Maxwell Stefan

Introduccion a la Ecuación de Maxwell-Stefan para Transferencia de Masa

Dr. Ing. Bernardo Yépez Silva Santisteban. EPG-UNALM 2006

Introducción a la Ecuación de Maxwell-Stefan para transferencia de masa

Wesselingh, J.A. and Krishna, R., Mass Transfer in Multicomponent Mixtures,

VSSD, Delft (2000)


J Ddcdzi

i= − ciα

ciβ

βαdiffusivity

kDzii=

∆

mass transfer coefficient

Fick’slaw

Ji : flux of i with respect to the mixture

J Dcz

k ci ii

i i= − = −∆∆

∆

∆z

∆c c ci i i= −β α

Mass Transfer as you have learned it

Introduccion a la Ecuación de Maxwell-Stefan


N Ddcdz

Nxi ii

i= − +differential equation

N k c Nxi i i i= − +∆difference equation123 {

Ni : flux with respect to an interface

diffusion flux

drift flux

Stefan or drift correction

N Nii

=∑flux of mixture

with Drift

− k s ci i i∆


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gas: c c c constant1 2+ = =

fluxes with respect to mixture

J Ddcdz1 1

1= −

J Ddcdz2 2

2= −

( )J J D

d c c

dz1 21 20+ = = −

++

0 1

D

x1

only one binary D, which is independent

of composition

Classic - in 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)



0.6

0.5

0.4

0.6

0.4

0.2

0.00 10 20

N2A

B

H2

CO2

A

B

B

A

mole fraction xi

reverse diffusion

timeh

Three Gases (2)





Two Cations

H+

Cl -

Na+

Cl-

cationpermeable membrane

highconcentration

lowconcentration

excess +chargeand electrical field

so 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)



He

298 K

100 kPa

Ar

298 K

101 kPa(for example)

main reason:viscous flow

retards He, accelerates Ar

∆p

N NHe Ar= −

2 Gases in a Porous Plug (2)





1kg

1 m

the potential difference is the work required to change the condition of the weighthere: ∆ψ ∆= = ≡mg z 981 9 81. .J ( Nm)

or, per mole ∆ψ ∆i iM g z=

1kg

Fi

the driving force is the negative potential gradient:

Fddz

M gii

i= − = −ψ

the force is downwards

Gravity - a simple Potential



Chemical Potential

xi γ i

mixture ( ) ii aRTTpconst ln, +=µchemical potential

activity

pure i (one mole)

( )µi const p T∗ = ,

iiii aRT ln−=µ−µ=µ∆ ∗

activity coefficientwork required: change in the chemical potential

iii xa γ=



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

in an ideal gas

+=µpp

RTTpconst ii ln),(

partial pressure

µ in an Ideal SolutionIntroduccion a la Ecuación de Maxwell-Stefan




momentum ‘in’ momentum ‘out’

forces

∑ invm )( &

change of momentum ∑∑∑ +−= Fvmvm

dtmvd

outin )()()(

&&

∑ outvm )( &

∑F

Momentum Balance



(2)(1)

1u2u

zH2 CO2

dzz +

Moving Through Each Other

species velocities



dzz +z

area A

volume Adz

zAp1 dzz

Ap +1

)( 1221 uuppforce

friction −∝

dzdp

forcedriving 1−=

Forces on Hydrogen (1)





dzda

aRT

dzad

RTdzd

F i

i

iii −=−=µ−= ln − RT

xdxdzi

i

in ideal solutionsfor a given T and p

Driving Force (per Mole of i)



( )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

Maxwell-Stefan Equation



the driving force on a species i in

a mixture

the sum of the friction forces between i and the

other species j

the friction exerted by j on i is proportional - to the fraction of j in the mixture and- to the difference in velocity between i and j.

Multicomponent Diffusion - in Words





engineer

scientist

great scientist

one day

one week

one month

∑≠

−ζ=∆ψ∆−

ijjijji uux

z)(,

( )∑≠

−ζ=µ−ij

jijjii uux

dzd

,

( )∑≠

−ζ=µ∇−ij

jijjii uuxrr

,

Why Simplify?



eddies & large scale convection

‘film ’: no eddies

phase boundary

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

the other phase

Film Theory



gases

liquids

membrane

in a solid particle

∆zd≈10

∆z ≈ −10 4 m

∆z ≈ −10 5 m

∆z = − −10 107 4K m

∆zd

Thickness of Films





Difference Form of Force

za

aRT

za

RTz

F i

i

iii ∆

∆−=∆

∆−=∆µ∆−= ln

zx

xRT i

i ∆∆−

in ideal solutions

for a given T and p



∆µ1

RT+1

0

-1

0 2

approximate

exact

‘approximate’ works out better in difference equations

α

β

1

1lna

a

( )αβ

αβ

+−

=∆

11

11

1

1

5.0 aa

aa

aa

α

β

1

1

a

a

− ∞

-2

Approximation



growing bubble of CO2 ∆z ≈ −10 5 m

mole fractionof CO2

x1 0003α = .

FRTx

dxdz

RTx

xz1

1

1

1

1= − ≈ − ∆∆

x1 0 001β = .

Forces in a Glass of Beer





1 kg of CO2 FRT

zx

x11

1

5

8

9

8314 30010

0 0020 002

24 10

5 10

≈ −

= − × × −+

= ×

≈ ×

−

∆∆

. ..

.N

mol CO

kgfkgCO

2

2500 000 tons

1025

particles

Supertanker!



J. C. MaxwellJ. C. Maxwell J. StefanJ. Stefan

The great-great-GrandfathersIntroduccion a la Ecuación de Maxwell-Stefan


A. FickA. Fick





Friction 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’



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

( ) sm

10104.01010106

300314.8 29

9323−

−− ≈×××××

×→

ÐRT

1212

,,

≡ζ

ÐRT

d122 13, =

A πη

Diffusion and Friction Coefficients

2,12,1 Ð

RT≡ζ

each others ‘inverse’we use both



2 components: 1 relative velocity

1 independent equation

3 components: 2 relative velocities

2 independent equations

n components: n - 1 relative velocities

n - 1 independent equations

One Equation Missing





only relative

velocities

∑≠

−ζ=ij

jijjii uuxF )(,

bootstrap

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

Bootstrap (1)



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)



∑ −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:

Fluxes





x1α

β1xx1

positivedirection

α β

binary:

infinitesimallayer

finite layer(approximate)

u1

( )212,1

21

1

uuÐRT

xdzda

aRT −=−

( )( )dzÐ

uux

ada

2,1

212

1

1 −=−

( )z

Ðk

kuu

xaa

∆=−=∆− 2,1

2,12,1

212

1

1

from Differential to Difference

α1a

β1a



876film

iu

0iu

α β0

icaverage concentration ic

species velocity(depends on position in

film)species velocity at the average composition

positive velocity

Average Velocity



)(1

2122,11

1 uuxka

a −=∆−cxc 1×

)(1

21122,11

11 NxNx

ckaa

x −=∆−

Differences with Fluxes





∑≠

−=∆−

ij ji

jij

i

i

k

uux

aa

,

)(

using velocities using fluxes

∑≠

−=∆−

ij ji

jiij

i

ii ck

NxNx

aa

x,

)(

for ideal solutions

ix∆−

Multicomponent Equations



100

10-2

10-4

10-6

ki j,1m s−

kÐ

zi ji j

,,=

∆

gases ≈ − −10 1 m s 1

liquids ≈ − −10 4 m s 1

Transfer Coefficient

s

in pores

in pores



x1α

x1

0α β

drops on a tray

gas: trace of NH3 (1)

bulk of N2 (2)

..as you already knew..

Stripping -dilute

12 ≈x

transport relation

bootstrap

flux

ckNxNx

x2,1

21121

−=∆−

02 =N

12,11 xckN ∆−=





0β

x1 05= .x1β

α

x1α

Stripping - concentrated

12,112

2,11 2 xckx

x

ckN ∆−=∆−=



Polarisationwater (1) permeates

u1

x2α

x2 ∆x2

salt (2) does not:

x xx

2 22

2= +α

∆

transport:

also

so

∆xx

2

2α

3

2

1

0 0 1 2

exact

− = −∆xx

uk

2

2

1

1 2

10

,

x u2 2 0small, =

∆xx

uk

uk

2

2

1

12

1

12

2

2α

=−

,

,uk

1

12, exact solution:∆xx

uk

2

2

1

12α

=

exp

,



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 Drop





NN

1

2

= ν

− = −∆xx N x N

k c12 1 1 2

12,

− = −∆xx N x N

k c21 2 2 1

1 2,

Nk c

x xx1

12

2 11= −

−,

ν∆ N

k c

x xx2

12

1 22= −

−,

ν∆

example ν = 2 x x1 2 0 5= = .

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

Fluxes from Vaporising Drop

Stefan (drift) corrections



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 Gasification



( )− = − =+

∆xx N x N

k c

x x N

k c12 1 1 2

1 2

2 1 1

12

2

, ,

N N2 12= −

( )Nk c

x xx1

12

2 11

2

210 10

07 2 0306= −

+= − ×

+ ×−

−,

. ..∆

( )N exact1 0046 0047= − −. : . mol m s2 1

( ) 122 smmol094.0:092.0 −−−−= exactN

Fluxes in Gasification

almost the same





Binary Distillation

0β

x1β

N1

α

x1αtransport relation

N2

x2α

x2β

hexane (1)

heptane (2)− = −∆x

x N x Nk c1

2 1 1 2

12,

bootstrap N N1 2= −

(equimolar exchange)

( )→ − =+

∆xx x N

k c11 2 1

12,

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



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 Bootstraps

N N1

1

2

2ν ν=



( ) ( )

( ) ( )

− = − + −

− = − + −

ddz

x u u x u u

ddz

x u u x u u

µ ζ ζ

µ ζ ζ

112 2 1 2 13 3 1 3

22 1 1 2 1 2 3 3 2 3

, ,

, ,

forces per mole of ‘1’

forces per mole of ‘2’

Ternary - per mole of i





( ) ( )

( ) ( )

− = − + −

− = − + −

xddz

x x u u x x u u

xddz

x x u u x x u u

11

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

22

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

µ ζ ζ

µ ζ ζ

, ,

, ,

forces per mole of mixture

these should cancel:

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

Ternary - per mole of Mixture



binary

ternary

quaternary

ckNxNx

ckNxNx

x3,1

3113

2,1

21121

−+−=∆−

ckNxNx

ckNxNx

x3,2

3223

1,2

12212

−+−=∆−

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