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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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Key Concepts

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

Cylindrical Enclosure
In radiative heat transfer, a cylindrical enclosure is a geometric configuration that consists of three main surfaces: the base, the top, and the side. These surfaces are denoted as \(A_1\), \(A_2\), and \(A_3\) respectively. When analyzing radiative interactions, especially in thermodynamic and mechanical engineering applications, understanding how these surfaces transfer heat to one another is crucial.

A cylinder is characterized by its length \(L\) and diameter \(D\), often combined into a dimensionless ratio \(K\), defined as \(K = \frac{L}{D}\). This ratio simplifies the analysis by relating the geometric proportions of the cylindrical enclosure. The heat transfer characteristics depend significantly on the arrangement and size of these surfaces, making \(K\) a vital parameter in the calculation of radiative heat transfer between surfaces.

In many practical scenarios, such as heat exchangers or industrial furnaces, these principles help engineers design systems for optimal heat dissipation or retention. Understanding how each surface interacts and transfers radiative heat plays a pivotal role in achieving efficient energy management.
View Factors
View factors, also known as configuration factors or shape factors, are fundamental in radiative heat transfer. They quantify the fraction of radiation energy leaving one surface that directly strikes another, within an enclosure. This concept is essential when determining how much energy is exchanged between surfaces in a non-participating medium.

To compute the view factor \(F_{ij}\) from surface \(A_i\) to \(A_j\), several geometric relationships are used. The cylindrical enclosure setup allows us to derive these factors based on the interaction between the base and the side surfaces. For instance, the view factor \(F_{13}\) from the base to the side surface can be expressed using the length-to-diameter ratio \(K\), as derived in the exercise.

  • \(F_{13} = 2K \tan^{-1}(\frac{1}{K})\)
Understanding and calculating these view factors is crucial since they directly influence design decisions in thermal system construction, helping assess how efficiently heat will be transferred between surfaces.
Reciprocity Theorem
The Reciprocity Theorem is a fundamental principle in radiative heat transfer, ensuring a balanced exchange of energy between surfaces. It states that the view factor from surface \(A_i\) to \(A_j\), multiplied by the area of \(A_i\), equals the view factor from \(A_j\) to \(A_i\), multiplied by the area of \(A_j\). Mathematically, this is represented as:
\[ A_i F_{ij} = A_j F_{ji} \]

In the context of the cylindrical enclosure problem, the Reciprocity Theorem allows us to relate the view factors \(F_{13}\) and \(F_{31}\) between the base and side surfaces. This relationship simplifies the calculation by using geometric properties of the surfaces involved.

  • Rearranging the theorem for the cylindrical surfaces gives: \(F_{13} = \frac{A_3}{A_1}F_{31}\).
This powerful theorem is instrumental in solving complex radiative exchanges in enclosures since it reduces the number of calculations by establishing interdependencies between different view factors.

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Most popular questions from this chapter

The base surface of a cubical furnace with a side length of \(3 \mathrm{~m}\) has an emissivity of \(0.80\) and is maintained at \(500 \mathrm{~K}\). If the top and side surfaces also have an emissivity of \(0.80\) and are maintained at \(900 \mathrm{~K}\), the net rate of radiation heat transfer from the top and side surfaces to the bottom surface is (a) \(194 \mathrm{~kW}\) (b) \(233 \mathrm{~kW}\) (c) \(288 \mathrm{~kW}\) (d) \(312 \mathrm{~kW}\) (e) \(242 \mathrm{~kW}\)

Why is the radiation analysis of enclosures that consist of black surfaces relatively easy? How is the rate of radiation heat transfer between two surfaces expressed in this case?

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.

What is the crossed-strings method? For what kind of geometries is the crossed-strings method applicable?

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) \(0.87\)

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