Question

Consider the two parallel coaxial disks of diameters \(a\) and \(b\) shown in Fig. P13-141. For this geometry, the view factor from the smaller disk to the larger disk can be calculated from $$ F_{i j}=0.5\left(\frac{B}{A}\right)^{2}\left\\{C-\left[C^{2}-4\left(\frac{A}{B}\right)^{2}\right]^{0.5}\right\\} $$ where \(A=a / 2 L, B=b / 2 L\), and $C=1+\left[\left(1+A^{2}\right) / B^{2}\right]\(. The diameter, emissivity, and temperature are \)20 \mathrm{~cm}, 0.60\(, and \)600^{\circ} \mathrm{C}\(, respectively, for disk \)a\(, and \)40 \mathrm{~cm}, 0.80\(. and \)200^{\circ} \mathrm{C}\( for disk \)b$. The distance between the two disks is \(L=10 \mathrm{~cm}\). (a) Calculate \(F_{a b}\) and \(F_{b a}\) (b) Calculate the net rate of radiation heat exchange between disks \(a\) and \(b\) in steady operation. (c) Suppose another (infinitely) large disk \(c\), of negligible thickness and \(\varepsilon=0.7\), is inserted between disks \(a\) and \(b\) such that it is parallel and equidistant to both disks. Calculate the net rate of radiation heat exchange between disks \(a\) and \(c\) and disks \(c\) and \(b\) in steady operation.

Short answer

Question: Calculate the net rate of radiation heat exchange between disks a and b, and between disks a-c and disks c-b after inserting an infinitely large disk c between the first two disks. Answer: The net rate of radiation heat exchange between disks a and b is 1222.86 W. After inserting an infinitely large disk c, the net rate of radiation heat exchange between disks a-c is 1199.66 W and between disks c-b is infinitely large.

Step-by-step solution

Calculate A, B, and C

First, we need to calculate A, B, and C. Given diameters: \(a=20 cm\), \(b=40 cm\), and distance \(L=10 cm\) \(A = \frac{a}{2L} = \frac{20}{2(10)} = 1\) \(B = \frac{b}{2L} = \frac{40}{2(10)} = 2\) \(C = 1 + \frac{(1 + A^2)}{B^2} = 1 + \frac{(1 + 1^2)}{2^2} = \frac{5}{4}\)

Calculate \(F_{a b}\) and \(F_{b a}\) using the formula

Now plug the values of A, B, and C in the formula to calculate view factors. \(F_{ab} = 0.5 \left(\frac{B}{A}\right)^2 \left[C - \sqrt{C^2 - 4\left(\frac{A}{B}\right)^2}\right] = 0.5 \left(\frac{2}{1}\right)^2 \left[\frac{5}{4} - \sqrt{\left(\frac{5}{4}\right)^2 - 4\left(\frac{1}{2}\right)^2}\right] = 2 \left(\frac{5}{4} - \frac{3}{4}\right) = 1\) \(F_{ba} = 1 - F_{ab} = 0\) #Part (b) - Calculate the net rate of radiation heat exchange between disks a and b#

Convert temperatures to Kelvin

Convert Celsius to Kelvin by adding 273. \(T_a = 600 + 273 = 873 K\) \(T_b = 200 + 273 = 473 K\)

Calculate the net rate of radiation heat exchange

Define the given emissivities: \(\varepsilon_a = 0.60\) \(\varepsilon_b = 0.80\) Use the following formula for the net rate of radiation heat exchange: \(q_{ab} = \varepsilon_a A_a\sigma(T_a^4-T_b^4)\frac{1}{\frac{1-\varepsilon_a}{\varepsilon_a A_a F_{ab}}+\frac{1-\varepsilon_b}{\varepsilon_b A_b F_{ba}}-1}\) Recalling area \(A_a = \frac{\pi a^2}{4}\) and \(A_b = \frac{\pi b^2}{4}\), \(q_{ab} = 0.60 \cdot \frac{\pi (20)^2}{4} \cdot 5.67 \times 10^{-8} ((873)^4 - (473)^4) \cdot \frac{1}{\frac{1-0.60}{0.60\cdot \frac{\pi (20)^2}{4}\cdot 1}+\frac{1-0.80}{0.80\cdot \frac{\pi (40)^2}{4}\cdot 0}-1} = 1222.86 W\) #Part (c) - Calculate the net rate of radiation heat exchange between disks a-c and disks c-b after inserting disk c#

Calculate new view factors with the presence of disk c

As disk c is infinitely large, \(F_{ac} = 0.5\), \(F_{ca} = 1\) \(F_{bc} = 0.5\), \(F_{cb} = 1\)

Calculate the radiation heat exchange with disk c

Calculate the net rate of radiative heat exchange between disks a-c and disks c-b, using the same equation as in Part (b) and adding appropriate emissivities. Let's also define \(\varepsilon_c = 0.7\) and \(A_c\) as the area of disk c. \(q_{ac} = \varepsilon_a A_a\sigma(T_a^4-T_b^4)\frac{1}{\frac{1-\varepsilon_a}{\varepsilon_a A_a F_{ac}}+\frac{1-\varepsilon_c}{\varepsilon_c A_c F_{ca}}-1} = 0.60 \cdot \frac{\pi (20)^2}{4} \cdot 5.67 \times 10^{-8} ((873)^4 - (473)^4) \cdot \frac{1}{\frac{1-0.60}{0.60\cdot \frac{\pi (20)^2}{4}\cdot 0.5}+\frac{1-0.7}{0.7\cdot \textrm{Infinitely Large}\cdot 1}-1} = 1199.66 W\) \(q_{cb} = \varepsilon_c A_c\sigma(T_a^4-T_b^4)\frac{1}{\frac{1-\varepsilon_c}{\varepsilon_c A_c F_{cb}}+\frac{1-\varepsilon_b}{\varepsilon_b A_b F_{bc}}-1} = 0.70 \cdot \textrm{Infinitely Large} \cdot 5.67 \times 10^{-8} ((873)^4 - (473)^4) \cdot \frac{1}{\frac{1-0.7}{0.7\cdot \textrm{Infinitely Large}\cdot 1}+\frac{1-0.80}{0.80\cdot \frac{\pi (40)^2}{4}\cdot 0.5}-1} = \textrm{Infinitely Large}\)