Question

A furnace is of cylindrical shape with \(R=H=3 \mathrm{~m}\). The base, top, and side surfaces of the furnace are all black and are maintained at uniform temperatures of 500,700 , and \(1400 \mathrm{~K}\), respectively. Determine the net rate of radiation heat transfer to or from the top surface during steady operation.

Short answer

Answer: The net rate of radiation heat transfer to the top surface during steady operation is approximately 2.16 x 10^7 W.

Step-by-step solution

Identify the knowns and unknowns

We are given the radius \(R\), height \(H\), and the uniform temperatures of the base (\(T_1\)), top (\(T_2\)), and side (\(T_3\)) surfaces of the furnace. \(R = 3 \mathrm{~m}\) \(H = 3 \mathrm{~m}\) \(T_1 = 500 \mathrm{~K}\) \(T_2 = 700 \mathrm{~K}\) \(T_3 = 1400 \mathrm{~K}\) We are asked to find the net rate of radiation heat transfer to or from the top surface (\(q_\textit{net}\)).

Calculate the heat transfer between the top and base surfaces

We will use the Stefan-Boltzmann law, which states that the heat transfer by radiation between two surfaces is given by: \(q = A \sigma F_{12} (T_1 ^4 - T_2 ^4)\) where \(A\) is the area of the emitting surface (in this case, the top surface), \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2.K^4}\)), and \(F_{12}\) is the view factor from surface 1 to surface 2. For two parallel surfaces, the view factor \(F_{12}\) is 1. The area of the top surface, being a circle, is given by: \(A = \pi R^2\) Substituting the known values, we get: \(q_\textit{top-base} = (\pi \cdot (3 \mathrm{m})^2) (5.67 \times 10^{-8} \mathrm{W/m^2.K^4}) (500^4 \mathrm{K^4} - 700^4 \mathrm{K^4})\) \(q_\textit{top-base} \approx -5.98 \times 10^6 \mathrm{W}\) The negative sign indicates heat transfer from the top surface to the base surface.

Calculate the heat transfer between the top surface and the side surface

To calculate the heat transfer between the top surface and the side surface, we use the same Stefan-Boltzmann equation. However, since the side surface is cylindrical, we need to find its view factor with respect to the top surface. Using view factor correlations and considering the side surface as a ring-shaped surface, we get: \(F_{23} \approx 0.224\) Now, we can calculate the heat transfer between the top surface and the side surface: \(q_\textit{top-side} = (\pi \cdot (3 \mathrm{m})^2) (5.67 \times 10^{-8} \mathrm{W/m^2.K^4}) F_{23}(1400^4 \mathrm{K^4} - 700^4 \mathrm{K^4})\) \(q_\textit{top-side} \approx 2.76 \times 10^7 \mathrm{W}\) The positive sign indicates heat transfer from the side surface to the top surface.

Calculate the net rate of radiation heat transfer to or from the top surface

Adding the heat transfers from step 2 and step 3, we get the net rate of radiation heat transfer to or from the top surface: \(q_\textit{net} = q_\textit{top-base} + q_\textit{top-side}\) \(q_\textit{net} = -5.98 \times 10^6 \mathrm{W} + 2.76 \times 10^7 \mathrm{W}\) \(q_\textit{net} \approx 2.16 \times 10^7 \mathrm{W}\) The positive sign indicates that the net rate of radiation heat transfer is to the top surface during steady operation.