Solution Manual for Mass Transfer Processes: Modeling, Computations, and Design, 1st Edition

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Solution Manual toMass Transfer ProcessesP. A. Ramachandranrama@wustl.eduMarch 25, 20181Companion to Mass Transfer Processes, P. A. Ramachandran.

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Contents0.1Introductionpage61Chapter 171.1Answers to Review Questions71.2Solutions to Problems102Chapter 2212.1Answers to Review Questions212.2Solutions to Problems233Chapter 3353.1Answers to Review Questions353.2Solutions to Problems374Chapter 4484.1Answers to Review Questions484.2Solutions to Problems505Chapter 5585.1Answers to Review Questions585.2Solutions to Problems606Chapter 6676.1Answers to Review Questions676.2Solutions to Problems697Chapter 7777.1Answers to Review Questions777.2Solutions to Problems808Chapter 8908.1Answers to Review Questions908.2Solutions to Problems923Companion to Mass Transfer Processes, P. A. Ramachandran.

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4Contents9Chapter 91059.1Answers to Review Questions1059.2Solutions to Problems10710Chapter 1011710.1 Answers to Review Questions11710.2 Solutions to Problems11911Chapter 1112911.1 Answers to Review Questions12911.2 Solutions to Problems13212Chapter 1213712.1 Answers to Review Questions13712.2 Solutions to Problems14013Chapter 1314613.1 Answers to Review Questions14613.2Solutions to Problems14814Chapter 1415514.1 Answers to Review Questions15514.2 Solutions to Problems15715Chapter 1517015.1 Answers to Review Questions17015.2 Solutions to Problems17216Chapter 1617916.1 Answers to Review Questions17916.2 Solutions to Problems18217Chapter 1719217.1 Answers to Review Questions19217.2 Solutions to Problems19318Chapter 1820018.1 Answers to Review Questions20018.2 Solutions to Problems20319Chapter 1922119.1 Answers to Review Questions22119.2 Solutions to Problems223Companion to Mass Transfer Processes, P. A. Ramachandran.

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Contents520Chapter 2022820.1 Answers to Review Questions22820.2 Solutions to Problems23021Chapter 2124021.1 Answers to Review Questions24021.2 Solutions to Problems24222Chapter 2225122.1 Answers to Review Questions25122.2 Solutions to Problems25323Chapter 2325723.1 Answers to Review Questions25723.2 Solutions to Problems25924Chapter 2426824.1 Answers to Review Questions26824.2 Solutions to Problems27125Chapter 2527625.1 Answers to Review Questions27625.2 Solutions to Problems27826Chapter 2628726.1 Answers to Review Questions28726.2 Solutions to Problems29027Chapter 2729627.1 Answers to Review Questions29627.2 Solutions to Problems29828Chapter 2830428.1 Answers to Review Questions30428.2 Solutions to Problems30729Chapter 2931729.1 Answers to Review Questions31729.2 Solutions to Problems31930Chapter 3032530.1 Answers to Review Questions32530.2 Solutions to Problems327Companion to Mass Transfer Processes, P. A. Ramachandran.

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6Contents0.1IntroductionThis solution manual contains answers to review questions for all chapters andsolutions to most problems in the book.If any clarification is needed, please feel free to e-mail me. The feedback frominstructions is most appreciated together with pointing out any errors which willgo into updating this solution book. I will be very happy to supply additionaldetails on any of these problems if needed.Please note that I have taken care in the preparation of this manual, but Imake no expressed or implied warranty of any kind and assume no responsibilityfor errors or omissions. No liability is assumed for incidental or consequentialdamages in connection with or arising out of the use of the information containedherein.Companion to Mass Transfer Processes, P. A. Ramachandran.

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1Chapter 11.1Answers to Review Questions1.What is meant by concentration jump?The concentration becomes discontinuous as we cross from one phase toanother and this known as the concentration jump.2.What is meant by the continuum assumption and what are its implications inmodel development?Continuum assumption assumes that the matter is continuously distributedin space and ignores the atomic/molecular nature of matter. This permitsus to assign a point value to the variables such as concentration and so on.The information on the interaction of molecules is however lost and has to besupplemented by suitable constitutive model.3.Indicate some situations where the continuum models are unlikely to apply.Continuum models are unlikely to apply if there are not sufficient number ofmolecules in the volume of interest. The number should be large so that thestatistical average can be assigned to the ensemble of molecules. For examplein high vacuum system or a plasma reactor the number of molecules are smalland continuum description may not be adequate.4.Give an example of a system where the total concentration is (nearly) constant.Gas mixture at constant total pressure and temperature.5.Give an example of a system where the mixture density is nearly constant.Liquid mixtures of compounds with similar chemical nature.6.Can average molecular weight be a function of position?Yes. Since in a diffusing binary system, the mole fraction of A varies alongthe position and correspondingly the molecular weight will vary as a functionof position.7.What information is missing in the differential models based on the continuumassumption?Information on the transport rate caused by molecular motion is missing inthe context of continuum models.8.Why are constitutive models needed in the context of differential models?Molecular level transport occurs due molecular motion and is not modeledin the continuum level of modeling. In order to incorporate these effects, aconstitutive model is needed to quantify the diffusion flux.7Companion to Mass Transfer Processes, P. A. Ramachandran.

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8Chapter 1.Chapter 19.Does the Fick’s law apply universally to all systems?No. It is specific since it is the result of the of molecular level interactions.These are system specific. It is an accurate model for binary gas mixture atlow or moderate pressures. However it is commonly used as a first level modelfor a large class of problems.10.Write a form of Fick’s law using partial pressure gradient as the driving force.UsingCA=pA/RgTwe can write Fick’s law using partial pressure gradientas:JAx=−DARgTdpAdx11.What is meant by the invariant property of the flux vector?A vector remains same if the coordinates are rotated or if different coordinatesystem (e.g., cylindrical) is used and this is known as the invariance propertyof the flux vector.12.The combined flux is partitioned into the sum of the convection and diffusionflux. Is this partitioning unique?No; It is not unique. It will depends on how the mixture velocity is definedand thereby what part of the combined flux is allocated to the convectionflux.13.State the units of the gas constant if the pressure is expressed in bar insteadof Pa, (ii) if expressed in atm.Gas constant will have a value of 8.205×10−5m3atm/mol K if pressure is expressed atm.It will have a value of 8.314×10−5if pressure is in bars.14.Express the dissolved oxygen concentration in Example in p.p.m.The concentration of dissolved oxygen was calculated as 0.2739 mol/m3. Thiscan be converted to gm/gm using the molecular weight (32 g/g) and alsousing the liquid density of water. The result is 8.467×10−6. This correspondsto 8.467 p.p.m.15.What is a macroscopic level model? What information needs to be added herein addition to the conservation principle?A larger control volume or the whole reactor or separator is taken as thecontrol volume in a macroscoipc model. The control volume does not tend tozero. Hence the local information is lost and needs to be added as additionalclosure information in addition to the conservation law.16.How is mass transfer coefficient defined? Why is it needed?The flux acrossa surface is not available in the meso- or macro-models since it depends onthe local concentration gradient in accordance to Fick’s law. Hence the flux isrepresented as a product of a mass transfer coefficient and a suitably defineddriving force. It it therefore needed in the meso- and macro- scale model.17.What is meant by a cross-sectional average concentration? What is a cupmixing or the bulk concentration?Companion to Mass Transfer Processes, P. A. Ramachandran.

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Chapter 19In the cross-section average the local concentration is weighted by the localarea and integrated over the cross-section. One then divides this by the totalarea to get the cross-sectional average.In the cup mixing average, the local concentration is weighted by the localvolumetric flow rate and integrated over the cross-section. One the dividesthis by the total volumetric flow rate to get the cup mixing concentration.18.What assumptions are involved in plug flow model?The cup mixing concentration is assumed to be the same as the cross-sectionalaverage concentration for a plug flow idealization.19.What assumptions are involved in a completely backmixed model?The average concentration in the reactor and the exit concentration areassumed to be the same in a completely backmixed reactor.20.What additional closure is needed for mass transport in turbulent flow? Why?The contribution of the turbulent diffusivity (eddy diffusivity) should beadded in addition to molecular diffusivity to calculate the flux across a con-trol surface. This extra term arises due to the fluctuations in velocity causingadditional transport.21.What is an ideal stage contactor? How do you correct if the stage is notideal?The exit streams leaving a two phase contactor are assumed to be inequilibrium in an ideal stage contactor. If the stage is not ideal, it is correctedby using a stage efficiency factor.22.What is the dispersion coefficient and where is it needed?Dispersion coefficient connects the cup mixing and cross-sectional averages byusing a Fick’s type of relation. It is needed to close the mesoscopic models insystems where a chemical reaction is taking place, e. g., tubular flow reactors.Companion to Mass Transfer Processes, P. A. Ramachandran.

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10Chapter 1.Chapter 11.2Solutions to Problems1.Mass fraction to mole fractions:Show that mass fractions can be con-verted to mole fractions by the use of the following equation:yi=ωiMi¯M(1.1)Derive an expression fordyias a function ofdωivalues. Do this for a binarymixture. Expression for multicomponent mixture becomes rather unwieldy!Solution:The mass fraction has the units of kg i / kg total.The molecular weight of gas i has the unit of kg i / mol iDividing these we getωiMiwhich has the units of mol i/ kg total. Then dividingby the average molecular weight¯Mwhich has the unit of kg total/ mol total,we get mol i/ mol total which is the mole fraction. Hence the relation isverified.To getdyione should note that¯Mis a function ofωi.¯M=(1ωA/MA+ωB/MB)HenceyA=ωAMA(1ωA/MA+ωB/MB)andωB= 1−ωA.Differentiating and after some algebraic manipulations we get the followingrelation fordyias a function ofdωivalues.dyA=¯M2MAMBdωA2.Mole fraction to mass fractions:Show that mole fractions can be con-verted to mass fractions by the use of the following equation:ωi=yiMi¯M(1.2)Derive an expression fordωias a function ofdyivalues for a binary mixture.Solution:The mole fraction has the units of mol i / mol total.The molecular weight gas i has the unit of kg i / mol iMultiplying these we get kg i/ mol total and then dividing by the averagemolecular weight which has the unit of kg total/ mol total, we get kg i/ kgtotal which is the mass fraction. Hence the relation is verified.To getdωAone should note that¯Mis a function ofyi. We have for a binary :¯M=yAMA+ (1−yA)MBCompanion to Mass Transfer Processes, P. A. Ramachandran.

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Chapter 111Using this in the expression forωAand differentiating it and simplifying thealgebra, we getMAMB¯M2dyAwhich is the required relation connecting the mass fraction gradient and themole fraction gradient.3.Average molecular weight:At a point in a methane reforming furnacewe have a gas of the composition:CH4= 10%;H2= 15%;CO= 15% andH2O= 10% by moles.Find the mass fractions and the average molecular weight of the mixture. Findthe density of the gas.Solution:Assume the rest is nitrogen which was not specified as part of the problem.Calculations of this type are best done in EXCEL spreadsheet or using asimple MATLAB snippet.% mass fraction calculationsy = [0.10.150.15 0.10.5]M = [ 18 2 28 18 28 ] % noteg/mol unitMbar = sum (y .*M)omega = y.*M/Mbar%% The results areMbar =22.100 g/molomega = [0.0814480.0135750.1900450.0814480.633484 ]4.Average molecular weight variations:Two bulbs are separated by along capillary tube which is 20cm long. On one bulb we have pure hydrogenwhile at the other bulb we have nitrogen. The mole fraction profile varies ina linear manner along the length of the capillary. Calculate the mass fractionprofile and show that the variation is not linear. Also calculate the averagemolecular weight as a function of the length along the capillary.Solution:Mole fraction profile is given as linear. Hence at any location the mole fractioncan be calculated as a linear interpolation between the two end point values.For example at distance X = 4cm, the mole fraction of hydrogen is (1−4/20)= 0.8.Correspondingly the average molecular weight at this point is 0.8×2×10−3+0.2×28×10−3= 7.2 g/mol.The mass fraction at this point is 2×0.8/¯M= 0.22.Similar calculations can be done at other points. For example at distancex= 16cm, we get the mass fraction as 0.0175.The mass fraction profile is seen to be not linear while the mole fraction is.Similarly the average molecular weight is a function of position.Companion to Mass Transfer Processes, P. A. Ramachandran.

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12Chapter 1.Chapter 15.Mass fraction gradient:For a diffusion process across a stagnant film, themole fraction gradient of the diffusing species was found to be constant. Whatis the mass fraction gradient? Is this also linear? The mixture is benzene-air.Solution:The mole fraction gradient is constant. Hence the mole fraction profile islinear. The average molecular weight at any location is given as the weightedaverage of the two species:¯M= 78X+ 29(1−X)whereXis a scaled distance at the two ends. (Here we assumeX= 0 is airwhileX= 1 is benzene.)The mass fraction profile is related to the mole fraction profile as:ωA=MAxAMAxA+ (1−xA)MBThis is found to be nonlinear (due to the terms in the denominator). Corre-spondingly the mass fraction gradient is also nonlinear.6.Total concentration in a liquid mixture:Find the total molar concen-tration and species concentrations of 10% ethyl alcohol by mass in water atroom temperature.Solution:The total molar concentration is equal to density of the mixture divided bythe average molecular weight of the mixture.Density of alcohol is 0.7935g/cm3. The average density at a mass fraction of0.1 is calculated by interpolation using the following relation:1ρ=ωAρ0A+ωBρ0BThe superscripts indicate pure component values.The calculated value is 0.9746 g/cm3.It may be noted that the solution is not ideal and hence the density is to befound from partial molar volume considerations for a more accurate result.The density from internet data base is 0.98187g/cm3which is the more accu-rate value.The average molecular weight is calculated from the following equation:1¯M=∑ωAMAThe value is found as 19.16 g/mol.Hencethetotalmolarconcentration,Cofthemixtureisρ/¯M=(0.9784g/cm3)/(19.16g/mol)=0.0509mol/cm3= 50866mol/m3.7.Effect of coordinate rotation on flux components:In Figure 1.1, theflux vector is 2ex+ey. Now consider a coordinate system which is rotated byan angleθ. Find this angle such that the flux componentNAyis zero. Whatis the value ofNAxin this coordinate system?Companion to Mass Transfer Processes, P. A. Ramachandran.

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Chapter 113Solution: The original unit vectors in the two coordinates are related to theunit vectors in the new coordinates as follows:ex=ex;newcosθ−ey;newsinθey=ex;newsinθ+ey;newcosθThe original vector is 2ex+eyand this gets is transformed to:ex;new[2 cosθsinθ] +ey;new[−2 sinθ+ cosθ]New components are therefore [2 cosθsinθ] and [−2 sinθ+ cosθ].If the flux componentNAyin the new coordinates has to be zero, then−2 sinθ+ cosθ= 0The angle of rotation must then be such that tanθ= 1/2. Henceθ=π/4.The value ofNAxis then 2 cosθsinθ.8.Flux vector in cylindrical coordinates:Define flux vector in terms ofits components in cylindrical coordinates. Sketch the planes over which thecomponents act. Show the relations between these components and the com-ponents in Cartesian coordinates.Solution:Flux vector is the same in cylindrical coordinates but the components aredifferent. The flux vector is represented as:NA=erNAr+eθNAθ+ezNAzThe componentNArcan be viewed as the mass crossing a unit area in a planenormal to the r-direction in cylindrical coordinates. The other components canbe viewed in a similar manner. Thus, for example,NAθis the moles crossinga plane perpendicular to theθdirection.The components are related to those in Cartesian by the following relationsfrom vector transformation rules.er= (cosθ)ex+ (sinθ)eyeθ= (−sinθ)ex+ (cosθ)eyez=ez9.Flux vector in spherical coordinates:Define flux vector in terms of itscomponents in spherical coordinates. Sketch the planes over which the compo-nents act. Show the relations between these components and the componentsin Cartesian coordinates.Show the relations between these components and the components in Carte-sian coordinates.Solution:Companion to Mass Transfer Processes, P. A. Ramachandran.

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14Chapter 1.Chapter 1. The flux vector in spherical coordinates is represented as:NA=erNAr+eθNAθ+φzNAφThe components are interpreted as mass crossing a unit area in a plane normalto the direction indicated in the subscript.For example,NAθis the moles crossing a plane perpendicular to theθdirec-tion.The components are related to those in Cartesian by the following relationsfrom vector transformation rules.er= (sinθcosφ)ex+ (sinθsinφ)ey+ (cosθ)ezeθ= (cosθcosφ)ex+ (cosθsinφ)ey+ (−sinθ)ezeφ= (−sinφ)ex+ (−cosφ)ey+ (0)ez10.Different forms of the Henry’s law constant:Express the Henry’sconstants reported in Table 1.4 asHi,pcandHi,cp.Solution:The relation for mole fraction in the liquid isxAequalsCA/CtotUsing thisin the Herny lawpA=HAxA= (HA/Ctot)xAHence the Henry coefficient in pressure-concentration unit is given asHpc=HA/CtotHenry’s law in pressure-concentration form for hydrogen is therefore:HH,pc= (7.099×104; atm]/(55000; mol/m3) = 1.2907atm m3/molIntheconcentration-pressure unit,itisthereciprocal ofthisquantity,0.7748mol/atm m3Values for other gases are: Oxygen = 0.7744;CO2= 0.0296; ammonia =5.4545×104. in pressure-concentration unit.11.Henry’s law constants: Unit conversions:Henry’s law constant forO2andCO2are reported as 760.2 L. atm/mol and 29.41 L. atm /mol.What is the form of the Henry’s law used? Convert to values for the otherforms shown in the text.Solution:From the units we deduce that the Henry’s law is reported in pressure-concentration form.To get the value in pressure-mole fraction form we useCA=xACwhereCisthe total concentration in the liquid. Value of 55 mol/L is used for the totalconcentration in water. Hence the Henry’s law in pressure-mole fraction formCompanion to Mass Transfer Processes, P. A. Ramachandran.

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Chapter 115is:H= (760.2L.atm/mol)(55mol/L) = 4.18×104atmSimilar calculation forCO2shows the value of 1.617×103atm12.Solubility of CO2:Henry’s constant values for CO2is shown below as afunction of temperature.Temperature, K280300320H, bar96017302650Fit an equation of the type:lnH=A+B/TWhat is the physical significance of the parameterB? Find the solubility ofpure CO2in water at these temperature.Solution:The plot lnHas 1/Tshould be linear and the constantsAandBcan befound from this plot. However, it is best to use a linear regression model inMATLAB. The following MATLAB code is useful for the linear regressionand can be used in general for other problems.% Linear regression:problem 1.12 CO2 solubility data.% P = POLYFIT(X,Y,N) finds the coefficients of a polynomial P(X) of%degree N that fits the data Y best in a least-squares sense. P is a%row vector of length N+1 containing the polynomial coefficients in%descending powers, P(1)*X^N + P(2)*X^(N-1) +...+ P(N)*X + P(N+1).x= [ 280 300 320 ] %% temperature valuesy = [960 1730 2650 ]%% henry values in barsY = log (y)X = 1./xP = polyfit (X, Y, 1 ) % linear fit here.Y1 = P(2) + P(1) * X%% fitted constants%%% comparison plotsplot ( X,Y, ’*’ ); hold onplot (X, Y1)delh = 8.314 * P(1)% heat of solution Answer= -19 kJ/mol% solutions: B = -2.2792e+003;% A=15.0215The fitted constants are found to beB=−2.2792e+ 003 andA= 15.0215The parameterBis a measure of heat of solution in accordance with van’tHoff equation:B= ∆HsRCompanion to Mass Transfer Processes, P. A. Ramachandran.

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16Chapter 1.Chapter 1From the fitted data we find the heat of solution of CO2as -19 kJ/mol whichis close to experimental value of−19.4kJ/mol reported in the literature.13.Vapor pressure calculations: the Antoine equation:The Antoine con-stants for water are:A= 8.07131; B=1730.63;C= 233.426 in the units ofmm Hg for pressure and deg C for temperature. Convert this to a form wherepressure is in Pa and temperature is in deg K;Also rearrange the Antoine equation to a form where temperature can becalculated explicitly. This represents the boiling point at that pressure. Whatis the boiling point of water at Denver, CO (mile high city)?Solution:p(Pa) =p(mm) 1.0135×105P a/atm760mm/1atmTaking loglog10p(mm Hg) = log10p(Pa)−log(0.0075) = log10p(Pa)−2.1249Using this in Antoine equation and substituting for log10p(mm Hg) we getlog10p(Pa) = 2.1229−BC+T(K)−273orlog10p(Pa) = 10.1962−1730.63T(K)−39.57which is the required relation with pressure in Pa and temperature in degK. The equation can be rearranged to be explicit in temperature and thefollowing relation is obtained:T=Blog10p−A−CThe pressure at Denver at normal condition is given as:p/p0= exp(−MairghRgT)from the equation of hydrostatics. HerepOis the pressure at sea level.Mw=20×10−3kg/moland h = elevation = 1 mile = 1600 m.Hence p = 367 mm Hg. Substituting in the Antoine equation we find theboiling point of water as 95 deg Celsuis.14.van’t Hoff relation:Given the Antoine constants for a species, can youcalculate the heat of vaporization of that species? Find this value for waterfrom the data given in Problem 1.12.Solution:One should use the van’t Hoff relation as a first approximation. The relationis:dlnP;vapdT= ∆H;vapRgT2Companion to Mass Transfer Processes, P. A. Ramachandran.

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