1. 1 CHAPTER 9 PERTURBATION SOLUTIONS 9.1 Introduction Objective Definitions Perturbation quantity Basic Problem To construct an approximate solution to a problem using an exact solution to a slightly different problem

2. 2 Regular Perturbation Problem Singular Perturbation Problem Parameter Perturbation Coordinate Perturbation Asymptotic Expansion Example (9.1) First term: n = 0, corresponds to the zero-order solution Second term: n = 1, represents the first-order solution, etc.

3. 3 9.2 Solution Procedure (1) Identification of the Perturbation Quantity: Non- dimensional formulation (2) Introduction of an Asymptotic Solution Example (9.2) Substitute eq. (9.2), into the governing equations and boundary conditions Equate terms of identical powers of (3) Formulation of theproblems

4. 4 Generate n-order problems (4) Solutions The n-order problems are solved consecutively 9.3 Examples of Perturbation Problems in Conduction Perturbation of the heat equation, boundary conditions or both (1) Cylinder with Non-circular Inner Radius Fig. 9.1 i R a  a  (9.3) Two-dimensional conduction

5. 5 For the departure of a is small from a circle of radius Basic problem: constant inner radius one-dimensional (2) Tapered Plate Example: Linearly tapered plate (9.4) For small departure from a rectangle Basic problem: rectangle

6. 6 (3) Cylinder with Eccentric Inner Radius e = eccentricity perturbation parameter (9.5) 2-D conduction For small departurefrom concentric circles Basic problem: concentric circle, one- dimensional

7. 7 (4) Surface Radiation Non-linear problems Small radiation compared to convection: Define (9.6) Basic problem: radiation is neglected (5) Conduction with Phase Change Non-linear problems Example of perturbation parameter: Stefan number = ratio of the sensible heat to latent heat, eq. (6.13)

8. 8 (9.7) Basic problem: represents a quasi-steady state model (6) Temperature Dependent Properties Non-linear problems Small variation in properties: Example: temperature dependent conductivity. Define (9.8)

9. 9 9.4 Perturbation Solutions: Examples Example 9.1: Transient Conduction with Surface Radiation and Convection Thin foil Surface area Initially at Cooling is by radiation and convection Ambient and surroundings are at zero kelvin Heat transfer coefficient is Surface emissivity is e Biot < 0.1

10. 10 Radiation is small compared to convection Identify appropriate perturbation parameter Obtain a second-order perturbation solution (1) Observations Use lumped-capacity method Perturbation solution for small radiation compared to convection Perturbation parameter: Ratio of radiation to convection (2) Formulation

11. 11 (i) Assumptions (1) Biot << 0.1 (2) Simplified radiation model, eq. (1.19) (3) Constant properties (4) No energy generation (5) Ambient and surroundings are at zero kelvin (ii) Governing Equations Conservation of energy, eq. (1.6) (1.6) (a)

12. 12 (b) (c) Substitute (b) and (c) into (a) (d) Initial condition (e) (iii) Identification of the Perturbation Parameter Dimensionless form: Define

13. 13 (f) Substitute into (d) and (e) (g) (h) Coefficient of the first term: Ratio of radiation to convection Define (9.9)

14. 14 Substitute (9.9) into (g) (9.10) (iv) Asymptotic Solution Assume (9.2) (v) Formulation of theproblems Substitute (9.2) into (9.10)

15. 15 Expand the first term and retain terms of order (i) Equate terms of identical powers of (j-0)(j-1)

16. 16 (j-2) Initial conditions: Substitute eq. (9.2) into (h) Equate terms of identical powers of (k-0) (k-1)

17. 17 (j-2) NOTE: Zero-order problem, (j-0): Foil exchanging heat by convection only The problem is linearized Governing equations: First order differential equations First-order problem, (j-1): Radiation is approximated by Second-order problem, (j-2): Radiation is approximated by Solutions to the n-order problems must proceed consecutively

18. 18 (3) Solutions Zero-order solution: Solution to (j-0) with initial condition (k-0) (l-0) First-order solution: Substitute (l-0) into (j-1) Solving and using initial condition (k-1) (l-1) Second-order solution: Substitute (l-0) and (l-1) into (j-2)

19. 19 Solving and using initial condition (k-2) (l-2) Substitute (l-0), (l-1) and (l-2) into eq. (9.2) (m) Rearrange (9.11)

20. 20 (4) Checking Limiting check: (i) Setin eq. (9.11) gives the correct solution to cooling with no radiation Initial condition check: Ateq. (9.11) gives (5) Comments Exact solution to eq. (9.10): separate variablesand and integrate (ii) Atthe foil should be ator Set in (9.11) gives

21. 21 Integrate (9.12) Good agreement with perturbation solution Does perturbation solution converge? Rewrite eq. (9.12) using binomial expansion

22. 22 (9.12a) Compare eq. (9.11) with eq. (9.12a): The first three terms are identical. This indicates convergence. Example 9.2: Variable Thermal Conductivity Steady state one-dimensional conduction in a wall of thickness L. One side is at, the opposite side is at Variable conductivity:

23. 23 (9.13) Obtain a second-order perturbation solution Identify an appropriate perturbation parameter (1) Observations Non-linearity due to Perturbation solution for (i) Assumptions (2) Formulation (1) Steady state (2) One-dimensional

24. 24 (3) (ii) Governing Equations (9.13) (iii) Boundary Conditions (1) (2)

25. 25 Dimensionless form: Define Substitute into eq. (9.14) Perturbation parameter (a) (iv) Identification of the Perturbation Parameter

26. 26 Equation (9.14) becomes (9.15) Eq. (9.15) is non-linear B.C. (1) (2) (v) Asymptotic Solution (9.2) (vi) Formulation of theproblems

27. 27 Substitute (9.2) into (9.15) Differentiate and expand

28. 28 Equate terms (b-0)(b-1)(b-2)

29. 29 Substitute eq. (9.2) into the first B.C. Equate terms of identical powers of (c-0) (c-1) (c-2) Second B.C.:

30. 30 Equate terms of identical powers of (d-0) (d-1) (d-2) NOTE (i) Zero-order problem, (b-0): One-dimensional, constant conductivity (ii) First-order problem, (b-1): Conductivity variation is approximated by the solution to the zero-order problem

31. 31 (iii) Second order problem, (b-2): Conductivity is approximated by the zero and first-order solutions (iv) (b-1) and (b-2) are linear (v) Solutions to the n-order problems must proceed consecutively (3) Solutions Zero-order solution: Solution to (b-0) with initial conditions (c-0) and (d-0) First-order solution: Substitute (e-0) into (b-1) (e-0)

32. 32 Second-order solution: Substitute (e-0) and (e-1) into (b-2) Integrate and use boundary conditions (c-1) and (d-1) (e-1) Integrate and apply boundary conditions (c-2) and (d-2) (e-2) Substituting (e-0), (e-1) and (e-2) into eq. (9.2) (9.15)

33. 33 (4) Checking Limiting check: Settingin eq. (9.16) gives the correct solution for constant conductivity Boundary Conditions check: Eq. (9.16) satisfies the boundary conditions at and (5) Comments Exact solution: By direct integration of eq. (9.15) and using the two B.C.

34. 34 (f) Accuracy: Table 9.1 Table 9.1 and at % Error 00.5 0.0 0.10.512 0.002 0.20.523 0.038 0.30.533 0.113 0.40.54 0.541 0.258 0.50.545 0.55 0.473 Excellent agreement

35. 35 Example 9.3: Fin with Convection and Radiation Heat transfer by convection and radiation Ambient fluid and surroundings are: Surface emissivity is e Constant area semi- infinite fin, radius The base is at Small radiation compared to convection

36. 36 Identify an appropriate perturbation parameter Obtain a second-order perturbation solution for the temperature distribution and fin heat transfer rate (1) Observations Perturbation solution: Small radiation compared to convection Perturbation parameter: Ratio of radiation to convection Non-linear problem NOTE: Exact solution to fin heat transfer rate: presented in Section 7.7

37. 37 (2) Formulation (i) Assumptions (1) Steady state (2) Biot << 0.1 (3) Simplified radiation model of eq. (1.19) (4) Constant properties (5) Ambient fluid and surroundings are at zero degree kelvin (ii) Governing Equations Setin eq. (7.16)

38. 38 (9.17) cross section area perimeter (iii) Boundary Conditions (1) (2) (iv) Identification of the Perturbation Parameter

39. 39 Dimensionless form: Define (a) Substitute (a) into eq. (9.17) (b) Introduce the Biot number (c) Coefficient ofterm: Ratio of radiation to convection at base.

40. 40 Define: (9.18) Substitute (c) and eq. (9.18) into (b) (9.19) Boundary conditions (1) (2) (v) Asymptotic Solution

41. 41 Assume (9.2) (vi) Formulation of theproblems Substitute eq. (9.2) into (9.19) (d) Expand the fourth power term

42. 42 Equate terms of identical powers of (e-0)(e-1)

43. 43 (e-2) Substitute eq. (9.2) into boundary condition (1) Equate terms of identical powers of (f-0) (f-1) (f-2)

44. 44 Boundary condition (2): Equate terms of identical powers of (g-0) (g-1) (g-2) NOTE: (i) Zero-order problem, (e-0): Fin exchanging heat by convection only

45. 45 (ii) First-order problem, (e-1):replaces (iii) Second-order problem, (e-2):replaces (iv) Non-linear Eq. (9.19) is replaced by linear problems (3) Solutions Zero-order solution: Solution to (e-0) with boundary conditions (f-0) and (g-0) is First-order solution: Substitute (h-0) into (e-1) (h-0)

46. 46 Solution is (Appendix A) (h-1) Second-order solution: Substitute (h-0) and (h-1) into (e-2) Solution is (Appendix A) (h-2)

47. 47 Substitute (h-0), (h-1) and (h-2) into eq. (9.2) (9.20) Fin heat transfer rate: Apply Fourier’s law at (i) or (j)

48. 48 Substitute eq. (9.20) into (j) (9.21) (4) Checking Limiting check (i) No radiation: Settingin eq. (9.21) gives the correct solution (5) Comments (ii) If any of the quantities A c, kandvanishes, q must also vanish. Eq. (9.21) satisfies this condition

49. 49 Exact solution: Obtained in Section 7.7. Set eq. (7.18): in Introduceand express in dimensionless form Accuracy: Table 9.2 compares the two solutions Excellent agreement Convergence: Binomial expansion of eq. (9.22) Table 9.2 0 1.0 0.2 0.99997 0.4 0.9998 0.6 0.9993 0.8 0.9985 1.0 0.9973 (9.22)

50. 50 (9.22a) Compare eq. (9.22a) with (9.21) (9.21) Example 9.4: Conduction with Phase Change: Stefan’s Problem Semi-infinite liquid region Initially at the fusion temperature Boundary atis suddenly fixed at

51. 51 Small Stefan number Obtain a second-order perturbation solution for the interface (1) Observations Use the Stefan number as the perturbation parameter Small Ste = small sensible heat compared to latent heat Non-linear problem Exact solution: Given in Section 6.7 Liquid phase remains at (2) Formulation

52. 52 (i) Assumptions (1) Constant properties (2) One-dimensional conduction (3) Semi-infinite region (ii) Governing Equations : The heat equation is (a) (iii) Boundary and Initial Conditions (1) (2)

53. 53 (4) Dimensionless form: Follow Section 6.5. Define: Substitute (b) into (a) and B.C. (1-4) (iv) Identification of the Perturbation Parameter (3) (b)

54. 54 (c) (1) (2) (3) (4) (v) Asymptotic Solution Assume:

55. 55 (d) (e) (vi) Formulation of theproblems Substitute(d) into (c) (f) Equate terms of identical powers of (g-0)

56. 56 (g-1) (g-2) First B.C. Equate terms of identical powers of (h-0) (h-1)

57. 57 (h-2) NOTE: B.C. (2) becomes (i) B.C. (2) and (3) are at which depends on That is, appears implicitly in these conditions. Use Taylor series expansion about to approximate and at

58. 58 Substitute (d) and (e) into (i)

59. 59 Expand, equate terms of identical powers ofand use (g-0) and (j-0) (j-1)

60. 60 (j-2) Using Taylor series expansion ofabout B.C. (3) becomes

61. 61 (k) Substitute (d) and (e) into (k) and use (g-0)

62. 62

63. 63 Expand and equate terms of identical powers of (l-0) (l-1)

64. 64 (l-2) Substitute (e) into initial condition (4) Equate terms of identical powers of

65. 65 (m-0) (m-1) (m-2) NOTE (i) Zero-order problem, (g-0), (h-0), (j-0), (l-0), (m-0): is the quasi-steady approximation of the Stefan problem (Example 6.1) (ii) First-order problem: Transient term is approximated by the solution to the zero order problem

66. 66 (3) Solutions Zero-order solution: Solution to (g-0) with conditions (h-0) and (j-0) is (n) Interface position is obtained from the solution to (l-0). Substitute (n) into (l-0) Solve and use initial condition (m-0)

67. 67 (o) The zero-order temperature solution (n) becomes (p) First-order solution: Substitute (p) into (g-1) Integrate twice, use conditions (h-1) and (j-1) (q)

68. 68 Interface: Use solution to (l-1). Substitute (o) and (q) into (l-1) Solve and use initial condition (m-1) (r) The first-order solution (q) becomes (s)

69. 69 Second-order solution. Substitute (s) into (g-2) Integrate twice, use conditions (h-2) and (j-2) (t) Interface: Use solution to (l-2). Substitute (o), (r) and (t) into (l-2)

70. 70 Integrate and use initial condition (m-2) (u) Second-order solution (t) becomes (v) The perturbation solution forbecomes

71. 71 (9.23) Interface solution: Substitute(o), (r) and (u) into (e) (w) Substitute (b) into (w) gives dimensional (9.24)

72. 72 (4) Checking Limiting check: (ii) For or, ( i.e.), interface remains stationary. Settingin eq. (9.24) gives Boundary and initial conditions check: Eq. (9.23) satisfiesEq. (w) satisfies initial condition (5) Comments (i)corresponds to. For this case the interface remains stationary. Setting in eq. (9.24) gives

73. 73 (i) Exact solution to Stefan’s problem is found in Section 6.7 (9.25) Whereis the root of eq. (6.24) (6.24) Table 9.3 0.1 0.22 1.0005 0.2 0.31 1.0022 0.3 0.37 1.0054 0.4 0.42 1.0097 0.5 0.46 1.0160 0.6 0.50 1.0241 0.7 0.54 1.0341 0.8 0.57 1.0463 (ii) Accuracy: Table 9.3.

74. 74 9.5 Useful Expansions Taylor series expansion (Example 9.4) Exponential expansion: Trigonometric expansion: (x) (y) If E does not appear explicitly in a governing equation or boundary condition an appropriate expansion must be used.

75. 75 Binomial expansion: (z)