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.
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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
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.
EKC 217: Introduction to Mass Transfer and Diffusion 7
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
EKC 217: Introduction to Mass Transfer and Diffusion
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.
EKC 217: Introduction to Mass Transfer and Diffusion
EKC 217: Introduction to Mass Transfer and Diffusion 10
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
EKC 217: Introduction to Mass Transfer and Diffusion
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.
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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.
EKC 217: Introduction to Mass Transfer and Diffusion
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.
EKC 217: Introduction to Mass Transfer and Diffusion
EKC 217: Introduction to Mass Transfer and Diffusion 16
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
EKC 217: Introduction to Mass Transfer and Diffusion
EKC 217: Introduction to Mass Transfer and Diffusion 18
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.
EKC 217: Introduction to Mass Transfer and Diffusion 19
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.
EKC 217: Introduction to Mass Transfer and Diffusion 20
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
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion 22
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
EKC 217: Introduction to Mass Transfer and Diffusion 23
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
EKC 217: Introduction to Mass Transfer and Diffusion 24
REVIEW: Different expressions of the concentrations in a binary mixtures of component
A and B.
EKC 217: Introduction to Mass Transfer and Diffusion 25
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]
EKC 217: Introduction to Mass Transfer and Diffusion 26
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:
EKC 217: Introduction to Mass Transfer and Diffusion 27
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:
EKC 217: Introduction to Mass Transfer and Diffusion 28
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
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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.
EKC 217: Introduction to Mass Transfer and Diffusion
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)
EKC 217: Introduction to Mass Transfer and Diffusion
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.
EKC 217: Introduction to Mass Transfer and Diffusion
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)
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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).
EKC 217: Introduction to Mass Transfer and Diffusion
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.
EKC 217: Introduction to Mass Transfer and Diffusion
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)
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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.
EKC 217: Introduction to Mass Transfer and Diffusion
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.
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
EKC 217: Introduction to Mass Transfer and Diffusion
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.
EKC 217: Introduction to Mass Transfer and Diffusion
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)
EKC 217: Introduction to Mass Transfer and Diffusion
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)
EKC 217: Introduction to Mass Transfer and Diffusion
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.
EKC 217: Introduction to Mass Transfer and Diffusion
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)
EKC 217: Introduction to Mass Transfer and Diffusion
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)
EKC 217: Introduction to Mass Transfer and Diffusion
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)
EKC 217: Introduction to Mass Transfer and Diffusion
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)
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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 .
EKC 217: Introduction to Mass Transfer and Diffusion
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
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]
EKC 217: Introduction to Mass Transfer and Diffusion
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)
EKC 217: Introduction to Mass Transfer and Diffusion
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)
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion
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
EKC 217: Introduction to Mass Transfer and Diffusion