Question

A 2.2-m-diameter spherical steel tank filled with iced water at $0^{\circ} \mathrm{C}$ is buried underground at a location where the thermal conductivity of the soil is \(k=0.55 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The distance between the tank center and the ground surface is \(2.4 \mathrm{~m}\). For a ground surface temperature of \(18^{\circ} \mathrm{C}\), determine the rate of heat transfer to the iced water in the tank. What would your answer be if the soil temperature were \(18^{\circ} \mathrm{C}\) and the ground surface were insulated?

Short answer

Answer: For the non-insulated ground surface case, the rate of heat transfer to the iced water in the tank is approximately 121.54 W. For the insulated ground surface case, the rate of heat transfer to the iced water in the tank would be 0 W.

Step-by-step solution

Identify the given information

We are given the following information: - Diameter of the spherical steel tank (\(D\)) = 2.2 meters - Temperature of iced water (\(T_{1}\)) = \(0^{\circ}\mathrm{C}\) - Distance between tank center and ground surface (\(L\)) = 2.4 meters - Thermal conductivity of the soil (\(k\)) = 0.55 W/m·K - Ground surface temperature (\(T_{2}\)) = \(18^{\circ}\mathrm{C}\)

Calculate the temperature difference

The temperature difference between the ground surface and the iced water is: \(\Delta T = T_{2} - T_{1} = 18 - 0 = 18\mathrm{^{\circ}C}\)

Find the radius of the tank and the distance from the tank's wall to the ground surface

The radius of the spherical tank (\(R\)) is half of its diameter: \(R = \frac{D}{2} = \frac{2.2}{2} = 1.1\mathrm{~m}\) Now, we can find the distance between the tank's wall and the ground surface (\(\delta L\)) by subtracting the radius of the tank from the distance between the center of the tank and the ground surface: \(\delta L = L - R = 2.4 - 1.1 = 1.3\mathrm{~m}\)

Calculate the rate of heat transfer using Fourier's Law of heat conduction

According to Fourier's Law, the rate of heat transfer (\(Q\)) is: \(Q = \frac{k\cdot A\cdot \Delta T}{\delta L}\) where the surface area of the tank (\(A\)) can be determined using the formula for the surface area of a sphere: \(A = 4\pi R^2 = 4\pi (1.1)^2 \approx 15.205\mathrm{~m^2}\) Now, plug in the values: \(Q = \frac{0.55\cdot 15.205 \cdot 18}{1.3} \approx 121.535 \mathrm{~W}\) The rate of heat transfer to the iced water in the tank is approximately 121.54 W.

Analyze the insulated ground surface case

If the ground surface were insulated, no heat would be gained or lost through the surface. Hence, the soil temperature would be constant at \(18^{\circ}\mathrm{C}\). As a result, there would be no temperature difference between the soil and the ground surface, and the rate of heat transfer would be zero. So, for the insulated ground surface case, the rate of heat transfer to the iced water in the tank would be 0 W.