Question 5.15: A frustum of a cone has its base uniformly heated as in Figu...
A frustum of a cone has its base uniformly heated as in Figure 5.13. The top is maintained at 550 K while the side is perfectly insulated on the outside. Surfaces 1 and 2 are diffuse-gray, while surface 3 is black. For energy transfer only by radiation, what is the temperature of side 1? How important is the value of ϵ_2?
By using the configuration factor for two parallel disks (factor 10 in Appendix C), F_{3–1} = 0.3249.
Then F_{3–2} = 1 – F_{3–1} = 0.6751. From reciprocity, A_1F_{1–3} = A_3F_{3–1} and A_2F_{2–3} = A_3F_{3–2}, so that F_{1–3 }= 0.1444 and F_{2–3} = 0.1310. Then F_{1–2} = 1 – F_{1–3} = 0.8556. From A_1F_{1–2} = A_{2}F_{2–1}, F_{2–1} = 0.3735. Finally, F_{2–2} = 1 – F_{2–1} – F_{2–3} = 0.4955. From Equation 5.29,
\sum\limits_{j=1}^{N}{\left\lgroup\frac{\delta _{kj}}{\epsilon _j} -F_{k-j}\frac{1-\epsilon _j}{\epsilon _j} \right\rgroup }\frac{Q_j}{A_j} =\sum\limits_{j=1}^{N}{(\delta _{kj}-F_{k-j})\sigma T_j^4} =\sum\limits_{j=1}^{N}{F_{k-j}\sigma (T_k^4-T_j^4)} (5.29)
and by noting that Q_2 = 0 and 1 – ϵ_3 = 0, the three equations are
\frac{3000}{0.6}=\sigma [T_1^4-0.8556T_2^4 -0.1444(550)^4]
-3000(0.3735)\frac{1-0.6}{0.6}=\sigma [-0.3735T_1^4+(1-0.4955)T_2^4-0.1310(550)^4]
-3000(0.3735)\frac{1-0.6}{0.6}=\sigma [-0.3735T_1^4+(1-0.4955)T_2^4-0.1310(550)^4]
These can be solved for T_1, T_2, and Q_3. (For this particular example, Q_3 can also be obtained from overall energy conservation, Q_3 = –Q_1.) The result is T_1 = 721.6 K (T2 = 667.4 K). Since Q_2 = 0, all the terms involving ϵ_2 are zero, so ϵ_2 does not appear in the simultaneous equations, and the emissivity of the insulated surface is not important. Physically, this results from the fact that, for no convection or conduction, all absorbed energy must be reemitted and hence J_2 = G_2 independent of ϵ_2. This example is now solved using Equation 5.35.
\sum\limits_{j=1}^{N}{[\delta _{kj}-(1-\epsilon _k)F_{k-j}]J_j=\epsilon _k\sigma T_k^4} 1\leq k\leq m (5.35a)
\sum\limits_{j=1}^{N}{[\delta _{kj}-F_{k-j}]J_j=\frac{Q_k}{A_k} } m+1\leq k\leq N (5.35b)
Because T_3 is specified and A_3 is black, Equation 5.35a gives J_3= σT_3^4 . Equation 5.35b is used at A_1 and A_2 since q_1 and q_2 are specified,
J_1-F_{1-2}J_2=F_{1-3}σT_3^4+q_1
-F_{2-1}J_1+(1-F_{2-2})J_2=F_{2-3}σT_3^4
This yields J_1 = 13,370 \ W/m^2 and J_2 = 11,250 \ W/m^2. From Equation 5.17
J_k=\sigma T_k^4-\frac{1-\epsilon _k}{\epsilon _k} q_k or \sigma T_k^4=\frac{1-\epsilon _k}{\epsilon _k} q_k+J_k (5.17a,b)
σT_1^4= J_1 q 1 +[(1−\epsilon_1)/\epsilon_1]q_1 , and similarly for σT_2^4. This gives the same temperatures as in the first part of this example.