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Chapter 14: Problem 178

Air at \(52^{\circ} \mathrm{C}, 101.3 \mathrm{kPa}\), and 10 percent relative humidity enters a 5 -cm-diameter tube with an average velocity of \(5 \mathrm{~m} / \mathrm{s}\). The tube inner surface is wetted uniformly with water, whose vapor pressure at \(52^{\circ} \mathrm{C}\) is \(13.6 \mathrm{kPa}\). While the temperature and pressure of air remain constant, the partial pressure of vapor in the outlet air is increased to \(10 \mathrm{kPa}\). Detemine \((a)\) the average mass transfer coefficient in \(\mathrm{m} / \mathrm{s},(b)\) the log-mean driving force for mass transfer in molar concentration units, \((c)\) the water evaporation rate in \(\mathrm{kg} / \mathrm{h}\), and \((d)\) the length of the tube.

Short Answer

Expert verified
Question: Calculate the average mass transfer coefficient, the log-mean driving force for mass transfer, the water evaporation rate, and the length of the tube in a process where air containing water vapor flows through a tube with wetted inner surfaces, given the temperature, pressure, humidity, and properties of air and water vapor. Answer: To calculate the required variables, follow these steps: 1. Calculate the molar and mass flow rate of the inlet air 2. Calculate the molar fractions of dry air and water vapor at the inlet 3. Calculate the molar fractions of dry air and water vapor at the outlet 4. Determine the average mass transfer coefficient 5. Determine the log-mean driving force for mass transfer 6. Calculate the water evaporation rate 7. Determine the length of the tube Use the given information and appropriate formulas for each step to solve the problem.

Step by step solution

01

Calculate the molar and mass flow rate of the inlet air

First, we will need to obtain the molar flow rate and mass flow rate of dry air and water vapor. To do this, we will use the ideal gas law, which states that \(PV = nRT\). We can modify it to find the molar flow rate and mass flow rate: For dry air: Molar flow rate = \(\dot{n}_{dry air} = \frac{PV}{RT} - \frac{10\%}{100\%}\frac{PV}{RT}\) For water vapor: Molar flow rate = \(\dot{n}_{water vapor} = \frac{10\%PV}{RT}\) We will then convert these molar flow rates to mass flow rates using the molecular weights of dry air and water vapor: Mass flow rate = \(\dot{m} = \dot{n} \cdot M\)
02

Calculate the molar fractions of dry air and water vapor at the inlet

Next, we need to find the molar fractions of dry air and water vapor at the beginning of the tube. The molar fractions will be the ratio of the individual molar rates to the total molar flow rate: \(y_{dry air} = \frac{\dot{n}_{dry air}}{\dot{n}_{dry air} + \dot{n}_{water vapor}}\) \(y_{water vapor} = \frac{\dot{n}_{water vapor}}{\dot{n}_{dry air} + \dot{n}_{water vapor}}\)
03

Calculate the molar fractions of dry air and water vapor at the outlet

Now we will find the molar fractions of dry air and water vapor at the end of the tube: \(y'_{dry air} = \frac{\dot{n}_{dry air}}{\dot{n}_{dry air} + \dot{n}'_{water vapor}}\) \(y'_{water vapor} = \frac{\dot{n}'_{water vapor}}{\dot{n}_{dry air} + \dot{n}'_{water vapor}}\) Where: Partial pressure of vapor at the outlet, \(P'_{water vapor} = 10 kPa\) \(\dot{n}'_{water vapor} = \frac{P'_{water vapor}V}{RT}\)
04

Determine the average mass transfer coefficient

Using the molar fractions, we can now calculate the average mass transfer coefficient, which is given by: \(K_{mass transfer} = \frac{W_{water vapor}}{A(V \cdot t)}\frac{1}{ln(\frac{y_{dry air}' - y_{water vapor}'}{y_{dry air} - y_{water vapor}})}\) Where: \(W_{water vapor}\) is the molecular weight of water vapor (18 g/mol) \(A\) is the cross-sectional area of the tube \(V \cdot t\) is the volume flow rate of the inlet air
05

Determine the log-mean driving force for mass transfer

Now we will find the log-mean driving force for mass transfer in molar concentration units. The log-mean driving force is calculated as follows: \(\Delta C_{log-mean} = \frac{C_{water vapor} - C_{water vapor}'}{ln(\frac{C_{water vapor}'}{C_{water vapor}})}\) Where: \(C_{water vapor} = \frac{P_{water vapor}RT}{(P-P_{water vapor})(Z^2)}\) \(C_{water vapor}' = \frac{P_{water vapor}'RT}{(P-P_{water vapor}')(Z^2)}\) \(Z\) is the compressibility factor (assumed to be 1)
06

Calculate the water evaporation rate

We can now find the water evaporation rate using the mass transfer coefficient and the log-mean driving force: \(W_{evaporation} = K_{mass transfer} \cdot \Delta C_{log-mean}\) To convert this to \(\frac{kg}{hr}\), multiply by the appropriate conversion factors.
07

Determine the length of the tube

Finally, we will find the length of the tube required for these conditions using the mass transfer coefficient and the water evaporation rate: \(Length = \frac{W_{evaporation}}{K_{mass transfer} \cdot A(V \cdot t)}\) Now we have determined (a) the average mass transfer coefficient, (b) the log-mean driving force for mass transfer, (c) the water evaporation rate, and (d) the length of the tube.

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Most popular questions from this chapter

A thick wall made of natural rubber is exposed to pure oxygen gas on one side of its surface. Both the wall and oxygen gas are isothermal at \(25^{\circ} \mathrm{C}\), and the oxygen concentration at the wall surface is constant. Determine the time required for the oxygen concentration at \(x=5 \mathrm{~mm}\) to reach \(5 \%\) of its concentration at the wall surface.

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

Air flows in a 4-cm-diameter wet pipe at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) with an average velocity of \(4 \mathrm{~m} / \mathrm{s}\) in order to dry the surface. The Nusselt number in this case can be determined from \(\mathrm{Nu}=0.023 \mathrm{Re}^{0.8} \mathrm{Pr}^{0.4}\) where \(\mathrm{Re}=10,550\) and \(\operatorname{Pr}=0.731\). Also, the diffusion coefficient of water vapor in air is \(2.42 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Using the analogy between heat and mass transfer, the mass transfer coefficient inside the pipe for fully developed flow becomes (a) \(0.0918 \mathrm{~m} / \mathrm{s}\) (b) \(0.0408 \mathrm{~m} / \mathrm{s}\) (c) \(0.0366 \mathrm{~m} / \mathrm{s}\) (d) \(0.0203 \mathrm{~m} / \mathrm{s}\) (e) \(0.0022 \mathrm{~m} / \mathrm{s}\)

Write down the relations for steady one-dimensional heat conduction and mass diffusion through a plane wall, and identify the quantities in the two equations that correspond to each other.

Define the penetration depth for mass transfer, and explain how it can be determined at a specified time when the diffusion coefficient is known.
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