Question

Two-phase gas-liquid oxygen is stored in a spherical tank of \(1 \mathrm{~m}\) 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 the 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.

Short answer

Answer: The heat transfer rate at the spherical tank surface is approximately -163.754 W, indicating that the heat is being transferred from the inner sphere to the outer sphere.

Step-by-step solution

Identify the known values

We are given the following information: - Inner sphere diameter: \(1 \mathrm{~m}\), which means the radius is \(0.5 \mathrm{~m}\) - Outer sphere diameter: \(1.6 \mathrm{~m}\), which means the radius is \(0.8 \mathrm{~m}\) - Both surfaces have an emissivity of \(0.01\) - Inner sphere temperature: normal boiling point of oxygen, which is \(90.19 \mathrm{~K}\) - Outer sphere temperature: \(273 \mathrm{~K}\)

Use the Stefan-Boltzmann Law to find the radiative heat transfer rate

Radiative heat transfer between two surfaces is given by the Stefan-Boltzmann Law: \(q = \sigma \epsilon A(T_1^4 - T_2^4)\) where: - \(\sigma\) is the Boltzmann constant, which is equal to \(5.67 \times 10^{-8} \mathrm{W/m^2K^4}\) - \(\epsilon\) is the emissivity of the surfaces - \(A\) is the surface area of the inner sphere - \(T_1\) and \(T_2\) are the temperatures of the inner and outer spheres, respectively In this case, we will use the formula to find the heat transfer rate at the inner sphere surface.

Calculate the surface area of the inner sphere

To find the surface area of the inner sphere, we can use the formula: \(A = 4 \pi r^2\) where \(r\) is the radius of the sphere. In this case, \(r = 0.5 \mathrm{~m}\), so: \(A = 4 \pi (0.5)^2 = 3.14 \mathrm{~m^2}\)

Calculate the heat transfer rate

Now we have all the values needed for the Stefan-Boltzmann Law. Let's plug them into the formula: $q = \sigma \epsilon A(T_1^4 - T_2^4) \\ = (5.67 \times 10^{-8})(0.01)(3.14)(90.19^4 - 273^4) \\ = -163.754 \mathrm{W}$ The heat transfer rate at the spherical tank surface is \(-163.754\mathrm{W}\). The negative sign indicates that the heat is being transferred from the inner sphere to the outer sphere, which is expected since the outer sphere is at a higher temperature.