1. 1 CHAPTER 4 TRANSIENT CONDUCTION Neglect spatial variation: 4.1.1 Criterion for Neglecting Spatial Temperature Variation Cooling of wire by surface convection: (4.1) = temperature drop across radius 4.1Simplified Model:Lumped-Capacity Method

2. 2 (1) radius  (2) conductivity k (3) heat transfer coefficient h Define Biot number Bi Factors affecting Neglect for small (4.2) Neglect if (4.3) < 0.1

3. 3 Irregular shape: Physical significance of Bi : Determining Long cylinder: Large plate of thickness L, convection on both sides: (4.4)

4. 4 Volume: V Determine: Transient temperature Assume: Conservation of energy: 4.1.2 Lumped-Capacity Analysis Surface area: Initial temperature: Convection at surface: Energy generation: (a) (1.6)

5. 5 (c), (d) and (e) into (b) (b) Assume that no energy is added Neglect radiation, is by convection (c) (d) (e) Energy generation Rate of energy change

6. 6 Separating variables and rearranging Equation (4.5) is the lumped-capacity equation for all geometries Limited to: (2) No radiation (3) Incompressible material Initial condition: (4.5) (1) Uniform Bi < 0.1 (4) (4.6)

7. 7 Integrate (4.5), use (4.6) NOTE (1) Temperature decay is exponential (2) Solution is independent of k Integration of (4.5): Assume: (5) Constant h and (6) Constant (3) Steady state temperature:set in (4.7): (4.7)

8. 8 Physical significance of (4.8) (4.8) Special Cases: Case (i): No energy generation. Set in (4.7) (4.9) Steady state: Case (ii): No convection. Set h =0 in (4.7)

9. 9 Physical significance:No Steady state (4.10) Case (iii): Initial and ambient temperatures are the same. Set in (4.7) (4.11) Steady state: setting in (4.11) gives (4.8)

10. 10 Bi > 0.1 Governing equation: PDE Solution by separation of variables Two HBC and one initial condition Example 4.1: Plate with Surface Convection Thickness: 2L Initial temperature: Assumef(x) is symmetrical L x L 0 Fig. 4.1  T h  T h HBC2 4.2Transient Conduction in Plates Heat exchange is by convection Determine

11. 11 (1) Observations Temperature is symmetrical Convection BC is NH (2) Origin and Coordinates (3) Formulation (i) Assumptions. (ii) Governing Equations. Let One-dimensional, constant k,h, (a) (4.12)

12. 12 (iii) Independent Variable with Two HBC: x -variable (iv) Boundary and Initial Conditions L x L 0 Fig. 4.1  T h  T h HBC2 (1), H IC (3), NH (4) Solution (i) Assumed Product Solution, H (2)

13. 13 (b) into (4.12) (b) (c) (d) (ii) Selecting the Sign of the Terms (e)

14. 14 (iii) Solutions to the ODE (f) yields (g) (h) (iv) Application of Boundary and Initial Conditions BC (1): A n = 0 (i)

15. 15 where Substituting into (b) IC (3): BC (2): gives (4.13) (j) (4.14) (k)

16. 16 (v) Orthogonality Eq. (3.6): in equation (k) are solutions to (e). Comparing (e) with eq. (3.5b) and p = w = 1 and q = 0 BC at x = 0 and x = L are homogeneous. w(x) = 1. Multiply both sides of (k) by integrate and apply orthogonality. Therefore,are orthogonal with respect to (4.15)

17. 17 Special case: Uniform initial temperature i T (5) Checking Dimensional check : Limiting check Plate is initially at  T Differential Equation Boundary and initial conditions (6) Comments (ii) Initial temperature f(x) need not be symmetrical (4.16) (i) Both separated equationshave term

18. 18 Example 4.2: Plate with Energy Generation, Specified Surface Temperature Thickness = L 4.3 Non-homogeneous Equations and Boundary Conditions Initial temperature: introduce At Surfaces at and Determine: Fig. 4.2 2 T 1 T L q x 0

19. 19 (1) Observations Asymmetry x-variable has two NHBC The heat equation is non-homogeneous (2) Origin and Coordinates (3) Formulation (i) Assumptions. (ii) Governing Equations: eq. (1.8) One-dimensional, constant (4.17)

20. 20 (iii) Independent Variable with Two HBC: x-variable (iv) Boundary and Initial Conditions Fig. 4.2 2 T 1 T L q x 0 (1) (2) (3) (4) Solution (a)

21. 21 Split (b). Let (a) into BC (1) (a) Into eq. (4.17) (b) (c) (d)

22. 22 (c-1) (d-1) (a) into BC (2) IC (3) Solution to (d) (c-2) (c-3) (e)

23. 23 (i) Assumed Product Solution (f) into (c), separating variables (f) (g) (ii) Selecting the Sign of the Two Terms (i) (h)

24. 24 (iii) Solutions to the Ordinary Differential Equations (j) for (k) (l) (iv) Application of Boundary and Initial Conditions BC (c-1)

25. 25 BC (c-2) Substituting into (f) and summing (m) (n) Solution to (o) NH initial condition (c-3): Return to BC (d-1) and (d-2) give and

26. 26 (v) Orthogonality (p) in (p) are solutions (i). Compare (i) with eq. (3.5a): Sturm- Liouville equation with Multiply both sides of (p) by apply orthogonality, gives integrate and (q)

27. 27 (o) into (q), evaluate integrals and use (m) Complete solution (r) (s)

28. 28 (5) Checking Dimensional check Limiting check: Temperature at steady state (6) Comments (i) Non-uniform initial temperature (ii) Non-homogeneous boundary conditions

29. 29 4.4 Transient Conduction in Cylinders Example 4.3: Cylinder with Energy Generation Long cylinder Surface cooling by convection  Th,  Th, r 0 q q Fig. 4.3 2 HBC Initial temperature Energy generation Determine (1) Observations Symmetry

30. 30 Heat equation is non-homogeneous (2) Origin and Coordinates (3) Formulation (i) Assumptions (4) Negligible end effect (ii) Governing Equations makes convection BC homogeneous (1) One-dimensional, (2) Uniform h and (3) Constant

31. 31 Heat equation (1.11) (iii) Independent Variable with 2 HBC: r-variable (iv) Boundary Conditions (4.18) or (1) finite  Th,  Th, r 0 q q Fig. 4.3 2 HBC

32. 32 (4) Solution (2) IC (3) (a) (a) into eq. (4.18) (b) Split (b)

33. 33 (a) into BC (1) Let (c) (d) (c-1) or finite

34. 34 BC (2): Initial condition Integrating (d) (c-2) (d-1) (d-2) (c-3) (e)

35. 35 (f) into (c), separating variables (i) Assumed Product Solution (f) (g) (h) (ii) Selecting the Sign of theTerms (i)

36. 36 (iii) Solutions to the ODE (j) For (k) (l) (iv) Application of Boundary and Initial Conditions BC (c-1) and (c-2)

37. 37 (k) and (l) into (f), summing Initial condition (c-3) and (m) (n) Return to solution (e) for BC (d-1) and (d-2) give 1 C and 2 C (o)

38. 38 (p) (v) Orthogonality Eq. (3.6) gives Functions in (p) are solutions to (i). Comparing (i) with eq. (3.5a) shows that it is a Sturm-Liouville equation with and HBC at and thereforeare orthogonal with respect to and

39. 39 Multiply both sidesof (p) by integrate and apply orthogonality (q) (o) into (q) and evaluate the integrals (r)

40. 40 Complete solution (4.19) (5) Checking Dimensional check Limiting check: (ii) Steady state (i) and

41. 41 (6) Comments Initial temperature (i) (ii) Two parameters: the Biot number Bi and 4.5 Transient Conduction in Spheres Example 4.4: Sphere with Surface Convection Determine: transient temperature Fig. 4.4 HBC2  Th, r o r 0 Initial temperature Convection at surface

42. 42 (1) Observations (2) Origin and Coordinates Convection BC is non-homogeneous One-dimensional, transient Define to give r -variable 2HBC (3) Formulation (i) Assumptions (1) One-dimensional (2) Constant (ii) Governing Equations

43. 43 Eq. (1.13) gives (iii) Independent Variable with 2 HBC: r-variable (iv) Boundary and Initial Conditions (a) (4.20) or (1) finite Fig. 4.4 HBC2  Th, r o r 0

44. 44 (4) Solution (i) Assumed Product Solution (b) into eq. (4.20), separate variables (2) IC (3) (b) (c) (d)

45. 45 (iii) Solutions to the ODE Use eqs. (2.32) and (2.33) (ii) Selecting the Sign of theTerms (e) (f) gives (g)

46. 46 Solution to (f) (iv) Application of Boundary and Initial Conditions BC (1) BC (2) (g) and (h) into (b) (h) k = 1, 2, … (i)

47. 47 (4.21) IC (3): (j) (v) Orthogonality The characteristic functions in (j) are it is a Sturm-Liouville problem with solutions to (e). Comparing (e) with eq. (3.5a) shows that

48. 48 are orthogonal with respect to Multiply (j) by orthogonality integrate and invoke (5) Checking Dimensional check (k)

49. 49 (ii) Steady state (6) Comments Limiting check: (i) Initial temperature = (i) Non-uniform initial temperature, (ii) Solution is expressed in terms of a single parameter: Biot number Bi

50. 50 Examples: Solar heating Reentry aerodynamic heating Reciprocating surface friction Periodic oscillation of temperature or flux Limitation: Linear equations 4.6 Temperature Dependent Boundary Conditions: Duhamel’s Superposition Integral

51. 51 Duhamel’s superposition integral Use solution of an auxiliary problem with constant boundary condition to construct the solution to the same problem with time dependent condition 4.6.1 Formulation of Duhamel’s Integral Initial temperature: boundary is at 01 T 02 T T 1  0102 TT  t 11.4.Fig

52. 52 Solution Decompose into two problems: One starts and the second atEach problem has a constant BC Solution to auxiliary problem, constant surface Temperature of magnitude unity. Thus (b) (a) Second problem starts at with surface temperature Solution to

53. 53 Adding (b) and (c) Generalize to arbitrarily boundary condition (c) (d) = surface temperature, or =ambient temperature, or =heat flux to many problems, each having a small step change, in B.C. Solution: superimpose solutions

54. 54 solution BC for first problem: F(0) starts at time t = 0, (e) Contribution of the i th problem, BC is Adding all solutions (f)

55. 55 or Integration by parts with (4.29) (4.28)

56. 56 NOTE: Duhamel’s method applies to linear equations Equations (4.28) and (4.29) apply to any coordinate system Define If equal to unity Auxiliary solution: is based on a constant BC Method applies to: (1) (2) Lumped-capacity models Integrals in (4.28) and (4.29) are with respect to the dummy variable

57. 57 4.6.2 Extension to Discontinuous Boundary Conditions and t a t )( 1 tF )( 2 tF )(tF Fig. 4.13 If is discontinuous, modify Duhamel’s integral. usein (4.28) & (4.29) modify (4.28) and (4.29) and with discontinuity at

58. 58 Result: For (4.28) add another solution, Integration by parts withgives (4.30)

59. 59 (4.31) Extension to several discontinuities

60. 60 4.6.3 Applications Example 4.5: Plate with Time Dependent Surface Temperature BC and IC: (1) 00  ),(tT(2) AttftLT  )(),( (3)  )0,(xT 0  i T Determine),(txT (1) Observations 0 x xAT  0  T 0)0,(  xT L Fig. 4.14 One BC depends on time 0  i T

61. 61 (2) Origin and Coordinates (3) Formulation (i)Assumptions (1) 1-D (ii) Governing equation (iii) Boundary and Initial conditions (a) (1) (2) constant

62. 62 0 x xAT  0  T 0)0,(  xT L Fig. 4.14 (2) (3) (4) Solution Eq.(a) is linear, applyDuhamel’s integral,eq.(4.28) solution to problem with BC (2) replaced by (4.28)

63. 63 Use resultof Example 4.2: set (b) where (c) Findand (d) and

64. 64 Substituting (b) and (d) intoeq. (4.28) (e) Evaluating the integrals in (e) (f)

65. 65 (5) Checking (i) A = 0 (ii)   t Dimensional check Limiting check: Example 4.6: Lumped-Capacity Method with Time Dependent Ambient Temperature Metal foil, Initial temperature Oven temperature starts at and changes with time as

66. 66 (1) Observations (a) Determine: Oven temperature is time dependent Use Duhamel’s method if lumped-capacity equation is linear appears in DE (2) Origin and Coordinates (3) Formulation

67. 67 (ii) Governing Equation Define (i)Assumptions (1) Bi < 0.1 (2) constant Set in (4.5) (a) (b) Oven temperature (c)

68. 68 Substitute into (a) Initial condition (4) Solution Apply Duhamel’s integral (d) (e) (4.28)

69. 69 Constant oven temperature equal to unity: Rewrite (a) = solution to auxiliary problem of constant oven temperature equal to unity and zero initial temperature (f) (g) (h) = time dependent oven temperature

70. 70 Initial condition (i) Integrate and use IC (j) Determine (k) Substituting (j) and (k) intoeq. (4.28), use

71. 71 (l) Performing the integration (5) Checking Dimensional check (m) Limiting check:

72. 72 Differential equation check Initial condition check: 4.7 Conduction in Semi-infinite Regions: The Similarity Method Limitations of separation of variables: 2 HBC separated by a finite distance Method fails for semi-infinite regions Alternate approach: Similarity method Combine two independent variables into oneand transform PDE to ODE Basic idea:

73. 73 (i) Semi-infinite region Limitations: (ii) Limitations on BC and IC Fig. 4.15 o T 0 x  i T Example: Semi-infinite plate, uniform initial Surface is suddenly temperature maintained at temperature (a) BC (1) (2) IC (3)

74. 74 Define: (b) Substitute into (a) BC and IC become (c) Fig. 4.15 o T 0 x  i T (1) (2) (3)

75. 75 Assume that x and t can be combined into a single variable (d) Try: Conditions onMust transform PDE, BC and IC in terms of and eliminate x and t. (4.32) Using (4.32), construct the derivatives in (c)

76. 76 (c) becomes (4.33)

77. 77 NOTE: (1) x and t are eliminated and replaced by (2) Solution to PDE (c) must satisfy 3 conditions. (3) Solution to ODE (4.33) can only satisfy two conditions. Transformation of BC and IC: Using transforms to

78. 78 BC and IC transform to: NOTE: One BC and IC coalesce. Fig. 4.15 o T 0 x  i T (1) (2) (3) Solution to (4.33): Separating variables Integrating

79. 79 or Integrating and using first BC (e) Define (f) Where

80. 80 and Equation (e) becomes BC (2) gives B = -1. (g) becomes (g) (4.33a) (4.33b) Derivative of (4.34)

81. 81