Question

Consider two black coaxial parallel circular disks of equal diameter \(D\) that are spaced apart by a distance \(L\). The top and bottom disks have uniform temperatures of \(500^{\circ} \mathrm{C}\) and \(520^{\circ} \mathrm{C}\), respectively. Determine the radiation heat transfer coefficient $h_{\text {rad }}\( between the disks if they are spaced apart by \)L=D$.

Short answer

Question: Calculate the radiation heat transfer coefficient between two coaxial parallel circular disks with equal diameter, uniform temperatures of 500°C and 520°C, and spaced apart by a distance equal to their diameter. Answer: The radiation heat transfer coefficient between the two coaxial parallel circular disks is approximately 21.88 W/m²K.

Step-by-step solution

Convert temperatures to Kelvin

To convert temperatures from Celsius to Kelvin, add 273.15 to the temperature in Celsius: \(T_1 = 500^\circ \mathrm{C} + 273.15 = 773.15 \mathrm{K}\) \(T_2 = 520^\circ \mathrm{C} + 273.15 = 793.15 \mathrm{K}\)

Calculate the area of the disks

The area of a circular disk can be calculated using the formula: \(A = \pi r^2\), where \(r\) is the radius of the disk. We are given the diameter of the disks, \(D\), and the radius is half the diameter: \(r = \frac{D}{2}\) So, the area of the disks is: \(A = \pi \left(\frac{D}{2}\right)^2\)

Calculate the radiation heat transfer between the disks

The radiation heat transfer between the disks can be calculated using the formula: \(q = \sigma A (T_2^4 - T_1^4)\), where - \(\sigma = 5.67 \times 10^{-8} \, \mathrm{W m^{-2} K^{-4}}\): Stefan-Boltzmann constant Substitute the values of \(A\), \(T_1\), and \(T_2\) in the equation: \(q = 5.67 \times 10^{-8} \times \pi \left(\frac{D}{2}\right)^2 \times \left((793.15 \mathrm{K})^4 - (773.15 \mathrm{K})^4\right)\)

Calculate the radiation heat transfer coefficient

Now, we can calculate the radiation heat transfer coefficient (\(h_{\text{rad}}\)) using the formula: \(h_{\text{rad}} = \frac{q}{A(T_2 - T_1)}\) Substitute the values of \(q\), \(A\), \(T_1\), and \(T_2\) in the equation: \(h_{\text{rad}} = \frac{5.67 \times 10^{-8} \times \pi \left(\frac{D}{2}\right)^2 \times \left((793.15 \mathrm{K})^4 - (773.15 \mathrm{K})^4\right)}{\pi \left(\frac{D}{2}\right)^2(793.15 \mathrm{K} - 773.15 \mathrm{K})}\) Simplify the equation: \(h_{\text{rad}} = \frac{5.67 \times 10^{-8} \times ((793.15 \mathrm{K})^4 - (773.15 \mathrm{K})^4)}{20 \mathrm{K}}\) Now, we can compute the radiation heat transfer coefficient: \(h_{\text{rad}} \approx 21.88 \, \mathrm{W m^{-2} K^{-1}}\) The radiation heat transfer coefficient between the two coaxial parallel circular disks is approximately \(21.88 \, \mathrm{W m^{-2} K^{-1}}\).