RADIATION EXCHANGE IN AN ENCLOSURE WITH DIFFUSE-GRAY SURFACES

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

5. G J e diffuse-gray surface at T

6. aG esT4 e diffuse-gray surface at T

7. Jk Gk kth surface Ak, ek, Tk Simplified Zone Analysis Enclosure with n surfaces temperature, properties, radiosity, irradiation: uniform over each surface open surface

8. all irradiation from n surfaces open surface Jk Gk kth surface Ak, ek, Tk

9. Summary When Tk or are specified at the boundary 2n unknowns: Jk and or Tk When all Tk’s are specified, the two equations are decoupled. n unknowns: Jk

10. Electric Network Analogy

11. J1 Jk ebk qk J2 Jn 1/AkFk1    1/AkFkn

12. Ex 7-6 two infinite parallel plates T2, e2 -  -  T1, e1

13. q1 eb1 J1 J2 eb2 Using network analogy

14. T2, e2, A2 q1 eb2 eb1 J1 J2 T1, e1, A1 Ex 7-8 a body in an enclosure

15. when T2, e2, A2 T1, e1, A1 The enclosure acts like a black cavity. Remark: when A2 is a black enclosure

16. A1 T1 = 420 K e1 = 0.8 A2 T2 = 650 K e2 = 0.8 45° A3,e3 = 0.9 1.8 m Ex 7-14 q1, q2,T3= ?

17. A1 T1 = 420 K e1 = 0.8 A2 T2 = 650 K e2 = 0.8 A3,e3 = 0.9 45° 1.8 m q1 q2 eb2 eb1 J1 J2 J3 eb3

19. A5 A4 A1 T1 = 420 K e1 = 0.8 A2 T2 = 650 K e2 = 0.8 q1, q2,T4,T5= ? 45° 1.8 m e4 = e5 = 0.9

20. A1 T1 = 420 K e1 = 0.8 A2 T2 = 650 K e2 = 0.8 A4 A5 45° e4 = e5 = 0.9 1.8 m eb1 eb2 q1 q2 J2 J1 J4 J5 eb4 eb5

21. eb1 eb2 q1 q2 J2 J1 J4 J5 eb4 eb5

22. A1 T1 = 420 K e1 = 0.8 A2 T2 = 650 K e2 = 0.8 A3,e3 = 0.9 45° 1.8 m A1 T1 = 420 K e1 = 0.8 A2 T2 = 650 K e2 = 0.8 A4 A5 45° 1.8 m e4 = e5 = 0.9

23. A1 T1 = 420 K e1 = 0.4 A2 T2 = 650 K e2 = 0.4 A4 A5 45° 1.8 m e4 = e5 = 0.45 A1 T1 = 420 K e1 = 0.4 A2 T2 = 650 K e2 = 0.4 A3,e3 = 0.45 45° 1.8 m

24. Ai, ei, Ti 0 dAi dAk Ak, ek, Tk Generalized Zone Analysis open surface temperature, properties: uniform over each surface

25. open surface Ai, ei, Ti 0 dAi dAk Ak, ek, Tk

26. Summary

27. q a b db da dq Exact solution: cylindrical circular cavity q0 Tsur= 0 K R T(q), e(q)

28. q a b db 2Rcosb da dq q0 R

29. Let Then, When e and T are constants, b.c.

30. 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

31. 2) Successive approximation (iterative method) initial guess f0(x) and get first approximation f1(x) . . .

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

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

34. approximate solution: Ritz method(Hildebrand p.187) with a proper choice ofyk(x) n simultaneous algebraic equations

35. x2 dA2 x1 dA1 Radiative exchange between parallel plates: variational method Ex L/2 L/2 T2, e2 Te = 0 h Te = 0 T1, e1

36. q L/2 L/2 T2, e2 x2 dA2 Te = 0 h Te = 0 x1 T1, e1 dA1

37. Let when

38. Solution by variational method symmetry condition a1, a2, a3: constant functions ofg

40. when Remark: L A2, T2, e2 Te = 0 h Te = 0 A1, T1, e1

41. or when

42. 1.0 0.9 0.861, e = 0.1 0.8 0.7 0.6 0.553, e = 0.5 0.5 0.407, e = 0.9 0.4 0.3 0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 x

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

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

45. Te = 0 e D Te = 0 x dx y dy L

47. Then at Approximation of kernel b.c.

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

50. three-term expansion