mass transfer & separation tectransfer & separation

februari 2008

MassMass transfer & separation technology 424302 2008 transfer & separation technology 424302 2008 -- APPENDIXAPPENDIX

Åbo Akademi - TkF - Värmeteknik - Biskopsgatan 8, 20500 Åbo MÖF-ST RZ 2008 1/42

MultiMulti--componentcomponent massmass transfertransferusingusing transparanciestransparancies that that accompanyaccompany ””MassMass Transfer in Transfer in MulticomponentMulticomponent mixturesmixtures””

by J.A. by J.A. WesselinghWesselingh & R. Krishna, & R. Krishna, DelftDelft University Press (2000)University Press (2000)

MassMass transfer and transfer and separation technologyseparation technologyMassMassööverfverfööring och separationsteknik ring och separationsteknik ((””MMÖÖFF--STST””))

Appendix. Appendix. MassMass transfer in transfer in multimulti--componentcomponent mixturesmixtures

Ron ZevenhovenÅbo Akademi University

Heat Engineering Laboratorytel. 3223 ; [email protected]

See alsoKrishna & WesselinghChem. Eng. Sci.52(6) 1997 861-911

februari 2008





A.1 A.1 ””OldOld--schoolschool”” massmass transfer transfer




J Ddcdzi

i= − ciα

ciβ

βdiffusivity

k Dzii=

Δ

mass transfer coefficient

Fick’slaw

Ji : flux of i with respect to the mixture

J D cz

k ci ii

i i= − = −ΔΔ

Δ

Δz

Δc c ci i i= −β α

Mass TransferMass Transfer as you have learned it

α




N D dcdz

Nxi ii

i= − +differential equation

N k c Nxi i i i= − +Δdifference equation123 {

Ni : flux with respect to an interface

diffusion flux

drift flux

Stefan or drift correction

N Nii

=∑flux of mixture

Diffusion with DriftDiffusion with Drift

−k s ci i iΔ




gas: c c c constant1 2+ = =

fluxes with respect to mixture

J D dcdz1 1

1= −

J D dcdz2 2

2= −

( )J J Dd c c

dz1 21 20+ = = −+

+

0 1

D

x1

only one binary D, which is independent

of composition

Classic Classic -- in Gasesin Gases




N2, CO2N2, H2

ideal gases, 100 kPa, 298 K

A B

beginning: xN2 = 0 46.xH2 = 0 54.

xN2 = 0 52.xCO2 = 0 48.

Question: Does N2 transfer (a) from A to B? (b) from B to A?(c) not at all? (d) or does it do (a), (b) and (c)?

Three Gases (1)Three Gases (1)

APPENDIX chapter to course material for ÅA TkFcourse 424302 / 2008

page 1 of 10




0.6

0.5

0.4

0.6

0.4

0.2

0.00 10 20

N2 A

B

H2

CO2

A

B

B

A

mol

e fra

ctio

n x i

reverse diffusion

timeh

Three Gases Three Gases (2)(2)




Two Two CationsCations

H+

Cl-

Na+

Cl-

cationpermeable membrane

highconcentration

lowconcentration

excess +chargeand electrical fieldso Na+ can move

against its concentration gradient!

H+ moves rapidly H+

Na+

1

3

2




He

298 K

100 kPa

Ar

298 K

100 kPa

!friction (He / plug) < friction (Ar / plug)

the plug, matrix or membrane is a

(pseudo)component

MM

Ar

He

N NHe Ar≈ −3

2 Gases in a Porous Plug (1)2 Gases in a Porous Plug (1)




He

298 K

100 kPa

Ar

298 K

101 kPa(for example)

main reason:viscous flow

retards He, accelerates Ar

Δp

N NHe Ar= −

2 Gases in a Porous Plug (2)2 Gases in a Porous Plug (2)

februari 2008





A.2 Driving A.2 Driving forcesforces




1kg

1 m

the potential difference is the work required to change the condition of the weight

here: Δψ Δ= = ≡mg z 981 981. .J ( Nm)

or, per mole Δψ Δi iM g z=

1kg

Fi

the driving force is the negative potential gradient:F d

dzM gi

ii= − = −

ψ

the force is downwards

Gravity Gravity -- a simple Potentiala simple Potential

Δz


page 2 of 10




Chemical PotentialChemical Potentialxi γ i

mixture ( ) ii aRTTpconst ln, +=μ

chemical potential

activity

pure i (one mole)

( )μi const p T∗ = ,

iiii aRT ln−=μ−μ=μΔ ∗

activity coefficientwork required: change in the chemical potential

iii xa γ=

februari 2008





chemical potential in an ideal solution ii xRTTpconst ln),( +=μ

in an ideal gas ⎟⎠⎞

⎜⎝⎛+=μ

ppRTTpconst i

i ln),(

partial pressure

μμ in an in an Ideal SolutionIdeal Solution

februari 2008





momentum ‘in’ momentum ‘out’

forces

∑ invm )( &

change of momentum ∑∑∑ +−= Fvmvm

dtmvd

outin )()()(&&

∑ outvm )( &

∑F

Momentum BalanceMomentum Balance

februari 2008





(2)(1)

1u2u

zH2 CO2

dzz+

Moving Through Each OtherMoving Through Each Other

species velocities

februari 2008





dzz+z

area A

volume Adz

zAp1 dzzAp+1

)( 1221 uuppforce

friction−∝⎟

⎠⎞

⎜⎝⎛

dzdp

forcedriving 1−=⎟

⎠⎞

⎜⎝⎛

Forces on Hydrogen (1)Forces on Hydrogen (1)

121221

21

21221212

2121

1

21211

===

−=−∝

−∝−=

−∝−

,,,

,

,1

11

ζDRT

DRTζt coefficienfriction with

)uu(xζ)uu(RTp F force Driving

moleper force )uu(RTpdzdp

pRT gives

RTpcwith

per volume force )uu(ppdzdp :balance Force Note:

often x2 ≈ 1and u2 = 0

Gases: u ~ 10-2 m/sLiquids:u ~10-4 m/s




dzda

aRT

dzadRT

dzdF i

i

iii −=−=

μ−=

ln−

RTx

dxdzi

i

in ideal solutionsfor a given T and p

Driving Force (per Mole of Driving Force (per Mole of ii))


page 3 of 10




( )F x u ui i j j i jj i

= −≠∑ζ ,driving force

on i

friction coefficient between i and j

mole fractionof j

(diffusive) species velocities

MaxwellMaxwell--Stefan EquationStefan Equation




in a solid particle

Δz d≈10

Δzd

gases

liquids

Δz ≈ −10 4 m

Δz ≈ −10 5 m

membrane

Δz = − −10 107 4K m

Film Theory, Thickness of FilmsFilm Theory, Thickness of Films

eddies & large scale convection

‘film’: no eddies

phase boundary

two thin, one dimensional ‘films’ next to the phase boundary

februari 2008





Difference Form of ForceDifference Form of Force

za

aRT

zaRT

zF i

i

iii Δ

Δ−=

ΔΔ

−=ΔμΔ

−=ln

zx

xRT i

i ΔΔ

−

in ideal solutions

for a given T and p




Δμ1

RT+1

0

-1

0 2

approximate

exact

‘approximate’ works out better in difference equations

⎟⎟⎠

⎞⎜⎜⎝

⎛

α

β

1

1lnaa

( )αβ

αβ

+−

=Δ

11

11

1

1

5.0 aaaa

aa

α

β

1

1

aa

−∞

-2

ApproximationApproximation




Δz ≈ −10 5 m

mole fractionof CO2

x1 0003α = .

F RTx

dxdz

RTx

xz1

1

1

1

1= − ≈ −ΔΔ

x1 0001β = .

Forces in a Forces in a Glass of BeerGlass of Beer

growing bubble of CO2

Note: u2 = 0

februari 2008





ExampleExample (3.1 from (3.1 from bookbook))


page 4 of 10

februari 2008





ExampleExample (3.1 from (3.1 from bookbook):): answeranswer

februari 2008





A.3 A.3 FrictionFriction




Friction Coefficients of SpheresFriction Coefficients of Spheres

( )ζ πη12 2 112 1 1

3 3 10, = A× ≈ × − − −d Nmol ms1

F1F1

A = ×6 1023

molecules mol-1

coefficient of a single sphere

spheres ‘1’

liquid ‘2’

SeePTG § 6.7:Stokes Law




Maxwell-Stefan diffusivity of large molecules in dilute liquids (not gases)

( ) sm10

104.01010106300314.8 2

99323

−−− ≈

××××××

→

Ð RT12

12,

,

≡ζ

Ð RTd12

2 13, = A πη

Diffusion and Friction CoefficientsDiffusion and Friction Coefficients

2,12,1 Ð

RT≡ζ

each others ‘inverse’

we use both




2 components: 1 relative velocity1 independent equation

3 components: 2 relative velocities2 independent equations

n components: n - 1 relative velocitiesn - 1 independent equations

One Equation MissingOne Equation Missing




only relative

velocities

∑≠

−ζ=ij

jijjii uuxF )(,bootstrap

‘floating’ transport relations: have to be ‘tied’ to surroundings

Bootstrap Bootstrap (1)(1)


page 5 of 10

februari 2008





N2CO2

N2H2

H+

Cl-Na+

Cl-

HeAr

no net volume flow plug does not move

membrane does not move

(almost) no charge transfer

Bootstraps (2)Bootstraps (2)

februari 2008





∑ −ζ=j

jijijiii uuxcxcxF )(,

∑ −ζ=j

jiijjii NxNxf )(,

force on i per unit volume of mixture

in practical problems we use fluxes:

N u c u cxi i i i i= =

flux form of MS-equation:

FluxesFluxes

Fi / V ; xi = ci / c




x1α

β1xx1

positivedirection

α β

binary:

infinitesimallayer

finite layer(approximate)

u1( )21

2,12

1

1

uuÐRTx

dzda

aRT

−=−

( )( )dzÐ

uuxa

da2,1

212

1

1 −=−

( )z

Ðk

kuux

aa

Δ=

−=

Δ− 2,1

2,12,1

212

1

1

From Differential to DifferenceFrom Differential to Difference

α1a

β1a

februari 2008





876film

iu

0iu

α β0

icaverage concentration ic

species velocity(depends on position in film)

species velocity at the average composition

positive velocity

Average VelocityAverage Velocity

februari 2008





)(1212

2,11

1 uuxka

a−=

Δ−

cxc 1×

)(12112

2,11

11 NxNx

ckaax −=

Δ−

Differences with FluxesDifferences with Fluxes

februari 2008





∑≠

−=

Δ−

ij ji

jij

i

i

kuu

xaa

,

)(

using velocities using fluxes

∑≠

−=

Δ−

ij ji

jiij

i

ii ck

NxNxaax

,

)(

for ideal solutions

ixΔ−

Multicomponent EquationsMulticomponent Equations


page 6 of 10




100

10-2

10-4

10-6

ki j,1m s−

kÐ

zi ji j

,,=

Δ

gases ≈ − −10 1 m s 1

liquids ≈ − −10 4 m s 1

Transfer Transfer CoefficientsCoefficients

in pores

in pores

februari 2008





changes are not very important

changes are not very important

Temperature EffectsTemperature EffectsMS-equation

terms) diffusion (thermal)u(uxζFj

jijji,i +−=∑small

driving force

dzdx

xRT

dzdF i

iT

ii −=⎟

⎠⎞

⎜⎝⎛ μ−=

difference form:

zx

xTRF i

ii Δ

Δ−=

at constant temperature

average film temperature

februari 2008





A.4 A.4 BinaryBinary examplesexamples




x1α

x1

0α β

drops on a tray

gas: trace of NH3 (1)

bulk of N2 (2)

..as you already knew..

Stripping Stripping -- dilutedilute

12 ≈x

transport relation

bootstrap

flux

ckNxNxx

2,1

21121

−=Δ−

02 =N

12,11 xckN Δ−=




0β

x1 05= .x1β

α

x1α

Stripping Stripping -- concentratedconcentrated

12,112

2,11 2 xckx

xck

N Δ−=Δ−=




heat

benzene (1), volatile

toluene (2)

x1α

x2α

x2β

x1β

y K x1 1 1= β

y K x2 2 2= β

vapour removed by convection

bootstrap:NN

yy

1

2

1

2

= = ν

Vaporising DropletVaporising Droplet


page 7 of 10




NN

1

2

= ν

− =−

Δx x N x Nk c1

2 1 1 2

12,

− =−

Δx x N x Nk c2

1 2 2 1

12,

Nk c

x xx1

12

2 11= −

−,

νΔ N

k cx x

x212

1 22= −

−,

νΔ

example ν = 2 x x1 2 0 5= = .

N k c x1 12 14= − , Δ ( )N k c x2 12 22= − − , Δ

Fluxes from Vaporising DropletFluxes from Vaporising Droplet

Stefan (drift) correctionsΔx1 < 0Δx2 > 0




O C CO2 2 2+ →

C

O2 1( )

CO( )2

1.0

0.60.4

0.0

both components are moving and have a high concentration

k12210, = − −ms 1

bootstrap:N N2 12= −

calculate N1 and N2

c = −10 mol m 3

Carbon GasificationCarbon Gasification




( )− =−

=+

Δx x N x Nk c

x x Nk c1

2 1 1 2

1 2

2 1 1

1 2

2

, ,

N N2 12= −

( )Nk c

x xx1

1 2

2 11

2

210 10

0 7 2 0 30 6= −

+= −

×+ ×

−−

,

. ..Δ

( )N exact1 0 046 0 047= − −. : . mol m s2 1

( ) 122 smmol094.0:092.0 −−−−= exactN

Fluxes in GasificationFluxes in Gasification

almost the same




Binary DistillationBinary Distillation

0β

x1β

N1

α

x1αtransport relation

N2

x2α

x2β

hexane (1)

heptane (2)− =

−Δx x N x N

k c12 1 1 2

12,

bootstrap N N1 2= −

(equimolar exchange)

( )→ − =+

Δxx x N

k c11 2 1

12,

N k c x1 12 1= − , Δ N k c x2 12 2= − , Δ




1

2

3

4

5

6

membrane stagnant

bulk stagnant (absorption)

trace stagnant (polarisation)

equimolar exchange (distillation)

interface determined (vaporisation)

reaction stoichiometry

uM = 0

02 =N

u1 = 0

N N1 2 0+ =

NN

yy

1

2

1

2

=

Some BootstrapsSome Bootstraps

N N1

1

2

2ν ν=

februari 2008





A.5 A.5 TernaryTernary examplesexamples


page 8 of 10




( ) ( )

( ) ( )

− = − + −

− = − + −

ddz

x u u x u u

ddz

x u u x u u

μζ ζ

μζ ζ

112 2 1 2 13 3 1 3

22 1 1 2 1 2 3 3 2 3

, ,

, ,

forces per mole of ‘1’

forcesper mole of ‘2’

Ternary Ternary -- per mole of per mole of ii




( ) ( )

( ) ( )

− = − + −

− = − + −

⎫

⎬⎪

⎭⎪

x ddz

x x u u x x u u

x ddz

x x u u x x u u

11

12 1 2 1 2 13 1 3 1 3

22

2 1 1 2 2 1 2 3 2 3 2 3

μζ ζ

μζ ζ

, ,

, ,

forces per mole of mixture

these should cancel:

ζ ζ2 1 12 2 1 12 2 1 12, , , , , ,≡ ≡ ≡Ð Ð k k

Ternary Ternary -- per mole of Mixtureper mole of Mixture




binary

ternary

quaternary

ckNxNx

ckNxNxx

3,1

3113

2,1

21121

−+

−=Δ−

ckNxNx

ckNxNxx

3,2

3223

1,2

12212

−+

−=Δ−

More ComponentsMore Components




CondensorCondensor vapour

liquid

cooling water

H O2

NH3

H2

0.6

0.4

0.2

0.0

NH3 (1) and H2O (2) condense on a tube

find the velocities in the gas film

H2 (3) does not condense

k k13 2 333 10, ,= = × − −ms 1

k1231 10, = × − −ms 1

Mix: NH3+H2O + H2




Condenser (2)Condenser (2)transport (MS) relations:

NH3 :

H2O :

bootstrapthree linear equations, three unknowns

exact solutions:

03 =N

30)103(4.03.0

30)101(3.04.02.0 3

313

21−− ×

−+

×−

=−NNNN

30)103(3.03.0

30)101(3.04.04.0 3

323

12−− ×

−+

×−

=NNNN

015.01 =N 122 smmol045.0 −−=N

013.01 =N 122 smmol049.0 −−=N




H O2

NH3

H2

mixture velocity

H2O moves down its gradient

NH3 dragged against its gradient

H2 does not move at all

Condenser (3)Condenser (3)


page 9 of 10




Ternary Distillation (1)Ternary Distillation (1)

in which direction does 2 move?

0.530.45

0.63

0.35

0.02 0.02

1

3

2

1 ethanol 3 water2 a trace of butanol

large friction between 1 2andk12

28 10, = × − −ms 1

k k13 2 3220 10, ,= = × −

bootstrap: equimolar exchange

vapour liquid

m s-1

u y u y u y1 1 2 2 3 3 0+ + =




0.018

0.020

0.0220.0214

0.018

0.020

0.0186

0.022y2α y2β

no motion

u22375 10= − × −.

u22158 10= − × −.

u220 81 10= + × −.

ButanolButanol -- which direction?which direction?




Ammonia reactionAmmonia reaction

N2 (1)

H2 (2)

NH3(3)

α β

catalytic surface

N H NH2 2 33 2+ ⇔

− =−

+−

Δx x N x Nk c

x N x Nk c1

2 1 1 2

12

3 1 1 3

13, ,

− =−

+−

Δx x N x Nk c

x N x Nk c2

1 2 2 1

12

3 2 2 3

2 3, ,

− =−

+−

Δx x N x Nk c

x N x Nk c3

1 3 3 1

13

2 3 3 2

2 3, ,

transport relations:

bootstrap: N N N1 2 3

1 3 2= =

−




− =−

+−Δx

xx u u

kx u u

k1

12

1 2

123

1 3

13, ,

L=−x u u

keffeff

eff

1

1,

a ternary can be approximated as a binary when

When is: 3 = 2?When is: 3 = 2?




1

2

3

one friction term dominates:

(example: mobile species in many membranes)equal velocity of two species:

(example: Na+ and Cl- in water)equal diffusivities (‘1’ in m- and p-xylene)

xk

xk

x u uk

2

12

3

133

1 3

13, , ,

<< →−

( )u u xk

xk

u u2 32

12

3

131 2= → +

⎛

⎝⎜

⎞

⎠⎟ −

, ,

( ) ( )k kx x u x u x u

k12 132 3 1 2 2 3 3

12, ,

,

= →+ − +

x xeff = 3

k keff1 13, ,=

xk

xk

xk

eff

eff1

2

12

3

13, , ,

= +⎛

⎝⎜

⎞

⎠⎟

x x xeff = +2 3

u x u x ux xeff =++

2 2 3 3

2 3




simplifying transport equation of N2 in ammonia formation:eliminate N2 and N3 with N N N N2 1 3 13 2= = −

N k c xeff1 1 1= − , Δ1 3 21

2 1

12

3 1

13kx x

kx x

keff, , ,

=−

++

with

similarly for H2 and NH3

Effective Binary in Reactive Effective Binary in Reactive SystemSystem

If all fluxes Ni are related via the same reaction stoichiometry→ pseudo - binary


page 10 of 10

mass transfer & separation tectransfer & separation

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