A spherical tank of diameter \(D=2 \mathrm{~m}\) that is filled with liquid nitrogen at \(100 \mathrm{~K}\) is kept in an evacuated cubic enclosure whose sides are \(3 \mathrm{~m}\) long. The emissivities of the spherical tank and the enclosure are \(\varepsilon_{1}=0.1\) and \(\varepsilon_{2}=0.8\), respectively. If the temperature of the cubic enclosure is measured to be \(240 \mathrm{~K}\), determine the net rate of radiation heat transfer to the liquid nitrogen. Answer: \(228 \mathrm{~W}\)

The surface area \(A\) of a sphere with diameter \(D\) is given by the formula: \(A = \pi D^2\). For a spherical tank with diameter \(D=2 \mathrm{~m}\), we compute the surface area as follows: \(A_1 = \pi \times (2 \mathrm{~m})^2 = 4\pi \mathrm{~m^2}\) For the cubic enclosure, we first compute the surface area of a single side, which is \(A_2 = (3 \mathrm{~m})^2 = 9 \mathrm{~m^2}\). Since a cube has six identical sides, the total surface area of the cube is then: \(A_\text{cube} = 6 \times A_2 = 6 \times 9 \mathrm{~m^2} = 54 \mathrm{~m^2}\)

To find the net rate of radiation heat transfer to the liquid nitrogen, we can apply the Stefan-Boltzmann law: \(Q = \sigma \varepsilon_{1} \varepsilon_{2} A_1 A_2 F_{1-2}(T_1^4 - T_2^4)\), where \(\sigma\) is the Stefan-Boltzmann constant, \(\varepsilon_i\) are the emissivities, \(A_i\) are the surface areas, \(F_{1-2}\) is the view factor between the tank and the enclosure, and \(T_i\) are the absolute temperatures. Given that the tank and the enclosure are concentric, the view factor \(F_{1-2}=1\). The temperature of the liquid nitrogen is \(T_1=100 \mathrm{~K}\), and the temperature of the cubic enclosure is \(T_2=240 \mathrm{~K}\). The emissivities are given as \(\varepsilon_{1}=0.1\) and \(\varepsilon_{2}=0.8\). The Stefan-Boltzmann constant \(\sigma = 5.67 \times 10^{-8} \mathrm{W/m^2K^4}\).

Now plug in the values into the Stefan-Boltzmann law: \(Q = 5.67 \times 10^{-8} \mathrm{W/m^2K^4} \cdot 0.1 \cdot 0.8 \cdot 4\pi \mathrm{~m^2} \cdot 54 \mathrm{~m^2} \cdot (100^4 - 240^4) \mathrm{K^4}\) Calculating the value, we find: \(Q = 228 \mathrm{~W}\) Thus, the net rate of radiation heat transfer to the liquid nitrogen is \(228 \mathrm{~W}\).