Question

Two parallel disks of diameter \(D=3 \mathrm{ft}\) separated by $L=2 \mathrm{ft}$ are located directly on top of each other. The disks are separated by a radiation shield whose emissivity is \(0.15\). Both disks are black and are maintained at temperatures of \(1350 \mathrm{R}\) and $650 \mathrm{R}$, respectively. The environment that the disks are in can be considered to be a blackbody at \(540 \mathrm{R}\). Determine the net rate of radiation heat transfer through the shield under steady conditions.

Short answer

Question: Calculate the net rate of radiation heat transfer through a radiation shield with an emissivity of 0.15 placed between two parallel black disks with a diameter of 3 ft, where the temperature of disk 1 is 1350 K and the temperature of disk 2 is 650 K.

Step-by-step solution

Calculate Area and Find Constants

The diameter of the disks (\(D\)) is given as 3 ft, so we calculate the area of one disk using the formula for the area of a circle, \(A = \pi r^2\): $$ A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{3}{2}\right)^2 = \frac{9}{4}\pi\ ft^2 $$ We are given the emissivity of the shield as \(0.15\) and the Stefan-Boltzmann constant, \(\sigma = 5.67 \times 10^{-8} W/m^2 K^4\).

Calculate Radiation Resistances

Since the disks are black, we have their emissivity at 1. Then we can write the expressions for radiation resistances between the disks, shield, and surroundings: $$ R_{1 \to disk} = \frac{1}{A (\epsilon_1 \sigma T_1^3 + \epsilon_2)} $$ Where \(\epsilon_1\) is the emissivity of disk 1 and \(\epsilon_2\) is the emissivity of the shield. $$ R_{disk \to 2} = \frac{1}{A (\epsilon_2 \sigma T_2^3 + 1)} $$ Substitute the values to calculate the resistances: $$ R_{1 \to disk} = \frac{1}{(\frac{9}{4}\pi) (1 \times 5.67 \times 10^{-8} \times (1350)^3 + 0.15)} $$ $$ R_{disk \to 2} = \frac{1}{(\frac{9}{4}\pi) (0.15 \times 5.67 \times 10^{-8} \times (650)^3 + 1)} $$

Find Total Radiation Resistance and Net Heat Transfer

Add the two resistances to find the total radiation resistance: $$ R_{total} = R_{1\to disk} + R_{disk \to 2} $$ Now, we can use the total radiation resistance to find the net heat transfer: $$ Q = \frac{T_1 - T_2}{R_{total}} $$ Substitute the given values and the total radiation resistance to find the net heat transfer: $$ Q = \frac{1350 - 650}{R_{total}} = \frac{700}{R_{total}} $$ After calculating the value of \(R_{total}\) and substituting, you will find the numerical value of the net rate of radiation heat transfer through the shield under steady conditions.