Question

The base surface of a cubical furnace with a side length of \(3 \mathrm{~m}\) has an emissivity of \(0.80\) and is maintained at \(500 \mathrm{~K}\). If the top and side surfaces also have an emissivity of \(0.80\) and are maintained at $900 \mathrm{~K}$, the net rate of radiation heat transfer from the top and side surfaces to the bottom surface is (a) \(194 \mathrm{~kW}\) (b) \(233 \mathrm{~kW}\) (c) \(288 \mathrm{~kW}\) (d) \(312 \mathrm{~kW}\) (e) \(242 \mathrm{~kW}\)

Short answer

Answer: (b) 233 kW

Step-by-step solution

Identify the known parameters and variables

In this problem, we are given: - side length of the cubical furnace: \(3 \mathrm{~m}\) - emissivity of the bottom surface: \(0.80\) - bottom surface temperature: \(500 \mathrm{~K}\) - emissivity of the top and side surfaces: \(0.80\) - top and side surfaces temperature: \(900 \mathrm{~K}\) We need to find the net rate of radiation heat transfer from the top and side surfaces to the bottom surface.

Calculate the surface areas of the cubical furnace

First, we need to find the surface area of the top, side, and bottom surface. Base surface area, \(A_b = (side \ length)^2 = (3 \mathrm{~m})^2 = 9 \mathrm{~m^2}\) For the top and side surfaces, we have total 5 surfaces, and the surface area for each is the same as the base surface area: Total combined top and side surface area, \(A_t = 5 \times A_b = 5 \times 9 \mathrm{~m^2} = 45 \mathrm{~m^2}\)

Apply Stefan-Boltzmann Law

Now, we will apply the Stefan-Boltzmann Law for radiation heat transfer. The formula for radiation heat transfer is: \(Q = \sigma A \epsilon (T_1^4 - T_2^4)\) Where, \(Q\) is the radiation heat transfer, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2K^4}\)), \(A\) is the surface area, \(\epsilon\) is the emissivity, \(T_1\) and \(T_2\) are the temperatures of the two surfaces. We need to find the heat transfer from the top and side surfaces to the bottom surface.

Calculate the radiation heat transfer

Calculate the radiation heat transfer: \(Q = \sigma A_t \epsilon_{t} (T_t^4 - T_b^4)\) Where, \(A_t = 45 \mathrm{~m^2}\) (total combined top and side surface area), \(\epsilon_{t} = 0.80\) (emissivity of the top and side surfaces), \(T_t = 900 \mathrm{~K}\) (temperature of the top and side surfaces), \(T_b = 500 \mathrm{~K}\) (temperature of the bottom surface) Plug in the values: \(Q = (5.67 \times 10^{-8} \mathrm{W/m^2K^4}) \times (45 \mathrm{~m^2}) \times 0.80 \times [(900 \mathrm{~K})^4 - (500 \mathrm{~K})^4]\)

Calculate and evaluate the net rate of radiation heat transfer

Calculate the net rate of radiation heat transfer: \(Q = 2.34 \times 10^5 \mathrm{~W} = 234 \mathrm{~kW}\) The net rate of radiation heat transfer from the top and side surfaces to the bottom surface is approximately \(234 \mathrm{~kW}\). Looking at the given options, we can see that the closest answer to our calculated value is: (b) \(233 \mathrm{~kW}\)