Question

A furnace is of cylindrical shape with a diameter of \(1.2 \mathrm{~m}\) and a length of \(1.2 \mathrm{~m}\). The top surface has an emissivity of \(0.60\) and is maintained at \(450 \mathrm{~K}\). The bottom surface has an emissivity of \(0.50\) and is maintained at \(800 \mathrm{~K}\). The side surface has an emissivity of \(0.40\). Heat is supplied from the base surface at a net rate of \(1400 \mathrm{~W}\). Determine the temperature of the side surface and the net rates of heat transfer between the top and the bottom surfaces and between the bottom and side surfaces.

Short answer

Based on the given information and calculations: 1. The side surface temperature of the furnace is approximately 455.34 K. 2. The net rate of heat transfer between the top and bottom surfaces is -14973.91 W (a negative value indicates heat is being absorbed by the top surface). 3. The net rate of heat transfer between the bottom and side surfaces is 12573.91 W (indicating heat is being released from the bottom surface to the side surface).

Step-by-step solution

Calculate the surface areas of the top, side, and bottom surfaces

To determine the temperature of the side surface and the net rates of heat transfer between top-bottom and bottom-side surfaces, first, we need to calculate the surface areas of the top, side, and bottom surfaces. The formula for the surface area of a cylinder will be used for the side surface. - Top Surface Area (A_top): It is a circle with a diameter of 1.2 m. A_top = π(D_top)^2 / 4 = π(1.2)^2 / 4 = 1.131 m^2 - Bottom Surface Area (A_bottom): It is also a circle with diameter 1.2 m. A_bottom = π(D_bottom)^2 / 4 = π(1.2)^2 / 4 = 1.131 m^2 - Side Surface Area (A_side): It is the lateral surface area of the cylindrical furnace with diameter 1.2 m and height 1.2 m. A_side = π * D * h = π(1.2)(1.2) = 4.523 m^2

Apply the Stefan-Boltzmann Law and find the heat transfer rates

Next, we will apply the Stefan-Boltzmann Law, which states that the rate of heat transfer by radiation is proportional to the emissivity, surface area, and temperature of the surface raised to the fourth power (T^4). The formula for radiative heat transfer is given as Q_rad = ε * σ * A * (T^4). Given the emissivities and temperatures, we can calculate the heat transfer rates between the surfaces. - Heat Transfer from Bottom Surface (Q1): Q1 = ε_bottom * σ * A_bottom * (T_bottom^4) = 0.50 * (5.67e-8) * 1.131 * (800^4) = 17033.81 W - Heat Transfer from Top Surface (Q2): Q2 = ε_top * σ * A_top * (T_top)^4 = 0.60 * (5.67e-8) * 1.131 * (450^4) = 2059.90 W

Calculate the side surface temperature

From the given information, we know that the heat transfer from the base is 1400 W. This heat must be balanced by the net heat transfer between the top and side surfaces. Let Q3 be the heat transfer from the side surface. So, Q2 + Q3 = 1400 W. Now, we can calculate the side surface temperature (T_side) using the following equation: Q3 = ε_side * σ * A_side * (T_side)^4 T_side^4 = Q3 / (ε_side * σ * A_side) T_side = (Q3 / (ε_side * σ * A_side))^(1/4) Plugging in the values, we get: Q3 = 1400 - Q2 = 1400 - 2059.90 = -659.90 W (The negative sign indicates heat is absorbed by the side surface) T_side = ((-659.90) / (0.40 * (5.67e-8) * 4.523))^(1/4) T_side ≈ 455.34 K

Find the net rates of heat transfer between the surfaces

Now that we have the temperature of the side surface (T_side), we can find the net rates of heat transfer between the top and the bottom surfaces (Q_top_bottom) and between the bottom and side surfaces (Q_bottom_side) using the following formulas: Q_top_bottom = Q2 - Q1 = 2059.90 - 17033.81 = -14973.91 W Q_bottom_side = Q1 - Q2 - 1400 = 17033.81 - 2059.90 - 1400 = 12573.91 W Therefore, the net rate of heat transfer between the top and bottom surfaces is -14973.91 W, and the net rate of heat transfer between the bottom and side surfaces is 12573.91 W.