A sphere of ice, \(5 \mathrm{~cm}\) in diameter, is exposed to \(50 \mathrm{~km} / \mathrm{h}\) wind with 10 percent relative humidity. Both the ice sphere and air are at \(-1^{\circ} \mathrm{C}\) and \(90 \mathrm{kPa}\). Predict the rate of evaporation of the ice in \(\mathrm{g} / \mathrm{h}\) by use of the following correlation for single spheres: Sh \(=\left[4.0+1.21(\mathrm{ReSc})^{2 / 3}\right]^{0.5}\). Data at \(-1^{\circ} \mathrm{C}\) and \(90 \mathrm{kPa}: D_{\text {air- } \mathrm{H}, \mathrm{O}}=2.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}^{3}\), kinematic viscosity (air) \(=1.32 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\), vapor pressure \(\left(\mathrm{H}_{2} \mathrm{O}\right)=\) \(0.56 \mathrm{kPa}\) and density (ice) \(=915 \mathrm{~kg} / \mathrm{m}^{3}\).

First, we need to calculate the Reynolds number (Re), which is the ratio of inertial forces to viscous forces and can be defined as: Re \(= \frac{u*d}{\nu}\) where: u = wind speed (\(50 \mathrm{~km/h}\)) d = diameter of the sphere (\(5 \mathrm{~cm}\)) \(\nu\) = kinematic viscosity of air (\(1.32 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\)) We need to convert the wind speed and diameter to meters/second and meters, respectively: u = \(50 \frac{\mathrm{km}}{\mathrm{h}} * \frac{1000\mathrm{m}}{1\mathrm{km}} * \frac{1\mathrm{h}}{3600\mathrm{s}} = 13.89 \frac{\mathrm{m}}{\mathrm{s}}\) d = \(5 \frac{\mathrm{cm}}{1} * \frac{1\mathrm{m}}{100\mathrm{cm}} = 0.05\mathrm{m}\) Now, substitute these values into the Re equation: Re \(= \frac{13.89 * 0.05}{1.32 \times 10^{-7}} = 5.27 \times 10^5\)

Next, we need to calculate the Schmidt number (Sc), which is the ratio of momentum diffusivity to mass diffusivity, and can be defined as: Sc \(= \frac{\nu}{D}\) where: \(D\) = mass diffusivity of air (\(2.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\)) Now, substitute the given value of \(\nu\) and \(D\) into the Sc equation: Sc \(= \frac{1.32 \times 10^{-7}}{2.5 \times 10^{-5}} = 0.00528\)

Now, we will use the given correlation formula to calculate the Sherwood number (Sh): Sh \(=\left[4.0+1.21(\mathrm{ReSc})^{2 /3}\right]^{0.5}\) Insert the calculated Re and Sc values: Sh \(=\left[4.0+1.21(5.27 \times 10^5*0.00528)^{2 / 3}\right]^{0.5} = 24.14\)

The mass transfer rate can be calculated using the Sherwood number: Rate \(= Sh * D * \frac{A \Delta P_{H_2O}}{RT}\) where: A = surface area of the sphere \(= 4 \pi r^2\) r = radius of the sphere (\(d/2\)) \(\Delta P_{H_2O}\) = vapor pressure of water \(= 0.56 \mathrm{kPa}\) R = gas constant \(= 8.314 \mathrm{J/(mol\cdot K)}\) T = temperature \(= -1^{\circ} \mathrm{C} = 272.15 \mathrm{K}\) Convert the vapor pressure to Pascals: \(\Delta P_{H_2O} = 0.56 \mathrm{kPa} * \frac{1000\mathrm{Pa}}{1\mathrm{kPa}} = 560\,\mathrm{Pa}\) Calculate the surface area: \(r = 0.05/2 = 0.025\,\mathrm{m}\) \(A = 4 \pi (0.025)^2 = 0.00785\,\mathrm{m}^2\) Now, insert all the values and solve for the mass transfer rate, noting that we use the molecular weight of water to convert from moles to grams: Rate \(= 24.14 * 2.5 \times 10^{-5} * \frac{0.00785 * 560}{8.314 * 272.15} * \frac{18.015\,\mathrm{g}}{1\,\mathrm{mol}} = 0.001928\,\mathrm{g/s}\)

Finally, convert the mass transfer rate from grams per second to grams per hour: Rate \(= 0.001928\,\mathrm{g/s} * \frac{3600\,\mathrm{s}}{1\,\mathrm{h}} = 6.94\,\mathrm{g/h}\) Thus, the evaporation rate of the ice sphere is approximately \(6.94\,\mathrm{g/h}\).