Mass Transfer Solutions

1.1a. Concentration of a gas mixture.A mixture of noble gases [helium (1), argon (2), krypton (3), and xenon (4)] is at a total pressure of200 kPa and a temperature of 400 K. If the mixture has equal mole fractions of each of the gases,determine:a) The composition of the mixture in terms of mass fractions.

Solution: Basis: 100 kmole of the mixture

b) The average molecular weight of the mixture.

Solution

c) The total molar concentration.

Solution

d) The mass density.

Solution

1.2a. Concentration of a liquid solution fed to a distillation column.A solution of carbon tetrachloride (1) and carbon disulfide (2) containing 50% by weight each is tobe continuously distilled at the rate of 4,000 kg/h Determine:a) The concentration of the mixture in terms of mole fractions.

Solution

Basis: 100 kg mixture

b) The average molecular weight of the mixture.

Solution

c) Calculate the feed rate in kmol/h.

Solution

1.3a. Concentration of liquified natural gas.A sample of liquified natural gas, LNG, from Alaska has the following molar composition: 93.5%CH4, 4.6% C2H6, 1.2% C3H8, and 0.7% CO2. Calculate:

a) Average molecular weight of the LNG mixture.

Solution

b) Weight fraction of CH4 in the mixture.

Solution

Basis: 100 kmoles of LNG

c) The LNG is heated to 300 K and 140 kPa, and vaporizes completely. Estimate the density of thegas mixture under these conditions.

Solution

1.4b. Concentration of a flue gas.A flue gas consists of carbon dioxide, oxygen, water vapor, and nitrogen. The molar fractions of CO2

and O2 in a sample of the gas are 12% and 6%, respectively. The weight fraction of H2O in the gas

is 6.17%. Estimate the density of this gas at 500 K and 110 kPa.

SolutionBasis: 100 kmole of gas mixture

Let x = molar fraction of water in the mixture (as a percent)

Initial estimate

From the given water weight fraction (0.0617):

1.5b. Material balances around an ammonia gas absorber.

A gas stream flows at the rate of 10.0 m3/s at 300 K and 102 kPa. It consists of an equimolarmixture of ammonia and air. The gas enters through the bottom of a packed bed gas absorberwhere it flows countercourrent to a stream of pure liquid water that absorbs 90% of all of theammonia, and virtually no air. The absorber vessel is cylindrical with an internal diameter of 2.5 m.a) Neglecting the evaporation of water, calculate the ammonia mol fraction in the gas leaving theabsorber.

Solution:Basis: 1 second

A = ammonia B = air

b) Calculate the outlet gas mass velocity (defined as mass flow rate per unit empty tubecross-sectional area).

Solution:

1.6b. Velocities and fluxes in a gas mixture.A gas mixture at a total pressure of 150 kPa and 295 K contains 20% H2, 40% O2, and 40% H2O

by volume. The absolute velocities of each species are -10 m/s, -2 m/s, and 12 m/s, respectively, all

in the direction of the z-axis.a) Determine the mass average velocity, v, and the molar average velocity, V, for the mixture.

Solution

Molar average velocity, V

Mass average velocity, vm

Basis: 1 kmole of gasmixture

b) Evaluate the four fluxes: jO2, nO2, JO2, NO2.

Solution:

1.7b. Properties of air saturated with water vapor.

Air, stored in a 30-m3 container at 340 K and 101.3 kPa is saturatedwith water vapor. Determine thefollowing properties of the gas mixture:a) Mole fraction of water vapor.b) Average molecular weight of the mixture.c) Total mass contained in the tank.d) Mass of water vapor in the tank.

Solution

a) Antoine equation for water vapor:

For a saturated mixture

b)

c)

d)

1.8c. Water balance around an industrial cooling tower.The cooling water flow rate to the condensers of a big coal-fired power plant is 8,970 kg/s. Thewater enters the condensers at 29 qC and leaves at 45 qC. From the condensers, the water flows toa cooling tower where it is cooled down back to 29 qC by countercourrent contact with air (seeFigure 1.11). The air enters the cooling tower at the rate of 6,500 kg/s of dry air, at a dry-bulbtemperature of 30 qC and a humidity of 0.016 kg of water/kg of dry air. It leaves the cooling towersaturated with water vapor at 38 qC.a) Calculate the water losses by evaporation in the cooling tower.

Solution

Consider the air leaving the tower saturated at 38 qC.

Antoine equation for water vapor:

E = water lost by evaporation in the air

b) To account for water losses in the cooling tower, part of the effluent from a nearby municipalwastewater treatment plant will be used as makeup water. This makeup water contains 500 mg/Lof dissolved solids. To avoid fouling of the condenser heat-transfer surfaces, the circulating wateris to contain no more than 2,000 mg/L of dissolved solids. Therefore, a small amount of thecirculating water must be deliberately discarded (blowdown). Windage losses from the tower areestimated at 0.2% of the recirculation rate. Estimate the makeup-water requirement.

SolutionW = windage losses xm = 500 ppm xc = 2000 ppm

M = makeup water rate B = blowdown rate

Initial estimates

Water balance

Solids balance:

1.9b. Water balance around a soap dryer.It is desired to dry 10 kg/min of soap continuously from 17% moisture by weight to 4% moisture in

a countercourrent stream of hot air. The air enters the dryer at the rate of 30.0 m3/min at 350 K,101.3 kPa, and initial water-vapor partial pressure of 1.6 kPa. The dryer operates at constanttemperature and pressure.a) Calculate the moisture content of the entering air, in kg of water/kg of dry air.

Solution

b) Calculate the flow rate of dry air, in kg/min.

Solution

c) Calculate the water-vapor partial pressure and relative humidity in the air leaving the dryer.

Solution

Calculate water vapor pressure at 350 K

Antoine equation for water vapor:

1.10b. Activated carbon adsorption; material balances.

A waste gas contains 0.3% toluene in air, and occupies a volume of 2,500 m3 at 298 K and 101.3kPa. In an effort to reduce the toluene content of this gas, it is exposed to 100 kg of activatedcarbon, initially free of toluene. The system is allowed to reach equilibrium at constant temperatureand pressure. Assuming that the air does not adsorb on the carbon, calculate the equilibriumconcentration of toluene in the gaseous phase, and the amount of toluene adsorbed by the carbon.The adsorption equilibrium for this system is given by the Freundlich isotherm(EAB Control CostManual, 3rd. ed., U. S. E. P. A., Research Triangle Park, NC, 1987.):

where W is the carbon equilibrium adsorptivity, in kg of toluene/kg of carbon, and p* is theequilibrium toluene partial pressure, in Pa, and must be between 0.7 and 345 Pa.

Solution

M = mass of carbonx = moles of toluene adsorbed

Initial estimates

1.11b. Activated carbon adsorption; material balances.It is desired to adsorb 99.5% of the toluene originally present in the waste gas of Problem 1.10.Estimate how much activated carbon should be used if the system is allowed to reach equilibriumat constant temperature and pressure.

Solution

1.12a, d. Estimation of gas diffusivity by the Wilke-Lee equation.E. M. Larson (MS thesis, Oregon State University, 1964) measured the diffusivity of chloroform

in air at 298 K and 1 atm and reported its value as 0.093 cm2/s. Estimate the diffusioncoefficient by the Wilke-Lee equation and compare it with the experimental value.

Solution

FromAppendixB

1.13a, d. Estimation of gas diffusivity by the Wilke-Lee equation.

a) Estimate the diffusivity of naphthalene (C10H8) in air at 303 K and 1 bar. Compare it with the

experimental value of 0.087 cm2/s reported in Appendix A. The normal boiling point of

naphthalene is 491.1 K, and its critical volume is 413 cm3/mol.

Solution

Experimental value

b) Estimate the diffusivity of pyridine (C5H5N) in hydrogen at 318 K and 1 atm. Compare it with the

experimental value of 0.437 cm2/s reported in Appendix A. The normal boiling point of pyridine is

388.4 K, and its critical volume is 254 cm3/mol.

Solution

Experimental value

c) Estimate the diffusivity of aniline (C6H7N) in air at 273 K and 1 atm. Compare it with the

experimental value of 0.061 cm2/s (Guilliland, E. R., Ind. Eng. Chem., 26:681, 1934). The

normal boiling point of aniline is 457.6 K, and its critical volume is 274 cm3/mol.

Solution

Experimental value

1.14d. Diffusivity of polar gases If one or both components of a binary gas mixture are polar, a modified Lennard-Jones relation isoften used. Brokaw (Ind. Eng. Chem. Process Design Develop., 8:240, 1969) has suggested analternative method for this case. Equation (1-49) is still used, but the collision integral is now given by

mp = dipole moment, debyes [1 debye = 3.162 10-25 (J-

m3)1/2]

a) Modify the Mathcad¨ routine of Figure 1.3 to implement Brokaw's method. Use the functionnameDABp(T, P, MA, MB, mA, mB, VA, VB, TbA, TbB)

Solution

b) Estimate the diffusion coefficient for a mixture of methyl chloride and sulfur dioxide at 1 bar and

323 K, and compare it to the experimental value of 0.078 cm2/s. The data required to use Brokaw'srelation are shown below (Reid, et al., 1987):

Parameter Methyl chloride Sulfur dioxideTb , K 249.1 263.2

Vb , cm3/mol 50.6 43.8

Pp, debyes 1.9 1.6

M 50.5 64.06

Solution

1.15d. Diffusivity of polar gases Evaluate the diffusion coefficient of hydrogen chloride in water at 373 K and 1 bar. The data requiredto use Brokaw's relation (see Problem 1.14) are shown below (Reid, et al., 1987):

Parameter Hydrogen chloride WaterTb , K 188.1 373.2

Vb , cm3/mol 30.6 18.9Pp, debyes 1.1 1.8

M 36.5 18

Solution

1.16d. Diffusivity of polar gases Evaluate the diffusion coefficient of hydrogen sulfide in sulfur dioxide at 298 K and 1.5 bar. The datarequired to use Brokaw's relation (seeProblem 1.14) are shown below (Reid, et al., 1987):

Parameter Hydrogen sulfide Sulfur dioxideTb , K 189.6 263.2Vb , cm3/mol 35.03 43.8Pp, debyes 0.9 1.6M 34.08 64.06

Solution

1.17a,d. Effective diffusivity in a multicomponent stagnant gas mixture.Calculate the effective diffusivity of nitrogen through a stagnant gas mixture at 373 K and 1.5 bar.The mixture composition is:

O2 15 mole %

CO 30%CO2 35%

N2 20%

Sol;ution

Calculate mole fractions on a nitrogen (1)-free basis:oxygen (2); carbon monoxide (3); carbon dioxide (4)

Calculate binary MS diffusivities from Wilke-Lee equation

1.18a,d. Mercury removal from flue gases by sorbent injection.Mercury is considered for possible regulation in the electric power industry under Title III of the 1990Clean Air Act Amendments. One promising approach for removing mercury from fossil-fired flue gasinvolves the direct injection of activated carbon into the gas. Meserole, et al. (J. Air & WasteManage. Assoc., 49:694-704, 1999) describe a theoretical model for estimating mercury removal bythe sorbent injection process. An important parameter of the model is the effective diffusivity ofmercuric chloride vapor traces in the flue gas. If the flue gas is at 1.013 bar and 408 K, and itscomposition (on a mercuric chloride-free basis) is 6% O2, 12% CO2, 7% H2O, and 75% N2,

estimate the effective diffusivity of mercuric chloride in the flue gas. Assume that only the HgCl2 is

adsorbed by the activated carbon. Meserole et al. reported an effective diffusivity value of 0.22

cm2/s.

Solution

HgCl2 (1) O2 (2) CO2 (3) H2O (4) N2 (5)

1.19a. Wilke-Chang method for liquid diffusivity.Estimate the liquid diffusivity of carbon tetrachloride in dilute solution into ethanol at 298 K.

Compare to the experimental value reported by Reid, et al. (1987) as 1.5 u 10-5 cm2/s. The critical

volume of carbon tetrachloride is 275.9 cm3/mol. The viscosity of liquid ethanol at 298 K is 1.08 cP.

Solution

1.20b. Diffusion in electrolyte solutions.When a salt dissociates in solution, ions rather than molecules diffuse. In the absence of anelectric potential, the diffusion of a single salt may be treated as molecular diffusion. For dilutesolutions of a single salt, the diffusion coefficient is given by the Nernst-Haskell equation (Harned,H. S., and B. B. Owen, "The Physical Chemistry of Electrolytic Solutions," ACS Monogr. 95, 1950):

a) Estimate the diffusion coefficient at 298 K for a very dilute solution of HCl in water.

Solution

b) Estimate the diffusion coefficient at 273 K for a very dilute solution of CuSO4 in water. The

viscosity of liquid water at 273 K is 1.79 cP.

1.21a. Oxygen diffusion in water: Hayduk and Minhas correlation.Estimate the diffusion coefficient of oxygen in liquid water at 298 K. Use the Hayduk andMinhas correlation for solutes in aqueous solutions. At this temperature, the viscosity of water

is 0.9 cP. The critical volume of oxygen is 73.4 cm3/mol. The experimental value of this

diffusivity was reported as 2.1 u 10Ð5 cm2/s (Cussler E. L., Diffusion, 2nd ed, CambridgeUniversity Press, Cambridge, UK, 1997).

Solution

1.22a, d. Liquid diffusivity: Hayduk and Minhas correlation.Estimate the diffusivity of carbon tetrachloride in a dilute solution in n-hexane at 298 K using theHayduk and Minhas correlation for nonaqueous solutions. Compare the estimate to the reported

value of 3.7 u 10Ð5 cm2/s. The following data are available (Reid, et al., 1987):

Solution

1.23b. Estimating molar volumes from liquid diffusion data.

The diffusivity of allyl alcohol (C3H6O) in dilute aqueous solution at 288 K is 0.9 u 10Ð5 cm2/s

(Reid, et al., 1987). Based on this result, and the Hayduk and Minhas correlation for aqueoussolutions, estimate the molar volume of allyl alcohol at its normal boiling point. Compare it to theresult obtained using the data on Table 1.2. The viscosity of water at 288 K is 1.15 cP.

Solution

Iniitial estimates

From Table 2.1

1.24b, d. Concentration dependence of binary liquid diffusivities.a) Estimate the diffusivity of ethanol in water at 298 K when the mol fraction of ethanol in solution is40%. Under these conditions (Hammond, B. R., and R. H. Stokes, Trans. Faraday Soc., 49, 890,1953):

The experimental value reported by Hammond and Stokes (1953) is 0.42 u 10-5

cm2/s.

SolutionEstimate the infinite dilution diffusivity of ethanol in water at 298 Kfrom Hayduk-Minhas for aqueous solutions

From Appendix A, the infinite dilution diffusivity of water in ethanolat 298 K is

b) Estimate the diffusivity of acetone in water at 298 K when the mol fraction of acetone in solutionis 35%. For this system at 298 K, the activity coefficient for acetone is given by Wilson equation(Smith, J. M., et al., Introduction to Chemical Engineering Thermodynamics, 5th ed, McGraw-HillCo., Inc., New York, NY, 1996):

Solution

Estimate the thermodynamic factor

From Appendix A

Estimate the infinite dilution diffusivity of acetone in water at 298 Kfrom Hayduk-Minhas for aqueous solutions

1.25b, d. Steady-state, one-dimensional, gas-phase flux calculation.A flat plate of solid carbon is being burned in the presence of pure oxygen according to thereaction

Molecular diffusion of gaseous reactant and products takes place through a gas film adjacentto the carbon surface; the thickness of this film is 1.0 mm. On the outside of the film, the gasconcentration is 40% CO, 20% O2, and 40% CO2. The reaction at the surface may be

assumed to be instantaneous, therefore, next to the carbon surface, there is virtually nooxygen.The temperature of the gas film is 600 K, and the pressure is 1 bar. Estimate the rate

of combustion of the carbon, in kg/m2-min.

Solution

CO (1), CO2 (2), O2 (3)

Calculate binary MS diffusivities from Wilke-Lee

Appendix C-2: Solution of the Maxwell-Stefan equations for amulticomponent mixture of ideal gases by orthogonal collocation (NC =

3).

Orthogonal collocation matrices

The pressure and temperature in the vapour phase are

The Maxwell-Stefan diffusion coefficicients are

The length of the diffusion path is

The density of the gas phase follows from the ideal gas law

Initial estimates of the fluxes

Initial estimates of the concentrations

Stoichiometric relations

(No oxygen)

1.26b. Steady-state, one-dimensional, liquid-phase flux calculation.A crystal of Glauber's salt (Na2SO4� 10H2O) dissolves in a large tank of pure water at 288 K.

Estimate the rate at which the crystal dissolves by calculating the flux of Na2SO4 from the crystal

surface to the bulk solution. Assume that molecular diffusion occurs through a liquid film uniformly0.085 mm thick surrounding the crystal. At the inner side of the film (adjacent to the crystalsurface) the solution is saturated with Na2SO4, while at the outer side of the film the solution is

virtually pure water. The solubility of Glauber's salt in water at 288 K is 36 g of crystal/100 g of

water and the density of the corresponding saturated solution is 1,240 kg/m3 (Perry and Chilton,1973). The diffusivity of Na2SO4 in dilute aqueous solution at 288 K can be estimated as

suggested in Problem 1-20. The density of pure liquid water at 288 K is 999.8 kg/m3; the viscosityis 1.153 cP.

SolutionA = Na2SO4 B = H2O

Calculate xA1 (saturated

solution)Basis: 100 g H2O (36 g of dissolved crystal)

(Pure water)

Calculate diffusivity

1.27c, d. Molecular diffusion through a gas-liquid interface.Ammonia, NH3, is being selectively removed from an air-NH3 mixture by absorption into water. In

this steady-state process, ammonia is transferred by molecular diffusion through a stagnant gaslayer 5 mm thick and then through a stagnant water layer 0.1 mm thick. The concentration ofammonia at the outer boundary of the gas layer is 3.42 mol percent and the concentration at thelower boundary of the water layer is esentially zero.The temperature of the system is 288 K and

the total pressure is 1 atm. The diffusivity of ammonia in air under these conditions is 0.215 cm2/s

and in liquid water is 1.77 u 10Ð5 cm2/s. Neglecting water evaporation, determine the rate of

diffusion of ammonia, in kg/m2-hr. Assume that the gas and liquid are in equilibrium at theinterface.

Solution:

Initial estimates:

1.28c. Steady-state molecular diffusion in gases.A mixture of ethanol and water vapor is being rectified in an adiabatic distillation column. Thealcohol is vaporized and transferred from the liquid to the vapor phase. Water vapor condenses(enough to suply the latent heat of vaporization needed by the alcohol being evaporated) and istransferred from the vapor to the liquid phase. Both components diffuse through a gas film 0.1 mmthick. The temperature is 368 K and the pressure is 1 atm. The mole fraction of ethanol is 0.8 onone side of the film and 0.2 on the other side of the film. Calculate the rate of diffusion of ethanol

and of water, in kg/m2-s. The latent heat of vaporization of the alcohol and water at 368 K can beestimated by the Pitzer acentric factor correlation (Reid, et al., 1987)

where Z is the acentric factor.

Solution A = ethanol B = water

Calculate ethanol heat of vaporization

Calculate water heat of vaporization

Estimate diffusivity from Wilke-Lee

1.29a, d. Analogy among molecular heat and mass transfer.It has been observed that for the system air-water vapor at near ambient conditions, Le = 1.0(Treybal, 1980). This observation, called the Lewis relation, has profound implications inhumidification operations, as will be seen later. Based on the Lewis relation, estimate the diffusivityof water vapor in air at 300 K and 1 atm. Compare your result with the value predicted by the Wilke-

Lee equation. For air at 300 K and 1 atm:Cp = 1.01 kJ/kg-K, k = 0.0262 W/m-K, m = 1.846 10-5

kg/m-s, and r = 1.18 kg/m3.

Solution

Estimate diffusivity from the Wilke-Lee equation

1.30b, d. Steady-state molecular diffusion in gases.Water evaporating from a pond at 300 K does so by molecular diffusion across an air film 1.5 mmthick. If the relative humidity of the air at the outer edge of the film is 20%, and the total pressure is 1bar, estimate the drop in the water level per day, assuming that conditions in the film remainconstant. The vapor pressure of water as a function of temperature can be accurately estimated fromthe Wagner equation (Reid, et al., 1987)

Solution

From Appendix A

1.31b, d. Steady-state molecular diffusion in a ternary gas system.Calculate the fluxes and concentration profiles for the ternary system hydrogen (1), nitrogen (2), andcarbon dioxide (3) under the following conditions. The temperature is 308 K and the pressure is 1

atm. The diffusion path length is 86 mm. At one end of the diffusion path the concentration is 20mole% H2, 40% N2, 40% CO2; at the other end, the concentration is 50% H2, 20% N2, 30% CO2.

The total molar flux is zero, N = 0. The MS diffusion coefficients are D12 = 83.8 mm2/s, D13 = 68.0

mm2/s, D23 = 16.8 mm2/s.

Solution

Appendix C-1: Solution of the Maxwell-Stefan equations for amulticomponent mixture of ideal gases by orthogonal collocation (NC =

3).

Orthogonal collocation matrices

The pressure and temperature in the vapour phase are

The Maxwell-Stefan diffusion coefficicients are

The length of the diffusion path is

The density of the gas phase follows from the ideal gas law

Initial estimates of the fluxesInitial estimates of the concentrations

2.1a. Mass-transfer coefficients in a gas absorber.A gas absorber is used to remove benzene (C6H6) vapors from air by scrubbing the gas mixture

with a nonvolatile oil at 300 K and 1 atm. At a certain point in the absorber, the benzene molefraction in the bulk of the gas phase is 0.02, while the corresponding interfacial benzene gas-

phase concentration is 0.0158. The benzene flux at that point is measured as 0.62 g/m2-s. a) Calculate the mass-transfer coefficient in the gas phase at that point in the equipment,expressing the driving force in terms of mole fractions .

Solution

b) Calculate the mass-transfer coefficient in the gas phase at that point in the equipment,

expressing the driving force in terms of molar concentrations, kmol/m3.

Solution

c) At the same place in the equipment, the benzene mole fraction in the bulk of the liquid phaseis 0.125, while the corresponding interfacial benzene liquid-phase concentration is 0.158.Calculate the mass-transfer coefficient in the liquid phase, expressing the driving force in terms ofmole fractions.Solution

2.2a. Mass-transfer coefficients from naphthalene sublimation data.In a laboratory experiment, air at 347 K and 1 atm is blown at high speed around a singlenaphthalene (C10H8) sphere, which sublimates partially. When the experiment begins, the diameter

of the sphere is 2.0 cm. At the end of the experiment, 14.32 min later, the diameter of the sphere is1.85 cm. a) Estimate the mass-transfer coefficient, based on the average surface area of the particle,expressing the driving force in terms of partial pressures. The density of solid naphthalene is 1.145

g/cm3, its vapor pressure at 347 K is 670 Pa (Perry and Chilton, 1973).

Solution

b) Calculate the mass-transfer coefficient, for the driving force in terms of molar concentrations.

Solution

2.3a. Mass-transfer coefficients from acetone evaporation data.In a laboratory experiment, air at 300 K and 1 atm is blown at high speed parallel to the surface of arectangular shallow pan that contains liquid acetone (C3H6O), which evaporates partially. The pan is

1 m long and 50 cms wide. It is connected to a reservoir containing liquid acetone whichautomatically replaces the acetone evaporated, maintaining a constant liquid level in the pan. Duringan experimental run, it was observed that 2.0 L of acetone evaporated in 5 min. Estimate the mass-

transfer coefficient. The density of liquid acetone at 300 K is 0.79 g/cm3; its vapor pressure is 27kPa (Perry and Chilton, 1973).

Solution

2.4b. Mass-transfer coefficients from wetted-wall experimental data.A wetted-wall experimental set-up consists of a glass pipe, 50 mm in diameter and 1.0 m long.Water at 308 K flows down the inner wall. Dry air enters the bottom of the pipe at the rate of 1.04

m3/min, measured at 308 K and 1 atm. It leaves the wetted section at 308 K and with a relativehumidity of 34%. With the help of equation (2-52), estimate the average mass-transfer coefficient,with the driving force in terms of molar fractions.

Solution

2.7c. Mass transfer in an annular space.a) In studying rates of diffusion of naphthalene into air, an investigator replaced a 30.5-cm section ofthe inner pipe of an annulus with a naphthalene rod. The annulus was composed of a 51-mm-ODbrass inner pipe surrounded by a 76-mm-ID brass pipe. While operating at a mass velocity within the

annulus of 12.2 kg of air/m2-s at 273 K and 1 atm, the investigator determined that the partialpressure of naphthalene in the exiting gas stream was 0.041 Pa. Under the conditions of theinvestigation, the Schmidt number of the gas was 2.57, the viscosity was 175 PP, and the vaporpressure of naphthalene was 1.03 Pa. Estimate the mass-transfer coefficient from the inner wall forthis set of conditions. Assume that equation (2-52) applies.

Solution

b) Monrad and Pelton (Trans. AIChE, 38, 593, 1942) presented the following correlation forheat-transfer coefficient in an annular space:

where do and di are the outside and inside diameters of the annulus, de is the equivalent

diameter defined as

Write down the analogous expression for mass transfer and use it to estimate the mass-transfercoefficient for the conditions of part a). Compare both results.

Solution

2.8c. The Chilton-Colburn analogy: flow across tube banks.Winding and Cheney (Ind. Eng. Chem., 40, 1087, 1948) passed air at 310 K and 1 atm through abank of rods of naphthalene. The rods were in a staggered arrangement, with the air flowing at rightangles to the axes of the rods. The bank consisted of 10 rows containing alternately five and four38-mm-OD tubes (d = 38 mm) spaced on 57-mm centers, with the rows 76 mm apart. The mass-transfer coefficient was determined by measuring the rate of sublimation of the naphthalene. Thedata could be correlated by:

where G' is the maximum mass velocity through the tube bank, in kg/m2-s, and kG is in kmol/m2-

s-Pa.a) Rewrite equation (2-68) in terms of the Colburn jD -factor. The diffusivity of naphthalene in air at

310 K and 1 atm is 0.074 cm2/s.

Solution

Properties of dilute mixtures of naphthalene in air at 310 K and 1 atm:

Dimensional constant in the given correlation:

b) Estimate the mass-transfer coefficient to be expected for evaporation of n-propyl alcohol intocarbon dioxide for the same geometrical arrangement when the carbon dioxide flows at a maximumvelocity of 10 m/s at 300 K and 1 atm. The vapor pressure of n-propyl alcohol at 300 K is 2.7 kPa.

Solution

Properties of dilute mixtures of propyl alcohol in carbon dioxide at 300 K and 1 atm:

c) Zakauskas (Adv. Heat Transfer, �, 93, 1972) proposed the following correlation for the heat-transfer coefficient in a staggered tube bank arrangement similar to that studied by Winding andCheney:

Use the mass-transfer expression analogous to equation (2-69) to estimate the mass-transfercoefficient of part b). Compare the results.

Solution

2.9b. Mass transfer from a flat plate.A 1-m square thin plate of solid naphthalene is oriented parallel to a stream of air flowing at 20m/s. The air is at 310 K and 101.3 kPa. The naphthalene remains at 290 K; at this temperaturethe vapor pressure of naphthalene is 26 Pa. Estimate the moles of naphthalene lost from the plateper hour, if the end effects can be ignored.

Solution

2.10b. Mass transfer from a flat plate.A thin plate of solid salt, NaCl, measuring 15 by 15 cm, is to be dragged through seawater at a

velocity of 0.6 m/s. The 291 K seawater has a salt concentration of 0.0309 g/cm3. Estimate therate at which the salt goes into solution if the edge effects can be ignored. Assume that the

kinematic viscosity at the average liquid film conditions is 1.02 u 10Ð6 m2/s, and the diffusivity is

1.25 u 10Ð9 m2/s. The solubility of NaCl in water at 291 K is 0.35 g/cm3, and the density of the

saturated solution is 1.22 g/cm3 (Perry and Chilton, 1973) .

Solution

Laminar flow

At the bulk of the solution, point 2:

At the interface, point 1:

2.11b. Mass transfer from a flat liquid surface.During the experiment described in Problem 2.3, the air velocity was measured at 6 m/s, parallel tothe longest side of the pan. Estimate the mass-transfer coefficient predicted by equation (2-28) or(2-29) and compare it to the value measured experimentally. Notice that, due to the high volatility ofacetone, the average acetone concentration in the gas film is relatively high. Therefore, propertiessuch as density and viscosity should be estimated carefully. The following data for acetone might

be needed: Tc = 508.1 K, Pc = 47.0 bar, M = 58, Vc = 209 cm3/mol, Zc = 0.232 (Reid, et al., 1987).

Solution

Average film properties:

Estimate the viscosity of the mixture from Lucas Method

Estimate the diffusivity from the Wilke-Lee equation

2.12b. Evaporation of a drop of water falling in air.Repeat Example 2.9 for a drop of water which is originally 2 mm in diameter.

Solution

2.13b. Dissolution of a solid sphere into a flowing liquid stream.Estimate the mass-transfer coefficient for the dissolution of sodium chloride from a cast sphere, 1.5cm in diameter, if placed in a flowing water stream. The velocity of the 291 K water stream is 1.0m/s.

Assume that the kinematic viscosity at the average liquid film conditions is 1.02 u 10Ð6 m2/s, and

the mass diffusivity is 1.25 u 10Ð9 m2/s. The solubility of NaCl in water at 291 K is 0.35 g/cm3, and

the density of the saturated solution is 1.22 g/cm3 (Perry and Chilton, 1973) .

Solution

From Prob. 2.10:

2.14b. Sublimation of a solid sphere into a gas stream. During the experiment described in Problem 2.2, the air velocity was measured at 10 m/s. Estimatethe mass-transfer coefficient predicted by equation (2-36) and compare it to the value measuredexperimentally. The following data for naphthalene might be needed: Tb = 491.1 K, Vc = 413

cm3/mol.

Solution

For air at 347 K and 1 atm:

Estimate DAB from the Wilke-Lee

equationLennard-Jones parameters for naphthalene

2.15b. Dissolution of a solid sphere into a flowing liquid stream. The crystal of Problem 1.26 is a sphere 2-cm in diameter. It is falling at terminal velocity under the

influence of gravity into a big tank of water at 288 K. The density of the crystal is 1,464 kg/m3 (Perryand Chilton, 1973). a) Estimate the crystal's terminal velocity.

Solution

b) Estimate the rate at which the crystal dissolves and compare it to the answer obtained in Problem1.26.

Solution

From Prob.1.26

From Prob.1.26:

2.16c. Mass transfer inside a circular pipe.Water flows through a thin tube, the walls of which are lightly coated with benzoic acid (C7H6O2).

The water flows slowly, at 298 K and 0.1 cm/s. The pipe is 1-cm in diameter. Under theseconditions, equation (2-63) applies. a) Show that a material balance on a length of pipe L leads to

where v is the average fluid velocity, and cA* is the equilibrium solubility concentration.

b) What is the average concentration of benzoic acid in the water after 2 m of pipe. The solubility of

benzoic acid in water at 298 K is 0.003 g/cm3, and the mass diffusivity is 1.0 u 10Ð5 cm2/s(Cussler, 1997).

Solution

2.17b. Mass transfer in a wetted-wall tower.Water flows down the inside wall of a 25-mm ID wetted-wall tower of the design of Figure 2.2, while

air flows upward through the core. Dry air enters at the rate of 7 kg/m2-s. Assume the air iseverywhere at its average conditions of 309 K and 1 atm, the water at 294 K, and the mass-transfercoefficient constant. Compute the average partial pressure of water in the air leaving if the tower is 1

m long.

Solution

For water at 294 K

2.18c. Mass transfer in an annular space.In studying the sublimation of naphthalene into an airstream, an investigator constructed a 3-m-longannular duct. The inner pipe was made from a 25-mm-OD, solid naphthalene rod; this wassurrounded by a 50-mm-ID naphthalene pipe. Air at 289 K and 1 atm flowed through the annular space at an average velocity of 15 m/s. Estimatethe partial pressure of naphthalene in the airstream exiting from the tube. At 289 K, naphthalene

has a vapor pressure of 5.2 Pa, and a diffusivity in air of 0.06 cm2/s. Use the results of Problem 2.7to estimate the mass-transfer coefficient for the inner surface; and equation (2-47), using theequivalent diameter defined in Problem 2.7, to estimate the coefficient from the outer surface.

Solution

In this situation, there will be a molar flux from the inner wall, NA1, with specific interfacial area, a1,

and a flux from the outer wall, NA2, with area a2. A material balance on a differential volume

element yields:

Define:

Then:

For the interior wall:

For the outer wall:

2.19c. Benzene evaporation on the outside surface of a single cylinder.Benzene is evaporating at the rate of 20 kg/hr over the surface of a porous 10-cm-diameter cylinder.Dry air at 325 K and 1 atm flows at right angle to the axis of the cylinder at a velocity of 2 m/s. Theliquid is at a temperature of 315 K where it exerts a vapor pressure of 26.7 kPa. Estimate the length

of the cylinder. For benzene, Tc = 562.2 K, Pc = 48.9 bar, M = 78, Vc = 259 cm3/mol, Zc = 0.271

(Reid, et al., 1987).

Solution

Calculate the average properties of the film

From the Wilke-Lee equation

From the Lucas Method

From Eq. 2-45:

2.20b. Mass transfer in a packed bed.Wilke and Hougan (Trans. AIChE, 41, 445, 1945) reported the mass transfer in beds of granularsolids. Air was blown through a bed of porous celite pellets wetted with water, and by evaporatingthis water under adiabatic conditions, they reported gas-film coefficients for packed beds. In one run,the following data were reported:effective particle diameter 5.71 mm

gas stream mass velocity 0.816 kg/m2-stemperature at the surface 311 Kpressure 97.7 kPa

kG 4.415 u 10Ð3 kmol/m2-s-atm

With the assumption that the properties of the gas mixture are the same as those of air, calculatethe gas-film mass-transfer coefficient using equation (2-55) and compare the result with the valuereported by Wilke and Hougan.

Solution


2.21b. Mass transfer and pressure drop in a packed bed.Air at 373 K and 2 atm is passed through a bed 10-cm in diameter composed of iodine spheres0.7-cm in diameter. The air flows at a rate of 2 m/s, based on the empty cross section of the bed.The porosity of the bed is 40%. a) How much iodine will evaporate from a bed 0.1 m long? The vapor pressure of iodine at 373 K is6 kPa.

Solution

From the Wilke-Lee equation:

b) Estimate the pressure drop through the bed.

Solution

2.22b. Volumetric mass-transfer coefficients in industrial towers.The interfacial surface area per unit volume, a, in many types of packing materials used inindustrial towers is virtually impossible to measure. Both a and the mass-transfer coefficientdepend on the physical geometry of the equipment and on the flow rates of the two contacting,inmiscible streams. Accordingly, they are normally correlated together as the volumetric mass-transfer coefficient, kca.

Empirical equations for the volumetric coefficients must be obtained experimentally for each type ofmass-transfer operation. Sherwood and Holloway (Trans. AIChE, 36, 21, 39, 1940) obtained thefollowing correlation for the liquid-film mass-transfer coefficient in packed absorption towers

The values of a and n to be used in equation (2-71) for various industrial packings are listed in thefollowing table, when SI units are used exclusively.

a) Consider the absorption of SO2 with water at 294 K in a tower packed with 25-mm Raschig

rings. If the liquid mass velocity is L' = 2.04 kg/m2-s, estimate the liquid-film mass-transfer

coefficient. The diffusivity of SO2 in water at 294 K is 1.7 u 10Ð9 m2/s.

Solution

For dimensional consistency, add the constants:

b) Whitney and Vivian (Chem. Eng. Progr., 45, 323, 1949) measured rates of absorption of SO2 in

water and found the following expression for 25-mm Raschig rings at 294 K

where kxa is in kmole/m2-s. For the conditions described in part a), estimate the liquid-film

mass-transfer coefficient using equation (2-72). Compare the results.

Solution

2.23b. Mass transfer in fluidized beds.Cavatorta, et al. (AIChE J., 45, 938, 1999) studied the electrochemical reduction of ferrycianide

ions, {Fe(CN)6}Ð3, to ferrocyanide, {Fe(CN)6}Ð4, in aqueous alkaline solutions. They studied

different arrangements of packed columns, including fluidized beds. The fluidized bed experimentswere performed in a 5-cm-ID circular column, 75-cm high. The bed was packed with 0.534-mm

spherical glass beads, with a particle density of 2.612 g/cm3. The properties of the aqueous

solutions were: density = 1,083 kg/m3, viscosity = 1.30 cP, diffusivity = 5.90 10Ð10 m2/s. Theyfound that the porosity of the fluidized bed, e, could be correlated with the superficial liquid velocitybased on the empty tube, vs, through

where vs is in cm/s.

a) Using equation (2-56), estimate the mass-transfer coefficient, kL, if the porosity of the bed is

60%.

Solution

b) Cavatorta et al. proposed the following correlation to estimate the mass-transfer coefficient fortheir fluidized bed experimental runs:

where Re is based on the empty tube velocity. Using this correlation, estimate the mass-transfercoefficient, kL, if the porosity of the bed is 60%. Compare your result to that of part a).

Solution

2.24b. Mass transfer in a hollow-fiber boiler feedwater deaerator.Consider the hollow-fiber BFW deaerator described in Example 2-13. If the water flow rateincreases to 60,000 kg/hr while everything else remains constant, calculate the fraction of theentering dissolved oxygen that can be removed.

Solution

2.25b. Mass transfer in a hollow-fiber boiler feedwater deaerator.a) Consider the hollow-fiber BFW deaerator described in Example 2-13. Assuming that only oxygendiffuses across the membrane, calculate the gas volume flow rate and composition at the lumenoutlet. The water enters the shell side at 298 K saturated with atmospheric oxygen, which means adissolved oxygen concentration of 8.38 mg/L.

Solution

b) Calculate the mass-transfer coefficient at the average conditions inside the lumen. Neglect thethickness of the fiber walls when estimating the gas velocity inside the lumen.

Solution

Calculate the average flow conditions inside the fibers

Calculate the average oxygen molar fraction in the gas

From Lucas method for pure N2


(From Example 2.13)

3.1a. Application of Raoult's law to a binary system.Repeat Example 3.1, but for a liquid concentration of 0.6 mole fraction of benzene and atemperature of 320 K.

Solution

3.2b. Application of Raoult's law to a binary system.a) Determine the composition of the liquid in equilibrium with a vapor containing 60 mole percentbenzene-40 mole percent toluene if the system exists in a vessel under 1 atm pressure. Predictthe equilibrium temperature.

Solution

Initial estimates

b) Determine the composition of the vapor in equilibrium with a liquid containing 60 mole percentbenzene-40 mole percent toluene if the system exists in a vessel under 1 atm pressure. Predictthe equilibrium temperature.

Solution

3.3a. Application of Raoult's law to a binary system.Normal heptane, n-C7H16, and normal octane, n-C8H18, form ideal solutions. At 373 K, normal

heptane has a vapor pressure of 106 kPa and normal octane of 47.1 kPa.a) What would be the composition of a heptane-octane solution that boils at 373 K under a 93 kPapressure?

Solution

b) What would be the composition of the vapor in equilibrium with the solution that is described in(a)?

Solution

3.4a. Henry's law: saturation of water with oxygen.A solution with oxygen dissolved in water containing 0.5 mg O2/100 g of H2O is brought in contact

with a large volume of atmospheric air at 283 K and a total pressure of 1 atm. The Henry's law

constant for the oxygen-water system at 283 K equals 3.27 u 104 atm/mole fraction.a) Will the solution gain or lose oxygen?b) What will be the concentration of oxygen in the final equilibrium solution?

Solution

At equilibrium:

Basis: 1 L water (1 kg water)

Equilibrium concentration, ce = 11.42 mg oxygen/L

Initial conditions:

The solution gains oxygen.

3.5c. Material balances combined with equilibrium relations.Repeat Example 3.3, but assuming that the ammonia, air, and water are brought into contact in a

closed container. There is 10 m3 of gas space over the liquid. Assuming that the gas-space volumeand the temperature remain constant until equilibrium is achieved, modify the Mathcad program inFigure 3.2 to calculate:a) the total pressure at equilibrium

Solution

Initial guesses

Answer: P = 1.755 atm

b) the equilibrium ammonia concentration in the gas and liquid phases.

Answer: yA = 0.145; xA=

0.162

3.6b. Mass-transfer resistances during absorption.In the absorption of component A (molecular weight = 60) from an airstream into an aqueoussolution, the bulk compositions of the two adjacent streams at a point in the apparatus were

analyzed to be pA,G = 0.1 atm, and cA,L = 1.0 kmol of A/m3 of solution. The total pressure was 2.0

atm; the density of the solution was 1,100 kg/m3. The Henry's constant for these conditions was

0.85 atm/mole fraction. The overall gas coefficient was KG = 0.27 kmol/m2-hr-atm. If 57% of the total

resistance to mass transfer resides in the gas film, determine

a) the gas-film coefficient, kG;

Solution

b) the liquid-film coefficient, kL;

Solution

Basis: 1 m3 of aqueous solution

c) the concentration on the liquid side of the interface, xA,i;

Solution

Initial estimates of interfacial concentrations:

d) the mass flux of A.

Solution

Check this result by calculating the gas-phase flux:

3.7b. Mass-transfer resistances during absorption.For a system in which component A is transferring from the gas phase to the liquid phase, theequilibrium relation is given by

where pA,i is the equilibrium partial pressure in atm and xA,i is the equilibrium liquid concentration

in molar fraction. At one point in the apparatus, the liquid stream contains 4.5 mole % and the gasstream contains 9.0 mole % A. The total pressure is 1 atm. The individual gas-film coefficient at

this point is kG = 3.0 mole/m2-s-atm. Fifty per cent of the overall resistance to mass transfer is

known to be encountered in the liquid phase. Evaluate

a) the overall mass-transfer coefficient, Ky;

Solution

b) the molar flux of A;

Solution

(yAe =

yA*)

c) The liquid interfacial concentration of A.

Solution

3.8d. Absorption of ammonia by water: use of F-type mass-transfer coefficients.Modify the Mathcad program in Figure 3.6 to repeat Example 3.5, but with yA,G = 0.70 and xA,L =

0.10. Everything else remains constant.

Solution

Initial guesses

3.9d. Absorption of ammonia by water: use of F-type mass-transfer coefficients.

Modify the Mathcad program in Figure 3.6 to repeat Example 3.5, but with FL = 0.0050 kmol/m2-s.

Everything else remains constant.

Solution

Initial guesses

3.10b. Mass-transfer resistances during absorption of ammonia.In the absorption of ammonia into water from an air-ammonia mixture a 300 K and 1 atm, the

individual film coefficients were estimated to be kL = 6.3 cm/hr and kG = 1.17 kmol/m2-hr-atm. The

equilibrium relationship for very dilute solutions of ammonia in water at 300 K and 1 atm is

Determine the following mass-transfer coefficients:

a) ky

Solution

b)kx

Solution

c) Ky

Solution

d) Fraction of the total resistance to mass transfer that resides in the gas phase.

Solution

3.11b. Mass-transfer resistances in hollow-fiber membrane contactors.For mass transfer across the hollow-fiber membrane contactors described in Example 2.13, theoverall mass-transfer coefficient based on the liquid concentrations, KL, is given by (Yang and

Cussler, AIChE J., 32, 1910, Nov. 1986)

where kL, kM, and kc are the individual mass-transfer coefficients in the liquid, across the

membrane, and in the gas, respectively; and H is Henry's law constant, the gas equilibriumconcentration divided by that in the liquid. The mass-transfer coefficient across a hydrophobicmembrane is from (Prasad and Sirkar, AIChE J., 34, 177, Feb. 1988)

where DAB = molecular diffusion coefficient in the gas filling the pores,

eM = membrane porosity,

tM = membrane tortuosity,

d = membrane thickness.

For the membrane modules of Example 2.13, eM = 0.4,tM = 2.2, and d = 25 10Ð6 m (Prasad

and Sirkar, 1988).

a) Calculate the corresponding value of kM.

Solution

For oxygen in nitrogen at 298 K and 1 atm:

b) Using the results of part (a), Example 2.13, and Problem 2.25, calculate KL, and estimate what

fraction of the total resistance to mass transfer resides in the liquid film.

Solution

From Example 2.13:

From Prob. 2.25:

Virtually all of the resistance resides in the liquid phase.

3.12c. Combined use of F- and k-type coefficients: absorption of low-solubility gases.During absorption of low-solubility gases, mass transfer from a highly concentrated gas mixture toa very dilute liquid solution frequently takes place. In that case, although it is appropriate to use ak-type mass-transfer coefficient in the liquid phase, an F-type coefficient must be used in the gasphase. Since dilute liquid solutions usually obey Henry's law, the interfacial concentrations duringabsorption of low-solubility gases are related through yA,i = mxA,i.

a) Show that, under the conditions described above, the gas interfacial concentration satisfies theequation

Solution

In the gas phase:

In the liquid phase:

For Henry's Law:

Then:

Rearranging:

b) In a certain apparatus used for the absorption of SO2 from air by means of water, at one point in

the equipment the gas contained 30% SO2 by volume and was in contact with a liquid containing

0.2% SO2 by mole. The temperature was 303 K and the total pressure 1 atm. Estimate the

interfacial concentrations and the local SO2 molar flux. The mass-transfer coefficients were

calculated as FG = 0.002 kmol/m2-s, kx = 0.160 kmol/m2-s. The equilibrium SO2 solubility data at

303 K are (Perry and Chilton, 1973):

kg SO2/100 kg water Partial pressure of SO2, mm Hg (torr)

0.0 00.5 421.0 851.5 1292.0 1762.5 224

Solution

Define: w = kg SO2/100 kg water

p = Partial pressure of SO2, mm Hg (torr)

Initial guess:

3.13d. Distillation of a mixture of methanol and water in a packed tower: use of F-typemass-transfer coefficients.At a different point in the packed distillation column of Example 3.6, the methanol content of thebulk of the gas phase is 76.2 mole %; that of the bulk of the liquid phase is 60 mole %. Thetemperature at that point in the tower is around 343 K. The packing characteristics and flow rates

at that point are such that FG = 1.542 u 10Ð3 kmol/m2-s, and FL = 8.650 u 10Ð3 kmol/m2-s.

Calculate the interfacial compositions and the local methanol flux. To calculate the latent heats ofvaporization at the new temperature, modify the values given in Example 3.6 using Watson'smethod (Smith, et al., 1996):

For water, Tc = 647.1 K; for methanol, Tc = 512.6

K

Solution

For methanol (A)

For water (B)

Parameters

Initial estimates

3.14b. Material balances: adsorption of benzene vapor on activated carbon.Activated carbon is used to recover benzene from a nitrogen-benzene vapor mixture. A nitrogen-benzene mixture at 306 K and 1 atm containing 1% benzene by volume is to be passed

countercurrently at the rate of 1.0 m3/s to a moving stream of activated carbon so as to remove85% of the benzene from the gas in a continuous process. The entering activated carbon contains

15 cm3 benzene vapor (at STP) adsorbed per gram of the carbon. The temperature and totalpressure are maintained at 306 K and 1 atm. Nitrogen is not adsorbed. The equilibrium adsorptionof benzene on this activated carbon at 306 K is reported as follows:

Benzene vapor adsorbed Partial pressure benzene, mm Hg

cm3 (STP)/g carbon15 0.5525 0.9540 1.6350 2.1865 3.2680 4.8890 6.22100 7.83

a) Plot the equilibrium data as X' = kg benzene/kg dry carbon, Y' = kg benzene/kg nitrogen for atotal pressure of 1 atm.

Solution

b) Calculate the minimum flow rate required of the entering activated carbon (remember that theentering carbon contains some adsorbed benzene).

Solution

On the XY diagram, locate the point (X2,Y2). Since the operating line is above the equilibrium

curve and the equilibrium curve is concave upwards, the minimum operating line is obtained bylocating, at the intersection of Y = Y1 with the equilibrium curve, X1max.

c) If the carbon flow rate is 20% above the minimum, what will be the concentration of benzeneadsorbed on the carbon leaving?

Solution

d) For the conditions of part (c), calculate the number of ideal stages required.

Solution

See stepwise construction on the XY graph

3.15b. Material balances: desorption of benzene vapor from activated carbon.The activated carbon leaving the adsorber of Problem 3.14 is regenerated by countercurrent contactwith steam at 380 K and 1 atm. The regenerated carbon is returned to the adsorber, while themixture of steam and desorbed benzene vapors is condensed. The condensate separates into anorganic and an aqueous phase and the two phases are separated by decantation. Due to the lowsolubility of benzene in water, most of the benzene will be concentrated in the organic phase, whilethe aqueous phase will contain only traces of benzene. The equilibrium adsorption data at 380 Kare as follows:

Benzene vapor adsorbed Partial pressure benzene, kPakg benzene/100 kg carbon2.9 1.05.5 2.012.0 5.017.1 8.020.0 10.025.7 15.030.0 20.0

a) Calculate the minimum steam flow rate required.

Solution Equilibrium curve

From Problem 3.14

From the XY diagram:

b) For a steam flow rate of twice the minimum, calculate the benzene concentration in the gasmixture leaving the desorber, and the number of ideal stages required.

Solution

3.16b. Material balances: adsorption of benzene vapor on activated carbon; cocurrentoperation.If the adsorption process described in Problem 3.14 took place cocurrently, calculate theminimum flow rate of activated carbon required.

Solution

Fom Problem 3.14:

From the XY diagram:

3.17b. Material balances in batch processes: drying of soap with air.It is desired to dry 10 kg of soap from 20% moisture by weight to no more than 6% moisture by

contact with hot air. The wet soap is placed in a vessel containing 8.06 m3 of air at 350 K, 1 atm,and a water-vapor partial pressure of 1.6 kPa. The system is allowed to reach equilibrium, andthen the air in the vessel is entirely replaced by fresh air of the original moisture content andtemperature. How many times must the process be repeated in order to reach the specified soapmoisture content of no more than 6%? When this soap is exposed to air at 350 K and 1 atm, theequilibrium distribution of moisture between the air and the soap is as follows:

Wt % moisture in soap Partial pressure of water, kPa

2.40 1.293.76 2.564.76 3.796.10 4.967.83 6.199.90 7.3312.63 8.4215.40 9.5819.02 10.60

Solution

Generate the XY diagram

From the XY diagram, at the exit of the fifth equilibrium stage, X = 0.06 and

3.18b. Material balances in batch processes: extraction of an aqueous nicotine solution

with kerosene.Nicotine in a water solution containing 2% nicotine is to be extracted with kerosene at 293 K.Water and kerosene are essentially insoluble. Determine the percentage extraction of nicotine if100 kg of the feed solution is extracted in a sequence of four batch ideal extractions using 49.0 kgof fresh, pure kerosene each. The equilibrium data are as follows (Claffey et al., Ind. Eng. Chem.,42, 166, 1950):

X', u 103 kg nicotine/kg water Y', u 103 kg nicotine/kg kerosene

1.01 0.812.46 1.965.02 4.567.51 6.869.98 9.1320.4 18.70

Solution

From the XY diagram, after 4 extractions, X = 0.00422

3.19b. Cross-flow cascade of ideal stages.The drying and liquid-liquid extraction operations described in Problems 3.17 and 3.18,respectively, are examples of a flow configuration called a cross-flow cascade. Figure 3.27 is aschematic diagram of a cross-flow cascade of ideal stages. Each stage is represented by a circle,and within each stage mass transfer occurs as if in cocurrent flow. The L phase flows from onestage to the next, being contacted in each stage by a fresh V phase. If the equilibrium-distributioncurve of the cross-flow cascade is everywhere straight and of slope m, it can be shown that(Treybal, 1980)

where S is the stripping factor, mVS/LS, constant for all stages, and N is the total number of

stages.

Solve Problem 3.18 using equation (3-60), and compare the results obtained by the two methods.

Solution

Initial estimate

3.20a. Cross-flow cascade of ideal stages: nicotine extraction.Consider the nicotine extraction of Problems 3.18 and 3.19. Calculate the number of ideal stagesrequired to achieve at least 95% extraction efficiency.

Solution

Use 8 ideal stages

3.21b. Kremser equations: absorption of hydrogen sulfide.

A scheme for the removal of H2S from a flow of 1.0 std m3/s of natural gas by scrubbing with

water at 298 K and 10 atm is being considered. The initial composition of the feed gas is 2.5 molepercent H2S. A final gas stream containing only 0.1 mole percent H2S is desired. The absorbing

water will enter the system free of H2S. At the given temperature and pressure, the system will

follow Henry's law, according to Yi = 48.3Xi, where Xi = moles H2S/mole of water; Yi = moles

H2S/mole of air.

a) For a countercurrent absorber, determine the flow rate of water that is required if 1.5 times the

minimum flow rate is used.

Solution

at SC

b) Determine the composition of the exiting liquid.

Solution

c) Calculate the number of ideal stages required.

Solution

3.22b. Absorption with chemical reaction: H2S scrubbing with MEA.

As shown in Problem 3-21, scrubbing of hydrogen sulfide from natural gas using water is notpractical since it requires large amounts of water due to the low solubility of H2S in water. If a 2N

solution of monoethanolamine (MEA) in water is used as the absorbent, however, the requiredliquid flow rate is reduced dramatically because the MEA reacts with the absorbed H2S in the

liquid phase, effectively increasing its solubility.For this solution strength and a temperature of 298 K, the solubility of H2S can be approximatedby (de Nevers, N., Air Pollution Control Engineering, 2nd ed., McGraw-Hill, Boston, MA, 2000):

Repeat the calculations of Problem 3.21, but using a 2N monoethanolamine solution asabsorbent.

Solution

3.23b. Kremser equations: absorption of sulfur dioxide.A flue gas flows at the rate of 10 kmol/s at 298 K and 1 atm with a SO2 content of 0.15 mole %.

Ninety percent of the sulfur dioxide is to be removed by absorption with pure water at 298 K. Thedesign water flow rate will be 50% higher than the minimum. Under these conditions, theequilibrium line is (Ben’tez, J., Process Engineering and Design for Air Pollution Control,Prentice Hall, Englewood Cliffs, NJ, 1993):

where Xi = moles SO2/mole of water; Yi = moles SO2/mole of air.

a) Calculate the water flow rate and the SO2 concentration in the water leaving the absorber.

Solution

b) Calculate the number of ideal stages required for the specified flow rates and percentageSO2 removal.

Solution

3.24b. Kremser equations: absorption of sulfur dioxide.An absorber is available to treat the flue gas of Problem 3.23 which is equivalent to 8.5equilibrium stages. a) Calculate the water flow rate to be used in this absorber if 90% of the SO2 is to be removed.

Calculate also the SO2 concentration in the water leaving the absorber.

Solution

Initial estimate

b) What is the percentage removal of SO2 that can be achieved with this absorber if the water flow

rate used is the same that was calculated in Problem 3.23 (a)?

Solution

Initial estimate

3.25b. Kremser equations: liquid extraction.An aqueous acetic acid solution flows at the rate of 1,000 kg/hr. The solution is 1.1% (by weight)acetic acid. It is desired to reduce the concentration of this solution to 0.037% (by weight) aceticacid by extraction with 3-heptanol at 298 K. The inlet 3-heptanol contains 0.02% (by weight) aceticacid. An extraction column is available which is equivalent to a countercurrent cascade of 15equilibrium stages. What solvent flow rate is required? Calculate the composition of the solventphase leaving the column. For this system, equilibrium is given byWt ratio acetic acid in solvent = 0.828 u Wt ratio acetic acid in water

Solution

Let the aqueous phase be the V-phase; the solvent phase is the L-phase.

Initial estimate:

3.26c. Countercurrent versus cross-flow extraction.A 1-butanol acid solution is to be extracted with pure water. The butanol solution contains4.5% (by weight) of acetic acid and flows at the rate of 400 kg/hr. A total water flow rate of1005 kg/hr is used. Operation is at 298 K and 1 atm. For practical purposes, 1-butanol andwater are inmiscible. At 298 K, the equilibrium data can be represented by YAi = 0.62 XAi,

where YAi is the weight ratio of acid in the aqueous phase and XAi is the weight ratio of acid in

the organic phase.a) If the outlet butanol stream is to contain 0.10% (by weight) acid, how many equilibriumstages are required for a countercurrent cascade?

Solution

b) If the water is split up equally among the same number of stages, but in a cross-flow cascade,what is the outlet 1-butanol concentration (see Problem 3.19)?

Solution

3.27c. Glucose sorption on an ion exchange resin.Ching and Ruthven (AIChE Symp. Ser., 81,p. 242, 1985) found that the equilibrium of glucose on anion exchange resin in the calcium form was linear for concentrations below 50 g/L. Their equilibriumexpression at 303 K is YAi = 1.961 XAi, where XAi is the glucose concentration in the resin.(g of

glucose per liter of resin) and YAi is the glucose concentration in solution.(g of glucose per liter of

solution). a) We wish to sorb glucose onto this ion exchange resin at 303 K in a countercurrent cascade ofideal stages. The concentration of the feed solution is 15 g/L. We want an outlet concentration of 1.0g/L. The inlet resin contains 0.25 g of glucose/L. The feed solution flows at the rate of 100 L/min,while the resin flows at the rate of 250 L/min. Find the number of equilibrium stages required.

Solution

b) If 5 equilibrium stages are added to the cascade of part a), calculate the resin flow required tomaintain the same degree of glucose sorption.

Solution

4.1a. Void fraction near the walls of packed beds.Consider a cylindrical vessel with a diameter of 305 mm packed with solid spheres with a diameterof 50 mm.a) From equation (4-1), calculate the asymptotic porosity of the bed.

Solution

Answer

b) Estimate the void fraction at a distance of 100 mm from the wall.

Answer

4.2b. Void fraction near the walls of packed beds.Because of the oscillatory nature of the void-fraction radial variation of packed beds, there are anumber of locations close to the wall where the local void fraction is exactly equal to the asymptoticvalue for the bed. For the bed described in Example 4.1, calculate the distance from the wall to thefirst five such locations.

Solution

The local porosity near the wall equals the asymptotic porosity when J0(Drd) = 0. From

Example 4.1,

Initial estimates of the roots can be obtained from Fig. 4.4

Initial estimate

Initial estimate

Initial estimate

Initial estimate

Initial estimate

4.3c. Void fraction near the walls of packed beds.(a) Show that the radial location of the maxima and minima of the function described by Equation(4-1) are the roots of the equation

Solution

b) For the packed bed of Example 4.1, calculate the radial location of the first five maxima, and ofthe first five minima; calculate the amplitude of the void fraction oscilations at those points.

Solution

Initial estimates of the roots can be obtained from Fig. 4.4

Initial estimate

Initial estimate

Repeating this procedure, the following results are obtained:Maxima: Minimar, mm (r*) Amplitude, % r,mm (r*) Amplitude, %20.8 (1.04) 37.2 11.3 (0.57) -56.739.6 (1.98) 21.1 30.2 (1.51) -27.358.3 (2.92) 13.5 49.0 (2.45) -16.777.1 (3.85) 9.2 67.7 (3.39) -11.195.8 (4.79) 6.4 86.4 (4.32) -7.6

c) Calculate the distance from the wall at which the absolute value of the porosity fluctuations hasbeen dampened to less than 10% of the asymptotic bed porosity.

SolutionFrom the results of part (b), this must happen at r* between 3.39 and 3.85

(d) What fraction of the cross-sectional area of the packed bed is characterized by porosityfluctuations which are within 10% of the asymptotic bed porosity?

SolutionFrom part (c)

(f) For the packed bed of Example 4.1, estimate the average void fraction by numerical integration ofequation (4-65) and estimate the ratio Hav/Hb.

Solution

4.4c. Void fraction near the walls of annular packed beds.Annular packed beds (APBs) involving the flow of fluids are used in many technical and engineeringapplications, such as in chemical reactors, heat exchangers, and fusion reactor blankets. It is wellknown that the wall in a packed bed affects the radial void fraction distribution. Since APBs have twowalls that can simultaneously affect the radial void fraction distribution, it is essential to include thisvariation in transport models. A correlation for this purpose was recently formulated (Mueller, G. E.,AIChE J., 45, 2458-60, Nov. 1999). The correlation is restricted to randomly packed beds in annularcylindrical containers of outside diameter Do, inside diameter Di, equivalent diameter De = Do Ð Di,

consisting of equal-sized spheres of diameter dp, with diameter aspect ratios of 4 ² De/dp ² 20. The

correlation is

Consider an APB with outside diameter of 140 mm, inside diameter of 40 mm, packed with identical10-mm diameter spheres.(a) Estimate the void fraction at a distance from the outer wall of 25 mm.

Solution

(b) Plot the void fraction, as predicted by Eq. (4-66), for r* from 0 to R*.

(c) Show that the average porosity for an APB is given by

(d) Estimate the average porosity for the APB described above.

Solution

4.5a. Minimum liquid mass velocity for proper wetting of packing.A 1.0-m diameter bed used for absorption of ammonia with pure water at 298 K is packed with25-mm plastic Intalox saddles. Calculate the minimum water flow rate, in kg/s, neded to ensureproper wetting of the packing surface.

Solution

For plastic packing, vL,min = 1.2 mm/s.

From Steam Tables

4.6a. Minimum liquid mass velocity for proper wetting of packing.Repeat Problem 4.5, but using ceramic instead of plastic Intalox saddles.

Solution

For ceramic packing, vL,min = 0.15 mm/s.

From Steam Tables

4.7b. Specific liquid holdup and void fraction in first-generation randompacking.Repeat Example 4.2, but using 25-mm ceramic Berl saddles as packing material.

Solution

From Table 4.1:

From Example 4.2:

4.8b. Specific liquid holdup and void fraction in structured packing.Repeat Example 4.2, but using Montz metal B1-200 structured packing (very similar to the one

shown in Figure 4.3). For this packing, a = 200 mÐ1, H = 0.979, Ch = 0.547 (Seader and Henley,

1998).

Solution

From Example 4.2:

4.9b. Specific liquid holdup and void fraction in first-generation random packing.A tower packed with 25-mm ceramic Raschig rings is to be used for absorbing benzene vapor from adilute mixture with an inert gas using a wash oil at 300 K. The viscosity of the oil is 2.0 cP and its

density is 840 kg/m3. The liquid mass velocity is L' = 2.71 kg/m2-s. Estimate the liquid holdup, thevoid fraction, and the hydraulic specific area of the packing.

Solution

From Table 4.1:

4.10b. Pressure drop in beds packed with first-generation random packings.Repeat Example 4.3, but using 15-mm ceramic Raschig rings as packing material. Assume that, forthis packing, Cp = 1.783.

Solution

Packed Column Design Program

This program calculates the diameter of a packedcolumn to satisfy a given pressure drop criterium,and estimates the volumetric mass-transfer coefficients.

Enter data related to the gas and liquid streams

Enter liquid flow rate, mL, in kg/s Enter gas flow rate, mG, in kg/s

Enter liquid density, in kg/m3 Enter gas density, kg/m3

Enter liquid viscosity, Pa-s Enter gas viscosity, Pa-s

Enter temperature, T, in K Enter total pressure, P, in Pa

Enter data related to the packing

Enter packing factor, Fp, in ft2/ft3 Enter specific area, a, m2/m3

Introduce a units conversion factor in Fp

Enter porosity, fraction Enter loading constant, Ch

Enter pressure drop constant, Cp Enter allowed pressure drop, in Pa/m

Calculate flow parameter, X

Calculate Y at flooding conditions

Calculate gas velocity at flooding, vGf

As a first estimate of the column diameter, D, design for 70% of flooding

Calculate gas volume flow rate, QG, in m3/s

Calculate liquid volume flow rate, QL, in m3/s

Calculate effective particle size, dp, in m

Iterate to find the tower diameter for the given pressure drop

Column diameter, in meters

Fractional approach to flooding

4.11b. Pressure drop and approach to flooding in structured packing.Repeat Example 4.3, but using Montz metal B1-200 structured packing (very similar to the one

shown in Figure 4.3). For this packing, Fp = 22 ft2/ft3, a = 200 mÐ1, H = 0.979, Ch = 0.547, Cp =

0.355 (Seader and Henley, 1998).

Solution

Packed Column Design Program

This program calculates the diameter of a packedcolumn to satisfy a given pressure drop criterium,and estimates the volumetric mass-transfer coefficients.

Enter data related to the gas and liquid streams

Enter liquid flow rate, mL, in kg/s Enter gas flow rate, mG, in kg/s

Enter liquid density, in kg/m3 Enter gas density, kg/m3

Enter liquid viscosity, Pa-s Enter gas viscosity, Pa-s

Enter temperature, T, in K Enter total pressure, P, in Pa

Enter data related to the packing

Enter packing factor, Fp, in ft2/ft3 Enter specific area, a, m2/m3


Enter porosity, fraction Enter loading constant, Ch

Enter pressure drop constant, Cp Enter allowed pressure drop, in Pa/m

Calculate flow parameter, X

Calculate Y at flooding conditions

Calculate gas velocity at flooding, vGf

As a first estimate of the column diameter, D, design for 70% of flooding

Calculate gas volume flow rate, QG, in m3/s

Calculate liquid volume flow rate, QL, in m3/s

Calculate effective particle size, dp, in m

Iterate to find the tower diameter for the given pressure drop

Column diameter, in meters

Fractional approach to flooding

4.12b, d. Pressure drop in beds packed with second-generation random packings.A packed tower is to be designed for the countercurrent contact of a benzene-nitrogen gas mixturewith kerosene to wash out the benzene from the gas. The gas enters the tower at the rate of 1.5

m3/s, measured at 110 kPa and 298 K, containing 5 mole % benzene. Esentially, all the benzene is

absorbed by the kerosene. The liquid enters the tower at the rate of 4.0 kg/s; the liquid density is

800 kg/m3, viscosity is 2.3 cP. The packing will be 50-mm metal Pall rings, and the tower diameterwill be chosen to produce a gas-pressure drop of 400 Pa/m of irrigated packing.(a) Calculate the tower diameter to be used, and the resulting fractional approach to flooding.(b) Assume that, for the diameter chosen, the irrigated packed height will be 5 m and that 1 m ofunirrigated packing will be placed over the liquid inlet to act as entrainment separator. The blower-motor combination to be used at the gas inlet will have an overall mechanical efficiency of 60%.Calculate the power required to blow the gas through the packing.

Solution

(a) Design for conditions at the bottom of the tower where the maximum flow of gas and liquid occur

Benzene entering with the gas:

Assuming that all of the benzene is absorbed:

From the Lucas method for mixtures of gases:

Using the Packed Column Design Program:

D = 0.913 m f = 0.825

(b) Calculate the pressure drop through the dry packing on top

From the Lucas method for mixtures of gases:

From equation (4-11):

(c) Estimate the volumetric mass-transfer coefficients for the gas and liquid phases. Assume that DL

= 5.0 10Ð10 m2/s.

From the wilke-Lee equation, DG = 0.0885 cm2/s

From Table 4.2, CL = 1.192, CV =

0.410Using the Packed Column Design Program:

kLah = 0.00675 sÐ1; kyah = 0.26 kmol/m3-s

4.13b, d. Pressure drop in beds packed with structured packings.Redesign the packed bed of Problem 4.12, but using Montz metal B1-200 structured packing (very

similar to the one shown in Figure 4.3). For this packing, Fp = 22 ft2/ft3, a = 200 mÐ1, e = 0.979, Ch

= 0.547, Cp = 0.355, CL = 0.971, CV = 0.390 (Seader and Henley, 1998).

Solution

Using the Packed Column Design Program:D = 0.85 m f = 0.859

kLah = 0.00861 sÐ1; kyah = 0.376 kmol/m3-s

4.14c, d. Air stripping of wastewater in a packed column.

A wastewater stream of 0.038 m3/s, containing 10 ppm (by weight) of benzene, is to be stripped

with air in a packed column operating at 298 K and 2 atm to reduce the benzene concentration to0.005 ppm. The packing specified is 50-mm plastic Pall rings. The air flow rate to be used is 5 times

the minimum. Henry's law constant for benzene in water at this temperature is 0.6 kPa-m3/mole(Davis and Cornwell, 1998). Calculate the tower diameter if the gas-pressure drop is not to exceed500 Pa/m of packed height. Estimate the corresponding mass-transfer coefficients. The diffusivity of

benzene vapor in air at 298 K and 1 atm is 0.096 cm2/s; the diffusivity of liquid benzene in water at

infinite dilution at 298 K is 1.02 u 10Ð5 cm2/s (Cussler, 1997).

Solution

Calculate m, the slope of the equilibrium curve:For water at 298 K,

Calculate the minimum air flow rate:Convert liquid concentrations from ppm to mole fractions

At these low concentrations the equilibrium and operating lines are straight, andy2 (max) = y2* = mx2

Using the Packed Column Design Program:D = 1.145 m f = 0.766 kLah = 0.032 sÐ1; kyah = 0.335 kmol/m3-s

4.15b. Stripping chloroform from water by sparging with air.Repeat Example 4.5, but using an air flow rate that is twice the minimum required.

Solution

Initial estimate of the column height, Z

Initial estimate of gas holdup

Calculate power required

4.16b. Stripping chloroform from water by sparging with air.Repeat Example 4.5, but using the same air flow rate used in Problem 4.15, and specifying achloroform removal efficiency of 99%. Solution

For 99% removal efficiency, xin/xout = 100

Z = 1.30 m

WT(Z) = 5.0 kW

4.17b. Stripping chlorine from water by sparging with air.A vessel 2.0 m in diameter and 2.0 m deep (measured from the gas sparger at the bottom to liquidoverflow at the top) is to be used for stripping chlorine from water by sparging with air. The waterwill flow continuously downward at the rate of 7.5 kg/s with an initial chlorine concentration of 5mg/L. Airflow will be 0.22 kg/s at 298 K. The sparger is in the form of a ring, 25 cm in diameter,containing 200 orifices, each 3.0 mm in diameter. Henry's law constant for chlorine in water at this

temperature is 0.11 kPa-m3/mole (Perry and Chilton, 1973). The diffusivity of chlorine at infinite

dilution in water at 298 K is 1.25 u 10Ð5 cm2/s (Cussler, 1997).(a) Assuming that all the resistance to mass transfer resides in the liquid phase, estimate thechlorine removal efficiency achieved.Solution

Iterating until Z = 2.0 m:

(b) Calculate power required

4.18c,d. Batch wastewater aeration using spargers.In the treatment of wastewater, undesirable gases are frequently stripped or desorbed from thewater, and oxygen is adsorbed into the water when bubbles of air are dispersed near the bottom ofaeration tanks or ponds. As the bubbles rise, solute can be transferred from the gas to the liquid orfrom the liquid to the gas depending upon the concentration driving force.

For batch aeration in a constant volume tank, an oxygen mass-balance can be written as

where cA* is the oxygen saturation concentration. Integrating between the time limits zero and t

and the corresponding dissolved oxygen concentration limits cA,0 and cA,t; assuming that cA*

remains essentially constant, and that all the resistance to mass transfer resides in the liquidphase:

In aeration tanks, where air is released at an increased liquid depth, the solubility of oxygen isinfluenced both by the increasing pressure of the air entering the aeration tank and by thedecreasing oxygen partial pressure in the air bubble as oxygen is absorbed. For these cases, theuse of a mean saturation value corresponding to the aeration tank middepth is suggested(Eckenfelder, W. W. Jr., Industrial Water Pollution Control, 3rd ed., McGraw-Hill, Boston, Ma,2000):

where cs = saturation dissolved oxygen concentration in fresh water exposed

to atmospheric air at 101.3 kPa containing 20.9% oxygen,Po = absolute pressure at the depth of air release,

Ps = atmospheric pressure,

Ot = molar oxygen percent in the air leaving the aeration tank.

The molar oxygen percent in the air leaving the aeration tank is related to the oxygen transferefficiency, Oeff, through

where

Consider a 567 m3 aeration pond aerated with 15 spargers, each using compressed air at a rate of0.01 kg/s. Each sparger is in the form of a ring, 100 cm in diameter, containing 20 orifices, each3.0 mm in diameter. The spargers will be located 5 m below the surface of the pond. The watertemperature is 298 K; atmospheric conditions are 298 K and 101.3 kPa. Under these conditions,cs = 8.38 mg/L (Davis and Cornwell, 1998).

(a) Estimate the volumetric mass-transfer coefficient for these conditions from equations (4-23)and (4-25).(b) Estimate the time required to raise the dissolved oxygen concentration from 0.5 mg/L to 6.0mg/L, and calculate the resulting oxygen transfer efficiency.(c) Estimate the power required to operate the 15 spargers, if the mechanical efficiency of thecompressor is 60%.

Solution

Initial estimate of transfer efficiency

(b) Calculate power required

4.19c,d. Batch wastewater aeration using spargers; effect of liquid depth.Consider the situation described in Problem 4.18. According to Eckenfelder (2000), for most typesof bubble-diffusion aeration systems the volumetric mass-transfer coefficient will vary with liquiddepth Z according to the relationship

where the exponent n has a value near 0.7 for most systems. For the aeration pond of Problem4.18, calculate kLa at values of Z = 3m, 4m, 6m, and 7 m. Estimate the corresponding value of n

from regression analysis of the results. Hint: Remember that the total volume of the pond mustremain constant, therefore the cross-sectional area of the pond must change as the water depthchanges.

Solution

Using the program developed in Prob 4.18, the following vector of results is generated

4.20c. Flooding conditions in a packed cooling tower.A cooling tower, 2 m in diameter, packed with 75-mm ceramic Hiflow rings, is fed with water at 316

K at a rate of 25 kg/m2-s. The water is contacted with air, at 300 K and 101.3 kPa essentially dry,drawn upward countercurrently to the water flow. Neglecting evaporation of the water and changes

in the air temperature, estimate the volumetric rate of airflow, in m3/s, which would flood the tower.

Solution

LMV = liquid mass velocity GMV = gass mass velocity


Initial estimate of GMV

4.21c,d. Design of a sieve-tray column for ethanol absorptionRepeat the calculations of Examples 4.6, 4.7, 4.8, and 4.9 for a column diameter corresponding to50% of flooding.

Solution

From the Sieve-Plate Design Program, the following results are obtained for I = 0.5

D = 1.176 m t = 0.6 m 'P = 590 Pa/tray

Froude No. = 1.22 (no excessive weeping) E = 0.0165

EOG = 0.8123 EMG = 0.8865 EMGE = 0.878

4.22c,d. Design of a sieve-tray column for aniline stripping.A sieve-tray tower is to be designed for stripping an aniline (C6H7N)-water solution with steam.

The circumstances at the top of the tower, which are to be used to establish the design, are:Temperature = 371.5 K Pressure = 100 kPaLiquid:Rate = 10.0 kg/s Composition = 7.00 mass % aniline

Density = 961 kg/m3 Viscosity = 0.3 cPSurface tension = 58 dyne/cm

Diffusivity = 4.27 10Ð5 cm2/s (est.) Foaming factor = 0.90Vapor:Rate = 5.0 kg/s Composition = 3.6 mole % aniline

Density = 0.670 kg/m3 Viscosity = 118 mP (est.)

Diffusivity = 0.116 cm2/s (est.)

The equilibrium data at this concentration indicates that m = 0.0636 (Treybal, 1980). (a) Design a suitable cross-flow sieve-tray for such a tower. Take do = 5.5 mm on an equilateral-

triangular pitch 12 mm between hole centers, punched in stainless steel sheet metal 2 mmthick. Use a weir height of 40 mm. Design for a 75% approach to the flood velocity. Reportdetails respecting tower diameter, tray spacing, weir length, gas-pressure drop, and entrainmentin the gas. Check for excessive weeping.

Solution

For aniline, the molecular weight = 93. The average molecular weight of the gas = 20.7.The composition of the liquid is calculated as 1.4 mole % aniline, the average molecular weight= 19.1.

From the Sieve-Plate Design Program, the following results are obtained for f = 0.75

D = 1.93 m t = 0.6 m DP = 435 Pa/tray

Froude No. = 0.93 (no excessive weeping) E = 0.045Weir length = 1.40 m

(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).

Solution

EOG = 0.716 EMG = 0.722 EMGE = 0.720

4.23c,d. Design of a sieve-tray column for aniline stripping.Repeat Problem 4.22, but for a 45% approach to flooding. Everything else remains the same as inProblem 4.22.Solution


D = 2.49 m t = 0.6 m DP = 399 Pa/tray

Froude No. = 0.491 ( excessive weeping) E = 0.0173Weir length = 1.81 m


Solution

EOG = 0.660 EMG = 0.666 EMGE = 0.665

4.24c,d. Design of a sieve-tray column for methanol stripping.A dilute aqueous solution of methanol is to be stripped with steam in a sieve-tray tower. Theconditions chosen for design are:Temperature = 368 K Pressure = 101.3 kPaLiquid:Rate = 0.25 kmol/s Composition = 15.0 mass % methanol

Density = 961 kg/m3 Viscosity = 0.3 cPSurface tension = 40 dyn/cm

Diffusivity = 5.70 10Ð5 cm2/s (est.) Foaming factor = 1.0Vapor:Rate = 0.1 kmol/s Composition = 18 mole % methanol

Viscosity = 125 mP (est.) Diffusivity = 0.213 cm2/s (est.)The equilibrium data at this concentration indicates that m = 2.5 (Perry and Chilton, 1973).

(a) Design a suitable cross-flow sieve-tray for such a tower. Take do = 6.0 mm on an equilateral-

triangular pitch 12 mm between hole centers, punched in stainless steel sheet metal 2 mm thick.Use a weir height of 50 mm. Design for 80% approach to the flood velocity. Report detailsrespecting tower diameter, tray spacing, weir length, gas-pressure drop, and entrainment in thegas. Check for excessive weeping.Solution


D = 1.174 m t = 0.6 m DP = 386 Pa/tray



Solution

EOG = 0.5987 EMG = 0.707 EMGE = 0.700

4.25c,d. Sieve-tray column for methanol stripping; effect of hole size.Repeat Problem 4.24, but changing the perforation size to 4.5 mm, keeping everything elseconstant.Solution


D = 1.174 m t = 0.6 m DP = 686 Pa/tray



Solution

EOG = 0.793 EMG = 0.936 EMGE = 0.918

4.26c,d. Design of sieve-tray column for butane absorption.A gas containing methane, propane, and n-butane is to be scrubbed countercurrently in a sieve-traytower with a hydrocarbon oil to absorb principally the butane. It is agreed to design a tray for thecircumstances existing at the bottom of the tower , where the conditions are:Temperature = 310 K Pressure = 350 kPaLiquid:Rate = 0.50 kmol/s Average molecular weight = 150


Diffusivity = 1.14 10Ð5 cm2/s (est.) Foaming factor = 0.9Vapor:Rate = 0.3 kmol/s Composition = 86% CH4, 12% C3H8, 2% C4H10

Viscosity = 113 mP (est.) Diffusivity = 0.035 cm2/s (est.)The system obeys Raoult's law; the vapor pressure of n-butane at 310 K is 3.472 bar (Reid et al.,1987).

(a) Design a suitable cross-flow sieve-tray for such a tower. According to Bennett and Kovak (2000),the optimal value of the ratio Ah/Aa is that which yields an orifice Froude number, Fro = 0.5. Design

for the optimal value of do on an equilateral-triangular pitch 12 mm between hole centers, punched in

stainless steel sheet metal 2 mm thick. Use a weir height of 50 mm. Design for a 75% approach tothe flood velocity. Report details respecting tower diameter, tray spacing, weir length, gas-pressuredrop, and entrainment in the gas.Solution

The slope of the equilibrium curve is approximately 1.0. The average molecular weight of the gas is20.2. By trial-and-error, using the Sieve-Plate Design Program, it was found that the orificediameter that results in a Froude number = 0.5 is do = 3.43 mm.


D = 2.73 m t = 0.6 m DP = 811 Pa/tray


(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).Solution

EOG = 0.833 EMG = 1.038 EMGE = 1.034

4.27c,d. Design of sieve-tray column for ammonia absorption.A process for making small amounts of hydrogen by cracking ammonia is being considered, andresidual uncracked ammonia is to be removed from the resulting gas. The gas will consist of H2

and N2 in the molar ratio 3:1, containing 3% NH3 by volume. The ammonia will be removed by

scrubbing the gas countercurrently with pure liquid water in a sieve-tray tower. Conditions at thebottom of the tower are:Temperature = 303 K Pressure = 200 kPaLiquid:Rate = 6.0 kg/s Average molecular weight = 18


Diffusivity = 2.42 10Ð5 cm2/s (est.) Foaming factor = 1.0Vapor:Rate = 0.7 kg/s

Viscosity = 113 mP (est.) Diffusivity = 0.230 cm2/s (est.)For dilute solutions, NH3-H2O follows Henry's law, and at 303 K the slope of the equilibrium curve

is m = 0.85 (Treybal, 1980).(a) Design a suitable cross-flow sieve-tray for such a tower. Take do = 4.75 mm on an equilateral-

triangular pitch 12.5 mm between hole centers, punched in stainless steel sheet metal 2 mm thick.Use a weir height of 40 mm. Design for an 80% approach to the flood velocity. Report detailsrespecting tower diameter, tray spacing, weir length, gas-pressure drop, and entrainment in thegas. Check for excessive weeping.Solution


D = 0.775 m t = 0.5 m DP = 624 Pa/tray



EOG = 0.806 EMG = 0.85 EMGE = 0.847

4.28c,d. Design of sieve-tray column for toluene-methylcyclohexane distillation.A sieve-tray tower is to be designed for distillation of a mixture of toluene and methylcyclohexane.The circumstances which are to be used to establish the design, are:Temperature = 380 K Pressure = 98.8 kPaLiquid:Rate = 4.8 mole/s Composition =48.0 mol % tolueneDensity = 726 kg/m3 Viscosity = 0.22 cPSurface tension = 16.9 dyne/cm

Diffusivity = 7.08 10Ð5 cm2/s (est.) Foaming factor = 0.80Vapor:Rate = 4.54 mole/s Composition =44.6 mole % tolueneDensity = 2.986 kg/m3 Viscosity = 337 mP (est.)

Diffusivity = 0.0386 cm2/s (est.)The equilibrium data at this concentration indicates that m = 1.152. (a) Design a suitable cross-flow sieve-tray for such a tower. Take do = 4.8 mm on an equilateral-

triangular pitch 12.7 mm between hole centers, punched in stainless steel sheet metal 2 mm thick.Use a weir height of 50 mm. Design for a 60% approach to the flood velocity. Report detailsrespecting tower diameter, tray spacing, weir length, gas-pressure drop, and entrainment in thegas. Check for excessive weeping.


Solution

a) From the Sieve-Plate Design Program, the following results are obtained for I = 0.60

D = 0.6 m t = 0.5 m DP = 295 Pa/tray



EOG = 0.611 EMG = 0.712 EMGE = 0.710

5.1a. Overall tray efficiencyConsider a tray absorber with a constant Murphree efficiency EMGE = 0.75, and an average

absorption factor A = 1.25.a) Estimate the overall tray efficiency.

Solution

b) If the absorber requires 5.34 equilibrium stages, calculate the number of real trays.

Solution

Use 8 trays

5.3a. Ammonia stripping from a wastewater in a tray tower.A tray tower providing six equilibrium stages is used for stripping ammonia from a wastewaterstream by means of countercurrent air at 1 atm and 300 K. Calculate the concentration ofammonia in the exit water if the inlet liquid concentration is 0.1 mole % ammonia, the inlet air isfree of ammonia, and 1.873 standard cubic meter of air are fed to the tower per kilogram ofwastewater. The equilibrium data for this system, in this range of concentrations and 300 K, canbe represented by yA,i= 1.414 xA,i (King, C. J., Separation Processes, McGraw-Hill, New York,

NY, 1971).

Solution

Initial estimate

5.4a. Ammonia stripping from a wastewater in a tray tower.The Murphree plate efficiency for the ammonia stripper of Problem 5.3 is constant at 0.581.Estimate the number of real trays required.

Solution

Use 9 trays

5.5b. Ammonia stripping from a wastewater in a tray tower.

If the air flow rate to the absorber of Problems 5.3 and 5.4 is reduced to 1.5 standard m3/kg of

water, calculate concentration of ammonia in the exit water if 9 real trays are used and theMurphree efficiency remains constant at 58.1%.

Solution

Initial estimate

5.6b. Absorption of an air pollutant in a tray tower.A heavy-oil stream at 320 K is used in an absorber to remove dilute quantities of pollutant A froman air stream. The heavy oil is then recycled back to the process where where A is stripped. Theprocess is being run on a pilot plan basis, and information for scale-up is desired. The currentabsorber is a 16-plate sieve-tray column. Pilot plat data are as follows:Liquid flow rate = 5.0 moles/hrGas flow rate = 2.5 moles/hryA,in = 0.04 yA,out = 0.0001 xA,in = 0.0001

Equilibrium for A is given as yA,i = 0.7xA,i. Find the overall column efficiency, and the Murphree

plate efficiency. assume that the liquid and gas flow rates are roughly constant.

Solution

5.7b. Absorption of ammonia in a laboratory-scale tray tower.An absorption column for laboratory use has been carefully constructed so that it has exactly 4equilibrium stages, and is being used to measure equilibrium data. Water is used as the solventto absorb ammonia from air. The system operates isothermally at 300 K and 1 atm. The inletwater is pure distilled water. The ratio of L/V = 1.2, inlet gas concentration is 0.01 mole fractionammonia; the measured outlet gas concentration is 0.0027 mole fraction ammonia. Assumingthat HenryÕs law applies, calculate the slope of the equilibrium line.

Solution

Initial estimate

5.8c,d. Absorption of ammonia in a seive-tray tower.A process for making small amounts of hydrogen by cracking ammonia is being consideredand residual, uncracked ammonia is to be removed from the resulting gas. The gas will consistof H2 and N2 in the molar ratio 3:1, containing 3% NH3 by volume, at a pressure of 2 bars and

a temperature of 303 K.There is available a sieve-tray tower, 0.75 m diameter, containing 6 cross-flow trays at 0.5 mtray spacing. The perforations are 4.75 mm in diameter, arranged in triangular pitch on 12.5mm centers, punched in sheet metal 2 mm thick. The weir height is 40 mm. Assumeisothermal scrubbing with pure water at 303 K. The water flow rate to be used should not

exceed 50% of the maximum recommended for cross-flow sieve trays which is 0.015 m3/s-mof tower diameter (Treybal, 1980). The gas flow rate should not exceed 80% of the floodingvalue.

a) Estimate the gas flow rate that can be processed in the column under the circumstancesdescribed above.b) Calculate the concentration of the gas leaving the absorber in part a).Data:

Liquid density = 996 kg/m3 Surface tension = 0.068 N/m

Foaming factor = 1.0 Liquid diffusivity = 2.42 u 10Ð9 m2/s

Gas viscosity = 1.122 u 10Ð5 kg/m-s Gas diffusivity = 0.23 cm2/sSlope of the equilibrium line, P = 0.85

Solution

a)

Run the sieve-tray design program of Appendix E with different valuesof gas flow rate until D = 0.75 m at f = 0.8. The results are:

mG = 0.657 kg/s Gas pressure drop = 615 Pa per tray

EMGE = 0.841 Fro = 1.042 E = 0.048

b)

Initial estimate

5.9b. Absorption of carbon dioxide in a bubble-cap tray tower.A plant manufacturing dry ice will burn coke in air to produce a flue gas which, when cleaned andcooled, will contain 15% CO2, 6% O2, and 79% N2. The gas will be blown into a bubble-cap tower

scrubber at 1.2 atm and 298 K, to be scrubbed countercurrently with a 30 wt% monoethanolamine(C2H7ON) aqueous solution entering at 298 K. The scrubbing liquid, which is recycled from a

stripper, will contain 0.058 mole CO2/mol solution. The gas leaving the scrubber is to contain 2%

CO2. A liquid-to-gas ratio of 1.2 times the minimum is specified. Assume isothermal operation. At

298 K and 1.2 atm, the equilibrium mole fraction of carbon dioxide over aqueous solutions ofmonoethanolamine (30 wt%) is given by

where xA,i is the mole fraction of CO2 in the liquid solution.

a) Calculate the kilograms of solution entering the tower per cubic meter of entering gas.

Solution

For the minimum liquid flow rate, x1max will be in equilibrium with yN1

Initial estimate

Calculate molecular weight of MEA solutionBasis: 100 gm of 30% MEA in water

b) Determine the number of theoretical trays required for part a)

Solution

Generate XY diagram

c) The monoethanolamine solution has a viscosity of 6.0 cP and a density of 1,012 kg/m3.Estimate the overall tray efficiency for the absorber, and the number of real trays required.Seader and Henley (1998) proposed the following empirical correlation to estimate the overallefficiency of absorbers and strippers using bubble-cap trays (it has also been used to obtainrough estimates for sieve-tray towers):

where: EO = overall fractional efficiency

m = slope of equilibrium curvemL = liquid viscosity, in cP

L = liquid density, in kg/m3

Hint: In this problem, the equilibrium-distribution curve is not a straight line, therefore m is notconstant. Estimate thhe average value of m at liquid concentrations along the operating lineand use theaverage in the correlation given.

Solution

5.11c,d. Absorption of carbon disulfide in a sieve-tray tower.Carbon disulfide, CS2, used as a solvent in a chemical plant, is evaporated from the product in a

dryer into an inert gas (essentially N2) in order to avoid an explosion hazard. The CS2-N2 mixture

is to be scrubbed with an absorbent hydrocarbon oil. The gas will flow at the rate of 0.4 m3/s at297 K and 1 atm. The partial pressure of CS2 in the original gas is 50 mm Hg, and the CS2

concentration in the outlet gas is not to exceed 0.5%. The oil enters the absorber essentially pureat a rate 1.5 times the minimum, and solutions of oil and CS2 follow RaoultÕs law. Design a

sieve-tray tower for this process. Design for a gas velocity which is 70% of the flooding velocity.Assuming isothermal operation, determine:a) Liquid flow rate, kg/s

Data (at 297 K):Oil average molecular weight = 254 Oil viscosity = 4 cP

Oil density = 810 kg/m3 Surface tension = 0.030 N/mFoaming factor = 0.9 CS2 vapor pressure = 346 mm Hg

Solution

Initial estimates

b) Tower diameter and plate spacing

Solution

Plate design conditions:f = 0.7 4.5-mm holes in triangular pitch; 12 mm between centersWeir height = 50 mm; Plate thickness = 2 mm

From the Mathcad program in Appendix E:D = 0.705 m; t = 0.5 m; Pressure drop = 414 Pa/trayFro = 0.712 E = 0.018 EMGE = 0.588

d) Number of real trays required

Solution

Approximation, assuming straight equilibrium and operating lines

Graphical solution: Generate operating line in xy diagram

Graphically, the number of ideal stages is also slightly over 5

Use 10 trays

e) Total gas-pressure drop.

Solution

5.12c. Adiabatic absorption of carbon disulfide in a sieve-tray tower.Determine the number of equilibrium trays for the absorber of Problem 5.11, assuming adiabaticoperation. The specific heats are:Substance J/mole-KOil 362.2CS2, liquid 76.2

N2 29.1

CS2, gas 46.9

the latent heat of vaporization of CS2 at 297 K is 27.91 kJ/mole. The vapor pressure of CS2 as a

function of temperature is given by

Solution

Base for enthalpies: liquid oil, liquid CS2, gaseous N2 at 297 K

Assume T1 = 298.0 K

Approximately 12 ideal stages

Gas outlet temperature = 298 K

5.13b. Steam-stripping of benzene in a sieve-plate column.A straw oil used to absorb benzene from coke-oven gas is to be steam-stripped in a sieve-platecolumn at atmospheric pressure to recover the disolved benzene, C6H6. Equilibrium conditions at

the operating temperature are approximated by HenryÕs law such that, when the oil phase contains10 mole % benzene, the equilibrium benzene partial pressure above the oil is 5.07 kPa. The oil maybe considered nonvolatile. It enters the stripper containing 8 mole % benzene, 75% of which is to beremoved. The steam leaving contains 3 mole % C6H6.a) How many theorethical stages are required?b) How many moles of steam are required per 100 moles of the oil-benzene mixture?c) If 85% of the benzene is to be recovered with the same steam and oil rates, how many theoreticalstages are required?

Solution a)

b)

Kremser equation

c)

(Graphical construction ahead)

Kremser equation

5.15c,d. Absorption of carbon disulfide in a random-packed tower.Design a tower packed with 50-mm ceramic Hiflow rings for the carbon disulfide scrubber ofProblem 5.11. Assume isothermal operation and use a liquid rate of 1.5 times the minimum anda gas-pressure drop not exceeding 175 Pa/m of packing. Calculate the tower diameter, packedheight, and total gas-pressure drop. Assume that Ch for the packing is 1.0.

Solution

From Prob. 5.11, the conditions at the bottom of the absorber are:

From Tables 4.1 and 4.2

Run the Packed Tower Design Program of Ap. DResults:

Estimate the packed height; assume dilute solutions

From Prob 5.11:

Calculate HtOG at the bottom of the tower

Calculate HtOG at the top of the tower

From Prob. 5.11, the conditions at the top of the absorber are:

The mass-tranfer coefficients remain fairly constant along the tower

5.16b,d. Absorption of sulfur dioxide in a random-packed tower.It is desired to remove 90% of the sulfur dioxide in a flue gas stream at 298 K and 1 atm bycountercurrent absorption with pure water at the same temperature, using a packed tower thatis 0.7 m in diameter. The tower is packed with 35-mm plastic NORPAC rings. The average gas-pressure drop is 200 Pa/m. Equilibrium is described by HenryÕs law with yi = 8.4xi.. If the

liquid flow is adjusted so that the driving force (y Ð y*) is constant, calculate the height of thepacked section. Flue gases usually contain less than 1 mole % of SO2, an air pollutant

regulated by law. Assume that the properties of the liquid are similar to those of pure water,and that the properties of the flue gases are similar to those of air.

Solution

For the driving force to be constant, the operating line is parallel to the equilibrium line, andthe absorption factor A = 1.0

It was found by trial and error with the Packed Column program that, for a pressure drop of200 Pa/m and D = 0.7 m:

5.17c. Benzene vapor recovery system.Benzene vapor in the gaseous effluent of an industrial process is scrubbed with a wash oil in acountercurrent packed absorber. The resulting benzene-wash-oil solution is then heated to 398K and stripped in a tray tower, using steam as the stripping medium. The stripped wash oil isthen cooled and recycled to the absorber. Some data relative to the operation follow:

Absorption:Benzene in entering gas = 1.0 mole %Operating pressure of absorber = 800 mm Hg

Oil circulation rate = 2 m3/1,000 m3 of gas at STPOil specific gravity = 0.88 Molecular weight = 260HenryÕs law constant = 0.095 at 293 K; = 0.130 at 300 KNtOG = 5 transfer units

Stripping:Pressure = 1 atm Steam at 1 atm, 398 KHenryÕs law constant = 3.08 at 398 KNumber of equilibrium stages = 5a) In the winter, it is possible to cool the recycled oil to 293 K, at which temperature theabsorber then operates. Under these conditions 72.0 kg of steam is used in the stripper per

1,000 m3 of gas at STP entering the absorber. Calculate the percent benzene recovery in thewinter.

Solution

Initial estimates:

b) In the summer it is impossible to cool the recycled wash oil to lower than 300 K with theavailable cooling water. Assuming that the absorber then operates at 300 K, with the same oiland steam rates, and that NtOG and equilibrium stages remain the same, what summer

recovery of benzene can be expected?

Solution

c) If the oil rate cannot be increased, but the steam rate in the summer is increased by 50%over the winter value, what summer recovery of benzene can be expected?

5.19c. Absorption of germanium tetrachloride used for optical fibers.Germanium tetrachloride (GeCl4) and silicon tetrachloride (SiCl4) are used in the production ofoptical fibers. Both chlorides are oxidized at high temperature and converted to glasslike particles.However, the GeCl4 oxidation is quite incomplete and it is necessary to scrub the unreacted GeCl4from its air carrier in a packed column operating at 298 K and 1 atm with a dilute caustic solution.At these conditions, the dissolved GeCl4 has no vapor pressure and mass transfer is controlled bythe gas phase. Thus, the equilibrium curve is a straight line of zero slope. The entering gas flows atthe rate of 23,850 kg/day of air containing 288 kg/day of GeCl4. The air also contains 540 kg/day ofCl2, which, when dissolved, also will have no vapor pressure.It is desired to absorb, at least, 99% of both GeCl4 and Cl2 in an existing 0.75-m-diameter columnthat is packed to a height of 3.0 m with 13-mm ceramic Raschig rings. The liquid rate should beset so that the column operates at 75% of flooding. Because the solutions are very dilute, it can beassumed that both gases are absorbed independenttly.Gas-phase mass-transfer coefficients for GeCl4 and Cl2 can be estimated from the followingempirical equations developed from experimental studies with 13-mm Raschig rings for liquid mass

velocities between 0.68Ð2.0 kg/m2-s (Shulman, H. L., et al., AIChE J., 17, 631, 1971):

where:ds = equivalent packing diameter (0.01774 m for

13-mm ceramic Raschig rings) = gas density, kg/m3

Gx,Gy mass velocities, kg/m2-s

For the two diffusing species, take DGeCl4 = 0.06 cm2/s; DCl2 = 013 cm2/s. For the packing,

Fp = 580 ftÐ1, e = 0.63.

Determine:a) Liquid flow rate, in kg/s.

Solution

Initial estimate

b) The percent absorption of GeCl4 and Cl2 based on the available 3.0 m of packing.

Solution (dilute solutions)

5.20c,d. Absorption of carbon disulfide in a structured-packed tower.Redesign the absorber of Problem 5.15 using metal Montz B1-300 structured packing. Thecharacteristics of this packing are (Seader and Henley, 1998):

Fp = 33 ftÐ1 a = 300 mÐ1 H = 0.93 Ch = 0.482

Cp = 0.295 CL = 1.165 CV = 0.422

Solution

From the Packed Tower Program in Appendix D:

Estimate the packed height; assume dilute solutions

From Prob 5.11:

Calculate HtOG at the bottom of the tower

Calculate HtOG at the top of the tower

From Prob. 5.11, the conditions at the top of the absorber are:

The mass-tranfer coefficients remain fairly constant along the tower

5.21c,d. Absorption of ammonia in a random-packed tower.

It is desired to reduce the ammonia content of 0.05 m3/s of an ammonia-air mixture (300 K and 1atm) from 5.0 to 0.04% by volume by water scrubbing. There is available a 0.3-m-diameter towerpacked with 25-mm ceramic Raschig rings to a depth of 3.5 m. Is the tower satisfactory, and if so,what water rate should be used? At 300 K, ammonia-water solutions follow HenryÕs law up to 5mole % ammonia in the liquid, with m = 1.414.

Solution

Calculate minimum water flow rate

Try

From Appendix D

Try

From Appendix D

Try

From Appendix D

Cubic spline interpolation

5.22b,d. Absorption of sulfur dioxide in a structured-packed tower.A tower packed with metal Montz B1-300 structured packing is to be designed to absorb SO2

from air by scrubbing with water. The entering gas, at an SO2-free flow rate of 37.44 moles/m2-s

of bed cross-sectional area, contains 20 mole % of SO2. Pure water enters at a flow rate of 1976

moles/m2-s of bed cross-sectional area. The exiting gas is to contain only 0.5 mole % SO2.

Assume that neither air nor water will transfer between the phases and that the tower operatesisothermally at 2 atm and 303 K. Equilibrium data for solubility of SO2 in water at 303 k and 1

atm have been fitted by least-squares to the equation (Seader and Henley, 1998):

a) Derive the following operating line equation for the absorber:

Solution

b) If the absorber is to process 1.0 m3/s (at 2 atm and 303 K) of the entering gas, calculatethe water flow rate, the tower diameter, and the gas-pressure drop per unit of packing heightat the bottom of the absorber.

Solution

From Ap. D, by trial-and-error:

c)

At the top of the column:

From Ap. D, by trial-and-error:

Average mass-transfer coefficients

Average height

Calculate NtG

Initial estimate

6.1a. Flash vaporization of a heptane-octane mixture.A liquid mixture containing 50 mole % n-heptane (A), 50 mole % n-octane (B), at 303 K, is to becontinuously flash-vaporized at a pressure of 1 atm to vaporize 30 mole % of the feed. What willbe the composition of the vapor and liquid and the temperature in the separator if it behaves asan ideal stage?

Solution

From the program in Figure 6.3,

6.2a. Flash vaporization of a heptane-octane mixture.A liquid mixture containing 50 mole % n-heptane (A), 50 mole % n-octane (B), at 303 K, is to becontinuously flash-vaporized at a temperature of 350 K to vaporize 30 mole % of the feed. Whatwill be the composition of the vapor and liquid and the pressure in the separator if it behaves as anideal stage?

Solution

From the program in Figure 6.3,

6.3d. Flash vaporization of a ternary mixture.Modify the Mathcad¨ program of Figure 6.3 for ternary mixtures. Test your program with the datapresented in Example 6.2 for a mixture of benzene (A), toluene (B), and o-xylene (C). Criticaltemperatures and pressures,and the parameters of the Wagner equation for estimating vaporpressure (equation 6-5) are included in the following table (Reid, et al., 1987).

Component Tc, K Pc, bar A B C Dbenzene 562.2 48.9 Ð 6.983 1.332 Ð 2.629 Ð 3.334toluene 591.8 41.0 Ð 7.286 1.381 Ð 2.834 Ð 2.792o-xylene 630.3 37.3 Ð 7.534 1.410 Ð 3.110 Ð 2.860

Solution

Parameters of the Wagner equation, A = benzene, B, toluene, C = o-xylene:

Initial estimates

6.4d. Flash vaporization of a ternary mixture.Consider the ternary mixture of Example 6.2 and Problem 6.3. Estimate the temperature, andcomposition of the liquid and vapor phases when 60% of the mixture has been vaporized at a constantpressure of 1 atm.

Solution

6.5d. Flash vaporization of a ternary mixture.Consider the ternary mixture of Example 6.2 and Problem 6.3. It is desired to recover in the vapor 75%of the benzene in the feed, and to recover in the liquid 70% of the o-xylene in the feed. Calculate thetemperature, pressure, fraction of the feed vaporized, and the concentration of the liquid and gas phases

Solution

6.6b.Batch distillation of a heptane-octane mixture.Repeat the calculations of Example 6.3, but for 80 mole % of the liquid distilled.

Solution

Batch distillation of a mixture of heptane and octane

Initial estimate

6.8a. Binary batch distillation with constant relative volatility.Consider the binary batch distillation of Example 6.3. For this system, n-heptane with n-octane at 1atm, the average relative volatility is D = 2.16 (Treybal, 1980). Using equation (6-102) derived in Problem6.7, compute the composition of the residue after 60 mole% of the feed is batch-distilled.

Solution

Initial estimate:

6.9b. Binary batch distillation with constant relative volatility.Consider the binary batch distillation of Example 6.3. For this system, n-heptane with n-octane at 1atm, the average relative volatility is D = 2.16 (Treybal, 1980). The mixture will be batch-distilled untilthe average concentration of the distillate is 65 mole % heptane. Using equation (6-102) derived in

Problem 6.7, compute the composition of the residue, and the fraction of the feed that is distilled.

Solution

Initial estimates:

6.10b. Mixtures of light hydrocarbons: m-value correlations.

What is the bubble point of a mixture that is 15 mole % isopentane, 30 mole % n-pentane, and 55mole % n-hexane? The pressure is 1.0 atm.

Solution

Initial estimate

6.11c. Mixtures of light hydrocarbons: m-value correlations.A solution has the following composition, expressed as mole percent: ethane, 0.25 %; propane, 25 %;isobutane, 18.5 %; n-butane, 56 %; isopentane, 0.25 %. In the following, the pressure is 10 bars. Useequation (6-103) and Table 6.4 to calculate equilibrium distribution coefficients.a) Calculate the bubble point

Solution

Initial estimate

b) Calculate the dew point

Solution

Initial estimate

c) The solution is flash-vaporized to vaporize 40 mol % of the feed. Calculate the composition of theproducts.

Solution

Initial estimates

Check!

6.12b. Binary batch distillation with constant relative volatility.A 30 mole % feed of benzene in toluene is to be distilled in a batch operation. A product having an

average composition of 45 mole % benzeneis to be produced. Calculate the amount of residue left,assuming that D = 2.5 and F = 100 moles. Solution

Initial estimates:

6.13b. Batch distillation of a mixture of isopropanol in water.A mixture of 40 mole % isopropanol in water is to be batch-distilled at 1 atm until 70 mole % of thecharge has been vaporized. Calculate the composition of the liquid residue remaining in the still pot,and the average composition of the collected distillate. VLE data for this system, in mole fraction ofisopropanol, at 1 atm are (Seader and Henley, 1998):T, K 366 357 355.1 354.3 353.6 353.2 353.3 354.5y 0.220 0.462 0.524 0.569 0.593 0.682 0.742 0.916x 0.012 0.084 0.198 0.350 0.453 0.679 0.769 0.944Composition of the azeotrope is x = y = 0.685; boiling point of the azeotrope = 353.2 K.

Solution

Initial estimate

6.15a. Continuous distillation of a binary mixture of constant relative volatility.For continuous distillation of a binary mixture of constant relative volatility, Fenske equation (6-58) canbe used to estimate the minimum number of equilibrium stages required for the given separation, Nmin.

Use Fenske equation to estimate Nmin for distillation of the benzene-toluene mixture of Example 6.4.

Assume that, for this system at 1 atm, the relative volatility is constant at D = 2.5.

Solution

6.16a. Continuous distillation of a binary mixture of constant relative volatility.For continuous distillation of a binary mixture of constant relative volatility, the minimum reflux ratio canbe determined analytically from the following equation (Treybal, 1980):

(6-105)

Use equation (6-105) to estimate Rmin for distillation of the benzene-toluene mixture of Example 6.4.

Assume that, for this system at 1 atm, the relative volatility is constant at D = 2.5.

Solution

Initial estimate

6.17b. Continuous rectification of a water-isopropanol mixture.A water-isopropanol mixture at its bubble point containing 10 mole % isopropanol is to becontinuously rectified at atmospheric pressure to produce a distillate containing 67.5 mole %isopropanol. Ninety-eight percent of the isopropanol in the feed must be recovered. VLE data aregiven in Problem 6.13. If a reflux ratio of 1.5 times the minimum is used, how many theoreticalstages will be required: (a) If a partial reboiler is used?

Solution

6.24c. A distillation-membrane hybrid for ethanol dehydration.Many industrially important liquid systems are difficult or impossible to separate by simple continuousdistillation because the phase behavior contains an azeotrope, a tangent pinch, or an overall low relativevolatility. One solution is to combine distillation with one or more complementary separation technologiesto form a hybrid. An example of such a combination is the dehydration of ethanol using a distillation-membrane hybrid.In a given application, 100 moles/s of a saturated liquid containing 37 mole % ethanol and 63 mole % watermust be separated to yield a product which is 99 mole % ethanol, and a residue containing 99 mole %water. The solution will be fed to a distillation column operating at atmospheric pressure, with a partialreboiler and a total condenser. The reflux ratio will be 1.5 the minimum. The distillate will enter a membranewith parameters aP = 70 and = 0.6. The membrane boosts the concentration so that the permeate

stream is the ethanol-rich product (xP = 0.99). The retentate stream is returned as a saturated liquid to the

column to the tray at the nearest liquid concentration.a) Calculate the molar flow rate of the product and of the residue.

Solution

Initial estimates:

b) Calculate the molar flow rate and composition of the distillate coming out of the column.

Solution

Initial estimates

6.26b. Fenske-Underwood-Gilliland method.A distillation column has a feed of 100 kmoles/h. The feed is 10 mole % LNK, 55 mole % LK, and 35mole % HK and is a saturated liquid. The reflux ratio is 1.2 the minimum. We desire 99.5 % recovery ofthe light key in the distillate. The mole fraction of the light key in the distillate should be 0.75. Use theFUG approach to estimate the number of ideal stages required and the optimal location of the feed-stage.Equilibrium data:DLNK = 4.0 DLK = 1.0 DHK = 0.75

Solution

Kilomoles in distillate:

Kilomoles in bottoms:

Determine minimum reflux: Saturated liquid, q = 1

Initial estimate

Determine number of ideal stages at R = 1.2 Rmin: Gilliland correlation

Initial estimate

Kirkbride equation:

Initial estimate:

Feed at stage 11

6.27b. Fenske-Underwood-Gilliland method.One hundred kmoles/h of a ternary bubble-point mixture to be separated by distillation has thefollowing composition:

Component Mole fraction Relative volatilityA 0.4 5B 0.2 3C 0.4 1a) For a distillate rate of 60 kmoles/h, five theoretical stages, and total reflux, calculate the distillateand bottoms composition by the Fenske equation.

Solution

Initial estimates:

b) Using the separation in part a) for components A and C, determine the minimum reflux ratio by theUnderwood equation.

Solution

Initial estimate:

Initial estimate:

Initial estimate:

c) For a reflux ratio of 1.2 times the minimum, determine the number of theoretical stagesrequired, and the optimum feed location.

Solution

Initial estimate

Kirkbride equation:

Initial estimate:

Feed at stage 5

6.35c Flash Calculations: the Rachford-Rice Method for Ideal Mixtures.a) Show that the problem of flash vaporization of a multicomponent ideal mixture can be reformulated assuggested by Rachford and Rice (Doherty and Malone, Conceptual Design of Distillation Systems,McGraw-Hill, New York, NY, 2001):

b) Solve Example 6.2 using the Rachford-Rice method.

Solution

Initial estimate:

7.2 Single extraction.

Appendix G-1: Single-Stage Extraction

Initial estimates:

7.3. Single-stage extraction: insoluble liquids�

7.4. Single-stage extraction: insoluble liquids�

7.5 Multistage crosscurrent extraction: insoluble liquids.

Rearranging equation (3-60)

Solvent flow rate per stage

Total solvent flow rate

7.6 Multistage crosscurrent extraction.

Appendix G-2: Multistage Crosscurrent Extraction

Data presented are from Example 7-3: acetone-water-chloroform system at 298 K

Initial estimates:

Initial guesses:

Initial guesses:

Composited extract

Check material balances

Acetone

Chloroform

Water

Mass Transfer Solutions

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