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Chapter 13: Problem 14

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 between the base and the side surface \(F_{13}\) in terms of \(K\) and \((b)\) the value of the view factor \(F_{13}\) for \(L=D\).

Short Answer

Expert verified
Question: Determine the view factor F13 between the base and side surfaces of a cylindrical enclosure in terms of length to diameter ratio K, and find the value of F13 for the case when length L equals diameter D. Answer: The view factor F13 in terms of length to diameter ratio K is given by the expression F13 = 2K * tan^(-1)(1/K). For the case when L = D, the view factor F13 is equal to π/2.

Step by step solution

01

Reviewing the concept of view factors

A view factor is a term used in radiative heat transfer to indicate the fraction of the total energy emitted by a surface that is intercepted by another surface. In this case, we are interested in finding the view factor \(F_{13}\) between the base surface (\(A_1\)) and the side surface (\(A_3\)) of a cylindrical enclosure.
02

The relation between view factors

The view factors between the different surfaces of the cylinder must obey the following relationship (Reciprocity Theorem): \(F_{13} = \frac{A_3}{A_1} F_{31}\) Using this relationship, we can find \(F_{13}\) if we determine \(F_{31}\).
03

The view factor between the side surface and the base surface

The view factor between the side surface (A3) and the base surface (A1) can be determined using the relation (Derived from "Hottel's crossed strings method" in cylindrical enclosures): \(F_{31}= \frac{1}{\pi} \tan^{-1}(\frac{D}{L})\)
04

Substitute the value of K

We are given the length to diameter ratio as: \(K = \frac{L}{D}\) Rearrange the equation to get the value of \(\frac{D}{L}\): \(\frac{D}{L} = \frac{1}{K}\) Now substitute the value of \(\frac{D}{L}\) in the equation of \(F_{31}\): \(F_{31}= \frac{1}{\pi} \tan^{-1}(\frac{1}{K})\)
05

Calculate F13 in terms of K

Substitute the value of F31 in the reciprocity theorem relation (Step 2): \(F_{13} = \frac{A_3}{A_1} F_{31} = \frac{2\pi DL}{\pi D^2} \cdot \frac{1}{\pi} \tan^{-1}(\frac{1}{K})\) Simplified: \(F_{13} = \frac{2L}{D} \tan^{-1}(\frac{1}{K}) = 2K \tan^{-1}(\frac{1}{K})\) Now, we have found the expression for \(F_{13}\) in terms of \(K\): \(F_{13} = 2K \tan^{-1}(\frac{1}{K})\)
06

Calculate F13 when L = D

For the cylindrical enclosure with L = D, we can determine the value of \(F_{13}\): When \(L=D\), \(K = \frac{L}{D} = 1\), therefore substitute this into our derived expression: \(F_{13} = 2(1) \tan^{-1}(\frac{1}{1}) = 2 \tan^{-1}(1)\) \(\tan^{-1}(1) = \frac{\pi}{4}\), on substituting this: We get \(F_{13} = 2 \times\frac{\pi}{4} = \frac{\pi}{2}\) for \(L = D\) Therefore, the view factor \(F_{13}\) for \(L=D\) is \(\frac{\pi}{2}\).

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Most popular questions from this chapter

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: \(228 \mathrm{~W}\)

What is latent heat? How is the latent heat loss from the human body affected by \((a)\) skin wettedness and \((b)\) relative humidity of the environment? How is the rate of evaporation from the body related to the rate of latent heat loss?

13-59 This question deals with steady-state radiation heat transfer between a sphere \(\left(r_{1}=30 \mathrm{~cm}\right)\) and a circular disk \(\left(r_{2}=120 \mathrm{~cm}\right)\), which are separated by a center-to- center distance \(h=60 \mathrm{~cm}\). When the normal to the center of disk passes through the center of the sphere, the radiation view factor is given by $$ F_{12}=0.5\left\\{1-\left[1+\left(\frac{r_{2}}{h}\right)^{2}\right]^{-0.5}\right\\} $$ Surface temperatures of the sphere and the disk are \(600^{\circ} \mathrm{C}\) and \(200^{\circ} \mathrm{C}\), respectively; and their emissivities are \(0.9\) and \(0.5\), respectively. (a) Calculate the view factors \(F_{12}\) and \(F_{21}\). (b) Calculate the net rate of radiation heat exchange between the sphere and the disk. (c) For the given radii and temperatures of the sphere and the disk, the following four possible modifications could increase the net rate of radiation heat exchange: paint each of the two surfaces to alter their emissivities, adjust the distance between them, and provide an (refractory) enclosure. Calculate the net rate of radiation heat exchange between the two bodies if the best values are selected for each of the above modifications.

Two-phase gas-liquid oxygen is stored in a spherical tank of \(1-m\) diameter, where it is maintained at its normal boiling point. The spherical tank is enclosed by a \(1.6-\mathrm{m}\) diameter concentric spherical surface at \(273 \mathrm{~K}\). Both spherical surfaces have an emissivity of \(0.01\), and the gap between the inner sphere and outer sphere is vacuumed. Assuming that the spherical tank surface has the same temperature as the oxygen, determine the heat transfer rate at the spherical tank surface.

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.
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