Question

Two long, parallel 20-cm-diameter cylinders are located \(30 \mathrm{~cm}\) apart from each other. Both cylinders are black and are maintained at temperatures \(425 \mathrm{~K}\) and \(275 \mathrm{~K}\). The surroundings can be treated as a blackbody at \(300 \mathrm{~K}\). For a 1 -m-long section of the cylinders, determine the rates of radiation heat transfer between the cylinders and between the hot cylinder and the surroundings.

Short answer

Answer: The rate of radiation heat transfer between the two cylinders is approximately 881 W, and the rate of radiation heat transfer between the hot cylinder and the surroundings is approximately 1243 W.

Step-by-step solution

Calculate the surface area of a 1-meter section of the cylinder

Given the diameter of the cylinders is \(20 \mathrm{~cm}\), we can find their radius (10 cm) and convert it to meters (0.1 m). The length of the cylinder section is given as 1 meter. The surface area of a cylinder can be calculated as: $$ A = 2 \pi r L $$ Using the given values: $$ A = 2 \pi (0.1 \mathrm{~m})(1 \mathrm{~m}) = 0.2 \pi \mathrm{~m^2}$$

Calculate the radiation heat transfer between the two cylinders

We'll use the given temperatures for the two cylinders (\(425 \mathrm{~K}\) and \(275 \mathrm{~K}\)) and the Stefan-Boltzmann constant (\(\sigma = 5.67 \times 10^{-8} \mathrm{W/m^2K^4}\)) in the radiation heat transfer formula: $$q = A \epsilon \sigma (T_1^4 - T_2^4)$$ Plugging in the values: $$q = 0.2 \pi (1)(5.67 \times 10^{-8})(425^4 - 275^4)$$ $$q \approx 881 \mathrm{~W}$$ The rate of radiation heat transfer between the two cylinders is approximately 881 W.

Calculate the radiation heat transfer between the hot cylinder and the surroundings

Now we'll use the temperature of the hot cylinder (\(425 \mathrm{~K}\)) and the temperature of the surroundings (\(300 \mathrm{~K}\)) in the radiation heat transfer formula: $$q = A \epsilon \sigma (T_1^4 - T_2^4)$$ Plugging in the values: $$q = 0.2 \pi (1)(5.67 \times 10^{-8})(425^4 - 300^4)$$ $$q \approx 1243 \mathrm{~W}$$ The rate of radiation heat transfer between the hot cylinder and the surroundings is approximately 1243 W.