Appendices 10.A & 10.B:An Educational

PresentationPresented By:

Joseph AshJordan Baldwin

Justin HirtAndrea Lance

History of Heat Conduction

Jean Baptiste Biot (1774-1862)

French Physicist Worked on analysis of

heat conduction Unsuccessful at dealing

with the problem of incorporating external convection effects in heat conduction analysis

History of Heat Conduction

Jean Baptiste Joseph Fourier (1768 – 1830) Read Biot’s work 1807 determined how to solve the

problem Fourier’s Law

Time rate of heat flow (Q) through a slab is proportional to the gradient oftemperature difference

History of Heat Conduction

Ernst Schmidt German scientist Pioneer in Engineering

Thermodynamics Published paper “Graphical Difference

Method for Unsteady Heat Conduction” First to measure velocity and

temperature field in free convection boundary layer and large heat transfer coefficients

Schmidt Number Analogy between heat and mass transfer that causes a dimensionlessquantity

Derivation of the Heat Conduction

EquationA first approximation of the equations that govern the conduction of heat in a solid rod.

Consider the following:

A uniform rod is insulated on both lateral ends. Heat can now only flow in the axial direction.

It is proven that heat per unit time will pass from the warmer section to the cooler one.

The amount of heat is proportional to the area, A, and to the temperature difference T2-T1, and is inversely proportional to the separation distance, d.

The final consideration can be expressed as the following:

is a proportionality factor called the thermal conductivity and is determined by material properties

Assumptions

The bar has a length L so x=0 and x=L Perfectly insulated Temperature, u, depends only on position, x,

and time, t Usually valid when the lateral dimensions are

small compared to the total length.

The differential equation governing the temperature of the bar is a physical balance between two rates:

Flux/Flow term Absorption term

Flux

The instantaneous rate of heat transfer from left to right across the cross sections x=x0 where x0 is arbitrary can be defined as:

The negative is needed in order to show a positive rate from left to right (hot to cold)

Flux

Similarly, the instantaneous rate of heat transfer from right to left across the cross section x=x0+Δx where Δx is small can be defined as:

Flux

The amount of heat entering the bar in a time span of Δt is found by subtracting the previous two equations and then multiplying the result by Δt:

Heat Absorption

The average change in temperature, Δu, can be written in terms of the heat introduced, Q Δt and the mass Δm of the element as:

where s = specific heat of the materialρ = density

Heat Absorption

The actual temperature change of the bar is simply the actual change in temperature at some intermediate point, so the above equation can also be written as:

This is the heat absorption equation.

Heat Equation

Equating the QΔt in the flux and absorption terms, we find the heat absorption equation to be:

If we divide the above equation by ΔxΔt and allow both Δx and Δt to both go to 0, we will obtain the heat conduction or diffusion equation:

where

and has the dimensions of length^2/time and called the thermal diffusivity

Boundary Conditions

Certain boundary conditions may apply to the specific heat conduction problem, for example: If one end is maintained at some constant

temperature value, then the boundary condition for that end is u = T.

If one end is perfectly insulated, then the boundary condition stipulates ux = 0.

Generalized Boundary Conditions Consider the end where x=0 and the rate of flow of

heat is proportional to the temperature at the end of the bar. Recall that the rate of flow will be given, from left to right, as

With this said, the rate of heat flow out of the bar from right to left will be

Therefore, the boundary condition at x=0 iswhere h1 is a proportionality constant

if h1=0, then it corresponds to an insulated endif h1 goes to infinity, then the end is held at 0

temp.

Generalized Boundary Conditions Similarly, if heat flow occurs at the end x = L, then the

boundary condition is as follows:

where, again, h2 is a nonzero proportionality factor

Initial Boundary Condition

Finally, the temperature distribution at one fixed instant – usually taken at t = 0, takes the form:

occurring throughout the bar

Generalizations

Sometimes, the thermal conductivity, density, specific heat, or area may change as the axial position changes. The rate of heat transfer under such conditions at x=x0 is now:

The heat equation then becomes a partial differential equation in the form:

or

Generalizations

Other ways for heat to enter or leave a bar must also be taken into consideration.

Assume G(x,t,u) is a rate per unit per time. Source

G(x,t,u) is added to the bar G(x,t,u) is positive, non-zero, linear, and u does not depend on t G(x,t,u) must be added to the left side of the heat equation

yielding the following differential equation

Generalizations

Similarly, Sink

G(x,t,u) is subtracted from the bar G(x,t,u) is positive, non-zero, linear, and u does not

depend on t G(x,t,u) then under this sink condition takes the

form:

Generalizations

Putting the source and sink equations together in the heat equation yields

which is commonly called the generalized heat conduction equation

Multi-dimensional space

Now consider a bar in which the temperature is a function of more than just the axial x-direction. Then the heat conduction equation can then be written: 2-D:

3-D:

Example 1: Section 10.6, Problem 9

Let an aluminum rod of length 20 cm be initially at the uniform temperature 25C. Suppose that at time t=0, the end x=0 is cooled to 0C while the end x=20 is heated to 60C, and both are thereafter maintained at those temperatures.

Find the temperature distribution in the rod at any time t

Example 1: Section 10.6, Problem 9Find the temperature distribution, u(x,t)

2uxx=ut, 0<x<20, t<0u(0,t)=0 u(20,t)=60, t<0u(x,0)=25, 0<x<20

From the initial equation we find that:L=20, T1=0, T2=60, f(x)=25

We look up the Thermal Diffusivity of aluminum→2=0.86

Example 1: Section 10.6, Problem 9Using Equations 16 and 17 found on page 614, we

find that

where

1112 sin, 2

222

n

Ltn

n LxnecT

LxTTtxu

L

n dxLxnT

LxTTxf

Lc

0 112 sin2

Example 1: Section 10.6, Problem 9Evaluating cn, we find that

nnc

nnnnnc

dxxnxc

n

n

L

n

50cos70

5sin12cos710

20sin0

2006025

202

2

0

Example 1: Section 10.6, Problem 9Now we can solve for u(x,t)

1

40086.0

1

2086.0

20sin50cos703,

20sin50cos700

20060,

2

2

222

n

tn

n

tn

xnennxtxu

xnennxtxu

Example 1: Section 10.6, Problem 9

Derivation of the Wave Equation

Applicable for:

• One space dimension, transverse vibrations on elastic string

• Endpoints at x = 0 and x = L along the x-axis

• Set in motion at t = 0 and then left undisturbed

Schematic of String in Tension

Equation DerivationSince there is no acceleration in the horizontal direction

However the vertical components must satisfy

where is the coordinate to the center of mass and the weight is neglected

Replacing T with V the and rearranging the equation becomes

0cos),()cos(),( txTtxxT

),(sin),()sin(),( txxutxTtxxT tt

x

),(),(),(

txux

txVtxxVtt

Derivation continued

Letting , the equation becomes

To express this in terms of only terms of u we note that

The resulting equation in terms of u is

and since H(t) is not dependant on x the resulting equation is

0x

),(),( txutxV ttx

),()(tan)(),( txutHtHtxV x

ttxx uHu )(

ttxx uHu

Derivation Continued

For small motions of the string, it is approximated that

using the substitution that

the wave equation takes its customary form of

TTH cos

/2 Ta

ttxx uua 2

Wave Equation Generalizations

The telegraph equation

where c and k are nonnegative constantscut arises from a viscous damping forceku arises from an elastic restoring forceF(x,t) arises from an external force

The differences between this telegraph equation and the customarywave equation are due to the consideration of internal elasticforces. This equation also governs flow of voltage or current in atransmission line, where the coefficients are related to the electricalparameters in the line.

),(2 txFuakucuu xxttt

Wave Equations in Additional DimensionsFor a vibrating system with more than on significant space

coordinate it may be necessary to consider the wave equation in more than one dimension.

For two dimensions the wave equation becomes

For three dimensions the wave equation becomes

ttyyxx uuua )(2

ttzzyyxx uuuua )(2

Example 2: Section 10.7, Problem 6Consider an elastic string of length L whose ends

are held fixed. The string is set in motion from its equilibrium position with an initial velocity g(x). Let L=10 and a=1. Find the string displacement for any time t.

,4,1

,4

LxL

Lx

xg

LxL

LxL

Lx

43

43

4

40

Example 2: Section 10.7, Problem 6From equations 35 and 36 on page 631, we find

that

where

1

sinsin,n

n Latn

Lxnktxu

L

n dxLxnxg

Lk

Lan

0sin2

Example 2: Section 10.7, Problem 6Solving for kn, we find:

4sin

43sin8

sin4

sin43sin42

sin4sinsin42

3

2

40

43

4 43

nnnaLk

nnnnL

ank

dxLxn

LxLdx

Lxndx

Lxn

Lx

Lk

Lan

n

n

L L

L

LLn

Example 2: Section 10.7, Problem 6Now we can solve for u(x,t)

133

133

13

10sin

10sin

4sin

43sin180,

sinsin4

sin43sin18,

sinsin4

sin43sin8,

n

n

n

tnxnnnn

txu

Latn

Lxnnn

nLtxu

Latn

Lxnnn

naLtxu

THE END