13-59 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 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.

Step by step solution

01

Calculate the view factors of F12 and F21

Given the formula for the view factor F12, $$ F_{12}=0.5\left\\{1-\left[1+\left(\frac{r_{2}}{h}\right)^{2}\right]^{-0.5}\right\\} $$ Use the given values for \(r_2 = 120 cm\) and \(h = 60 cm\): $$ F_{12} = 0.5\left\\{1-\left[1+\left(\frac{120}{60}\right)^{2}\right]^{-0.5}\right\\} $$ Calculate \(F_{12}\): $$ F_{12} ≈ 0.240 $$ Now to find \(F_{21}\), we can use the reciprocity theorem: $$ A_1 F_{12} = A_2 F_{21} $$ Where \(A_1\) is the surface area of the sphere and \(A_2\) is the surface area of the disk. Calculate \(A_1\) and \(A_2\): $$ A_1 = 4\pi r_1^2 = 4\pi(30)^2 ≈ 11310 cm^2 $$ $$ A_2 = \pi r_2^2 = \pi(120)^2 ≈ 45239 cm^2 $$ Now, we can solve for \(F_{21}\): $$ F_{21} = \frac{A_1 F_{12}}{A_2} ≈ \frac{11310 * 0.240}{45239} ≈ 0.06 $$ So, \(F_{12} ≈ 0.240\) and \(F_{21} ≈ 0.06\).

02

Calculate the net rate of radiation heat exchange between the sphere and the disk

The net rate of radiation heat exchange between the sphere and the disk can be calculated by: $$ Q_{12} = A_1 F_{12} \sigma \epsilon_{12} (T_1^4 - T_2^4) $$ Where \(\sigma\) is the Stefan-Boltzmann constant (\(\approx 5.67 * 10^{-8} W/(m^2 K^4)\)), \(\epsilon_{12}\) is the combined emissivity between the sphere and the disk, and \(T_1\) and \(T_2\) are the surface temperatures of the sphere and the disk in Kelvin, respectively. First, we need to calculate the combined emissivity \(\epsilon_{12}\): $$ \frac{1}{\epsilon_{12}} = \frac{1 - \epsilon_1}{\epsilon_1} + \frac{1 - \epsilon_2}{\epsilon_2} $$ Using the given emissivities of the sphere and disk which are \(0.9\) and \(0.5\), respectively: $$ \frac{1}{\epsilon_{12}} = \frac{1 - 0.9}{0.9} + \frac{1 - 0.5}{0.5} $$ Calculate \(\epsilon_{12}\): $$ \epsilon_{12} ≈ 0.357 $$ Now, we need to convert the surface temperatures of the sphere and the disk to Kelvin: $$ T_1 = 600 + 273 = 873 K $$ $$ T_2 = 200 + 273 = 473 K $$ Finally, we can calculate the net rate of radiation heat exchange between the sphere and the disk: $$ Q_{12} = 11310 * 0.240 * 5.67 * 10^{-8} * 0.357 * (873^4 - 473^4) W $$ Calculate \(Q_{12}\): $$ Q_{12} ≈ 33298 W $$

03

Part (c) Optimal conditions for heat exchange

We are given four potential modifications for increasing the net rate of radiation heat exchange between the sphere and the disk: adjusting emissivities, adjusting distance, and adding an enclosure. In this step, we will discuss what would be the optimal conditions for each modification without calculating the exact result. 1. Emissivities: To increase the net rate of radiation heat exchange, we would want to maximize the emissivities of both the sphere and the disk. The ideal value would be \(\epsilon_1 = \epsilon_2 = 1\). 2. Distance: Decreasing the distance between the two surfaces would increase the view factor, and hence increase the net rate of radiation heat exchange. However, this might not always be practical due to physical constraints. 3. Enclosure: Adding an enclosure would improve the radiation transfer by reflecting the radiated energy back to the surfaces, but it would depend on the emissivity of the enclosure material and the geometry of the enclosure. The exact influence of the enclosure would need to be determined case by case. Considering the above optimal conditions for each modification, the net rate of radiation heat exchange between the two bodies would be higher than the previously calculated value of \(33298 W\).

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