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TRANSIENT CONDUCTION

TRANSIENT CONDUCTION. Lumped Thermal Capacitance Method Analytical Method: Separation of Variables Semi-Infinite Solid: Similarity Solution Numerical Method: Finite Difference Method. volume V. surface area A s. Lumped Thermal Capacitance Method. negligible spatial effect.

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TRANSIENT CONDUCTION

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

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

  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 tt

  6. T2 T2 A B B A + ↑ V I - T1 T1 Seebeck effect (1821) Peltier effect (1834) Seebeck effect and Peltier effect

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

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

  11. Step 1: Heating Step 2: Cooling Te = ?°C Ti,o = 25°C Tc = 150°C Tt = 37°C curing cooling heating 300 s t = 0 t =tc t = te t = tt Example 5.3 2L = 3 mm epoxy e = 0.8 aluminumT(0) = Ti,o = 25°C T(tt) = 37°C Find: Total time tt required for the two-step process Assumption: Thermal resistance of epoxy is negligible.

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

  13. Te = ?°C Ti,o = 25°C Tc = 150°C Tt = 37°C curing cooling heating 300 s t = 0 t = tc t = te t = tt Numerical solutions Heating process tc = 124 s Curing process Te= 175°C Cooling process tt = 989 s

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

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

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

  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. When Approximate solution at x = 0, Bi = 1.0 Fo = 0.1 Fo = 1 q1 1.0393 0.5339 q2 -0.0469 -1.22 10-5 q3 0.0007 4.7 10-20

  24. When See Table 5.1 Approximate solution

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

  26. or Radial systems with convection ro r i.c. b.c.

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

  28. Since

  29. initial condition

  30. Approximate solution Total energy transfer (net out-going) Since V = pL, dV = 2prdrL

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

  32. oil

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

  34. 3) 4) The energy transfer to the pipe wall over the 8-min interval

  35. Ts x1 = d(t1) x2 = d(t2) 1 1 Semi-Infinite Solid: Similarity Solution i.c. b.c. Ti x similarity solution similarity variable

  36. Let Scaling analysis Ts x Ti x = d(t)

  37. i.c. b.c. merge into one Ts x Ti x = d(t)

  38. or Similarity solution integrating factor

  39. error function: erfc: complimentary error function

  40. Ti x Ts x2 = d(t2) x1 = d(t1)

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