Question

Consider a 3-m \(\times 3-\mathrm{m} \times 3-\mathrm{m}\) cubical furnace whose top and side surfaces closely approximate black surfaces at a temperature of \(1200 \mathrm{~K}\). The base surface has an emissivity of \(\varepsilon=0.7\), and is maintained at \(800 \mathrm{~K}\). Determine the net rate of radiation heat transfer to the base surface from the top and side surfaces.

Short answer

Answer: The net rate of radiation heat transfer to the base surface from the top and side surfaces is approximately \(1.0107\times10^8~\mathrm{W}\).

Step-by-step solution

Calculate the area of the top and side surfaces

The area of each side of the cube is \(A = L\times W\). Since the cube has the dimensions of 3 m x 3 m x 3 m and the top and side surfaces have the same dimensions as the base surface, the area of each of these surfaces is: $$A = 3~\text{m} \times 3~\text{m} = 9~\mathrm{m}^2$$

Determine the radiation heat transfer rate for the top and side surfaces

We can now use the Stefan-Boltzmann Law to determine the radiation heat transfer rate for the top and side surfaces. The Stefan-Boltzmann Law states that the power emitted by a black body is given by: $$P = \sigma A T^4$$ where \(\sigma\) is the Stefan-Boltzmann constant (\(\sigma = 5.67\times10^{-8}~\frac{\text{W}}{\text{m}^2 \text{K}^4}\)), \(A\) is the area of the emitting surface, and \(T\) is the temperature of the surface in Kelvin. We have the temperature of the top and side surfaces, which closely approximate black surfaces, as 1200 K. Therefore: $$P = (5.67\times10^{-8}~\frac{\text{W}}{\text{m}^2 \text{K}^4}) (9~\mathrm{m^2})(1200~\mathrm{K})^4 = 2.15 \times 10^7~\mathrm{W}$$ We have to calculate this for the top and 4 side surfaces, multiplying this 5 times: $$P_{total} = 5 \times 2.15 \times 10^7~\mathrm{W} = 1.075 \times 10^8~\mathrm{W}$$

Calculate the radiation heat transfer rate for the base surface

For the base surface with emissivity \(\varepsilon=0.7\) and temperature \(T_{base} = 800~\text{K}\), we can write the Stefan-Boltzmann Law as: $$P_{base} = \varepsilon\sigma A T_{base}^4$$ Substitute the emissivity, area, and temperature values in the above formula: $$P_{base} = (0.7)(5.67\times10^{-8}~\frac{\text{W}}{\text{m}^2 \text{K}^4}) (9~\mathrm{m^2})(800~\mathrm{K})^4 = 6.43\times10^6~\mathrm{W}$$

Determine the net rate of radiation heat transfer to the base surface

To find the net rate of radiation heat transfer to the base surface, we can subtract the base surface radiation from the total radiation emitted by the top and side surfaces. $$Q_{net} = P_{total}- P_{base} = 1.075 \times 10^8~\mathrm{W} - 6.43\times10^6~\mathrm{W} = 1.0107\times10^8~\mathrm{W}$$

Final Answer:

The net rate of radiation heat transfer to the base surface from the top and side surfaces is approximately \(1.0107\times10^8~\mathrm{W}\).