Mass Transfer (Topic 2)

Mass Transfer and Diffusion

Introduction to Mass Transfer

When a system contains two or more components whose concentrationsWhen a system contains two or more components whose concentrationsvary from point to point, there is a natural tendency for mass to betransferred, minimizing the concentration differences within a system.The transport of one constituent from a region of higher concentration toThe transport of one constituent from a region of higher concentration toa lower concentration is called mass transfer.

The transfer of mass within a fluid mixture or across a phase boundary is a process that plays a major role in many industrial processes. Example of such processes are:

• Dispersion of gasses from stacks• Removal of pollutants from plant discharge by means of absorption

St ippin f s s f m st t• Stripping of gases from wastewater• Neurton diffusion within nuclear reactors• Air conditioning

@K. Hidajat, Department of Chemical & Biomolecular Engineering National University of Singapore 1

Objectives

Your objectives in studying this section are to be able to:

1. Understand mass transfer between phases.

2. Calculate interfacial mass-transfer rates in terms of localmass-transfer coefficient for each phase.

3 D fi d h i t ll t f3. Define and use, where appropriate, overall mass transfercoefficients



Steady-state ordinary molecular diffusion

• Mass transfer by ordinary molecular diffusion occurs because of aconcentration difference or gradient; that is, a species diffusesof decreasing concentration.f ng n n n.

• The mass transfer rate is proportional to the area normal to thedirection of mass transfer and not to the volume of the mixture.Thus the rate can be expressed as fluxThus, the rate can be expressed as flux.

• Mass transfer stops when the concentration is uniform.

F k’ L f D ffFick’s Law of Diffusion

For binary mixture of A and B,

andz z

A BA AB B BA

dc dcJ D J Ddz dz

(1)



Many alternative forms of equations (1) are used, depending on thechoice of driving force or potential in the gradient. For example,g p g p ,

z

AA AB

dxJ cDdz

(2)

The fluxes relative to the fixed position for two components A and Bcan be derived as,

c dc A AA A B AB

c dcN N N Dc dz

(3)

B Bc dcN N N D (4) B BB A B BA

c dcN N N Dc dz

(4)

Adding these gives,d dA B

AB BAdc dcD Ddz dz

(5)

A BJ J


A B


If cA + cB = constant,AB BAD D

Equimolar Counterdiffusion (EMD), NA = – NB

A ABdx cDN J cD x x 1 2

2 1A A AB A AN J cD x x

dz z z

(6)

Unimolecular Diffusion (UMD) NB = 0Unimolecular Diffusion (UMD), NB = 0

21 2

1ln1 1

AB A ABA A A

cD x cDN x x

(7)

2 1 1 2 1 LM1 1A Az z x z z x


Convective Mass Transfer

The basic mass transfer equation in words is:

Mass transfer rate = area mass transfer coefficient driving force

orA c An Ak c (7)A c A ( )

mass transfer coefficientmass transfer coefficient

Note: there are various theories on describing this mass transfergcoefficient in standard textbooks on mass transfer.



Equimolar Counter Diffusion

Gases:

1 2( ) ABA c A A c

DN k c c k

1 2( ) ABA G A A G

DN k p p kRT

PD

Liquids

1 2( ) ABA y A A y

PDN k y y kRT

q

1 2( ) ABA L A A L

DN k c c k

cD

1 2( ) ABA x A A x

cDN k x x k



Diffusion through Stagnant Film

Gases:

LM

1 2( )

( )AB

A c A A cB

DN k c c kc

LM

1 2( )

( )AB

A G A A GB

PDN k p p kRT p

Li id

LM

2

1 2( )( )

ABA y A A y

B

P DN k y y kRT p

Liquids

LM1 2( )

( )AB

A L A A LB

DN k c c kx

LM1 2( )

( )AB

A x A A xB

cDN k x x kx



To convert from one type of mass transfer coefficient to another:

k ck k k x ck x Liquids: ( ) ( )x L L x B L B

B B

k ck k k x ck xM

c c xk

LM LM

LM LM

Liquids: ( ) ( )

( ) ( )( )

B ycG y B G

p kPk Pk k p kRT P

LM

LM

( )Gases: ( )


Convective Mass Transfer Between Phases


Convective Mass Transfer Between Phases

Two-Film Theory:

i iG G i iLiquid phaseGas phase

pAb

Gas phase Liquid phase

pAb

cAi

pAipAi

cAi

cAbcAb

Interface

GL

Ab

c f pAt the phase interface, cAi and pAi are in equilibrium,

Interface

Ai Aic f p

( ) ( )A G Ab Ai c Ai AbN k p p k c c


Convective Mass Transfer Between PhasesInterfacial compositions:

kc Ab Ai

G Ab Ai

k p pk c c

pA

pAb Equilibrium curve

pAipAi

cAc c@K. Hidajat, Department of Chemical & Biomolecular Engineering

National University of Singapore 12

AcAb cAi

Overall Mass Transfer Coefficients

Imaginary Bulk gas phasecomposition pointed to measurable pAb

compositionpAi

variable

Bulk liquid phase

cAi*Ac

Bulk liquid phase

compositioncAbcAb

*ApAp

*( ) for liquid phaseA Abc cDriving force: *( ) for gaseous phaseAb Ap p



Equilibrium curve

pAb

curve

* *N K c c K p p

pAi

A L A Ab G Ab AN K c c K p p

* *Ab A Ab Ai Ai Ap p p p p p

*Ap

*Ai A

xAi Ab

p pmc c

mx

cAb cAi*Ac

Ai Abc c

*Ab A Ab Ai x Ai Ab

A A x A

p p p p m c c

N N m NK k k

1 1 x

G G L

mK k k


G G LK k k G G LK k k


Equilibrium curve

pAb

curve

myIn a similar manner, we can find

pAi

* *A L A Ab G Ab AN K c c K p p

* *A Ab A Ai Ai Abc c c c c c

*Ap

A Ab A Ai Ai Abc c c c c c

*

Ab Aiy

p pm

c c

cAb cAi*Ac

A Aic c

* /A Ab Ab Ai y Ai Ab

A A A

c c p p m c c

N N NK m k k

1 1 1

L G LK m k k


L y G LK m k k L y G LK m k k


Note:

for pAi = HAcAi

1 1 H1 1 A

G G L

HK k kG G L

1 1 1

L A G LK H k k


Mass Transfer Resistance

The resistance to mass transfer is defined as the reciprocal of the masstransfer coefficient:

1 GK

represents the resistance to mass transfer in the gas phase

1 LK represents the resistance to mass transfer in the liquid phase

It is important to know if one of the 2 resistances is controlling the masstransfer. If so, the rate of mass transfer can be increased by promotingturbulence in and/or increasing the dispersion of the controlling phaseturbulence in and/or increasing the dispersion of the controlling phase.

Recall the relationship between overall and film mass transfer coefficients, andRecall the relationship between overall and film mass transfer coefficients, andthat the 1/K represents the mass transfer resistance.


Mass Transfer ResistanceIf mx is small (i.e. the equilibrium curve is very flat), the term mx/kL is notsignificant, therefore:

1 1

G GK k

and the major resistance to diffusional mass transfer lies inthe gas phase and the mass transfer is said to be gas-phasecontrolled.

In this case, solute A can be interpreted as being very soluble in the liquid: atequilibrium, a small concentration of A in the gas will bring about a very largeconcentration in the liquid.

If my is large (i.e. the equilibrium curve is very steep), the term 1/mykG is notsignificant, therefore:

d h j i diff i l f li i1 1

L LK k

and the major resistance to diffusional mass transfer lies inthe liquid phase and the mass transfer is said to be liquid-phase controlled.

Solute A is relatively insoluble in the liquid: a very large concentration of A in thegas phase is required to provide even a small change of concentration in the liquid.


Mass Transfer Between Two PhasesExample:

In a dilute concentration region equilibrium data for SO distributedIn a dilute concentration region, equilibrium data for SO2 distributedbetween air and water can be approximated by

pA = 25xA

where the partial pressure of SO2 is expressed in atmospheres. For anabsorption column operating at 10 atm, the bulk vapour and liquidconcentrations at one point in the column are yA = 0.01 and xA = 0.0. Theconcentrations at one point in the column are yA 0.01 and xA 0.0. Themass transfer coefficient for this process are

210 kgmol/m h mole fractionxk 28 kgmol/m h mole fractionyk

Assuming equimolar counter transfer, (a) find the overall liquid phase masstransfer coefficient, (b) determine the interfacial compositions, xAi and yAi,and (c) calculate the molar flux, NA


Mass Transfer Between Two PhasesSolution:

(a) p = 25x but y = p /P y = 2 5x(a) pA = 25xA, but yA = pA/P yA = 2.5xA.

1 1 1

x y y xK m k k

x y y x

Upon substituting my = 2.5 and the mass transfer coefficients intothe above equation, we obtain

1 1 1

2

1 1 1(2.5)(8) 10

6.67kgmol/m h mole fractionx

x

K

K

gx

(b) Using the rate ratio line,

10k y y 10 1.258

x Ab Ai

y Ab Ai

k y yk x x


Mass Transfer Between Two Phases

yA1.25x Ab Aik y y

k x x

0.01 y Ab Aik x x

yAi

yA = 2.5xA

0.0067yA = 1.25xA + 0.01

xAi0 0.01 0.02 xA

0.00267(c) The mass flux

2kgmol/m h( ) 10(0.00267 0) 0.0267A x Ai AbN k x x


g( ) ( )A x Ai Ab

Mass Transfer Between Two PhasesRepeat part (b) of the previous example for bulk concentrations yA = 0.04and xA = 0.01. Assuming transfer of component A through a stagnant film.

Solution

The determination of interfacial compositions for transfer through at t fil i th t t i l d d b dstagnant film requires that a trial-and-error procedure be used.

To begin we assume a counter diffusion to find the interfacial compositions.10k y y 10 1.258

x Ab Ai

y Ab Ai

k y yk x x

1.25 0.0525Ai Aiy x xAi = 0.0141.25 0.0525Ai Aiy x

Equilibrium: yAi = 2.5xAi

Ai

yAi = 0.035

For diffusion through a stagnant film:

LM/(1 )/(1 )

x A Ab Aik x y yk y x x


LM/(1 )y A Ab Aik y x x

Mass Transfer Between Two Phases

LM(1 ) (1 ) (1 0.014)(1 0.01)(1 ) 0.988

1 0.0141 lnln

Ai AbA

Ai

x xxx

LM

lnln 1 0.011(1 ) (1 ) (1 0.035)(1 0.04)(1 ) 0.962

Ai

Ab

Ai AbA

xy yy

LM(1 ) 0.962

1 0.0351 lnln 1 0.041

AAi

Ab

yyy

/(1 ) 10 / 0 988k x LM

LM

/(1 ) 10 / 0.988Therefore, 1.217/(1 ) 8 / 0.962

x A

y A

k xk y

As before plot a line from the bulk concentrations with a slope equals to1.217 to intersect with the equilibrium curve.

xAi = 0.01405 yAi = 0.0364

Use the new values for the log mean concentration differences.Consequently, the interfacial conditions are xAi = 0.01405 and yAi =0 0364


0.0364.

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Mass Transfer (Topic 2)

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