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)

[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...

Thermal Radiation Heat Transfer

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

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

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

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

$-F_{4-1}J_1-F_{4-2}J_2+J_4-F_{4-5}J_5=q_4$

$-F_{5-1}J_1-F_{5-2}J_2-F_{5-4}J_4+J_5=q_5$

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

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

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

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

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