Question

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) \(=915 \mathrm{~kg} / \mathrm{m}^{3}\).

Short answer

Answer: The evaporation rate of the ice sphere is approximately 6.94 grams per hour.

Step-by-step solution

Calculate the Reynolds number (Re)

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 \(= \frac{13.89 * 0.05}{1.32 \times 10^{-7}} = 5.27 \times 10^5\)

Calculate Schmidt number (Sc)

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 \(= \frac{1.32 \times 10^{-7}}{2.5 \times 10^{-5}} = 0.00528\)

Calculate Sherwood number (Sh)

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 \(=\left[4.0+1.21(5.27 \times 10^5*0.00528)^{2 / 3}\right]^{0.5} = 24.14\)

Calculate mass transfer rate

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 \(= 24.14 * 2.5 \times 10^{-5} * \frac{0.00785 * 560}{8.314 * 272.15} * \frac{18.015\,\mathrm{g}}{1\,\mathrm{mol}} = 0.001928\,\mathrm{g/s}\)

Convert mass transfer rate to grams per hour

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 \(6.94\,\mathrm{g/h}\).

Key concepts

Reynolds Number

The Reynolds Number is a dimensionless number that helps us understand the flow characteristics of a fluid, like air or water, around an object, here a sphere of ice. It is defined as the ratio of inertial forces to viscous forces and gives us insight into whether the flow will be laminar (smooth) or turbulent. For our problem with the ice sphere, we calculate the Reynolds Number using the formula:

• Re = \(\frac{u \cdot d}{u}\)

where:

• \(u\) is the wind speed, converted from km/h to m/s,
• \(d\) is the sphere's diameter, converted to meters,
• \(u\) is the kinematic viscosity of air.

By substituting the values, we get a Reynolds Number of approximately \(5.27 \times 10^5\), indicating that the airflow around the sphere is likely turbulent.

Schmidt Number

In the context of mass transfer between phases, the Schmidt Number is used to describe fluid flow. It compares the ratio of momentum diffusivity (viscous effects) to mass diffusivity (how substances spread). The Schmidt Number tells us about how effectively a fluid can transport mass compared to momentum.For our ice sphere exercise, we define the Schmidt Number as:

• Sc = \(\frac{u}{D}\)

where:

• \(u\) is the kinematic viscosity of air,
• \(D\) is the mass diffusivity of air.

By calculating based on the given values, we find that Sc is approximately 0.00528. This low value indicates that the momentum diffuses through the fluid more effectively compared to mass.

Sherwood Number

The Sherwood Number is another dimensionless number crucial in mass transfer calculations, similar to the Nusselt number in heat transfer. It represents the mass transfer coefficient in relation to the convective mass transfer rate.The formula provided in the problem states:

• Sh = \(\left[4.0 + 1.21(ReSc)^{2/3}\right]^{0.5}\)

Here, we use the previously calculated values for the Reynolds Number (Re) and Schmidt Number (Sc) to find the Sherwood Number. Upon substituting these values, the Sherwood Number comes out to be approximately 24.14. This represents an enhanced mass transfer due to the combined effects of flow and diffusion.

Mass Transfer Rate

Finally, we focus on determining the Mass Transfer Rate, which essentially tells us how fast mass is moving from one place to another—in this case, how fast the ice is evaporating. The mass transfer rate incorporates the Sherwood Number to calculate how quickly the water vaporizes from the ice sphere.The formula to calculate the mass transfer rate is:

• Rate = Sh \(\cdot D \cdot \frac{A \cdot \Delta P_{H_2O}}{RT}\)

where:

• \(A\) is the sphere's surface area,
• \(\Delta P_{H_2O}\) is the vapor pressure difference,
• \(R\) is the gas constant, and
• \(T\) is the absolute temperature.

By further converting vapor pressure to Pascals, calculating the sphere's surface area, and using the ideal gas law, we find the mass transfer rate is approximately 0.001928 g/s. To make it more relatable, converting it gives about 6.94 g/h. This is how fast the ice would evaporate under the given conditions.