Question

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.

Short answer

Question: Calculate the view factors, net rate of radiation heat exchange, and the new net rate of radiation heat exchange after implementing the given modifications. Solution: Step 1: Find the view factors using the given equation and reciprocal relation. - Given: \(r_1 = 30 cm\), \(r_2 = 120 cm\), \(h = 60 cm\) - Calculate \(F_{12}\) using the given equation. - Calculate \(F_{21}\) using the reciprocal relation and finding the areas of the sphere and the circular disk. Step 2: Find the net rate of radiation heat exchange between the sphere and the disk. - Given: \(T_1 = 600^\circ C\), \(T_2 = 200^\circ C\), \(\varepsilon_1 = 0.9\), \(\varepsilon_2 = 0.5\) - Convert temperatures to Kelvin. - Calculate the net rate of radiation heat exchange \(q_{net}\) using the given equation and temperatures in Kelvin. Step 3: Determine the modifications and calculate the new net rate of radiation heat exchange. - Modification 1: Find the new emissivities that maximize the exchange (\(\varepsilon_1'\) and \(\varepsilon_2'\)). - Modification 2: Find the optimal distance (\(h'\)) for the new emissivities. - Modification 3: Assume minimal heat loss to the environment with a refractory enclosure. - Calculate the new net rate of radiation heat exchange (\(q_{net}'\)) using the new emissivities, distance, and same equation with minimal heat loss.

Step-by-step solution

Calculate the view factors \(F_{12}\) and \(F_{21}\)

To calculate the view factor \(F_{12}\), we will use the given equation: $$ F_{12}=0.5\left\\{1-\left[1+\left(\frac{r_{2}}{h}\right)^{2}\right]^{-0.5}\right\\} $$ We are given \(r_2 = 120 cm\) and \(h = 60 cm\). Substituting these values into the equation, we can find \(F_{12}\). After finding \(F_{12}\), we can calculate the view factor \(F_{21}\) using the reciprocal relation: $$ F_{21} = \frac{A_1}{A_2}F_{12} $$ where \(A_1\) and \(A_2\) are the areas of the sphere and the circular disk, respectively. We can find these areas using the formulas: \(A_1 = 4 \pi r_1^2\) for the sphere, \(A_2 = \pi r_2^2\) for the circular disk. We are given \(r_1 = 30 cm\). Substituting these values into the area formulas and finding the areas, we can find \(F_{21}\) with the above equation.

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

To calculate the net rate of radiation heat exchange between the sphere and the disk, we can use the following equation: $$ q_{net} = F_{12}\sigma(T_1^4 - T_2^4)\frac{A_1\varepsilon_1\varepsilon_2}{1 - \varepsilon_1(1 - \varepsilon_2)} $$ where \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} W/m^2K^4)\), \(T_1\) and \(T_2\) are the temperatures of the sphere and the disk (\(600^\circ C\) and \(200^\circ C\)), and \(\varepsilon_1\) and \(\varepsilon_2\) are their emissivities (\(0.9\) and \(0.5\)). First, we need to convert the temperatures from Celsius to Kelvin by adding \(273.15\): \(T_1 = 600 + 273.15 = 873.15 K\), \(T_2 = 200 + 273.15 = 473.15 K\). Now, we can substitute the values of \(F_{12}\), \(T_1\), \(T_2\), \(\varepsilon_1\), and \(\varepsilon_2\) into the equation and find the net rate of radiation heat exchange \(q_{net}\).

Determine the modifications to increase the net rate of radiation heat exchange and calculate the new net rate

The given potential modifications to increase the net rate of radiation heat exchange are: 1. Change the emissivities of the surfaces by painting them. 2. Adjust the distance between the sphere and the circular disk. 3. Provide a refractory enclosure to reduce heat losses to the environment. For the first modification, we will find the new emissivities (\(\varepsilon_1'\) and \(\varepsilon_2'\)) that maximize the net rate of radiation heat exchange. We can assume that these new emissivities are close to \(1\) for both surfaces. For the second modification, we will find the optimal distance \(h'\) for the new emissivities \(\varepsilon_1'\) and \(\varepsilon_2'\). For the third modification, we can assume that the refractory enclosure reduces the heat losses to the environment and all the heat is exchanged between the sphere and the circular disk. Using the new emissivities, distance and the assumption of minimal heat loss to the environment, we can calculate the new net rate of radiation heat exchange \(q_{net}'\) by using the same equation: $$ q_{net}' = F_{12}'\sigma(T_1^4 - T_2^4)\frac{A_1\varepsilon_1'\varepsilon_2'}{1 - \varepsilon_1'(1 - \varepsilon_2')} $$ where \(F_{12}'\) is the view factor for the new distance \(h'\), which can be found using the same view factor equation with \(h'\) replacing \(h\).