Question

Heat transfer by radiation occurs between two large parallel plates, which are maintained at temperatures T1T_{1} and T2T_{2}, with T1>T2T_{1}>T_{2}. To reduce the rate of heat transfer between the plates, it is proposed that they be separated by a thin shield that has different emissivities on opposite surfaces. In particular, one surface has the emissivity εs<0.5\varepsilon_{s}<0.5, while the opposite surface has an emissivity of 2εs2 \varepsilon_{s}. (a) How should the shield be oriented to provide the larger reduction in heat transfer between the plates? That is, should the surface of emissivity εs\varepsilon_{s} or that of emissivity 2εs2 \varepsilon_{s} be oriented toward the plate at T1T_{1}? (b) What orientation will result in the larger value of the shield temperature TsT_{s}?

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We are given following data:\texttt{We are given following data:}

T1>T2\begin{align*} T_{1}>T_{2}\\ \end{align*}

$\texttt{Thermal Resistance for the first case (R1)(R_{1}) per unit area:}$

R1=1ϵ1ϵ1+1F1s+1ϵsϵs+12ϵs2ϵs+1Fs2+1ϵ2ϵ2    (1)\begin{align*} R_{1} &=\dfrac{1-\epsilon_{1}}{\epsilon_{1}}+\dfrac{1}{F_{1s}}+\dfrac{1-\epsilon_{s}}{\epsilon_{s}}+\dfrac{1-2\epsilon_{s}}{2\epsilon_{s}}+\dfrac{1}{F_{s2}}+\dfrac{1-\epsilon_{2}}{\epsilon_{2}}\ \ \ \ (1)\\ \end{align*}

$\texttt{Thermal Resistance for the second case (R2)(R_{2}) per unit area:}$

R2=1ϵ1ϵ1+1F1s+12ϵs2ϵs+1ϵsϵs+1Fs2+1ϵ2ϵ2     (2)\begin{align*} R_{2} &=\dfrac{1-\epsilon_{1}}{\epsilon_{1}}+\dfrac{1}{F_{1s}}+\dfrac{1-2\epsilon_{s}}{2\epsilon_{s}}+\dfrac{1-\epsilon_{s}}{\epsilon_{s}}+\dfrac{1}{F_{s2}}+\dfrac{1-\epsilon_{2}}{\epsilon_{2}}\ \ \ \ \ (2)\\ \end{align*}

$\texttt{We can see from equation (1) and equation (2) that R1=R2R_{1}=R_{2}}$

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