Two-phase gas-liquid oxygen is stored in a spherical tank of \(1-m\) diameter, where it is maintained at its normal boiling point. The spherical tank is enclosed by a \(1.6-\mathrm{m}\) diameter concentric spherical surface at \(273 \mathrm{~K}\). Both spherical surfaces have an emissivity of \(0.01\), and the gap between the inner sphere and outer sphere is vacuumed. Assuming that the spherical tank surface has the same temperature as the oxygen, determine the heat transfer rate at the spherical tank surface.

The normal boiling point of oxygen is \(90.2 \mathrm{~K}\). Thus, the temperature of the oxygen at its normal boiling point is: \(T_{1} = 90.2 \mathrm{~K}\)

We are given the diameters of both the spherical tank (\(1\) m) and the concentric spherical surface (\(1.6\) m), so we can calculate their radii: \(r_{1}=0.5\) m and \(r_{2}=0.8\) m. The surface area of a sphere is given by the formula \(A = 4 \pi r^2\). Therefore, we can find the surface areas for both the inner (spherical tank) and outer (concentric spherical surface) spheres: \(A_{1} = 4 \pi r_{1}^2 = 4 \pi (0.5)^2 = \pi \mathrm{~m}^2\) \(A_{2} = 4 \pi r_{2}^2 = 4 \pi (0.8)^2 = 2.56 \pi \mathrm{~m}^2\)

In a vacuum, heat transfer occurs only through radiation. The radiative heat transfer equation is given by: \(q = \dfrac{F_{12}\sigma (T_{1}^4 - T_{2}^4)}{1/\varepsilon_{1} - 1 + (A_{1}/A_{2})(1/\varepsilon_{2} - 1)}\), where \(q\) - heat transfer rate, \(F_{12}\) - view factor between surface 1 and surface 2, which is equal to \(1\) for concentric spheres, \(\sigma\) - Stefan-Boltzmann constant \((5.67 \times 10^{-8} \mathrm{W/m}^2 \mathrm{K}^4)\), \(T_{1}\) - temperature of the inner sphere (temperature of the oxygen), \(T_{2}\) - temperature of the outer sphere (\(273 \mathrm{~K}\)), \(\varepsilon_{1}\) and \(\varepsilon_{2}\) - emissivities of surface 1 (spherical tank) and surface 2 (concentric spherical surface) (\((0.01)\) in both cases), and \(A_{1}\) and \(A_{2}\) - surface areas of surfaces 1 and 2. Plugging the given values into the radiative heat transfer equation: \(q = \dfrac{1 \times (5.67 \times 10^{-8}) \times [(90.2)^4 - (273)^4]}{1/0.01 - 1 + (\pi / 2.56 \pi)(1/0.01 - 1)}\) After calculating, we get: \(q \approx -13.15 \mathrm{~W}\) The negative sign indicates that the heat transfer is happening from the outer sphere to the inner sphere. Therefore, the heat transfer rate at the spherical tank surface is approximately \(13.15 \mathrm{~W}\).