Question

An ice skating rink is located in a building where the air is at $T_{\text {air }}=20^{\circ} \mathrm{C}\( and the walls are at \)T_{w}=25^{\circ} \mathrm{C}$. The convection heat transfer coefficient between the ice and the surrounding air is \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissivity of ice is \(\varepsilon=0.95\). The latent heat of fusion of ice is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\), and its density is $920 \mathrm{~kg} / \mathrm{m}^{3}$. (a) Calculate the refrigeration load of the system necessary to maintain the ice at \(T_{s}=0^{\circ} \mathrm{C}\) for an ice rink of \(12 \mathrm{~m}\) by \(40 \mathrm{~m}\). (b) How long would it take to melt \(\delta=3 \mathrm{~mm}\) of ice from the surface of the rink if no cooling is supplied and the surface is considered insulated on the back side?

Short answer

In this exercise, we calculated the refrigeration load necessary to maintain the ice at \(0^{\circ}\mathrm{C}\), which is approximately \(101694\text{ W}\). We also calculated that it would take about \(4350\text{ s}\) or \(1.2\text{ hours}\) to melt \(3\text{ mm}\) of ice if no cooling is supplied.

Step-by-step solution

(Step 1: Calculate the total surface area of the ice rink)

To calculate the refrigeration load and the melting time, we first need to know the total surface area of the ice rink. The given ice rink dimensions are \(12\text{ m}\) by \(40\text{ m}\), and the surface area can be calculated with the formula \(A=L \cdot W\). \( A= 12\text{ m} \times 40 \text{ m} = 480 m^2 \)

(Step 2: Calculate the convection heat transfer)

Next, we need to calculate the convection heat transfer between the ice and the air. We will use the following convection heat transfer equation: \(Q_c = h \cdot A \cdot (T_{air} -T_s)\). \(Q_c = 10\frac{\text{W}}{\text{m}^2\text{K}} \times 480\text{ m}^2 \times (20^{\circ}\text{C}-0^{\circ}\text{C})\) \(Q_c = 96000\text{ W}\)

(Step 3: Calculate the radiation heat transfer)

Now we need to calculate the radiation heat transfer from the ice to the room. We'll use the Stefan-Boltzmann equation for radiation heat transfer: \(Q_r = \varepsilon \cdot A \cdot \sigma \cdot (T_{w}^4 - T_{s}^4)\), where \(\sigma = 5.67 \times 10^{-8}\text{ W}/\text{m}^2\text{K}^4\) is the Stefan-Boltzmann constant. First, we need to convert the temperatures to Kelvin: \(T_{w,K} = 273+25 = 298\text{ K}\) and \(T_{s,K} = 273+0 = 273\text{ K}\) \(Q_r = 0.95 \times 480\text{ m}^2 \times 5.67 \times 10^{-8}\frac{\text{W}}{\text{m}^2\text{K}^4} \times (298^4 \text{ K}^4 - 273^4 \text{ K}^4) \) \(Q_r \approx 5694\text{ W}\)

(Step 4: Calculate the total heat transfer and refrigeration load)

Add the convection and radiation heat transfers to find the total heat transfer: \( Q=Q_c + Q_r = 96000\text{ W} + 5694 \text{W} \approx 101694\text{ W}\) So, the refrigeration load of the system is approximately \(101694 \text{ W}\).

(Step 5: Calculate the volume of the ice that melts)

We are given that \(\delta = 3\text{ mm}\) of ice would melt, and we are to find the time it will take to melt that much ice. First, we need to determine the volume of the ice that melts: \(V_{melt}=\delta \cdot A = 3 \times 10^{-3}\text{ m} \times 480\text{ m}^2 = 1.44\text{ m}^3\)

(Step 6: Calculate the mass of the ice that melts)

Next, we find the mass of the ice that melts using its density. The density of ice is \(920\frac{\text{kg}}{\text{m}^3}\). Therefore, mass can be calculated using: \(m_{melt} = V_{melt} \times \rho_{ice} = 1.44\text{ m}^{3} \times 920\frac{\text{kg}}{\text{m}^{3}} \approx 1325\text{ kg}\)

(Step 7: Calculate the total heat required to melt the ice)

Now, we can calculate the total heat required to melt the ice using the latent heat of fusion: \(Q_{melt} = m_{melt} \cdot h_{if} = 1325\text{ kg} \times 333.7\frac{\text{kJ}}{\text{kg}} = 442387.5\text{ kJ}\)

(Step 8: Calculate the time required to melt the ice)

Finally, we will calculate the time required to melt the ice using the total heat transfer per second (converted from Watts to kJ/s) and the heat required to melt the ice: \(t = \frac{Q_{melt}}{Q} = \frac{442387.5\text{ kJ}}{101694\frac{\text{W}}{\text{s}} \times \frac{1\text{ kJ}}{10^3\text{ W}}} \approx 4350\text{ s}\) The time required to melt \(3\text{ mm}\) of ice from the surface of the rink if no cooling is supplied is approximately \(4350\text{ s}\) or \(1.2\text{ hours}\).