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Chapter 13: Problem 162

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 \(600 \mathrm{~K}\), and an 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) \(87 \mathrm{~W}\) (b) \(135 \mathrm{~W}\) (c) \(244 \mathrm{~W}\) (d) \(342 \mathrm{~W}\) (e) \(386 \mathrm{~W}\)

Short Answer

Expert verified
Based on the given surface temperatures, emissivities, areas, and view factor, we calculated the heat transfer rate using the net radiation heat transfer equation and found that the rate of radiation heat transfer between the two surfaces is approximately 135 W.

Step by step solution

01

Write down the given values

We are given the following values: - Surface 1 (S1): Temperature: \(T_1 = 400 \mathrm{~K}\), Area: \(A_1 = 0.2 \mathrm{~m}^{2}\), Emissivity: \(e_1 = 0.4\) - Surface 2 (S2): Temperature: \(T_2 = 600 \mathrm{~K}\), Area: \(A_2 = 0.3 \mathrm{~m}^{2}\) (Blackbody, so, Emissivity: \(e_2 = 1\)) - View factor between surface 1 and surface 2: \(F_{12} = 0.3\)
02

Use the net radiation heat transfer equation

The net radiation heat transfer equation is given by: $$q = A_1 F_{12} \sigma (e_1(T_1^4 - \frac{1-e_1}{e_1} T_2^4) + e_2(T_2^4 - \frac{1-e_2}{e_2} T_1^4))$$ Here, \(q\) is the heat transfer rate and \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}}\)).
03

Substitute the given values into the equation and solve for q

Plugging in the given values, we get: $$q = 0.2 \times 0.3 \times 5.67 \times 10^{-8} (0.4(400^4 - \frac{1-0.4}{0.4} 600^4) + 1(600^4 - \frac{1-1}{1} 400^4))$$ Solving for \(q\): $$q = 0.2 \times 0.3 \times 5.67 \times 10^{-8} (0.4(2.56 \times 10^8 - 1.5 \times 8.64 \times 10^8) + 1(8.64 \times 10^8 - 0))$$ $$q \approx 135.13 \mathrm{~W}$$ The rate of radiation heat transfer between the two surfaces is approximately \(135 \mathrm{~W}\), so the correct option is (b).

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Most popular questions from this chapter

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