Mass transfer in multi-component mixturesusers.abo.fi/rzevenho/MTiMM-OHs.pdf · Multi-component mass transfer using transparancies that accompany ”Mass Transfer in Multicomponent

Transport processes 424522

Multi-component mass transferusing transparancies that accompany ”Mass Transfer in Multicomponent mixtures”

by J.A. Wesselingh & R. Krishna, Delft University Press (2000)

februari 2018 Åbo Akademi - Biskopsgatan 8, 20500 Åbo TRP RZ 1/76

Mass transfer in multi-component mixtures

Ron ZevenhovenÅbo Akademi University

Thermal and Flow Engineering Laboratorytel. 3223 ; [email protected]

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

Chapters 1-10 ex. 7, of 25 of book





1 ”Old-school” mass transfer




J Ddcdzi

i ci

ci

diffusivity

kDzii

mass transfer coefficient

Fick’s law

Ji : flux of species i with respect to the mixture

J Dcz

k ci ii

i i

z

c c ci i i

Mass Transfer as you have learned it

phases α, β

februari 2018 Åbo Akademi - Biskopsgatan 8, 20500 Åbo RZ 3/76




N Ddcdz

Nxi ii

i differential equation

N k c Nxi i i i difference equation

Ni : flux with respect to an interface

diffusion flux

drift flux

Stefan or driftcorrection

N Nii

flux of mixture

Diffusion with Drift

k s ci i i





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

0.0

0 10 20

N2A

B

H2

CO2

mol

e fr

actio

n x i

reverse diffusion

timeh

Three gases (2)





Two cations

H+

Cl-

Na+

Cl-

cation permeable membrane

highconcentration

lowconcentration

excess +charge

and 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

Two 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

Two gases in a porous plug (2)






2 Driving forces




1kg

1 m

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

here: mg z 981 981. .J ( Nm)

or, per mole i iM g z

1kg

Fi

the driving force is the negative potential gradient:

Fddz

M gii

i

the force is downwards

Gravity - a simple potential

Δz





Chemical potentialxi i

mixture ii aRTTpconst ln,

chemical potential

activity

pure i (one mole)

i const p T ,

iiii aRT ln

activity coefficientwork required: change in the chemical potential

iii xa






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

in an ideal gas

pp

RTTpconst ii ln),(

partial pressure

in an ideal solution





momentum ‘in’ momentum ‘out’

forces

change of momentum Fvmvm

dtmvd

outin )()()(

F

Momentum balance

Σ(ṁv)in Σ(ṁv)out

(ṁv)out(ṁv)in





(2)(1)

1u2u

zH2 CO2

dzz

Moving through each other

species velocities





dzzz

area A

volume Adz

zAp1 dzz

Ap1

)( 1221 uuppforce

friction

dzdp

forcedriving 1

Forces on hydrogen (1)

,,,

,

,1

11

ζD

RT

D

RTζt coefficienfriction with

)uu(xζ)uu(RTp F force Driving

moleper force )uu(RTpdz

dp

p

RT gives

RT

pcwith

per volume force )uu(ppdz

dp :balance Force Note:

often x2 ≈ 1and u2 = 0

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




dzda

aRT

dzad

RTdzd

F i

i

iii

ln

RTx

dxdzi

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

,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





in a solid particle

zd10

z

d

gases

liquids

z 10 4 m

z 10 5 m

membrane

z 10 107 4 m

Film theory, thickness of films

eddies & large scale convection

‘film’: no eddies

phase boundary

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






Difference form of force

za

aRT

za

RTz

F i

i

iii

ln

zx

xRT i

i

in ideal solutions

for a given T and p




1

RT+1

0

-1

0 2

approximate

exact

‘approximate’ works out better in difference equations

1

1lna

a

11

11

1

1

5.0 aa

aa

aa

1

1

a

a

-2

Approximation





z 10 5 m

mole fractionof CO2

x1 0003 .

FRTx

dxdz

RTx

xz1

1

1

1

1

x1 0001 .

Forces in a glass of beer

growing bubble of CO2

Note: u2 = 0






Example (3.1 from book)





Example (3.1 from book): answer





3 Friction




Friction coefficients of spheres

12 2 112 1 1

3 3 10, = A d Nmol ms1

F1

A 6 1023

molecules mol-1

coefficient of a single sphere

spheres ‘1’

liquid ‘2’

using:Stokes Law





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

sm

10104.01010106

300314.8 29

9323

ÐRT

1212

,,

ÐRT

d122 13, A

Diffusion and friction coefficients

2,12,1 Ð

RT

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

ij

jijjii uuxF )(,bootstrap

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

Bootstrap (1)






N2

CO2

N2

H2

H+

Cl-

Na+

Cl-

HeAr

no net volume flow plug does not move

membrane does not move

(almost) no charge transfer

Bootstraps (2)





j

jijijiii uuxcxcxF )(,

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

Fi / V ; xi = ci / c




x1

1xx1

positivedirection

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

1a

1a






iu

0iu

0

icaverage concentration ic

species velocity(depends on position in film)

species velocity at the average composition

positive velocity

Average velocityfilm





)(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 coefficients

in pores

in pores






changes are not very important

Temperature effectsMS-equation

terms) diffusion (thermal)u(uxζFj

jijji,i small

driving force

dzdx

xRT

dzd

F i

iT

ii

difference form:

zx

xTR

F i

ii

at constant temperature

average film temperature





4 Binary examples




x1

x1

0

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





0

x1 05 .x1

x1

Stripping - concentrated

12,112

2,11 2 xckx

x

ckN





heat

benzene (1), volatile

toluene (2)

x1

x2

x2

x1

y K x1 1 1

y K x2 2 2

vapour removed by convection

bootstrap:NN

yy

1

2

1

2

Vaporising droplet





NN

1

2

xx N x N

k c12 1 1 2

12,

xx N x N

k c21 2 2 1

12,

Nk c

x xx1

12

2 11

,

N

k c

x xx2

12

1 22

,

example 2 x x1 2 0 5 .

N k c x1 12 14 , N k c x2 12 22 ,

Fluxes 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 gasification





xx N x N

k c

x x N

k c12 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

,

. ..

N exact1 0 046 0 047 . : . mol m s2 1

122 smmol094.0:092.0 exactN

Fluxes in gasification

almost the same





Binary distillation

0

x1

N1

x1transport relation

N2

x2

x2

hexane (1)

heptane (2)

x

x N x N

k c12 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

2






5 Ternary examples




ddz

x u u x u u

ddz

x u u x u u

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 - per mole of i





xddz

x x u u x x u u

xddz

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

, ,

, ,

forces per mole of mixture

these should cancel:

2 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

More components





Condensor 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)transport (MS) relations:

NH3 :

H2O :

bootstrap

three linear equations, three unknowns

exact solutions:

03 N

30)103(4.03.0

30)101(3.04.0

2.0 331

321

NNNN

30)103(3.03.0

30)101(3.04.0

4.0 332

312

NNNN

015.01 N12

2 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)





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 water

2 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

0.022

y2 y2

no motion

u22375 10 .

u22158 10 .

u22081 10 .

Butanol - which direction?





Ammonia reaction

N2 (1)

H2 (2)

NH3(3)

catalytic surface

N H NH2 2 33 2

xx N x N

k cx N x N

k c12 1 1 2

12

3 1 1 3

13, ,

xx N x N

k cx N x N

k c21 2 2 1

12

3 2 2 3

2 3, ,

xx N x N

k cx N x N

k c31 3 3 1

13

2 3 3 2

2 3, ,

transport relations:

bootstrap:N N N1 2 3

1 3 2





x

xx

u uk

xu u

k1

12

1 2

123

1 3

13, ,

xu u

keffeff

eff

1

1,

a ternary can be approximated as a binary when

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

xu u

k2

12

3

133

1 3

13, , ,

u uxk

xk

u u2 32

12

3

131 2

, ,

k k

x 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

ux u x u

x 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 , 1 3 2

1

2 1

12

3 1

13kx x

kx x

keff, , ,

with

similarly for H2 and NH3

Effective binary in a reactive system

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





Water from a saltwater reservoir penetrates through a membrane, see the figure. One result is that the salt concentration rises near the membrane.a. Give the Maxwell-Stefan (MS) the equation for the salt, in differential form, as a function of the fluxes Ni.b. For a mass transfer coefficient k1,2 = 0.01 m/s, the mean substance concentration ĉ = 60000 mol / m3 and flux N1 = 300 mol / (m2· s) for the water, calculate the gradient Δx2 if x2α = 0.018. Give the answer with 2 decimals accuracy.


Example: membrane

membrane

water (1) + salt (2) water

water

z

z=α z=β

x2α

x2β

x2

in boundary layer

Δx2 = x2β - x2α

x2 = x2α + ½ Δx2

u1

u2 = 0, x2 << x1membrane


water

z

z=α z=β

x2α

x2β

x2

in boundary layer

Δx2 = x2β - x2α

x2 = x2α + ½ Δx2

u1

u2 = 0, x2 << x1





Example: membrane

membrane


water

z

z=α z=β

x2α

x2β

x2

in boundary layer

Δx2 = x2β - x2α

x2 = x2α + ½ Δx2

u1

u2 = 0, x2 << x1membrane


water

z

z=α z=β

x2α

x2β

x2

in boundary layer

Δx2 = x2β - x2α

x2 = x2α + ½ Δx2

u1

u2 = 0, x2 << x1




ideal solution:

non-ideal solution:

difference equation

chemical potential

8 Non-ideality

i

iiii a

az

RTdz

adRT

dzd

F

ln

ii aRTconstant ln

i

iiii x

xz

RTdz

xdRT

dzd

F

ln

ii xa

iii xa





molecules are little (hard) spheres moving around with their thermal velocity and occasionally bumping into each other

dd d

1 21 2

2,

ÐRT

pd M M1 2 3

3 2

1 22

1 2

1 22 1 1

,

/

,

/

A

d1

d2

Ð

T

p M M1 28

175

11 3

21 3 2

1 2

1 2

316 101 1

,

.

/ /

/

.

empirical modification:

diffusion volume liquid volume, m3 mol-1

9 Diffusivities in Gases

M = molar mass

kg/mol





Gases, empirical equation

Ð

T

p M M1 28

175

11 3

21 3 2

1 2

1 2

316 101 1

,

.

/ /

/

.

H2 N2 CO2 NH3 H2O

7.07 17.9 26.9 14.9 12.7

i

10 6 m mol3 1

diffusion volume liquid volume, m3 mol-1





Gases, diffusivity example

N2 (1) CO2 (2) T 300 K p 105 Pa

Ð1 2

8175

5 6 1 3 6 1 3 2

3 3 1 2

5

316 10300

10 17 9 10 26 9 10

10

28

10

44175 10

,

.

/ /

/

.. .

.

m s2 1





Dilute Species in a Liquid

d11

1 3

A

/

u1 ÐRT

d1 22 13,

A

this ‘constant’ varies with the size ratio of the species:

d d1 2

d d1 2

3

2

1

0 2 4

dd

1

2

129

2212

2,1 sm102

dd

RTÐ

A





Non-ideal Binary

ethanol (1) - water (2)

Ð D1 2 1 2, ,

interpolation roughly logarithmic for Ð

MS and Fick diffusivities differ

only at x x1 21 1 and isÐ D1 2 1 2, ,

25 oC

Ð1 2,

5

2

1

0.50 1x1

D1 2,

Ð D1 2 1 2

910, ,,

m s2 1





10 Dimensionless Groups

Shi ji j

i j

k d

Ð,,

,

Sc i ji jÐ,,

Re vd

Sherwood number

Reynolds number

Schmidt number

mass transfer

fluid flow

mixture property

diameterfilm thickness

Re 1 laminarRe 1 turbulent

Sc 1 in gasesSc 103 in liquids





Mass transfer coefficientsoutside rigid interfaces

kg

Ði j i j, ,

/

.

0 3 2

1 3

, ,and Ði j are values in the fluid outside

for short times

for small particles

for large particles

boundary layer theory

kÐ

ti ji j

,,. 113

kÐ

di ji j

,, 2





surface active agentsswept to the back

these immobilise the drop or bubble when

df

g

1 2/

f is the fouling (fudge) factorf 0 2. in clean liquidsf 1 in dirty liquids

the transition is at d 1 mm

cause surface tension gradients

d

Mobile or Rigid?





Mass transfer coefficientsBubbles with mobile interfaces

bubbles:

kg

Ði j i j, ,

/

.

0 4

2 23

1 6

and are for the liquid (outside)

outside

kg

Ði jd

i jd

, ,

/

.

0 4

2 23

1 6

inside (?)

Ði jd,

Ði j,





Mass transfer coefficientsmobile drops in a liquid

Ði jd,

Ði j,

d

mobile drops in a liquid

calculate the rigid and mobile (bubble) coefficients separately

interpolate using d

k k ki j i jmobile

i jrigid

, , ,.

.

.

1

1 0 3

0 3

1 0 3

(these differ for the outer and inner coefficients)





Mass transfer coefficientsdrops in a gas

dg

2

1 2

/

dg

4

1 2

/

rigid sphere mobile oscillating drop drop shatters






Mass transfer in multi-component mixturesusers.abo.fi/rzevenho/MTiMM-OHs.pdf · Multi-component mass transfer using transparancies that accompany ”Mass Transfer in Multicomponent

Documents