Chapter 13: Problem 140

Consider a cubical furnace with a side length of \(3 \mathrm{~m}\). The top surface is maintained at \(700 \mathrm{~K}\). The base surface has an emissivity of \(0.90\) and is maintained at \(950 \mathrm{~K}\). The side surface is black and is maintained at \(450 \mathrm{~K}\). Heat is supplied from the base surface at a rate of \(340 \mathrm{~kW}\). Determine the emissivity of the top surface and the net rates of heat transfer between the top and the bottom surfaces, and between the bottom and side surfaces.

Short Answer

Expert verified

Answer: To determine the emissivity of the top surface and the net rates of heat transfer between the surfaces, follow the steps provided in the solution. The four steps are: 1. Calculate area and radiosity for each surface. 2. Calculate net radiative heat transfer between surfaces. 3. Find emissivity for the top surface. 4. Find net rates of heat transfer. By following these steps, we will be able to determine the required values for the emissivity of the top surface and the net rates of heat transfer between the top and bottom surfaces, and between the bottom and side surfaces of the cubical furnace.

Step by step solution

First, we need to calculate the area, \(A\), of each surface of the cubical furnace: \(A = \text{side length} \times \text{side length} = 3\text{m} \times 3\text{m} = 9\text{ m}^2\). Next, let's calculate the radiosity, \(J_i\), for each surface using the Stefan-Boltzmann Law (\(J_i = \varepsilon_i \sigma T_i^4 + \rho_i \sigma T_{surroundings}^4\)) and the given temperatures, where \(\varepsilon_i\) is the emissivity of the surface, \(\rho_i\) is the reflectivity of the surface (\(\rho_i = 1 - \varepsilon_i\)), \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \text{W/m}^2\text{K}^4\)), and \(T_i\) is the temperature in Kelvin: Top surface: \(J_1 = \varepsilon_1 \sigma T_1^4 + \rho_1 \sigma T_{surroundings}^4\) Base surface: \(J_2 = \varepsilon_2 \sigma T_2^4 + \rho_2 \sigma T_{surroundings}^4\) Side surface: Since it's black, the emissivity is \(1\), and we can use the simpler Stefan-Boltzmann Law: \(J_3 = \sigma T_3^4\) #Step 2: Calculating net radiative heat transfer between surfaces#

Now we will calculate the net radiative heat transfer between the surfaces, using the formula \(q_{net} = A F_{i \rightarrow j}(J_i - J_j)\), where \(F_{i \rightarrow j}\) represents the view factor between surfaces \(i\) and \(j\) and \(q_{net}\) is the net radiative heat transfer between the two surfaces: Top and base surfaces: \(q_{net, 1 \rightarrow 2} = A F_{1 \rightarrow 2}(J_1 - J_2)\) Within a cube, the surface to surface view factor is \(1/4\) which can be used to calculate the net radiative heat transfer: Bottom and side surfaces: \(q_{net, 2 \rightarrow 3} = A F_{2 \rightarrow 3}(J_2 - J_3)\) #Step 3: Finding emissivity for the top surface#

We are given that the heat supplied from the base surface is \(340 \text{ kW}\). This heat is transferred from the base surface to the top and side surfaces. Therefore, the net radiative heat transfer from the base surface should be equal to the heat supplied: \(q_{net, 1 \rightarrow 2} + q_{net, 2 \rightarrow 3} = 340000\text{ W}\) or equivalently: \(A F_{1 \rightarrow 2}(J_1 - J_2) + A F_{2 \rightarrow 3}(J_2 - J_3) = 340000\text{ W}\) Using the known values and solving this equation for the unknown variable, \(\varepsilon_1\), we obtain the emissivity of the top surface. #Step 4: Finding net rates of heat transfer#

Finally, we use the obtained emissivity value for the top surface, and the known values for the base and side surfaces, to calculate the net radiative heat transfer rates between the top and bottom surfaces, and between the bottom and side surfaces: \(q_{net, 1 \rightarrow 2} = A F_{1 \rightarrow 2}(J_1 - J_2)\) \(q_{net, 2 \rightarrow 3} = A F_{2 \rightarrow 3}(J_2 - J_3)\)