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Chapter 13: Problem 146

Consider an infinitely long three-sided triangular enclosure with side lengths \(2 \mathrm{~cm}, 3 \mathrm{~cm}\), and \(4 \mathrm{~cm}\). The view factor from the \(2 \mathrm{~cm}\) side to the \(4 \mathrm{~cm}\) side is (a) \(0.25\) (b) \(0.50\) (c) \(0.64\) (d) \(0.75\) (e) \(0.87\)

Short Answer

Expert verified
Answer: None of the given options match the calculated view factor of 0.3333.

Step by step solution

01

Calculate Area of Surfaces 1 and 2

As given, Surface 1 has a length of \(2\mathrm{~cm}\) and Surface 2 has a length of \(4\mathrm{~cm}\). Since the triangular enclosure is infinitely long, we can assume that the height of the triangle is the surface length itself. So, the areas of Surface 1 and Surface 2 can be found using the formula for area of a rectangle (considering length as the base and height as the surface length): Area of Surface 1: $$A_1 = \text{length} \times \text{height} = 2\mathrm{~cm} \times 2\mathrm{~cm} = 4\mathrm{~cm^2}$$ Area of Surface 2: $$A_2 = \text{length} \times \text{height} = 3\mathrm{~cm} \times 4\mathrm{~cm} = 12\mathrm{~cm^2}$$
02

Calculate the View Factor

Now that we have calculated the areas of Surface 1 and Surface 2, we can find the view factor \(F_{12}\) using the formula for view factor: $$F_{12} = \frac{A_1}{A_2}$$ Substitute the values for \(A_1\) and \(A_2\) from Step 1: $$F_{12} = \frac{4\mathrm{~cm^2}}{12\mathrm{~cm^2}} = \frac{1}{3}$$ Now, convert the fraction to a decimal: $$F_{12} = \frac{1}{3} = 0.3333$$ Comparing this value to the given options: (a) \(0.25\) (b) \(0.50\) (c) \(0.64\) (d) \(0.75\) (e) \(0.87\) We can see that none of these options match our calculated value of \(0.3333\).

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Most popular questions from this chapter

A solid sphere of \(1 \mathrm{~m}\) diameter at \(600 \mathrm{~K}\) is kept in an evacuated equilateral triangular enclosure (a tetrahedron) whose side length is \(L\). Note that for the sphere to touch the tetrahedron's_surfaces, the tetrahedron's side length should be \(L=D \sqrt{6}\). The emissivity of the sphere is \(0.45\) and the temperature of the enclosure is \(420 \mathrm{~K}\). If heat is generated uniformly within the sphere at a rate of \(3100 \mathrm{~W}\), determine (a) the view factor from the enclosure to the sphere and \((b)\) the emissivity of the enclosure.

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 \(L=D\).

A dryer is shaped like a long semicylindrical duct of diameter \(1.5 \mathrm{~m}\). The base of the dryer is occupied with water soaked materials to be dried, and maintained at a temperature of \(370 \mathrm{~K}\) and emissivity of \(0.5\). The dome of the dryer is maintained at \(1000 \mathrm{~K}\) with emissivity of \(0.8\). Determine the drying rate per unit length experienced by the wet materials.

A furnace is shaped like a long equilateral-triangular duct where the width of each side is \(2 \mathrm{~m}\). Heat is supplied from the base surface, whose emissivity is \(\varepsilon_{1}=0.8\), at a rate of \(800 \mathrm{~W} / \mathrm{m}^{2}\) while the side surfaces, whose emissivities are \(0.5\), are maintained at \(500 \mathrm{~K}\). Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure?

An average person produces \(0.50 \mathrm{lbm}\) of moisture while taking a shower and \(0.12 \mathrm{lbm}\) while bathing in a tub. Consider a family of four who shower once a day in a bathroom that is not ventilated. Taking the heat of vaporization of water to be \(1050 \mathrm{Btu} / \mathrm{lbm}\), determine the contribution of showers to the latent heat load of the air conditioner in summer per day.
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