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CBE 3233 Chemical &Biomolecular Separation
Qing Song Spring 2013
Lecture 3 Mass Transfer
1
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Mass Transfer and Diffusion
Mass transfer by ordinary molecular diffusionoccurs because of a concentration gradient
Net movement of a species in a mixture from onelocation to another.2
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Applications of Mass TransferOperation Extract
Phase
RaffinatePhase
Distillation
Gas
adsorption/dehumidification
Membrane separationsAdsorption
Liquid extraction
Leaching
Crystallization
Drying
Vapor
Gas
Gas or liquid
Gas or liquidExtract
Liquid
Mother liquor
Gas (usually
air)
Liquid
Liquid
Gas or liquid
SolidRaffinate
Solid
Crystals
Wet solid
Transfer of material from one homogeneous phase to another
Driving force for transfer is a concentration difference 3
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Mass TransferAbsorption: solute transfer through the gas to the gas-liquid interface, across the interface, and into the liquid
Distillation: separate a liquid mixture by vaporization into
individual components.
A solute gas is absorbed from an inert gas into a liquid in
which the solute is more or less soluble.
Washing of ammonia from mixture of NH3 and air,
Remove of CO2 and H2S from natural gas.
The low boiler (more volatile) diffuses through the liquid
phase to the interface and away from the interface into
the vapor. The high boiler (less volatile) diffuses in the
reverse direction.
Example: ethanol and water, crude oil into gasoline,
kerosene, fuel oil, and lubricating stock. 4
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Mechanisms of Mass TransferMolecular diffusion: random, spontaneous,
Eddy (turbulent) diffusion by random,
macroscopic fluid motion
Both molecular and eddy diffusion may
involve the movement of different species in
opposing directions.
Bulk flow of the mixture in a direction
parallel to the direction of diffusion.
5
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Molar concentration:Mole fraction:
(liquids, solid (gases)
RT
p
V
n
Mc AA
A
AA
Mass Transfer Fundamentals
6
RT
p
V
n
Mc ii
i
ii
c
cx i
i
c
cx A
A
c
cy i
i
c
cy A
A
molar average velocity,c
cn
i
ii 1
M
v
v
velocity of a particular species relative to molar averagevelocity is the diffusion velocity
MiiD v
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Flux DefinitionsMolar flux iii cN With respect to a fixed reference frame
Molar fluxWith respect to a molar average velocity vM
Miii cJ
7
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Flux: A vector quantity denoting amount of a particular
Species that passes per given time through a unit area
normal to the vector, given by Ficks First Law, for basic
molecular diffusion
Molar flux
or, in the z-direction,
For a general relation in a non-isothermal, isobaric
system,
AABA cD J
dz
dcDJ AABzA ,
dz
dycDJ AABzA ,
Theory of Diffusion
dz
dwDj A
ABA
Mass flux 8
ii cDJ
dz
dcDJ izi ,
dz
dycDJ izi ,
dz
dwDj i
i
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Units of Above ParametersJi: molar flux of component i, kgmol/m
2.h (kmol/m2.h) or lb
mol/ft2.h
D: diffusivity m2/h
Ci: mole concentration kgmol/m3, kmol/m3, or lbmol/ft3
Z: distance in direction of diffusion, m or ft
i: mass density of component i, kg/m3, or lb/ft3
xi: mole fraction of i
wi: mass fraction of i
9
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10
Units of Above Parameters for Binary SystemJA: molar flux of component i, kgmol/m
2.h (kmol/m2.h) or lb
mol/ft2.h
DAB: diffusivity m2/h
CA: mole concentration kgmol/m3, kmol/m3, or lbmol/ft3
Z: distance in direction of diffusion, m or ft
A: mass density of component A, kg/m3, or lb/ft3
xA: mole fraction of A
wA: mass fraction of A
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Similarities between Mass , Heat and Momentum Transfer
Mass
Heat
Momentum
dzdcDJ AABzA ,
Ficks first law
dz
dTkq Fouriers law
dz
vdx
zx
Newtowns law
11
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Since mass is transferred by two means: concentration differences and convection differences from densitydifferences
Flux Relationships
12
MiiMiiiMiii cNcccJ
Mii
Miii cdz
dcDcJN earranging:
Diffusion flux Convective flux
n
i
i
n
i
iiiin
i
i
iin
i
i
n
i
ii
iii Nc
cJc
c
cJ
c
c
cJN11
11
1
n
iiiii
NxJN1
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13
Flux Relationships for Binary System
dzdxcDJ AABA
BAAA
ABBAAAA NNxdzdxcDNNxJN
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Summary of Molar and Mass fluxes
14
JA, and NA are equivalent statement of theFick rate equation. Any one of theseequations is adequate to describe moleculardiffusion.
Molar fluxes, JA and NA, are used to describemass-transfer operations in which chemicalreactions are involved. They are often used todescribe the mass transfer in diffusion cellsfor measuring diffusion coefficients.
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Molecular Diffusion
15
Unit area
NA NB
NA is positive NB is negativeNet flux=N=NA+NB
A(Water) B(Ethanol)
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16
Equimolar Counter Diffusion (EMD)NxJN AAA
BA
AA
ABA NNc
c
dz
dcDN
BABBBAB NNcc
dz
dcDN
dzdcD
dzdcD
cccNNNN BBA
A
AB
BA
BABA
similarly,
Adding them together,
dz
dcD
dz
dcD B
BA
A
AB
BA JJ 0 BA NNN
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17
In the Absence of Bulk flowdz
dxcDJN AABAA
dz
dxcDJN B
ABBB
(3-16)(3-17)
Total concentration c, T, P are constants,z=z1, xA=xA1, z=z2, xA=xA2
A
AB
A dxdzcDN
A
A
x
x
A
z
z AB
A dxdzcDN
11
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18
11 AAAB
A xxzzcD
N
Integrate between z1 and z
Equimolar Counter Diffusion (EMD)
11
AA
AB
A xxzz
cDJ
BB
AB
Bxx
zz
cDJ
1
1
(3-18)
(3-19)
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19
Unimolecular Diffusion (UMD)N=NA, NB=0
Only one component such as A is being transferred,
Example: ammonia (NH3) (A) & air (B) absorbed by water
AA
A
ABA Nxdz
dxcDN (3-26)
dz
dxcDNx AABAA )1(
dzdx
x
cD
NA
A
AB
A 1 (3-27)
Integrating eq.(3-27)
At quasi-steady-state conditions and constant molar activity
)1ln(
1
1
11
1
1
1A
x
x A
AB
A
Ax
xA
AB
A
A
x
x A
ABz
zxd
N
cD
x
xd
N
cD
x
dx
N
cDdz
A
A
A
A
A
A
1
11
1ln)1(ln
1
A
A
A
ABx
xA
A
AB
x
x
N
cDx
N
cDzz
A
A(3-31)
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AB
A
A
A
cD
zzN
x
x 1
11
1ln
AB
AcD
zzN
A
A ex
x 1
11
1
AB
A
cD
zzN
AA exx
1
111
Unimolecular Diffusion (UMD)
(3-32)
1
2
12 1
1
lnA
AAB
A x
x
zz
cD
N
1
2
21
1
2
12
1
1ln
1
1ln
11
1
A
A
AA
A
A
AA
LMA
x
x
xx
x
x
xx
x
z=z2, eq.(3-21) becomes
LMA
AAAB
Ax
xx
zz
cDN
1
21
12
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Mass Transfer Coefficients
21
AcA ckN
AC
Solute A
CA1
Rate equation
Steady state diffusion of solute A
through a membrane as shown in
figure. After the solute diffuses
through the membrane, it is swept
from the external surface by a gas
stream. A mass transfer coefficient
for transfer of component A into the
free stream is defined in terms of
diffusion at the interface by
AA
z
A
AA
Ac
cc
z
cD
cc
Jk
1
0
1
kc: mass transfer coefficients
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22
Equimolar Counter Diffusion (EMD)
AA
ABA
ccD
J 1JA=-JB 11
AA
AB
A xxzz
cDJ
AB
c
D
k
'
NA=-NB 11
AAAB
A xxzz
cDN
AA
ABA
ccDN 1
AAcA cckN 1'
For liquids AALAAcA cckcckN 1'
1
'
AAxAALA xxkxxckN 1'
1
'And
For gases AAGAAc
A ppkppRT
kN 1'
1
'
RTP
Vnc t
tAAPyp
Dalton's law )( 1'
1
'
AAyAAGtA yykyykPN
nRTVPt
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Unimolecular Diffusion (UMD)NA=constant, NB=0
LM
A
AAAB
LMA
AAAB
A
c
c
ccD
x
xx
zz
cDN
11
11
1
cc
cc
c
c
c
c
c
c
A
A
AA
LM
A
1
1
1
1ln
11
1
LMA
AB
ccc
Dk
1
AALMA
c
AAcAcc
cc
kcckN
1
'
11
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AALAAcA cckcckN 11 AAxLAALA xxkxxckN 11
For liquids
or
Unimolecular Diffusion (UMD)
Introducing ideal gas law and make use of dalton's
law
AGAyAAGtAAGA yykyykPppkN 111
or
AALMB
Gt
AA
LMB
ct
A ppp
kPpp
pRT
kPN 1
'
1
'
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Units of Mass Transfer Coefficients
25
EMD UMD
Gases:
Units
AB
cDk '
RT
Dk ABG
'
RT
DPk ABty
'
Liquids:
AB
LDk '
AB
x
cDk '
LMB
ABtc
PDPk
LMB
ABt
G
pRT
DPk
MB
ABL
xDk
LMB
ABt
ypRT
DPk
2
mole(time)(area)(mol/vol)
mole(time)(area)(pressure
mole(time)(area)(mol fra.)
mole(time)(area)(mol/vol)
mole(time)(area)(mol fra.)
LMB
AB
x
x
cDk
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Conversion between Mass Transfer Coefficients
LMBLLMBxLLx xCkxkkM
Ckk '''
LMBLMB
xCC
P
kPkPkPk
RT
Pk yLMBGLMByG
c '''
Gases:
Liquids:
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Example 1
27
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28
Film Theory
Resistance to the diffusion is equivalent to that in astagnant film of a certain thickness
A thin boundary layernear the interface wherethe fluid is in laminar flowAll resistance in the filmTransport or Mass transferis by molecular diffusion
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29
Penetration Theory
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Diffusion of a short distance Unsteady-state molecular transport Mass flux at interface of gas and liquid is:
Surface elements will be renewed byeddies from the turbulent core Instantaneous mass transfer, with solutepenetrating into eddy after exposure tosurface
AbAiABA cctD
N
30
Penetration Theory
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Total solute transferred is
with average mass transfer rate
With distribution in element ages at thesurface, rate of surface renewal is constantand given a factor s, so mass transfer is
2/1
exp
0 2exp
tD
ccdtNAB
AbAi
t
A
2/1
exp
2
tDccN ABAbAiA
AbAiABA ccsDN 31
Penetration Theory
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32
For absorption, the solute may diffuse through a gasphase and then diffuse through and be absorbed in anadjacent and immiscible liquid phase.
The two phases are in direct contact with each other,and the interfacial area between the phases is usuallynot well defined. A concentration gradient must exist to cause this mass
transfer through the resistances in each phase.
Mass Transfer Between Phases
Ammonia (NH3) (A) +air (B) absorbed by water (C)
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distance from interface
NA
yAGyAi
interface
gas-phase mixtureof A in gas Gliquid-phase solutionof A in liquid L.
xAixAL
WhereyAG = concentration of A in the bulk gas phaseyAi = concentration of gas A at the interfacexAi =concentration of liquid A at the interfacexAL = concentration of A in the bulk liquid phase
Mass Transfer Between Phases
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3434
Mass Transfer Between Phases
ALAixAiAGyA xxkyykN ''
Equimolar counterdiffusion (EMD)
Whereky = Gas phase mass transfer coefficient (kg mol/s.m2.mol frac)
kx = Liquid phase mass transfer coefficient (kg mol/s.m2.mol frac)
Diffusion of A through stagnant or
nondiffusing B (UMD)
ALAixAiAGyA xxkyykN
LMA
y
y
y
kk
1
'
LMA
x
x
x
kk
1
'
Where
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The interface composition (xAi and yAi ) can be determined by drawing
he line PMwith the slope (-kx/ky) intersecting the equilibrium line
Mass Transfer Between Phases
y*A xAL~ x*A~yAG
yAG
yAi
y*AxAL xAi x*A
M
D
E
P equilibriumline
slope = my
slope = mx
AiA
AiAG
yxx
yym
*
AGAi
AAi
x xx
yym
*
AiAL
AiAG
y
x
xxyy
kkslope
'
'
Fig.: Concentration driving force and interface concentrations in inter
phase mass transfer (equimolar counter diffusion EMD)
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yAG
yAi
y*AxAL xAi x*A
M
D
E
P equilibriumline
slope = my
slope = mx
AiA
AiAGy
xxyym *
AGAi
AAi
xxx
yym
*
ALAiAiAG
LMAy
LMAx
xx
yy
yk
xk
slope
1
1
'
'
Fig.: Concentration driving force and interface concentrations in inter
phase mass transfer (A diffusing through stagnant B).
Mass Transfer Between Phases
y*A xAL~ x*A~yAG
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Overall mass-transfer coefficientSimilar to overall heat-transfer coefficient
Mass Transfer Between PhasesUnimolecular diffusion (UMD)
ALAxAAGyA xxKyyKN **
ALAixAiAGyA xxkyykN
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Overall Mass Transfer Coefficient
38
ALAixAiAGAAiAiAGAAG xxmyyyyyyyy **
mx
is the slope of ME
x
A
x
y
A
y
A
k
Nm
k
N
K
N
x
x
yy km
kK
111
ALAiAiAGy
ALAiAiAALA xxyym
xxxxxx 1**
x
A
yy
A
x
A
k
N
km
N
K
N
xyyx kkmK
111
Similarly
my is the slope of DM
(3-242)
(3-243)
AGAi
AAi
x xx
yym
*
AiA
AiAG
y
xx
yym
*
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39
Mass Transfer Resistance
39
Resistance in gas phase
Total resistance in both phasesy
y
K
k
1
1
Resistance in liquid phase
Total resistance in both phasesx
x
K
k
1
1
Solute A is very soluble, mx is small
Solute A is insoluble, my is large
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40
Two Resistance Theory
Three resistance: the gas phase, the liquid phase, andthe interface.The interface resistance is negligible and anyfluctuations in yAi, and xAi are small, yAi and xAi areequilibrium values given by the systems equilibrium-distribution curve
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Individual Mass Transfer Coefficient
41
iAgAGzA ppkN ,,, LAiALzA cckN ,,,
Combining both and rearrange to
iALA
iAGA
G
L
cc
pp
k
k
,,
,,
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42
Overall Mass Transfer Coefficient*
, AGAGA ppKN
iAiAmcp ,, LAA mcp ,
* *, AGA mcp
zA
LAiA
zA
iAGA
G Nccm
Npp
K ,
,,
,
,,1
At low concentrations, we have linear
equilibrium relations
LGGk
m
kK
11
Where : PAG the bulk composition in the gas phase
PA* is the partial pressure of A in equilibrium with the bulkconcentration of CALKG the overall mass transfer coefficient based on partial pressure
(time)(interfacial area)(pressure)
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4343
Overall Mass Transfer CoefficientLAALA ccKN ,*
iAiAmcp ,, LAA mcp ,
* *, AGA mcp At low concentrations, we have linear
equilibrium relations
LGL kmkK
111
Where : CA* the bulk composition in the liquid in equilibrium with
the bulk gas phaseCAL is the bulk concentration in the liquid
KL the overall mass transfer coefficient based on a liquid driving force
(time)(interfacial area)(mole A/volume)
zA
LAiA
zA
iAGA
zA
ALA
LN
cc
mN
pp
N
CC
K ,
,,
,
,,
,
*1
oncentration Driving Force for the Two-resistance Theory
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44
*
AP
*
AC
APtotal
AGP
ALP
AC
AGC
ALC
total
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45
Ratio of resistances in individual phase to
total resistance
Ratio of resistances in individual phase to
total resistance
L
L
totalA
filmliquidA
K
k
C
C
/1
/1
,
,
G
G
totalA
filmgasA
K
k
P
P
/1
/1
,
,
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Mass Transfer Resistance
46
Resistance in gas phase
Total resistance in both phases G
G
K
k
1
1
Resistance in liquid phase
Total resistance in both phasesL
L
K
k
1
1
Solute A is very soluble, mx is small
Solute A is insoluble, my is large
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47
Wetted Wall Column
Wetted Wall Column (glass tube)
Header
Header
Water
Moist air
Moist air
Experimental Setup
Blower
Thomas
Meter
Wet and Dry Bulb
Temperature
Thermometer
Hygrometer
Area for mass transfer is defined
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Film TheoryResistance to the diffusion is equivalent to that in astagnant film of a certain thickness
Aicc 0z
Abcc z
AbAiABA ccD
J
AB
c
Dk
atat
is not directly measureable contains all of the fluiddynamics in the system and willchange with the Re
0z z
cAb
Thin layer
(stagnant)
able to calculate (Re)from experimental data
48
Example : Interface Composition in Interphase Mass Transfer.
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49
The solute A is being absorbed from a gas mixture of A and B in a
wetted-wall tower with the liquid flowing as a film downward along the
wall. At certain point in the tower the bulk gas concentration
yAG = 0.380 molfraction and the bulk liquid concentration isxAL = 0.1. The tower is operating at298 K and 1.013 x 10
5 Paand the
equilibrium data are as follows:
xA yA
0 0
0.05 0.022
0.1 0.052
0.15 0.087
0.2 0.131
0.25 0.187
0.3 0.265
0.35 0.385
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The solute A diffuse through stagnant B in the gas
phase and then through a nondiffusing liquid.Using correlations for dilute solutions in wetted-wall
towers, the film mass-transfer coefficient for A in the
gas phase is predicted as
ky = 1.465 x 10-3 kg mol A/s.m2 mol frac. And for the
liquid phase as
kx = 1.967 x 10-3 kg mol A/s.m2 mol frac.
Calculate the interface concentrations yAi and xAi
and the flux NA.
SolutionFirst we plot the data
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yA
xA
00
0.1
0.1
0.2
0.2
0.3
0.3
0.4
0.4
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Now we need to find the point P on the graphSo :Since the correlations are fordilute solutions, (1-yA)iM and (1-
xA)iM are approximately 1.0 and the coefficients are the sameas ky and kx .
Point P is plotted at yAG = 0.380 and xAL = 0.1.For the first trial (1-yA)iM and (1-xA)iM are assumed as 1.0 andthe slope of line PM is,
A line through point P with a slope of1.342 is plotted in thefigure intersecting the equilibrium line at M1, whereyAi = 0.183 and xAi = 0.247.
342.10.1/10465.1
0.1/10967.1
)1/(
)1/(3
3
'
'
LMAy
LMAx
yk
xkslope
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53
For the second trial we use yAi and xAi fromthe first trial to calculate the new slope.Substituting into corresponding Eqs.,
)]1/()1ln[(
)1()1()1(
AGAi
AGAiLMA
yy
yyy
715.0)]38.01/()183.01ln[()380.01()183.01(
)]1/()1ln[(
)1()1()1(
AiAL
AiALLMA
xx
xxx
825.0)]247.01/()1.01ln[(
)247.01()1.01(
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54
Substituting into Eq. (10.4-9) to obtain the new slope,
A line through point P with a slope of1.163 is
plotted and intersects the equilibrium line at M,
where yAi = 0.197 and xAi = 0.257. Using these newvalues for the third trial, the following values are
calculated:
163.1715.0/10465.1
825.0/10967.1
)1/(
)1/(
3
3
'
'
LMAy
LMAx
yk
xkslope
709.0)]38.01/()197.01ln[(
)380.01()197.01()1(
LMAy
820.0)]257.01/()1.01ln[(
)257.01()1.01()1(
LMAx
Please notice the valuesof xAi and yAi didntchange much from thefirst trial and that meanswe are on the right wayof solving the problem(refining the answer)
8200/10961)1/(3'k
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55
This slope of1.160 is essentially the same as theslope of1.163 for the second trial.Hence, the final values are yAi= 0.197 and xAi =
0.257 and are shown as point M. To calculate the
flux,
Note that the flux NA through each phase is the
same as in other phase, which should be the case at
steady state.
160.1709.0/10465.1
820.0/10967.1
)1/(
)1/(3
3
'
LMAy
LMAx
yk
xkslope
24
3'
./1078.3
)197.0380.0(709.0
10967.1)(
)1(
mskgmol
yyy
kN AiAG
LMA
y
A
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Fig.: Location of interface concentrationsfor example.
yAG
yAi
y*A
xAL xAi x*A
M
D
E
P
M1
00
0.1
0.1
0.2
0.2
0.3
0.3
0.4
0.4
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Example 3
57
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Example 4
62
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Diffusion Coefficients
69
Diffusion can be considered by a mechanisticapproach in which a consideration of atommovement is important or by a continuum approachsuch as with Ficks first law, where no considerationis given to the actual mechanism by which atomtransfer occurs.Gases: molecules are far apart, intermolecular forcescan often be disregarded or considered only duringcollisions.A molecule is assumed to travel along a straight lineuntil it collides with another molecule, after which itsspeed and direction are altered.
Diffusion Coefficients for Gases
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The diffusion coefficient can be derived foran idea gas by using the simplified discussionpresented by Sherwood et al (1975).From kinetic theory, the diffusion coefficientis assumed to be directly proportional to themean molecular velocity , and the meanfree path U
UD
Diffusion Coefficients for Gases
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For an idea gas the motion of the molecule is assumedto be totally random, with the mean free path , beinginversely proportional to both the average cross-sectional area of the molecules, A, and the numberdensity, n, of all molecules in a specified volume.Thenumber density of an idea gas varies directly withpressure and inversely with the temperature,
PA
T
nA
1
Mean molecular velocity is related to the temperatureand molecular weight of the molecule by theexpression
PA
T
M
T
M
TD
BA
AB
21
21
23
' 11
BAavg
ABMMPA
TKD
Aavg: average cross-sectional areaof both types of moleculesK a costant
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Semi-empirical Equations
231312175.100143.0
BVAVAB
BAAB
PM
TDD
Equation of Fuller, Schettler, and Giddings
DAB is in cm2/sP: atmT: K V =summation of atomic and structural diffusionvolumes from Table 3.1, which includes diffusionvolumes of simple molecules