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 \(900 \mathrm{~K}\), 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) \(22.5 \mathrm{~kW}\) (b) \(38.6 \mathrm{~kW}\) (c) \(60.7 \mathrm{~kW}\) (d) \(89.8 \mathrm{~kW}\) (e) \(151 \mathrm{~kW}\)

When analyzing this problem, we must first find out the information about view factors. The given view factor \(F_{bt}\) from the base surface to the top surface is \(0.2\). By conservation of energy and view factors, we also need to find the view factor \(F_{bs}\) of the base surface to the side surface: \(F_{bs} = 1 - F_{bt} = 1 - 0.2 = 0.8\) Now that we have found \(F_{bs}\), let's move on to the next step.

Next, we need to calculate the radiosity (radiant energy) of each surface. Since the surfaces closely approximate black surfaces, we can assume that they are blackbodies. Thus, we can use the equation for blackbody emissive power: \(E_b = \sigma T^4\) Where \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W/m^2K^4}\)) and \(T\) is the absolute temperature in Kelvin. We will apply this equation to each surface of the furnace: \(E_{b1} = \sigma T_1^4\) (Base surface) \(E_{b2} = \sigma T_2^4\) (Top surface) \(E_{b3} = \sigma T_3^4\) (Side surface) Plugging in the values, we get: \(E_{b1} = (5.67 \times 10^{-8})(400)^4 = 5.791 \times 10^4 \mathrm{W/m^2}\) \(E_{b2} = (5.67 \times 10^{-8})(600)^4 = 2.6145 \times 10^5 \mathrm{W/m^2}\) \(E_{b3} = (5.67 \times 10^{-8})(900)^4 = 1.1634 \times 10^6 \mathrm{W/m^2}\)

Now we will calculate the net radiation heat transfer \(Q_{bs}\) between the base and side surfaces using the following equation: \(Q_{bs} = F_{bs}(A_1E_{b1} - A_3E_{b3})\) Where \(A_1\) and \(A_3\) are the surface areas of the base and side surfaces, respectively. We can determine these surface areas using the diameter of the cylinder, \(2 \mathrm{m}\). The base surface is a circle with radius \(1 \mathrm{m}\) and the side surface is a cylinder with height \(2 \mathrm{m}\). \(A_1 = \pi r^2 = \pi(1)^2 = \pi \mathrm{m^2}\) \(A_3 = 2\pi rh = 2\pi(1)(2) = 4\pi \mathrm{m^2}\) Now we can plug the values into the equation for the net radiation heat transfer: \(Q_{bs} = 0.8[(\pi)(5.791 \times 10^4) - (4\pi)(1.1634 \times 10^6)]\) \(Q_{bs} = 0.8[1.819 \times 10^5 \mathrm{W} - 4.6536 \times 10^6 \mathrm{W}]\) \(Q_{bs} = 0.8[-4.4717 \times 10^6 \mathrm{W}] = -3.57736 \times 10^6 \mathrm{W}\) But, since we know that a negative net radiation heat transfer is not possible, this indicates that radiation is coming from the side surface to the base surface. Taking the absolute value, we have: \(Q_{bs} = 3.57736 \times 10^6 \mathrm{W} = 35.7736 \times 10^3 \mathrm{W}\) = \(35.7736 \mathrm{kW}\) Comparing this result against the given options, we get the closest answer to be: (b) \(38.6 \mathrm{kW}\) So, the net radiation heat transfer between the base and the side surfaces of the furnace is approximately \(38.6 \mathrm{kW}\).