The second half of Example 5.14 is now calculated by using Equation 5.35. A1 and A2 have specified T, while A4 and A5 have specified q. Then Equation 5.35 yields

Question

The second half of Example 5.14 is now calculated by using Equation 5.35. [latex]A_1[/latex] and [latex]A_2[/latex] have specified T, while [latex]A_4[/latex] and [latex]A_5[/latex] have specified q. Then Equation 5.35 yields

[latex]\sum\limits_{j=1}^{N}{[\delta _{kj}-(1-\epsilon _k)F_{k-j}]J_j}=\epsilon _k\sigma T^4_k [/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)

Accepted Answer

[latex]J_1-(1-\epsilon _1)[F_{1-2}J_2+F_{1-4}J_4+F_{1-5}J_5] =\epsilon _1\sigma T_1^4[/latex]

[latex]J_2-(1-\epsilon _2)[F_{2-1}J_1+F_{2-4}J_4+F_{2-5}J_5] =\epsilon _2\sigma T_2^4[/latex]

[latex]-F_{4-1}J_1-F_{4-2}J_2+J_4-F_{4-5}J_5=q_4[/latex]

[latex]-F_{5-1}J_1-F_{5-2}J_2-F_{5-4}J_4+J_5=q_5[/latex]

The solution yields J in [latex]W/m^2: J_1 = 3277, J_2 = 9741, J_4 = 8370,[/latex] and [latex]J_5 = 11,048. [/latex] Then from Equation 5.15,

[latex]\frac{Q_k}{A_k}=q_k=\frac{\epsilon _k}{1-\epsilon _k}(\sigma T^4_k-J_k) [/latex] (5.15)

[latex]q_1=[\epsilon _1/(1-\epsilon _1)](\sigma T^4_1-J_1)[/latex] and similarly for [latex]q_2. [/latex] From Equation 5.17,

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

[latex]\sigma T_4^4=[\epsilon _4/(1-\epsilon _4)]q_4+J_4[/latex] and similarly for [latex] T_5[/latex] . This provides the same results as for Example 5.14.

Question 5.16: The second half of Example 5.14 is now calculated by using E...

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