What are the integral equations for radiation exchange between two parallel opposed plates finite in one dimension and infinite in the other, as in Figure 5.20? Each plate has a specified temperature variation that depends only on the x or y coordinate, and the environment is at Te.

Question

What are the integral equations for radiation exchange between two parallel opposed plates finite in one dimension and infinite in the other, as in Figure 5.20? Each plate has a specified temperature variation that depends only on the x or y coordinate, and the environment is at [latex]T_e.[/latex]

Accepted Answer

From the discussion in Example 4.2, the configuration factors between the infinitely long parallel strips [latex]dA_1[/latex] and [latex]dA_2[/latex] are

[latex]dF_{d1-d2}=\frac{1}{2}d\left(\sin \beta \right) =\frac{1}{2}\frac{a^2}{[(y-x)^2+a^2]^{3/2}}dy [/latex]
[latex]dF_{d2-d1} =\frac{1}{2}\frac{a^2}{[(y-x)^2+a^2]^{3/2}}dx [/latex]

The distribution of heat flux added to each plate is found by applying Equation 5.60

[latex]\frac{q_k(r_k)}{\epsilon _k}-\sum\limits_{j=1}^{N}{\frac{1-\epsilon _j}{\epsilon _j}\int_{A_j}^{}{q_j(r_j)dF_{dk-dj}(r_j,r_k)} } [/latex]
[latex]=\sigma T_k^4(r_k)-\sum\limits_{j=1}^{N}{\int_{A_j}^{}{\sigma T_j^4(r_j)}dF_{dk-dj}(r_j,r_k) } [/latex] (5.60)

to each plate. As discussed before Example 5.20, reduced temperatures are used to account for [latex]T_e.[/latex] With [latex]T_1=(T_{w1}^4-T_e^4)^{1/4} [/latex] and [latex]T_2=(T_{w2}^4-T_e^4)^{1/4} [/latex] ,the governing equations are

[latex]\frac{q_1(x)}{\epsilon _1} -\frac{1-\epsilon _2}{\epsilon _2}\int_{-L/2}^{L/2}{} q_2(y) \frac{1}{2}\frac{a^2}{[(y-x)^2+a^2]^{3/2}}dy=\sigma T_1^4(x) -\int_{-L/2}^{L/2}{} \sigma T_2^4(y) \frac{1}{2}\frac{a^2}{[(y-x)^2+a^2]^{3/2}}dy[/latex] (5.70a)
[latex]\frac{q_1(y)}{\epsilon _2} -\frac{1-\epsilon _1}{\epsilon _1}\int_{-L/2}^{L/2}{} q_1(x) \frac{1}{2}\frac{a^2}{[(y-x)^2+a^2]^{3/2}}dx=\sigma T_2^4(y) -\int_{-L/2}^{L/2}{} \sigma T_1^4(x) \frac{1}{2}\frac{a^2}{[(y-x)^2+a^2]^{3/2}}dx[/latex] (5.70b)

An alternative formulation using Equation 5.63

[latex]J_k(r_k)=\epsilon _k\sigma T_k^4(r_k)+(1-\epsilon _k)\sum\limits_{j=1}^{N}{\int_{A_j}^{}{J_j}(r_j)dF_{dk-dj} (r_j,r_k)} [/latex] (5.63)

yields two equations for [latex] J_1(x)[/latex] and [latex] J_2(y),[/latex]

[latex]J_1(x) -\left(1-\epsilon _1\right) \int_{-L/2}^{L/2}{} J_2(y) \frac{1}{2}\frac{a^2}{[(y-x)^2+a^2]^{3/2}}dy=\epsilon _1\sigma T_1^4(x)[/latex] (5.71a)
[latex]J_2(y) -\left(1-\epsilon _2\right) \int_{-L/2}^{L/2}{} J_1(x) \frac{1}{2}\frac{a^2}{[(y-x)^2+a^2]^{3/2}}dx=J _2\sigma T_2^4(y)[/latex] (5.71b)

After the [latex]J_1(x)[/latex] and [latex]J_2(y)[/latex] are found, the desired [latex]q_1(x)[/latex] and [latex]q_1(y)[/latex] are obtained from Equation 5.65a as

[latex]q_k(r_k)=\frac{\epsilon _k}{1-\epsilon _k}[\sigma T_k^4\left(r_k\right)-J_k(r_k) ] [/latex] [latex] 1 ≤k ≤m [/latex] (5.65a)

[latex]q_1(x)=\frac{\epsilon _1}{1-\epsilon _1} [\sigma T_1^4(x)-J_1(x)][/latex] (5.72a)
[latex]q_2(y)=\frac{\epsilon _2}{1-\epsilon _2} [\sigma T_2^4(y)-J_2(y)][/latex] (5.72b)

Question 5.21: What are the integral equations for radiation exchange betwe...

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