#### Question

Heat transfer by radiation occurs between two large parallel plates, which are maintained at temperatures $T_{1}$ and $T_{2}$, with $T_{1}>T_{2}$. 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 $\varepsilon_{s}<0.5$, while the opposite surface has an emissivity of $2 \varepsilon_{s}$. (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 $\varepsilon_{s}$ or that of emissivity $2 \varepsilon_{s}$ be oriented toward the plate at $T_{1}$? (b) What orientation will result in the larger value of the shield temperature $T_{s}$?

#### Solution

Verified#### Step 1

1 of 6$\texttt{We are given following data:}$

$\begin{align*} T_{1}>T_{2}\\ \end{align*}$

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

$\begin{align*} R_{1} &=\dfrac{1-\epsilon_{1}}{\epsilon_{1}}+\dfrac{1}{F_{1s}}+\dfrac{1-\epsilon_{s}}{\epsilon_{s}}+\dfrac{1-2\epsilon_{s}}{2\epsilon_{s}}+\dfrac{1}{F_{s2}}+\dfrac{1-\epsilon_{2}}{\epsilon_{2}}\ \ \ \ (1)\\ \end{align*}$

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

$\begin{align*} R_{2} &=\dfrac{1-\epsilon_{1}}{\epsilon_{1}}+\dfrac{1}{F_{1s}}+\dfrac{1-2\epsilon_{s}}{2\epsilon_{s}}+\dfrac{1-\epsilon_{s}}{\epsilon_{s}}+\dfrac{1}{F_{s2}}+\dfrac{1-\epsilon_{2}}{\epsilon_{2}}\ \ \ \ \ (2)\\ \end{align*}$

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