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Chapter 13: Problem 29

How does radiosity for a surface differ from the emitted energy? For what kind of surfaces are these two quantities identical?

Short Answer

Expert verified
Answer: Radiosity is the total amount of energy leaving a surface per unit area and time, including both emitted and reflected energy. Emitted energy is the amount of energy generated and released by a surface per unit area and time, depending on the surface's emissivity and temperature. Radiosity and emitted energy are identical for surfaces with ideal black-body characteristics, which have an emissivity of 1 and absorb all incident energy, converting it into emitted energy with no reflected energy.

Step by step solution

01

1. Definition of Radiosity

Radiosity (B) is the total amount of energy leaving a surface per unit area and time. It is the sum of both emitted energy and reflected energy. It takes into account all the energy interactions between the given surface and other objects in the scene.
02

2. Definition of Emitted Energy

Emitted energy (E) is the amount of energy generated and released by a surface per unit area and time. It depends on the surface's emissivity and temperature. It is important to note that not all surfaces emit energy; only those with non-zero emissivity emit energy.
03

3. Relationship between Radiosity and Emitted Energy

Radiosity and emitted energy are related for a given surface as: B = E + R, where R is the reflected energy (energy arriving at the surface from other surfaces and reflected back).
04

4. Identical Scenarios for Radiosity and Emitted Energy

Radiosity and emitted energy are identical for surfaces with ideal black-body characteristics. These surfaces have an emissivity of 1, meaning they absorb all the incident energy and convert it into emitted energy. In this case, there is no reflected energy (R = 0), so B = E.

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Most popular questions from this chapter

Consider a cylindrical enclosure with \(A_{1}, A_{2}\), and \(A_{3}\) representing the internal base, top, and side surfaces, respectively. Using the length to diameter ratio, \(K=L / D\), determine (a) the expression for the view factor from the side surface to itself \(F_{33}\) in terms of \(K\) and \((b)\) the value of the view factor \(F_{33}\) for \(L=D\).

Two gray surfaces that form an 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 has a temperature of \(600 \mathrm{~K}\), an area of \(0.3 \mathrm{~m}^{2}\), and a total emissivity of \(0.6\). If the view factor \(F_{12}\) is \(0.3\), the rate of radiation heat transfer between the two surfaces is (a) \(135 \mathrm{~W}\) (b) \(223 \mathrm{~W}\) (c) \(296 \mathrm{~W}\) (d) \(342 \mathrm{~W}\) (e) \(422 \mathrm{~W}\)

Consider an equimolar mixture of \(\mathrm{CO}_{2}\) and \(\mathrm{O}_{2}\) gases at \(800 \mathrm{~K}\) and a total pressure of \(0.5\) atm. For a path length of \(1.2 \mathrm{~m}\), determine the emissivity of the gas.

A flow-through combustion chamber consists of 15 -cm diameter long tubes immersed in water. Compressed air is routed to the tube, and fuel is sprayed into the compressed air. The combustion gases consist of 70 percent \(\mathrm{N}_{2}\), 9 percent \(\mathrm{H}_{2} \mathrm{O}, 15\) percent \(\mathrm{O}_{2}\), and 6 percent \(\mathrm{CO}_{2}\), and are maintained at \(1 \mathrm{~atm}\) and \(1500 \mathrm{~K}\). The tube surfaces are near black, with an emissivity of \(0.9\). If the tubes are to be maintained at a temperature of \(600 \mathrm{~K}\), determine the rate of heat transfer from combustion gases to tube wall by radiation per \(m\) length of tube.

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