1. Appendices 10.A & 10.B: An Educational Presentation
Presented By:
Joseph Ash
Jordan Baldwin
Justin Hirt
Andrea Lance

2. History of Heat Conduction
Jean Baptiste Biot
( )
French Physicist
Worked on analysis of heat conduction
Unsuccessful at dealing with the problem of incorporating external convection effects in heat conduction analysis

3. 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 of
temperature difference

4. 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 dimensionless
quantity

5. Derivation of the Heat Conduction Equation
A first approximation of the equations that govern the conduction of heat in a solid rod.

6. 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.

7. The final consideration can be expressed as the following:
is a proportionality factor called the thermal conductivity and is determined by material properties

8. 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.

9. The differential equation governing the temperature of the bar is a physical balance between two rates:
Flux/Flow term
Absorption term

10. 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)

11. 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:

12. 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:

13. 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 material
ρ = density

14. 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.

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

16. 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

17. 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.

18. 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 is
where h1 is a proportionality constant
if h1=0, then it corresponds to an insulated end
if h1 goes to infinity, then the end is held at 0 temp.

19. 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

20. Initial Boundary Condition
Finally, the temperature distribution at one fixed instant – usually taken at t = 0, takes the form:
occurring throughout the bar

21. 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

22. 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

23. 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:

24. Generalizations
Putting the source and sink equations together in the heat equation yields
which is commonly called the generalized heat conduction equation

25. 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:

26. Example 1: Section 10.6, Problem 9
Let an aluminum rod of length 20 cm be initially at the uniform temperature 25C. Suppose that at time t=0, the end x=0 is cooled to 0C while the end x=20 is heated to 60C, and both are thereafter maintained at those temperatures.
Find the temperature distribution in the rod at any time t

27. Example 1: Section 10.6, Problem 9
Find the temperature distribution, u(x,t)
2uxx=ut, 0<x<20, t<0
u(0,t)=0 u(20,t)=60, t<0
u(x,0)=25, <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

28. Example 1: Section 10.6, Problem 9
Using Equations 16 and 17 found on page 614, we find that
where

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

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

31. Example 1: Section 10.6, Problem 9

32. 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

33. Schematic of String in Tension

34. Equation Derivation
Since 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

35. 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

36. Derivation Continued
For small motions of the string, it is approximated that
using the substitution that
the wave equation takes its customary form of

37. Wave Equation Generalizations
The telegraph equation
where c and k are nonnegative constants
cut arises from a viscous damping force
ku arises from an elastic restoring force
F(x,t) arises from an external force
The differences between this telegraph equation and the customary
wave equation are due to the consideration of internal elastic
forces. This equation also governs flow of voltage or current in a
transmission line, where the coefficients are related to the electrical
parameters in the line.

38. Wave Equations in Additional Dimensions
For 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

39. Example 2: Section 10.7, Problem 6
Consider 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.

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

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

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

43. THE END