1 introduction to mass transfer and diffusion.pdf

TOPIC 1: DIFFUSION

1

EKC 217:

MASS TRANSFER

COURSE OUTCOME 2

Course Strategies:

Lectures covering:

Laws on diffusion and their applications as related to

mass transfer in practical applications such as diffusion

with bulk flow and through stagnant media.

The use of dimensionless co-relations of Sherwood

number with Reynolds Number and Schmidt Number.

Course Activities:

Lectures; in-class examples; tutorials and assignments

CO1: Perform calculations on diffusion and mass transfer problems.

EKC 217: Introduction to Mass Transfer and Diffusion

Course Learning Objectives

3

By end of this topic, student should be able to:

1) Understand the concepts of mass transfer and

diffusion.

2) Carry out calculations on diffusion and mass

transfer problems.


What is MASS TRANSFER?

4

A term used to indicate the transference/movement of a

component (at molecular level) in a mixture from a region

where its concentration is high to a region where the

concentration is lower.

Among some of the familiar phenomena that involves mass

transfer are:

1) Liquid in an open pail

evaporates into still air

because of the difference

in concentration of water

vapor at the water surface

and the surrounding air.

Liquid to gas


5

2) A piece of sugar added

into a cup of coffee

eventually dissolves by

itself and diffuses to the

surrounding solutions.

3) A piece of solid CO2

(dry ice) also gets smaller

and smaller in time as the

CO2 molecules diffuse

into the air.

Solid to liquid

Solid to gas


EKC 217: Introduction to Mass Transfer and Diffusion 6

Mass transfer occurs by two basic mechanisms:

Molecular diffusion - random and spontaneous

microscopic movement of individual molecules in a gas,

liquid or solid.

Eddy (turbulent) diffusion – due to turbulent flow of fluid.

In a binary mixture, molecular diffusion occurs because of

one or more different potentials or driving force, including

differences (gradients) of concentration, pressure (pressure

diffusion), temperature (thermal diffusion) and external force

fields (forced diffusion).

This chapter will only focus on molecular diffusion caused by

CONCENTRATION GRADIENTS as it is the most common

type of molecular diffusion in commercial separation process.


The rate at which the process takes place is dependent both on

the driving force (concentration difference) and on the mass

transfer resistance.

In most of these applications, mass transfer takes place across

a phase boundary where the concentrations on either side of the

interface are related by the phase equilibrium relationship.

Where a chemical reaction takes place during the course of

the mass transfer process, the overall transfer rate depends on

both the chemical kinetics of the reaction and the mass transfer

resistance.

Classification of Mass Transfer Operations 8

Based on the three various phases available,

i.e.: solid, liquid and gas, six possibilities of

contact are available:

1. Gas-Gas

2. Gas-Liquid

3. Gas-Solid

4. Liquid-Liquid

5. Liquid-Solid

6. Solid-Solid


9

1) Gas-Gas

The phase contact between gas-gas is not practically

realized as most gasses are completely soluble in each

other.

2) Gas-Liquid

This is on of the most common phase contact in the

industries.

Among some of the industrial process that has this two

phase contact are distillation, absorption,

desorption/stripping.



a) Distillation

The main function is to separate a

liquid mixture, of miscible and

volatile substances into individual

components by vaporization;

Utilizes the differences in volatility

for separation while the vapor phase

is created from the liquid by

application of heat;

Example of industrial process:

Separation of a mixture of alcohol

and water into its component and

the separation of a mixture of

benzene and toluene into its

component (Figure 1). Figure 1

Benzene

Toluene

Mixture of

benzene &

toluene

11

b) Gas Absorption

In gas absorption, a soluble vapor

is absorbed by means of a liquid in

which the solute gas is more or less

soluble using selective absorbent.

For instance, if a mixture of air

and ammonia is in contact with

liquid water, a large portion of

ammonia but essentially no air will

dissolve in the liquid

(Figure 2).

Through this, the air/ammonia

mixture can be separated.

The process need more than one

column in order to achieve 100%

separation.

Absorption

column

Air (outlet) Water (inlet)

Air & ammonia

(inlet)

Water & ammonia

(outlet)

Figure 2


12

c) Gas Desorption/Stripping

Similar to gas absorption, but the only difference is purely

in the direction of solute transfer;

For example, if air is brought into contact with an

ammonia-water solution, some of the ammonia will leave

the liquid phase and enters the gas phase.

3) Gas-Solid

If a solid is moistened with a volatile liquid and is exposed

to a relatively dry gas, the liquid leaves the solid and

diffuses into the gas, an operation generally known as

DRYING.


13

4) Liquid-Liquid

Separations involving the contact of two insoluble

liquid phase are known as liquid-liquid extraction.

A simple and familiar example is the separation of acetone

from acetone-water solution using carbon tetrachloride. If the

acetone-water solution is shaken in a separatory funnel with

carbon tetrachloride and the liquids are allowed to settle, a

large portion of the acetone will be found in the carbon

tetrachloride-rich phase and will thus have been separated

from the water.

Acetone-water

solution

Carbon tetrachloride

solution

Water

Carbon

tetrachloride-acetone

solution


14

5) Liquid-Solid

Crystallization is one of the process that involves liquid and solid

phase. It is used to obtain materials in attractive and uniform

crystals of good purity, separating a solute from the melt or a

solution and leaving impurities behind .

Leaching is a process whereby liquid solvent is used to dilute

selective component in a solid mixture. For example, the leaching of

gold from its ore by cyanide solution and cotton seed oil from the

seeds by hexane.

6) Solid-Solid

E.g.: diffusion of carbon into iron during

case-hardening, doping of semi-

conductors for transistors, migration

of doped molecules in semiconductors

at high temperature.


DIFFUSION

15

Diffusion involves movement of a component (at molecular level) in a

mixture from a region where its concentration is high to a region

where the concentration is lower.

The rate of diffusion is conveniently described in terms of a molar

flux (mole per unit area per unit time), the area being measured in a

direction normal to the diffusion.

In a binary mixture, molecular diffusion occurs because of one or

more different potentials or driving force, including differences

(gradients) of concentration, pressure (pressure diffusion),

temperature (thermal diffusion) and external force fields (forced

diffusion).

We will only focus on molecular diffusion caused by concentration

gradients as it is the most common type of molecular diffusion in

commercial separation process.



In all mass transfer operations, diffusion occurs in at least one

phase and often in both phases.

For example, in gas absorption, solute diffuses through the gas

phase to the interface between the phases and through the liquid

phase from the interface.

This chapter is only restricted to binary mixture and steady state

is assumed.

GAS LIQUID

17

MOLECULAR DIFFUSION

Water molecules

Dye molecules

A container is filled

with dye solution Clear water is added

on top, dye solution is

undisturbed

Water and dye

molecules move

across the

horizontal plane

Uniform mixture of dye

and water is formed



A container is filled with dye solution.

Clear water is then carefully added on top, so that the dye solution on the bottom is undisturbed.

At the first instance, a sharp boundary between the two layers. However, after a short while, the upper layer becomes colored and the lower layer becomes less colored.

The color change process is through diffusion of the dye molecules.

By taking the horizontal plane across the solution, on the average, a fraction of molecules in the solution below the plane will cross over to the region above, and the same fraction will cross in the opposite direction.


Since the concentration of dye molecules is higher in

the lower region than the upper, there will be a net rate

of mass transfer of dye molecules will take place from

the lower region to the upper region.

After a longer period, the concentration of dye will be

uniform throughout the solution.


Based on these observations, we can conclude that:

Mass transfer by ordinary molecular diffusion occurs because of a concentration difference or gradient, i.e.: a species diffuses in the direction of decreasing concentration.

The mass transfer rate is proportional to the area normal to the direction of mass transfer and not the volume of the mixture. Therefore, the rate of diffusion can be expressed as a FLUX.

Mass transfer stops when the concentration is uniform.

These observations were then quantified by Fick in 1855, and it is known as FICK’S LAW.

21


FICK’S LAW

Fick’s Law for steady state diffusion of a binary mixture of A and B may be written as:

(for component A) ------ (1.1)

where:

JA = molar flux of component A in the z direction due to

molecular diffusion (mole per unit area per unit time,

mole A/m2.s)

DAB = Diffusivity/Diffusion coefficient of molecule A in B (m2/s)

cA = Concentration of component A (mole per unit volume,

mole A/m3)

z = Distance in the direction of diffusion (m)

dz

dcDJ A

ABA


Similarly, for component B :

(for component B) ------ (1.2)

Fick’s Law for steady state diffusion of a binary mixture of A and B may also be written in many other ways, such as:

1) Gradients of mole fraction:

------ (1.3)

where:

c = Total molar concentration of component A and B

xA = Molar fraction of component A

NOTE: cxA = cA

dz

dcDJ B

BAB

dz

dxcDJ A

ABA


2) In term of partial pressure:

(since yA = pA/pT)

where:

pA = partial pressure of A

pT = total pressure of the system

Therefore:

------ (1.4)

RT

py

RT

p

p

p

RT

pc T

AT

T

AAA

dz

dy

RT

pD

dz

dcDJ AT

ABA

ABA


REVIEW: Different expressions of the concentrations in a binary mixtures of component

A and B.


Exercise:

The composition of air is often given in terms of only the two

principal species in the gas mixture:

Oxygen, O2 yO2 = 0.21

Nitrogen, N2 yN2 = 0.79

Determine the mass fraction of both oxygen and nitrogen and

the mean molecular weight of the air when it is maintained at

25C (298 K) and 1 atm (1.013 x 105 Pa). The molecular weight of

oxygen is 0.032 kg/mol and of nitrogen is 0.028 kg/mol.

[Ans: O2 = 0.23, N2 = 0.77,

mean molecular weight of mixture = 0.0288 kg/mol]


Example 1: Molecular diffusion of He in N2

26

dz

dcDJ He

HeNHe 2

A mixture of He and N2 gas is contained in a pipe at 298 K

and 1 atm total pressure, which is constant throughout. At

one end of the pipe at point 1 the partial pressure, pA1 of He

is 0.6 atm and the other end 0.2 m pA2 = 0.20 atm. Calculate

the flux of He at steady state if DAB of the He-N2 is 0.687 x

10-4 m2/s. The universal gas constant, R is given as 82.057

cm3.atm/gmol.K

Solution:

Beginning from the Fick’s Law:


2

12

2

1

He

He

c

cHeHeN

z

zHe dcDdzJ

12

212

zz

ccDJ

HeHeHeN

He

V

n

RT

pc HeHe

He 11

Rearranging Eq. (1.1) and integrating:

From the perfect gas law, pHeV = nHeRT, thus:


sHe/m kgmole 10 x 5.63

)02.0)(298)(10 x 06.82(

)2.06.0)(10x 687.0(

26-

3-

4

HeJ

Substituting into the earlier equation:

Putting in all the known values:

)( 12

212

zzRT

ppDJ

HeHeHeN

He


29

Correlation between Diffusivity

For diffusion of A and B in a gas at constant temperature and

pressure:

------ (1.5)

The relationship between DAB and DBA is easily determined

for ideal gases, since the molar density (concentration) does not

depend on the composition:

------ (1.6) 0 dcdcdc BA

RT

Pccc BA

30

Choosing the reference plane for which there is zero volume

flow, the sum of the molar diffusion fluxes of A and B can be

set to zero:

------ (1.7)

The subscript z is often dropped when the direction is obvious.

Writing Fick’s Law for A and B for constant total

concentration, c :

------ (1.8)

0 BzAz JJ

and dz

dcDJ

dz

dcDJ B

BABA

ABA


31

Replacing Eq. (1.8) into (1.7):

------ (1.9)

Since dcA = -dcB , therefore:

------ (1.10)

0dz

dcD

dz

dcD B

BAA

AB

BAAB DD


Diffusion process together with convection

32

Up to now, we have considered Fick’s Law for diffusion in a

stationary fluid; i.e.: there has been no net movement or

convection flow of the entire phase of the binary mixture A and

B.

Many practical problems such as the

evaporation of water from a lake

under the influence of the wind or

the mixing of two fluids as they flow

in a pipe, involve diffusion in moving

medium, i.e. bulk motion is cause by

the external force.


33

The diffusion flux, JA occurred because of the concentration gradient. The

rate at which moles of A passed a fixed point to the right, which will be

taken as a positive flux, is JA kg mole A/m2·s.

This flux can be converted to a velocity of diffusion of A to the right by:

where Ad is the diffusion velocity of A in m/s.

3

2

m

mole kg

s

m s)/m mole kg( AAdA cAJ ------ (1.11)

What is CONVECTIVE MASS TRANSFER?

-Mass transfer between a moving fluid and a surface or between

immiscible moving fluids separated by a mobile interface (gas/liquid or

liquid/liquid contactor)

Mass transfer = diffusion + bulk motion of medium (convective)


34

Now let’s consider what happens when the whole fluid is moving in bulk

or convective flow to the right.

The molar average velocity of the whole fluid relative to a stationary

point is M (m/s). Component A is still diffusing to the right, but now

its diffusion velocity, Ad , is measured relative to the moving fluid.

To a stationary observer, A is moving faster than the bulk of the phase,

since its diffusion velocity, Ad , is added to that of the bulk phase, M .

Expressed mathematically, the velocity of A relative to the stationary

point is the sum of the diffusion velocity and the average or convective

velocity:

------ (1.12) MAdA

where A is the velocity of A relative to a stationary point.


35

Expressed pictorially:

Multiplying Eq. (1.12) by cA:

Each of the three terms represents a flux. The first term, cAA, can

be represented by the flux NA kg mole A/m2·s. This is the total flux

of A relative to the stationary point. The second term is JA, the

diffusion flux relative to the moving fluid. The third term is the

convective flux of A relative to the stationary point.

A

Ad M

A = actual velocity of A

Ad = diffusional velocity of A

M = velocity of bulk

observer sees actual movement as A

MAAdAAA ccc ------ (1.13)


36

Hence, Eq. (1.13) becomes:

------ (1.14)

Let N be the total convective flux of the whole stream relative to the

stationary point. Then,

------ (1.15)

Solving for M (bulk velocity) gives:

------ (1.16)

Substituting Eq. (1.16) into (1.14):

------ (1.17)

MAAA cJN

BAM NNcN

c

NN BAM

)( BAA

AA NNc

cJN


37

Since JA is Fick’s law:

------ (1.18)

Equation (1.18) is the final general equation for diffusion plus

convection to use when the flux NA is used, which is relative to the

stationary point. A similar equation can be written for NB:

------ (1.19)

To solve Eq. (1.18) or (1.19), the relation between the flux NA and NB

must be known. Eq. (1.18) and (1.19) hold for diffusion in gas, liquid or

solid.

For equimolar counter diffusion, NA = -NB and the convective term in

Eq. (1.18) becomes zero. Then, NA = JA = -NB = -JB .

)( BAAA

ABA NNc

c

dz

dxcDN

)( BABB

BAB NNc

c

dz

dxcDN


38

Eq. (1.18) and (1.19) can also be used in different forms. For

example, since N = cM and cxA = cA, thus:

------ (1.20)

Therefore, the appropriate equation used to solve a problem

would entirely depends on the information given in the

problem.

Equation (1.18) or (1.20) is the basic equation for mass transfer

in a non-turbulent fluid phase. It accounts for the amount of

component A carried by the convective bulk flow of the fluid

and the amount of A being transferred by molecular diffusion.

dz

dcDcN A

ABMAA


39

There are several types of situation covered by Eq. (1.18) or

(1.20).

Among two of them that will be discussed in the syllabus are:

1) Equimolar counterdiffusion

2) Unimolecular diffusion (diffusion of a single component

through stationary second component).


Equimolar counterdiffusion 40

In equimolar counterdiffusion,

the molar fluxes of A and B is

equal, but in opposite direction

or the net volumetric and molar

flows are zero.

A typical example of this case is

the diffusion of A and B in the

vapor phase for distillation that

have constant molar overflow.


41

Since the net volumetric and molar flows are zero, thus Eq.

(1.18) can be used with the convective term is set to zero, as

shown below:

dz

dxcDJ A

ABA ------ (1.21)

Eq. (1.21) is then integrated over a film thickness of zT,

assuming a constant flux, JA:

TA

Ai

z

A

x

xAAB dzJdxcD

0------ (1.22)


42

Integrating Eq. (1.22) and rearranging gives:

------ (1.23)

The concentration gradient for A is linear in the film, and the

gradient for B has the same magnitude but the opposite sign.

)(or )( AAi

T

ABAAAi

T

ABA cc

z

DJxx

z

cDJ


43

Example 2:

Ammonia gas (A) is diffusing through a uniform tube 0.10 m

long containing N2 gas (B) at 1.0132 x 105 Pa pressure and 298 K.

At a point 1, pA1 = 1.013 x 104 Pa and at a point 2, pA2 = 0.507 x 104

Pa. The diffusivity DAB = 0.230 x 10-4 m2/s. Calculate the flux JA

at steady state and repeat for JB.

Solution:

Given: P = 1.0132 x 105 Pa T = 298 K

z2 – z1 = 0.10 m

Substitute known values into the following equation:

)( 12

21

zzRT

ppDJ AAAB

A


44

sA/m kgmole 10 x 4.70

)010.0)(298(8314

10x507.010x013.110x23.0

27-

444

AJ

For component B:

pB1 = P - pA1 = 1.0132 x 105 – 1.013 x 104 = 9.119 x 104 Pa

pB2 = P – pA2 = 1.0132 x 105 – 0.507 x 104 = 9.625 x 104 Pa

Hence,

sB/m kgmole 10 x 4.70-

)010.0)(298(8314

10x625.910x119.910x23.0

27-

444

BJ

The negative value for JB means the flux goes from point 2 to point 1.


Unimolecular Diffusion 45

In unimolecular diffusion,

mass transfer of

component A occurs

through stagnant

component B, NB = 0.

Therefore, the total flux to

or away from the interface,

N is the same as NA.


46

A typical example of this case is the evaporation of a liquid

with the diffusion of the vapor from the interface into a gas

stream.

Based on the definition of unimolecular diffusion, Eq. (1.18)

becomes:

------ (1.24) AA

AABA Nx

dz

dxcDN


47

Rearranging:

------ (1.25)

Integrating:

------ (1.26)

Or:

------ (1.27)

dxx

dzcD

N

dz

dxcDxN

AAB

AAABAA

1

1or )1(

Ai

Ax

xA

z

AB

A

x

x

x

dxdz

cD

N A

Ai

T

1

1ln

10

Ai

A

T

ABA

x

x

z

cDN

1

1ln



48

MOLECULAR DIFFUSION IN GASES

In the gas phase,

e.g.: if ammonia (A)

were being absorbed from

air (B) into water, only

ammonia diffuses since air

does not dissolve

appreciably in water.

Special case for A diffusing through stagnant B:

49

)( BAAA

ABA NNc

c

dz

dxcDN

Since NB = 0, therefore:

)( AAA

ABA Nc

c

dz

dxcDN

------ (1.28)

------ (1.29)

From Eq. (1.18):

Thus, NB = 0 and NA = constant.


50

TAAT Pxp

RT

Pc ,

Hence:

A

T

AAABA N

P

p

dz

dp

RT

DN

Rearranging:

dz

dp

RT

D

P

pN AAB

T

AA

1

Integrating:

2

1

2

1 /1

A

A

p

pTA

Az

z

ABA

Pp

dp

RT

DdzN

------ (1.30)

------ (1.31)

------ (1.32)

------ (1.33)


51

1

2

12

ln)( AT

ATTABA

pP

pP

zzRT

PDN

After integration, Eq. (1.33) becomes:

------ (1.34)

Since:

PT – pA2 = pB2 ,

PT – pA1 = pB1 ,

pB2 – pB1 = pA1 – pA2 , then:

1

2

12

21

12

ln)( B

B

BB

AATABA

p

p

pp

pp

zzRT

PDN

------ (1.35)

The logarithmic mean of pB1 and pB2 is given by:

)/ln( 12

12

BB

BBBM

pp

ppp

------ (1.36)


52

)()(

21

12

AA

BM

TABA pp

pzzRT

PDN

Substituting Eq. (1.36) into Eq. (1.35) gives:

------ (1.37)

)( 12

21

zzRT

ppDJ AAAB

A

Compare with the earlier equation for equimolar counterdiffusion:

Therefore, in the present case, PT/pBM can be regarded as

correction factor.


53

In addition, for gases, Eq. (1.14): can also be

expressed using mole fraction in vapor phase (yA), since:

MAAA cJN

M

MAMA

Nyc

and

where:

M = molar density (kgmole/m3)

= 1/22.41 kgmole/m3 (at standard conditions, 0C & 1 atm)

yA = mole fraction of component A in vapor phase

N = total convective flux of the whole stream relative to

the stationary point (kgmole/m2·s)

M = molar average velocity (ms-1)

cA = molar concentration of component A (kgmole/m3)


54

Eq. (1.14) becomes:

dz

dyDNyN A

MABAA ------ (1.38)

Since N = NA + NB , and when only component A is being

transferred (i.e.: NB = 0), the total flux to or away from the

interface N is the same as NA, then Eq. (1.38) becomes:

dz

dyDNyN A

MABAAA ------ (1.39)


55

Rearranging and integrating:

dz

dyDyN A

MABAA )1( ------ (1.40)

2

1

2

1 )1(

A

A

y

yA

AMAB

z

zA

y

dyDdzN ------ (1.41)

1

2

12 1

1ln

A

AMABA

y

y

zz

DN

------ (1.42)


56

Similarly,

yB1 = 1 – yA1

yB2 = 1 – yA2

yB2 – yB1 = yA1 – yA2

Then,

1

2

12

21

12

lnB

B

BB

AAMABA

y

y

yy

yy

zz

DN

------ (1.43)

The logarithmic mean of

yB1 and yB2 is given by: )/ln( 12

12

BB

BBBM

yy

yyy

------ (1.44)

)()(

21

12

AA

BM

MABA yy

yzz

DN

Finally, by substituting Eq. (1.44) into Eq. (1.43) gives:

------ (1.45)


57

Example 3:

Water in the bottom of a narrow metal

tube is held at a constant temperature of

293 K. The total pressure of air (assume

dry) is 1.01325 x 105 Pa (1.0 atm) and the

temperature is 293 K (20C). Water

evaporates and diffuses through the air in

the tube and the diffusion path z2 –z1 is

0.1524 m (0.5 ft) long. The diagram is

similar to the shown figure. Calculate

the rate of evaporation at steady state in

lb mol/ft2· h and kgmole/m2· s. The

diffusivity of water vapor at 293 K and 1

atm pressure is 0.250 x 10-4 m2/s. Assume

that the system is isothermal. Use SI and

English units.

T = 293 K

0.5 ft

Water (A)

Air (B)

pA1 = 0.0231 atm

pA2 = 0

NA


58

Solution:

The diffusivity is converted to ft2/h by using the conversion factor (refer

Appendix 1, McCabe, Smith and Harriott).

/hft969.0)10x875.3)(10x 250.0( 24-4

ABD

Using Appendix 7 (McCabe, Smith and Harriott), the vapor pressure of

water at 20C (68 F) is 0.3402 lbf/in2 or 2345.6 N/m2.

air) (pure 0 and atm0231.010x01325.1

6.2345251 AA pp

T = 460 +68 = 528R = 293 K

R = 82.057 cm3· atm/gmole· K = 0.730 ft3 · atm/lbmole· R


59

To calculate the value of pBM (from Eq. 1.36):

pB1 = PT – pA1 = 1.00 – 0.0231 = 0.9769 atm

pB2 = PT – pA2 = 1.00 – 0 = 1.00 atm

Therefore:

Pa10x001.1atm988.0)9769.0/00.1ln(

9769.000.1

)/ln(

5

12

12

BB

BBBM

pp

ppp

Since pB1 is close to pB2 , the linear mean (pB1 +pB2)/2 could be

used and would be very close to pBM .


60

Substituting in Eq. (1.37) with z2 – z1 = 0.5 ft (0.1524 m), thus:

hlbmole/ft10x175.1

)988.0)(5.0)(528(730.0

)00231.0)(0.1(969.0

)()(

24

21

12

AA

BM

TABA pp

pzzRT

PDN

skgmole/m10x595.1

)10x001.1)(1524.0)(293(8314

)010x341.2)(10x01325.1)(10x250.0(

)()(

27

5

354

21

12

AA

BM

TABA pp

pzzRT

PDN


EKC 217: Introduction to Mass Transfer and

Diffusion 61

Exercise:

a) For the diffusion of solute A through a layer of gas to an

absorbing liquid with yAi = 0.20 and yA = 0.10, calculate the

rate transfer for unimolecular diffusion compared to that for

equimolar counter diffusion.

b) What is the value of yA halfway through the layer for

unimolecular diffusion?

[Ans: yA = 0.1515]


62

MOLECULAR DIFFUSION IN LIQUIDS

Diffusion of solutes in liquid is very important in many

industrial processes especially in such separation operations

such as:

1) Gas absorption

2) Distillation

3) Liquid-liquid extraction or solvent extraction

Rate of molecular diffusion in liquids is considerably slower

than in gases.

The molecules in a liquid are very close together compared to

a gas. Therefore, the molecules of the diffusing solute A will

collide with molecules of liquid B more often and diffuse

more slowly than in gases.

63

In diffusion in liquids, an important difference from diffusion

in gases is that the diffusivities are often dependent on the

concentration of the diffusing components.

Similar to those for gases, equations for diffusion in liquids

can be classified in two cases:

1) Steady-state equimolar counterdiffusion:

Starting from Eq. (1.18):

and knowing NA = -NB , then:

)( BAAA

ABA NNc

c

dz

dxcDN

)()( AAi

T

ABAAi

T

avABA cc

z

Dxx

z

cDJ ------ (1.46)


64

2

2

2

1

1

MM

Mc

av

av

------ (1.47)

cav is defined as follows:

where:

cav = average total concentration of A + B (kgmole/m3)

M1 = average molecular weight of the solution at point 1

(kg mass/kgmole)

1 = average density of the solution at point 1 (kg/m3)

EKC 217: Introduction to Mass Transfer and

Diffusion

65

2) Steady-state diffusion of A through non-diffusing B:

NA = constant, NB = 0

)()(

21

12

AA

avBM

ABA xx

Mxzz

DN

where:

)/ln( 12

12

BB

BBBM

xx

xxx

------ (1.48)

Note that xA1 + xA2 = xB1 + xB2 = 1.0

For dilute solution, xBM is close to 1.0 and c is essentially constant.

Then, Eq. (1.48) simplifies to:

)()(

21

12

AAAB

A cczz

DN

------ (1.49)


66

Example 4:

Calculate the rate of diffusion of acetic acid (A) across a film of

nondiffusing water (B) solution 1 mm thick at 17C when the

concentrations on opposite sides of the film are 9 and 3 wt %,

respectively. The diffusivity of acetic acid in the solution is

0.95 x 10-9 m2/s.

Solution:

Given:

(z2 – z1) = 0.001 m

MA = 60.03 kg/kmole

MB = 18.02 kg/kmole

At 17 C: Density of the 9% solution = 1012 kg/m3

Density of the 3% solution = 1003.2 kg/m3


67

Consider basis of solution = 1 kg,

At point 1:

acid aceticfraction mole 0.0288

0.0520

0.0015

02.18/91.003.60/09.0

03.60/09.01

Ax

aterfraction w mole9712.00288.011 Bx

kg/kmole 21.190.0520

1 solution, theofweight Molecular 1 M

3

1

1 kmole/m 7.5221.19

1012

M


68

Similarly, at point 2:

acid aceticfraction mole 0.0092 0.0543

0.0005

02.18/97.003.60/03.0

03.60/03.02

Ax

aterfraction w mole9908.00092.012 Bx

kg/kmole 42.180.0543

1 solution, theofweight Molecular 2 M

3

2

2 kmole/m 5.5442.18

2.1003

M

Then,

3kmole/m6.532

5.547.52

avM


69

980.0)9712.0/9908.0ln(

9712.09908.0

)/ln( 12

12

BB

BBBM

xx

xxx

Finally, substitute all known values in Eq. (1.48):

skmole/m 10 x 1.018

)0092.00288.0(6.53)980.0)(001.0(

10x95.0

26-

9

AN


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