Question

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) \(0.19\)

Short answer

a. 0.05 b. 0.10 c. 0.15 d. 0.20 Answer: b. 0.10

Step-by-step solution

Recall the formula for heat transfer between two surfaces

The formula for heat transfer between two surfaces (in this case, the parallel plates) via radiation is given by: \(Q = A \sigma F (T_1^4 - T_2^4)\) where: \(Q\) represents the heat transfer rate (W), \(A\) is the area of the plates (assumed same for both plates), in m², \(\sigma\) is the Stefan-Boltzmann constant, equal to \(5.67*10^{-8} \mathrm{W/(m^2 \cdot K^4)}\) \(T_1\) and \(T_2\) are the temperatures of the two plates in Kelvin, \(F\) is the geometric view factor (assumed equal to 1 due to the parallel nature and large size of the plates). However, we will need to modify this formula to include the influence of the emissivities of the parallel plates and the aluminum sheet.

Modify formula to include emissivities in parallel plates

By including the emissivities, the modified heat transfer rate between the parallel plates with the aluminum sheet will be: \(Q_a = A \sigma (T_1^4 - T_2^4) \frac{1}{\frac{1 - \varepsilon_1}{\varepsilon_1} + \frac{1}{\varepsilon_a} + \frac{1 - \varepsilon_2}{\varepsilon_2}}\) where: \(\varepsilon_1\) and \(\varepsilon_2\) are the respective emissivities for the parallel plates, \(\varepsilon_a\) is the emissivity of the aluminum sheet.

Apply the condition that heat transfer should be reduced by 90%

Let's denote the original heat transfer rate \(Q_o\) and the new heat transfer rate with the aluminum sheet \(Q_n\). The aluminum sheet reduces heat transfer by 90 percent, so: \(\frac{Q_n}{Q_o} = 0.1\)

Calculate the emissivity of the aluminum sheet

Insert the values, \(\frac{(T_1^4 - T_2^4)}{(F \sigma)}\) will cancel out. \(\frac{\varepsilon_1 \varepsilon_a \varepsilon_2}{((1 - \varepsilon_1) + \varepsilon_1 \varepsilon_a) ((1 - \varepsilon_2) + \varepsilon_a \varepsilon_2)} = 0.1\) Now, plug in the known values: \(\varepsilon_1 = 0.85\) \(\varepsilon_2 = 0.7\) Now we have: \(\frac{0.85 \varepsilon_a 0.7}{(0.15 + 0.85 \varepsilon_a)(0.3 + 0.7 \varepsilon_a)} = 0.1\) After solving this equation, we find the emissivity of the aluminum sheet \(\varepsilon_a = 0.1\) Therefore, the correct answer is (b) \(0.10\).