7. Short Introductions To: Mass Transfer; Separation .
Introduction to Process Engineering (PTG)#7/8Processteknikens grunder (”PTG”)Introduction to Process Engineeringv.20081/567. Short introductions to:Mass transfer;Separation processes;Particulate technology &multi-phase flowRon ZevenhovenÅbo Akademi UniversityHeat Engineering Laboratory / Värmetekniktel. 3223 ; ron.zevenhoven@abo.fiTkF VT rz08Introduction to Process Engineering (PTG)#7/82/567.1 Mass transfer processes;Diffusion, convection;Mass transfer coefficientTkF VT rz08
Mass transfer process examplesDrying of a solid using heat or dry gasAdsorption of gas or liquid using a solidDistillation using heat to create a vapour phaseGas absorption using a liquid absorbentExtraction of liquid or solid using a solvent thatcreates immiscible phases Crystallisation using coolingto create a solid phaseNote the ”support phase”:solvent, sorbent, heat, .3/56 Picture: http://www.petrogas.org/absorption.htmIntroduction to Process Engineering (PTG)#7/8TkF VT rz08Introduction to Process Engineering (PTG)#7/8Mass transfer mechanisms /14/56Diffusion: Fick’s Law* (for binary systems)Spreading of a substance from a region with highconcentration to regions with lower concentration;more correct for ”concentration” is ”chemical potential”Molecular diffusion coefficient Đ; for species A in medium B Đ ĐAB& ′′ dm D dcmA dtdxccdiffusiontimexxsurface* Note analogy with Fourier’s Law for heat conductionTkF VT rz08
Introduction to Process Engineering (PTG)#7/85/56Mass transfer mechanisms /2Diffusion (forced or free) convectionFlow as a result of a pressure difference, gravity, .If needed,needed, Đ Đmol Đturb can be used to include turbulent eddy diffusionflow, vflow, vdc v cdxv dx / dt& ′′ DmccdiffusiontimexxTkF VT rz08Introduction to Process Engineering (PTG)#7/86/56Diffusion Transfer of matter as a Mass flux Φ”mA,x (kg / sresult of a concentrationper m2) through a surface(x0) perpen-dicular to the(or density) difference,transport direction (x) asmore accuratelya result of a gradient inchemical potentialmass concentration (ρA).difference (”gradient”) Main cause: Brownian More conventional forgases: use concentrationmotion of moleculescA (mol/m3) ρA/MA pA/RTΦ"molA ,x - D Adc AdxPicture: SSJ84TkF VT rz08
Introduction to Process Engineering (PTG)#7/8Diffusion coefficients (some values)ABwaterCO2waterN22DAB m /s-91.17 10-92.01 10-9waterO22.60 10airNH319.6 10airairCO2H2O vapour-6-613.6 10-626 107/56Temp C1822250025(Ö96 p. 40)Ambient conditions: Đ 10-5 m2/s in gasesĐ 10-8 . 10-9 m2/s in liquidsĐ 10-11 . 10-13 m2/s in solidsTkF VT rz08Introduction to Process Engineering (PTG)#7/88/56Diffusion and convectionA m2 An example:Flow of a fluid from atube into a regionwhere c c0 for acertain speciesconvection, v Steady state, mass balance givesdiffusionc0xc0cdc0 D A v A c ; c c0 @ x 0dxxvx vx withc c 0. exp - Pe (Péclet number)D D vx ρvx ηηPe Re Sc with Sc (Schmidt number)Dη ρDρDTkF VT rz08
Introduction to Process Engineering (PTG)#7/89/56Mass transfer and boundary layers For mass transferbetween two phases Aand B, concentrationgradients usually onlyexist near the physicalboundary thatseparates A and BΦ molAcA,bulkAcA,interfaceABAcA,interfaceBboundary layermedium AcA,bulkBboundary layermedium B Thus, the driving forcesare active only inboundary layers at the Equilibriumbetweenseparating surfacecA,interfaceAphaseboundaryand cA,interfaceBTkF VT rz08Introduction to Process Engineering (PTG)#7/810/5610/56Mass heat transfer analogy /1Phase Phase12T1c1µ1c2,ic1c1,iTiµiT2c2µ2c2,ic1,ic2A temperatureA chemicalprofile, heatpotential profile,transfer 1 2 mass transfer 1 2Concentration profiles,mass transfer 1 2TkF VT rz08
Introduction to Process Engineering (PTG)#7/8Mass transfer coefficientinterface a1 (L)2 (G)xiC1.iΦAxC1yC211/5611/56 Mass flow species A:ṅA ΦA mol/s Mass transfer rate per area:ṄA ṅA/a Φ”A mol/(m2·s) Mass transfer coefficients, k,(unit: m/s) for both sides of theinterface:ṄA kx ·(c1,i-c1) ky·(c2-c2,i) Interface concentrations can beeliminated using equilibriumconstant K c1,i /c2,i c1*/c2 c1/c2*c1* c1 at equilibrium with c2, etc.yiC2.iL,G exampleTkF VT rz08Introduction to Process Engineering (PTG)#7/8The film model12/5612/56 A mass transfer coefficient canbe linked to the film model:Φ”A,mol k·(cA0-cA1) - ĐA·dcA/dy ĐA·(cA0-cA1)/δcwhich gives k ĐA/δc Thus, the boundary layer thicknesscan be estimated if k and ĐA areknown. The mass transfer limitations areconcentrated in a well-defined region.See heat transfermass transferNusselt numberSherwood numberPicture: SSJ84Nu hD/λ D/δTSh kD/Đ D/δcTkF VT rz08
Introduction to Process Engineering (PTG)#7/8 13/5613/56Mass heat transfer analogy /2Heat transferNu f(Re, Pr, L/D, Gr,.) Convection around a sphere: Nu 2 0.6Re½Pr⅓Transfer from a wall and a 5turbulent flow: 2000 Re 10and Pr 0.7Nu 0.027Re0.8Pr0.33(η/ηwall)1/7General: Nu CRemPrn, where m 0.33 . 0.8, n 0.33Mass transferSh f(Re, Sc, L/D, Gr,.)Convection around a sphere:Sh 2 0.6Re½Sc⅓Transfer from a wall and aturbulent flow: 2000 Re 105and Sc 0.7Sh 0.027Re0.8Sc0.33General: Sh CRemScn,where m 0.33 . 0.8, n 0.33 Chilton-Colburn analogies, heat and mass transfer values jH, jD:jD ShRe-1Sc-⅓jH NuRe-1Pr-⅓jH jD CRem-1 ½ƒ ƒ Fanning friction factor for pipe flowTkF VT rz08Introduction to Process Engineering (PTG)#7/814/5614/567.2 Phase equilibrium(gas-gas, gas-liquid, liquid-liquid)Henry’s Law, Raoult’s LawTkF VT rz08
Introduction to Process Engineering (PTG)#7/8Mass transfer and equilibrium15/5615/56Drying of wet gas in an glycol absorberwet gasdry gasEquilibrium determined bythermodynamicscH2O,eqRate determined bytransport processes andequipment designcH2Oin liqcH2OtimeglycolTkF VT rz08Introduction to Process Engineering (PTG)#7/816/5616/56Air above a lakeat equilibrium : K H2O yH2O 1and: someair dissolvesin the lake !y H2Ox H2OyO2 0.21yN2 0.79Equilibriumat surfaceConcentrationjumpxH2O 1TxO2 0.01xN2 0.01zx, y, Tx fraction in liquid,liquid, y fraction in gas,z position coordinate,coordinate, T temperatureTkF VT rz08
Gas-liquid phase equilibrium:17/5617/56Raoult’s law Assume a liquid mixture of components A, B,C, . at a temperature T. ”A” occupies a large fraction, say xA 5 % At temperature T, saturation pressure of puresubstance A (i.e. A vapour above liquid A) is pA0 For the mixture, the vapour pressure of Aequals pA yA.ptot xA.pA0 For a two-component mixtureof A and B: pB xB.pB0 (1-xA).pB0 ,using xA xB 1Picture: http://www.tannerm.com/raoult.htmIntroduction to Process Engineering (PTG)#7/8TkF VT rz08 Gas-liquid phase equilibrium:18/5618/56Henry’s lawAssume a liquid mixture of components A, B,C, . at a temperature T.”A” occupies a small fraction, say xA 5 %For the mixture, the vapour pressure of Aequals pA yA.ptot HcA.xAwith Henry constant Hc(unit: Pa, bar, .)yA /xA HcA/ptot β, yA β.xAdistribution coefficient β (mol/mol) Hc is a function of temperature, butindependent of pressure at ptot 5 bar.Picture: 34.gifIntroduction to Process Engineering (PTG)#7/8TkF VT rz08
Introduction to Process Engineering (PTG)#7/8Example: Water-ammonia vapour /1T 40 CPtot ?GASLIQUIDxH2O 0.3 mol/molxNH3 0.7 mol/mol19/5619/56 A mixture of ammonia and water 40 C, total pressure unknown Liquid composition:70 %-mol NH3 30 %-mol H2O What is the composition of thegas, yNH3, yH2O ? At equilibrium, no driving forcesor temperature gradients,xNH3 xH2O 1, yNH3 yH2O 1but xNH3 yNH3, xH2O yH2O !!!Relative volatility of NH3 with respect to water:α (yNH3 / xNH3) / (yH2O / xH2O) KNH3 / KH2O at equilibriumTkF VT rz08Introduction to Process Engineering (PTG)#7/8Example: Water-ammonia vapour /220/5620/56 Gas above an NH3/H2O liquid mixture withxNH3 0.7 and xH2O 0.3, T 40 C. Questions: pressure & equilibrium composition of gas ? From tabelised data for saturation pressures, at 40 C :p H2O 7.348 kPa; p NH3 1554.33 kPa x-values for liquid 5 % : use Raoult’s Law :pH2O xH2O· p H2O 0.3 · 7.348 kPa 2.22 kPapNH3 xNH3· p NH3 0.7 · 1554.33 kPa 1088.3 kPa Ptotal pH2O pNH3 1090.25 kPa 10.9025 bar yH2O 2.22 kPa / 1090.25 kPa 0.002 0.2 %-vyNH3 1088.3 kPa / 1090.25 kPa 0.998 99.8 %-vSource: ÇB98TkF VT rz08
Introduction to Process Engineering (PTG)#7/8Example: Water in air above a lake21/5621/56 Assume a lake with T 17 C, ptot 92 kPa atwater surface level. At the surface, the water will be saturated withwater, which means pH2O p H2O 1920 Pa. Fraction of water in the air at the water surfaceat equilbrium with the air above is thenyH2O pH2O/ptot 1.92 kPa / 92 kPa 2.09 %note: % %-v (volume %)Source: ÇB98TkF VT rz08Introduction to Process Engineering (PTG)#7/8Example: Air dissolved in lake water22/5622/56 Assume the same lake, T 17 C, ptot 92 kPa at watersurface level. At the surface pH2O p H2O 1920 Pa. A small amount ( 5 %-vol) of air will be dissolved inthe water: use Henry’s Law to calculate the equilibrium: At T 290 K, Hc AIR,water 6200 MPa pAIR ptot – pH2O 90.008 kPa xAIR,AIR, water side pAIR,air side / Hc AIR, water 0.090008 MPa / 6200 MPa 1.45 10-5 0.00145 %-v This means 1.45 moles air (molar mass 29 kg/kmol) in100000 moles water (molar mass 18 kg/kmol), whichmeans 23.4 mg air / kg waterSource: ÇB98TkF VT rz08
#7/823/5623/56Introduction to Process Engineering (PTG)Two-component phase diagram (G/L)critical pointspure A, Bx A 1- x By A 1- y BP, T, x – diagram for abinary gas-liquid systemPicture: T68TkF VT rz08#7/824/5624/56Introduction to Process Engineering (PTG)Binary vapour-liquid equilibriumConstant pressure; the x-y diagramyTemperaturexPicture: T68TkF VT rz08
Introduction to Process Engineering (PTG)#7/8Relative volatility, αα Picture: WK9225/5625/56 Raoult (for not-small xi):yA xApA , yB xBpB xB 1-xA, yB 1-yA yB/yA α ·xB/xAα relative volatilityα pB /pA α (T) result:(1-yA)/yA α ·(1-xA)/xAyA α·xA / (1 (α-1)·xA)TkF VT rz08Introduction to Process Engineering (PTG)#7/826/5626/567.3 Separation processes(gas-gas, gas-liquid, liquid-liquid):equilibrium stagesTkF VT rz08
#7/827/5627/56Introduction to Process Engineering (PTG)Separation of mixtures: 1 stageFor example: separating phenol fromwater (L) by adding benzene (V) in aseparation funnel.x phenol conc.conc. in L, y phenol conc.conc. in VEquilibrium constant: K y1 / x1Separation factor: S K·K·V/LFraction of phenol separated S / (S 1) for 1 stageTkF VT rz08#7/8Introduction to Process Engineering (PTG)Cu - ore2,0releasefrom oreDiluted H2SO40,1extraction2,2Organic solvent0,2recovery50Concentrated H2SO440electrolysisCuNumbers are concentrations kg leFour-stageprocessing for Curecovery from Cucontaining ore(liquid / liquid)Source: WK92TkF VT rz08
Introduction to Process Engineering (PTG)#7/8Phase equilibrium stages ntThe extraction and the recovery process arecombinations of mixing tanks and settling tanksPicture: WK92TkF VT rz08Introduction to Process Engineering (PTG)#7/8Phase equilibrium stages nt Cu is transferred from feed stream L to solvent (”support phase”)phase”) V; theextraction unit can be described as a series of equilibrium stages.stages. Equilibrium constant for Cu in feed stream andsolvent stream is K yCu/xCu Thus, for an equilibrium stage n: yn Kxn Streams L and V are often roughly constantL1 L2 . Ln L; V1 V2 . . Vn V separation factor S K·K·V/LLnxnLn 1xn 1equilibriumstage nequilibriumstage n 1Vnynequilibriumstage n-1Vn-1yn-1TkF VT rz08
Introduction to Process Engineering (PTG)#7/8Example: phase equilibrium stage31/5631/56Picture: WK92 For example: phase equilibrium constant K 5;L 1000 kg/s, V 800 kg/s S KV/L 4.Cu in feed x0 0.2 %-wt 0.002 kg/kg For clean solvent, y2 0, then Vy1 / Lx1 S Mass balance gives Lx0 (S 1)w-rLx0 Vy2 Lx0 Lx1 Vy1 Lx1(1 KV/L) Lx1(1 S) Lx0 2 kg Cu/s w Lx1 Lx0 / (1 S) 0.4 kg Cu/sTkF VT rz08Introduction to Process Engineering (PTG)#7/832/5632/567.4 Separation processes(gas-gas, gas-liquid, liquid-liquid):continuous distillation;packed tower columnsTkF VT rz08
#7/8Introduction to Process Engineering (PTG)Continuousdistillation33/5633/56Pictures: T68TkF VT rz08Introduction to Process Engineering (PTG)#7/8Distillation, principle34/5634/56 Liquid withcomposition a1 boils attemperature T2, givingvapour withcomposition a’2 whichis enriched incomponent A Taking out and coolingvapour a’2 gives liquida4 at temperature T4 In equilibrium withliquid a4 is vapour a 4,again further enrichedin component APicture: A83TkF VT rz08
Introduction to Process Engineering (PTG)#7/8Continuous distillation (binary) Separating a mixture of2 components A and B,with A more volatile Liquid for absorptionsection produced bycondensing some topproduct 35/5635/56Top product:more volatilecomponentTop section:rectifying sectionabsorbing the lessvolatile componentGas for strippingsection produced byboiling some bottomproductBottom section:stripping sectiondesorbing the morevolatile componentRoughly equimolarexchange:exchange: 1 mol Aliquid gas 1mol Bgas liquidBottom product:less volatilecomponentTkF VT rz08Introduction to Process Engineering (PTG)#7/836/5636/56Packed tower columns ech.comhttp://www.sulzerchemtech.comTkF VT rz08
Introduction to Process Engineering (PTG)#7/8Packed tower columns /237/5637/56 Mass transfer from gas to liquid or vice versa wherethe liquid forms a (thin) film on the surface of packingmaterial elements, creating a large contact surface ”a”(m2 / m3 apparatus) For relatively smallamounts of materialtransferred (say, 2% ofthe streams) the processmay be consideredisotherm (vaporisation andcondensation have a heateffect!) and streams Vand L may beconsidered constant.Picture: W92TkF VT rz08#7/8Introduction to Process Engineering (PTG)38/5638/567.5 Particle technology;Multi-phase flows”Consider a particle”Prof. Brian Scarlett1938-2004TkF VT rz08
#7/839/5639/56Frequency distribitionXn dN/dX n 0,1,2,3Introduction to Process Engineering (PTG)Particle (or droplet) size distributiondN/dXXdN/dX dL/dXN numberL lengthS surfaceV volumeX2dN/dX dS/dXX3dN/dX dV/dXParticle size, X Different distributions for number,length, surface and volume ! Different particle size analysers givedifferent distributions: some measurelength, others measure surface, etc.TkF VT rz08Introduction to Process Engineering (PTG)#7/8Particle shape40/5640/56Shape factor,“sphericity” φφ 4.836 (volume)2/3surface surface of sphere with same volumesurface of particleTkF VT rz08
Introduction to Process Engineering (PTG)#7/841/5641/56Particles in fluid flowsTurbulence as seenby Da VinciA smooth (a) androughened (b) ball enteringwater at 25 CParticles andturbulent eddiesPicture: CR93TkF VT rz08Introduction to Process Engineering (PTG)#7/8Eulerian vs. Lagrangianparticle representation42/5642/56time t4time t1time t3time t1time t2time t2time t4Focussing on a control volume (Euler), left,or focussing on particle trajectories (Lagrange), rightEuler-Euler (for fluid and particulate phase) andEuler-Lagrange methods are both widely used Source: ZH00TkF VT rz08
Aerosols Aerosol: A suspension ofsolid or liquid particles in agas. Aerosols are stable forat least a few seconds and insome cases may last a yearor more. The term ”aerosol”includes both the particlesand the gas, which is usuallyair. Particle size ranges from0.001 to over 100 µm.For example smoke is adispersion of solid particlesor droplets in air. Sol: particles dispersed in aliquid, for example inkSource: ZH00Picture:http://kapstadt.org/en/photos of cape town/people of south africa/43/5643/56Picture: http://www.aerosol-soc.org.uk/images/Stack plume.jpgIntroduction to Process Engineering (PTG)#7/8TkF VT rz08Introduction to Process Engineering (PTG)#7/844/5644/56Flow of particle swarmsDrag coefficient for sphere in swarm, CD*,corrected for effect of neighbour particlesCD * CD f ( ε ) CD ε nε voidage, porosityRichardson - Zaki hindrance factor:Small particles,low Re: ƒ(ε) ε-4.7Re 0.20.2 11 500 500n4.654.35 Re-0.034.45 Re-0.012.39Source: ZH00TkF VT rz08
Introduction to Process Engineering (PTG)#7/845/5645/56Flow in packed bedsDarcy’s Law:u K Δpη fluidLwith permeability KLKozeny - Carman equation:u ε 3 ( Δp )5(1 ε)S 2v η fluidLSv specific surface surface/volumeSource: ZH00Sv 6 /dp for a sphere with diameter dpTkF VT rz08Introduction to Process Engineering (PTG)#7/8Fluidised beds46/5646/56Source: ZH00gas bubble in agas/solidfluidised bedTkF VT rz08
#7/8Introduction to Process Engineering (PTG)mechanic, pneumatic, hydraulicPneumatic conveyor / drier2Conveyor beltIntroduction to Process Engineering (PTG)#7/834 pneumatic conveying regimes :- Solid Dense Phase- Discontinuous Dense Phase- Continuous Dense Phase- Dilute PhasePicture 1: http://www.bateman.co.za/pro-eng-pne.htm1TkF VT rz08Flow of powders in/from silosa. Mass flowc. Expanded flowe. RatholePicture 3:http://www.mactenn.com/pnu overview.htmlConveying systems:Picture 2: nveyor/coal-baltimore-lg.jpg47/5647/5648/5648/56b. Funnel flowd. “Pipe”f. ArchingSource: ZH00TkF VT rz08
Introduction to Process Engineering (PTG)#7/8A gas cyclone49/5649/56AdvantagesSimple, cheap andcompactLarge capacityDisadvantagesLarge pressure dropLow efficiency“Catch” removal problemsNo removal below 5 μmProblems above 400 CSee also hydro-cyclones andother cyclones for liquid-solid,liquid-liquid and liquid-gas separationsSource: ZH00TkF VT rz08#7/850/5650/564 process steps:1. Particle charging2. Particle movement relativeto the gas flow3. Particle collection atdeposition surface4. Particle removal fromdeposition surface (oftendiscontinuous)Note: the electric propertiesof the particles to beremoved should besuitable, otherwise usea filter systemTypically quite large, mainlyused at power plants forfly-ash removal from flue gasPictures: http://www.eas.asu.edu/ holbert/wise/electrostaticprecip.htmlIntroduction to Process Engineering (PTG)Electrostatic precipitators (ESPs)TkF VT rz08
Inside out / outside in operationSource: ZH00Picture: terkop.jpgBaghouse filters51/5651/56Picture: bag.gifIntroduction to Process Engineering (PTG)#7/8TkF VT rz08Introduction to Process Engineering (PTG)#7/8Liquid filtration52/5652/56ΔΔϕϕRotary drum filter Horizontal belt filterSource: IGH91, ZH00 Constant pressure filtrationN rotation /min (rpm), drumradius R (m),length L (m),submerged angle Δ ϕ (0 . π)volume element ΔA R L Δ ϕis submerged for a timeΔt Δ ϕ / (2π N) (min)TkF VT rz08
Sedimentation of suspensionsFeed53/5653/56EffluentClear zoneFeed zoneThickening zoneDry solids concentrationSludge dischargeContinuous thickenerBatch sedimentation testPicture: http://www.geocities.com/fitzgerrell/work img/thickener.jpgIntroduction to Process Engineering (PTG)#7/8Source: IGH91, ZH00TkF VT rz08Solid-solid separations Used for separatingunwanted materialsor size fractions Equipment examples:––––Sieves, screensCyclones, centrifugesHydraulic separatorsSink-float, frothflotation separators– Magnetic, electrostaticseparatorsScreens54/5654/56Picture http://www.key.net/p product.cfm?productid 30Picture: on to Process Engineering (PTG)#7/8Froth flotationTkF VT rz08
Crystallisers Solid product crystals can beproduced from gases, liquid mel
TkF VT rz08 Mass transfer and equilibrium Drying of wet gas in an glycol absorber c H2O wet gas dry gas time c H2O in liq c H2O,eq Equilibrium determined by thermodynamics Rate determined by transport processes and equipment design glycol #7/8 16/56 Introduction to Process Engineering (PTG) TkF VT rz08 Air above a lake x fraction in liquid, y .
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Chemical Reactions, and Stoichiometry Section 2.1: The Atomic Mass The atomic mass is the mass of 1 atom. Atoms are very small and their mass is a . molecular or ionic compound. The formula mass is expressed in a.m.u. Molar mass is the sum of atomic masses of all atoms in a mole of pure substance. The molar mass is expressed in g/mol.
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existing mass families (Place Mass) or create in place masses (Create Mass). Using either tool the first thing you'll see is a dialog box indicating that Revit has activated the Show Mass mode. Mass visibility is controlled in the following ways: The Show Mass button on the View toolbar toggles mass visibility on and off in all views
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1. Define key messages that are imbedded in the introductions 2. Explain why starting a rollout of the standards with introductions is important. 3. Describe actions that you can consider in your district as you begin to roll out the standards. 1. What are standards? 2. To whom do standards apply? 3. Why were the standards revised? 4.
