Question

Two concentric spheres are maintained at uniform temperatures \(T_{1}=45^{\circ} \mathrm{C}\) and \(T_{2}=280^{\circ} \mathrm{C}\) and have emissivities \(\varepsilon_{1}=0.25\) and \(\varepsilon_{2}=0.7\), respectively. If the ratio of the diameters is \(D_{1} / D_{2}=0.30\), the net rate of radiation heat transfer between the two spheres per unit surface area of the inner sphere is (a) \(86 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(1169 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(1181 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(2510 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(3306 \mathrm{~W} / \mathrm{m}^{2}\)

Short answer

Answer: (c) 1181 W/m²

Step-by-step solution

Convert temperatures to Kelvin

Given temperatures are \(T_1=45^\circ \mathrm{C}\) and \(T_2=280^\circ \mathrm{C}\). To convert them to Kelvin, we need to add 273 to both values: $$T_{1K} = T_1 + 273 = 45 + 273 = 318\, \mathrm{K}$$ $$T_{2K} = T_2 + 273 = 280 + 273 = 553\, \mathrm{K}$$

Calculate the surface areas

Given the ratio of the diameters \(D_1 / D_2 = 0.30\). Since the surface area of a sphere with diameter \(D\) is \(A = \pi D^2\), we can substitute the relation of diameters to the surface areas: $$\frac{A_1}{A_2} = \left(\frac{D_1}{D_2}\right)^2 = 0.30^2 = 0.09$$ Let \(A_2 = k\), then \(A_1 = 0.09 k\). Here, \(k\) is just an arbitrary parameter representing the actual surface area of sphere 2.

Apply the formula

Now, let's plug the given values into the formula for the net rate of radiation heat transfer: $$ H_{net} = \frac{\varepsilon_1 \sigma T_1^4 - \varepsilon_2 \sigma T_2^4}{\frac{1 - \varepsilon_1}{\varepsilon_1 A_1} + \frac{1 - \varepsilon_2}{\varepsilon_2 A_2}} = \frac{0.25 \sigma (318)^4 - 0.7\sigma (553)^4}{\frac{1 - 0.25}{0.25 (0.09k)} + \frac{1 - 0.7}{0.7 k}} $$

Simplify and find the answer

Let's simplify the equation to find the value of \(H_{net}\): $$ H_{net} = \frac{0.25\sigma (318)^4 - 0.7\sigma (553)^4}{\frac{0.75}{0.0225k} + \frac{0.3}{0.7 k}} = \frac{0.25\sigma (318)^4 - 0.7\sigma (553)^4}{\frac{1}{k}}k = 0.25\sigma (318)^4 - 0.7\sigma (553)^4 $$ Now, evaluate the value of H_net using \(\sigma = 5.67 \times 10^{-8} Wm^{-2}K^{-4}\): $$ H_{net} = 0.25(5.67 \times 10^{-8})(318)^4 - 0.7(5.67 \times 10^{-8})(553)^4 \approx 1181 \, \mathrm{W/m^2} $$ Hence, the correct option is (c) \(1181 \, \mathrm{W/m^2}\).