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Consider the parallel rectangles shown schematically. Show that the view factor \(F_{12}\) can be expressed as $$ F_{12}=\frac{1}{2 A_{1}}\left[A_{(1,4)} F_{(1,4)(2,3)}-A_{1} F_{13}-A_{4} F_{42}\right] $$ where all view factors on the right-hand side of the equation can be evaluated from Figure \(13.4\) (see Table 13.2) for aligned parallel rectangles.

Short Answer

Expert verified
To show that the view factor \(F_{12}\) for parallel rectangles can be expressed as \(F_{12}=\frac{1}{2 A_{1}}\left[A_{(1,4)} F_{(1,4)(2,3)}-A_{1} F_{13}-A_{4} F_{42}\right]\), we used the reciprocity theorem for view factors and evaluated the individual view factors \(F_{13}\), \(F_{42}\), and \(A_{(1,4)}F_{(1,4)(2,3)}\) using reference Figure 13.4 and Table 13.2 for aligned parallel rectangles. After substituting the calculated view factors into the given equation and simplifying, we found the value of \(F_{12}\), demonstrating that it can be expressed in terms of known view factors as stated.

Step by step solution

01

Recall the reciprocity theorem for view factors

The reciprocity theorem states that, for two surfaces A1 and A2, their view factors are related as follows: \(A_1F_{12} = A_2F_{21}\). This property will be used to define our relationships between view factors. Step 2: Evaluate view factors F13 and F42
02

Evaluate view factors F13 and F42

In order to find F13 and F42, we will look at the aligned parallel rectangles shown in Figure 13.4 and refer to Table 13.2, which provides values for view factors of aligned parallel rectangles. By substituting the appropriate values into the equation, we can find the values of F13 and F42. Step 3: Evaluate view factors for A(1,4)F(1,4)(2,3)
03

Evaluate view factors for A(1,4)F(1,4)(2,3)

To find the view factor for A(1,4)F(1,4)(2,3), we will again refer to the aligned parallel rectangles shown in Figure 13.4 and Table 13.2. Substitute the appropriate values into the equation and solve for the value of A(1,4)F(1,4)(2,3). Step 4: Apply the given equation for F12
04

Apply the given equation for F12

With the view factors F13, F42, and A(1,4)F(1,4)(2,3) obtained from the previous steps, substitute their values into the given equation: \(F_{12} = \frac{1}{2A_1}[A_{(1,4)}F_{(1,4)(2,3)} - A_1F_{13} - A_4F_{42}]\) Step 5: Simplify the equation to find F12
05

Simplify the equation to find F12

After substituting the calculated view factors from the previous steps, simplify the equation to find the value of F12. By following these steps and using the reference figures and tables, we have shown that the view factor F12 can be expressed in terms of other known view factors as given in the equation.

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Key Concepts

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

Reciprocity Theorem
The reciprocity theorem is a fundamental principle in the study of radiative heat transfer, particularly when dealing with view factors between surfaces. The theorem states that for two surfaces, A1 and A2, with respective view factors F12 and F21, the product of each surface area and its view factor to the other is equal. Mathematically, this relationship is expressed as:
\[ A_1F_{12} = A_2F_{21} \]
This means that if the amount of radiation leaving surface A1 and striking surface A2 is known, one can easily calculate the radiation leaving surface A2 and striking surface A1. The reciprocity theorem simplifies the calculation of view factors, which are essential for understanding and predicting radiative heat transfer between different surfaces. Moreover, it ensures the consistency and accuracy of radiative heat transfer analyses, making it an invaluable tool for engineers and scientists in thermal systems design.
In the context of the given problem, the reciprocity theorem allows us to link the view factors of different surfaces and thereby solve for unknown view factors based on known ones. Simplifying complex geometric configurations into a series of reciprocal relationships reduces the problem's complexity and makes it solvable.
Radiative Heat Transfer
Radiative heat transfer is one of the three fundamental modes of heat transfer, alongside conduction and convection. It's unique in that it does not require a medium and can occur in a vacuum. Radiative heat transfer is the energy exchange via electromagnetic waves, primarily in the infrared spectrum for thermal applications. This form of transfer is dictated by the Stefan-Boltzmann law, which states that the energy radiated from a surface is proportional to the fourth power of the surface's absolute temperature.

Key to understanding radiative heat exchange between surfaces is the concept of view factors, also known as configuration factors or shape factors. They quantify the fraction of radiation leaving one surface that directly strikes another. In mathematical terms, for two surfaces A1 and A2, the view factor F12 is the fraction of radiation leaving surface A1 that strikes surface A2 directly. Accurate calculation of these factors is crucial for predicting thermal radiation heat transfer, which plays a significant role in the thermal management of various systems, from electronic devices to spacecraft.

For students solving textbook problems on radiative heat transfer, like the one at hand, a thorough grasp of how view factors are determined and used is essential. Through the formulations and relationships between view factors, students can navigate through complex geometrical arrangements to find solutions to heat transfer challenges.
Aligned Parallel Rectangles
Aligned parallel rectangles are a common geometric configuration encountered in the calculation of view factors for radiative heat transfer analyses. Such configurations are simpler compared to other irregular geometries, and their view factors can often be found in reference tables or can be calculated using well-defined equations.

For this particular exercise, we consider two parallel and aligned rectangles where specific view factors need to be determined. The advantage of working with aligned parallel rectangles lies in the symmetry and simplicity, which allows for the application of basic principles, like the reciprocity theorem, without complex integrations or approximations. Reference figures and tables that provide view factors for such configurations are essential tools and offer a reliable method to find accurate view factor values.

When dealing with these kinds of problems, students are advised to clearly identify the surfaces involved and to understand the directional nature of view factors. As the solution suggests, by stepwise evaluation of view factors for each set of rectangles and applying given formulas, the complex problem becomes manageable. Moreover, having the ability to interpret reference figures and tables, like Figure 13.4 and Table 13.2 in this case, is a practical skill that aids in solving many real-world engineering problems involving radiative heat exchange between surfaces.

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Most popular questions from this chapter

A long, cylindrical heating element of \(20-\mathrm{mm}\) diameter operating at \(700 \mathrm{~K}\) in vacuum is located \(40 \mathrm{~mm}\) from an insulated wall of low thermal conductivity. (a) Assuming both the element and the wall are black, estimate the maximum temperature reached by the wall when the surroundings are at \(300 \mathrm{~K}\). (b) Calculate and plot the steady-state wall temperature distribution over the range \(-100 \mathrm{~mm} \leq\) \(x \leq 100 \mathrm{~mm}\).

A cylindrical cavity of diameter \(D\) and depth \(L\) is machined in a metal block, and conditions are such that the base and side surfaces of the cavity are maintained at \(T_{1}=1000 \mathrm{~K}\) and \(T_{2}=700 \mathrm{~K}\), respectively. Approximating the surfaces as black, determine the emissive power of the cavity if \(L=20 \mathrm{~mm}\) and \(D=10 \mathrm{~mm}\).

Derive expressions for the view factor \(F_{12}\) associated with the following arrangements. Express your results in terms of \(A_{1}, A_{2}\) and any appropriate hypothetical surface area, as well as the view factor for coaxial parallel disks (Table 13.2, Figure 13.5). (a) A circular disk and a coaxial, ring-shaped disk. (b) A circular disk and a coaxial, right-circular cone. Sketch the variation of \(F_{12}\) with \(\theta\) for \(0 \leq \theta \leq \pi / 2\), and explain the key features.

Liquid oxygen is stored in a thin-walled, spherical container \(0.8 \mathrm{~m}\) in diameter, which is enclosed within a second thin-walled, spherical container \(1.2 \mathrm{~m}\) in diameter. The opaque, diffuse, gray container surfaces have an emissivity of \(0.05\) and are separated by an evacuated space. If the outer surface is at \(280 \mathrm{~K}\) and the inner surface is at \(95 \mathrm{~K}\), what is the mass rate of oxygen lost due to evaporation? (The latent heat of vaporization of oxygen is \(2.13 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).)

A heat flux gage of 4-mm diameter is positioned normal to and \(1 \mathrm{~m}\) from the 5 -mm-diameter aperture of a blackbody furnace at \(1000 \mathrm{~K}\). The diffuse, gray cover shield \((\varepsilon=0.2)\) of the furnace has an outer diameter of \(100 \mathrm{~mm}\) and its temperature is \(350 \mathrm{~K}\). The furnace and gage are located in a large room whose walls have an emissivity of \(0.8\) and are at \(300 \mathrm{~K}\). (a) What is the irradiation on the gage, \(G_{g}\) (W/m²), considering only emission from the aperture of the furnace? (b) What is the irradiation on the gage due to radiation from the cover and aperture?

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