Question

A radiation shield that has the same emissivity \(\varepsilon_{3}\) on both sides is placed between two large parallel plates, which are maintained at uniform temperatures of \(T_{1}=650 \mathrm{~K}\) and \(T_{2}=400 \mathrm{~K}\) and have emissivities of \(\varepsilon_{1}=0.6\) and \(\varepsilon_{2}=0.9\), respectively. Determine the emissivity of the radiation shield if the radiation heat transfer between the plates is to be reduced to 15 percent of that without the radiation shield.

Short answer

Answer: The emissivity of the radiation shield is approximately 0.322.

Step-by-step solution

Determine the radiation heat transfer rate without the shield

According to the Stefan-Boltzmann law, the radiation heat transfer rate between two large parallel plates can be calculated as follows: \(Q = \sigma(1 / \frac{1 - \varepsilon_1}{\varepsilon_1} + \frac{1}{A} + \frac{1 - \varepsilon_2}{\varepsilon_2})^{-1}(T_1^4 - T_2^4)\) where \(Q\) is the heat transfer rate, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 × 10^{-8} \, W/m^2K^4\)), A is the area, and T are the temperatures. Since there is no shield between the plates, we can simplify the equation as follows: \(Q_0 = \sigma (1 / \frac{1 - \varepsilon_1}{\varepsilon_1} + \frac{1 - \varepsilon_2}{\varepsilon_2})^{-1}(T_1^4 - T_2^4)\) Substitute the given values into the equation: \(Q_0 = 5.67 × 10^{-8}(1 / \frac{1 - 0.6}{0.6} + \frac{1 - 0.9}{0.9})^{-1}(650^4 - 400^4)\) Calculate the heat transfer rate: \(Q_0 \approx 7035 \, W\)

Determine the condition for the radiation heat transfer rate with the shield

The condition given is that the radiation heat transfer between the plates should be reduced to 15 percent of that without the shield. Mathematically, this can be represented as: \(Q_{with\,shield} = 0.15 × Q_{without\,shield}\) Substitute the value of \(Q_{without\,shield}\) from Step 1: \(Q_{with\,shield} = 0.15 × 7035 \, W\) Calculate the desired heat transfer rate: \(Q_{with\,shield} \approx 1055.25 \, W\)

Determine the emissivity of the radiation shield

When the shield is in place, we have to account for the reflections on both sides of the shield. The equation for the radiation heat transfer rate becomes: \(Q_{with\,shield} = \sigma(1 / \frac{1 - \varepsilon_1}{\varepsilon_1} + \frac{1}{A} + \frac{1 - \varepsilon_3}{\varepsilon_3} + \frac{1}{A} + \frac{1 - \varepsilon_2}{\varepsilon_2})^{-1}(T_1^4 - T_2^4)\) Since we know the desired heat transfer rate (\(Q_{with\,shield}\)), we can solve this equation for the unknown emissivity of the shield (\(\varepsilon_3\)): \(\varepsilon_3 = \frac{1}{ (\frac{1 - \varepsilon_1}{\varepsilon_1} + \frac{1}{A} + \frac{(1 - \varepsilon_2)Q_{with\,shield}}{\varepsilon_2\sigma (T_1^4 - T_2^4)} + \frac{1 - \varepsilon_1}{\varepsilon_1}) - (\frac{1 - \varepsilon_1}{\varepsilon_1} + \frac{1 - \varepsilon_2}{\varepsilon_2})}\) Substitute the known values and calculate the emissivity of the radiation shield: \(\varepsilon_3 = \frac{1}{ (\frac{1 - 0.6}{0.6} + \frac{1}{A} + \frac{(1 - 0.9)1055.25}{0.9(5.67 × 10^{-8})(650^4 - 400^4)} + \frac{1 - 0.6}{0.6}) - (\frac{1 - 0.6}{0.6} + \frac{1 - 0.9}{0.9})}\) \(\varepsilon_3 \approx 0.322\) The emissivity of the radiation shield is approximately 0.322.