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Chapter 13: Problem 151

Consider two concentric spheres forming an enclosure with diameters of \(12 \mathrm{~cm}\) and \(18 \mathrm{~cm}\) and surface temperatures \(300 \mathrm{~K}\) and \(500 \mathrm{~K}\), respectively. Assuming that the surfaces are black, the net radiation exchange between the two spheres is (a) \(21 \mathrm{~W}\) (b) \(140 \mathrm{~W}\) (c) \(160 \mathrm{~W}\) (d) \(1275 \mathrm{~W}\) (e) \(3084 \mathrm{~W}\)

Short Answer

Expert verified
Question: Calculate the net radiation exchange between two concentric black spheres with diameters of 12 cm and 18 cm and surface temperatures of 300 K and 500 K, respectively. Answer: The net radiation exchange between the two concentric black spheres is approximately 5814.7 W. Unfortunately, none of the given options matches this value, which might indicate a mistake in the provided options or the question itself.

Step by step solution

01

Write down the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power radiated per unit area of a black body is given by: \(P/A = \sigma T^4\) where - \(P\) is the radiated power, - \(A\) is the surface area, - \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{Wm^{-2}K^{-4}}\)), - \(T\) is the temperature of the body in Kelvin.
02

Calculate the surface area of each sphere

To calculate the surface area of each sphere, we can use the formula: \(A = 4\pi r^2\) where \(r\) is the radius of the sphere. The diameters are \(12\mathrm{~cm}\) and \(18\mathrm{~cm}\), so the radii are \(6\mathrm{~cm}\) and \(9\mathrm{~cm}\). Convert these values to meters. \(r_1 = 0.06\mathrm{~m}\) \(r_2 = 0.09\mathrm{~m}\) Now, we can calculate the surface areas: \(A_1 = 4\pi (0.06)^2 = 0.0452\mathrm{~m^2}\) \(A_2 = 4\pi (0.09)^2 = 0.1018\mathrm{~m^2}\)
03

Determine the radiated power of each sphere

By using the Stefan-Boltzmann Law, we can now determine the radiated power of each sphere: \(P_1 = A_1 \sigma (300)^4 = 0.0452 \times 5.67 \times 10^{-8} \times (300)^4 = 304.775\mathrm{~W}\) \(P_2 = A_2 \sigma (500)^4 = 0.1018 \times 5.67 \times 10^{-8} \times (500)^4 = 6119.475\mathrm{~W}\)
04

Calculate the net radiation exchange between the spheres

The net radiation exchange is the difference between the radiated power of the outer sphere and the radiated power of the inner sphere: \(P_\mathrm{net} = P_2 - P_1 = 6119.475 - 304.775 = 5814.7\mathrm{~W}\)
05

Compare to the given options

Comparing our result with the given options, we can see that none of the options matches our calculated value of \(5814.7\mathrm{~W}\). Therefore, it seems that there might be a mistake in the provided options or the question itself.

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Most popular questions from this chapter

What is operative temperature? How is it related to the mean ambient and radiant temperatures? How does it differ from effective temperature?

Why is the radiation analysis of enclosures that consist of black surfaces relatively easy? How is the rate of radiation heat transfer between two surfaces expressed in this case?

Consider a surface at \(0^{\circ} \mathrm{C}\) that may be assumed to be a blackbody in an environment at \(25^{\circ} \mathrm{C}\). If \(300 \mathrm{~W} / \mathrm{m}^{2}\) of radiation is incident on the surface, the radiosity of this black surface is (a) \(0 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(15 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(132 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(300 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(315 \mathrm{~W} / \mathrm{m}^{2}\)

A spherical tank, with an inner diameter of \(D_{1}=\) \(3 \mathrm{~m}\), is filled with a solution undergoing an exothermic reaction that heats the surface to a uniform temperature of \(120^{\circ} \mathrm{C}\). To prevent thermal burn hazards, the tank is enclosed with a concentric outer cover that provides an evacuated gap of \(5 \mathrm{~cm}\) in the enclosure. Both spherical surfaces have the same emissivity of \(0.5\), and the outer surface is exposed to natural convection with a heat transfer coefficient of \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and radiation heat transfer with the surrounding at a temperature of \(30^{\circ} \mathrm{C}\). Determine whether or not the vacuumed gap is sufficient to keep the outer surface temperature below \(45^{\circ} \mathrm{C}\) to prevent thermal burns. If not, propose a solution to keep the outer surface temperature below \(45^{\circ} \mathrm{C}\).

What does the view factor represent? When is the view factor from a surface to itself not zero?
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