numerical study of laser interaction with solid materials

57
Kurdistan Iraqi Region Ministry of Higher Education University of Sulaimani College of Science Physics Department Numerical Study of Laser Interaction with Solid Materials Prepared by Chia H.Qadr Shiraz Q .Ghafur Hewar A.Abdul Supervised by Dr. Omed Ghareb Abdullah 2010 - 2011

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Numerical Study of Laser Interaction with Solid Materials. University of Sulaimani, College of Science, Physics Department, 2011

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Page 1: Numerical Study of Laser Interaction With Solid Materials

Kurdistan Iraqi Region Ministry of Higher Education University of Sulaimani College of Science Physics Department

Numerical Study of Laser Interaction with

Solid Materials

Prepared by

Chia HQadr Shiraz Q Ghafur Hewar AAbdul

Supervised by

Dr Omed Ghareb Abdullah

2010 - 2011

ii

Acknowledgments

First of all we would like to say Alhamdulillah for giving us the

strength and health to do this project work until it done and not forgotten

to our family for providing with everything that are related to this project

work and their advice They also supported us and encouraged us to

complete this task so that we will not procrastinate in doing it

Then we would like to thank our Supervisor Dr Omed Ghareb

Abdullah for guiding us throughout this project We had some difficulties

in doing this task but he taught us patiently until we knew what to do He

tried and tried to teach us until we understand what we supposed to do with

the project work

Last but not least those who were helping us in doing this project by

sharing ideas They were helpful that when we combined and discussed

together we had this task done

Chia Shiraz amp Hewar

iii

Contents

Chapter One Basic Concepts

11 Introduction

12 Definition of the Laser

13 Active laser medium or gain medium

14 A Survey of Laser Types

141 Gas Lasers

142 Solid Lasers

143 Molecular Lasers

144 Free-Electron Lasers

15 Pulsed operation

16 Heat and heat capacity

17 Thermal conductivity

18 Derivation in one dimension

19 Aim of present work

Chapter Two Theoretical Aspects

21 Introduction

22 One dimension laser heating equation

23 Numerical solution of Initial value problems

24 Finite Difference Method

241 First Order Forward Difference

iv

242 First Order Backward Difference

242 First Order Central Difference

25 Procedures

Chapter three Results and Discussion

31 Introduction

32 Numerical solution with constant laser power density and

constant thermal properties

33 Evaluation of function 119920119920(119957119957) of laser flux density

34 Numerical solution with variable laser power density (

119920119920 = 119920119920 (119957119957) ) and constant thermal properties

35 Evaluation the Thermal Conductivity as functions of

temperature

36 Evaluation the Specific heat as functions of temperature

37 Evaluation the Density as functions of temperature

38 Numerical solution with variable laser power density (

119920119920 = 119920119920 (119957119957) ) and variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 =

119914119914(119931119931)120646120646 = 120646120646(119931119931)

39 Laser interaction with copper material

310 Conclusions

References

v

Abstract

In recent years much effort has gone into the understanding of the

interaction of short laser pulses with matter The present works have

typically involved studying the interaction of high intensity laser pulses with

high-density solid target In this study the NDYAG pulsed laser with

maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for

laser energy with time as well as a function of laser intensity with time are

presented in this study

The finite difference method was used to calculate the temperature

distribution as a function of laser depth penetration in lead and copper

materials

The best polynomial fits for thermal conductivity specific heat capacity

and density of metals as a function of temperature was obtained using

Matlab software At the first all these properties were assumed to be

constants and then the influence of varying these properties with

temperature was tacked in to account The temperature gradient of lead

shows to be greater than that of copper this may be due to the high thermal

conductivity and high specific heat capacity of copper with that of lead

1

Chapter One

Basic concepts

11 Introduction

Laser is a mechanism for emitting light with in electromagnetic radiation

region of the spectrum with different output intensity Max Plank published

work in 1900 that provided the understanding that light is a form of

electromagnetic radiation without this understanding the laser would have

been invented The principle of the laser was first known in 1917 when Albert

Einstein describe the theory of stimulated emission and Theodor Maiman in

1960 invent the first laser using a lasing medium of ruby that was stimulated

by using high energy flash of intense light

We have four types of laser according to their gain medium which are

(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)

So laser is provided a controlled source of atomic and electronic excitations

involving non equilibrium phenomena that lend themselves to processing of

novel material and structure because laser used in wide range application in

our life such as welding cutting drilling industrial and medical field Maiman

and other developer of laser weapons sighting system and powerful laser for

use in surgery and other areas where moderated powerful pinpoint source of

heat was needed And today laser are used in corrective eye surgery and

providing apprecise source of heat for cutting and cauterizing tissue

12 Definition of the Laser

The word laser is an acronym for Light Amplification by Stimulated Emission

of Radiation The laser makes use of processes that increase or amplify light

signals after those signals have been generated by other means These

processes include (1) stimulated emission a natural effect that was deduced

2

by considerations relating to thermodynamic equilibrium and (2) optical

feedback (present in most lasers) that is usually provided by mirrors

Thus in its simplest form a laser consists of a gain or amplifying medium

(where stimulated emission occurs) and a set of mirrors to feed the light back

into the amplifier for continued growth of the developing beam as seen in

Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

more intense directional monochromatic and coherent Most lasers consist

of a column of active material with a partly reflecting mirror at one end and a

fully reflecting mirror at the other The active material can be solid (ruby

crystal) liquid or gas (HeNe COR2R etc)

Fig(11) Simplified schematic of typical laser

13 Active laser medium or gain medium

Laser medium is the heart of the laser system and is responsible for

producing gain and subsequent generation of laser It can be a crystal solid

liquid semiconductor or gas medium and can be pumped to a higher energy

state The material should be of controlled purity size and shape and should

have the suitable energy levels to support population inversion In other

words it must have a metastable state to support stimulated emission Most

lasers are based on 3 or 4 level energy level systems which depends on the

lasing medium These systems are shown in Figs (12) and (13)

3

In case of a three-level laser the material is pumped from level 1 to level 3

which decays rapidly to level 2 through spontaneous emission Level 2 is a

metastable level and promotes stimulated emission from level 2 to level 1

Fig(12) Energy states of Three-level active medium

On the other hand in a four-level laser the material is pumped to level 4

which is a fast decaying level and the atoms decay rapidly to level 3 which is

a metastable level The stimulated emission takes place from level 3 to level 2

from where the atoms decay back to level 1 Four level lasers is an

improvement on a system based on three level systems In this case the laser

transition takes place between the third and second excited states Since

lower laser level 2 is a fast decaying level which ensures that it rapidly gets

empty and as such always supports the population inversion condition

Fig(13) Energy states of Four-level active medium

4

14 A Survey of Laser Types

Laser technology is available to us since 1960rsquos and since then has been

quite well developed Currently there is a great variety of lasers of different

output power operating voltages sizes etc The major classes of lasers

currently used are Gas Solid Molecular and Free Electron lasers Below we

will cover some most popular representative types of lasers of each class and

describe specific principles of operation construction and main highlights

141 Gas Lasers

1 Helium-Neon Laser

The most common and inexpensive gas laser the helium-neon laser is

usually constructed to operate in the red at 6328 nm It can also be

constructed to produce laser action in the green at 5435 nm and in the

infrared at 1523 nm

One of the excited levels of helium at 2061 eV is very close to a level in

neon at 2066 eV so close in fact that upon collision of a helium and a neon

atom the energy can be transferred from the helium to the neon atom

Fig (14) The components of a Hilium-Neon Laser

5

Fig(15) The lasing action of He-Ne laser

Helium-Neon lasers are common in the introductory physics laboratories

but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

brightness equal to sunshine on a clear day (01 wattcmP

2P) and is just as

dangerous to stare at directly

2- Carbon Dioxide Laser

The carbon dioxide gas laser is capable of continuous output powers above

10 kilowatts It is also capable of extremely high power pulse operation It

exhibits laser action at several infrared frequencies but none in the visible

spectrum Operating in a manner similar to the helium-neon laser it employs

an electric discharge for pumping using a percentage of nitrogen gas as a

pumping gas The COR2R laser is the most efficient laser capable of operating at

more than 30 efficiency

The carbon dioxide laser finds many applications in industry particularly for

welding and Cutting

6

3- Argon Laser

The argon ion laser can be operated as a continuous gas laser at about 25

different wavelengths in the visible between (4089 - 6861) nm but is best

known for its most efficient transitions in the green at 488 nm and 5145 nm

Operating at much higher powers than the Helium-Neon gas laser it is not

uncommon to achieve (30 ndash 100) watts of continuous power using several

transitions This output is produced in hot plasma and takes extremely high

power typically (9 ndash 12) kW so these are large and expensive devices

142 Solid Lasers

1 Ruby Laser

The ruby laser is the first type of laser actually constructed first

demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

aluminum oxide with a small amount (about 005) of Chromium which gives

it its characteristic pink or red color by absorbing green and blue light

The ruby laser is used as a pulsed laser producing red light at 6943 nm

After receiving a pumping flash from the flash tube the laser light emerges for

as long as the excited atoms persist in the ruby rod which is typically about a

millisecond

A pulsed ruby laser was used for the famous laser ranging experiment which

was conducted with a corner reflector placed on the Moon by the Apollo

astronauts This determined the distance to the Moon with an accuracy of

about 15 cm

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 2: Numerical Study of Laser Interaction With Solid Materials

ii

Acknowledgments

First of all we would like to say Alhamdulillah for giving us the

strength and health to do this project work until it done and not forgotten

to our family for providing with everything that are related to this project

work and their advice They also supported us and encouraged us to

complete this task so that we will not procrastinate in doing it

Then we would like to thank our Supervisor Dr Omed Ghareb

Abdullah for guiding us throughout this project We had some difficulties

in doing this task but he taught us patiently until we knew what to do He

tried and tried to teach us until we understand what we supposed to do with

the project work

Last but not least those who were helping us in doing this project by

sharing ideas They were helpful that when we combined and discussed

together we had this task done

Chia Shiraz amp Hewar

iii

Contents

Chapter One Basic Concepts

11 Introduction

12 Definition of the Laser

13 Active laser medium or gain medium

14 A Survey of Laser Types

141 Gas Lasers

142 Solid Lasers

143 Molecular Lasers

144 Free-Electron Lasers

15 Pulsed operation

16 Heat and heat capacity

17 Thermal conductivity

18 Derivation in one dimension

19 Aim of present work

Chapter Two Theoretical Aspects

21 Introduction

22 One dimension laser heating equation

23 Numerical solution of Initial value problems

24 Finite Difference Method

241 First Order Forward Difference

iv

242 First Order Backward Difference

242 First Order Central Difference

25 Procedures

Chapter three Results and Discussion

31 Introduction

32 Numerical solution with constant laser power density and

constant thermal properties

33 Evaluation of function 119920119920(119957119957) of laser flux density

34 Numerical solution with variable laser power density (

119920119920 = 119920119920 (119957119957) ) and constant thermal properties

35 Evaluation the Thermal Conductivity as functions of

temperature

36 Evaluation the Specific heat as functions of temperature

37 Evaluation the Density as functions of temperature

38 Numerical solution with variable laser power density (

119920119920 = 119920119920 (119957119957) ) and variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 =

119914119914(119931119931)120646120646 = 120646120646(119931119931)

39 Laser interaction with copper material

310 Conclusions

References

v

Abstract

In recent years much effort has gone into the understanding of the

interaction of short laser pulses with matter The present works have

typically involved studying the interaction of high intensity laser pulses with

high-density solid target In this study the NDYAG pulsed laser with

maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for

laser energy with time as well as a function of laser intensity with time are

presented in this study

The finite difference method was used to calculate the temperature

distribution as a function of laser depth penetration in lead and copper

materials

The best polynomial fits for thermal conductivity specific heat capacity

and density of metals as a function of temperature was obtained using

Matlab software At the first all these properties were assumed to be

constants and then the influence of varying these properties with

temperature was tacked in to account The temperature gradient of lead

shows to be greater than that of copper this may be due to the high thermal

conductivity and high specific heat capacity of copper with that of lead

1

Chapter One

Basic concepts

11 Introduction

Laser is a mechanism for emitting light with in electromagnetic radiation

region of the spectrum with different output intensity Max Plank published

work in 1900 that provided the understanding that light is a form of

electromagnetic radiation without this understanding the laser would have

been invented The principle of the laser was first known in 1917 when Albert

Einstein describe the theory of stimulated emission and Theodor Maiman in

1960 invent the first laser using a lasing medium of ruby that was stimulated

by using high energy flash of intense light

We have four types of laser according to their gain medium which are

(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)

So laser is provided a controlled source of atomic and electronic excitations

involving non equilibrium phenomena that lend themselves to processing of

novel material and structure because laser used in wide range application in

our life such as welding cutting drilling industrial and medical field Maiman

and other developer of laser weapons sighting system and powerful laser for

use in surgery and other areas where moderated powerful pinpoint source of

heat was needed And today laser are used in corrective eye surgery and

providing apprecise source of heat for cutting and cauterizing tissue

12 Definition of the Laser

The word laser is an acronym for Light Amplification by Stimulated Emission

of Radiation The laser makes use of processes that increase or amplify light

signals after those signals have been generated by other means These

processes include (1) stimulated emission a natural effect that was deduced

2

by considerations relating to thermodynamic equilibrium and (2) optical

feedback (present in most lasers) that is usually provided by mirrors

Thus in its simplest form a laser consists of a gain or amplifying medium

(where stimulated emission occurs) and a set of mirrors to feed the light back

into the amplifier for continued growth of the developing beam as seen in

Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

more intense directional monochromatic and coherent Most lasers consist

of a column of active material with a partly reflecting mirror at one end and a

fully reflecting mirror at the other The active material can be solid (ruby

crystal) liquid or gas (HeNe COR2R etc)

Fig(11) Simplified schematic of typical laser

13 Active laser medium or gain medium

Laser medium is the heart of the laser system and is responsible for

producing gain and subsequent generation of laser It can be a crystal solid

liquid semiconductor or gas medium and can be pumped to a higher energy

state The material should be of controlled purity size and shape and should

have the suitable energy levels to support population inversion In other

words it must have a metastable state to support stimulated emission Most

lasers are based on 3 or 4 level energy level systems which depends on the

lasing medium These systems are shown in Figs (12) and (13)

3

In case of a three-level laser the material is pumped from level 1 to level 3

which decays rapidly to level 2 through spontaneous emission Level 2 is a

metastable level and promotes stimulated emission from level 2 to level 1

Fig(12) Energy states of Three-level active medium

On the other hand in a four-level laser the material is pumped to level 4

which is a fast decaying level and the atoms decay rapidly to level 3 which is

a metastable level The stimulated emission takes place from level 3 to level 2

from where the atoms decay back to level 1 Four level lasers is an

improvement on a system based on three level systems In this case the laser

transition takes place between the third and second excited states Since

lower laser level 2 is a fast decaying level which ensures that it rapidly gets

empty and as such always supports the population inversion condition

Fig(13) Energy states of Four-level active medium

4

14 A Survey of Laser Types

Laser technology is available to us since 1960rsquos and since then has been

quite well developed Currently there is a great variety of lasers of different

output power operating voltages sizes etc The major classes of lasers

currently used are Gas Solid Molecular and Free Electron lasers Below we

will cover some most popular representative types of lasers of each class and

describe specific principles of operation construction and main highlights

141 Gas Lasers

1 Helium-Neon Laser

The most common and inexpensive gas laser the helium-neon laser is

usually constructed to operate in the red at 6328 nm It can also be

constructed to produce laser action in the green at 5435 nm and in the

infrared at 1523 nm

One of the excited levels of helium at 2061 eV is very close to a level in

neon at 2066 eV so close in fact that upon collision of a helium and a neon

atom the energy can be transferred from the helium to the neon atom

Fig (14) The components of a Hilium-Neon Laser

5

Fig(15) The lasing action of He-Ne laser

Helium-Neon lasers are common in the introductory physics laboratories

but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

brightness equal to sunshine on a clear day (01 wattcmP

2P) and is just as

dangerous to stare at directly

2- Carbon Dioxide Laser

The carbon dioxide gas laser is capable of continuous output powers above

10 kilowatts It is also capable of extremely high power pulse operation It

exhibits laser action at several infrared frequencies but none in the visible

spectrum Operating in a manner similar to the helium-neon laser it employs

an electric discharge for pumping using a percentage of nitrogen gas as a

pumping gas The COR2R laser is the most efficient laser capable of operating at

more than 30 efficiency

The carbon dioxide laser finds many applications in industry particularly for

welding and Cutting

6

3- Argon Laser

The argon ion laser can be operated as a continuous gas laser at about 25

different wavelengths in the visible between (4089 - 6861) nm but is best

known for its most efficient transitions in the green at 488 nm and 5145 nm

Operating at much higher powers than the Helium-Neon gas laser it is not

uncommon to achieve (30 ndash 100) watts of continuous power using several

transitions This output is produced in hot plasma and takes extremely high

power typically (9 ndash 12) kW so these are large and expensive devices

142 Solid Lasers

1 Ruby Laser

The ruby laser is the first type of laser actually constructed first

demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

aluminum oxide with a small amount (about 005) of Chromium which gives

it its characteristic pink or red color by absorbing green and blue light

The ruby laser is used as a pulsed laser producing red light at 6943 nm

After receiving a pumping flash from the flash tube the laser light emerges for

as long as the excited atoms persist in the ruby rod which is typically about a

millisecond

A pulsed ruby laser was used for the famous laser ranging experiment which

was conducted with a corner reflector placed on the Moon by the Apollo

astronauts This determined the distance to the Moon with an accuracy of

about 15 cm

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 3: Numerical Study of Laser Interaction With Solid Materials

iii

Contents

Chapter One Basic Concepts

11 Introduction

12 Definition of the Laser

13 Active laser medium or gain medium

14 A Survey of Laser Types

141 Gas Lasers

142 Solid Lasers

143 Molecular Lasers

144 Free-Electron Lasers

15 Pulsed operation

16 Heat and heat capacity

17 Thermal conductivity

18 Derivation in one dimension

19 Aim of present work

Chapter Two Theoretical Aspects

21 Introduction

22 One dimension laser heating equation

23 Numerical solution of Initial value problems

24 Finite Difference Method

241 First Order Forward Difference

iv

242 First Order Backward Difference

242 First Order Central Difference

25 Procedures

Chapter three Results and Discussion

31 Introduction

32 Numerical solution with constant laser power density and

constant thermal properties

33 Evaluation of function 119920119920(119957119957) of laser flux density

34 Numerical solution with variable laser power density (

119920119920 = 119920119920 (119957119957) ) and constant thermal properties

35 Evaluation the Thermal Conductivity as functions of

temperature

36 Evaluation the Specific heat as functions of temperature

37 Evaluation the Density as functions of temperature

38 Numerical solution with variable laser power density (

119920119920 = 119920119920 (119957119957) ) and variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 =

119914119914(119931119931)120646120646 = 120646120646(119931119931)

39 Laser interaction with copper material

310 Conclusions

References

v

Abstract

In recent years much effort has gone into the understanding of the

interaction of short laser pulses with matter The present works have

typically involved studying the interaction of high intensity laser pulses with

high-density solid target In this study the NDYAG pulsed laser with

maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for

laser energy with time as well as a function of laser intensity with time are

presented in this study

The finite difference method was used to calculate the temperature

distribution as a function of laser depth penetration in lead and copper

materials

The best polynomial fits for thermal conductivity specific heat capacity

and density of metals as a function of temperature was obtained using

Matlab software At the first all these properties were assumed to be

constants and then the influence of varying these properties with

temperature was tacked in to account The temperature gradient of lead

shows to be greater than that of copper this may be due to the high thermal

conductivity and high specific heat capacity of copper with that of lead

1

Chapter One

Basic concepts

11 Introduction

Laser is a mechanism for emitting light with in electromagnetic radiation

region of the spectrum with different output intensity Max Plank published

work in 1900 that provided the understanding that light is a form of

electromagnetic radiation without this understanding the laser would have

been invented The principle of the laser was first known in 1917 when Albert

Einstein describe the theory of stimulated emission and Theodor Maiman in

1960 invent the first laser using a lasing medium of ruby that was stimulated

by using high energy flash of intense light

We have four types of laser according to their gain medium which are

(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)

So laser is provided a controlled source of atomic and electronic excitations

involving non equilibrium phenomena that lend themselves to processing of

novel material and structure because laser used in wide range application in

our life such as welding cutting drilling industrial and medical field Maiman

and other developer of laser weapons sighting system and powerful laser for

use in surgery and other areas where moderated powerful pinpoint source of

heat was needed And today laser are used in corrective eye surgery and

providing apprecise source of heat for cutting and cauterizing tissue

12 Definition of the Laser

The word laser is an acronym for Light Amplification by Stimulated Emission

of Radiation The laser makes use of processes that increase or amplify light

signals after those signals have been generated by other means These

processes include (1) stimulated emission a natural effect that was deduced

2

by considerations relating to thermodynamic equilibrium and (2) optical

feedback (present in most lasers) that is usually provided by mirrors

Thus in its simplest form a laser consists of a gain or amplifying medium

(where stimulated emission occurs) and a set of mirrors to feed the light back

into the amplifier for continued growth of the developing beam as seen in

Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

more intense directional monochromatic and coherent Most lasers consist

of a column of active material with a partly reflecting mirror at one end and a

fully reflecting mirror at the other The active material can be solid (ruby

crystal) liquid or gas (HeNe COR2R etc)

Fig(11) Simplified schematic of typical laser

13 Active laser medium or gain medium

Laser medium is the heart of the laser system and is responsible for

producing gain and subsequent generation of laser It can be a crystal solid

liquid semiconductor or gas medium and can be pumped to a higher energy

state The material should be of controlled purity size and shape and should

have the suitable energy levels to support population inversion In other

words it must have a metastable state to support stimulated emission Most

lasers are based on 3 or 4 level energy level systems which depends on the

lasing medium These systems are shown in Figs (12) and (13)

3

In case of a three-level laser the material is pumped from level 1 to level 3

which decays rapidly to level 2 through spontaneous emission Level 2 is a

metastable level and promotes stimulated emission from level 2 to level 1

Fig(12) Energy states of Three-level active medium

On the other hand in a four-level laser the material is pumped to level 4

which is a fast decaying level and the atoms decay rapidly to level 3 which is

a metastable level The stimulated emission takes place from level 3 to level 2

from where the atoms decay back to level 1 Four level lasers is an

improvement on a system based on three level systems In this case the laser

transition takes place between the third and second excited states Since

lower laser level 2 is a fast decaying level which ensures that it rapidly gets

empty and as such always supports the population inversion condition

Fig(13) Energy states of Four-level active medium

4

14 A Survey of Laser Types

Laser technology is available to us since 1960rsquos and since then has been

quite well developed Currently there is a great variety of lasers of different

output power operating voltages sizes etc The major classes of lasers

currently used are Gas Solid Molecular and Free Electron lasers Below we

will cover some most popular representative types of lasers of each class and

describe specific principles of operation construction and main highlights

141 Gas Lasers

1 Helium-Neon Laser

The most common and inexpensive gas laser the helium-neon laser is

usually constructed to operate in the red at 6328 nm It can also be

constructed to produce laser action in the green at 5435 nm and in the

infrared at 1523 nm

One of the excited levels of helium at 2061 eV is very close to a level in

neon at 2066 eV so close in fact that upon collision of a helium and a neon

atom the energy can be transferred from the helium to the neon atom

Fig (14) The components of a Hilium-Neon Laser

5

Fig(15) The lasing action of He-Ne laser

Helium-Neon lasers are common in the introductory physics laboratories

but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

brightness equal to sunshine on a clear day (01 wattcmP

2P) and is just as

dangerous to stare at directly

2- Carbon Dioxide Laser

The carbon dioxide gas laser is capable of continuous output powers above

10 kilowatts It is also capable of extremely high power pulse operation It

exhibits laser action at several infrared frequencies but none in the visible

spectrum Operating in a manner similar to the helium-neon laser it employs

an electric discharge for pumping using a percentage of nitrogen gas as a

pumping gas The COR2R laser is the most efficient laser capable of operating at

more than 30 efficiency

The carbon dioxide laser finds many applications in industry particularly for

welding and Cutting

6

3- Argon Laser

The argon ion laser can be operated as a continuous gas laser at about 25

different wavelengths in the visible between (4089 - 6861) nm but is best

known for its most efficient transitions in the green at 488 nm and 5145 nm

Operating at much higher powers than the Helium-Neon gas laser it is not

uncommon to achieve (30 ndash 100) watts of continuous power using several

transitions This output is produced in hot plasma and takes extremely high

power typically (9 ndash 12) kW so these are large and expensive devices

142 Solid Lasers

1 Ruby Laser

The ruby laser is the first type of laser actually constructed first

demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

aluminum oxide with a small amount (about 005) of Chromium which gives

it its characteristic pink or red color by absorbing green and blue light

The ruby laser is used as a pulsed laser producing red light at 6943 nm

After receiving a pumping flash from the flash tube the laser light emerges for

as long as the excited atoms persist in the ruby rod which is typically about a

millisecond

A pulsed ruby laser was used for the famous laser ranging experiment which

was conducted with a corner reflector placed on the Moon by the Apollo

astronauts This determined the distance to the Moon with an accuracy of

about 15 cm

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 4: Numerical Study of Laser Interaction With Solid Materials

iv

242 First Order Backward Difference

242 First Order Central Difference

25 Procedures

Chapter three Results and Discussion

31 Introduction

32 Numerical solution with constant laser power density and

constant thermal properties

33 Evaluation of function 119920119920(119957119957) of laser flux density

34 Numerical solution with variable laser power density (

119920119920 = 119920119920 (119957119957) ) and constant thermal properties

35 Evaluation the Thermal Conductivity as functions of

temperature

36 Evaluation the Specific heat as functions of temperature

37 Evaluation the Density as functions of temperature

38 Numerical solution with variable laser power density (

119920119920 = 119920119920 (119957119957) ) and variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 =

119914119914(119931119931)120646120646 = 120646120646(119931119931)

39 Laser interaction with copper material

310 Conclusions

References

v

Abstract

In recent years much effort has gone into the understanding of the

interaction of short laser pulses with matter The present works have

typically involved studying the interaction of high intensity laser pulses with

high-density solid target In this study the NDYAG pulsed laser with

maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for

laser energy with time as well as a function of laser intensity with time are

presented in this study

The finite difference method was used to calculate the temperature

distribution as a function of laser depth penetration in lead and copper

materials

The best polynomial fits for thermal conductivity specific heat capacity

and density of metals as a function of temperature was obtained using

Matlab software At the first all these properties were assumed to be

constants and then the influence of varying these properties with

temperature was tacked in to account The temperature gradient of lead

shows to be greater than that of copper this may be due to the high thermal

conductivity and high specific heat capacity of copper with that of lead

1

Chapter One

Basic concepts

11 Introduction

Laser is a mechanism for emitting light with in electromagnetic radiation

region of the spectrum with different output intensity Max Plank published

work in 1900 that provided the understanding that light is a form of

electromagnetic radiation without this understanding the laser would have

been invented The principle of the laser was first known in 1917 when Albert

Einstein describe the theory of stimulated emission and Theodor Maiman in

1960 invent the first laser using a lasing medium of ruby that was stimulated

by using high energy flash of intense light

We have four types of laser according to their gain medium which are

(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)

So laser is provided a controlled source of atomic and electronic excitations

involving non equilibrium phenomena that lend themselves to processing of

novel material and structure because laser used in wide range application in

our life such as welding cutting drilling industrial and medical field Maiman

and other developer of laser weapons sighting system and powerful laser for

use in surgery and other areas where moderated powerful pinpoint source of

heat was needed And today laser are used in corrective eye surgery and

providing apprecise source of heat for cutting and cauterizing tissue

12 Definition of the Laser

The word laser is an acronym for Light Amplification by Stimulated Emission

of Radiation The laser makes use of processes that increase or amplify light

signals after those signals have been generated by other means These

processes include (1) stimulated emission a natural effect that was deduced

2

by considerations relating to thermodynamic equilibrium and (2) optical

feedback (present in most lasers) that is usually provided by mirrors

Thus in its simplest form a laser consists of a gain or amplifying medium

(where stimulated emission occurs) and a set of mirrors to feed the light back

into the amplifier for continued growth of the developing beam as seen in

Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

more intense directional monochromatic and coherent Most lasers consist

of a column of active material with a partly reflecting mirror at one end and a

fully reflecting mirror at the other The active material can be solid (ruby

crystal) liquid or gas (HeNe COR2R etc)

Fig(11) Simplified schematic of typical laser

13 Active laser medium or gain medium

Laser medium is the heart of the laser system and is responsible for

producing gain and subsequent generation of laser It can be a crystal solid

liquid semiconductor or gas medium and can be pumped to a higher energy

state The material should be of controlled purity size and shape and should

have the suitable energy levels to support population inversion In other

words it must have a metastable state to support stimulated emission Most

lasers are based on 3 or 4 level energy level systems which depends on the

lasing medium These systems are shown in Figs (12) and (13)

3

In case of a three-level laser the material is pumped from level 1 to level 3

which decays rapidly to level 2 through spontaneous emission Level 2 is a

metastable level and promotes stimulated emission from level 2 to level 1

Fig(12) Energy states of Three-level active medium

On the other hand in a four-level laser the material is pumped to level 4

which is a fast decaying level and the atoms decay rapidly to level 3 which is

a metastable level The stimulated emission takes place from level 3 to level 2

from where the atoms decay back to level 1 Four level lasers is an

improvement on a system based on three level systems In this case the laser

transition takes place between the third and second excited states Since

lower laser level 2 is a fast decaying level which ensures that it rapidly gets

empty and as such always supports the population inversion condition

Fig(13) Energy states of Four-level active medium

4

14 A Survey of Laser Types

Laser technology is available to us since 1960rsquos and since then has been

quite well developed Currently there is a great variety of lasers of different

output power operating voltages sizes etc The major classes of lasers

currently used are Gas Solid Molecular and Free Electron lasers Below we

will cover some most popular representative types of lasers of each class and

describe specific principles of operation construction and main highlights

141 Gas Lasers

1 Helium-Neon Laser

The most common and inexpensive gas laser the helium-neon laser is

usually constructed to operate in the red at 6328 nm It can also be

constructed to produce laser action in the green at 5435 nm and in the

infrared at 1523 nm

One of the excited levels of helium at 2061 eV is very close to a level in

neon at 2066 eV so close in fact that upon collision of a helium and a neon

atom the energy can be transferred from the helium to the neon atom

Fig (14) The components of a Hilium-Neon Laser

5

Fig(15) The lasing action of He-Ne laser

Helium-Neon lasers are common in the introductory physics laboratories

but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

brightness equal to sunshine on a clear day (01 wattcmP

2P) and is just as

dangerous to stare at directly

2- Carbon Dioxide Laser

The carbon dioxide gas laser is capable of continuous output powers above

10 kilowatts It is also capable of extremely high power pulse operation It

exhibits laser action at several infrared frequencies but none in the visible

spectrum Operating in a manner similar to the helium-neon laser it employs

an electric discharge for pumping using a percentage of nitrogen gas as a

pumping gas The COR2R laser is the most efficient laser capable of operating at

more than 30 efficiency

The carbon dioxide laser finds many applications in industry particularly for

welding and Cutting

6

3- Argon Laser

The argon ion laser can be operated as a continuous gas laser at about 25

different wavelengths in the visible between (4089 - 6861) nm but is best

known for its most efficient transitions in the green at 488 nm and 5145 nm

Operating at much higher powers than the Helium-Neon gas laser it is not

uncommon to achieve (30 ndash 100) watts of continuous power using several

transitions This output is produced in hot plasma and takes extremely high

power typically (9 ndash 12) kW so these are large and expensive devices

142 Solid Lasers

1 Ruby Laser

The ruby laser is the first type of laser actually constructed first

demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

aluminum oxide with a small amount (about 005) of Chromium which gives

it its characteristic pink or red color by absorbing green and blue light

The ruby laser is used as a pulsed laser producing red light at 6943 nm

After receiving a pumping flash from the flash tube the laser light emerges for

as long as the excited atoms persist in the ruby rod which is typically about a

millisecond

A pulsed ruby laser was used for the famous laser ranging experiment which

was conducted with a corner reflector placed on the Moon by the Apollo

astronauts This determined the distance to the Moon with an accuracy of

about 15 cm

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 5: Numerical Study of Laser Interaction With Solid Materials

v

Abstract

In recent years much effort has gone into the understanding of the

interaction of short laser pulses with matter The present works have

typically involved studying the interaction of high intensity laser pulses with

high-density solid target In this study the NDYAG pulsed laser with

maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for

laser energy with time as well as a function of laser intensity with time are

presented in this study

The finite difference method was used to calculate the temperature

distribution as a function of laser depth penetration in lead and copper

materials

The best polynomial fits for thermal conductivity specific heat capacity

and density of metals as a function of temperature was obtained using

Matlab software At the first all these properties were assumed to be

constants and then the influence of varying these properties with

temperature was tacked in to account The temperature gradient of lead

shows to be greater than that of copper this may be due to the high thermal

conductivity and high specific heat capacity of copper with that of lead

1

Chapter One

Basic concepts

11 Introduction

Laser is a mechanism for emitting light with in electromagnetic radiation

region of the spectrum with different output intensity Max Plank published

work in 1900 that provided the understanding that light is a form of

electromagnetic radiation without this understanding the laser would have

been invented The principle of the laser was first known in 1917 when Albert

Einstein describe the theory of stimulated emission and Theodor Maiman in

1960 invent the first laser using a lasing medium of ruby that was stimulated

by using high energy flash of intense light

We have four types of laser according to their gain medium which are

(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)

So laser is provided a controlled source of atomic and electronic excitations

involving non equilibrium phenomena that lend themselves to processing of

novel material and structure because laser used in wide range application in

our life such as welding cutting drilling industrial and medical field Maiman

and other developer of laser weapons sighting system and powerful laser for

use in surgery and other areas where moderated powerful pinpoint source of

heat was needed And today laser are used in corrective eye surgery and

providing apprecise source of heat for cutting and cauterizing tissue

12 Definition of the Laser

The word laser is an acronym for Light Amplification by Stimulated Emission

of Radiation The laser makes use of processes that increase or amplify light

signals after those signals have been generated by other means These

processes include (1) stimulated emission a natural effect that was deduced

2

by considerations relating to thermodynamic equilibrium and (2) optical

feedback (present in most lasers) that is usually provided by mirrors

Thus in its simplest form a laser consists of a gain or amplifying medium

(where stimulated emission occurs) and a set of mirrors to feed the light back

into the amplifier for continued growth of the developing beam as seen in

Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

more intense directional monochromatic and coherent Most lasers consist

of a column of active material with a partly reflecting mirror at one end and a

fully reflecting mirror at the other The active material can be solid (ruby

crystal) liquid or gas (HeNe COR2R etc)

Fig(11) Simplified schematic of typical laser

13 Active laser medium or gain medium

Laser medium is the heart of the laser system and is responsible for

producing gain and subsequent generation of laser It can be a crystal solid

liquid semiconductor or gas medium and can be pumped to a higher energy

state The material should be of controlled purity size and shape and should

have the suitable energy levels to support population inversion In other

words it must have a metastable state to support stimulated emission Most

lasers are based on 3 or 4 level energy level systems which depends on the

lasing medium These systems are shown in Figs (12) and (13)

3

In case of a three-level laser the material is pumped from level 1 to level 3

which decays rapidly to level 2 through spontaneous emission Level 2 is a

metastable level and promotes stimulated emission from level 2 to level 1

Fig(12) Energy states of Three-level active medium

On the other hand in a four-level laser the material is pumped to level 4

which is a fast decaying level and the atoms decay rapidly to level 3 which is

a metastable level The stimulated emission takes place from level 3 to level 2

from where the atoms decay back to level 1 Four level lasers is an

improvement on a system based on three level systems In this case the laser

transition takes place between the third and second excited states Since

lower laser level 2 is a fast decaying level which ensures that it rapidly gets

empty and as such always supports the population inversion condition

Fig(13) Energy states of Four-level active medium

4

14 A Survey of Laser Types

Laser technology is available to us since 1960rsquos and since then has been

quite well developed Currently there is a great variety of lasers of different

output power operating voltages sizes etc The major classes of lasers

currently used are Gas Solid Molecular and Free Electron lasers Below we

will cover some most popular representative types of lasers of each class and

describe specific principles of operation construction and main highlights

141 Gas Lasers

1 Helium-Neon Laser

The most common and inexpensive gas laser the helium-neon laser is

usually constructed to operate in the red at 6328 nm It can also be

constructed to produce laser action in the green at 5435 nm and in the

infrared at 1523 nm

One of the excited levels of helium at 2061 eV is very close to a level in

neon at 2066 eV so close in fact that upon collision of a helium and a neon

atom the energy can be transferred from the helium to the neon atom

Fig (14) The components of a Hilium-Neon Laser

5

Fig(15) The lasing action of He-Ne laser

Helium-Neon lasers are common in the introductory physics laboratories

but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

brightness equal to sunshine on a clear day (01 wattcmP

2P) and is just as

dangerous to stare at directly

2- Carbon Dioxide Laser

The carbon dioxide gas laser is capable of continuous output powers above

10 kilowatts It is also capable of extremely high power pulse operation It

exhibits laser action at several infrared frequencies but none in the visible

spectrum Operating in a manner similar to the helium-neon laser it employs

an electric discharge for pumping using a percentage of nitrogen gas as a

pumping gas The COR2R laser is the most efficient laser capable of operating at

more than 30 efficiency

The carbon dioxide laser finds many applications in industry particularly for

welding and Cutting

6

3- Argon Laser

The argon ion laser can be operated as a continuous gas laser at about 25

different wavelengths in the visible between (4089 - 6861) nm but is best

known for its most efficient transitions in the green at 488 nm and 5145 nm

Operating at much higher powers than the Helium-Neon gas laser it is not

uncommon to achieve (30 ndash 100) watts of continuous power using several

transitions This output is produced in hot plasma and takes extremely high

power typically (9 ndash 12) kW so these are large and expensive devices

142 Solid Lasers

1 Ruby Laser

The ruby laser is the first type of laser actually constructed first

demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

aluminum oxide with a small amount (about 005) of Chromium which gives

it its characteristic pink or red color by absorbing green and blue light

The ruby laser is used as a pulsed laser producing red light at 6943 nm

After receiving a pumping flash from the flash tube the laser light emerges for

as long as the excited atoms persist in the ruby rod which is typically about a

millisecond

A pulsed ruby laser was used for the famous laser ranging experiment which

was conducted with a corner reflector placed on the Moon by the Apollo

astronauts This determined the distance to the Moon with an accuracy of

about 15 cm

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 6: Numerical Study of Laser Interaction With Solid Materials

1

Chapter One

Basic concepts

11 Introduction

Laser is a mechanism for emitting light with in electromagnetic radiation

region of the spectrum with different output intensity Max Plank published

work in 1900 that provided the understanding that light is a form of

electromagnetic radiation without this understanding the laser would have

been invented The principle of the laser was first known in 1917 when Albert

Einstein describe the theory of stimulated emission and Theodor Maiman in

1960 invent the first laser using a lasing medium of ruby that was stimulated

by using high energy flash of intense light

We have four types of laser according to their gain medium which are

(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)

So laser is provided a controlled source of atomic and electronic excitations

involving non equilibrium phenomena that lend themselves to processing of

novel material and structure because laser used in wide range application in

our life such as welding cutting drilling industrial and medical field Maiman

and other developer of laser weapons sighting system and powerful laser for

use in surgery and other areas where moderated powerful pinpoint source of

heat was needed And today laser are used in corrective eye surgery and

providing apprecise source of heat for cutting and cauterizing tissue

12 Definition of the Laser

The word laser is an acronym for Light Amplification by Stimulated Emission

of Radiation The laser makes use of processes that increase or amplify light

signals after those signals have been generated by other means These

processes include (1) stimulated emission a natural effect that was deduced

2

by considerations relating to thermodynamic equilibrium and (2) optical

feedback (present in most lasers) that is usually provided by mirrors

Thus in its simplest form a laser consists of a gain or amplifying medium

(where stimulated emission occurs) and a set of mirrors to feed the light back

into the amplifier for continued growth of the developing beam as seen in

Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

more intense directional monochromatic and coherent Most lasers consist

of a column of active material with a partly reflecting mirror at one end and a

fully reflecting mirror at the other The active material can be solid (ruby

crystal) liquid or gas (HeNe COR2R etc)

Fig(11) Simplified schematic of typical laser

13 Active laser medium or gain medium

Laser medium is the heart of the laser system and is responsible for

producing gain and subsequent generation of laser It can be a crystal solid

liquid semiconductor or gas medium and can be pumped to a higher energy

state The material should be of controlled purity size and shape and should

have the suitable energy levels to support population inversion In other

words it must have a metastable state to support stimulated emission Most

lasers are based on 3 or 4 level energy level systems which depends on the

lasing medium These systems are shown in Figs (12) and (13)

3

In case of a three-level laser the material is pumped from level 1 to level 3

which decays rapidly to level 2 through spontaneous emission Level 2 is a

metastable level and promotes stimulated emission from level 2 to level 1

Fig(12) Energy states of Three-level active medium

On the other hand in a four-level laser the material is pumped to level 4

which is a fast decaying level and the atoms decay rapidly to level 3 which is

a metastable level The stimulated emission takes place from level 3 to level 2

from where the atoms decay back to level 1 Four level lasers is an

improvement on a system based on three level systems In this case the laser

transition takes place between the third and second excited states Since

lower laser level 2 is a fast decaying level which ensures that it rapidly gets

empty and as such always supports the population inversion condition

Fig(13) Energy states of Four-level active medium

4

14 A Survey of Laser Types

Laser technology is available to us since 1960rsquos and since then has been

quite well developed Currently there is a great variety of lasers of different

output power operating voltages sizes etc The major classes of lasers

currently used are Gas Solid Molecular and Free Electron lasers Below we

will cover some most popular representative types of lasers of each class and

describe specific principles of operation construction and main highlights

141 Gas Lasers

1 Helium-Neon Laser

The most common and inexpensive gas laser the helium-neon laser is

usually constructed to operate in the red at 6328 nm It can also be

constructed to produce laser action in the green at 5435 nm and in the

infrared at 1523 nm

One of the excited levels of helium at 2061 eV is very close to a level in

neon at 2066 eV so close in fact that upon collision of a helium and a neon

atom the energy can be transferred from the helium to the neon atom

Fig (14) The components of a Hilium-Neon Laser

5

Fig(15) The lasing action of He-Ne laser

Helium-Neon lasers are common in the introductory physics laboratories

but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

brightness equal to sunshine on a clear day (01 wattcmP

2P) and is just as

dangerous to stare at directly

2- Carbon Dioxide Laser

The carbon dioxide gas laser is capable of continuous output powers above

10 kilowatts It is also capable of extremely high power pulse operation It

exhibits laser action at several infrared frequencies but none in the visible

spectrum Operating in a manner similar to the helium-neon laser it employs

an electric discharge for pumping using a percentage of nitrogen gas as a

pumping gas The COR2R laser is the most efficient laser capable of operating at

more than 30 efficiency

The carbon dioxide laser finds many applications in industry particularly for

welding and Cutting

6

3- Argon Laser

The argon ion laser can be operated as a continuous gas laser at about 25

different wavelengths in the visible between (4089 - 6861) nm but is best

known for its most efficient transitions in the green at 488 nm and 5145 nm

Operating at much higher powers than the Helium-Neon gas laser it is not

uncommon to achieve (30 ndash 100) watts of continuous power using several

transitions This output is produced in hot plasma and takes extremely high

power typically (9 ndash 12) kW so these are large and expensive devices

142 Solid Lasers

1 Ruby Laser

The ruby laser is the first type of laser actually constructed first

demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

aluminum oxide with a small amount (about 005) of Chromium which gives

it its characteristic pink or red color by absorbing green and blue light

The ruby laser is used as a pulsed laser producing red light at 6943 nm

After receiving a pumping flash from the flash tube the laser light emerges for

as long as the excited atoms persist in the ruby rod which is typically about a

millisecond

A pulsed ruby laser was used for the famous laser ranging experiment which

was conducted with a corner reflector placed on the Moon by the Apollo

astronauts This determined the distance to the Moon with an accuracy of

about 15 cm

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 7: Numerical Study of Laser Interaction With Solid Materials

2

by considerations relating to thermodynamic equilibrium and (2) optical

feedback (present in most lasers) that is usually provided by mirrors

Thus in its simplest form a laser consists of a gain or amplifying medium

(where stimulated emission occurs) and a set of mirrors to feed the light back

into the amplifier for continued growth of the developing beam as seen in

Fig(11) Laser light differs from ordinary light in four ways Briefly it is much

more intense directional monochromatic and coherent Most lasers consist

of a column of active material with a partly reflecting mirror at one end and a

fully reflecting mirror at the other The active material can be solid (ruby

crystal) liquid or gas (HeNe COR2R etc)

Fig(11) Simplified schematic of typical laser

13 Active laser medium or gain medium

Laser medium is the heart of the laser system and is responsible for

producing gain and subsequent generation of laser It can be a crystal solid

liquid semiconductor or gas medium and can be pumped to a higher energy

state The material should be of controlled purity size and shape and should

have the suitable energy levels to support population inversion In other

words it must have a metastable state to support stimulated emission Most

lasers are based on 3 or 4 level energy level systems which depends on the

lasing medium These systems are shown in Figs (12) and (13)

3

In case of a three-level laser the material is pumped from level 1 to level 3

which decays rapidly to level 2 through spontaneous emission Level 2 is a

metastable level and promotes stimulated emission from level 2 to level 1

Fig(12) Energy states of Three-level active medium

On the other hand in a four-level laser the material is pumped to level 4

which is a fast decaying level and the atoms decay rapidly to level 3 which is

a metastable level The stimulated emission takes place from level 3 to level 2

from where the atoms decay back to level 1 Four level lasers is an

improvement on a system based on three level systems In this case the laser

transition takes place between the third and second excited states Since

lower laser level 2 is a fast decaying level which ensures that it rapidly gets

empty and as such always supports the population inversion condition

Fig(13) Energy states of Four-level active medium

4

14 A Survey of Laser Types

Laser technology is available to us since 1960rsquos and since then has been

quite well developed Currently there is a great variety of lasers of different

output power operating voltages sizes etc The major classes of lasers

currently used are Gas Solid Molecular and Free Electron lasers Below we

will cover some most popular representative types of lasers of each class and

describe specific principles of operation construction and main highlights

141 Gas Lasers

1 Helium-Neon Laser

The most common and inexpensive gas laser the helium-neon laser is

usually constructed to operate in the red at 6328 nm It can also be

constructed to produce laser action in the green at 5435 nm and in the

infrared at 1523 nm

One of the excited levels of helium at 2061 eV is very close to a level in

neon at 2066 eV so close in fact that upon collision of a helium and a neon

atom the energy can be transferred from the helium to the neon atom

Fig (14) The components of a Hilium-Neon Laser

5

Fig(15) The lasing action of He-Ne laser

Helium-Neon lasers are common in the introductory physics laboratories

but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

brightness equal to sunshine on a clear day (01 wattcmP

2P) and is just as

dangerous to stare at directly

2- Carbon Dioxide Laser

The carbon dioxide gas laser is capable of continuous output powers above

10 kilowatts It is also capable of extremely high power pulse operation It

exhibits laser action at several infrared frequencies but none in the visible

spectrum Operating in a manner similar to the helium-neon laser it employs

an electric discharge for pumping using a percentage of nitrogen gas as a

pumping gas The COR2R laser is the most efficient laser capable of operating at

more than 30 efficiency

The carbon dioxide laser finds many applications in industry particularly for

welding and Cutting

6

3- Argon Laser

The argon ion laser can be operated as a continuous gas laser at about 25

different wavelengths in the visible between (4089 - 6861) nm but is best

known for its most efficient transitions in the green at 488 nm and 5145 nm

Operating at much higher powers than the Helium-Neon gas laser it is not

uncommon to achieve (30 ndash 100) watts of continuous power using several

transitions This output is produced in hot plasma and takes extremely high

power typically (9 ndash 12) kW so these are large and expensive devices

142 Solid Lasers

1 Ruby Laser

The ruby laser is the first type of laser actually constructed first

demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

aluminum oxide with a small amount (about 005) of Chromium which gives

it its characteristic pink or red color by absorbing green and blue light

The ruby laser is used as a pulsed laser producing red light at 6943 nm

After receiving a pumping flash from the flash tube the laser light emerges for

as long as the excited atoms persist in the ruby rod which is typically about a

millisecond

A pulsed ruby laser was used for the famous laser ranging experiment which

was conducted with a corner reflector placed on the Moon by the Apollo

astronauts This determined the distance to the Moon with an accuracy of

about 15 cm

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 8: Numerical Study of Laser Interaction With Solid Materials

3

In case of a three-level laser the material is pumped from level 1 to level 3

which decays rapidly to level 2 through spontaneous emission Level 2 is a

metastable level and promotes stimulated emission from level 2 to level 1

Fig(12) Energy states of Three-level active medium

On the other hand in a four-level laser the material is pumped to level 4

which is a fast decaying level and the atoms decay rapidly to level 3 which is

a metastable level The stimulated emission takes place from level 3 to level 2

from where the atoms decay back to level 1 Four level lasers is an

improvement on a system based on three level systems In this case the laser

transition takes place between the third and second excited states Since

lower laser level 2 is a fast decaying level which ensures that it rapidly gets

empty and as such always supports the population inversion condition

Fig(13) Energy states of Four-level active medium

4

14 A Survey of Laser Types

Laser technology is available to us since 1960rsquos and since then has been

quite well developed Currently there is a great variety of lasers of different

output power operating voltages sizes etc The major classes of lasers

currently used are Gas Solid Molecular and Free Electron lasers Below we

will cover some most popular representative types of lasers of each class and

describe specific principles of operation construction and main highlights

141 Gas Lasers

1 Helium-Neon Laser

The most common and inexpensive gas laser the helium-neon laser is

usually constructed to operate in the red at 6328 nm It can also be

constructed to produce laser action in the green at 5435 nm and in the

infrared at 1523 nm

One of the excited levels of helium at 2061 eV is very close to a level in

neon at 2066 eV so close in fact that upon collision of a helium and a neon

atom the energy can be transferred from the helium to the neon atom

Fig (14) The components of a Hilium-Neon Laser

5

Fig(15) The lasing action of He-Ne laser

Helium-Neon lasers are common in the introductory physics laboratories

but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

brightness equal to sunshine on a clear day (01 wattcmP

2P) and is just as

dangerous to stare at directly

2- Carbon Dioxide Laser

The carbon dioxide gas laser is capable of continuous output powers above

10 kilowatts It is also capable of extremely high power pulse operation It

exhibits laser action at several infrared frequencies but none in the visible

spectrum Operating in a manner similar to the helium-neon laser it employs

an electric discharge for pumping using a percentage of nitrogen gas as a

pumping gas The COR2R laser is the most efficient laser capable of operating at

more than 30 efficiency

The carbon dioxide laser finds many applications in industry particularly for

welding and Cutting

6

3- Argon Laser

The argon ion laser can be operated as a continuous gas laser at about 25

different wavelengths in the visible between (4089 - 6861) nm but is best

known for its most efficient transitions in the green at 488 nm and 5145 nm

Operating at much higher powers than the Helium-Neon gas laser it is not

uncommon to achieve (30 ndash 100) watts of continuous power using several

transitions This output is produced in hot plasma and takes extremely high

power typically (9 ndash 12) kW so these are large and expensive devices

142 Solid Lasers

1 Ruby Laser

The ruby laser is the first type of laser actually constructed first

demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

aluminum oxide with a small amount (about 005) of Chromium which gives

it its characteristic pink or red color by absorbing green and blue light

The ruby laser is used as a pulsed laser producing red light at 6943 nm

After receiving a pumping flash from the flash tube the laser light emerges for

as long as the excited atoms persist in the ruby rod which is typically about a

millisecond

A pulsed ruby laser was used for the famous laser ranging experiment which

was conducted with a corner reflector placed on the Moon by the Apollo

astronauts This determined the distance to the Moon with an accuracy of

about 15 cm

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 9: Numerical Study of Laser Interaction With Solid Materials

4

14 A Survey of Laser Types

Laser technology is available to us since 1960rsquos and since then has been

quite well developed Currently there is a great variety of lasers of different

output power operating voltages sizes etc The major classes of lasers

currently used are Gas Solid Molecular and Free Electron lasers Below we

will cover some most popular representative types of lasers of each class and

describe specific principles of operation construction and main highlights

141 Gas Lasers

1 Helium-Neon Laser

The most common and inexpensive gas laser the helium-neon laser is

usually constructed to operate in the red at 6328 nm It can also be

constructed to produce laser action in the green at 5435 nm and in the

infrared at 1523 nm

One of the excited levels of helium at 2061 eV is very close to a level in

neon at 2066 eV so close in fact that upon collision of a helium and a neon

atom the energy can be transferred from the helium to the neon atom

Fig (14) The components of a Hilium-Neon Laser

5

Fig(15) The lasing action of He-Ne laser

Helium-Neon lasers are common in the introductory physics laboratories

but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

brightness equal to sunshine on a clear day (01 wattcmP

2P) and is just as

dangerous to stare at directly

2- Carbon Dioxide Laser

The carbon dioxide gas laser is capable of continuous output powers above

10 kilowatts It is also capable of extremely high power pulse operation It

exhibits laser action at several infrared frequencies but none in the visible

spectrum Operating in a manner similar to the helium-neon laser it employs

an electric discharge for pumping using a percentage of nitrogen gas as a

pumping gas The COR2R laser is the most efficient laser capable of operating at

more than 30 efficiency

The carbon dioxide laser finds many applications in industry particularly for

welding and Cutting

6

3- Argon Laser

The argon ion laser can be operated as a continuous gas laser at about 25

different wavelengths in the visible between (4089 - 6861) nm but is best

known for its most efficient transitions in the green at 488 nm and 5145 nm

Operating at much higher powers than the Helium-Neon gas laser it is not

uncommon to achieve (30 ndash 100) watts of continuous power using several

transitions This output is produced in hot plasma and takes extremely high

power typically (9 ndash 12) kW so these are large and expensive devices

142 Solid Lasers

1 Ruby Laser

The ruby laser is the first type of laser actually constructed first

demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

aluminum oxide with a small amount (about 005) of Chromium which gives

it its characteristic pink or red color by absorbing green and blue light

The ruby laser is used as a pulsed laser producing red light at 6943 nm

After receiving a pumping flash from the flash tube the laser light emerges for

as long as the excited atoms persist in the ruby rod which is typically about a

millisecond

A pulsed ruby laser was used for the famous laser ranging experiment which

was conducted with a corner reflector placed on the Moon by the Apollo

astronauts This determined the distance to the Moon with an accuracy of

about 15 cm

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 10: Numerical Study of Laser Interaction With Solid Materials

5

Fig(15) The lasing action of He-Ne laser

Helium-Neon lasers are common in the introductory physics laboratories

but they can still be quite dangerous An unfocused 1-mW HeNe laser has a

brightness equal to sunshine on a clear day (01 wattcmP

2P) and is just as

dangerous to stare at directly

2- Carbon Dioxide Laser

The carbon dioxide gas laser is capable of continuous output powers above

10 kilowatts It is also capable of extremely high power pulse operation It

exhibits laser action at several infrared frequencies but none in the visible

spectrum Operating in a manner similar to the helium-neon laser it employs

an electric discharge for pumping using a percentage of nitrogen gas as a

pumping gas The COR2R laser is the most efficient laser capable of operating at

more than 30 efficiency

The carbon dioxide laser finds many applications in industry particularly for

welding and Cutting

6

3- Argon Laser

The argon ion laser can be operated as a continuous gas laser at about 25

different wavelengths in the visible between (4089 - 6861) nm but is best

known for its most efficient transitions in the green at 488 nm and 5145 nm

Operating at much higher powers than the Helium-Neon gas laser it is not

uncommon to achieve (30 ndash 100) watts of continuous power using several

transitions This output is produced in hot plasma and takes extremely high

power typically (9 ndash 12) kW so these are large and expensive devices

142 Solid Lasers

1 Ruby Laser

The ruby laser is the first type of laser actually constructed first

demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

aluminum oxide with a small amount (about 005) of Chromium which gives

it its characteristic pink or red color by absorbing green and blue light

The ruby laser is used as a pulsed laser producing red light at 6943 nm

After receiving a pumping flash from the flash tube the laser light emerges for

as long as the excited atoms persist in the ruby rod which is typically about a

millisecond

A pulsed ruby laser was used for the famous laser ranging experiment which

was conducted with a corner reflector placed on the Moon by the Apollo

astronauts This determined the distance to the Moon with an accuracy of

about 15 cm

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 11: Numerical Study of Laser Interaction With Solid Materials

6

3- Argon Laser

The argon ion laser can be operated as a continuous gas laser at about 25

different wavelengths in the visible between (4089 - 6861) nm but is best

known for its most efficient transitions in the green at 488 nm and 5145 nm

Operating at much higher powers than the Helium-Neon gas laser it is not

uncommon to achieve (30 ndash 100) watts of continuous power using several

transitions This output is produced in hot plasma and takes extremely high

power typically (9 ndash 12) kW so these are large and expensive devices

142 Solid Lasers

1 Ruby Laser

The ruby laser is the first type of laser actually constructed first

demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is

aluminum oxide with a small amount (about 005) of Chromium which gives

it its characteristic pink or red color by absorbing green and blue light

The ruby laser is used as a pulsed laser producing red light at 6943 nm

After receiving a pumping flash from the flash tube the laser light emerges for

as long as the excited atoms persist in the ruby rod which is typically about a

millisecond

A pulsed ruby laser was used for the famous laser ranging experiment which

was conducted with a corner reflector placed on the Moon by the Apollo

astronauts This determined the distance to the Moon with an accuracy of

about 15 cm

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 12: Numerical Study of Laser Interaction With Solid Materials

7

Fig (16) Principle of operation of a Ruby laser

2- Neodymium-YAG Laser

An example of a solid-state laser the neodymium-YAG uses the NdP

3+P ion to

dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet

geometry which makes population inversion possible Neodymium-YAG lasers

have become very important because they can be used to produce high

powers Such lasers have been constructed to produce over a kilowatt of

continuous laser power at 1065 nm and can achieve extremely high powers in

a pulsed mode

Neodymium-YAG lasers are used in pulse mode in laser oscillators for the

production of a series of very short pulses for research with femtosecond time

resolution

Fig(17) Construction of a Neodymium-YAG laser

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 13: Numerical Study of Laser Interaction With Solid Materials

8

3- Neodymium-Glass Lasers

Neodymium glass lasers have emerged as the design choice for research in

laser-initiated thermonuclear fusion These pulsed lasers generate pulses as

short as 10-12 seconds with peak powers of 109 kilowatts

143 Molecular Lasers

Eximer Lasers

Eximer is a shortened form of excited dimer denoting the fact that the

lasing medium in this type of laser is an excited diatomic molecule These

lasers typically produce ultraviolet pulses They are under investigation for use

in communicating with submarines by conversion to blue-green light and

pulsing from overhead satellites through sea water to submarines below

The eximers used are typically those formed by rare gases and halogens in

electron excited Gas discharges Molecules like XeF are stable only in their

excited states and quickly dissociate when they make the transition to their

ground state This makes possible large population inversions because the

ground state is depleted by this dissociation However the excited states are

very short-lived compared to other laser metastable states and lasers like the

XeF eximer laser require high pumping rates

Eximer lasers typically produce high power pulse outputs in the blue or

ultraviolet after excitation by fast electron-beam discharges

The rare-gas xenon and the highly active fluorine seem unlikely to form a

molecule but they do in the hot plasma environment of an electron-beam

initiated gas discharge They are only stable in their excited states if stable

can be used for molecules which undergo radioactive decay in 1 to 10

nanoseconds This is long enough to achieve pulsed laser action in the blue-

green over a band from 450 to 510 nm peaking at 486 nm Very high power

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 14: Numerical Study of Laser Interaction With Solid Materials

9

pulses can be achieved because the stimulated emission cross-sections of the

laser transitions are relatively low allowing a large population inversion to

build up The power is also enhanced by the fact that the ground state of XeF

quickly dissociates so that there is little absorption to quench the laser pulse

action

144 Free-Electron Lasers

The radiation from a free-electron laser is produced from free electrons

which are forced to oscillate in a regular fashion by an applied field They are

therefore more like synchrotron light sources or microwave tubes than like

other lasers They are able to produce highly coherent collimated radiation

over a wide range of frequencies The magnetic field arrangement which

produces the alternating field is commonly called a wiggler magnet

Fig(18) Principle of operation of Free-Electron laser

The free-electron laser is a highly tunable device which has been used to

generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of

this range they are the highest power source Applications of free-electron

lasers are envisioned in isotope separation plasma heating for nuclear fusion

long-range high resolution radar and particle acceleration in accelerators

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 15: Numerical Study of Laser Interaction With Solid Materials

10

15 Pulsed operation

Pulsed operation of lasers refers to any laser not classified as continuous

wave so that the optical power appears in pulses of some duration at some

repetition rate This encompasses a wide range of technologies addressing a

number of different motivations Some lasers are pulsed simply because they

cannot be run in continuous mode

In other cases the application requires the production of pulses having as

large an energy as possible Since the pulse energy is equal to the average

power divided by the repitition rate this goal can sometimes be satisfied by

lowering the rate of pulses so that more energy can be built up in between

pulses In laser ablation for example a small volume of material at the surface

of a work piece can be evaporated if it is heated in a very short time whereas

supplying the energy gradually would allow for the heat to be absorbed into

the bulk of the piece never attaining a sufficiently high temperature at a

particular point

Other applications rely on the peak pulse power (rather than the energy in

the pulse) especially in order to obtain nonlinear optical effects For a given

pulse energy this requires creating pulses of the shortest possible duration

utilizing techniques such as Q-switching

16 Heat and heat capacity

When a sample is heated meaning it receives thermal energy from an

external source some of the introduced heat is converted into kinetic energy

the rest to other forms of internal energy specific to the material The amount

converted into kinetic energy causes the temperature of the material to rise

The amount of the temperature increase depends on how much heat was

added the size of the sample the original temperature of the sample and on

how the heat was added The two obvious choices on how to add the heat are

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 16: Numerical Study of Laser Interaction With Solid Materials

11

to add it holding volume constant or to add it holding pressure constant

(There may be other choices but they will not concern us)

Lets assume for the moment that we are going to add heat to our sample

holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the

subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549

be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the

amount of material and the temperature In the limit where 119876119876119889119889 goes to zero

(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative

lim119876119876119889119889rarr0

119876119876119889119889∆120549120549119889119889

= 120597120597119876119876120597120597120549120549119889119889

= 119862119862119889119889 (11)

We have given this derivative the symbol 119862119862119889119889 and we call it the heat

capacity at constant volume Usually one quotes the molar heat capacity

119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899

(12)

We can rearrange Equation (11) as follows

119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)

Then we can integrate this equation to find the heat involved in a finite

change at constant volume

119876119876119889119889 = 119862119862119889119889

1205491205492

1205491205491

119889119889120549120549 (14)

If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes

out of the integral and the heat at constant volume becomes

119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)

Let us now go through the same sequence of steps except holding pressure

constant instead of volume Our initial definition of the heat capacity at

constant pressure 119862119862119875119875 R Rbecomes

lim119876119876119875119875rarr0

119876119876119875119875∆120549120549119875119875

= 120597120597119876119876120597120597120549120549119875119875

= 119862119862119875119875 (16)

The analogous molar heat capacity is

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 17: Numerical Study of Laser Interaction With Solid Materials

12

119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899

(17)

Equation (16) rearranges to

119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)

which integrates to give

119876119876119875119875 = 119862119862119875119875

1205491205492

1205491205491

119889119889120549120549 (19)

When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes

119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)

Very frequently the temperature range is large enough that 119862119862119875119875 cannot be

regarded as constant In these cases the heat capacity is fit to a polynomial (or

similar function) in 120549120549 For example some tables give the heat capacity as

119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)

where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-

dependent heat capacity the heat at constant pressure would integrate as

follows

119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)

1205491205492

1205491205491

119889119889120549120549 (112)

119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492

2 minus 12054912054912 + 119899119899

1205741205743

12054912054923 minus 1205491205491

3 (113)

Occasionally one finds a different form for the temperature dependent heat

capacity in the literature

119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)

When you do calculations with temperature dependent heat capacities you

must check to see which form is being used for 119862119862119875119875 We are using the

convention that 119876119876 will always designate heat absorbed by the system 119876119876 can

be positive or negative and the sign indicates which way heat is flowing If 119876119876 is

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 18: Numerical Study of Laser Interaction With Solid Materials

13

positive then heat was indeed absorbed by the system On the other hand if

119876119876 is negative it means that the system gave up heat to the surroundings

17 Thermal conductivity

In physics thermal conductivity 119896119896 is the property of a material that indicates

its ability to conduct heat It appears primarily in Fouriers Law for heat conduction

Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)

The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through

a piece of material The reciprocal of thermal conductivity is thermal

resistivity

18 Derivation in one dimension

The heat equation is derived from Fouriers law and conservation of energy

(Cannon 1984) By Fouriers law the flow rate of heat energy through a

surface is proportional to the negative temperature gradient across the

surface

119902119902 = minus119896119896 120571120571120549120549 (115)

where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one

dimension the gradient is an ordinary spatial derivative and so Fouriers law is

119902119902 = minus119896119896 120549120549119909119909 (116)

where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal

energy per unit volume in the material 120549120549119876119876 is proportional to the change in

temperature 120549120549120549120549 That is

∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)

where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the

material Choosing zero energy at absolute zero temperature this can be

rewritten as

∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 19: Numerical Study of Laser Interaction With Solid Materials

14

The increase in internal energy in a small spatial region of the material

(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is

given by

119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909

119909119909minus∆119909119909

= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591

119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(119)

Where the fundamental theorem of calculus was used Additionally with no

work done and absent any heat sources or sinks the change in internal energy

in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat

across the boundaries By Fouriers law this is

119896119896 120597120597120549120549120597120597119909119909

(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909

(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905

119905119905minus∆119905119905

= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591

119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

(120)

again by the fundamental theorem of calculus By conservation of energy

119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909

119909119909minus∆119909119909

119905119905+∆119905119905

119905119905minus∆119905119905

= 0 (121)

This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]

Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0

Which can be rewritten as

120549120549119905119905 =119896119896119888119888119875119875 120588120588

120549120549119909119909119909119909 (122)

or

120597120597120549120549120597120597119905119905

=119896119896119888119888119875119875 120588120588

12059712059721205491205491205971205971199091199092 (123)

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 20: Numerical Study of Laser Interaction With Solid Materials

15

which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal

diffusivity and is often denoted 120572120572

19 Aim of present work

The goal of this study is to estimate the solution of partial differential

equation that governs the laser-solid interaction using numerical methods

The solution will been restricted into one dimensional situation in which we

assume that both the laser power density and thermal properties are

functions of time and temperature respectively In this project we attempt to

investigate the laser interaction with both lead and copper materials by

predicting the temperature gradient with the depth of the metals

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 21: Numerical Study of Laser Interaction With Solid Materials

16

Chapter Two

Theoretical Aspects

21 Introduction

When a laser interacts with a solid surface a variety of processes can

occur We are mainly interested in the interaction of pulsed lasers with a

solid surface in first instance a metal When such a laser interacts with a

copper surface the laser energy will be transformed into heat The

temperature of the solid material will increase leading to melting and

evaporation of the solid material

The evaporated material (vapour atoms) will expand Depending on the

applications this can happen in vacuum (or very low pressure) or in a

background gas (helium argon air)

22 One dimension laser heating equation

In general the one dimension laser heating processes of opaque solid slab is

represented as

120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597

( 119870119870 119879119879120597120597 ) (21)

With boundary conditions and initial condition which represent the pre-

vaporization stage

minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907

minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 22: Numerical Study of Laser Interaction With Solid Materials

17

119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897

where

119870119870 represents the thermal conductivity

120588120588 represents the density

119862119862 represents the specific heat

119879119879 represents the temperature

119879119879infin represents the ambient temperature

119879119879119907119907 represents the front surface vaporization

120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab

Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes

119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)

With the same boundary conditions as in equation (22)

where 119889119889119889119889 = 119870119870120588120588119862119862

which represents the thermal diffusion

But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation

equation (21) with this assuming implies

119879119879119905119905 = 1

120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)

With the same boundary and initial conditions in equation (22) Where 119870119870

represents the derivative of K with respect the temperature

23 Numerical solution of Initial value problems

An immense number of analytical solutions for conduction heat-transfer

problems have been accumulated in literature over the past 100 years Even so

in many practical situations the geometry or boundary conditions are such that an

analytical solution has not been obtained at all or if the solution has been

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 23: Numerical Study of Laser Interaction With Solid Materials

18

developed it involves such a complex series solution that numerical evaluation

becomes exceedingly difficult For such situation the most fruitful approach to

the problem is numerical techniques the basic principles of which we shall

outline in this section

One way to guarantee accuracy in the solution of an initial values problems

(IVP) is to solve the problem twice using step sizes h and h2 and compare

answers at the mesh points corresponding to the larger step size But this requires

a significant amount of computation for the smaller step size and must be

repeated if it is determined that the agreement is not good enough

24 Finite Difference Method

The finite difference method is one of several techniques for obtaining

numerical solutions to differential equations In all numerical solutions the

continuous partial differential equation (PDE) is replaced with a discrete

approximation In this context the word discrete means that the numerical

solution is known only at a finite number of points in the physical domain The

number of those points can be selected by the user of the numerical method In

general increasing the number of points not only increases the resolution but

also the accuracy of the numerical solution

The discrete approximation results in a set of algebraic equations that are

evaluated for the values of the discrete unknowns

The mesh is the set of locations where the discrete solution is computed

These points are called nodes and if one were to draw lines between adjacent

nodes in the domain the resulting image would resemble a net or mesh Two key

parameters of the mesh are ∆120597120597 the local distance between adjacent points in

space and ∆119905119905 the local distance between adjacent time steps For the simple

examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 24: Numerical Study of Laser Interaction With Solid Materials

19

The core idea of the finite-difference method is to replace continuous

derivatives with so-called difference formulas that involve only the discrete

values associated with positions on the mesh

Applying the finite-difference method to a differential equation involves

replacing all derivatives with difference formulas In the heat equation there are

derivatives with respect to time and derivatives with respect to space Using

different combinations of mesh points in the difference formulas results in

different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the

numerical solution obtained with any useful scheme will approach the true

solution to the original differential equation However the rate at which the

numerical solution approaches the true solution varies with the scheme

241 First Order Forward Difference

Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894

empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971

+1205751205751205971205972

2 1205971205972empty1205971205971205971205972

1205971205971

+1205751205751205971205973

3 1205971205973empty1205971205971205971205973

1205971205971

+ ⋯ (25)

where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie

consider the value of empty at the location of the 120597120597119894119894+1 mesh line

empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (26)

Solve for (120597120597empty120597120597120597120597)120597120597119894119894

120597120597empty120597120597120597120597120597120597119894119894

=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (27)

Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right

hand side have been reduced by one

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 25: Numerical Study of Laser Interaction With Solid Materials

20

Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)

and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (28)

The mean value theorem can be used to replace the higher order derivatives

∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ =∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (29)

where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus

120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894

∆120597120597+∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (210)

120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894

∆120597120597asymp∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585 (211)

The term on the right hand side of previous equation is called the truncation

error of the finite difference approximation

In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also

unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the

truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in

analytical form) the big 119978119978 notation can be used to express the dependence of

the truncation error on the mesh spacing Note that the right hand side of last

equation contains the mesh parameter ∆120597120597 which is chosen by the person using

the finite difference simulation Since this is the only parameter under the users

control that determines the error the truncation error is simply written

∆1205971205972

2 1205971205972empty1205971205971205971205972

120585120585= 119978119978(∆1205971205972) (212)

The equals sign in this expression is true in the order of magnitude sense In

other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 26: Numerical Study of Laser Interaction With Solid Materials

21

equality Rather the expression means that the left hand side is a product of an

unknown constant and ∆1205971205972 Although the expression does not give us the exact

magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term

approaches zero as ∆120597120597 is reduced

Using big 119978119978 notation Equation (28) can be written

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597+ 119978119978(∆120597120597) (213)

This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because

it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a

truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under

our control because we can choose the mesh size ∆120597120597 The part of the truncation

error that is not under our control is |120597120597empty120597120597120597120597|120585120585

242 First Order Backward Difference

An alternative first order finite difference formula is obtained if the Taylor series

like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh

variables in place of all the unknowns one obtains

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (214)

Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894

to get

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894

∆120597120597minus∆1205971205972

1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (215)

Or using big 119978119978 notation

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894 minus empty119894119894minus1

∆120597120597+ 119978119978(∆120597120597) (216)

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 27: Numerical Study of Laser Interaction With Solid Materials

22

This is called the backward difference formula because it involves the values of

empty at 120597120597119894119894 and 120597120597119894119894minus1

The order of magnitude of the truncation error for the backward difference

approximation is the same as that of the forward difference approximation Can

we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller

truncation error The answer is yes

242 First Order Central Difference

Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1

empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

+∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (217)

empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+∆1205971205972

2 1205971205972empty1205971205971205971205972

120597120597119894119894

minus∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (218)

Subtracting Equation (10) from Equation (9) yields

empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894

+ 2∆1205971205973

3 1205971205973empty1205971205971205971205973

120597120597119894119894

+ ⋯ (219)

Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597minus∆1205971205972

3 1205971205973empty1205971205971205971205973

120597120597119894119894

minus ⋯ (220)

or

120597120597empty120597120597120597120597120597120597119894119894

=empty119894119894+1 minus empty119894119894minus1

2∆120597120597+ 119978119978(∆1205971205972) (221)

This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good

approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1

the truncation error for the central difference approximation goes to zero much

faster than the truncation error in forward and backward equations

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 28: Numerical Study of Laser Interaction With Solid Materials

23

25 Procedures

The simple case in this investigation was assuming the constant thermal

properties of the material First we assumed all the thermal properties of the

materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor

point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we

assume the constant energy after that the pulse of special shapes was selected

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) was investigated using Matlab program as shown in Appendix

The equation of thermal conductivity and specific heat capacity of metal as a

function of temperature was obtained by best fitting of polynomials using

tabulated data in references

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 29: Numerical Study of Laser Interaction With Solid Materials

24

Chapter Three

Results and Discursion

31 Introduction

The development of laser has been an exciting chapter in the history of

science and engineering It has produced a new type of advice with potential for

application in an extremely wide variety of fields Mach basic development in

lasers were occurred during last 35 years The lasers interaction with metal and

vaporize of metals due to itrsquos ability for welding cutting and drilling applicable

The status of laser development and application were still rather rudimentary

The light emitted by laser is electro magnetic radiation this radiation has a wave

nature the waves consists of vibrating electric and magnetic fields many studies

have tried to find and solve models of laser interactions Some researchers

proposed the mathematical model related to the laser - plasma interaction and

the others have developed an analytical model to study the temperature

distribution in Infrared optical materials heated by laser pulses Also an attempt

have made to study the interaction of nanosecond pulsed lasers with material

from point of view using experimental technique and theoretical approach of

dimensional analysis

In this study we have evaluate the solution of partial difference equation

(PDE) that represent the laser interaction with solid situation in one dimension

assuming that the power density of laser and thermal properties are functions

with time and temperature respectively

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 30: Numerical Study of Laser Interaction With Solid Materials

25

32 Numerical solution with constant laser power density and constant

thermal properties

First we have taken the lead metal (Pb) with thermal properties

119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870

119862119862 = 014016119869119869119892119892119870119870

120588120588 = 10751 1198921198921198981198981198981198983

119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870

119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870

and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860

represent the area under laser influence

The numerical solution of equation (23) with boundary and initial conditions

in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties

of lead metal by explicit method using Matlab program give us the results as

shown in Fig (31)

Fig(31) Depth dependence of the temperature with the laser power density

1198681198680 = 76 times 106 119882119882119898119898119898119898 2

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 31: Numerical Study of Laser Interaction With Solid Materials

26

33 Evaluation of function 119920119920(119957119957) of laser flux density

From following data that represent the energy (119869119869) with time (millie second)

Time 0 001 01 02 03 04 05 06 07 08

Energy 0 002 017 022 024 02 012 007 002 0

By using Matlab program the best polynomial with deduced from above data

was

119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984

minus 10069 1198981198985 (31)

As shown in Fig (32)

Fig(32) Laser energy as a function of time

Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The

normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the

maximum value (119864119864119898119898119907119907119898119898 )

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)

(119864119864119898119898119907119907119898119898 ) (32)

The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 32: Numerical Study of Laser Interaction With Solid Materials

27

Fig(33) Normalized laser energy as a function of time

The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must

equal to 3 (total laser energy) ie

119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (33)

Therefore there exist a real number 119875119875 such that

119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (34)

that implies 119875119875 = 68241 and

119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

08

00

119899119899119898119898 = 3 (35)

The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =

08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore

119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808

00

(36)

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 33: Numerical Study of Laser Interaction With Solid Materials

28

Where 119863119863119898119898 put to balance the units of equation (36)

But integral

119868119868 = 119864119864119860119860

(37)

and from equations (35) (36) and (37) we have

119868119868 (119898119898)11989911989911989811989808

00

= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

0800 119899119899119898119898

119860119860 119863119863119898119898 (38)

Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation

(38)

There fore

119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899

119860119860 119863119863119898119898 (39)

As shown in Fig(34) Matlab program was used to obtain the best polynomial

that agrees with result data

119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983

+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)

Fig(34) Time dependence of laser intensity

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 34: Numerical Study of Laser Interaction With Solid Materials

29

34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

constant thermal properties

With all constant thermal properties of lead metal as in article (23) and

119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in

equation (23) with boundary and initial condition as in equation (22) and the

depth penetration is shown in Fig(35)

Fig(35) Depth dependence of the temperature when laser intensity function

of time and constant thermal properties of Lead

35 Evaluation the Thermal Conductivity as functions of temperature

The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of

temperature for Lead material was obtained by Matlab program using the

experimental data tabulated in researches

119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920

times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 35: Numerical Study of Laser Interaction With Solid Materials

30

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (36)

Fig(36) The best fitting of thermal conductivity of Lead as a function of

temperature

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 36: Numerical Study of Laser Interaction With Solid Materials

31

36 Evaluation the Specific heat as functions of temperature

The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345

The best polynomial fitted for these data was

119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792

+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730

times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (37)

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 37: Numerical Study of Laser Interaction With Solid Materials

32

Fig(37) The best fitting of specific heat capacity of Lead as a function of

temperature

37 Evaluation the Density as functions of temperature

The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature

was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

300 11330 400 11230 500 11130 600 11010 800 10430

1000 10190 1200 9940

The best polynomial of this data was

120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793

minus 45158 times 10minus12 1198791198794 (313)

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 38: Numerical Study of Laser Interaction With Solid Materials

33

The density of Lead as a function of temperature and the best polynomial fitting

are shown in Fig (38)

Fig(38) The best fitting of density of Lead as a function of temperature

38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and

variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)

We have deduced the solution of equation (24) with initial and boundary

condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)

and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)

respectively then by using Matlab program the depth penetration is shown in

Fig (39)

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 39: Numerical Study of Laser Interaction With Solid Materials

34

Fig(39) Depth dependence of the temperature for pulse laser on Lead

material

39 Laser interaction with copper material

The same time dependence of laser intensity as shown in Fig(34) with

thermal properties of copper was used to calculate the temperature distribution as

a function of depth penetration

The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for

copper material was obtained from the experimental data tabulated in literary

The Matlab program used to obtain the best polynomial equation that agrees

with the above data

119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792

minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844

times 10minus17 1198791198795 (314)

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 40: Numerical Study of Laser Interaction With Solid Materials

35

119879119879 ( 119870119870) 119870119870 119869119869

119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5

100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332

The previous thermal conductivity data and the best fitting of the data are

shown in Fig (310)

Fig(310) The best fitting of thermal conductivity of Copper as a function of

temperature

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 41: Numerical Study of Laser Interaction With Solid Materials

36

The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper

material was obtained from the following experimental data tacked from

literatures

119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)

100 0254

200 0357

273 0384

298 0387

400 0397

600 0416

800 0435

1000 0454

1100 0464

1200 0474

1300 0483

The best polynomial fitted for these data was

119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792

+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895

times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)

The specific heat capacity data and the best polynomial fitting of the data are

shown in Fig (311)

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 42: Numerical Study of Laser Interaction With Solid Materials

37

Fig(311) The best fitting of specific heat capacity of Copper as a function of

temperature

The density of copper 120588120588(119879119879) as a function of temperature tacked from

literature was used to find the best polynomial fitting

119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)

100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686

1000 8576 1100 8519 1200 8458 1300 8396

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 43: Numerical Study of Laser Interaction With Solid Materials

38

The best polynomial of this data was

120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793

minus 76765 times 10minus14 1198791198794 (316)

The density of copper as a function of temperature and the best polynomial

fitting are shown in Fig (312)

Fig(312) The best fitting of density of copper as a function of temperature

The depth penetration of laser energy for copper metal was calculated using

the polynomial equations of thermal conductivity specific heat capacity and

density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature

(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a

function of time the result was shown in Fig (313)

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 44: Numerical Study of Laser Interaction With Solid Materials

39

The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862

for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of

copper metal so the depth penetration of laser energy of lead metal was smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

Fig(313) Depth dependence of the temperature for pulse laser on Copper

material

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 45: Numerical Study of Laser Interaction With Solid Materials

40

310 Conclusions

The Depth dependence of temperature for lead metal was investigated in two

case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the

thermal properties (thermal conductivity specific heat) and density of metal are

also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity

vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity

specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =

120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the

penetration depth in the first case is smaller than that of the second case about

(190) times

The temperature distribution as a function of depth dependence for copper

metal was also investigated in the case when the laser intensity vary with time

119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and

density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)

The depth penetration of laser energy of lead metal was found to be smaller

than that of copper metal this may be due to the high thermal conductivity and

high specific heat capacity of copper with that of lead metal

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 46: Numerical Study of Laser Interaction With Solid Materials

41

References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp

Sons Inc Hoboken New Jersey Canada (2003)

[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)

[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd

[9]

Conference of Pure amp Applied Sciences (11-12) March p47 (2009)

[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)

[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)

[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)

[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)

[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)

httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml

[10] httpwwwworldoflaserscomlaserprincipleshtm

[11] httpenwikipediaorgwikiLaserPulsed_operation

[12] httpenwikipediaorgwikiThermal_conductivity

[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension

[14] httpwebh01uaacbeplasmapageslaser-ablationhtml

[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 47: Numerical Study of Laser Interaction With Solid Materials

42

Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 48: Numerical Study of Laser Interaction With Solid Materials

43

E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off

This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off

This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off

This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 49: Numerical Study of Laser Interaction With Solid Materials

44

title(Dencity as a function of temperature) hold off

This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 50: Numerical Study of Laser Interaction With Solid Materials

45

elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 51: Numerical Study of Laser Interaction With Solid Materials

46

r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 52: Numerical Study of Laser Interaction With Solid Materials

47

for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 53: Numerical Study of Laser Interaction With Solid Materials

48

for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 54: Numerical Study of Laser Interaction With Solid Materials

49

6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 55: Numerical Study of Laser Interaction With Solid Materials

50

u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 56: Numerical Study of Laser Interaction With Solid Materials

51

alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser
Page 57: Numerical Study of Laser Interaction With Solid Materials

52

715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))

  • 01 Title
    • 0TDr Omed Ghareb Abdullah
      • Ch1-Laser
        • 16 Heat and heat capacity
        • 18 Derivation in one dimension
          • Ch2 laser2
          • Ch3 laser2
          • Appendix-Laser