Question

A 3-m-internal-diameter spherical tank made of 1 -cm-thick stainless steel is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located outdoors at \(25^{\circ} \mathrm{C}\). Assuming the entire steel tank to be at \(0^{\circ} \mathrm{C}\) and thus the thermal resistance of the tank to be negligible, 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 -hour period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\). The emissivity of the outer surface of the tank is \(0.75\), and the convection heat transfer coefficient on the outer surface can be taken to be \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume the average surrounding surface temperature for radiation exchange to be \(15^{\circ} \mathrm{C}\).

Short answer

The rate of heat transfer to the iced water in the tank is approximately 6750π + 657.38 watts. (b) How much ice at 0°C melts during a 24-hour period? The amount of ice at 0°C that melts during a 24-hour period is approximately ((6750π + 657.38) × 86400) / (333.7 × 10^3) kg.

Step-by-step solution

Calculate the surface area of the tank

The surface area of a sphere is given by: \(A = 4\pi r^{2}\) Where: \(A\) = surface area \(r\) = radius of the sphere Given the internal diameter of the tank is 3m: \(r = \frac{3}{2} \mathrm{m}\) Plugging the value of \(r\) into the formula, we get: \(A = 4\pi (\frac{3}{2})^2\) \(A = 9\pi \mathrm{m}^2\)

Calculate the convective and radiative heat transfer

The convection heat transfer (\(Q_{conv}\)) is given by: \(Q_{conv} = h \cdot A \cdot (T_{\infty} - T_{s})\) Where: \(h\) = convection heat transfer coefficient (\(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)) \(A\) = surface area of the tank (from step 1) \(T_{\infty}\) = ambient temperature (\(25^{\circ} \mathrm{C}\)) \(T_s\) = surface temperature of the tank (assumed to be \(0^{\circ} \mathrm{C}\)) Plugging the values into the formula, we get: \(Q_{conv} = 30 \cdot 9\pi \cdot (25 - 0)\) \(Q_{conv} = 6750\pi \mathrm{~W}\) The radiative heat transfer (\(Q_{rad}\)) is given by: \(Q_{rad} = \epsilon \cdot \sigma \cdot A(T_{surr}^4 - T_{s}^4)\) Where: \(\epsilon\) = emissivity of the outer surface of the tank (0.75) \(\sigma\) = Stefan-Boltzmann constant (\(5.6703 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^4\)) \(A\) = surface area of the tank (from step 1) \(T_{surr}\) = average surrounding surface temperature for radiation exchange (\(15^{\circ} \mathrm{C}\) + 273.15) \(T_s\) = surface temperature of the tank (assumed to be \(0^{\circ} \mathrm{C}\) + 273.15) Plugging the values into the formula, we get: \(Q_{rad} = 0.75 \cdot 5.6703 \times 10^{-8} \cdot 9\pi \cdot ((15 + 273.15)^4 - (0 + 273.15)^4)\) \(Q_{rad} \approx 657.38 \mathrm{~W}\) Now we sum the convective and radiative heat transfer to find the total rate of heat transfer (\(Q_{total}\)): \(Q_{total} = Q_{conv} + Q_{rad}\) \(Q_{total} \approx 6750\pi + 657.38 \mathrm{~W}\) \((a)\) The rate of heat transfer to the iced water in the tank is \(\approx 6750\pi + 657.38 \mathrm{~W}\).

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

In order to calculate the amount of ice melted during a 24-hour period, we need to find the energy received by the ice during the time period. Energy received (\(E_{total}\)) is equal to the rate of heat transfer multiplied by the time period. Given the time period of 24-hours, we convert it to seconds: \(24 \mathrm{~hours} = 24 \times 60 \times 60 \mathrm{~s} = 86400 \mathrm{~s}\) Therefore: \(E_{total} = Q_{total} \cdot t\) \(E_{total} \approx (6750\pi + 657.38) \cdot 86400 \mathrm{~J}\) Now, we can find the mass of ice that melts by using the heat of fusion of water (\(h_{if} = 333.7 \mathrm{~kJ/kg}\)). We convert heat of fusion to J/kg: \(h_{if} = 333.7 \times 10^3 \mathrm{~J/kg}\) The mass of ice melted (\(m_{melted}\)) is given by: \(m_{melted} = \frac{E_{total}}{h_{if}}\) \(m_{melted} \approx \frac{(6750\pi + 657.38) \cdot 86400}{333.7 \times 10^3} \mathrm{~kg}\) \((b)\) The amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24-hour period is \(\approx \frac{(6750\pi + 657.38) \cdot 86400}{333.7 \times 10^3} \mathrm{~kg}\).

Key concepts

Convection Heat Transfer

Convection heat transfer happens when heat is transferred between a solid surface and a fluid (like air or water) in contact with it. This process is crucial in many industrial and daily applications. In our exercise, as the spherical tank is exposed outdoors, the surrounding air at 25°C warms the tank's surface at 0°C. This temperature difference causes convection to occur.

• The convection heat transfer coefficient, denoted by \(h\), measures the efficiency of heat transfer during convection. In our scenario, it is given as 30 W/m²·K.
• The formula used for convection heat transfer is \(Q_{conv} = h \cdot A \cdot (T_{\infty} - T_s)\). Here, \(Q_{conv}\) is the heat transfer per unit time, \(A\) is the surface area, \(T_{\infty}\) is the surrounding air temperature, and \(T_s\) is the surface temperature of the tank.

By substituting the known values, we calculate the heat transfer due to convection, which is measured in watts (W). This value tells us how much heat is being transferred from the air to the tank per second.

Radiative Heat Transfer

Radiative heat transfer is the emission of electromagnetic waves, usually in the form of infrared radiation, which carries energy away from an object. Unlike conduction and convection, radiation does not require a medium, which means it can occur in a vacuum as well. In the case of the spherical tank, radiative heat transfer occurs between the outer surface and the surrounding environment.

• Emissivity \(\epsilon\) indicates how effective a surface is at emitting thermal radiation. The tank's surface has an emissivity of 0.75, making it a relatively good emitter.
• The radiative heat transfer is calculated using the Stefan-Boltzmann Law: \(Q_{rad} = \epsilon \cdot \sigma \cdot A((T_{surr})^4 - (T_s)^4)\), where \(\sigma\) is the Stefan-Boltzmann constant \((5.6703 \times 10^{-8} W/m^2 \cdot K^4)\), and temperatures must be in Kelvin.

By applying this formula, we find the energy transferred through radiation, which contributes to the total heat gaining entry into the tank.

Heat of Fusion

Heat of fusion refers to the amount of energy required to change a substance from solid to liquid at its melting point, without changing its temperature. This concept is vital in understanding phase changes, such as melting ice.

• For water, the heat of fusion \(h_{if}\) is 333.7 kJ/kg. This means that 333.7 kilojoules of energy are needed to melt one kilogram of ice at 0°C.
• To find how much ice will melt in the tank, we calculate the total energy absorbed by the ice due to heat transfer over 24 hours. This value is obtained by multiplying the total heat transfer rate \(Q_{total}\) by the time in seconds (86400 seconds in 24 hours).
• The mass of ice melted \(m_{melted}\) is determined by dividing the total energy by the heat of fusion.

Understanding heat of fusion helps us calculate the physical changes occurring inside the tank over time, based on the heat entering the system.