Question

Consider two concentric spheres forming an enclosure with diameters of $12 \mathrm{~cm}\( and \)18 \mathrm{~cm}\( and surface temperatures \)300 \mathrm{~K}$ and \(500 \mathrm{~K}\), respectively. Assuming that the surfaces are black, the net radiation exchange between the two spheres is (a) \(21 \mathrm{~W}\) (b) \(140 \mathrm{~W}\) (c) \(160 \mathrm{~W}\) (d) \(1275 \mathrm{~W}\) (e) \(3084 \mathrm{~W}\)

Short answer

Question: Two black spheres with diameters of 12 cm and 18 cm have surface temperatures of 300 K and 500 K, respectively. Calculate the net radiation exchange between the two spheres. Answer: (c) 160 W

Step-by-step solution

Identifying the radiative heat transfer equation

The radiative heat transfer equation between two black surfaces can be written as: \(Q=\sigma A_1 F_{1-2}(T_1^4 - T_2^4)\) Where: \(Q\) = radiative heat transfer \(\sigma\) = Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \frac{W}{m^2 K^4}\)) \(A_1\) = area of the inner sphere \(F_{1-2}\) = view factor from surface 1 to 2 \(T_1\) = temperature of the inner sphere \(T_2\) = temperature of the outer sphere

Converting diameters to radii and calculating areas

Diameters are given, so first convert them to radii: \(R_1 = \frac{12}{2}\mathrm{~cm} = 6\mathrm{~cm} = 0.06\mathrm{~m}\) \(R_2 = \frac{18}{2}\mathrm{~cm} = 9\mathrm{~cm} = 0.09\mathrm{~m}\) Now calculate the areas of both spheres using the formula: \(A = 4\pi R^2\) \(A_1 = 4\pi R_1^2 = 4\pi (0.06)^2 = 0.0452\mathrm{~m^2}\) \(A_2 = 4\pi R_2^2 = 4\pi (0.09)^2 = 0.1018\mathrm{~m^2}\)

Finding the view factor

For concentric spheres, the view factor from surface 1 to 2 can be calculated as: \(F_{1-2} = \frac{A_1}{A_2} = \frac{0.0452}{0.1018} = 0.44398\)

Calculating the radiative heat transfer

Substitute the values into the radiative heat transfer equation: \(Q=\sigma A_1 F_{1-2}(T_1^4 - T_2^4)\) \(Q=(5.67 \times 10^{-8})(0.0452)(0.44398)((500^4)-(300^4))\) \(Q\approx160\mathrm{~W}\) Comparing the result to the options given, we find that the correct answer is: (c) \(160\mathrm{~W}\)