Question

Two parallel concentric disks, \(20 \mathrm{~cm}\) and \(40 \mathrm{~cm}\) in diameter, are separated by a distance of \(10 \mathrm{~cm}\). The smaller disk \((\varepsilon=0.80)\) is at a temperature of \(300^{\circ} \mathrm{C}\). The larger disk \((\varepsilon=0.60)\) is at a temperature of $800^{\circ} \mathrm{C}$. (a) Calculate the radiation view factors. (b) Determine the rate of radiation heat exchange between the two disks. (c) Suppose that the space between the two disks is completely surrounded by a reflective surface. Estimate the rate of radiation heat exchange between the two disks.

Short answer

Based on the given problem, we calculated the radiation view factors for both parallel concentric disks as 1/3. Using the Stefan-Boltzmann law, we determined that the rate of radiation heat exchange between the two disks is -1319 W, indicating heat transfer from disk 2 to disk 1. When a reflective surface is introduced between the two disks, the rate of radiation heat exchange is approximately -1575 W, with heat still being transferred from disk 2 to disk 1.

Step-by-step solution

Calculate the radiation view factors

To find the radiation view factors, we will use the formula for parallel concentric disks: $$F_{1 \to 2} = \frac{r_1^2}{r_2^2 - r_1^2}$$ Where \(F_{1 \to 2}\) is the radiation view factor from disk 1 to disk 2, \(r_1\) is the radius of the smaller disk, and \(r_2\) is the radius of the larger disk. Given the diameters of the disks: \(r_1 = \frac{20 \mathrm{~cm}}{2} = 10 \mathrm{~cm}\) \(r_2 = \frac{40 \mathrm{~cm}}{2} = 20 \mathrm{~cm}\) Now let's calculate the radiation view factor from disk 1 to disk 2: $$F_{1 \to 2} = \frac{(10 \mathrm{~cm})^2}{(20 \mathrm{~cm})^2 - (10 \mathrm{~cm})^2} = \frac{100 \mathrm{~cm^2}}{300 \mathrm{~cm^2}} = \frac{1}{3}$$ Similarly, the radiation view factor from disk 2 to disk 1 is: $$F_{2 \to 1} = F_{1 \to 2} = \frac{1}{3}$$

Determine the rate of radiation heat exchange between the two disks

To calculate the rate of radiation heat exchange, we will use the Stefan-Boltzmann law: $$q = \varepsilon_1 \varepsilon_2 \sigma (T_1^4 - T_2^4) \cdot A$$ Where \(q\) is the radiation heat exchange, \(\varepsilon_1\) and \(\varepsilon_2\) are the emissivities of the disks, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2 K^4}\)), \(T_1\) and \(T_2\) are the temperatures of the disks in Kelvin, and \(A\) is the area of the smaller disk. First, we need to convert the temperatures to Kelvin: \(T_1 = 300^{\circ} \mathrm{C} + 273.15 = 573.15 \mathrm{K}\) \(T_2 = 800^{\circ} \mathrm{C} + 273.15 = 1073.15 \mathrm{K}\) Next, we'll determine the area of the smaller disk: \(A = \pi (r_1)^2 = \pi (10 \mathrm{~cm})^2 = 100 \pi \mathrm{~cm^2}\) Now, we can calculate the rate of radiation heat exchange: $$q = (0.80)(0.60)[5.67 \times 10^{-8} \mathrm{W/m^2 K^4}][(573.15 \mathrm{K})^4 - (1073.15 \mathrm{K})^4] \cdot (100 \pi \mathrm{~cm^2})$$ $$q \approx -1319 \mathrm{W}$$ Since the value is negative, it shows that heat is being transferred from disk 2 to disk 1.

Estimate the rate of radiation heat exchange with a reflective surface

When a reflective surface is introduced between the two disks, we can assume that the emissivity of the reflective surface is 0, meaning there is no heat transfer through the surface. Therefore, the heat exchange between the two disks will only occur through the view factors, which remain unchanged. With the reflective surface, the rate of radiation heat exchange, \(q'\), can be calculated as: $$q' = \frac{F_{1 \to 2} \varepsilon_1 \varepsilon_2 \sigma (T_1^4 - T_2^4) \cdot A}{1 - (1 - \varepsilon_1)(1 - \varepsilon_2)F_{1 \to 2} F_{2 \to 1}}$$ Plugging in the values, we get: $$q' = \frac{\frac{1}{3}(0.80)(0.60)[5.67 \times 10^{-8} \mathrm{W/m^2 K^4}][(573.15 \mathrm{K})^4 - (1073.15 \mathrm{K})^4] \cdot (100 \pi \mathrm{~cm^2})}{1 - (1 - 0.80)(1 - 0.60)\frac{1}{3} \frac{1}{3}}$$ $$q' \approx -1575 \mathrm{W}$$ With the reflective surface, the rate of radiation heat exchange between the two disks is approximately -1575 W, with heat being transferred from disk 2 to disk 1.