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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 from the side surface to itself \(F_{33}\) in terms of \(K\) and \((b)\) the value of the view factor \(F_{33}\) for \(L=D\).

Short Answer

Expert verified
Answer: The expression for the view factor from the side surface to itself in terms of the length to diameter ratio (K) is \(F_{33} = \frac{1-K}{1+2K}\). When the length equals the diameter (K = 1), the view factor is 0.

Step by step solution

01

Understanding the problem

The view factor between two surfaces in the enclosure can be understood as the fraction of the total emitting or receiving power at one surface that reaches or leaves the other surface directly. In our case, we need to determine the view factor from the side surface to itself, denoted by \(F_{33}\).
02

View factor relation

First, let's write down the view factor relations for the case of the cylindrical enclosure with three surfaces. The relations are given by: $$F_{11}+F_{12}+F_{13}=1$$ $$F_{21}+F_{22}+F_{23}=1$$ $$F_{31}+F_{32}+F_{33}=1$$
03

Reciprocity theorem

We also know that a view factor obeys the reciprocity theorem, which can be stated as: $$A_{1}F_{12} = A_{2}F_{21}$$ $$A_{1}F_{13} = A_{3}F_{31}$$ $$A_{2}F_{23} = A_{3}F_{32}$$ Now, let's use these relations in our problem.
04

View factor expression

Given that \(K=L / D\), we can use this ratio to express the areas of the cylindrical enclosure as: $$A_1 = A_2 = \frac{\pi D^2}{4}$$ $$A_3 = \pi D L = \pi D^2 K$$ Now we need to find the value of \(F_{33}\). We can use the relation for \(F_{31}\): $$F_{31} = \frac{A_3}{A_1}F_{13}$$ Since \(F_{33}+F_{32}+F_{31}=1\), we can write the expression for the view factor \(F_{33}\) as: $$F_{33} = 1-F_{31}-F_{32}=1-\frac{A_3}{A_1}F_{13}-\frac{A_2}{A_3}F_{23}$$ From the reciprocity theorem, we can further simplify the expression, because \(F_{13}=\frac{A_3}{A_1}F_{31}\) and \(F_{23}=\frac{A_3}{A_2}F_{32}\): $$F_{33} = 1-\frac{A_3}{A_1}F_{31}-\frac{A_3}{A_1}(1-F_{31}-F_{33})$$ $$F_{33}\left(\frac{2A_3}{A_1}+1\right) = 1-\frac{A_3}{A_1}$$ Now we can express the view factor \(F_{33}\) in terms of \(K\): $$F_{33} = \frac{1-K}{1+2K}$$
05

Evaluate the value of \(F_{33}\) for \(L = D\)

Now, we need to find the value of \(F_{33}\) when the length equals the diameter, or \(K=1\). Plug \(K=1\) into the expression for \(F_{33}\): $$F_{33} = \frac{1-1}{1+2(1)} = \frac{0}{3} = 0$$ Thus, the value of the view factor \(F_{33}\) for \(L = D\) is 0.

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Key Concepts

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

View Factors
A view factor, also known as a configuration factor, is a concept in radiative heat transfer. It represents the fraction of radiation leaving one surface that directly reaches another surface. For any two surfaces in an enclosure, understanding view factors helps in determining the amount of radiative heat exchange between them.

View factors are crucial in analyses involving enclosures because they account for geometrical dependencies in radiation heat transfer. Here are a few important points about view factors:
  • View factors are dimensionless and range from 0 to 1.
  • They depend solely on the geometry of the surfaces, not their temperatures or emissivities.
  • The sum of view factors from a surface to all other surfaces, including itself, is always equal to 1. This is known as the summation rule.
Understanding view factors is essential in analyzing problems where radiation is a significant mode of heat transfer, like in a cylindrical enclosure. They allow engineers to calculate how much radiation a surface sees, influencing thermal analysis and design.
Cylindrical Enclosure
A cylindrical enclosure is a common geometric arrangement in heat transfer problems, particularly in calculating radiative heat exchange. In such an enclosure, typically, there are three distinct surfaces: the two circular bases and the cylindrical side.

The cylindrical geometry poses unique challenges and opportunities in heat transfer because of its curved surface, which can alter how radiation is exchanged between surfaces. Here’s what you need to consider about a cylindrical enclosure:
  • The length-to-diameter ratio, represented by the symbol \(K = \frac{L}{D}\), is an important parameter that influences the view factors between the surfaces of the cylinder.
  • The symmetry in the geometry can simplify the calculation of view factors, but also requires careful analysis to ensure accurate results.
  • The side surface of the cylinder can have a view factor with itself, which might be zero depending on the ratio \(K\).
When analyzing such systems, understanding how the view factors interact within the enclosure helps in solving for the thermal behavior of the system. The cylindrical nature affects not just the view factors but also the thermal macro-dynamics of the system.
Reciprocity Theorem
The reciprocity theorem is a vital principle in understanding heat transfer involving view factors. It provides a relationship between the view factors and surface areas of enclosures, stating that the product of the view factor and the area of one surface is equal to the product of the view factor and the area of another surface. Mathematically, it is expressed as:

\[A_i F_{ij} = A_j F_{ji}\]

This equation relates the fraction of radiation leaving surface \(A_i\) and arriving at surface \(A_j\) to the opposite process. Key points about the reciprocity theorem include:
  • It relies on surface properties and not on temperatures.
  • Helps simplify and solve for unknown view factors in complex geometries.
  • Ensures the conservation of energy principle holds in radiative heat exchange.
The reciprocity theorem was a critical tool in the original problem of determining the view factor \(F_{33}\) in the cylindrical enclosure. Without it, finding relationships between various view factors could be significantly difficult, affecting the analysis and interpretation of heat transfer problems.

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Most popular questions from this chapter

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

Two square plates, with the sides \(a\) and \(b\) (and \(b>a\) ), are coaxial and parallel to each other, as shown in Fig. P13-132, and they are separated by a center-to-center distance of \(L\). The radiation view factor from the smaller to the larger plate, \(F_{a b}\), is given by $$ F_{a b}=\frac{1}{2 A}\left\\{\left[(B+A)^{2}+4\right]^{0.5}-\left[(B-A)^{2}+4\right]^{0.5}\right\\} $$ where, \(A=a / L\) and \(B=b / L\). (a) Calculate the view factors \(F_{a b}\) and \(F_{b a}\) for \(a=20 \mathrm{~cm}\), \(b=60 \mathrm{~cm}\), and \(L=40 \mathrm{~cm}\). (b) Calculate the net rate of radiation heat exchange between the two plates described above if \(T_{a}=800^{\circ} \mathrm{C}\), \(T_{b}=200^{\circ} \mathrm{C}, \varepsilon_{a}=0.8\), and \(\varepsilon_{b}=0.4\). (c) A large square plate (with the side \(c=2.0 \mathrm{~m}, \varepsilon_{c}=0.1\), and negligible thickness) is inserted symmetrically between the two plates such that it is parallel to and equidistant from them. For the data given above, calculate the temperature of this third plate when steady operating conditions are established.

Consider two rectangular surfaces perpendicular to each other with a common edge which is \(1.6 \mathrm{~m}\) long. The horizontal surface is \(0.8 \mathrm{~m}\) wide and the vertical surface is \(1.2 \mathrm{~m}\) high. The horizontal surface has an emissivity of \(0.75\) and is maintained at \(400 \mathrm{~K}\). The vertical surface is black and is maintained at \(550 \mathrm{~K}\). The back sides of the surfaces are insulated. The surrounding surfaces are at \(290 \mathrm{~K}\), and can be considered to have an emissivity of \(0.85\). Determine the net rate of radiation heat transfers between the two surfaces, and between the horizontal surface and the surroundings.

Two aligned parallel rectangles with dimensions \(6 \mathrm{~m} \times\) \(8 \mathrm{~m}\) are spaced apart by a distance of \(2 \mathrm{~m}\). If the two parallel rectangles are experiencing radiation heat transfer as black surfaces, determine the percentage of change in radiation heat transfer rate when the rectangles are moved \(8 \mathrm{~m}\) apart.

A spherical tank, with an inner diameter of \(D_{1}=\) \(3 \mathrm{~m}\), is filled with a solution undergoing an exothermic reaction that heats the surface to a uniform temperature of \(120^{\circ} \mathrm{C}\). To prevent thermal burn hazards, the tank is enclosed with a concentric outer cover that provides an evacuated gap of \(5 \mathrm{~cm}\) in the enclosure. Both spherical surfaces have the same emissivity of \(0.5\), and the outer surface is exposed to natural convection with a heat transfer coefficient of \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and radiation heat transfer with the surrounding at a temperature of \(30^{\circ} \mathrm{C}\). Determine whether or not the vacuumed gap is sufficient to keep the outer surface temperature below \(45^{\circ} \mathrm{C}\) to prevent thermal burns. If not, propose a solution to keep the outer surface temperature below \(45^{\circ} \mathrm{C}\).

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