As an example with a more complex geometry, consider the long rectangular enclosure shown in cross-section in Figure 5.11 (the end walls are neglected). There is a thin baffle along a diagonal dividing the interior into two triangular enclosures. Heat fluxes are specified along two sides,

Question

As an example with a more complex geometry, consider the long rectangular enclosure shown in cross-section in Figure 5.11 (the end walls are neglected). There is a thin baffle along a diagonal dividing the interior into two triangular enclosures. Heat fluxes are specified along two sides, and the other two sides are each cooled to a specified uniform temperature. It is required to find the temperatures [latex]T_1[/latex] and [latex]T_2[/latex] of the heated walls and the heat fluxes [latex]q_5[/latex] and [latex]q_6[/latex] that must be removed through the cooled walls to maintain their temperatures. For simplicity the surfaces are not subdivided into smaller areas. Subdivision improves the results (see Example 5.14).

Accepted Answer

Since [latex]T_3, T_4, q_3,[/latex] and [latex]q_4[/latex] are also unknown, there are eight unknowns, so eight simultaneous equations are required. Two are the constraints at the baffle wall that yield [latex]T_3 = T_4[/latex] and [latex]q_3 = –q_4.[/latex]
The other six equations are found from Equation 5.29 and are written here in abbreviated form using the notation [latex]F_{mn} ≡F_{m−n} ,ϑ ≡T^4 ,E_n ≡(1− ε_n )/ε_n :[/latex]

[latex](q_1/\epsilon _1)-q_2E_2F_{12}=\sigma (\vartheta _1-F_{12}\vartheta _2-F_{13}\vartheta _3)[/latex]
[latex]-q_1E_1F_{21}+(q_2/\epsilon _2)-q_3E_2F_{23}=\sigma (-F_{21}\vartheta _1+\vartheta _2-F_{23}\vartheta _3)[/latex]
[latex]-q_1E_1F_{31}-q_2E_2F_{32}+(q_3/\epsilon _3)=\sigma (-F_{31}\vartheta _1+F_{32}\vartheta _2-\vartheta _3)[/latex]
[latex](q_5/\epsilon _5)-q_6E_6F_{56}-q_4E_4F_{54}=\sigma (\vartheta _5-F_{56}\vartheta _6-F_{54}\vartheta _4)[/latex]
[latex]-q_5E_5F_{65}+(q_6/\epsilon _6)-q_4E_4F_{64}=\sigma (-F_{65}\vartheta _5+\vartheta _6-F_{64}\vartheta _4)[/latex]
[latex]-q_5E_5F_{45}-q_6E_6F_{46}+(q_4/\epsilon _4)=\sigma (-F_{45}\vartheta _5-F_{46}\vartheta _6+\vartheta _4)[/latex]

For each three-sided enclosure, the configuration factors are found from Example 4.13. Typical values are [latex]F_{1–2} = F_{5–6} = 0.33632[/latex] and [latex]F_{1–3} = F_{5–4} = 0.66369[/latex]. A matrix solver can be used to substitute the values and solve the equations to obtain [latex]T_1 = 817.0 K, T_2 = 821.7 K, T_3 = 709.1 K, q_5 = –559.0 W/m^2, \ and \ q_6 = –2890.1 \ W/m^2.[/latex] Most of the energy is leaving through the more strongly cooled side [latex]A_6[/latex]. The effect of various parameters can easily be examined. If [latex]\epsilon _1 = \epsilon _2[/latex] are changed, the only results affected are [latex] T_1[/latex] and [latex] T_2.[/latex] This is because the same total energy must pass through surface 4, so [latex] T_4 [/latex] remains the same in order to transfer the energy to sides 5 and 6.For [latex]\epsilon _1 = \epsilon _2 = 1.0[/latex], the [latex] T_1 \ and \ T_2[/latex] are slightly decreased to [latex]T_1 = 812.9[/latex] K and [latex]T_2 = 815.8[/latex] K. When [latex]\epsilon _1 = \epsilon _2 = 0.50,[/latex] these temperatures increase to [latex]T_1 = 824.9 \ K \ and \ T_2 = 833.1 \ K [/latex]. The results should be checked by determining whether [latex]\sum\limits_{k=0}^{N}{q_k}A_k=0. [/latex] In this case (0.55 × 1500.0) + (0.75 × 2200.0) – (0.55 × 559.0) – (0.75 × 2890.1) =–0.025 W/m of length, which is within the roundoff accuracy.

Question 5.13: As an example with a more complex geometry, consider the lon...

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