(a) The thermal circuit diagram is shown in Figure (b). Here \dot{S}_{ 2} = 0 .
(b) The radiation from surface 1 to 2, in the presence of a reradiating surface 3, is given by -Q_{1}=Q_{r,1}=Q_{r,1-2}=\frac{E_{b,1}(T_{1})-E_{b,2}(T_{2})}{\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}} \right)_{1}+\frac{1}{A_{r,1}F_{1-2}+ \frac{1}{\frac{1}{A_{r,1}F_{1-3}}+\frac{1}{A_{r,2}F_{2-3}} } }+\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}} \right)_{2} } . We need to determine the three view factors. Before any numerical evaluation, we note that
F_{1-3} = 1 − F_{1-2} summation rule
F_{2-3} = 1 − F_{2-1} = 1 −\frac{A_{r,1}}{A_{r,2}} F_{1-2} summation and reciprocity rules
Then we only need to evaluate F_{1-2}. This is done using the relation or the graphical results given in Figure (a).
R_{1}^{\ast }=\frac{D_{1}}{2l}=\frac{0.05(m)}{2\times 0.25(m)} =0.1 , R_{2}^{\ast }=\frac{D_{2}}{2l}=\frac{0.30(m)}{2\times 0.25(m)} =0.6
F_{1-2} = 0.26 for \frac{1}{R_{1}^{\ast }} = 10 and R_{2}^{\ast } = 0.6 Figure (a).
Then
F_{1-3} = 1 − 0.26 = 0.74
F_{2-3}=1-\frac{\pi D_{1}^{2}/4}{\pi D_{2}^{2}/4} F_{1-2}=1-\frac{(0.05)^{2}(m)^{2}}{(0.30)^{2}(m)^{2}} \times 0.26=0.993
Next, from -Q_{1}=Q_{r,1}=Q_{r,1-2}=\frac{E_{b,1}(T_{1})-E_{b,2}(T_{2})}{\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}} \right)_{1}+\frac{1}{A_{r,1}F_{1-2}+ \frac{1}{\frac{1}{A_{r,1}F_{1-3}}+\frac{1}{A_{r,2}F_{2-3}} } }+\left(\frac{1-\epsilon _{r}}{A_{r}\epsilon _{r}} \right)_{2} } and [(R_{r,F})_{1-2}]_{app} \equiv \frac{1}{A_{r,1}F_{1-2}+ \frac{1}{\frac{1}{A_{r,1}F_{1-3}}+\frac{1}{A_{r,2}F_{2-3}} } } , we have
Q_{r,2-1}=\frac{\sigma _{SB}(T_{2}^{4}-T_{1}^{4})}{(R_{r,\epsilon })_{1}+[(R_{r,F})_{1-2}]_{app}+(R_{r,\epsilon })_{2}}
(R_{r,\epsilon })_{1}=\frac{1-\epsilon _{r,1}}{\pi R_{1}^{2}\epsilon _{r,1}} =\frac{1-0.8}{\pi (0.05)^{2}(m)^{2}\times 0.8} =1.273m^{2}
(R_{r,\epsilon })_{2}=\frac{1-\epsilon _{r,2}}{\pi R_{1}^{2}\epsilon _{r,2}} =\frac{1-0.9}{\pi (0.30)^{2}(m)^{2}\times 0.9} =1.572m^{-2}
[(R_{r,F})_{1-2}]_{app}
=\frac{1}{\pi R_{1}^{2}F_{1-2}+\frac{1}{\frac{1}{\pi R_{1}^{2}F_{1-3}}+\frac{1}{\pi R_{2}^{2}F_{2-3}} } }
=\frac{1}{\pi (0.025)^{2}(m)^{2}\times 0.26+\frac{1}{\frac{1}{\pi (0.025)^{2}(m)^{2}\times 0.74}+\frac{1}{\pi (0.15)^{3}(m)^{2}\times 0.993} } }
=\frac{1}{5.060\times 10^{-4}+\frac{1}{\frac{1}{1.453\times 10^{-3}}+\frac{1}{7.020\times 10^{-2}} } }
=\frac{1}{5.060\times 10^{-4}+1.424\times 10^{-3} } =5.183\times 10^{2}m^{-2}
Now substituting for the surface-grayness and view-factor resistances in the expression for Q_{r,2-1}, we have
Q_{r,2-1}=\frac{5.670\times 10^{-8}(W/m^{2}-K)[(1,200)^{4}(K)^{4}-(573.15)^{4}(K)^{4}]}{(1.273\times 10^{2}+1.572+5.183\times 10^{2})(m^{-2})}
=\frac{1.115\times 10^{5}(W/m^{2})}{6.471\times 10^{2}(1/m^{2})}=1.723\times 10^{2} W
(c) The ratio of [(R_{r,F} )_{1-2}]_{app} to (R_{r,F} )_{1-2} is found noting that
(R_{r,F} )_{1-2}=\frac{1}{\pi R_{1}F_{1-2}} =\frac{1}{5.111\times 10^{-4}(m^{3})} =1,957m^{-2}
Then
\frac{[(R_{r,F} )_{1-2}]_{app}}{(R_{r,F})_{1-2}}=\frac{518.2(1/m^{2})}{1,957(1/m^{2})} =0.2648
