cn3132 ii lecture 01 mass transfer models
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CN3132Separation Processes (II)
Lecture 01:Mass Transfer Models
Dr.ZHAODanDepartmentofChemicalandBiomolecularEngineering
4EngineeringDrive4,Blk E5,#0216
Tel:(65)65164679
Wankat 3rd: 15.1; 15.2.1; 15.2.4; 15.3.1; 15.3.2Treybal: Chapter 2
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2Course Outline MassTransfer (Lecture0103,week1)
Modelsformasstransfer Twofilmtheory Individualandoverallmasstransfercoefficients
RateBasedMethod (Lecture0409,week23) Transferunitsconceptsinratebaseddesign Applicationofratebaseddesignforcontinuouscontactoperationof
absorptionanddistillation Designofpackedcolumn
Humidification (Lecture1014,week45) Humidity,adiabaticsaturation,wetbulbtemperature Humidificationanddehumidificationprocesses Psychrometricchart Designofcoolingtower
Adsorption (Lecture1517,week6) Definitions Sorbenttypes Isotherms Chromatography
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3Schedules
Lectures Asusual,Monday9:0010:35am;Thursday9:009:45am,LT6 AmakeupclassscheduledthisSaturday(11Oct)9:0010:35amLT6
Tutorials Asusual,totally5tutorials,startingnextweek(13Oct2014)
Consultation Fridays14:0016:00pminmyoffice(E50216) Appointmentonotherdays IVLE,Webcast,Email
Midterm Test Time:Thursday,13Nov2014,9:009:45am Venue:tobeannounced Openbooktest Bringincalculatorsandstationery
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4How to excel in CN3132?(1) Attend lectures!
WhatWebcast cando: Makeupforemergentabsence
Reviewlectures
Prepareexams
Saveyouseveralsleepinghours
WhatWebcast canNOTdo: Liveshow!
Feelpeerpressure
Interactwithlecturers
Selffulfillment
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How to excel in CN3132?(2) Solve problems!!
Readthetextbook Derivetheequationsatleastoncebyyourself Workonthehomeworkbeforeyoucometothetutorials Balancebetweengroupstudyandindividualstudy
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6Recap DesignConceptforSeparation
Equilibrium Gibbsphaserule:F=C P+2 Relativevolatility
FlashDistillation Equilibriumline Operatingline Graphicalsolution
MultiComponentFlashDistillation Trialanderror
BinaryMultiStageDistillation Topoperatingline Bottomoperatingline Feedline q McCabeThielemethod Numberofstages Optimumfeedlocation Minimumrefluxratio
BinaryAbsorptionandStripping Equilibriumline Operatingline Molefractionvs.moleratio Kremser equations MinL/Gratio(absorption) MaxL/Gratio (stripping)
MultiComponentAbsorption Identifythekeycomponent
Extraction(ImmiscibleSystems) Analogytostripping
Extraction(PartiallyMiscible) Triangulardiagrams Mixingpoint Inverseleverarmrule Equilibriumline Operablerangeoffeedcomposition Correlationcurve HunterNashmethod
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7Thermodynamics vs. Kinetics(Equilibrium vs. Rate)
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8Staged Column vs. Packed Column
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9Mass Transfer
Whenasystemcontainstwoormorecomponentswhoseconcentrationsvaryfrompointtopoint,thereisanaturaltendencyformasstobetransferred,minimizingtheconcentrationdifferenceswithinasystem.Thetransportofoneconstituentfromaregionofhigherconcentrationtothatofalowerconcentration iscalledmasstransfer.
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Models for Mass Transfer:(1) Molecular Movement
Allmoleculesmoveandcollidebecauseofthermalenergy Molecularcollisionsresultinmasstransferbydiffusion Moleculestendtodistributethroughoutthevolumeavailable Atequilibriumthereisanequalnumberofdensity
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ForabinarymixtureofAandB,
JAz:molecularfluxofAinBalongzdirection[mole/(m2s)] DAB:moleculardiffusivity(m2/s) dcA/dz:concentrationgradientofAalongzdirection(mole/m4) Minussign:diffusiondirectionisoppositetoconcentrationgradient
Models for Mass Transfer:(2) Ficks 1st Law of Diffusion
AAz AB
dcJ Ddz
= BBz BA dcJ D dz= AdolfEugen Fick (18291901)
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Other Forms of Ficks Law
c:mixtureconcentration(mole/m3)
xA:molefractionofA
R:idealgasconstant8.314[J/(Kmol)]
T:temperature(K)
dpA/dz:pressuregradientofAalongzdirection(Pa/m)
AAz AB
dxJ cDdz
= BBz BA dxJ cD dz=
AB AAz
D dpJRT dz
= BA BBz D dpJ RT dz=
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Fickian Binary Gas Diffusivities
DAB:moleculardiffusivity(m2/s)
T:temperature(K)
MW:averagemolecularweight
ptot:totalabsolutepressure(Pa)
:averagediameterofthesphericalmolecules()
3/2 1/2
2
(1/ )AB
tot
T MWDp
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Fickian Binary Liquid Diffusivities
16 1/20
0.6
1.173 10 [ ( )]B BAB
B A
MW TDV
=
DAB:moleculardiffusivity(m2/s)
B:solventinteractionparameter MWB:molecularweightofB
T:temperature(K)
B:solventviscosity(Pas) VA:molarvolumeofsolute
(m3/kmol)
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Convection vs. Diffusion
ForabinarymixtureofAandB,thefluxesrelativetothefixedpositionforeachcomponentscanbederivedas:
NA:fluxofA NB:fluxofB cA:concentrationofA cB:concentrationofB c:totalconcentration
( )A AA A B ABc dcN N N Dc dz
= + ( )B BB A B BAc dcN N N Dc dz= +
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Equimolar Counterdiffusion (EMD)
Inequimolar counterdiffusion,themolarfluxesorAandBareequal,butoppositeindirection,andthetotalpressureisconstantthroughout,soN=NA+NB=0
A A AB AA Az AB AB
dc dx D dpN J D cDdz dz RT dz
= = = = 2 2 2 2
1 1 1 1
A A A
A A A
z c x pAB AB AB
A A AA A Az c x p
D cD Ddz dc dx dpN N RTN
= = = 1 2 1 2 1 2
2 1 2 1 2 1
( ) ( ) ( )( ) ( ) ( )
AB AB ABA A A A A A A
D cD DN c c x x p pz z z z RT z z
= = =
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Constant Molar Overflow (CMO) in Distillation
Theheatofvaporizationpermoleisconstant Withineachsectiontheliquidandthevaporflowrates
remainconstantinthewholesection
EMDapplies17
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Unimolecular Diffusion (UMD) (1)
SteadystatediffusionofAthroughstagnantB,soNB=0
Ammonia
+
Air
Water
A A A A AB AA A AB A A AB A
c dc dx p D dpN N D x N cD Nc dz dz p RT dz
= = = 2 2 2 2
1 1 1 11
A A A
A A A
z c x pAB A AB A AB A
A A A A A Az c x p
cD dc cD dx pD dpdzN c c N x RTN p p
= = = 2 2 2
2 1 1 2 1 1 2 1 1
1ln ln ln( ) ( ) 1 ( )
AB A AB A AB AA
A A A
cD c c cD x pD p pNz z c c z z x RT z z p p
= = =
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Unimolecular Diffusion (UMD) (2)
logarithmicmean: ( , )ln lnlmy xM x yy x=
1 2 1 22 1 2 1
( ) ( ) [UMD]( )(1 ) ( )( )
AB ABA A A A A
A lm A lm
cD pDN x x p pz z x RT z z p p
= =
1 2 1 2 1 22 1 2 1 2 1
( ) ( ) ( ) [EMD]( ) ( ) ( )
AB AB ABA A A A A A A
D cD DN c c x x p pz z z z RT z z
= = =
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Example Question
Oxygen(A)isdiffusingthroughcarbondioxide(B)understeadystateconditions,withtheCO2 nondiffusing.Thetotalpressureis1x105 Pa,andthetemperatureis0C.Thediffusionpathis2.0mm.Thepartialpressuresofoxygenatthe2endsare13,000and6,500Parespectively.Thediffusivityofthemixtureis1.87x105 m2/s.CalculatethemolarfluxofO2 inthemixture.GivenR=8.314[J/(Kmol)]
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Solution
ThisisacaseofcomponentAdiffusinginanotherNondiffusingcomponentB.Theequationtobeusedis:
Given: p=1x105 Pa T=0C=273K DAB =1.87x105 m2/s R=8.314J/(Kmol) (z2 z1)=2.0mm=2x103 m pA2 =6500Pa pA1 =13000Pa
2
2 1 1
ln( )
AB AA
A
pD p pNRT z z p p
=
NA =2.97102 gmol/(m2s)
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