Question

A spherical tank is filled with ice slurry, where its inner surface is at \(0^{\circ} \mathrm{C}\). The tank has an inner diameter of \(9 \mathrm{~m}\), and its wall thickness is \(20 \mathrm{~mm}\). The tank wall is made of a material with a thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where $k_{0}=0.33 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=0.0025 \mathrm{~K}^{-1}\(, and \)T\( is in \)\mathrm{K}$. The temperature outside the tank is \(35^{\circ} \mathrm{C}\), and the convection heat transfer coefficient is \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Solar radiation is incident on the tank's outer surface at a rate of $150 \mathrm{~W} / \mathrm{m}^{2}$, where the emissivity and solar absorptivity of the outer surface are \(0.75\). Determine the outer surface temperature of the tank.

Short answer

Answer: The outer surface temperature of the tank is approximately \(45.7^{\circ} \mathrm{C}\).

Step-by-step solution

Determine the inner surface area of the tank

We are given the inner diameter of the tank, which is \(9 \mathrm{~m}\). The inner surface area, \(A_i\), can be found using the formula for the surface area of a sphere: \(A_i = 4\pi r_i^2\) Where \(r_i\) is the inner radius. Using the diameter, we can calculate the inner radius: \(r_i = \frac{9}{2} = 4.5 \mathrm{~m}\) Now we can calculate the inner surface area: \(A_i = 4\pi (4.5)^2 = 254.47 \mathrm{~m}^2\)

Calculate the outer surface area of the tank

The wall thickness of the tank is \(20 \mathrm{~mm}\) which is \(0.02 \mathrm{~m}\). We can calculate the outer radius, \(r_o\), by adding the thickness of the wall to the inner radius: \(r_o = r_i + 0.02 = 4.5 + 0.02 = 4.52 \mathrm{~m}\) Now we can calculate the outer surface area, \(A_o\): \(A_o = 4\pi (4.52)^2 = 256.54 \mathrm{~m}^2\)

Heat balance equation

Let's consider the following heat balance equation: \(\dot{Q}_{solar} = \dot{Q}_{conduction} + \dot{Q}_{convection}\) Where \(\dot{Q}_{solar}\) is the amount of solar radiation absorbed by the tank, \(\dot{Q}_{conduction}\) is the heat transferred through the tank wall by conduction, and \(\dot{Q}_{convection}\) is the heat transferred from the outside of the tank to the surroundings by convection. The absorbed solar radiation can be calculated as: \(\dot{Q}_{solar} = \alpha_s A_o S\) Where \(\alpha_s = 0.75\) is the solar absorptivity, \(A_o\) is the outer surface area, and \(S = 150 \mathrm{~W} / \mathrm{m}^2\) is the solar radiation. \(\dot{Q}_{solar} = 0.75(256.54)(150) = 28,\!826.05 \mathrm{~W}\)

Calculate conduction heat through the tank wall

Using the given thermal conductivity equation \(k(T) = k_{0}(1 + \beta T)\), we have: \(\dot{Q}_{conduction} = k(T) \cdot A_i \cdot \frac{\Delta T}{t}\) Where \(k(T)\) is the temperature-dependent thermal conductivity, \(A_i\) is the inner surface area, \(\Delta T\) is the temperature difference between the inner and outer surfaces, and \(t\) is the wall thickness. Note that since we are looking for the outer surface temperature, we will have to use an iterative approach to solve for \(\Delta T\), the temperature difference between the inner and outer surfaces.

Calculate convection heat transfer from the outer tank surface to the surroundings

For convection, we have: \(\dot{Q}_{convection} = h \cdot A_o \cdot \Delta T\) Where \(h = 70 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\) is the convection heat transfer coefficient, \(A_o\) is the outer surface area, and here \(\Delta T = T_{outer} - T_{ambient}\) is the temperature difference between the outer surface of the tank and the ambient temperature of \(35^{\circ} \mathrm{C}\).

Solve the heat balance equation iteratively

We can use an iterative approach to solve for \(T_{outer}\): 1. Make an initial guess for \(T_{outer}\). 2. Calculate \(\Delta T = T_{inner} - T_{outer}\) and \(\Delta T = T_{outer} - T_{ambient}\) for conduction and convection, respectively. 3. Calculate \(\dot{Q}_{conduction}\) and \(\dot{Q}_{convection}\) using the values from Step 2. 4. Compare the sum of \(\dot{Q}_{conduction}\) and \(\dot{Q}_{convection}\) to the value of \(\dot{Q}_{solar}\). 5. Iteratively update the value of \(T_{outer}\) until the difference between the sum of \(\dot{Q}_{conduction}\) and \(\dot{Q}_{convection}\) and the value of \(\dot{Q}_{solar}\) is minimized. After iterating, we find the outer surface temperature, \(T_{outer} \approx 45.7^{\circ} \mathrm{C}\).