Question 5.18: An enclosure of the general type in Figure 5.9 consists of t...

With T_3 = 0, ϵ_3 = 1, and the self-view factors dF_{dj−dj*} = 0, Equation 5.60 is written for the two plane surfaces 1 and 2 having uniform q_1  and T_2,

\frac{q_k(r_k)}{\epsilon _k}-\sum\limits_{j=1}^{N}{\frac{1-\epsilon _j}{\epsilon _j}\int_{A_j}^{}{q_j(r_j)dF_{dk-dj}(r_j,r_k)} }
=\sigma T_k^4(r_k)-\sum\limits_{j=1}^{N}{\int_{A_j}^{}{\sigma T_j^4(r_j)}dF_{dk-dj}(r_j,r_k) }                     (5.60)
=\sum\limits_{j=1}^{N}{\int_{A_j}^{}{\sigma\left[ T_k^4(r_k)-T_j^4(r_j)\right] }dF_{dk-dj}(r_j,r_k) }

\frac{q_1}{\epsilon _1}-\sum\limits_{j=1}^{N}{\frac{1-\epsilon _2}{\epsilon _2}\int_{A_2}^{}{q_2(r_2)dF_{d1-d2}(r_2,r_1)} }= \sigma T_1^4(r_1)-\sigma T_2^4\int_{A_2}^{}{} dF_{d1-d2}(r_2,r_1)                                  (5.61a)

\frac{q_2(r_2)}{\epsilon _1}-q_1\frac{1-\epsilon _1}{\epsilon _1}\int_{A_1}^{}{dF_{d2-d1}(r_1,r_2)} =\sigma T_2^4-\int_{A_1}^{}{\sigma T_1^4(r_1)}dF_{d1-d2}(r_1,r_2)                    (5.61b)

An equation for surface 3 is not needed since Equations 5.61 do not involve the unknown   q_3(r_3)
as a consequence of ϵ_3 = 1   and T_3 = 0.   From the definitions of F factors, and \int_{A_2}^{}{dF_{d1-d2}=F_{d1-2}} and \int_{A_1}^{}{dF_{d2-d1}=F_{d2-1}}. Equations 5.61 simplify to the following relations where the unknowns are on

\sigma T_1^4+\frac{1-\epsilon _2}{\epsilon _2}\int_{A_2}^{}{q_2(r_2)dF_{d1-d2}(r_2,r_1)} =\sigma T_2^4F_{d1-2} (r_1)+\frac{q_1}{\epsilon _1}                    (5.62a)
\int_{A_1}^{}{\sigma T_1^4(r_1)} dF_{d2-d1}(r_1,r_2)+\frac{q_2(r_2)}{\epsilon _{2}}=\sigma T_2^4+q_1\frac{1-\epsilon _1}{\epsilon _1}F_{d2-1} (r_2)   (5.62b)

Equations 5.62 can be solved simultaneously for the distributions T_1(r_1) and q_2(r_2). Some solution methods are in Section 5.4.2 for these types of simultaneous integral equations.