1. Heat flux through the wall

2. between two semi-infinite solids
Interfacial contact
between two semi-infinite solids
TA,i
kA, rA, cA Ts
kB, rB, cB
TB,i
Boundary and interfacial conditions

3. The interface temperature is not function of time.
Ex) A: man, B: wood (pine) or steel (AISI 1302)
Assume
wood:
steel:

4. Objects with Constant Surface Temperatures
Utilization of solution to convection boundary condition
As Bi → ∞,
Ts,2 → T∞

5. Semi-Infinite Solid
In dimensionless form
Plane wall, Cylinder, and Sphere for Bi → ∞

6. Summary of transient heat transfer results for constant surface temperature cases

8. Objects with Constant Surface Heat Fluxes
Summary of transient heat transfer results for constant surface heat flux cases

10. Example 5.7 Cancer treatment by laser heating using nanoshells
1) Prior to treatment, antibodies are attached to the nanoscale particles.
2) Doped particles are then injected into the patient’s bloodstream and distributed throughout the body.
3) The antibodies are attracted to malignant sites, and therefore carry and adhere the nanoshells only to cancerous tissue.
4) A laser beam penetrates through the tissue between the skin and the cancer, is absorbed by the nanoshells, and, in turn, heats and destroys the cancerous tissues.

11. Known:
Size of a small sphere
Thermal conductivity (k), reflectivity (r), and extinction coefficient (k) of tissue
Depth of sphere below the surface of the skin

12. laser heat flux
Find:
Heat transfer rate from the tumor to the surrounding healthy tissue for a steady-state treatment temperature of Tt,ss = 55ºC at the surface of the tumor.
Laser power needed to sustain the tumor surface temperature at Tt,ss = 55ºC.
Time for the tumor to reach Tt = 52ºC when heat transfer to the surrounding tissue is neglected. Water property can be used.
Time for the tumor to reach Tt = 52ºC when heat transfer to the surrounding is considered and the thermal mass of the tumor is neglected.

13. laser heat flux
Assumptions:
1D conduction in the radial direction.
Constant properties.
Healthy tissue can be treated as an infinite medium.
The treated tumor absorbs all irradiation incident from the laser.
Lumped capacitance behavior for the tumor.
Neglect potential nanoscale heat transfer effects.
Neglect the effect of perfusion.

14. q
(Case 12 of Table 4.1)
1. Steady-state heat loss q from the tumor

15. laser heat flux
projected area of the tumor:
2. Laser power Pl,
Energy balance : heat transfer rate from tumor = absorbed laser energy

16. laser heat flux
3. Time for the tumor to reach Tt = 52ºC when heat transfer to the surrounding tissue is neglected.

17. 4. Time for the tumor to reach Tt = 52ºC when heat transfer to the surrounding is considered and thermal mass of the tumor is neglected.
Heat transfer between a sphere and an exterior infinite medium subjected to constant heat flux
By trial and error,

18. Periodic Heating Oscillating surface temperature
thermal penetration depth
(reduction of temperature amplitude by 90% relative to that of surface)
Quasi-steady state temperature distribution
Surface heat flux

19. Sinusoidal heating by a strip
C1: depends on thermal contact resistance at interface
between heater strip and underlying material

20. Finite Difference Method
Numerical Method
Finite Difference Method Dt
p-1 p
p+1
in time Dx Dy
(m,n)
(m,n+1)
(m-1,n)
(m+1,n)
(m,n-1)
in space
truncation error: O(Dt)
first order accuracy in time

21. Explicit Method (Euler Method) : forward difference
stability criterion:
If the system is one-dimensional in x, or or
stability criterion:

22. one-dimensional in x, t
p = 5
p = 4
p = 3
p = 2
p = 1
p = 0 x
m = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

23. Boundary node subjected to convection1 1 2 3
qconv,in
qcond,out
See Table 5.3
(p. 306)
stability criterion: or

25. Example 5.9 m + 1 m - 1 m qcond,in qcond,out L = 10 mm qcond,in
Fuel element:
Steady operation
Sudden change to
Symmetry adiabat
Coolant 1 2 3 4 5
m + 1
m - 1 m 5 4
qcond,in
qcond,out
L = 10 mm
qcond,in
qconv,out
Dx = 2 mm
Find:
Temperature distribution at 1.5 s after a change in operating
power by using the explicit finite difference method

26. m + 1 m - 1 m qcond,in qcond,out Dx Thus, Symmetry adiabat1 2 3 4 5
For node 0, set 5 4
qcond,in
qconv,out
Dx/2
For node 5, or

27. Dt: stability criterionor
Thus,
choose Dt = 0.3 s
Then,

28. nodal equations
Initial distribution: steady-state solution with

29. Calculated nodal temperatures

30. Comments:
Expanding the finite difference solution, the new steady-state
condition may be determined.

31. Implicit Method (fully) : backward difference
stability criterion : no restriction
qconv,in
qcond,out 1
If the system is one-dimensional in x,
Boundary node subjected to convection

32. Explicit Method
one-dimensional in x, t
p = 5
p = 4
p = 3
p = 2
p = 1
p = 0 x
m = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

33. Implicit Method (fully)
one-dimensional in x, t
p = 5
p = 4
p = 3
p = 2
p = 1
p = 0 x
m = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

34. Crank – Nicolson Method
Second order accuracy in time, but serious stability problem

35. backward difference :
forward difference:
averaging:
stability criterion: or
explicit method:

36. Example 5.10 very thick slab of copper 1 2 3 4 m + 1 m - 1 m1 2 3 4
m + 1
m - 1 m
Dx = 75 mm 1
Dx/2
qcond,in
qrad,in
qcond,out
qcond,out
T(x,0) = 20°C
Find:
Using the explicit FDM, determine temperature at the surface and 150 mm from the surface after 2 min, T(0, 2 min), T(150 mm, 2 min)
Repeat the calculations using the implicit FDM.
Determine the same temperatures analytically.

37. Determination of nodal points
T(x,2 min) 20 x
150
300
Table A.1, copper (300 K) :

38. 2 min → p = 5 Explicit FDM m + 1 m - 1 m qcond,in qcond,out Dx = 75 mm
qrad,in
qcond,out
Dx/2
node 0: or
interior nodes:
time step: stability criterion
Table A.1, copper (300 K) :
2 min → p = 5

39. finite-difference equations
and
After 2 min,
and

40. Improvement of the accuracy
domain length: 600 mm
After 2 min,
and
When Dt = 24 s,
and

41. Implicit FDM
node 0: or
Arbitrarily choosing,
interior nodes:
A set of nine equations must be solved simultaneously for
each time increment.
The equations are in the form [A][T]=[C].

42. [A][T] = [C]

43. After 2 min,
and

44. Analytical Solution
Comparison

45. Implicit method with Dx = 18
Implicit method with Dx = mm (37 nodalpoints) and Dt = 6 s (Fo = 2.0)
exact:

46. Conduction (Chap.2 – Chap5.)
2. Introduction
Fourier law, Newton’s law of cooling
Heat diffusion equation & boundary conditions
3. One-Dimensional Steady-State Conduction
Without heat generation: electric network analogy
With heat generation
Extended surfaces
4. Two-Dimensional Steady-State Conduction
Analytical method: Separation of variables
Conduction shape factor
Numerical method
5. Transient Conduction
Lumped capacitance method: Bi and Fo numbers
Analytical method: separation of variables
Semi-infinite solid: similarity solution