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: \(k=\) \(0.02588 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7282, v=1.608 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \mu_{\infty}=\) \(\left.1.872 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \mu_{\mathrm{s}}=1.729 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)\)

Step by step solution

01

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), \(v=1.608 \times 10^{-5}\,\text{m}^2/\mathrm{s}\): \(R e=\frac{6.944 \cdot 0.5}{1.608 \times 10^{-5}}=215800\)

02

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: \(Nu=2+0.4 \cdot 215800^{1 / 2} \cdot 0.7282^{1 / 3}=192.17\)

03

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: \(h=\frac{192.17 \cdot 0.02588}{0.5}=9.926\,\text{W/m}^{2}\text{.K}\)

04

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: \(q=9.926 \cdot 0.785 \cdot 30=232.975\,\text{W}\)

05

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 \(\boxed{4.78 \,\text{kg/h}}\).

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