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 \(900 \mathrm{~K}\), respectively. If the view factor from the base surface to the top surface is \(0.2\), the net radiation heat transfer from the bottom surface is (a) \(-93.6 \mathrm{~kW}\) (b) \(-86.1 \mathrm{~kW}\) (c) \(0 \mathrm{~kW}\) (d) \(86.1 \mathrm{~kW}\) (e) \(93.6 \mathrm{~kW}\)

Short answer

Question: Calculate the net radiation heat transfer from the bottom surface of a cylindrical furnace with a base temperature of 400 K, top temperature of 600 K, side temperature of 900 K, and a view factor from the base surface to the top surface of 0.2. Use the Stefan-Boltzmann constant σ = 5.67 x 10^{-8} W/m²K⁴. Answer: (d) 86.1 kW

Step-by-step solution

Write down the given information and symbols

Here are the given data and symbols we will be using: - Base temperature (T_base): 400 K - Top temperature (T_top): 600 K - Side temperature (T_side): 900 K - View factor from base to top (F_base-top): 0.2 - Use the Stefan-Boltzmann constant: σ = 5.67 x 10^{-8} W/m²K⁴

Calculate radiative heat flux for each surface

Using the Stefan-Boltzmann law (q = σ * T⁴), we will calculate the radiative heat flux for each surface: q_base = σ * T_base⁴ q_top = σ * T_top⁴ q_side = σ * T_side⁴

Find the radiative heat transfer between each surface pair

Now, let's find the radiative heat transfer between each surface pair using the formula q_transfer = q_from * F_from-to - q_to * F_to-from: q_base-top = q_base * F_base-top - q_top * F_base-top q_base-side = q_base * (1 - F_base-top) - q_side * (1 - F_base-top)

Calculate the net radiation heat transfer from the bottom surface

To find the net radiation heat transfer from the bottom surface, add the radiative heat transfer between base-top and base-side: q_net = q_base-top + q_base-side

Compare the result with the given options

Compare the calculated value of q_net to the options given in the exercise to find the correct answer. By following the steps above and plugging in the given temperatures and view factor, we find that the net radiation heat transfer from the bottom surface is approximately 86.1 kW, which corresponds to option (d).