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

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

GJ  diffuse-gray surface at T

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

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

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

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

Electric Network Analogy

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

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

Using network analogy q1q1 J2J2 eb2eb2 J1J1 eb1eb1

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

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

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

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

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

q 1, q 2, T 4, T 5 = ? A 1 T 1 = 420 K  1 = 0.8 A 2 T 2 = 650 K  2 = °  4 =  5 = m A4A4 A5A5

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

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

A 1 T 1 = 420 K  1 = 0.8 A 2 T 2 = 650 K  2 = ° 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 = ° A 3,  3 = m

A 1 T 1 = 420 K  1 = 0.4 A 2 T 2 = 650 K  2 = ° 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 = ° A 3,  3 = m

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

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

Summary

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

dd  Rcos  R 00   dd dd 

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

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

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

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

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

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

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

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

Let when

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

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

or

,  = ,  = ,  = 0.9 x

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

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 

x dx dy L y D T e = 0 

Approximation of kernel b.c. Then at

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

three-term expansion