A thin aluminum sheet with an emissivity of \(0.15\) on both sides is placed between two very large parallel plates, which are maintained at uniform temperatures \(T_{1}=900 \mathrm{~K}\) and \(T_{2}=650 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.8\), respectively. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates and compare the result with that without the shield.

We need to use the equation for effective emissivity between surfaces, given by: \(\frac{1}{\varepsilon_{\text{effective,1-2}}} = \frac{1-\varepsilon_1}{\varepsilon_1 \mathcal{A}_1} +\frac{\mathcal{A}_1}{\mathcal{A}_2}\frac{1-\varepsilon_2}{\varepsilon_2}\) In this case, the area of both plates is the same, so \(\mathcal{A}_1=\mathcal{A}_2\), and the equation simplifies to: \(\varepsilon_{\text{effective,1-2}} = \frac{\varepsilon_1 \varepsilon_2}{\varepsilon_1+\varepsilon_2-\varepsilon_1 \varepsilon_2}\) Now we can plug in the given emissivity values to find the effective emissivity between the plates without the shield: \(\varepsilon_{\text{effective,1-2}} = \frac{(0.5)(0.8)}{0.5+0.8-(0.5)(0.8)}\) We get: \(\varepsilon_{\text{effective,1-2}} = 0.615\) With the shield in place, we have to modify our equation to account for the additional surface: \(\frac{1}{\varepsilon_{\text{effective,1-3}}} = \frac{1-\varepsilon_1}{\varepsilon_1 \mathcal{A}_1} +\frac{\mathcal{A}_1}{\mathcal{A}_{s}}\frac{1-\varepsilon_s}{\varepsilon_s}+\frac{\mathcal{A}_{s}}{\mathcal{A}_2}\frac{1-\varepsilon_2}{\varepsilon_2}\) For this case, the areas of the plates are also the same as the shield's (\(\mathcal{A}_1=\mathcal{A}_{s}=\mathcal{A}_2\)), and we can plug in the given emissivity values: \(\varepsilon_{\text{effective,1-3}} = \frac{(0.5)(0.15)(0.8)}{[(0.5)(0.15)+0.15+0.15^2+(0.15)(0.8)]}\) We get: \(\varepsilon_{\text{effective,1-3}} = 0.162\)

Now we can use the formula for radiation heat transfer per unit surface area of the plates: \(q = \varepsilon_{\text{effective}}\sigma (T_1^4-T_2^4)\) Plug in the given temperatures and the calculated effective emissivities to find the heat transfer for both cases: \(q_{\text{Without Shield}} = 0.615\sigma (900^4-650^4)\) \(q_{\text{With Shield}} = 0.162\sigma (900^4-650^4)\)

Divide the heat transfer rate with and without the shield to find the ratio: \(\text{Heat Transfer Ratio} = \frac{q_{\text{With Shield}}}{q_{\text{Without Shield}}} = \frac{0.162}{0.615}\) We get the heat transfer ratio as: \(\text{Heat Transfer Ratio} \approx 0.263\) So, by comparing the results, we find that with the shield in place, the rate of radiation heat transfer between the two plates is only about 26.3% of the heat transfer without the shield.