Question

Two gray surfaces that form an enclosure exchange heat with one another by thermal radiation. Surface 1 has a temperature of \(400 \mathrm{~K}\), an area of \(0.2 \mathrm{~m}^{2}\), and a total emissivity of \(0.4\). Surface 2 has a temperature of \(600 \mathrm{~K}\), an area of \(0.3 \mathrm{~m}^{2}\), and a total emissivity of \(0.6\). If the view factor \(F_{12}\) is \(0.3\), the rate of radiation heat transfer between the two surfaces is (a) \(135 \mathrm{~W}\) (b) \(223 \mathrm{~W}\) (c) \(296 \mathrm{~W}\) (d) \(342 \mathrm{~W}\) (e) \(422 \mathrm{~W}\)

Short answer

Answer: The rate of radiation heat transfer between the two gray surfaces in the given enclosure is approximately 296 W.

Step-by-step solution

Use the net radiation heat transfer equation

The net radiation heat transfer between two surfaces can be expressed as: $$Q_{net} = \frac{\sigma F_{12} A_1 A_2 (\epsilon_2 T_2^4 - \epsilon_1 T_1^4)}{(1 - \epsilon_1) A_1 + (1 - \epsilon_2) A_2}$$ where \(Q_{net}\) is the net radiation heat transfer, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{~W/m^2 K^4}\)), \(F_{12}\) is the view factor, \(A_i\) and \(\epsilon_i\) are the area and emissivity of surface \(i\), and \(T_i\) is the temperature of surface \(i\).

Substitute the values provided

Write the equation with given values: $$Q_{net} = \frac{(5.67 \times 10^{-8} \mathrm{~W/m^2 K^4})(0.3)(0.2 \mathrm{\, m^2})(0.3 \mathrm{\, m^2})(0.6 \times 600^4 \mathrm{\, K^4} - 0.4 \times 400^4 \mathrm{\, K^4})}{(1 - 0.4) (0.2 \mathrm{\, m^2}) + (1 - 0.6) (0.3 \mathrm{\, m^2})}$$

Perform calculations

Calculate the values and simplify: $$Q_{net} \approx \frac{(5.67 \times 10^{-8})(0.3)(0.2)(0.3)(132.475 \times 10^{8})}{0.32}$$ $$Q_{net} \approx 296.35 \mathrm{~W}$$

Choose the closest multiple-choice answer

From the given options, the closest value to our calculated heat transfer rate is \(296 \mathrm{~W}\). So, the correct answer is (c).