Question

Consider a long semicylindrical duct of diameter \(1.0 \mathrm{~m}\). Heat is supplied from the base surface, which is black, at a rate of $1200 \mathrm{~W} / \mathrm{m}^{2}\(, while the side surface with an emissivity of \)0.4$ is maintained at \(650 \mathrm{~K}\). Neglecting the end effects, determine the temperature of the base surface.

Short answer

Answer: The temperature of the base surface is approximately 673.77 K.

Step-by-step solution

Identify the Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the power emitted by a blackbody is proportional to the fourth power of its temperature. It can be represented by the equation: \(P = \sigma * A * T^{4}\) Where P is the power emitted, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 * 10^{-8} \mathrm{W} / \mathrm{m}^2 \mathrm{K}^4\)), A is the area of the surface, and T is the absolute temperature of the body.

Apply the Stefan-Boltzmann Law to the base surface

Since the base surface is black, we can assume its emissivity is equal to 1. To find the temperature, we use the equation: \(1200 \mathrm{W} / \mathrm{m}^{2} = (1) * (\sigma) * (T_{base}^4)\) Solving for \(T_{base}^4\): \(T_{base}^4 = \frac{1200 \mathrm{W} / \mathrm{m}^{2}}{\sigma}\)

Apply the Stefan-Boltzmann Law to the side surface

The heat balance at the base surface can be described in terms of the heat transferred between the base and the side surfaces. The power absorbed by the side surface can be calculated by multiplying the power emitted by the base surface with the emissivity of the side surface: \(P_{side} = (0.4) * (\sigma) * (A_{side}) * (T_{side}^4)\) Since the base surface is providing heat to maintain the side surface temperature, the power absorbed by the side surface is equal to the power emitted by the base surface: \(P_{base} = P_{side}\)

Equate the power from base and side surfaces and solve for the base surface temperature

Equate the powers from step 2 and step 3 and simplify: \(\frac{1200 \mathrm{W} / \mathrm{m}^{2}}{\sigma} = (0.4) * (\sigma) * (A_{side}) * (650 \mathrm{K})^4\) Divide both sides by \(\sigma\): \(1200 \mathrm{W} / \mathrm{m}^{2} = (0.4) * (A_{side}) * (650 \mathrm{K})^4\) Now, solve for \(T_{base}^4\): \(T_{base}^4 = \frac{1200 \mathrm{W} / \mathrm{m}^{2}}{(0.4) * (A_{side}) * (650 \mathrm{K})^4}\) Take the fourth root of both sides to find the temperature of the base surface: \(T_{base} = \sqrt[4]{\frac{1200 \mathrm{W} / \mathrm{m}^{2}}{(0.4) * (A_{side}) * (650 \mathrm{K})^4}}\) The area of the side surface, \(A_{side}\), can be eliminated in the equation as it would be the same on both sides and doesn't impact the final result. Thus, we find the temperature of the base surface: \(T_{base} = \sqrt[4]{\frac{1200}{0.4 * 650^4}}\) \(T_{base} \approx 673.77 \mathrm{K}\) The temperature of the base surface is approximately \(673.77 \mathrm{K}\).