Question

A sphere of ice, \(5 \mathrm{~cm}\) in diameter, is exposed to $65 \mathrm{~km} / \mathrm{h}$ wind with 15 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: $\mathrm{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 {ais } \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}$.

Short answer

Answer: The predicted rate of evaporation of the ice sphere is 30.36 g/h.

Step-by-step solution

Calculate the Reynolds number (Re)

First, we need to compute the Reynolds number, defined as the ratio of inertial forces to viscous forces. It is given by the formula: \( \mathrm{Re} = \frac{vd}{\nu} \) where \(v\) is the wind velocity, \(d\) is the diameter of the sphere, and \(\nu\) is the kinematic viscosity of air. Given values: \(v=65 \mathrm{~km/h}=18.06 \mathrm{~m/s}\) \(d=5 \mathrm{~cm} = 0.05 \mathrm{~m}\), \(\nu=1.32 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) Plugging the values, we get: \( \mathrm{Re}=\frac{18.06 \times 0.05}{1.32 \times 10^{-7}}=68541.67 \)

Calculate the Schmidt number (Sc)

Next, we need to compute the Schmidt number, which is a dimensionless number representing the ratio of momentum diffusivity to mass diffusivity. It is given by the formula: \( \mathrm{Sc} = \frac{\nu}{D}\) where \(D\) is the mass diffusivity of the air-water vapor system. Given values: \(D = 2.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) \(\nu = 1.32 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) Plugging the values, we get: \( \mathrm{Sc} = \frac{1.32 \times 10^{-7}}{2.5 \times 10^{-5}} = 0.00528 \)

Evaluate the Sherwood number (Sh) using the correlation

We have the correlation for single spheres: \( \mathrm{Sh} = \left[ 4.0 + 1.21(\mathrm{ReSc})^{2/3}\right]^{0.5}\) We can substitute the Reynolds and Schmidt numbers computed earlier: \( \mathrm{Sh} = \left[4.0+1.21(68541.67 \cdot 0.00528)^{2/3}\right]^{0.5} = 6.85\)

Calculate the mass transfer coefficient (k)

The mass transfer coefficient can be found using the relationship: \( \mathrm{Sh} = \frac{k \cdot d}{D} \) Where \(k\) is the mass transfer coefficient we want to find. Rearranging the equation, we get: \( k = \mathrm{Sh}\frac{D}{d} \) Using the obtained value of \(\mathrm{Sh}\) and the given values for \(d\) and \(D\), we find: \( k = 6.85\frac{2.5 \times 10^{-5}}{0.05} = 3.425 \times 10^{-6} \mathrm{~m/s} \)

Determine the rate of evaporation

To find the rate of evaporation, we first find the driving force for mass transfer, which is the vapor pressure difference: \(\Delta P_{H_2O} = P_{H_2O}^* - \phi P_{H_2O}\) Here, \(P_{H_2O}^*\) is the saturation vapor pressure, \(\phi\) is the relative humidity and \(P_{H_2O}\) is the current vapor pressure. The given values are: \(P_{H_2O}^* = 0.56 \mathrm{kPa}\) \(\phi = 15 \% = 0.15\) Plugging the values, we get: \(\Delta P_{H_2O} = 0.56 - 0.15(0.56) = 0.476 \mathrm{kPa}\) Now we can find the rate of evaporation through the mass transfer coefficient \(k\) and the driving force (mass transfer area is \(A=\pi d^2\)): \(\dot{m} = k A \frac{\Delta P_{H_2O}}{R T}\) Use values previously calculated and given: \(T = -1^\circ\mathrm{C} = 272.15 \mathrm{K}\), R (specific gas constant for \(\mathrm{H_2O}\)) = \(461.5 \mathrm{J/(kg \cdot K)}\), \(\rho = 915 \mathrm{kg/m^3}\): \(\dot{m} = 3.425 \times 10^{-6} \cdot \pi (0.05)^2 \frac{0.476 \times 10^3}{461.5 \cdot 272.15} = 8.43566 \times 10^{-6} \mathrm{kg/s}\) Convert this value to grams per hour: \(\dot{m} = 8.43566 \times 10^{-6} \cdot 1000 \cdot 3600 = 30.36 \mathrm{g/h}\) The predicted rate of evaporation of the ice sphere is \(30.36 \mathrm{g/h}\).