1. RADIATION EXCHANGE IN AN ENCLOSURE WITH DIFFUSE-GRAY SURFACES Net radiation method Simplified zone analysis Electric network analogy Generalized zone analysis Methods for solving integral equations

2. Net Radiation Method irradiation radiosity G J diffuse-gray surface at T net radiative heat flux 

5. GJ  diffuse-gray surface at T

6. GG  T 4 diffuse-gray surface at T 

7. Simplified Zone Analysis GkGk JkJk k th surface A k,  k, T k Enclosure with n surfaces open surface temperature, properties, radiosity, irradiation: uniform over each surface

8. all irradiation from n surfaces GkGk JkJk k th surface A k,  k, T k open surface

9. Summary When all T k ’s are specified, the two equations are decoupled. When T k or are specified at the boundary 2n unknowns: J k and or T k n unknowns: J k

10. Electric Network Analogy

11. 1/A k F kn 1/A k F k1  JkJk e bk J1J1 qkqk J2J2 JnJn

12. two infinite parallel plates -- T 1,  1  -- T 2,  2  Ex 7-6

13. Using network analogy q1q1 J2J2 eb2eb2 J1J1 eb1eb1

14. a body in an enclosure T 1,  1, A 1 T 2,  2, A 2 q1q1 J2J2 eb2eb2 J1J1 eb1eb1 Ex 7-8

15. The enclosure acts like a black cavity. T 1,  1, A 1 T 2,  2, A 2 when Remark: when A 2 is a black enclosure

16. q 1, q 2, T 3 = ? A 1 T 1 = 420 K  1 = 0.8 A 2 T 2 = 650 K  2 = 0.8 45° A 3,  3 = 0.9 1.8 m Ex 7-14 q 1 = ? q 2 = ? T 3 = ?

17. q1q1 q2q2 A 1 T 1 = 420 K  1 = 0.8 A 2 T 2 = 650 K  2 = 0.8 45° A 3,  3 = 0.9 1.8 m J2J2 eb2eb2 J1J1 eb1eb1 eb3 eb3 J3J3

19. T 1 = 420 K  1 = 0.8 T 2 = 650 K  2 = 0.8 45° A 3,  3 = 0.9 1.8 m q 1 = 8077 W q 2 = 2317 W T 3 = 649 K q = 5760 W

20. q 1, q 2, T 4, T 5 = ? A 1 T 1 = 420 K  1 = 0.8 A 2 T 2 = 650 K  2 = 0.8 45°  4 =  5 = 0.9 1.8 m A4A4 A5A5

21. A 1 T 1 = 420 K  1 = 0.8 A 2 T 2 = 650 K  2 = 0.8 45° 1.8 m  4 =  5 = 0.9 A4A4 A5A5 q1q1 J2J2 eb2eb2 J1J1 eb1eb1 eb4 eb4 q2q2 J4J4 eb5 eb5 J5J5

22. q1q1 J2J2 eb2eb2 J1J1 eb1eb1 eb4 eb4 q2q2 J4J4 eb5 eb5 J5J5

23. A 1 T 1 = 420 K  1 = 0.8 A 2 T 2 = 650 K  2 = 0.8 45° 1.8 m  4 =  5 = 0.9 A4A4 A5A5 A 1 T 1 = 420 K  1 = 0.8 A 2 T 2 = 650 K  2 = 0.8 45° A 3,  3 = 0.9 1.8 m

24. A 1 T 1 = 420 K  1 = 0.4 A 2 T 2 = 650 K  2 = 0.4 45° 1.8 m  4 =  5 = 0.45 A4A4 A5A5 A 1 T 1 = 420 K  1 = 0.4 A 2 T 2 = 650 K  2 = 0.4 45° A 3,  3 = 0.45 1.8 m

25. Generalized Zone Analysis open surface dA k dA i A k,  k, T k A i,  i, T i 0 temperature, properties: uniform over each surface

26. open surface 0 dA k dA i A k,  k, T k A i,  i, T i

27. Summary

28. Exact solution: cylindrical circular cavity dd R 00   dd dd  T(  ),  (  ) T sur = 0 K

29. dd  Rcos  R 00   dd dd 

30. Let Then, When  and T are constants, b.c.

31. Methods for Solving Integral Equations (Hildebrand “Methods of Applied Mathematics”) Fredholm integral equation of the second kind 1) Reduction to sets of algebraic equations integral  summation (finite difference)  algebraic linear equations ▪ trapezoidal rule ▪ Simpson’s rule ▪ Gaussian quadrature

32. 2) Successive approximation (iterative method) initial guess  0 (x) and get first approximation  1 (x)...

33. M : maximum value of the kernel K(x,  ) if << 1 rapidly converge (proof: Hildebrand pp.421-424) as when

34. 3) Variational method (Courant & Hilbert “Methods of Mathematical Physics”p.205-) extremum of I :  (x) satisfies the integral equation.

35. approximate solution: Ritz method (Hildebrand p.187) with a proper choice of  k (x) n simultaneous algebraic equations

36. Radiative exchange between parallel plates: variational method T 2,   h L/2 T 1,   T e = 0 x2x2 dA 2 x1x1 dA 1 T e = 0 Ex

37. T 2,   x1x1 x2x2 dA 2 dA 1 h L/2 T 1,   T e = 0 

38. Let when

39. Solution by variational method symmetry condition a 1, a 2, a 3 : constant functions of 

41. Remark: A 2, T 2,   h L A 1, T 1,   T e = 0 when

42. or

43. 0.10.20.30.40.50 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.861,  = 0.1 0.553,  = 0.5 0.407,  = 0.9 x

44. 4) Approximation of kernel integral equation → 2 nd order O.D.E. accuracy increases when integral equation → 4 th order O.D.E.

45. Circular tube with uniform heat flux: approximation of kernel x dx L dy y D T e = 0 Tube surface temperature T = ? Ex 7-23 

46. x dx dy L y D T e = 0 

48. Approximation of kernel b.c. Then at

50. 5) Taylor series expansion substitute into integral equation decreases very rapidly as increases

51. three-term expansion