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Chapter 13: Problem 156

Consider two infinitely long concentric cylinders with diameters 20 and \(25 \mathrm{~cm}\). The inner surface is maintained at \(700 \mathrm{~K}\) and has an emissivity of \(0.40\), while the outer surface is black. If the rate of radiation heat transfer from the inner surface to the outer surface is \(2400 \mathrm{~W}\) per unit area of the inner surface, the temperature of the outer surface is (a) \(605 \mathrm{~K}\) (b) \(538 \mathrm{~K}\) (c) \(517 \mathrm{~K}\) (d) \(451 \mathrm{~K}\) (e) \(415 \mathrm{~K}\)

Short Answer

Expert verified
Answer: (b) 538 K

Step by step solution

01

Identify the given information

We have two concentric cylinders. The inner cylinder has diameter 20 cm, temperature 700 K, and emissivity 0.40. The outer cylinder has an unknown temperature and is a blackbody (emissivity=1). The rate of radiation heat transfer from the inner surface to the outer surface is 2400 W per unit area.
02

Apply the Stefan-Boltzmann Law

The Stefan-Boltzmann law states that the heat transfer rate per unit area Q/A is proportional to the temperature difference between the two surfaces raised to the fourth power. So, we have: Q/A = ε_inner * σ * (T1^4 - T2^4) where ε_inner is the emissivity of the inner surface, σ is the Stefan-Boltzmann constant (5.67 * 10^-8 W/m².K⁴), T1 is the temperature of the inner surface, and T2 is the temperature of the outer surface.
03

Insert given values and solve for T2

We know the Q/A, ε_inner, and T1 values, so we can plug those values in and solve for T2: 2400 = 0.40 * (5.67 * 10^-8) * (700^4 - T2^4) Now we need to isolate T2^4. First, divide both sides by (0.40 * (5.67 * 10^-8)): 2400 / (0.40 * (5.67 * 10^-8)) = 700^4 - T2^4 Next, subtract 700^4 from both sides: 700^4 - 2400 / (0.40 * (5.67 * 10^-8)) = T2^4 Now, we need to find the fourth root of the value on the right side of the equation to determine the temperature of the outer surface: T2 = (700^4 - 2400 / (0.40 * (5.67 * 10^-8)))^(1/4) T2 ≈ 538 K
04

Select the correct answer

We have calculated the temperature of the outer surface to be 538 K, so the correct answer is: (b) 538 K

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