Question

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.

Short answer

In summary, a thin aluminum sheet with emissivity of 0.15 on both sides is placed between two large parallel plates with temperatures T1 and T2, and emissivities ε1 and ε2. The aluminum sheet acts as a shield, reducing the net rate of radiation heat transfer per unit surface area between the two plates. The net rate of radiation heat transfer without the shield is 12615.53 W/m², whereas with the shield, it is reduced to 3123.11 W/m².

Step-by-step solution

Define the formula for radiative heat transfer between two surfaces

The formula for radiative heat transfer between two surfaces \(A_1\) and \(A_2\) with temperatures \(T_1\) and \(T_2\) is given by: $$Q=\frac{\sigma (T_1^4 - T_2^4)}{\frac{1-\varepsilon_1}{\varepsilon_1 A_1} + \frac{1}{A_1 F_{1-2}} + \frac{1-\varepsilon_2}{\varepsilon_2 A_2}}$$ where \(\sigma\) is the Stefan-Boltzmann constant, and \(F_{1-2}\) is the view factors between the two surfaces. We can assume \(A_1 = A_2 = A\) for our case, as the plates are very large parallel plates. Since we want to find the net rate of radiation heat transfer per unit surface area, we can divide the above equation by \(A\), which gives: $$q=\frac{\sigma (T_1^4 - T_2^4)}{\frac{1-\varepsilon_1}{\varepsilon_1} + \frac{1}{ F_{1-2}} + \frac{1-\varepsilon_2}{\varepsilon_2}}$$

Calculate the case without the shield

In this case, we don't have the aluminum sheet between the two plates, therefore, we can directly plug the values into the radiative heat transfer formula: $$q_{\text{without shield}}=\frac{\sigma (T_1^4 - T_2^4)}{\frac{1-\varepsilon_1}{\varepsilon_1} + \frac{1}{1} + \frac{1-\varepsilon_2}{\varepsilon_2}}$$ Substituting the given values of temperatures, emissivities and the Stefan-Boltzmann constant(\(\sigma=5.67 \times 10^{-8} \frac{W}{m^2 K^4}\)): $$q_{\text{without shield}}=\frac{(5.67 \times 10^{-8})(900^4 - 650^4)}{\frac{1-0.5}{0.5} + 1 + \frac{1-0.8}{0.8}} \Rightarrow q_{\text{without shield}} = 12615.53 \frac{W}{m^2}$$

Define the formula for the case with the shield

Now, we need to calculate the case with the shield. Since the shield is placed between the plates, we have one additional surface with emissivity 0.15 to consider. We can generalize the formula for radiative heat transfer as follows: $$q_{\text{with shield}}=\frac{\sigma (T_1^4 - T_2^4)}{\frac{1-\varepsilon_1}{\varepsilon_1} + \frac{1}{F_{1-2}} + \frac{1-\varepsilon_3}{\varepsilon_3} + \frac{1}{ F_{2-3}} + \frac{1-\varepsilon_2}{\varepsilon_2}}$$ Since we have very large parallel plates, we can assume \(F_{1-2} = F_{2-3} = 1\).

Calculate the case with the shield

Plugging the values of temperatures, emissivities, and the Stefan-Boltzmann constant(\(\sigma\)): $$q_{\text{with shield}}=\frac{(5.67 \times 10^{-8})(900^4 - 650^4)}{\frac{1-0.5}{0.5} + \frac{1}{1} + \frac{1-0.15}{0.15}+\frac{1}{1}+\frac{1-0.8}{0.8}} \Rightarrow q_{\text{with shield}} = 3123.11 \frac{W}{m^2}$$

Compare the results

Now we have calculated the values for the case without the shield (\(q_{\text{without shield}} = 12615.53 \frac{W}{m^2}\)) and the case with the shield (\(q_{\text{with shield}} = 3123.11 \frac{W}{m^2}\)). As we can see, the net rate of radiation heat transfer without the shield is higher than the case with the shield. The shield acts as an insulator, reducing the heat transfer between the two plates by a significant amount.