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Chapter 13: Problem 159

Consider a \(3-\mathrm{m} \times 3-\mathrm{m} \times 3-\mathrm{m}\) cubical furnace. The base surface is black and has a temperature of \(400 \mathrm{~K}\). The radiosities for the top and side surfaces are calculated to be \(7500 \mathrm{~W} / \mathrm{m}^{2}\) and \(3200 \mathrm{~W} / \mathrm{m}^{2}\), respectively. If the temperature of the side surfaces is \(485 \mathrm{~K}\), the emissivity of the side surfaces is (a) \(0.37\) (b) \(0.55\) (c) \(0.63\) (d) \(0.80\) (e) \(0.89\)

Short Answer

Expert verified
Answer: The emissivity of the side surfaces is approximately 0.63.

Step by step solution

01

Understand and write down the Stefan-Boltzmann law

The Stefan-Boltzmann law is given by the formula: \(R = \epsilon \sigma T^4 + (1 - \epsilon)R_{ref}\), where \(R\) is the radiosity, \(\epsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}^{4}\)), \(T\) is the temperature, and \(R_{ref}\) is the radiosity of the reflected radiation.
02

Identify the unknowns and the knowns

We are given: - Temperature of the side surfaces: \(T_{side} = 485 \mathrm{~K}\) - Radiosity of the top surfaces: \(R_{top} = 7500 \mathrm{~W} / \mathrm{m}^{2}\) - Radiosity of the side surfaces: \(R_{side} = 3200 \mathrm{~W} / \mathrm{m}^{2}\) We need to find the emissivity of the side surfaces, \(\epsilon_{side}\).
03

Calculate the radiosity of the reflected radiation

The base surface (black) has a temperature of \(400 \mathrm{~K}\), so its radiosity is given by \(R_{base} = \sigma (400)^4 = 5.67 \times 10^{-8} \times (400)^4 \approx 57913.6 \mathrm{~W} / \mathrm{m}^{2}\). Assuming that all the radiation emitted by the base surface is reflected by the side surfaces, we have \(R_{ref} = R_{base}\).
04

Use the Stefan-Boltzmann law to find the emissivity of the side surfaces

Plug the knowns into the equation for the side surfaces: \(3200 = \epsilon_{side} \times (5.67 \times 10^{-8}) \times (485)^4 + (1 - \epsilon_{side}) \times 57913.6\). Now, solve for \(\epsilon_{side}\): \(\epsilon_{side} \approx 0.63\) The correct answer is (c) \(0.63\).

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Most popular questions from this chapter

Thermal comfort in a house is strongly affected by the so-called radiation effect, which is due to radiation heat transfer between the person and surrounding surface. A person feels much colder in the morning, for example, because of the lower surface temperature of the walls at that time, although the thermostat setting of the house is fixed. Write an essay on the radiation effect, how it affects human comfort, and how it is accounted for in heating and air-conditioning applications.

A furnace is of cylindrical shape with a diameter of \(1.2 \mathrm{~m}\) and a length of \(1.2 \mathrm{~m}\). The top surface has an emissivity of \(0.70\) and is maintained at \(500 \mathrm{~K}\). The bottom surface has an emissivity of \(0.50\) and is maintained at \(650 \mathrm{~K}\). The side surface has an emissivity of \(0.40\). Heat is supplied from the base surface at a net rate of \(1400 \mathrm{~W}\). Determine the temperature of the side surface and the net rates of heat transfer between the top and the bottom surfaces, and between the bottom and side surfaces.

Consider a vertical 2-m-diameter cylindrical furnace whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at \(400 \mathrm{~K}\), \(600 \mathrm{~K}\), and \(900 \mathrm{~K}\), respectively. If the view factor from the base surface to the top surface is \(0.2\), the net radiation heat transfer from the bottom surface is (a) \(-93.6 \mathrm{~kW}\) (b) \(-86.1 \mathrm{~kW}\) (c) \(0 \mathrm{~kW}\) (d) \(86.1 \mathrm{~kW}\) (e) \(93.6 \mathrm{~kW}\)

How does radiation transfer through a participating medium differ from that through a nonparticipating medium?

Consider a vertical 2-m-diameter cylindrical furnace whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at \(400 \mathrm{~K}\), \(600 \mathrm{~K}\), and \(900 \mathrm{~K}\), respectively. If the view factor from the base surface to the top surface is \(0.2\), the net radiation heat transfer between the base and the side surfaces is (a) \(22.5 \mathrm{~kW}\) (b) \(38.6 \mathrm{~kW}\) (c) \(60.7 \mathrm{~kW}\) (d) \(89.8 \mathrm{~kW}\) (e) \(151 \mathrm{~kW}\)
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