Question

The surfaces of a two-surface 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 is black, has a temperature of \(800 \mathrm{~K}\), and has area of \(0.3 \mathrm{~m}^{2}\). If the view factor \(F_{12}\) is \(0.3\), the rate of radiation heat transfer between the two surfaces is (a) \(340 \mathrm{~W}\) (b) \(560 \mathrm{~W}\) (c) \(780 \mathrm{~W}\) (d) \(900 \mathrm{~W}\) (e) \(1160 \mathrm{~W}\)

Short answer

Answer: The rate of radiation heat transfer between the two surfaces is approximately 560 W.

Step-by-step solution

Identify the given information

Temperature of Surface 1 (T1): 400K Area of Surface 1 (A1): 0.2 m² Total emissivity of Surface 1 (ε1): 0.4 Temperature of Surface 2 (T2): 800K Area of Surface 2 (A2): 0.3 m² Surface 2 is black, so its emissivity (ε2) is 1. View factor between the surfaces (F12): 0.3

Apply the Stefan-Boltzmann Constant

The Stefan-Boltzmann Constant (σ) is a universal constant with a value of \(5.67 \times 10^{-8} \mathrm{~W}/\mathrm{m}^2 \mathrm{~K}^4\).

Calculate the radiation heat transfer

Now, we can plug all the given information and constants into the formula to calculate the rate of radiation heat transfer: $$ Q = 0.3 * 0.2 * ( (0.4 * (5.67 * 10^{-8}) * (400^4 - 800^4)) + ((1 - 0.4) * 1 * (5.67 * 10^{-8}) * (800^4 - 400^4)) )$$ After solving the equation, we get the value of Q. $$ Q \approx 560 \mathrm{~W} $$ The radiation heat transfer rate between the two surfaces is approximately \(560 \mathrm{~W}\), which corresponds to option (b).