bruce mayer, pe registered electrical & mechanical engineer bmayer@chabotcollege

[email protected] • ENGR-45_Lec-07_Diffusion_Fick-2.ppt1

Bruce Mayer, PE Engineering-45: Materials of Engineering

Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 45

Solid Solid StateState

Diffusion-Diffusion-22



Learning Goals - DiffusionLearning Goals - Diffusion

How Diffusion Proceeds How Diffusion Can be Used in Material

Processing How to Predict The Rate Of Diffusion

Be Predicted For Some Simple Cases• Fick’s first and SECOND Laws

How Diffusion Depends On Structure And Temperature



Recall Fick’s FIRST Law.Recall Fick’s FIRST Law. Fick’s 1st Law

Position, x

Cu flux Ni flux

x

C

dx

dCDJ

• Where– J Flux in kg/m2•s

or at/m2•s– dC/dx = Concentration

GRADIENT in units of kg/m4 or at/m4

– D Proportionality Constant (Diffusion Coefficient) in m2/s

In the SteadyState Case J = const• So dC/dx = const

– For all x & t

Thus for ANY two points j & k

jk

jk

xx

CC

x

C

dx

dC

Con

cen

., C



NONSteady-State DiffusionNONSteady-State Diffusion In The Steady Case

tfC

tfJ

In The NONSteady, or Transient, Case the Physical Conditions Require

txfdxdCCJ ,,,

Liquid Source Vapor Bubble Saturation Transient (Calculated)

0.0

0.2

0.4

0.6

0.8

1.0

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

Radial Position Inside Bubble, r (mm)

Vap

or

Satu

rati

on

Fra

cti

on

, v

Pv(r,t) (t=0.01)

Pv(r,t) (t=0.04 s)

Pv(r,t) (t=0.10 s)

Pv(r,t) (t=0.15 s)

Pv(r,t) (t=0.2 s)

Pv(r,t) (t=0.3 s)

Pv(r,t) (t=0.35 s)

file = BubPv(t)2.xls

• Bubble Diameter = 4 mm

• Dv = 5 mm2/s

Increasing Time

222 /

01

sin121, ov rtDno

n

n

v errn

rrntr

• In The Above Concen-vs-Position Plot Note how, at x 1.5 mm, Both C and dC/dx CHANGEwith Time



NONSteady State Diffusion MathNONSteady State Diffusion Math Consider the

Situation at Right Box Dimensions

• Width = x

• Height = 1 m

• Depth = 1 m– Into the slide

Box Volume, V = x•1•1 = x

Now if x is small

xCCC

C rightleft

2

avg

• Can Approximate C(x) as

Concentration,C, in the Box

J(right)J(left)

x

The Amount of Matl in the box, M

33

mm

kgVC M



NONSteady State Diffusion contNONSteady State Diffusion cont or

Material ENTERING the Box in time t

rightleft JJ

For NONsteady Conditions


J(right)J(left)

dx

So Matl ACCUMULATES in the Box

xCxC M 11

tJJ

tAreaJM

leftleft

leftin

11

Material LEAVING the Box in time t

tJJ

tAreaJM

rightright

rightout

11



NONSteady State Diffusion cont.2NONSteady State Diffusion cont.2 So the NET Matl

Accumulation

Adding (or Subtracting) Matl From the Box CHANGES C(x)

xxx

rightleft

x

CD

x

CD

t

Cx

ortJJCx11

With V = 1•1•x


J(right)J(left)

x

tJJM rightleft

CVMV

MC

V

M

V

MCCC oldnew

Partials Req’d asC = C(x,t)



NONSteady State Diffusion cont.3NONSteady State Diffusion cont.3 In Summary for

CONSTANT D

Now, And this is CRITICAL, by TAYLOR’S SERIES

xxx x

C

x

CD

t

Cx Concentration,

C, in the Box

J(right)J(left)

x

x

x

xCxCxC

xxxx

/

so

xxx

x

x

C

x

Cx

x

xCD

t

Cx



NONSteady State Diffusion cont.4NONSteady State Diffusion cont.4 After Canceling

Now for very short t


J(right)J(left)

x

Finally Fick’s SECOND LAW for Constant Diffusion Coefficient Conditions

xx

x

x

CD

x

xCD

t

C2

2

2

2

2

2

0lim

x

CD

x

CD

t

C

t

C

xt

2

2

x

CD

t

C



Comments of Fick’s 2nd LawComments of Fick’s 2nd Law The Formal

Statement

This Leads to the GENERAL, and much more Complicated, Version of the 2nd Law


J(right)J(left)

x

2

2

x

CD

t

C

This Assumes That D is Constant, i.e.;

txCDD , In many Cases

Changes in C also Change D

x

CD

xt

C



Example – NonSS DiffusionExample – NonSS Diffusion Example: Cu Diffusing into a Long Al Bar

The Copper Concentration vs x & t

pre-existing conc., Co of copper atoms

Surface conc., Cs of Cu atoms bar

Co

Cs

position, x

C(x,t)

tot1

t2t3

The General Soln is Gauss’s Error Function, “erf”

Dt

xerf

CC

CtxC

oS

o

21

,



Comments on the erfComments on the erf Gauss's Defining

Eqn

dyezerfz y

0

22

z is just a NUMBER

• Thus the erf is a (hard to evaluate) DEFINITE Integral

Treat the erf as any other special Fcn

Some Special Fcns with Which you are Familiar: sin, cos, ln, tanh• These Fcns used to

be listed in printed Tables, but are now built into Calculators and MATLAB

See Text Tab 5.1 for Table of erf(z)



Comments on the erf cont.Comments on the erf cont. 1-erf(z) appears So

Often in Physics That it is Given its Own Name, The COMPLEMENTARY Error Function:

dyezerfcz y

0

221

Recall The erfc

Diffusion solution

Notice the Denom in this Eqn

Dt

xerfc

CC

CtxC

oS

o

4

,

dLDt 4 This Qty has SI

Units of meters, and is called the “Diffusion Length”• The Natural Scaling

Factor in the efrc



SemiInfite Slab Diffusion vs Time

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

x (arbitrary units)

Rel

ativ

e C

on

cen

tati

on

ErfC t1; Ld = 0.2236

ErfC t2; Ld = 0.3873

ErfC t3; Ld = 0.6708

ErfC t4; Ld = 1.1619

ErfC t5; Ld = 2.0125

file = erfc_0401.xls

TIME

PARAMETERS• D = 0.05 (arb Units)



Example Example D = f(T) D = f(T) Given Cu Diffusing

into an Al Bar At given point in the

bar, x0, The Copper Concentration reaches the Desired value after 10hrs at 600 °C• The Processing

Recipe

Get a New Firing Furnace that is Only rated to 1000 °F = 538 °C• To Be Safe, Set the

New Fnce to 500 °C

Need to Find the NEW Processing TIME for 500 °C to yield the desired C(x0) ChrsxC 600@100



Example Example D = f(T) cont D = f(T) cont Recall the erf

Diffusion Eqn

For this Eqn to be True, need Equal Denoms in the erf

Dt

xerf

CC

CtxC

oS

o

41

,

Since CS and Co have NOT changed, Need

oS

oC

oS

oC

CC

CtxC

CC

ChrsxC

50006000 ,10,

Since by the erf

tD

xerf

hrsD

xerf

500

0

600

0

41

1041

500

600

500600

10

10

D

hrsDt

or

tDhrsD



Example Example D = f(T) cont.2 D = f(T) cont.2 Now Need to

Find D(T) As With Xtal Pt-

Defects, D Follows an Arrhenius Rln

– Qd Arrhenius Activation Energy in J/mol or eV/at

– R Gas Constant = 8.31 J/mol-K = 8.62x10-5 eV/at-K

– T Temperature in K

Find D0 and Qd from Tab 5.2 in Text• For Cu in Al

– D0 = 6.5x10-5 m2/s

– Qd = 136 kJ/mol

• Where– D0 Temperature

INdependent Exponential PreFactor in m2/s

RT

QDD dexp0



Example Example D = f(T) cont.3 D = f(T) cont.3 Thus D(T) for

Cu in Al Thus for the new

500 °C Recipe

In this Case• D600 = 4.692x10-13 m2/s

• D500 = 4.152x10-14 m2/s

Now Recall the Problem Solution

TKmolJ

molJD

/31.8

/136000exp105.6 5

500

600 10

D

hrsDt

hrshrs

t 0.113152.4

1092.46

This is 10x LONGER than Before; Should have bought a 600C fnce



Find D Arrhenius ParametersFind D Arrhenius Parameters Recall The D(T) Rln

Applied to the D(T) Relation

Take the Natural Log of this Eqn

TR

QDD d 1

lnln 0

RT

QDD dexp0

This takes the form of the slope-intercept Line Eqn:

mxby

0ln

1

lny

Db

Tx

R-Qm

D

d



Find D(T) Parameters contFind D(T) Parameters cont And, Since TWO

Points Define a Line If We Know D(T1) and D(T2) We can calc• D0

• Qd

Quick Example• D(T) For Cu in Au at

Upper Right

Slope, m = y/x

x

y

x = (1.1-0.8)x1000/K

• = 0.0003 K-1

y = ln(3.55x10-16) − ln(4x10-13) = −7.023



Find D(T) Parameters cont.2Find D(T) Parameters cont.2 By The Linear Form

in the (x,y) format• x1 = 0.0008

• y1 = ln(4x10-13) = −28.55

So b

J/mol194500

0003.0023.731.8

d

d

d

d

Q

Q

xyRmRQ

orRQm

Now, the intercept, b

11 mxyb

orbmxy

Pick (D,1/T) pt as• (4x10-13,0.8)

819.9

0008.00003.0

023.755.28

b

b

Finally D0

smD

eeD b

250

819.90

1044.5



Diff vs. Structure & PropertiesDiff vs. Structure & Properties Faster Diffusion for

• Open crystal structures

• Lower melting Temp materials

• Materials with secondary bonding

• Smaller diffusing atoms

• Cations• Lower density

materials

Slower Diffusion for• Close-packed

structures• Higher melting Temp

materials• Materials with

covalent bonding• Larger diffusing

atoms• Anions• Higher density

materials



Diffusion SummarizedDiffusion Summarized Phenomenon: Mass Transport In Solids Mechanisms

• Vacancy InterChange by KickOut

• Interstitial “squeezing”

Governing Equations• Fick's First Law

• Fick's Second Law

Diffusion coefficient, D• Affect of Temperature

• Qd & D0

– How to Determine them from D(T) Data



WhiteBoard WorkWhiteBoard Work Problem 5.28

• Ni Transient Diffusion into CuProb 5.28: Ni Diffusion into Cu

0%

1%

2%

3%

4%

5%

0.0 0.5 1.0 1.5 2.0 2.5 3.0

x into Cu (mm)

CN

i (W

t% N

i in

Cu

)

file = erfc_0401.xls

500hrs @ 1000 ° 500hrs @ T2

Same Concentration

bruce mayer, pe registered electrical & mechanical engineer bmayer@chabotcollege

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