A triangular enclosure has long sides normal to the cross section shown in Figure 5.12. The triangular end walls can be neglected in the radiative exchange. Specified quantities are the temperatures of two sides and the heat flux added to the third. The $q_1, q_2,$ and $T_3$ are required. The solution is obtained first with side 3 being a single area. Then this side is divided into two equal parts to
refine the calculation.

Question

A triangular enclosure has long sides normal to the cross section shown in Figure 5.12. The triangular end walls can be neglected in the radiative exchange. Specified quantities are the temperatures of two sides and the heat flux added to the third. The [latex]q_1, q_2,[/latex] and [latex]T_3[/latex] are required. The solution is obtained first with side 3 being a single area. Then this side is divided into two equal parts to
refine the calculation.

Accepted Answer

Equation 5.29 is used for analysis. From Example 4.13,

[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)

[latex] F_{1−2}= (A_1+A_2 -A_3 )/2A_1 =0.2929 =F_{2−1} .[/latex] Then [latex] F_{1-3}=1-F_{1-2}= 0.7071 =F_{2−3}.[/latex] From symmetry, [latex] F_{3–1} = F_{3–2} = 0.5. [/latex] The self-view factors are zero. The first three equations from Example 5.13 are used. If [latex] ϵ_1 = ϵ_2 = 0.80[/latex] and [latex]ϵ_3 = 0.90,[/latex] the results are [latex]q_1 = –6346 \ W/m^2, q_2 = 1820 \ W/m^2,[/latex] and [latex]T_3 = 649.1 \ K.[/latex] Thus energy must be added to maintain the high temperature of [latex]A_2.[/latex] The energy added to [latex]A_2[/latex] combined with that from [latex]A_3[/latex] flows out of [latex]A_1[/latex] at a lower temperature. Now consider what happens if the emissivities are reduced by half, [latex]ϵ_1=ϵ_2=0.40[/latex] and [latex]ϵ_3 = 0.45[/latex]. The energy supplied to [latex] A_3[/latex] has greater difficulty in being transferred to the other walls, and the temperature rises to [latex] T_3 = 733.9 \ K.[/latex] Since [latex]T_3[/latex] is now larger than both [latex]T_1[/latex] and [latex]T_2,[/latex] energy is transferred out from both [latex]A_1[/latex] and [latex]A_2: q_1 = –4101 \ W/m^2[/latex] and [latex]q_2 = –424.6 \ W/m^2.[/latex] To refine the calculation, [latex]A_3[/latex] is divided into two equal parts [latex]A_4 [/latex] and [latex]A_5. [/latex] From the geometry, [latex]F_{1–4} = 0.5 = F_{1–5} + F_{1–2} = F_{2–5}. [/latex] With [latex]F_{1–2}[/latex] known, this gives [latex]F_{1–5} = 0.2071 = F_{2–4}.[/latex] As in the previous examples, four equations are written from Equation 5.29. With [latex]T_1, T_2, q_4, [/latex] and [latex]q_5[/latex] known, these are solved for [latex]q_1, q_2, T_4, [/latex] and [latex]T_5. [/latex] For [latex]ϵ_1 = ϵ_2 = 0.80 [/latex] and [latex]ϵ_3 = 0.90,[/latex] this gives [/latex]q_1 = –6049 \ W/m^2,[/latex]
[latex]q_2 = 1524 \ W/m^2, T_4 = 626.3 \ K,[/latex] and [latex]T_5 = 669.7 \ K. [/latex] This reveals the temperature variation along [latex] A_3[/latex] as compared with the uniform value obtained in the first part of this example. The [latex] q_2[/latex] is somewhat smaller because [latex]A_2 [/latex] is adjacent to the higher-temperature portion of [latex] A_3[/latex]. By further subdividing [latex]A_3, [/latex] its temperature distribution would be obtained. For [latex]ϵ_1 = ϵ_2 = 0.40[/latex] and [latex]ϵ_3 = 0.45, T_4 = 726.8[/latex] K and [latex]T_5 = 740.8[/latex] K, compared with [latex]T_3 = 733.9[/latex] K in the previous calculation. The q are changed somewhat to [latex]q_1 = –4038 \ W/m^2[/latex] and [latex]q_2 = –487.2 \ W/m^2.[/latex] The q values in this example have all been verified to satisfy overall energy conservation (the results given have been rounded off).

Question 5.14: A triangular enclosure has long sides normal to the cross se...

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