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Two infinitely long parallel plates of width \(w\) are located at \(w\) distance apart, as shown in Fig. P13-24. Using the Hottel's crossed-strings method, determine the view factor \(F_{12}\).

Short Answer

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Question: Determine the view factor between two infinitely long parallel plates with width w and separated by a distance w using Hottel's crossed strings method. Answer: The view factor, \(F_{12}\), between the two infinitely long parallel plates in the given scenario is 0.5 or 50%.

Step by step solution

01

Understand Hottel's crossed-strings method

Hottel's crossed-strings method is a graphical technique used to determine view factors between two surfaces. The method involves drawing a string from the corner of one surface to the adjacent corner of the other surface and calculating the length of the string and angles formed by the string at the corners. The lengths of the strings and angles will be used to determine the view factor between the surfaces.
02

Visualize the scenario

Visualize two infinitely long parallel plates with width \(w\) separated by a distance \(w\). Draw a string from the corner A of plate 1 to the corner B of plate 2, and another string from corner C of plate 2 to corner D of plate 1.
03

Determine the string lengths

In the scenario described, there are two strings: AB and CD. These strings are the same length since the plates are parallel and equidistant. Utilizing the Pythagorean theorem, the length of the strings can be determined: Length (AB) = Length (CD) = \(\sqrt{w^2 + w^2}\) = \(\sqrt{2w^2}\) = \(w\sqrt{2}\)
04

Determine the angles

To find the angles of both strings, denote the angles at A and D as \(\theta_1\) and \(\theta_2\) respectively. Using the tan function, these angles can be determined: Angle A: \(\theta_1 = \tan^{-1} \frac{w}{w} = \tan^{-1}(1)\) Angle D: \(\theta_2 = \tan^{-1} \frac{w}{w} = \tan^{-1}(1)\) The angles at B and C are also the same: Angle B: \(\tan^{-1} \frac{w}{w} = \tan^{-1}(1)\) Angle C: \(\tan^{-1} \frac{w}{w} = \tan^{-1}(1)\)
05

Determine the view factor using the cross-string method

Apply Hottel's crossed-string formula to calculate the view factor \(F_{12}\): For two parallel infinite plates of equal width, the formula becomes: \(F_{12} = \frac{(\theta_1 + \theta_2)}{\pi}\) Using the angles determined earlier: \(F_{12} = \frac{\tan^{-1}(1) + \tan^{-1}(1)}{\pi} = \frac{\pi/4 + \pi/4}{\pi} = \frac{\pi/2}{\pi} = \frac{1}{2}\) Thus, the view factor \(F_{12}\) is equal to 0.5 or 50% for this specific scenario. It implies that 50% of the radiant energy leaving plate 1 reaches plate 2.

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Most popular questions from this chapter

Consider two diffuse surfaces \(A_{1}\) and \(A_{2}\) oriented on a spherical surface as shown in the figure. Determine \((a)\) the expression for the view factor \(F_{12}\) in terms of \(A_{2}\) and \(L\), and (b) the value of the view factor \(F_{12}\) when \(A_{2}=0.02 \mathrm{~m}^{2}\) and \(L=1 \mathrm{~m}\).

Give examples of radiation effects that affect human comfort.

Consider two coaxial parallel circular disks of equal diameter $D=1 \mathrm{~m}\( spaced apart by \)1 \mathrm{~m}$, and two aligned parallel square plates \((1 \mathrm{~m} \times 1 \mathrm{~m})\) are also spaced apart by $1 \mathrm{~m}\(. Determine the view factors \)F_{12}$ between the circular disks and the square plates. Which of the two geometries has the higher view factor value?

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 the 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.

A row of tubes, equally spaced at a distance that is twice the diameter of the tubes, is positioned between two large parallel plates. The surface temperature of the tubes is constant at \(10^{\circ} \mathrm{C}\) and the top and bottom plates are at constant temperatures of \(100^{\circ} \mathrm{C}\) and \(350^{\circ} \mathrm{C}\), respectively. If the surfaces behave as blackbody, determine the net radiation heat flux leaving the bottom plate.

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