Question

=Consider a vertical 2 -m-diameter cylindrical furnace whose surfaces closely approximate black surfaces. The base, top, and side surfaces of the furnace are maintained at \(400 \mathrm{~K}, 600 \mathrm{~K}\), and \(1200 \mathrm{~K}\), respectively. If the view factor from the base surface to the top surface is \(0.2\), the net radiation heat transfer between the base and the side surfaces is (a) \(73 \mathrm{~kW}\) (b) \(126 \mathrm{~kW}\) (c) \(215 \mathrm{~kW}\) (d) \(292 \mathrm{~kW}\) (e) \(344 \mathrm{~kW}\)

Short answer

Answer: (b) 126 kW

Step-by-step solution

Recall the formula for net radiation heat transfer between two surfaces

The formula to calculate the net radiation heat transfer between two black surfaces i and j is: \(Q_{i \rightarrow j} = A_i F_{i \rightarrow j} \epsilon_{ij} \sigma [T_i^4 - T_j^4]\) Where \(A_i\) is the area of surface i, \(F_{i \rightarrow j}\) is the view factor from surface i to surface j, \(\epsilon_{ij}\) is the effective emissivity between surfaces i and j, \(\sigma\) is the Stefan-Boltzmann constant(\(5.67 \times 10^{-8} \mathrm{W/m^2 K^4}\)), and \(T_i\) and \(T_j\) are the absolute temperatures of surfaces i and j, respectively. Since the surfaces are black, \(\epsilon_{ij} = 1\).

Calculate the area of the base surface

The base and top surfaces are circles with diameter 2 m. The area of a circle is given by the formula: \(A = \pi r^2\) The radius (r) can be calculated by dividing the diameter by 2: \(r = \frac{2}{2} = 1 \mathrm{m}\) Now, we can calculate the area (A): \(A = \pi (1 \mathrm{m})^2 = \pi \mathrm{m^2}\)

Find the view factor from the base surface to the side surface

Given that the view factor from the base surface to the top surface (\(F_{b \rightarrow t}\)) is 0.2, we can find the view factor from the base surface to the side surface (\(F_{b \rightarrow s}\)) using the knowledge that the sum of the view factors from one surface to all other surfaces in an enclosure must be equal to 1: \(F_{b \rightarrow s} = 1 - F_{b \rightarrow t} = 1 - 0.2 = 0.8\)

Calculate the net radiation heat transfer between the base and side surfaces

Now that we have all the necessary information, we can apply the formula and calculate the net radiation heat transfer: \(Q_{b \rightarrow s} = A_b F_{b \rightarrow s} \sigma [T_b^4 - T_s^4]\) Substitute the values into the equation: \(Q_{b \rightarrow s} = \pi \mathrm{m^2} (0.8) (5.67 \times 10^{-8} \mathrm{W/m^2 K^4}) [(400 \mathrm{K})^4 - (1200 \mathrm{K})^4]\) Evaluate the expression: \(Q_{b \rightarrow s} = \pi \mathrm{m^2} (0.8) (5.67 \times 10^{-8} \mathrm{W/m^2 K^4}) [(2.56 \times 10^7 \mathrm{K^4}) - (2.0736 \times 10^9 \mathrm{K^4})]\) \(Q_{b \rightarrow s} = -126.3 \mathrm{kW}\) Since the heat transfer is negative, this means that the heat is actually transferred from the side surface to the base surface. The magnitude of the heat transfer is 126.3 kW. The correct option is: (b) \(126 \mathrm{~kW}\)