Two very large parallel plates are maintained at uniform temperatures \(T_{1}=750 \mathrm{~K}\) and \(T_{2}=500 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.85\) and \(\varepsilon_{2}=0.7\), respectively. If a thin aluminum sheet with the same emissivity on both sides is to be placed between the plates in order to reduce the net rate of radiation heat transfer between the plates by 90 percent, the emissivity of the aluminum sheet must be (a) \(0.07\) (b) \(0.10\) (c) \(0.13\) (d) \(0.16\) (e) \(0.19\)

#tag_title#Step 3: Calculate the net radiation heat transfer with the aluminum sheet#tag_content#Since we want to reduce the net radiation heat transfer by 90%, we can write the desired net radiation heat transfer as: \(Q_{12_{reduced}} = 0.1 * Q_{12}\) Now, we can also express the net radiation heat transfer with the aluminum sheet in terms of the equivalent emissivity of the three surfaces: \(Q_{12_{reduced}} = A\sigma \varepsilon_e (T_1^4 - T_2^4)\) #tag_title#Step 4: Find the required emissivity of the aluminum sheet#tag_content#To find the required emissivity of the aluminum sheet, we need to substitute the expressions for \(Q_{12_{reduced}}\) and \(\varepsilon_e\) in terms of \(Q_{12}\) and the emissivities of the original plates. Then, we can solve for the emissivity of the aluminum sheet: \(0.1 * Q_{12} = A\sigma (\frac{1}{\frac{1 - \varepsilon_1}{\varepsilon_1} + \frac{1 - \varepsilon_2}{\varepsilon_2} + \frac{1 - \varepsilon_{Al}}{\varepsilon_{Al}}}) (T_1^4 - T_2^4)\) Where \(\varepsilon_{Al}\) is the emissivity of the aluminum sheet. We can then solve for \(\varepsilon_{Al}\): \(\varepsilon_{Al} = \frac{1}{\frac{10 - \varepsilon_1 / \varepsilon_1}{\varepsilon_1} + \frac{10 - \varepsilon_2 / \varepsilon_2}{\varepsilon_2} - 9}\) Substituting the given values for the emissivities and temperatures, we can find the required emissivity of the aluminum sheet. #Short Answer# The required emissivity of the aluminum sheet to reduce the net radiation heat transfer between the two plates by 90% is approximately 0.087.