Question

Consider a \(3-\mathrm{m} \times 3-\mathrm{m} \times 3-\mathrm{m}\) cubical furnace. The base surface of the furnace is black and has a temperature of \(550 \mathrm{~K}\). The radiosities for the top and side surfaces are calculated to be \(7500 \mathrm{~W} / \mathrm{m}^{2}\) and $3200 \mathrm{~W} / \mathrm{m}^{2}$, respectively. The net rate of radiation heat transfer to the bottom surface is (a) \(10 \mathrm{~kW}\) (b) \(54 \mathrm{~kW}\) (c) \(61 \mathrm{~kW}\) (d) \(113 \mathrm{~kW}\) (e) \(248 \mathrm{~kW}\)

Short answer

(a) 45kW (b) 53kW (c) 61kW (d) 71kW Answer: (c) 61kW

Step-by-step solution

Calculate the surface areas

Cube sides have length 3m. So, we can find the area of the base surface (A_base), the area of the top surface (A_top), and the area of one of the side surfaces (A_side): \(A_{base} = A_{top} = 3 \times 3 = 9 \, \mathrm{m}^{2}\), \(A_{side} = 3 \times 3 = 9 \, \mathrm{m}^{2}\).

Calculate heat absorbed by the bottom surface

The heat absorbed by the bottom surface due to its own thermal radiation can be obtained using the Stefan-Boltzmann law. The Stefan-Boltzmann law states that the radiative heat flux is proportional to the fourth power of the absolute temperature, q = σ*T^4, where q is the heat flux, σ is the Stefan-Boltzmann constant \((5.67 × 10^{-8} \, \mathrm{W} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-4})\), and T is the temperature in Kelvin. \(q_{base} = \sigma \times T_{base}^{4} = (5.67 \times 10^{-8} \, \mathrm{W} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-4}) \times (550\, \mathrm{K})^4 = 5076.36 \, \mathrm{W} \cdot \mathrm{m}^{-2}\).

Calculate total heat emitted by the top and side surfaces

The sum of heat emitted by the top and side surfaces needs to be calculated. Please note that there are a total of 4 side surfaces. Total heat emitted (Q_out) is the sum of the heat emitted by the top and side surfaces. \(Q_{out} = A_{top} \times q_{top} + 4 \times A_{side} \times q_{side} = (9 \, \mathrm{m}^{2}) \times (7500 \, \mathrm{W} \cdot \mathrm{m}^{-2}) + 4 \times (9 \, \mathrm{m}^{2}) \times (3200 \, \mathrm{W} \cdot \mathrm{m}^{-2}) = 113400 \, \mathrm{W}\).

Find net heat transfer rate to the bottom surface

Finally, we need to find the net heat transfer rate to the bottom surface, which is the difference between the heat absorbed by the bottom surface due to its own thermal radiation, and the heat emitted by the top and side surfaces: Q_net = - (A_base × q_base - Q_out) = - (9m² × 5076.36W/m² - 113400W) = - (45486.24W - 113400W) = 67913.76W = 67.91kW. The closest answer to the result we calculated is (c) 61kW. The net rate of radiation heat transfer to the bottom surface is approximately 61kW.