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?

Question

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 [latex]ϵ_2?[/latex]

Accepted Answer

By using the configuration factor for two parallel disks (factor 10 in Appendix C), [latex]F_{3–1} = 0.3249.[/latex]
Then [latex]F_{3–2} = 1 – F_{3–1} = 0.6751.[/latex] From reciprocity, [latex]A_1F_{1–3} = A_3F_{3–1}[/latex] and [latex]A_2F_{2–3} = A_3F_{3–2}[/latex], so that [latex]F_{1–3 }= 0.1444[/latex] and [latex]F_{2–3} = 0.1310.[/latex] Then [latex]F_{1–2} = 1 – F_{1–3} = 0.8556.[/latex] From [latex]A_1F_{1–2} = A_{2}F_{2–1}, F_{2–1} = 0.3735.[/latex] Finally, [latex]F_{2–2} = 1 – F_{2–1} – F_{2–3} = 0.4955.[/latex] From Equation 5.29,

[latex]\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)} [/latex] (5.29)

and by noting that [latex]Q_2 = 0[/latex] and [latex]1 – ϵ_3 = 0,[/latex] the three equations are

[latex]\frac{3000}{0.6}=\sigma [T_1^4-0.8556T_2^4 -0.1444(550)^4] [/latex]
[latex]-3000(0.3735)\frac{1-0.6}{0.6}=\sigma [-0.3735T_1^4+(1-0.4955)T_2^4-0.1310(550)^4] [/latex]
[latex]-3000(0.3735)\frac{1-0.6}{0.6}=\sigma [-0.3735T_1^4+(1-0.4955)T_2^4-0.1310(550)^4][/latex]

These can be solved for [latex]T_1, T_2[/latex], and [latex]Q_3.[/latex] (For this particular example, [latex]Q_3[/latex] can also be obtained from overall energy conservation, [latex]Q_3 = –Q_1.) [/latex] The result is [latex]T_1 = 721.6 K (T2 = 667.4 K).[/latex] Since [latex]Q_2 = 0,[/latex] all the terms involving [latex]ϵ_2[/latex] are zero, so [latex]ϵ_2[/latex] 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 [latex]J_2 = G_2[/latex] independent of [latex]ϵ_2[/latex]. This example is now solved using Equation 5.35.

[latex]\sum\limits_{j=1}^{N}{[\delta _{kj}-(1-\epsilon _k)F_{k-j}]J_j=\epsilon _k\sigma T_k^4} [/latex] [latex]1\leq k\leq m[/latex] (5.35a)
[latex]\sum\limits_{j=1}^{N}{[\delta _{kj}-F_{k-j}]J_j=\frac{Q_k}{A_k} } [/latex] [latex]m+1\leq k\leq N[/latex] (5.35b)

Because [latex]T_3[/latex] is specified and [latex]A_3[/latex] is black, Equation 5.35a gives [latex]J_3= σT_3^4[/latex] . Equation 5.35b is used at [latex]A_1[/latex] and [latex]A_2[/latex] since [latex]q_1[/latex] and [latex]q_2[/latex] are specified,

[latex]J_1-F_{1-2}J_2=F_{1-3}σT_3^4+q_1[/latex]

[latex]-F_{2-1}J_1+(1-F_{2-2})J_2=F_{2-3}σT_3^4[/latex]

This yields [latex]J_1 = 13,370 \ W/m^2[/latex] and [latex]J_2 = 11,250 \ W/m^2.[/latex] From Equation 5.17

[latex]J_k=\sigma T_k^4-\frac{1-\epsilon _k}{\epsilon _k} q_k[/latex] or [latex]\sigma T_k^4=\frac{1-\epsilon _k}{\epsilon _k} q_k+J_k[/latex] (5.17a,b)

[latex]σT_1^4= J_1 q 1 +[(1−\epsilon_1)/\epsilon_1]q_1 ,[/latex] and similarly for [latex]σT_2^4[/latex]. This gives the same temperatures as in the first part of this example.

Question 5.15: A frustum of a cone has its base uniformly heated as in Figu...

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