Question

Consider a \(4-\mathrm{m} \times 4-\mathrm{m} \times 4-\mathrm{m}\) cubical furnace whose floor and ceiling are black and whose side surfaces are reradiating. The floor and the ceiling of the furnace are maintained at temperatures of \(550 \mathrm{~K}\) and \(1100 \mathrm{~K}\), respectively. Determine the net rate of radiation heat transfer between the floor and the ceiling of the furnace.

Short answer

Answer: The net rate of radiation heat transfer between the floor and the ceiling of the furnace is approximately 2.93 x 10^6 W.

Step-by-step solution

Determine the reradiating surface behavior

A reradiating surface is a surface that reflects all the radiation it receives and emits no radiation of its own. Here, the floor and ceiling of the furnace are black, meaning they emit and absorb radiation efficiently, while the side surfaces are reradiating and do not contribute to the overall heat transfer.

Calculate the view factor

The view factor between two surfaces is the proportion of the radiation leaving one surface that is intercepted by the other surface. For this problem, the view factor between the floor and ceiling of a cubical furnace is 1 since all the radiation leaving the floor is absorbed by the ceiling and vice versa.

Apply Stefan-Boltzmann law

To calculate the net rate of radiation heat transfer between the floor and the ceiling, we can use the Stefan-Boltzmann law, which is given by the formula: \(q = \sigma F_{1 \to 2} A_1 (T_1^4 - T_2^4)\) Where \(q\) is the net rate of radiation heat transfer, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2 K^4}\)), \(F_{1 \to 2}\) is the view factor between surface 1 and surface 2, \(A_1\) is the area of surface 1, and \(T_1\) and \(T_2\) are the absolute temperatures of surface 1 and surface 2, respectively.

Calculate the net rate of radiation heat transfer

Given the temperatures of the floor and ceiling (\(550 \mathrm{~K}\), \(1100 \mathrm{~K}\)) and the furnace's dimensions (\(4 \mathrm{m} \times 4 \mathrm{m} \times 4 \mathrm{m}\)), we can calculate the net rate of radiation heat transfer: \(q = (5.67 \times 10^{-8} \mathrm{W/m^2 K^4}) (1)(4^2 \mathrm{m}^2)((1100 \mathrm{~K})^4 - (550 \mathrm{~K})^4) \) \(q \approx 2.93 \times 10^6 \mathrm{W}\) So, the net rate of radiation heat transfer between the floor and the ceiling of the furnace is approximately \(2.93 \times 10^6 \mathrm{W}\).