Question

A \(0.55\)-m-internal-diameter spherical tank made of 1 -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 outdoors at $30^{\circ} \mathrm{C}\( and is subjected to winds at \)8 \mathrm{~km} / \mathrm{h}$. Assuming the entire steel tank to be at \(0^{\circ} \mathrm{C}\) and thus its thermal resistance 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-\mathrm{h}$ period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\). Disregard any heat transfer by radiation.

Short answer

(a) The rate of heat transfer to the iced water in the tank is 2311.34 Watts. (b) The amount of ice at 0 degrees Celsius that melts during a 24-hour period is 604.13 kg.

Step-by-step solution

Find the properties of air at the film temperature

First, we have to find the properties of air. As the outdoor temperature is 30°C and the tank is at 0°C, the film temperature can be estimated as: $$T_f = \frac{T_\infty + T_s}{2} = \frac{30 + 0}{2} = 15 °C$$ From standard air property tables, we find that at 15°C: - Density \((\rho)\): 1.225 kg/m³ - Thermal conductivity \((k_a)\): 0.0242 W/(m·K) - Dynamic viscosity \((\mu)\): 1.81 x 10⁻⁵ kg/(m·s) - Specific heat at constant pressure \((C_p)\): 1006 J/(kg·K) - Kinematic viscosity \((\nu)\): 1.47 x 10⁻⁵ m²/s

Calculate wind velocity and Reynolds number

We need to convert wind velocity from km/h to m/s and find the Reynolds number to determine the flow regime. $$V = \frac{8 \mathrm{~km} / \mathrm{h}}{3.6} = 2.22 \mathrm{~m} / \mathrm{s}$$ The Reynolds number for flow over a sphere can be calculated as: $$Re = \frac{VD}{\nu} = \frac{2.22 * 0.55}{1.47 * 10^{-5}} = 83014$$ This is greater than 60000, which indicates a turbulent flow regime.

Calculate the Nusselt number using the Seider-Tate equation

To find the convection coefficient, we can now apply the empirical correlation (Seider-Tate equation) for turbulent flow over a sphere: $$Nu = 2 + \left [ 0.4(Re^{1/2})(Pr^{1/3}) \right ] + \left [ 0.06(Re^{2/3})(Pr^{1/3})(\frac{Pr}{Pr_s})^{1/4} \right ]$$ For air at the film temperature 15°C, the values are as follows: - Prandtl number \((Pr) = \frac{C_p\mu}{k_a} = \frac{1006 * 1.81 * 10^{-5}}{0.0242} = 0.712\) - Prandtl number at the surface temperature \((Pr_s) = 0.708\) (since air properties do not change much between 0°C and 15°C) Now substituting these values, we can find the Nusselt number: $$Nu = 2 + [0.4(83014^{1/2})(0.712^{1/3})] + [0.06(83014^{2/3})(0.712^{1/3})(\frac{0.712}{0.708})^{1/4}] = 214.34$$

Calculate the convection coefficient and the heat transfer rate

The convection coefficient \((h_c)\) is given by: $$h_c = \frac{Nu * k_a}{D} = \frac{214.34 * 0.0242}{0.55} = 9.37 \mathrm{~W} / (\mathrm{m}^2 \cdot \mathrm{K}) $$ Applying Newton's Law of Cooling for the convective heat transfer, we can find the rate of heat transfer: $$Q = h_c A\Delta T = h_c * 4\pi r^2 (T_\infty - T_s) = 9.37 * 4 * \pi * (0.275)^2 * (30 - 0) = 2311.34 \mathrm{~W}$$ Thus, the rate of heat transfer to the iced water in the tank is 2311.34 Watts. (a)

Calculate the melted ice during a 24-hour period

The heat of fusion of water \((h_{if})\) is given as 333.7 kJ/kg. Now, the energy required to melt a mass (\(m\)) of ice per day can be calculated by: $$E = m * h_{if} = 2311.34 \mathrm{~W} * 86400 \mathrm{~s}$$ Rearranging and solving for mass, we get: $$m = \frac{2311.34 * 86400}{333.7 * 1000} = 604.13 \mathrm{~kg}$$ Therefore, 604.13 kg of ice at 0 degrees Celsius melts during a 24-hour period. (b)