Question

An 8-m-internal-diameter spherical tank made of \(1.5\)-cm-thick stainless steel \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located in a room whose temperature is \(25^{\circ} \mathrm{C}\). The walls of the room are also at $25^{\circ} \mathrm{C}$. The outer surface of the tank is black (emissivity \(\varepsilon=1\) ), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat transfer coefficients at the inner and the outer surfaces of the tank are $80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( and \)10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(, respectively. Determine \)(a)$ the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -h period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\).

Short answer

After analyzing the given exercise of a spherical tank containing iced water at \(0^{\circ} \mathrm{C}\) and surrounded by a room at \(25^{\circ}\mathrm{C}\), we have determined that the objective is to determine the rate of heat transfer to the iced water and the amount of ice that melts during a 24-hour period. We used a step-by-step solution including finding the heat transfer through the stainless steel, determining the overall heat transfer coefficient, calculating the rate of heat transfer to the iced water, and finally, calculating the amount of ice that melts during a 24-hour period.

Step-by-step solution

Find the heat transfer through the stainless steel

To find the heat transfer through the stainless steel, we have to calculate the conductive heat transfer across the thickness of the stainless steel. The formula for conductive heat transfer is given by: $$ q_{cond} = k * \frac{A * (T_{in} - T_{out})}{d} $$ where \(k\) is the thermal conductivity of the material in \(\mathrm{W/(m \cdot K)}\), \(A\) is the surface area in \(m^2\), \(T_{in}\) and \(T_{out}\) are the inner and the outer surface temperatures in degrees Celsius, and \(d\) is the thickness in meters. We can find the surface area of the tank using the formula for a sphere: $$ A = 4 \pi r^2 $$ where \(r\) is the radius of the tank in meters. Then, we can proceed to calculate the conductive heat transfer.

Calculate the overall heat transfer coefficient

The overall heat transfer coefficient (\(U\)) can be found using the formula: $$ \frac{1}{U*A} = \frac{1}{h_{in}*A} + \frac{d}{k*A} + \frac{1}{h_{out}*A} $$ In this formula, \(h_{in}\) and \(h_{out}\) are the convection heat transfer coefficients on the inner and outer surfaces of the tank, respectively. Solving for \(U\), we get: $$ U = \frac{1}{\frac{1}{h_{in}} + \frac{d}{k} + \frac{1}{h_{out}}} $$ Now, we can substitute the given values and calculate the overall heat transfer coefficient.

Calculate the rate of heat transfer to the iced water

To find the rate of heat transfer to the iced water, we can use the overall heat transfer coefficient and the temperatures of the room and the iced water: $$ q_{total} = U * A * (T_{room} - T_{iced}) $$ Substituting the values, we can calculate the rate of heat transfer to the iced water.

Calculate the amount of ice that melts during a 24-hour period

To find the amount of ice that melts during a 24-hour period, we will use the latent heat of fusion of ice, which is given as \(h_{if}\) in \(\mathrm{kJ/kg}\). First, we need to convert the rate of heat transfer from \(\mathrm{W}\) to \(\mathrm{kJ/h}\): $$ q_{total} (\mathrm{kJ/h}) = \frac{q_{total} \times 3600}{1000} $$ Next, we can use the latent heat of fusion to find the mass of ice melted during a 24-hour period: $$ m_{melt} = \frac{q_{total} (\mathrm{kJ/h}) * 24 \, \mathrm{h}}{h_{if}} $$ Calculating this will give us the amount of ice that melts during a 24-hour period.