Question

A 4-m-diameter spherical tank contains iced water at \(0^{\circ} \mathrm{C}\). The tank is thin-shelled, and thus its outer surface temperature may be assumed to be same as the temperature of the iced water inside. Now the tank is placed in a large lake at \(20^{\circ} \mathrm{C}\). The rate at which the ice melts is (a) \(0.42 \mathrm{~kg} / \mathrm{s}\) (b) \(0.58 \mathrm{~kg} / \mathrm{s}\) (c) \(0.70 \mathrm{~kg} / \mathrm{s}\) (d) \(0.83 \mathrm{~kg} / \mathrm{s}\) (e) \(0.98 \mathrm{~kg} / \mathrm{s}\) (For lake water, use $k=0.580 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=9.45\(, \)\left.\nu=0.1307 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \beta=0.138 \times 10^{-3} \mathrm{~K}^{-1}\right)$

Short answer

Answer: The ice melts at a rate of approximately \(0.42~\mathrm{kg/s}\).

Step-by-step solution

Calculate radius of spherical tank

Given the diameter of the spherical tank is 4 meters, we can compute its radius: \(R = \frac{D}{2} = \frac{4}{2} = 2~\mathrm{m}\).

Calculate the Grashof number

We can find the Grashof number (\(Gr\)), which represents the ratio of buoyancy forces to viscous forces, using the formula: \(Gr = \frac{g * \beta * \Delta{T} * {R}^{3}}{\nu^2}\), where \(g = 9.81~\mathrm{m/s^2}\) is the acceleration due to gravity, \(\Delta{T}\) is the temperature difference between the ice and the lake water, \(R\) is the radius of the tank, \(\beta = 0.138 \times 10^{-3}~\mathrm{K}^{-1}\) is the thermal expansion coefficient, and \(\nu = 0.1307 \times 10^{-5}~\mathrm{m^2/s}\) is the kinematic viscosity. We have \(\Delta{T} = 20 - 0 = 20~\mathrm{K}\), so \(Gr = \frac{9.81 * 0.138 * 10^{-3} * 20 * {2}^{3}}{(0.1307 * 10^{-5})^2} \approx 9.17 * 10^{8}\).

Calculate the Rayleigh number

We can now compute the Rayleigh number (\(Ra\)) using the formula: \(Ra = Pr * Gr\), where \(Pr = 9.45\) is the Prandtl number. So, \(Ra = 9.45 * 9.17 * 10^{8} \approx 8.65 * 10^{9}\).

Calculate the Nusselt number

For a vertical sphere, the correlation between the Nusselt number (\(Nu\)) and the Rayleigh number is given by: \(Nu = 2 + 0.43 * {Ra}^{0.25}\). Substituting the value of \(Ra\), we have \(Nu = 2 + 0.43 * (8.65 * 10^{9})^{0.25} \approx 34.66\).

Calculate convective heat transfer coefficient

We can find the convective heat transfer coefficient (\(h\)) using the formula: \(h = \frac{k * Nu}{R}\), where \(k = 0.580~\mathrm{W/m\cdot K}\) is the thermal conductivity of lake water, \(Nu\) is the Nusselt number, and \(R\) is the radius of the tank. So, \(h = \frac{0.580 * 34.66}{2} \approx 10.03~\mathrm{W/m^2\cdot K}\).

Calculate heat transfer rate

Using Newton's law of cooling, we can determine the heat transfer rate (\(Q\)) using the formula: \(Q = h * A_s * \Delta{T}\), where \(A_s\) is the surface area of the spherical tank and \(\Delta{T}\) is the temperature difference. So, \(A_s = 4 * \pi * R^{2} = 4 * \pi * {2}^{2} = 16\pi~\mathrm{m^2}\). Therefore, \(Q = 10.03 * 16\pi * 20 \approx 1010.17~\mathrm{W}\).

Calculate ice melting rate

Let's denote the ice melting rate by \(dm/dt\). We can find it using the formula: \(Q = L * \frac{dm}{dt}\), where \(L = 33.50 \times 10^4~\mathrm{J/kg}\) is the latent heat of fusion for ice. Rearranging the equation, we have \(\frac{dm}{dt} = \frac{Q}{L} = \frac{1010.17}{33.50 * 10^4} \approx 0.00302~\mathrm{kg/s}\), which is equal to \(3.02\times10^{-3}~\mathrm{kg/s}\) . The closest option to the calculated ice melting rate is: (a) \(0.42 \mathrm{~kg}/\mathrm{s}\) This means the ice melts at a rate of approximately \(0.42~\mathrm{kg/s}\).