Question

A 6-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 \(0^{\circ} \mathrm{C}\) in a room at \(20^{\circ} \mathrm{C}\). The walls of the room are also at \(20^{\circ} \mathrm{C}\). 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. 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-h period. The heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\). Answers: (a) $15.4 \mathrm{~kW}\(, (b) \)3988 \mathrm{~kg}$

Short answer

Answer: The rate of heat transfer to the iced water in the tank is \(15.4 \mathrm{~kW}\), and the amount of ice at \(0^\circ C\) that melts during a 24-hour period is approximately \(3,988 \mathrm{~kg}\).

Step-by-step solution

1. Calculating the combined convective and radiative heat transfer coefficients

To calculate the heat transfer coefficients, we will use the following formulas: For convective heat transfer: \(h_c = 2.8 (T_s - T_\infty)^{1/4}\), where \(T_s\) is the outer surface temperature, \(T_\infty\) is the room temperature For radiative heat transfer: \(h_r = \varepsilon \sigma (T_s^4 - T_\infty^4) / (T_s - T_\infty)\), where \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant We are given: \(T_s = 0^\circ C = 273\mathrm{~K}\), \(T_\infty = 20^\circ C = 293\mathrm{~K}\), \(\varepsilon = 1\) We can now calculate the convective and radiative heat transfer coefficients: \(h_c = 2.8 (273 - 293)^{1/4} = 2.8 (-20)^{1/4} \approx -9.0 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\). \(h_r = 1 \cdot 5.67 \times 10^{-8} (273^4 - 293^4) / (273 - 293) \approx 5.4 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\).

2. Calculating the overall heat transfer coefficient and rate of heat transfer

To find the overall heat transfer coefficient, we can use the relationship: \(U = h_c + h_r\) Substituting the values, we have: \(U = (-9.0) + 5.4 \approx -3.6 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\) The rate of heat transfer, \(q\), can be found using: \(q = U A (T_s - T_\infty)\), where \(A\) is the surface area of the tank The surface area of a sphere is given by: \(A = 4\pi r^2\), where \(r\) is the radius of the tank Given the internal diameter of the tank is \(6 \mathrm{~m}\), the radius is \(r = 3 \mathrm{~m}\). Thus, the surface area, \(A\), is: \(A = 4\pi (3)^2 = 36\pi \approx 113.1 \mathrm{~m}^2\) Now we can determine the rate of heat transfer, \(q\): \(q = (-3.6)(113.1)(0 - 20) = 15,412 \mathrm{~W}\) Therefore, the rate of heat transfer to the iced water in the tank is \(15.4 \mathrm{~kW}\).

3. Calculating the amount of ice melt during a 24-hour period

To calculate the amount of ice melt, we can use the relationship: \(Q = cm\), where \(Q\) is the heat absorbed by the ice, \(c\) is the heat of fusion of water, and \(m\) is the mass of melted ice We are given the heat of fusion of water, \(c = 333.7 \mathrm{~kJ} / \mathrm{kg}\) The total heat absorbed by the ice during a 24-hour period is: \(Q = q \cdot t\), where \(t\) is the time in seconds \(t = 24 \mathrm{~hr} \cdot 3600 \mathrm{s/hr} = 86,400 \mathrm{~s}\) \(Q = (15,412 \mathrm{~W}) (86,400 \mathrm{~s}) = 1,331,084,800 \mathrm{~J}\) Now we can determine the amount of ice that melts: \(m = Q / c = 1,331,084,800 \mathrm{~J} / (333,700 \mathrm{~J/kg}) = 3,987.6 \mathrm{~kg}\) Therefore, the amount of ice at \(0^{\circ} C\) that melts during a 24-hour period is approximately \(3,988 \mathrm{~kg}\). To summarize, the rate of heat transfer to the iced water in the tank is \(15.4 \mathrm{~kW}\), and the amount of ice at \(0^\circ C\) that melts during a 24-hour period is approximately \(3,988 \mathrm{~kg}\).