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MASS TRANSFERForCHEMICAL ENGINEERING

MASS TRANSFERSYLLABUSFick’s laws, molecular diffusion in fluids, mass transfer coefficients, film, penetrationand surface renewal theories; momentum, heat and mass transfer analogies; stage-wiseand continuous contacting and stage efficiencies; HTU & NTU concepts; design andoperation of equipment for distillation, absorption, leaching, liquid-liquid extraction,drying, humidification, dehumidification and adsorption.ANALYSIS OF GATE PAPERSExam 01210129912 Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

CONTENTSTopicsPage €™s First LawRelation among Molar FluxesDiffusivitySteady State DiffusionTransient gnificant Parameters in Convective Mass TransferApplication of Dimensionless Analysis in Mass TransferAnalysis among Mass, Heat and Momentum TransferConvective Mass Transfer CorrelationsMass Transfer between PhasesTheories of Mass TransferGate .93.103.113.123.13IntroductionT-XY DiagramRelative VolatilityRaoult’s LawConstant Boiling MixturesDifferential DistillationFlash VaporizationFractionationNumber of Plates AnalysisTypes of ReboilersImportant TermsTray EfficiencyEffect of PressureGate 4.24.34.44.5IntroductionSelection of SolventCounter Current AbsorptionRate of AbsorptionCalculation of Tower Height62626263653.4.MASS TRANSFER COEFFICIENTSDISTILLATION Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

4.6Calculation for Ideal Number of Plates by AbsorptionFactor MethodHETP (Height Equivalent to Theoretical Plate)Gate Questions6667705.15.25.35.45.5IntroductionVapor Pressure CurveDry Bulb TemperatureWet Bulb TemperatureDew Point TemperatureGate nMechanism of Moisture ContentResistances in DryingImportant TermsRate of DryingEffect of Variables on DryingGate Questions88888888899092IntroductionEquilibrium Relations in ExtractionTypes of System in ExtractionStagesGate oductionFactors Affecting Rate of LeachingContacting PatternSuper Critical Fluid Extraction1111111111119.19.29.3IntroductionTypes of AdsorptionAdsorption IsothermsGate Questions113113113115ADSORPTION Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

1DIFFUSION1.1 INTRODUCTIONTransfer of mass from one point to another in asingle phase or from one phase to another phasewith the help of a driving force is known as masstransfer. The driving force is required for anytransfer; in mass transfer concentrationdifferenceis driving force.Types of mass transfer:1. Molecular Diffusion: The transfer of massor diffusion of species takes place in astagnant medium.2. Turbulent Diffusion: If the diffusion ofspecies takes place in a turbulentmedium, then the mechanism is known asturbulent or eddy diffusion.Flux: The net rate at which a species in solutionpasses through unit area, which is normal to thedirection of diffusion in unit time.Molar Flux: The molar flux of species definedwith reference to fixed spatial coordinates, NA isNA CA VA . (1.1)This could be written in terms of diffusionvelocity of A, (i.e.VA-V) and molar averagevelocity of mixture, V asNA CA (VA-V) CAV (1.2)By definition,Ξ£ Ci ViV 𝐢𝐢Therefore, equation (1.2) becomesΞ£C ViNA CA (VA-V) CA 𝐢𝐢i CA (VA-V) XAΞ£Ci ViFor binary systems containing two componentsA and B,NA CA (VA-V) XA (CAVA CBVB) CA (VA-V) XA (NA NB)NA CA (VA-V) XA N . (1.3)The first term on the right hand side of thisequation is diffusional molar flux of A, and thesecond term is flux due to bulk motion.Mass Flux: The mass flux of species defined withreference to fixed spatial coordinates, nA isnA ρA vA . (1.4)This could be written in terms of diffusionvelocity of A, (i.e.vA-v) and mass average velocityof mixture, v asnA ρA (vA-v) ρAv (1.5)By definition,Ξ£ ρi viv 𝜌𝜌Therefore, equation (1.5) becomesΣρi vinA ρA (vA-v) ρA𝜌𝜌 ρA (vA-v) YAΣρi vi1.2 FICK’S FIRST LAW:The molar flux of a species with respect toobserver moving with molar average velocityis directly proportional to the gradient of theconcentration of that species.dCJA Ξ± 𝑑𝑑𝑑𝑑AIf we remove the proportionality, then theconstant of proportionality must beintroduced, which is known as diffusivity ordiffusion coefficient; denoted by DAB for abinary mixture or Dij for a multicomponentmixture. This coefficient tells us how easilyspecies I diffuse in the solution or for a binarymixture how easily A diffuses into B. dCA/dZis the concentration gradient in the Zdirection.

dCJA -DAB 𝑑𝑑𝑑𝑑AWhere, minus sign represents the transfer ofmass taking place from a region of higherconcentration to a region of lowerconcentration.dXJA -CDAB 𝑑𝑑𝑑𝑑AUsing this expression, Equation (1.3)could be written as,dXNA -CDAB 𝑑𝑑𝑑𝑑A XA N . (1.6)1.3 RELATION AMONG MOLAR FLUXES:For a binary system containing A and B, fromEquation (1.6),NA JA XA NOr, JA NA-XA N . (1.7)Similarly, JB NB-XB N . (1.8)Addition of equation (1.7) & (1.8) gives,JA JB NA NB– (XA XB) N . (1.9)By definition,N NA NBand XA XB 1Therefore equation (1.9)becomes,JA JB 0JA - JBdXdXCDAB 𝑑𝑑𝑑𝑑A -CDBA 𝑑𝑑𝑑𝑑B (1.10)From XA XB 1dXA dXBTherefore equation (1.10) becomes,DAB - DBA . (1.11)1.4 DIFFUSIVITYFick’s law proportionality, DAB, is known asmass diffusivity (simply as diffusivity) or asthe diffusion coefficient. DAB has thedimension of L2/t, identical to thefundamental dimensions of the othertransport properties: Kinematic viscosity,πœ‡πœ‡π›Ύπ›Ύ in momentum transfer, and thermal𝜌𝜌diffusivity, 𝛼𝛼 𝐾𝐾𝜌𝜌 𝐢𝐢𝑝𝑝in heat transfer.Diffusivity is normally reported in cm2/sec;the SI unit being m2/sec. Diffusivity dependson pressure, temperature, and compositionof the system. Diffusivities of gases at lowdensity are almost composition independent,increase with the temperature and varyinversely with pressure. Liquid and soliddiffusivities are strongly concentrationdependent and increase with temperature.General range of values of diffusivity:Gases: 5 x 10-6 - 1 x 10-5 m2/secLiquids: 10-6 - 10-9 m2/secSolids: 5 x 10-14 - 1 x 10-10 m2/sec1.4.1 DIFFUSIVITY IN GASES:Pressure dependence of diffusivity is given by1DAB 𝑃𝑃And temperature dependency is according toDAB 𝑇𝑇 3/2Diffusivity of a component in a mixture ofcomponents can be calculated using thediffusivities for the various binary pairs involvedin the mixture. The relation given by Wilke is1𝐷𝐷1 ‘šπ‘š 𝑦𝑦 ′′𝑦𝑦𝑦𝑦𝑛𝑛′23 𝐷𝐷𝐷𝐷𝐷𝐷1 21 31 𝑛𝑛Where, D1-mixtureis the diffusivity for component1 in the gas mixture; D1-n is the diffusivity for thebinary pair, component 1 diffusing throughcomponent n; and y’nis the mole fraction ofcomponent n in the gas mixture evaluated on acomponent –1 – free basis, that is

𝑦𝑦2β€² 𝑦𝑦2𝑦𝑦2 𝑦𝑦3 . . 𝑦𝑦𝑛𝑛1.4.2 DIFFUSIVITY IN LIQUIDS:Temperature dependency is according toDAB 𝑇𝑇1.5 STEADY STATE DIFFUSIONIn this section, steady-state molecular masstransfer through simple systems in which theconcentration and molar flux are functions of asingle space coordinate will be considered.In a binary system, containing A and B, thismolar flux in the direction of z, as given byEqn (1.6) isdXNA -CDAB 𝑑𝑑𝑑𝑑A XA (NA NB) . (1.11)1.5.1 DIFFUSION OF A THROUGH NONDIFFUSING B:Liquid A is evaporating into gas B. Solubility ofgas B in liquid A is negligible or B is nondiffusing. A stream of gaseous mixture A and Bhaving concentration xA2 flows slowly over thetop of the tube to maintain the mole fraction of Aat xA2. Entire system is kept at constanttemperature and pressure. Gases A and B areassumed to be ideal.NA JA XA NNA JA XA (NA NB)NB 0,dXNA XA NA - CDAB 𝑑𝑑𝑑𝑑AdX(1- XA )NA - CDAB 𝑑𝑑𝑑𝑑A𝑍𝑍𝑋𝑋dXANA 0 𝑑𝑑𝑑𝑑 - CDAB 𝑋𝑋 𝐴𝐴2 (1 XA)𝐴𝐴11 𝑋𝑋NA Z CDABln 1 𝑋𝑋𝐴𝐴2NA NA NA ·π·π΄π΄π΄π΄π‘π‘1 𝑋𝑋𝐴𝐴1ln 1 𝑋𝑋𝐴𝐴2𝑋𝑋ln ·π΄π΄π΄π΄ 𝑋𝑋𝐡𝐡2 𝑋𝑋𝐡𝐡1𝑍𝑍Where,𝑋𝑋𝐡𝐡2 𝑋𝑋𝐡𝐡1𝑋𝑋ln 𝐡𝐡2 𝑋𝑋𝐡𝐡1𝑋𝑋𝐡𝐡𝐡𝐡𝐡𝐡 𝑋𝑋𝐡𝐡2 𝑋𝑋𝐡𝐡1𝑋𝑋ln 𝐡𝐡2 𝑋𝑋𝐡𝐡1logarithmic mean of B.is called𝑋𝑋𝐴𝐴1 𝑋𝑋𝐴𝐴2 𝑋𝑋𝐡𝐡2 𝑋𝑋𝐡𝐡1NA 𝐢𝐢𝐷𝐷𝐴𝐴𝐴𝐴 𝑋𝑋𝐴𝐴1 ΅π΅ . (1.12)When the difference between Concentration𝑋𝑋𝐡𝐡2 &𝑋𝑋𝐡𝐡1 is very less then we take arithmeticmean not logarithmic mean.Equation (1.12) can be written as in the form oflogarithmic average concentration of B,Where,NA 𝐢𝐢𝐷𝐷𝐴𝐴𝐴𝐴 𝐢𝐢𝐴𝐴1 ΅π΅πΆπΆπ΅π΅π΅π΅π΅π΅ (1.13)𝐢𝐢𝐡𝐡2 𝐢𝐢𝐡𝐡1𝐢𝐢ln 𝐡𝐡2 𝐢𝐢𝐡𝐡1is calledlogarithmic average concentration of B.For an ideal gas, 𝐢𝐢 P𝑅𝑅𝑅𝑅Therefore, for an ideal gas mixture equation.(1.13) becomesNA 𝑃𝑃𝐷𝐷𝐴𝐴𝐴𝐴 𝑃𝑃𝐴𝐴1 ΅π΅πΏπΏπΏπΏ . (1.14)

Example 1.1 Oxygen is diffusing in a mixtureof oxygen-nitrogen at 1 std-atm, 25ΒΊC.Concentration of oxygen at planes 2 mmapart are 10 and 20 volume % respectively.Nitrogen is non-diffusing.(a) Derive the appropriate expression tocalculate the flux of oxygen. Define units ofeach term clearly.(b) Calculate the flux of oxygen. Diffusivity ofoxygen in nitrogen 1.89 * 10-5 m / s.Solution: Let us denote oxygen as A and nitrogenas B. Flux of A (i.e.) NA is made up of twocomponents, namely that resulting from the bulkmotion of A (i.e.), NXA and that resulting frommolecular diffusion JA:NA JA XA NFrom Fick’s law of Diffusion,dCJA -DAB 𝑑𝑑𝑑𝑑AdCNA XA N - DAB 𝑑𝑑𝑑𝑑A (1)Since N NA NB and XA CA/C equation (1)becomesNA (NA NB )CA𝐢𝐢dC- DAB 𝑑𝑑𝑑𝑑A . (2)Since B is non-diffusing NB 0. Also, the totalconcentration C remains constant. Therefore,equation (2) becomes,𝑍𝑍 0 𝑑𝑑𝑑𝑑𝐢𝐢 𝐷𝐷𝐴𝐴𝐴𝐴𝐢𝐢 𝐢𝐢 𝐴𝐴2dC𝐴𝐴1 (NA C NA CA )𝑁𝑁𝐴𝐴 ‘§π‘§πΆπΆ 𝐢𝐢ln 𝐢𝐢 𝐢𝐢𝐴𝐴2𝐴𝐴1𝐢𝐢 𝐢𝐢ln 𝐢𝐢 𝐢𝐢𝐴𝐴2 . (3)𝐴𝐴1Replacing concentration in terms of pressuresusing Ideal gas law, equation (3) becomes𝑁𝑁𝐴𝐴 𝑃𝑃𝑑𝑑 ‘ƒπ‘ƒ 𝑃𝑃ln 𝑃𝑃𝑑𝑑 𝑃𝑃𝐴𝐴2 . (4)𝑑𝑑𝐴𝐴1Where,𝐷𝐷𝐴𝐴𝐴𝐴 molecular diffusivity of A in B𝑃𝑃𝑑𝑑 total pressure of systemR universal gas constantT temperature of system in absolute scalez distance between two planes across thedirection of diffusion𝑃𝑃𝐴𝐴1 partial pressure of A at plane 1, and𝑃𝑃𝐴𝐴2 partial pressure of A at plane 2Given:𝐷𝐷𝐴𝐴𝐴𝐴 1.89 * 10-5m2/sec𝑃𝑃𝑑𝑑 1 atm 1.01325 * 105 N/m 2T 25ΒΊC 273 25 298 Kz 2 mm 0.002 m𝑃𝑃𝐴𝐴1 0.2 *1 0.2 atm 20265 N/m2 (From Idealgas law and additive pressure rule)𝑃𝑃𝐴𝐴2 0.1 * 1 0.1 atm 10132.5 N/m2Substituting these in equation (4),𝑁𝑁𝐴𝐴 (1.89 10 5) (1.01325 105 )(8314 298 0.002)ln(101325 10132.5)(101325 20265)𝑁𝑁𝐴𝐴 4.55 10 5 kmol/m2.s1.5.2EQUIMOLAR COUNTER DIFFUSION:A physical situation which is encountered in thedistillation of two constituents whose molarlatent heats of vaporization are essentially equal,stipulates that the flux of one gaseouscomponent is equal to but acting in the oppositedirection from the other gaseous component;that is, NA -NB.The molar flux NA, for a binary system at constanttemperature and pressure is described by,CdCNA (NA NB ) 𝐢𝐢A - DAB 𝑑𝑑𝑑𝑑A (1.15)

With the substitution of NA -NB, Equation (1.15)becomes,dCNA - DAB 𝑑𝑑𝑑𝑑A (1.16)For steady state diffusion Equation (1.16) maybe integrated, using the boundary conditions: atz z1 𝐢𝐢𝐴𝐴 𝐢𝐢𝐴𝐴1and z z2𝐢𝐢𝐴𝐴 𝐢𝐢𝐴𝐴2Giving,𝑍𝑍𝐢𝐢NA 𝑍𝑍 2 𝑑𝑑𝑑𝑑 - DAB 𝐢𝐢 𝐴𝐴2 𝑑𝑑𝐢𝐢𝐴𝐴1𝑁𝑁𝐴𝐴 (𝑍𝑍𝐷𝐷𝐴𝐴𝐴𝐴2 𝑍𝑍1𝐴𝐴1)(𝐢𝐢𝐴𝐴2 𝐢𝐢𝐴𝐴1 ) . (1.17)For an ideal gas, 𝐢𝐢 P𝑅𝑅𝑅𝑅Therefore, for an ideal gas mixture equation.(1.17) becomes𝐷𝐷𝑁𝑁𝐴𝐴 𝑅𝑅𝑅𝑅(𝑍𝑍 𝐴𝐴𝐴𝐴(𝑃𝑃𝐴𝐴2 𝑃𝑃𝐴𝐴1 ) . (1.17) 𝑍𝑍 )21This is the equation of molar flux for steady-stateequimolar counter diffusion.Concentration profile in this equimolar counterdiffusion may be obtained from,𝑑𝑑(𝑁𝑁 ) 0𝑑𝑑𝑑𝑑 𝐴𝐴(Since𝑁𝑁𝐴𝐴 is constant over the diffusion path)And from equation (1.16)Therefore,Or,𝑑𝑑2 CA𝑑𝑑𝑑𝑑 2dCANA - DAB 𝑑𝑑𝑑𝑑𝑑𝑑dCA DAB 0𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 0 . (1.18)This equation may be solved using the boundaryconditions to give,𝐢𝐢𝐴𝐴 𝐢𝐢𝐴𝐴1𝐢𝐢𝐴𝐴 𝐢𝐢𝐴𝐴2𝑍𝑍 𝑍𝑍1 𝑍𝑍1 𝑍𝑍2 (1.19)Equation, (1.19) indicates a linear concentrationprofile for equimolar counter diffusion.Example 1.2 Methane diffuses at steady statethrough a tube containing helium. At point 1the partial pressure of methane is 𝑷𝑷𝑨𝑨𝑨𝑨 55kPa and at point 2, 0.03 m apart 𝑷𝑷𝑨𝑨𝑨𝑨 15 KPa.The total pressure is 101.32 kPa, and thetemperature is 298 K. At this pressure andtemperature, the value of diffusivity is 6.75 *10-5 m2/sec.(i). Calculate the flux of CH4at steady state forequimolar counter diffusion.(ii). Calculate the partial pressure at a point0.02 m apart from point 1.Solution: For steady state equi-molar counterdiffusion, molar flux is given by,𝐷𝐷𝑁𝑁𝐴𝐴 𝑅𝑅𝑅𝑅(𝑍𝑍 𝐴𝐴𝐴𝐴(𝑃𝑃𝐴𝐴2 𝑃𝑃𝐴𝐴1 ) (1) 𝑍𝑍 )Therefore,2𝑁𝑁𝐴𝐴 1(6.75 10 5 )(55 15)(8314 298 0.003)𝑁𝑁𝐴𝐴 3.633 10 5π‘˜π‘˜π‘˜π‘˜π‘˜π‘˜π‘˜π‘˜π‘šπ‘š2 . 𝑠𝑠𝑠𝑠𝑠𝑠And from (1), partial pressure at 0.02 m frompoint 1 is:3.633 10 5 (6.75 10 5 )(55 𝑃𝑃𝐴𝐴 )(8314 298 0.002)𝑃𝑃𝐴𝐴 28.33 𝐾𝐾𝐾𝐾𝐾𝐾Example 1.3 For Equimolar counter diffusionfrom a sphere to a surrounding stationaryinfinite medium, the mass flux 𝑡𝑡𝑨𝑨𝑨𝑨 of thediffusing component A at the interface is

given by 𝑡𝑡𝑨𝑨𝑨𝑨 𝑫𝑫𝑨𝑨𝑹𝑹(π‘ͺπ‘ͺ𝑨𝑨𝑨𝑨 π‘ͺπ‘ͺ𝑨𝑨𝑨𝑨 ) where 𝑫𝑫𝑨𝑨 is thediffusivity, R the radius of sphere andπ‘ͺπ‘ͺ𝑨𝑨𝑨𝑨 and π‘ͺπ‘ͺ𝑨𝑨𝑨𝑨 the molar concentrations of A atthe interface and at a point far away from thesphere. Show that the Sherwood number,based on the diameter of the sphere is equalto 2.Solution: The mass flux 𝑁𝑁𝐴𝐴𝐴𝐴 is given by𝑁𝑁𝐴𝐴𝐴𝐴 𝐷𝐷𝐴𝐴𝑅𝑅(𝐢𝐢𝐴𝐴𝐴𝐴 𝐢𝐢𝐴𝐴𝐴𝐴 )molecular motions into the large body ofstagnant gas that surrounds the droplet.Consider a naphthalene ball, as shown in figure.At any moment, when the radius of the ball is r1,the flux of naphthalene ball at any distance rfrom the center is given by,dXNA XA (NA NB) - CDAB 𝑑𝑑𝑑𝑑A . (1)The mass flux 𝑁𝑁𝐴𝐴𝐴𝐴 of the diffusing component A atthe interface is also given by,𝑁𝑁𝐴𝐴𝐴𝐴 𝐾𝐾𝑐𝑐 (𝐢𝐢𝐴𝐴𝐴𝐴 𝐢𝐢𝐴𝐴𝐴𝐴 ) (2)From the equations (1) and (2), we get𝐾𝐾𝑐