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The number of view factors that need to be evaluated directly for a 10 -surface enclosure is (a) 1 (b) 10 (c) 22 (d) 34 (e) 45

Short Answer

Expert verified
Answer: (e) 45

Step by step solution

01

Understand the view factor formula for a closed enclosure

For a closed enclosure with 'n' surfaces, the total number of view factors can be calculated using the following formula: Number of view factors = n(n-1)/2 In this problem, we have a 10-surface enclosure, so n = 10.
02

Calculate the number of view factors for the given enclosure

Now, we can use the view factor formula to determine the number of view factors for a 10-surface enclosure: Number of view factors = 10(10-1)/2 This simplifies to: Number of view factors = 10(9)/2 = 45
03

Compare to the given options and find the correct answer

After calculating the number of view factors for the given enclosure, we can now compare it to the provided options: (a) 1 (b) 10 (c) 22 (d) 34 (e) 45 As we calculated in Step 2, there are indeed 45 view factors that need to be directly evaluated for the given 10-surface enclosure. Therefore, the correct answer is (e) 45.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radiative Heat Transfer
Radiative heat transfer is a mode of heat exchange that involves the transfer of energy via electromagnetic waves. Unlike conduction and convection, radiative transfer does not require a medium; it can occur across the vacuum of space. Materials can emit, absorb, reflect, or transmit radiative energy, and these properties determine how they interact with thermal radiation. For students dealing with heat transfer problems, it's crucial to understand that all bodies above absolute zero emit thermal radiation, and the amount of radiative heat transfer depends on the temperature difference between the surfaces and their emissive properties.

In our exercise regarding a 10-surface enclosure, each surface can exchange radiation with the others, and the complexity of calculating this exchange can increase with the number of surfaces. This detail is particularly important because it affects the comfort, design, and energy efficiency in real-life applications such as buildings, furnaces, and electronic devices.
Enclosure Analysis
Enclosure analysis forms an integral part of radiative heat transfer calculations, wherein the interactions between various surfaces are evaluated. A 'view factor', also known as a 'configuration factor' or 'angle factor', quantifies the proportion of radiated energy leaving one surface that directly reaches another surface. View factors depend on the geometry of the enclosure and the relative orientations of the surfaces. They are utilized in determining the heat exchange within an enclosure with multiple surfaces.

To simplify the understanding for students, consider that an enclosure analysis is akin to analyzing how much each person in a room can see of every other person. This concept can be translated into the fraction of heat exchangeable between surfaces. In the exercise example, a 10-surface enclosure equates to calculating how each of the 10 surfaces 'views' the other 9. Thus, it is essential to evaluate this complex web of interactions for accurate heat transfer analyses in enclosures, impacting a broad range of engineering fields such as aerospace, automotive, and architectural design.
Heat Transfer Calculations
Heat transfer calculations are fundamental to engineering, as these calculations help engineers to design systems for heating and cooling, and ensure the thermal comfort of environments. These calculations can range from simple to complex, depending on the modes of heat transfer involved—conduction, convection, and radiation—and the number of interacting surfaces within a system.

In the textbook exercise, we encountered the formula \( \frac{{n(n-1)}}{2} \) for determining the number of view factors in a closed enclosure. For students, mastering heat transfer calculations requires understanding the interplay between surface geometry, material properties, and environmental conditions. Accuracy in these calculations ensures effective thermal management, be it for preventing electronic components from overheating or optimizing the climate control systems of a building. Remember, even small errors in these calculations can have significant impacts on system performance and energy consumption.

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Most popular questions from this chapter

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.

Consider a 15-cm-diameter sphere placed within a cubical enclosure with a side length of \(15 \mathrm{~cm}\). The view factor from any of the square-cube surfaces to the sphere is (a) \(0.09\) (b) \(0.26\) (c) \(0.52\) (d) \(0.78\) (e) 1

A 2-m-internal-diameter double-walled spherical tank is used to store iced water at \(0^{\circ} \mathrm{C}\). Each wall is \(0.5 \mathrm{~cm}\) thick, and the \(1.5\)-cm-thick air space between the two walls of the tank is evacuated in order to minimize heat transfer. The surfaces surrounding the evacuated space are polished so that each surface has an emissivity of \(0.15\). The temperature of the outer wall of the tank is measured to be \(20^{\circ} \mathrm{C}\). Assuming the inner wall of the steel tank to be at \(0^{\circ} \mathrm{C}\), determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24-h period.

Two very large parallel plates are maintained at uniform temperatures of \(T_{1}=600 \mathrm{~K}\) and \(T_{2}=400 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.9\), respectively. Determine the net rate of radiation heat transfer between the two surfaces per unit area of the plates.

A spherical tank of diameter \(D=2 \mathrm{~m}\) that is filled with liquid nitrogen at \(100 \mathrm{~K}\) is kept in an evacuated cubic enclosure whose sides are \(3 \mathrm{~m}\) long. The emissivities of the spherical tank and the enclosure are \(\varepsilon_{1}=0.1\) and \(\varepsilon_{2}=0.8\), respectively. If the temperature of the cubic enclosure is measured to be \(240 \mathrm{~K}\), determine the net rate of radiation heat transfer to the liquid nitrogen. Answer: \(228 \mathrm{~W}\)

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