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Chapter 13: Problem 14

Consider two diffuse surfaces \(A_{1}\) and \(A_{2}\) on the inside of a spherical enclosure of radius \(R\). Using the following methods, derive an expression for the view factor \(F_{12}\) in terms of \(A_{2}\) and \(R\). (a) Find \(F_{12}\) by beginning with the expression \(F_{i j}=\) \(q_{i \rightarrow j} / A_{i} J_{i} .\) (b) Find \(F_{12}\) using the view factor integral, Equation 13.1.

Short Answer

Expert verified
In summary, the view factor \(F_{12}\) for two diffuse surfaces \(A_1\) and \(A_2\) inside a spherical enclosure of radius \(R\) can be found using two different methods: (a) using the expression \(F_{ij}=\frac{q_{i\rightarrow j}}{A_i J_i}\) and (b) using the view factor integral (Equation 13.1). In both cases, we derived that \(F_{12} = \frac{A_2}{4\pi R^2 J_1}\) for part (a) and \(F_{12} =\frac{A_2}{4\pi R^2}\) for part (b) when surface 1 is emitting unit energy.

Step by step solution

01

(a) Using the expression \(F_{ij} = \frac{q_{i\rightarrow j}}{A_i J_i}\)

First, let's analyze the given expression. \(F_{ij}\) represents the view factor from surface \(i\) to surface \(j\). The fraction \(\frac{q_{i\rightarrow j}}{A_i}\) represents the energy transfer between the two surfaces. Hence, to find the view factor \(F_{12}\), we need to calculate the energy transfer \(q_{1\rightarrow 2}\), as well as the area \(A_1\) and the total energy leaving surface 1, which is \(J_1\). Due to diffuse nature of the surfaces, the energy transfer \(q_{1\rightarrow 2}\) can be defined as: \(q_{1\rightarrow 2}=A_1 F_{12} J_1\) The view factor \(F_{12}\) now can be obtained by rearranging the terms: \(F_{12}=\frac{q_{1\rightarrow 2}}{A_1 J_1}\) As the surfaces are inside a spherical enclosure of radius \(R\), the energy incident on surface 2 can be written as: \(q_{1\rightarrow2}=\frac{A_1A_2}{4\pi R^2}\) Substituting the above expression for \(q_{1\rightarrow2}\) in the equation for \(F_{12}\), we get: \(F_{12}=\frac{A_1A_2}{4\pi R^2 A_1 J_1}\) Now canceling \(A_1\) on the numerator and denominator, we get the expression for view factor: \(F_{12}=\frac{A_2}{4\pi R^2 J_1}\)
02

(b) Using the view factor integral, Equation 13.1

The integral expression of view factor (Equation 13.1) is given as: \(F_{12} = \frac{1}{A_1} \int_{A_1} \int_{A_2} \frac{\cos\theta_1 \cos\theta_2 dA_1 dA_2}{\pi r^2}\) Where \(\theta_1\) and \(\theta_2\) are the angles between the normal vectors of surfaces 1 and 2, respectively, and \(r\) is the distance between the elemental areas \(dA_1\) and \(dA_2\). For the spherical enclosure, observe that the \(\cos\theta_1\) in the integral refers to the cosine of the angle between the normal vector of the surface patch \(dA_1\) and the vector from the center of surface 1 to the center of surface 2. Similarly, \(\cos\theta_2\) refers to the cosine of the angle between the normal vector of the surface patch \(dA_2\) and the same center-to-center vector. As the surfaces are diffuse, \(\cos\theta_1 = \cos\theta_2 = 1\). The integral expression of view factor (Equation 13.1) now simplifies as: \(F_{12} = \frac{1}{A_1} \int_{A_1} \int_{A_2} \frac{dA_1 dA_2}{\pi r^2}\) Now integrating the above expression, we find that: \(F_{12} = \frac{A_2}{4\pi R^2}\) The view factor \(F_{12}\) derived in part (b) coincides with the result derived in part (a) when surface 1 is emitting unit energy.

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Key Concepts

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

View Factor
The concept of the view factor is crucial in radiative heat transfer, especially when dealing with complex geometries like a spherical enclosure. The view factor, denoted as \( F_{ij} \), represents the fraction of radiation leaving surface \( i \) that strikes surface \( j \). This quantity is influenced by the geometry and relative orientation of the surfaces. For a spherical enclosure, the symmetry simplifies calculations greatly, especially when surfaces are diffuse.

**Radiative Exchange:** The view factor helps quantify the energy exchange between surfaces by accounting for their geometric orientations and distances. It is defined by the equation:

\[F_{ij} = \frac{q_{i\rightarrow j}}{A_i J_i}\]

where \( q_{i\rightarrow j} \) is the energy transferred from surface \( i \) to \( j \), \( A_i \) is the area of surface \( i \), and \( J_i \) is the energy leaving surface \( i \) per unit area.

**Geometric Factor:** In a spherical enclosure, for two areas \( A_1 \) and \( A_2 \), the symmetry leads to simplied expressions for view factor calculations:
  • View factors depend significantly on distance and orientation.
  • In spherical geometries, symmetry simplifies the mathematics since each point on the sphere contributes uniformly to radiation exchange.
Diffuse Surfaces
Diffusivity is a property of surfaces that affects how they emit or reflect radiation. When dealing with diffuse surfaces, the radiation is emitted uniformly in all directions. This characteristic simplifies the mathematical treatment of radiation since the reflectance or emission does not depend on direction.

**Uniform Emission:** In heat transfer problems, using diffuse surface assumptions can simplify view factor calculations because they remove directional dependencies. Here鈥檚 what鈥檚 important:
  • Diffuse surfaces emit energy equally in all directions.
  • This uniformity means the angle between the surface normal and the radiation does not change the emission intensity.

**Simplification in Calculations:** In the spherical enclosure, diffuse surfaces allow us to approximate the radiation exchange simply by considering geometrical relations without worrying about directional biases. When calculating the view factor using integrals, the cosine terms \( \cos \theta_1 \) and \( \cos \theta_2 \) simplify because for diffuse surfaces, they both become \( 1 \). This is a major mathematical convenience, leading to straightforward solutions.
Spherical Enclosure
A spherical enclosure provides a unique geometry where radiative transfer calculations become elegant due to its symmetry. In problems involving spherical enclosures, radiative heat exchange analysis needs to consider the curvature and enclosed volume of the system.

**Enclosure Impact:** The main aspects to consider when dealing with spherical enclosures include:
  • The symmetry significantly simplifies the integration involved in view factor calculations.
  • Due to the curvature, the view of any point on the surface potentially encompasses the entire opposite hemisphere.

**Integrating Over a Sphere:** With spherical geometry, the view factor between any two surfaces simplifies as the curvature of the sphere is homogeneous. Thus, the result is dependent on enclosed area size and radial distance:
  • View factor integrals are calculated over the full solid angle for the surface.
  • In a typical setup with surface \( A_1 \), and \( A_2 \) situated within a sphere of radius \( R \), the calculations adapt to treat distance and area uniformly.
This uniformity and symmetry make spherical enclosures a common model in theoretical studies and practical applications for radiative heat transfer.

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Most popular questions from this chapter

To enhance heat rejection from a spacecraft, an engineer proposes to attach an array of rectangular fins to the outer surface of the spacecraft and to coat all surfaces with a material that approximates blackbody behavior. Consider the U-shaped region between adjoining fins and subdivide the surface into components associated with the base (1) and the side (2). Obtain an expression for the rate per unit length at which radiation is transferred from the surfaces to deep space, which may be approximated as a blackbody at absolute zero temperature. The fins and the base may be assumed to be isothermal at a temperature \(T\). Comment on your result. Does the engineer's proposal have merit?

A double-glazed window consists of two panes of glass, each of thickness \(I=6 \mathrm{~mm}\). The inside room temperature is \(T_{t}=20^{\circ} \mathrm{C}\) with \(h_{i}=7.7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the outside temperature is \(T_{o}=-10^{\circ} \mathrm{C}\) with \(h_{a}=25 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\). The gap between the glass sheets is of thickness \(L=5 \mathrm{~mm}\) and is filled with a gas. The glass surfaces may be treated with a low-emissivity coating to reduce their emissivity from \(\varepsilon=0.95\) to \(\varepsilon=0.05\). Determine the heat flux through the window for case 1: \(\varepsilon_{1}=\varepsilon_{2}=0.95\), case \(2: \varepsilon_{1}=\varepsilon_{2}=0.05\), and case 3: \(\varepsilon_{\mathrm{L}}=0.05, \varepsilon_{2}=0.95\). Consider either air of argon of thermal conductivity \(k_{A R}=17.7 \times 10^{-3}\) W/m. \(\mathrm{K}\) to be within the gap. Radiation heat transfer occurring at the external surfaces of the two glass sheets is negligible, as is free convection between the glass sheets.

Consider the attic of a home located in a hot climate. The floor of the attic is characterized by a width of \(L_{1}=10 \mathrm{~m}\) while the roof makes an angle of \(\theta=30^{\circ}\) from the horizontal direction, as shown in the schematic. The homeowner wishes to reduce the heat load to the home by adhering bright aluminum foil \(\left(\varepsilon_{f}=0.07\right)\) onto the surfaces of the attic space. Prior to installation of the foil, the surfaces are of emissivity \(s_{e}=0.85\). (a) Consider installation on the bottom of the attic roof only. Determine the ratio of the radiation heat transfer after to before the installation of the foil. (b) Determine the ratio of the radiation heat transfer after to before installation if the foil is installed only on the top of the attic floor. (c) Determine the ratio of the radiation heat transfer if the foil is installed on both the roof bottom and the floor top.

A fumace having a spherical cavity of \(0.5-\mathrm{m}\) diameter contains a gas mixture at 1 atm and \(1400 \mathrm{~K}\). The mixture consists of \(\mathrm{CO}_{2}\), with a partial pressure of \(0.25 \mathrm{~atm}\) and nitrogen with a partial pressure of \(0.75 \mathrm{arm}\). If the cavity wall is black, what is the cooling rate needed to maintain its temperature at \(500 \mathrm{~K}\) ?

A right-circular cone and a right-circular cylinder of the same diameter and length \(\left(A_{2}\right)\) are positioned coaxially at a distance \(L_{0}\) from the circular disk \(\left(A_{1}\right)\) shown schematically. The inner base and lateral surfaces of the cylinder may be treated as a single surface, \(A_{2}\). The hypothetical area corresponding to the opening of the cone and cylinder is identified as \(A_{3}\). (a) Show that, for both arrangements, \(F_{21}=\left(A_{1} / A_{2}\right) F_{13}\) and \(F_{22}=1-\left(A_{3} / A_{2}\right)\), where \(F_{13}\) is the view factor between two coaxial, parallel disks (Table 13.2). (b) For \(L=L_{o}=50 \mathrm{~mm}\) and \(D_{1}=D_{3}=50 \mathrm{~mm}\), calculate \(F_{21}\) and \(F_{22}\) for the conical and cylindrical configurations and compare their relative magnitudes. Explain any similarities and differences. (c) Do the relative magnitudes of \(F_{21}\) and \(F_{22}\) change for the conical and cylindrical configurations as \(L\) increases and all other parameters remain fixed? In the limit of very large \(L\), what do you expect will happen? Sketch the variations of \(F_{21}\) and \(F_{22}\) with \(L\), and explain the key features.
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