Question

A 3 -m-internal-diameter spherical tank made of \(1-\mathrm{cm}\) 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 -h 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 \(15^{\circ} \mathrm{C}\). Answers: (a) \(23.1 \mathrm{~kW}\), (b) \(5980 \mathrm{~kg}\)

Short answer

Question: Determine the rate of heat transfer to the iced water and the amount of ice that melts during a 24-hour period for a spherical tank made of stainless steel containing iced water at \(0^\circ\, \mathrm{C}\) and located outdoors at \(25^\circ\, \mathrm{C}\). Consider the internal diameter of the tank to be 3 m, the thickness of the steel to be 1 cm, the emissivity of the outer surface to be 0.75, and the convection heat transfer coefficient on the outer surface to be \(30\,\mathrm{W/m^2}\cdot\mathrm{K}\).

Step-by-step solution

(1) Calculate Surface Area of the Spherical Tank

To calculate the outer surface area of the tank, we can use the formula \(A = 4 \pi r^2\), where \(r\) is the radius of the sphere. Since the internal diameter of the tank is 3 m, the internal radius is 1.5 m, and the tank has a 1 cm (0.01 m) steel thickness. So, the outer radius is: \(r_o = 1.5 + 0.01\) m. The surface area of the spherical tank is: \(A = 4 \pi r_o^2\).

(2) Calculate Temperature Difference for Convection and Radiation

We need to calculate the temperature difference for convection, which is the difference between the tank's temperature and the outside air temperature: \(\Delta T_c = 25 - 0 = 25 ^\circ \mathrm{C}\) For radiation, we need to convert the tank's and surrounding surface temperature to Kelvin and calculate the temperature difference: \(T_{tank} = 273\,\mathrm{K}\) \(T_{surrounding} = 15 + 273 = 288\,\mathrm{K}\) \(\Delta T_r = T_{surrounding} - T_{tank} = 288 - 273 = 15\,\mathrm{K}\)

(3)Calculate Heat Transfer Rates for Convection and Radiation

Now we can calculate the heat transfer rates for convection \(Q_c\) and radiation \(Q_r\). For convection: \(Q_c = h_c A \Delta T_c\) where \(h_c\) is the convection heat transfer coefficient, given as \(30\,\mathrm{W/m^2}\cdot\mathrm{K}\). For radiation: \(Q_r = \sigma \varepsilon A (T_{surrounding}^4 - T_{tank}^4)\) where \(\sigma\) is the Stefan-Boltzmann constant \(= 5.67 \times 10^{-8} \,\mathrm{W/m^2}\cdot\mathrm{K^4}\), and \(\varepsilon\) is the emissivity of the outer surface, given as 0.75.

(4) Calculate Total Heat Transfer Rate

To determine the total heat transfer rate, sum up the heat transfer rates from convection and radiation: \(Q_{total} = Q_c + Q_r\)

(5) Calculate Amount of Melted Ice

To find out how much ice melts during a 24-hour (24*3600 s) period, we will use the heat of fusion of water, given by \(h_{if} = 333.7\, \mathrm{kJ/kg}\). We can find the amount of melted ice, \(m\), using the formula: \(m = \frac{Q_{total}\times 24\times 3600}{h_{if}\times 10^3}\) Now, we can follow these steps to find the answers.