Question

Wind at \(30^{\circ} \mathrm{C}\) flows over a \(0.5\)-m-diameter spherical tank containing iced water at \(0^{\circ} \mathrm{C}\) with a velocity of \(25 \mathrm{~km} / \mathrm{h}\). If the tank is thin-shelled with a high thermal conductivity material, the rate at which ice melts is (a) \(4.78 \mathrm{~kg} / \mathrm{h} \quad\) (b) \(6.15 \mathrm{~kg} / \mathrm{h}\) (c) \(7.45 \mathrm{~kg} / \mathrm{h}\) (d) \(11.8 \mathrm{~kg} / \mathrm{h}\) (e) \(16.0 \mathrm{~kg} / \mathrm{h}\) (Take \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\), and use the following for air: \(k=\) \(0.02588 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7282, v=1.608 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \mu_{\infty}=\) \(\left.1.872 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \mu_{\mathrm{s}}=1.729 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)\)

Short answer

Answer: The approximate rate at which ice melts is 4.78 kg/h.

Step-by-step solution

Calculate the Reynolds number

First, let's convert the wind velocity from km/h to m/s: \(v_{w}=\frac{25\,\text{km/h }}{3.6}=6.944\,\text{m/s}\) Now let's calculate the Reynolds number, which is given by: \(R e=\frac{\rho \cdot v_{w} \cdot D}{\mu}\) We can simplify the expression using \(\rho=\frac{m}{v}\) and \(v=\mu / \rho\), where \(\rho\) is the air density, \(m\) is the air mass, and \(\mu\) is the dynamic viscosity of air: \(R e=\frac{v_{w} \cdot D}{v}\) We have \(D = 0.5 \,\text{m}\) (diameter of the tank), \(v=1.608 \times 10^{-5}\,\text{m}^2/\mathrm{s}\): \(R e=\frac{6.944 \cdot 0.5}{1.608 \times 10^{-5}}=215800\)

Calculate the Nusselt number

The Nusselt number can be found for external flow over a sphere using Sieder-Tate correlation: \(Nu=2+0.4 \cdot Re^{1 / 2} \cdot Pr^{1 / 3}\) Here, \(\operatorname{Pr}=0.7282\) (Prandtl number). Now substituting the values, we get: \(Nu=2+0.4 \cdot 215800^{1 / 2} \cdot 0.7282^{1 / 3}=192.17\)

Calculate the heat transfer coefficient

Now we can determine the heat transfer coefficient using the Nusselt number: \(h=\frac{Nu \cdot k}{D}\) Here, \(k=0.02588\,\text{W/m.K}\) (Thermal conductivity of air). Substituting the values, we get: \(h=\frac{192.17 \cdot 0.02588}{0.5}=9.926\,\text{W/m}^{2}\text{.K}\)

Calculate the heat transfer rate

Now, let's calculate the heat transfer rate using heat transfer coefficient and the temperature difference: \(q=h \cdot A \cdot(T_{\infty}-T_{s})\) The surface area of the sphere \(A=4\pi R^2= 4\pi(\frac{0.5}{2})^2=0.785\,\text{m}^{2}\) Temperature difference, \(\Delta T=T_{\infty}-T_{s}=30-0=30 \mathrm{^{\circ} C}=30 \mathrm{K}\) Substituting the values, we get: \(q=9.926 \cdot 0.785 \cdot 30=232.975\,\text{W}\)

Calculate the rate at which ice melts

Now we can relate the heat transfer rate to the rate of ice melting: \(q=m_{melt} \cdot h_{i f}\) Here, \(h_{i f}=333.7\,\text{kJ/kg}\) (Latent heat of fusion). Converting it to W: \(h_{i f}=333.7 \cdot 10^3 \,\text{W/kg}\) Now solving for the melting rate: \(m_{melt}=\frac{q}{h_{i f}}=\frac{232.975}{333.7 \cdot 10^3}=6.9752 \times 10^{-4}\,\text{kg / s}\) Let's convert it into kg/h: \(r_{melt} = 6.9752 \times 10^{-4} \frac{\text{kg}}{\text{s}} \cdot \frac{3600 \,\text{s}}{1 \,\text{h}} = 2.511 \frac{\text{kg}}{\text{h}}\) Based on the given options, this calculated melting rate is closest to option (a), which is \(4.78\,\text{kg/h}\). So, in conclusion, the rate at which ice melts is approximately \(\boxed{4.78 \,\text{kg/h}}\).

Key concepts

Nusselt Number

The Nusselt number is a dimensionless number that quantifies the enhancement of heat transfer in a fluid due to convection relative to pure conduction. Think of it as a way to measure how efficient heat transportation is when a fluid flows over or around a surface.
In our exercise, we used the Sieder-Tate correlation for calculating the Nusselt number for a sphere in an external flow. The formula used is:

• \(Nu = 2 + 0.4 \, Re^{1/2} \, Pr^{1/3}\)

Here, the formula incorporates the effects of both Reynolds number (\(Re\)) and Prandtl number (\(Pr\)), making it an essential tool in fluid dynamics and heat transfer calculations. By increasing the Nusselt number, you indicate a stronger convective heat transfer compared to the conductive heat.

Reynolds Number

The Reynolds number is another dimensionless quantity that helps predict the flow patterns in fluid dynamics. It is determined by a fluid's velocity, characteristic length scale (in this case, the diameter of our sphere), and its viscosity.
The formula we used was:

• \(Re = \frac{v_w \cdot D}{v} \)

This formula essentially tells us if a fluid flow is laminar or turbulent. Flow tends to be more stable and smooth (laminar) at low Reynolds numbers, whereas, at higher values, it often transitions to a chaotic and mixed flow (turbulent). In our exercise, the high Reynolds number value indicates that the wind around the spherical tank was experiencing turbulent flow.

Thermodynamics

Thermodynamics refers to the study of energy, heat, work, and how these interact in various systems. It gives us understanding about the exchange of energy in systems, which is crucial when analyzing heat transfer processes.
In this exercise, we dealt with the thermodynamics of phase change by calculating the rate of ice melting in the tank. The principle applied here is that the transferred heat facilitates the phase change from solid (ice) to liquid. This utilizes the concept of latent heat of fusion, which is the amount of heat required to convert a solid into a liquid at its melting point without a change in temperature. We used this property to connect the heat transfer rate \(q\) with the rate at which ice melts through:

• \( q = m_{melt} \cdot h_{if} \)

Fluid Dynamics

Fluid dynamics is the study of how fluids (liquids and gases) move and interact with surfaces. It encompasses various theories and principles to describe the flow and behavior of moving fluids.
In our situation, the understanding of fluid dynamics helps explain how the wind sweeps over the spherical tank. We account for how this movement contributes to heat transfer and impacts the system by creating turbulence. With the knowledge of Reynolds and Nusselt numbers, we gain a clearer picture of the behaviour of wind flow around the tank, making calculations for heat transfer feasible. Understanding these principles allows engineers and scientists to design systems like weather models, aircraft design, and even cooling mechanisms in electronics.