Question 5.14: A triangular enclosure has long sides normal to the cross se...
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.
Equation 5.29 is used for analysis. From Example 4.13,
\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)
F_{1−2}= (A_1+A_2 -A_3 )/2A_1 =0.2929 =F_{2−1} . Then F_{1-3}=1-F_{1-2}= 0.7071 =F_{2−3}. From symmetry, F_{3–1} = F_{3–2} = 0.5. The self-view factors are zero. The first three equations from Example 5.13 are used. If ϵ_1 = ϵ_2 = 0.80 and ϵ_3 = 0.90, the results are q_1 = –6346 \ W/m^2, q_2 = 1820 \ W/m^2, and T_3 = 649.1 \ K. Thus energy must be added to maintain the high temperature of A_2. The energy added to A_2 combined with that from A_3 flows out of A_1 at a lower temperature. Now consider what happens if the emissivities are reduced by half, ϵ_1=ϵ_2=0.40 and ϵ_3 = 0.45. The energy supplied to A_3 has greater difficulty in being transferred to the other walls, and the temperature rises to T_3 = 733.9 \ K. Since T_3 is now larger than both T_1 and T_2, energy is transferred out from both A_1 and A_2: q_1 = –4101 \ W/m^2 and q_2 = –424.6 \ W/m^2. To refine the calculation, A_3 is divided into two equal parts A_4 and A_5. From the geometry, F_{1–4} = 0.5 = F_{1–5} + F_{1–2} = F_{2–5}. With F_{1–2} known, this gives F_{1–5} = 0.2071 = F_{2–4}. As in the previous examples, four equations are written from Equation 5.29. With T_1, T_2, q_4, and q_5 known, these are solved for q_1, q_2, T_4, and T_5. For ϵ_1 = ϵ_2 = 0.80 and ϵ_3 = 0.90, this gives [/latex]q_1 = –6049 \ W/m^2,[/latex]
q_2 = 1524 \ W/m^2, T_4 = 626.3 \ K, and T_5 = 669.7 \ K. This reveals the temperature variation along A_3 as compared with the uniform value obtained in the first part of this example. The q_2 is somewhat smaller because A_2 is adjacent to the higher-temperature portion of A_3. By further subdividing A_3, its temperature distribution would be obtained. For ϵ_1 = ϵ_2 = 0.40 and ϵ_3 = 0.45, T_4 = 726.8 K and T_5 = 740.8 K, compared with T_3 = 733.9 K in the previous calculation. The q are changed somewhat to q_1 = –4038 \ W/m^2 and q_2 = –487.2 \ W/m^2. The q values in this example have all been verified to satisfy overall energy conservation (the results given have been rounded off).