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Chapter 14: Problem 176

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

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

Step by step solution

01

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 \(= \frac{13.89 * 0.05}{1.32 \times 10^{-7}} = 5.27 \times 10^5\)
02

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 \(= \frac{1.32 \times 10^{-7}}{2.5 \times 10^{-5}} = 0.00528\)
03

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 \(=\left[4.0+1.21(5.27 \times 10^5*0.00528)^{2 / 3}\right]^{0.5} = 24.14\)
04

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 \(= 24.14 * 2.5 \times 10^{-5} * \frac{0.00785 * 560}{8.314 * 272.15} * \frac{18.015\,\mathrm{g}}{1\,\mathrm{mol}} = 0.001928\,\mathrm{g/s}\)
05

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攊n 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.

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Most popular questions from this chapter

Determine the mole fraction of the water vapor at the surface of a lake whose temperature is \(15^{\circ} \mathrm{C}\) and compare it to the mole fraction of water in the lake. Take the atmospheric pressure at lake level to be \(92 \mathrm{kPa}\).

A tank with a 2-cm-thick shell contains hydrogen gas at the atmospheric conditions of \(25^{\circ} \mathrm{C}\) and \(90 \mathrm{kPa}\). The charging valve of the tank has an internal diameter of \(3 \mathrm{~cm}\) and extends \(8 \mathrm{~cm}\) above the tank. If the lid of the tank is left open so that hydrogen and air can undergo equimolar counterdiffusion through the 10 -cm- long passageway, determine the mass flow rate of hydrogen lost to the atmosphere through the valve at the initial stages of the process.

Air flows through a wet pipe at \(298 \mathrm{~K}\) and 1 atm, and the diffusion coefficient of water vapor in air is \(2.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). If the heat transfer coefficient is determined to be \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the mass transfer coefficient is (a) \(0.0326 \mathrm{~m} / \mathrm{s}\) (b) \(0.0387 \mathrm{~m} / \mathrm{s}\) (c) \(0.0517 \mathrm{~m} / \mathrm{s}\) (d) \(0.0583 \mathrm{~m} / \mathrm{s}\) (e) \(0.0707 \mathrm{~m} / \mathrm{s}\)

What is the relation \((f / 2) \mathrm{Re}=\mathrm{Nu}=\mathrm{Sh}\) known as? Under what conditions is it valid? What is the practical importance of it? \(\mathrm{St}_{\text {mass }} \mathrm{Sc}^{2 / 3}\) and what are the names of the variables in it? Under what conditions is it valid? What is the importance of it in engineering?

Consider a 15-cm-internal-diameter, 10-m-long circular duct whose interior surface is wet. The duct is to be dried by forcing dry air at \(1 \mathrm{~atm}\) and \(15^{\circ} \mathrm{C}\) through it at an average velocity of \(3 \mathrm{~m} / \mathrm{s}\). The duct passes through a chilled room, and it remains at an average temperature of \(15^{\circ} \mathrm{C}\) at all times. Determine the mass transfer coefficient in the duct.
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