1. TRANSIENT CONDUCTION Lumped Thermal Capacitance Method Analytical Method: Separation of Variables Semi-Infinite Solid: Similarity Solution Numerical Method: Finite Difference Method

2. Lumped Thermal Capacitance Method negligible spatial effect volume V surface area A s

3. excess temperature: initial condition:

4. thermal time constant : convection resistance : lumped thermal capacitance Thermocouple?

5. Transient temperature response of lumped capacitance solids for different thermal time constant  t

6. Seebeck effect and Peltier effect AB T2T2 T1T1 V + - Seebeck effect (1821) AB ↑ I Peltier effect (1834) T2T2 T1T1

7. volume V surface area A s total energy transfer in time t

8. Validation of Lumped Capacitance Method Bi : Biot number

9. Transient temperature distributions for different Biot numbers in a plane wall symmetrically cooled by convection

10. When spatial effect is negligible. L c : characteristic length Fo : Fourier number(dimensionless time)

11. Find: Total time t t required for the two-step process Assumption: Thermal resistance of epoxy is negligible. Example 5.3 Step 1: HeatingStep 2: Cooling t = 0 T i,o = 25°C t = t c t = t e t = t t 300 s T c = 150°C T e = ?°C T t = 37°C heating cooling curing aluminum T(0) = T i,o = 25°C T(t t ) = 37°C epoxy  = 0.8 2L = 3 mm

12. Epoxy Biot numbers for the heating and cooling processes Aluminum: Thus, lumped capacitance approximation can be applied. q conv q rad T(t)T(t)

13. Heating process Curing process Cooling process t c = 124 s T e = 175°C t t = 989 s t = 0 T i,o = 25°C t = t c t = t e t = t t 300 s T c = 150°C T e = ?°C T t = 37°C heating cooling curing Numerical solutions

14. Total time for the two-step process : Intermediate times :

15. Plane wall with convection Separation of Variables LL x i.c. b.c. Analytical Method

16. Dimensional analysis dimensionless variables Fo : Fourier number, Bi : Biot number L: m [L]

17. Equation in dimensionless form

18. initial condition boundary conditions

19. Drop out * for convenience afterwards

20. boundary conditions

21. b.c.

22. initial condition

23. Approximate solution at x = 0, Bi = 1.0 Fo = 0.1Fo = 1  1 1.03930.5339  2 -0.0469-1.22 10 -5  3 0.00074.7 10 -20 When

24. See Table 5.1 Approximate solution When

26. total energy transfer (net out-going) maximum amount of energy transfer

27. Radial systems with convection i.c. b.c. r or roro

28. dimensionless variables Drop out * for convenience afterwards i.c. b.c. or

31. Since

32. initial condition

33. Approximate solution Total energy transfer (net out-going) Since V =  L, dV = 2  rdrL

34. Assumption: Pipe wall can be approximated as plane wall, since L << D. Example 5.4 oil insulation Steel, AISI 1010 Find: 1) Biot and Fourier numbers after 8 min 2) Temperature of exterior pipe surface after 8 min, T(0, 8 min) 3) Heat flux to the wall at 8 min, (8 min) 4) Energy transferred to pipe per unit length after 8 min,

35. oil

36. 1) Bi and Fo at t = 8 min With Bi = 0.313, the lumped capacitance method is inappropriate. However, since Fo > 0.2, approximate solution can be applicable. 2) T(0, 8 min) AISI 1010: from Table 5.1

37. 4) The energy transfer to the pipe wall over the 8-min interval 3)

38. i.c. b.c. TiTi x TsTs similarity solution similarity variable 1 1 x 1 =  (t 1 )x 2 =  (t 2 ) Semi-Infinite Solid: Similarity Solution

39. Scaling analysis TsTs TiTi x x =  (t) Let

42. merge into one i.c. b.c. TsTs TiTi x x =  (t)

43. Similarity solution integrating factor or

44. erfc : complimentary error function error function:

45. TsTs TiTi x 1 =  (t 1 ) x 2 =  (t 2 ) x