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Introduction to Mass Transfer

Mass transfer is the movement of mass from one place to another due to differences in

chemical potential of a compound in the two places.In a single phasethis means transfer of a component from a region where its

concentration is high to a region where its concentration is lower. At equilibrium the

concentration is the same everywhere.

If the transfer is across a phase boundaryit can be from a low to a high

concentration. Imagine solutions of salt in oil and water with an equal concentration

of salt in bothif you mixed them you would expect a mass transfer of salt into the

water in which it is more soluble. At equilibrium the concentrations will be different

in the two phases.

Examples1. Dissolution of a crystal

Initially there is a high solute concentration close to the crystal surface and a low

solute concentration in the bulk of the liquid. Through time the concentration in the

bulk rises. Figure 1 shows how the concentration of the bulk solution might change

from the time that the crystals are added. One curve represents the case where the

crystals never completely dissolve. Which curve? Why do they not completely

dissolve? What does the dotted line represent?

Figure 1 Crystal dissolution

2. Evaporation of water into air

Figure 2 shows water evaporating from two shallow trays into a flow of air. Which

tray will dry up first, A or B? What factors affect the rate of drying?

Figure 2 Water evaporation into a flow of air.

Time

Bulk

Concentration

1 cm depth2 cm depth

A B

0.5 m2 1.0 m2

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3. Gas Phase Catalytic Reaction

Consider a gas phase catalytic reaction in which A reacts to form B. There are three

steps in series (three resistances):

Mass transfer of A through the boundary layer.

Chemical reaction on the surface. Mass transfer of B through the boundary layer.

Figure 3 illustrates the concentration profiles. Notice that the concentrations where

the reaction takes place (at the surface) are different from the ones you would measure

in the bulk gas.

Figure 3 Concentration profiles for gas phase catalytic reaction

4. Separation of components in a mixture by a Mass Transfer Operation

These operations involve concentration gradients within phases and transfer across

phase boundaries.

Gas / Liquid

Absorption

Evaporation

Distillation

Gas / Solid

Drying

Adsorption

Liquid / Solid

Crystallisation

Ion exchange

Liquid / Liquid

Solvent extraction

Measures of Concentration for Gas and Liquid Phases

For ideal gases: PTV=RT

And Daltons Law: pA=yAPT

pAis the partial pressure of A.

Solid non-porous

Catalyst

Bulk

Gas

CA

CB

Boundary

Layer

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yAis the mole fraction of A.

V is molar volume, i.e. m3per mole of total gas.

V

1gives molar concentration, cTi.e. moles per m

3.

For a binary ideal gas mixture with components A and B:

RT

p

RT

Py

V

ycyc ATAATAA

And similarly for B.

For liquids: xi= ci/cTxiis the mole fraction of component i.

Molecular Diffusion in a Binary Gas Mixture

Transfer of a component within a phase can be by:

Molecular diffusion: random movement of molecules. This predominates in a

stagnant fluid or in laminar flow. Eddy diffusion: transport by natural or forced convection currents or

turbulence.

In this course we only deal with molecular diffusion. Figure 4 shows diffusion from

point 1 to 2 in a single phase.

Figure 4 Diffusion in a single phase

There are two important idealised cases to be considered:

Transfer of A and B at equal molar rates but in opposite directions. This is

equimolar counterdiffusion of A and B. There is no net flow of material

across any plane.

e.g. Catalytic Reaction, A B

Binary distillation of eg. ethanol and water.

No transfer of B. Diffusion of A occurs through a stagnant/stationary gas B.there is a net flow of material in the direction of decreasing cA.

e.g. Evaporation of water vapour into air or condensation of water vapour

from air.Absorption of CO2from air using a liquid absorbent.

cA1

cA2

distance

Transfer of A

Perhaps transfer of B

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It is important to grasp the difference between molecular diffusion and bulk flow.

Look atFigure 5a. Initially pure A is separated from pure B by a permeable barrier.

The pressure is the same everywhere. A will diffuse into B, and B will diffuse into A.

Imagine that you stand on either side of the barrier. Face left. Would you feel a wind

on your face? Face right. Would you feel a wind now?The diffusion of A is exactly balanced by the diffusion of B.

Now look atFigure 5b. Ammonia in air is brought into contact with water. The

ammonia will start to dissolve. Assume that the air does not dissolve and the water

does not evaporate. Stand on the air side of the interface. Face left then right. Would

you feel a wind on your face in either direction? Why?

There is a bulk movement of gas towards the water. The air cannot move so while it

is blown towards the water it must diffuse back at the same rate. For the diffusion of

air to take place the concentration of air must be higher at the interface than in the

bulk gas.

Figure 5 Illustration of counter-diffusion and movement of one species

We will use J to represent diffusional flux and N to represent the total movement of a

species by diffusion and bulk movement.

Ficks 1st Law

The basic law describing molecular diffusion is Ficks 1stLaw:

dx

dcDJ

J is the molar flux, mol/m2s

D is diffusivity, m2/s

cis concentration, mol/m3

x is distance, m

Note that it is convenient to work in molar terms.

In a mixture of A and B, the diffusion of A is given by:dx

dcDJ AABA

And for B:dx

dcDJ BBAB

For a constant pressure ideal gas system, cA+ cB = cT = constant.

Hence:dx

dc

dx

dc BA

And if the pressure is the same everywhere there is not net transfer of molecules soJA= -JB.

Ammonia

in air

A BWater

(a) (b)

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And hence DAB= DBA. From now on we will use D to represent both DABand DBA.

For an ideal gas:dx

dyDc

dx

dp

RT

D

dx

dcDJ AT

AAA

Equimolar Counterdiffusion

Figure 6 Equimolar counterdiffusion in a single phase binary system

The fluxes of A and B are equal and opposite

JA = - JB =dx

dcD A

2

10

A

A

c

c

A

x

A dcDdxJ

Integrate to give: 2121 AAT

AAA yy

x

Dccc

x

DJ

or for an ideal gas: 2121 AAT

AAA yyRTx

DPpp

RTx

DJ

Since there is no bulk flow with equimolar counterdiffusion, N = J for any species.

21 AAAA ccx

DJN

21 AABB ccx

DJN

Figure 7 shows the concentration profiles for A and B. Note that they are linear.

Concentration

Distance, x

cA1 x

x cA2

x0

Transfer

of A

Transfer

of B

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Figure 7 Concentration profiles for equimolar counterdiffusion

Summary of fluxes

A B A + B

Diffusion, Jdx

dcD A dx

dcD A 0

Bulk Flow 0 0 0

Total, Ndy

dcD A

dy

dcD A 0

Diffusion Through A Stationary Gas

Again:dx

dcDJJ ABA

But the total movement of B, NB= 0 since B is stationary.

So JBmust be opposed by a bulk flow of B =dx

dcD A

NB= diffusional flux of B + bulk flow of B 0dx

dcD

dx

dcD AA

Accompanying a bulk flow of B =dx

dcD A

there must be a proportional bulk flow of A =dx

dc

c

cD A

B

A

NA= diffusional flux of A + bulk flow of Adx

dc

c

cD

dx

dc

c

cD

dx

dcD A

B

TA

B

AA

The ratio cT/cB is called the drift factor (>1) ( 1 for systems dilute in A, for

which cBcT)

CT

CA1CB2

CA2CB1

00 x

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Stefans Law,dx

dc

c

cDN A

B

TA

Integrating,

2

10

A

A

c

c AT

A

x

T

A

cc

dcdx

Dc

N

integrate to give:

1

2

1

2

1

1lnln

A

AT

AT

ATTA

y

y

x

Dc

cc

cc

x

DcN

or for an ideal gas:

1

2

1

2

1

1lnln

A

AT

AT

ATTA

y

y

RTx

DP

pp

pp

RTx

DPN

Figure 8 shows the concentration profiles for A and B. Note that they are non-linear.

Figure 8 Concentration profiles for diffusion of A through stagnant B

Summary of fluxes

A B A + B

Diffusion, Jdx

dcD A

dx

dcD A 0

Bulk Flowdx

dc

c

cD A

B

A dx

dcD A

dx

dc

c

cD A

B

T

Total, Ndx

dc

c

cD A

B

T 0dx

dc

c

cD A

B

T

ExampleA layer of benzene (C6H6) 5 mm deep lies at the bottom of an open cylindrical tank 5m

in diameter. The tank temperature is 295K and atmospheric pressure is 101.3 kN/m2. At

this temperature, the diffusivity of benzene in air is 8 x 10-6m2/s and the saturated

vapour pressure of benzene is 13.3 kPa. If diffusion of benzene may be assumed to take

CTCA1

CB2

CA2CB1

0

0 x

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place through a stagnant gas film, 3mm thick, how long will it take for the benzene to

evaporate?

The density of liquid benzene is 880 kg/m3at 295K.

At the point of contact, the gas and liquid will be assumed to be in equilibrium and sothe benzene partial pressure will equal its vapour pressure, i.e.13.3 kPa On the other

side of the stagnant gas film the partial pressure of the benzene will be zero.

This is a case of diffusion of A (benzene) through stagnant B (air). It is probably

simplest to use the form:

1

2lnAT

ATTA

pp

pp

RTx

DPN or

1

2lnBzT

BzTTBz

pp

pp

RTx

DPN

pT= 101.3 kN/m2

pBz1= 13.3 kN/m2

pBz2= 0.0 kN/m

2

56

Bz 10x1.553.133.101

03.101ln

0.003x295x8.314

10x8N

kmol/m2s

The area for mass transfer is the area of the tank = x 2.52.

The rate of transfer is NBzx Area = 3.04 x 10-4kmol/s

The mass of benzene initially in the tank (from area, depth and density) is 86.4 kg

(i.e. 1.107 kmol).

The time taken for this to evaporate will be: 1.107/3.04 x 10-4= 3640 s

Alternatively the mass transfer rate can be found from:

1

2lnBzT

BzTTBz

cc

cc

x

DcN

cT= nT/V = PT/ RT =295x314.8

3.101= 0.0413 kmol/m3

cBz1=295x314.8

3.13= 0.0054 kmol/m3

cBz2= 0.0 kmol/m3

NBz=

0054.00413.0

00413.0ln

0.003

0413.010x8 6 x= 1.55 x 10-5kmol/m2s as before.

Experimental Determination of Vapour Diffusivities

The following method was developed by Winkelmann, see C&R vol 1 p581.

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Figure 9 Winkelmanns method for determination of vapour diffusivities

The partial pressure of vaporising component, A, just above the liquid surface will

equal its saturated vapour pressure; and it will be zero at the top of the vertical tube,

where A is diluted and carried off by the gas phase component B.

The gas stream flow is selected to ensure sufficient dilution of A so that c Acan be

taken to be zero, but without causing eddy currents in the vertical tube.

While in the previous example we were dealing with diffusion over a fixed distance

(3mm), in this case the distance, L, increases as vaporisation proceeds.

If cA*is the concentration of A in the vapour corresponding to its saturated vapour

pressure pA*then:

*1

2 0lnlnAT

TT

AT

ATTA

cc

c

L

Dc

cc

cc

x

DcN

Since the rate of diffusion must also equal the rate at which material is lost from the

liquid phase:

*

0ln

AT

TTLA

cc

c

L

Dc

dt

dL

MN

M = molar mass of A

If we take the diffusion distance at t = 0 to be L0, this integrates to:

tcc

cMDcLL

AT

T

L

T

*

2

0

2 ln2

In principle we should be able to plot (L

2

L02

) against t, to get a straight line,through the origin. The gradient would allow us to determine D.

While L0is difficult to determine accurately, (LL0) is easier to find and the

equation above can be organised as follows:

tcc

cMDcLLLLL

AT

T

L

T

*000ln

22

*

0

*

0

0

ln

2

2

ln

2

AT

T

L

T

AT

T

L

T

cc

cMDc

L

cc

cMDc

LL

LL

t

Distance, L, over

which diffusion

occurs

Gas

Stream, B

Liquid, A

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A plot of0L-L

tagainst ( LL0) will give a straight line with a gradient:

*

ln2

1

AT

T

L

T

cccMDc

from which D can be found. (see example in C&R vol 1 p582the algebra in the

equations is different but the numbers still work.)

Prediction of Vapour Diffusivities

If experimental values of diffusivity are not known, various theoretical models (based

on the kinetic theory of gases) can estimate diffusivity. For example, a method due to

Gilliland based on the Stefan-Maxwell hard sphere model:

BABATAB

MMVVP

TD

1110x3.423/13/1

5.14

DAB = diffusivity for system A/B (m2/s)

PT = total pressure (N/m2)

T = absolute temperature (K)

MA,MB= molecular masses (kg/kmol) of A and B

VA,VB= molar volumes (m3/kmol) of liquid A and B at their boiling

points

If the molar volumes are not known, they can be built up from tables of contributions

associated with the atoms within its structure. See C&R vol 1 p585.

If the diffusivity of a mixture is known at one condition, the above equation can be

used to estimate its value at other conditions. For example if the diffusivity of

benzene in air is 8 x 10-6m2/s at 295K and 101.3 kPa, find DBzat

350K and 101.3 kPa (10.34 x 10-6 m2/s)

295K and 200 kPa (4.052 x 10-6 m2/s)

Combining the above equation with, for example, the equation for diffusion through astationary gas,

1

2lnAT

ATTA

pp

pp

RTx

DPN

we can predict how mass transfer rates will be affected by changes in temperature and

pressure.

Diffusion in the Liquid Phase

Liquid phase diffusion is governed by essentially the same laws as for gas phase

diffusion, but:

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In the gas phase the average distance between molecular collisions >> molecular

size.

In the liquid phase the average distance between molecular collisions < molecular

size.

Diffusivities in liquid phase are very much less than those in gas phase.

Liquid phase diffusivities are much more dependent on concentration than thosein gas phase.

In the liquid phase, DABDBA.

For dilute systems with diffusion of A through stagnant B, the bulk flow will be small

so:

dx

dcDJN AABAA which on integration becomes:

21 AAAB

A ccx

DN

Guidance to estimation of liquid phase diffusivities is provided in C&R vol 1, p597

and in Sinnott, p332.

Mass Transfer In The Vicinity Of A Phase Boundary

Figure 10 shows schematically what might be happening near a phase boundary. If it

helps, think of phase 1 as Coke and phase 2 as air. A represents CO2coming out of

the Coke into air. Molecular diffusion (a slow process) predominates in the laminar

sub-layer. Eddy diffusion (a fast process) predominates in the turbulent bulk.

Figure 10 Laminar and turbulent regions near an interface

A simpler picture is to assume that all of the mass transfer resistance within a phase

lies in a thin stagnant film through which the transfer mechanism is molecular

diffusion. This is illustrated inFigure 11.

1 2 3

1 = Laminar sub-layer

2 = Buffer layer

3 = Turbulent bulk

Phase 1

Liquid?

Phase 2

Gas?

Flux of A

Interphase

boundary

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Figure 11 Simplified picture for mass transfer at an interface: linear

gradient

For equimolar counterdiffusion through the boundary layer

AoAiA ccL

DN

or AoAiDA cchN where hD= D/L = the film mass transfer coefficient with units m

3/m2s (or m/s)

For mass transfer through stagnant B in the boundary layer

Figure 12 Simplified picture for mass transfer at an interface: non-linear

gradient

AiT

AoTTA

cc

cc

L

DcN ln

or AoAiDA cchN

where

AiT

AoT

AoAi

TD

cc

cc

cc

c

L

Dh ln

)(= the film mass transfer coefficient.

Show that for dilute systems this simplifies to hD= D/L, the same as for equimolar

counterdiffusion. (Hint: divide the terms inside the log by cTand remember that

ln(1 - x ) = -x if x is a small number.)

cA0

cAi

L

cAi

cAo

L

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Mass Transfer across a Phase Boundary

The Whitman two-film theory models the mass transfer across a phase boundary by

assuming a stagnant film in each phase as shown inFigure 13.

Figure 13 Mass transfer through two stagnant films at an interface

Assumptions in the theory are: There is a discrete interface between the phases, and all material transferring

must pass through it.

Mass transfer resistance is confined to a thin layer of fluid on each side of theinterface.

There is no resistance to mass transfer at the interface itself.

The phases are in equilibrium at the interface. (The two concentrations, c Ai1

and cAi2, will be related by some phase equilibrium or solubility relationship.)

The system is at steady state. (Flux of A to the interface through phase 1equals its flux away through the interface through phase 2.)

Flux through the film in phase 1 = flux through the film in phase 2

NA= hD1(cAo1- cAi1) = hD2(cAi2- cAo2)

221

2

11 AoAi

D

D

AoAi cch

hcc

On a graph of cA1versus cA2, this equation will give a straight line which will pass

through cAo1, cAo2and have a gradient of1

2

D

D

h

h . We can then find the interface

concentrations (remembering that they are in equilibrium) usingFigure 14.

L2

cAi2

cAi1

cAo1

cAo2

L1

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Figure 14 Graphical determination of interface concentrations

Figure 15 shows the concentrations if hD2is much larger than hD1. There is a low

concentration difference in phase 2 (cAi2 cAo2) and a large concentration difference

in phase 1. A large h value means a small resistance and so a small driving force is

necessary to give the flux.

Figure 15 Graphical determination of interface concentrations with a small

resistance in phase 2

Figure 16 shows the concentrations if hD2is much smaller than hD1. There is a largeconcentration difference in phase 2 and a small concentration difference in phase 1.

cA1 Equilibrium

Line

cAo1

cAi1

cAi2cAo2 cA2

Bulk

Conditions

Interface

Conditions

cA1 Equilibrium

Line

cAo1

cAi1

cAi2cAo2 cA2

Bulk

Conditions

Interface

Conditions

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Figure 16 Graphical determination of interface concentrations with a small

resistance in phase 1

Overall Mass Transfer Coefficients

Up until now we have used film mass transfer coefficients:

NA= hD1(cAo1- cAi1) = hD2(cAi2- cAo2)

Because the interfacial concentrations, cAi1and cAi2, cannot readily be measured or

estimated, it is convenient to define overall mass transfer coefficients K1, K2, such

that:

NA = K1(cAo1- cAe1)

NA = K2(cAe2- cAo2)

K1= overall mass transfer coefficient based on phase 1

K2= overall mass transfer coefficient based on phase 2

cAe1= phase 1 composition that would be in equilibrium with cAo2in phase 2

cAe2= phase 2 composition that would be in equilibrium with cAo1in phase 1

Figure 17 shows the driving forces.

Figure 17 Mass transfer driving forces for overall coefficients

For a straight equilibrium line, i.e Henrys Law:

cA1 Equilibrium

Line

cAo1

cAi1

cAi2cAo2 cA2

Bulk

Conditions

Interface

Conditions

cA1

Equilibrium

Line

cAo1

cAo2cA2

Bulk

Conditions

cAe1

cAe2

Phase 2

driving force

Phase

1

driving

force

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cAi1= H cAi2 and cAe1= H cAo2 and cAo1= H cAe2

Using film coefficients: NA= hD1(cAo1- cAi1) = hD2(cAi2- cAo2)

Using overall coefficients: NA= K1(cAo1- cAe1) = K2(cAe2- cAo2)

Hence: K1(cAo1- cAe1) = hD1(cAo1- cAi1)

11

11

11

11

AiAo

AeAo

D cc

cc

hK

Adding and subtracting cAi1:

11

11

111

11

111

1111

11

1111

AiAo

AiAo

DAiAo

AeAi

DAiAo

AiAiAeAo

D cc

cc

hcc

cc

hcc

cccc

hK

also:22

11

21

11

AoAi

AiAo

DD cc

cc

hh

combining these:122

11

21

111

DAoAi

AeAi

D hcc

cc

hK

But: )( 2211 AoAiAeAi ccHcc

So:211

11

DD h

H

hK

Similarly:

212

111

DD hHhK

and:21

1

K

H

K

If hD1>> hD2 or H is large (fairly insoluble gas) K2hD2

If hD1

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cA1= 4.98 cA2

where cA1and cA2are the concentrations (kmol/m3) of propionic acid in the aqueous

and organic phases respectively.

a) Find the overall mass transfer coefficients K1and K2, based on the aqueous andorganic phases respectively.

b) Find the film mass transfer coefficients, hD1and hD2.

c) Find the interfacial concentration of propionic acid in each phase.

First we need to work out the molar concentration of propionic acid in the bulk of

each phase.

Molar concentration of acid in the aqueous phase =74

998x03.0= 0.405 kmol/m3

Molar concentration of acid in the organic phase =

74

879x0.004= 0.0475 kmol/m3

Using the equilibrium relationship we can find cAe1and cAe2

cAe1= 4.98 x 0.0475 = 0.237 kmol/m3

0.405 = 4.98 x cAe2 cAe2= 0.0813 kmol/m3

Now NA = K1(cAo1- cAe1)

K1= 1.2 x 10-6/(0.4050.237) = 7.1 x 10-6 m3/m2s

and: NA = K2( cAe2- cAo2)

K2= 1.2 x 10-6/(0.08130.0475) = 3.55 x 10-5 m3/m2s

And211

11

DD h

H

hK

i.e. Total resistance = Aqueous phase resistance + Organic phase resistance

Since 60% of the resistance is in the aqueous phase, then11

1

100

601

KhD

so hD1= 1.19 x 10-6m3/m2s

and:12

1

100

40

Kh

H

D

so hD2= 8.9 x 10-5 m3/m2s

Finally we can obtain the interfacial concentrations using:

NA = hD1(cAo1- cAi1)

1.2 x 10-6 = 1.19 x 10-5(0.405cAi1)

cAi1= 0.304 kmol/m3

Similarly we could use NA = hD2(cAi2- cAo2) to find cAi2. Alternatively we can use

the equilibrium relationship:cAi1= 4.98 cAi2

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cAi2= 0.062 kmol/m3

Correlations for Film Mass Transfer Coefficients

Mechanisms of mass transfer are very similar to those of convective heat transfer.

For convective heat transfer in pipes we frequently used the Colburn equation for film

heat transfer coefficients:

Nu = 0.023 Re0.8Pr0.33

Or its alternative forms:

jh= StPr0.67= 0.023 Re-0.2

where St is the Stanton number (for heat transfer)

St = Nu/(Re Pr) = h/(cpu)

For mass transfer from a liquid flowing down the inside of a tube into a gas passing

upwards, Gilliland and Sherwood proposed:

Sh = 0.023 Re0.83Sc0.44

Sherwood Number, Sh =

T

BMD

c

c

D

dh where cBMis the mean value of CB. For

equimolar counterdiffusion and dilute systems, Sherwood Number, Sh =

D

dhD .

Schmidt Number, Sc =

D

A very closely related form is:

jd= StSc0.67 = 0.023 Re-0.2

where St is the Stanton number (for mass transfer)

St = Sh/(Re Sc) = (hD/u) (cBM/cT)

There is a mistake on the SI sheet. It gives Nu in the definition of St for mass transfer instead Sh.

Comparing the jdand jh factor expressions we see that mass transfer coefficients canbe estimated in terms of heat transfer coefficients and vice versa:

67.0Pr

Scc

c

c

hh

BM

T

p

D

C&R vol 1, Section 10.8. p646 gives similar experimental correlations for other

important geometries, e.g:

Plane surfaces

Single particles

Fixed beds of particles Fluidised beds of particles

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Laboratory Determination of Liquid-Liquid Mass TransferEffects

Effects of parameters such as liquid density, viscosity, temperature and suspended

solids concentration on liquid-liquid mass transfer can be investigated using a Lewis

Cell shown inFigure 18.

Figure 18 Lewis cell for finding liquid-liquid mass transfer coefficients

Constant flows of feed solution (containing solute A) and extracting solvent are

established so that the system reaches steady state.

Ideally each phase should be perfectly mixed (so that the exit stream has the same

concentration as the liquid in the cell) but the interface should remain flat and smoothso that the area is well-defined. The stirrers are surrounded by wire mesh to reduce

turbulence and avoid vortex formation. Flat disc stirrers may be useful.

ExampleA Lewis cell,Figure 19,is being used in a laboratory study of the transfer of acetone

from an aqueous solution (phase 1) to liquid chloroform (phase 2). The apparatus is

maintained under steady state conditions at 25oC. The cell is a vertical cylinder with

an internal diameter of 15cm.

Figure 19 Lewis cell extracting acetone from water into chloroform

Interface

Lean solventRich solvent

(Extract)

Feed

Solution

Raffinate

Phase 1

Phase 2

Stirrer

Aqueous Feed

(1)

Aqueous Raffinate

(2)

Chloroform Feed (3) Chloroform Extract (4)

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The aqueous feed enters at 24 ml/min and the chloroform enters at 60 ml/min. The

acetone concentration in the fresh aqueous feed is 2.2 mass%, while that in the

chloroform extract is 0.18 mass%. The chloroform feed to the cell contains no

acetone.

It may be assumed that neither solvent is soluble in the other and that the small

quantities of acetone transferred have no significant effect on the volumetric flow or

density of either phase. The densities of the aqueous and organic phases at 25oC are

997 kg/m3and 1482 kg/m3respectively. The molar mass of acetone is 58.1 kg/kmol.

The equilibrium relationship for the distribution of acetone at low concentration

between water and chloroform at 25oC is:

cA1= 2.07 cA2

a) Find the acetone concentration in the aqueous raffinate (kmol/m3). In liquid-liquid

extraction, raffinate is the stream that has been refined.

b) Find the overall mass transfer coefficients, K1and K2.

Chloroform phase flow = 1.482 x 10-3 kg/s

Aqueous phase flow = 0.399 x 10-3 kg/s

Material balance on acetone:

Acetone in (2) + acetone in (4) = acetone in (1) + acetone in (3)

Acetone in (2) = 8.78 x 10-6 + 0 - 2.67 x 10-6 = 6.11 x 10-6 kg/s

Mass % acetone in (2) = 1.53%

Molar concentration of acetone in (2) = 58.1

997x0.0153= 0.263 kmol/m

3

Stream Acetone

(kg/s)

Acetone

(mass%)

Acetone

(kmol/m3)

(1) 8.78x10- 2.20 0.378

(2) 6.11x10- 1.53 0.263=cAo1(3) 0.0 0.00 0.000

(4) 2.67x10- 0.18 0.046=cAo2

b) Acetone transferred = 4.6 x 10-8 kmol/s

Interfacial area = 215.04 = 0.01767 m2

Flux of acetone, NA = 4.6 x 10-8/0.01767 = 2.60 x 10-6 kmol/m2s

NA = K1(cAo1- cAe1)

K1 = 1.55 x 10-5 m3/m2s

NA = K2(cAe2- cAo2)

K2 = 3.21 x 10-5 m3/m2s

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Alternatives to Film Theory for Interfacial Mass Transfer

See C&R vol 1, Section 10.5.2 p602.

These broadly involve two new concepts:

Surface Renewal at the interface by interchange of fluid elements between

interface and bulk.

Penetration of transferring species into these elements while at the interface.

Surface RenewalAs shown inFigure 20,fluid elements in Phase 2 are brought to the interface by

eddies. Elements are in contact with Phase 1 for a short interval of time and then mix

back into the bulk.

An element coming from the bulk will contain a uniform concentration of transferring

species (cAo2). On reaching the interface, the side of the element at the interface will

rapidly reach equilibrium (cAi2) with the interfacial concentration of Phase 1. Whilethe element is at the surface, A will penetrate further into the element, thus

transferring A into Phase 2.

Figure 20 Surface renewalfluid elements move to and from the interface

PenetrationIf the fluid element is considered as an infinitely thick slab, the concentration of A

(cA) within it will vary with time (t) and position (x) according to the unsteady statediffusion equation:

2

2

dx

cdD

dt

dc AA

with boundary conditions:

cA= cAo at t = 0 for all x

cA= cAi at x = 0 for all t > 0

cA= cAo at x = for all t

The solution to this equation develops with time as shown inFigure 21.

Phase1 Phase 2

FluidElement

x

1

2

3

Transfer of A

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Figure 21 Developing profile in an infinitely thick slab

Solution of the above equation at any point in time, t, gives the flux through the

interface (i.e. x=0) into the element at any time t,

t

DccN AoAixtA )()(

0,

Regular Surface RenewalIn the original formulation of the surface renewal or penetration concepts, Higbie

postulated that all fluid elements spend the same time, , at the surface, so that totaltransfer per unit area

0

2/1

0

1)()(

t

Dcc

t

Dcc AoAiAoAi

D

cc AoAi )(2

Dividing this by the time, , gives the average flux,

DccN AoAiA )(2

Note that:

The driving force is (cAicAo) as expected.

The rate is proportional to D0.5 rather than D as in all our earlier treatment.

The rate increases with decreasing , e.g. through increased agitation, as we

would expect

Random Surface RenewalIn 1951 Dankwerts proposed that instead of all fluid elements spending the same time

at the interface, any element would be replaced randomly irrespective of the time

already spent at the interface.

If the rate of production of fresh surface per unit total area of surface is s, then:

DsccN AoAiA )(

Note that:

The driving force is (cAicAo) as expected.

The rate is again proportional to D0.5 rather than D.

The rate increases with increasing s, e.g. through increased agitation, as we wouldexpect.

cAi

cAo

cA

x

Increasing Time

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Comparisons Between Heat and Mass Transfer

Heat and mass transfer are closely related phenomena. Table 1brings together some

obvious comparisons. If you think of others, please let me know.

Table 1 Properties of heat and mass transfer

Mass Transfer Heat Transfer

Transfer is proportional to a

gradient dx

dci dx

dT

Diffusion and conductionFicks Law:

dx

dcDJ

Fouriers Law:

dx

dTkQf

Film coefficients NA= hD1(cAo1- cAi1) Q = hA(T1T2)

At interface cA1in equilibrium with cA2 T1= T2

Total resistance is sum of

individual resistances211

11

DD h

H

hK 21

111

hhU

Correlations for film

coefficients

Sh = 0.023 Re . Sc . Nu = 0.023 Re . Pr .

Bibliography

1. "C&R vol. 1" refers to Chemical Engineering, J.M.Coulson, J.F.Richardson et al.,

vol. 1, 6thed., Butterworth, 1999

2. "Sinnott" refers to Chemical Engineering, R.K.Sinnott vol 6, 3rded., Butterworth,1999