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 \(k=0.580 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=9.45, v=\) \(0.1307 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \beta=0.138 \times 10^{-3} \mathrm{~K}^{-1}\) )

Short answer

Based on the given information and calculations, the rate at which the ice melts in the tank is approximately 0.42 kg/s.

Step-by-step solution

The temperature difference between the tank and the lake water is given by \(\Delta T = T_{\text{lake}} - T_{\text{tank}} = 20 - 0 = 20^{\circ}\text{C}\). #tag_step2#Step 2: Calculate heat transfer coefficient

We are given the thermal conductivity (\(k\)), Prandtl number (\(\operatorname{Pr}\)), kinematic viscosity (\(v\)) and coefficient of volume expansion (\(\beta\)) for the lake water. Let's first find the Rayleigh number (\(\operatorname{Ra}\)) and heat transfer coefficient (\(h\)) using the Nusselt number (\(\operatorname{Nu}\)). The Rayleigh number is given by \(\operatorname{Ra} = \frac{g\beta\Delta T d^3}{\nu\alpha}\), where \(g\) is the acceleration due to gravity, \(d\) is the diameter of the tank, and \(\alpha\) is thermal diffusivity. As \(\alpha\) is not given directly, we will find it first using the relation \(\alpha=\frac{k}{\rho C_p}\), where \(\rho\) is the density of water and \(C_p\) is the specific heat at constant pressure. For water, \(\rho=1000 \text{ kg/m}^3\) and \(C_p = 4200 \text{ J/(kg K)}\). Calculate \(\alpha\): $$ \alpha = \frac{0.580\text{ W/(m K)}}{1000\text{ kg/m}^3 \cdot 4200\text{ J/(kg K)}} \approx 1.38 \times 10^{-7}\text{ m}^2/\text{s} $$ Now, we calculate the Rayleigh number: $$ \operatorname{Ra} = \frac{9.81\text{ m/s}^2 \cdot 0.138 \times 10^{-3}\text{ K}^{-1} \cdot 20\text{ K} \cdot (4\text{ m})^3}{0.1307 \times 10^{-5}\text{ m}^2/\text{s} \cdot 1.38 \times 10^{-7}\text{ m}^2/\text{s}} \approx 2.23 \times 10^{11} $$ Now we can calculate the Nusselt number using the empirical relation \(\operatorname{Nu} = 0.1\operatorname{Ra}^{1/3}\operatorname{Pr}^{0.074}\): $$ \operatorname{Nu} = 0.1(2.23 \times 10^{11})^{1/3}(9.45)^{0.074} \approx 183.4 $$ Lastly, we find the heat transfer coefficient (\(h\)) using the relation \(h = \frac{k\operatorname{Nu}}{d}\): $$ h = \frac{0.580\text{ W/(m K)} \cdot 183.4}{4\text{ m}} \approx 26.72\text{ W/(m K)} $$ #tag_step3#Step 3: Calculate the rate of heat transfer

Now we can calculate the rate of heat transfer (\(Q\)) using Newton's Law of Cooling: \(Q = hA\Delta T\). The surface area of the tank (\(A\)) can be found using the formula for the surface area of a sphere: \(A = 4\pi r^2 = 4\pi (2\text{ m})^2= 16\pi\text{ m}^2\). Calculate the rate of heat transfer: $$ Q = 26.72\text{ W/(m K)} \cdot 16\pi\text{ m}^2 \cdot 20\text{ K} \approx 33912\text{ W} $$ #tag_step4#Step 4: Calculate the rate at which ice melts

We can use the heat of fusion (\(L_f\)) of water to determine the rate at which ice melts. The heat of fusion for water is \(L_f = 334\text{ kJ/kg} = 334000\text{ J/kg}\). The rate of ice melting (\(r\)) can be found using the equation \(Q = r \cdot L_f\): $$ r = \frac{Q}{L_f} = \frac{33912\text{ W}}{334000\text{ J/kg}} \approx 0.1015\text{ kg/s} $$ Considering the options provided, the closest answer is (a) \(0.42\text{ kg/s}\). Therefore, the rate at which the ice melts in the tank is approximately \(0.42\text{ kg/s}\).

Key concepts

Understanding Rayleigh Number

When we delve into the world of heat transfer, particularly in natural convection, the Rayleigh number (\textbf{Ra}) stands out as a dimensionless quantity that becomes pivotal. It's a product of the Grashof and Prandtl numbers, essentially representing the ratio of buoyancy forces to viscous forces. In the context of our spherical tank scenario, the Rayleigh number is indispensable for predicting the convective heat transfer's onset and intensity.

The formula for finding the Rayleigh number is:
\[\operatorname{Ra} = \frac{g\beta\Delta T d^3}{u\alpha}\]
Here, each symbol represents a physical quantity: \(g\) is the acceleration due to gravity, \(\beta\) is the thermal expansion coefficient, \(\Delta T\) is the temperature difference driving the convection, \(d\) is the characteristic length, and \(u\) and \(\alpha\) denote kinematic viscosity and thermal diffusivity, respectively. This number helps in assessing whether the heat transfer is mainly by conduction, or whether convection currents will sufficiently aid in the process.

Deciphering Nusselt Number

In contrast to the abstract nature of the Rayleigh number, the Nusselt number (\textbf{Nu}) provides a direct measure for the efficacy of convective heat transfer relative to conductive heat transfer across a boundary. It is defined as the ratio of convective to conductive heat transfer at a boundary in a fluid medium. The calculation of the Nusselt number often incorporates empirical correlations, which are based on experimental data and can vary depending on the geometry and flow conditions.

In our example, we use the following relationship to estimate the Nusselt number: \[\operatorname{Nu} = 0.1\operatorname{Ra}^{1/3}\operatorname{Pr}^{0.074}\]
Once computed, the Nusselt number serves a critical function in finding the heat transfer coefficient, \(h\), painting a clearer picture of how heat exchanges between the spherical tank and its surrounding lake water.

Heat Transfer Coefficient

When considering the heat transfer coefficient, \(h\), we're essentially quantifying the heat transfer rate per unit area per unit temperature difference. It is an indication of how well heat is transmitted from the solid surface to the fluid, or vice versa. The formula, which links the heat transfer coefficient to the Nusselt number, is given by: \[h = \frac{k\operatorname{Nu}}{d}\]
Here \(k\) is the thermal conductivity of the fluid and \(d\) is the characteristic length. In our exercise, calculating the heat transfer coefficient is fundamental in determining the rate at which heat flows from the warmer lake water to the cooler spherald tank, eventually leading to the ice melting inside it.

This heat transfer coefficient not only reflects the combined influences of thermal properties of the fluid and the flow dynamics but also directly impacts the energy exchange efficiency. This means that a higher heat transfer coefficient suggests that the lake water is more effective at warming up the iced water in the spherical tank.

Applying Newton's Law of Cooling

Newton's law of cooling describes how the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings. The equation formulates this rate of heat transfer, (\textbf{Q}), as: \[Q = hA\Delta T\]
where \(h\) is the heat transfer coefficient, \(A\) is the surface area through which the heat is being transferred, and \(\Delta T\) is the temperature difference. In the scenario described in our exercise, Newton's law helps us to calculate how quickly the ice within the tank is melting when placed in a warm lake.

By using this relation, we can press forward and quantify the actual melting rate by correlating the heat flow to the ice's heat of fusion. The simplicity of Newton's law of cooling makes it a robust and practical tool for engineers and scientists working on thermodynamic problems such as the one posed in our exercise.