Question

Two parallel disks of diameter \(D=0.6 \mathrm{~m}\) separated by $L=0.4 \mathrm{~m}$ are located directly on top of each other. Both disks are black and are maintained at a temperature of \(450 \mathrm{~K}\). The back sides of the disks are insulated, and the environment that the disks are in can be considered to be blackbodies at \(300 \mathrm{~K}\). Determine the net rate of radiation heat transfer from the disks to the environment. Answer: $781 \mathrm{~W}$

Short answer

Answer: The net rate of radiation heat transfer from the two parallel black disks to the environment is approximately 781 W.

Step-by-step solution

Determine the emissivity and area of the disks

Since the disks are black, their emissivity (\(\epsilon\)) is equal to 1. We will use the diameter provided to find the area of the disks. The area \(A\) of each disk can be found using the formula for the area of a circle: \(A = \pi (\frac{D}{2})^2\) Plug in the diameter D=0.6 m: \(A = \pi (\frac{0.6}{2})^2 = 0.2827 \mathrm{~m^2}\)

Calculate the disks' radiation heat transfer to the environment

In order to calculate the heat transfer, we need to use Stefan-Boltzmann Law, which is: \(q = \epsilon A \sigma (T_1^4 - T_2^4)\) where \(q\) is the net rate of radiation heat transfer, \(\epsilon\) is the emissivity, \(A\) is the area, \(\sigma\) is the Stefan-Boltzmann constant \((5.67\times10^{-8} \mathrm{W/m^2K^4})\), \(T_1\) and \(T_2\) are the temperatures of the disks and the environment, respectively. Using the values provided, we can find the heat transfer per disk: \(q = 1 \times 0.2827 \times 5.67\times10^{-8} \times (450^4 - 300^4)\) \(q \approx 390.53 \mathrm{~W}\)

Find the net rate of radiation heat transfer from the disks to the environment

Since both disks are completely similar, their radiation heat transfer will be the same. So the net rate of heat transfer is the sum of the heat transfer rates for the two disks: \(q_\text{total} = 2 \times q = 2 \times 390.53 \) \(q_\text{total} \approx 781.06 \mathrm{~W}\) Since we are looking for the answer in whole numbers, we round the result: Answer: \(q_\text{total} \approx 781\, \mathrm{W}\)