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LECTURE 3: INTERPHASE MASS TRANSFER AND DIFFUSION COEFFICIENT
Dr Aradhana Srivastava,
Associate Professor and Group Leader,
Chemical engineering Group,
BITS Pilani, Hyderabad Campus
2
INTERPHASE MASS TRANSFER
So far we only discussed mass transfer of species in single phase (gas or liquid)
Many mass-transfer operations involve the contact of two insoluble phases to permit mass transfer
In such cases, the thermodynamic equilibrium between phases is important
Mass transfer is derived by the deviation from the equilibrium state If equilibrium between the phases is
established, there would be no net mass transfer
3
EQUILIBRIUM RELATION
There are equations that describes the equilibrium relation between the concentration of certain component in the liquid and gas phase at certain temperature and pressure
Raoult’s Law(gas-liq mixture)
Henry’s Law (gas-liq dilute solution)
Distribution-law (liquid-liquid)
AAAAA PxPyp
pressurevapor
A
tcoefficienactivity
AA
pressuretotal
A
pressurepartial
A PxPyp
AAA HxPyp
2,1, liquidAliquidA Kcc
4
DIFFUSION BETWEEN PHASES Consider the absorption of ammonia
from a mixture with air using liquid water, in a wetted wall column
The ammonia-air mixture enters the column from the bottom and flow upward and water flow downward on the inner wall of the column The concentration of the ammonia in the
mixture decreases as it flow upward the concentration of the ammonia in water
increases as it flow downward under steady state conditions, the
concentration at any point of the column does not change with time
NH3-Air
Water
NH3-H2O
5
TWO-RESISTANCE THEORY
The ammonia (solute) diffuses from the gas phase to the liquid through an interfaceThere is a concentration gradient
in the direction of mass transfer in each phase
On concentrations on the interface (yA,i and xA,i, are assumed to be in equilibrium)
This simply means the mass transfer resistance is only in the fluid phase and no resistance across the interface
Water
Inte
rface
Bulk NH3-Air
yA, G
yA,i
xA,i
xA,G
6
PHASE TRANSFER COEFFICIENTS
NNH3 can be expressed in terms of k-type coefficients:
yA,i= f(xA,i)
A y A ,G A ,i x A ,i A ,LN k y y k x x Water
Inte
rface
Bulk NH3-Air
yA, G
yA,i
xA,i
xA,G
A ,G A ,ix
y A ,L A ,i
y yk
k x x
yA,G
xA,i
yA,i
xA,L
Equi
libriu
m C
urveSlope=
-kx /k
y
A ,i
A ,i
y
y is the interface concentration in the gas phase side
x is the interface concentration in the liquid phase side
k
is the gas phase mass trans
fer
Wher
coe i
:
e
ffic
x
en
k is the liquid phase mass transfer coeffic
t
ient
7
OVERALL-MASS TRANSFER COEFFICIENT
The interface concentration is hard to accurately measure
The flux can be estimated in terms of overall mass transfer coefficient as follow
Water
Inte
rface
Bulk NH3-Air
yA, G
yA,i
xA,i
xA,G
* *A y A ,G A x A A ,LN K y y K x x
*A A ,L
*A A ,G
y is in equilibrium with x
x is in equilibrium with
re :
y
Whe
yA,G
xA*
yA*
xA,L
Equi
libriu
m C
urve
Slope=-k
x /ky
yA,i
xA,i
8
OVERALL AND INDIVIDUAL PHASE COEFFICIENT
If the equilibrium relation is linearDilute solution where Henry’s Law applies
A ,i A ,iy m x
* *A A ,L A ,G Ay m x ; y m x
*A y A ,G A
* *A ,i A ,LA ,G A A ,G A ,i A ,i A A ,G A ,i
y A A A A A
N K y y
m x xy y y y y y y y1
K N N N N N
y y x
1 1 m
K k k
x y x
1 1 1
K mk k Similarly
y
y
1
kResistance in gas phase1Total Resistance in both phases
K
x
X
1
kResistance in liquid phase1Total Resistance in both phases
K
9
EFFECT OF THE GAS SOLUBILITY IN THE MASS TRANSFER COEFFICIENTS
For highly soluble gas (the slope of the equilibrium line, m, is small)The major resistance is in the
gas phase
For low solubility gas (the slope of the equilibrium line, m, is large)The major resistance is in the
liquid phase
High solubility
yA
XA
Low so
lubi
lity
y y x y
1 1 m 1
K k k k
x y x x
1 1 1 1
K mk k k
10
EXAMPLEIn an experimental study of the absorption of NH3 by water in a wetted-wall column, the value of KG was found to be 2.75×10-6 kmol/(m2-s-kPa). At one point in the column, the composition of the gas and liquid phases were 8.0 and 0.115 mol% NH3, respectively. The temperature was 300 K and the total pressure was 1 atm. 85% of the total resistance to mass transfer was found to be in the gas phase. At 300° K, NH3-water solution follows Henry’s law up to 5 mol% NH3 in the liquid with m = 1.65 when the total pressure is 1 atm. Calculate:
1.Flux of NH3
2. Individual film coefficients3. Interfacial Compositions (yA,i and xA,i)
11
SOLUTION
GivenT=300 K; P=1 atmKG= 2.75x10-6 kmol/m2-s-kPayA,G=0.080; xA,L=0.00115 .
.y y
1 0.85
k K
*A y A ,G AN K y y
* 3A A ,Ly mx 1.65 * 0.00115 1.886 x10
* 4 5 2A y A ,G AN K y y 2.786 x10 0.08 0.001886 2.18 x10 kmol / m s
-6 -4 2y GK =K P = 2.75x10 x 101.3 = 2.786x10 kmol/m -s
12
SOLUTION
.
. .
.
y y
1 0.85
k K
5A
A ,i A ,G 4y
N 2.18 x10y y 0.08 0.01362
k 3.28 x10
4y 4 2
y
K 2.786 x10k 3.28 x10 kmol / m .s
0.85 0.85
y y x
1 1 m
K k k
x y y y y y
m 1 1 1 0.85 0.15
k K k K K K
A y A ,G A ,iN k y y
A i,i
3,A
0.013628.
y3x
m05 x10
1.64
6y 3 2
x
mK 1.64 x 2.75 x10k 3.05 x10 kmol / m .s
0.15 0.15
PROBLEMThe solubility of gaseous substance (Mol Wt. 26) in
water is given by Henry’s law: pA = 105 xA, pA in mm of Hg. Convert the equilibrium relation to the following forms: (a) yA = m xA if the total pressure is 10 bar; (b) pA =m’CA, where CA is in gmol/litre. Also write down the equilibrium relation using the mole ratio unit. Assume that solution is dilute and has a density equal to that of water (= 1000 kg/m3)
SOLUTION
PROBLEMIn a certain equipment used for absorption of SO2 from air by
water at one section, the gas and liquid phase concentration of the solute are 10 mole % and 4 mass % respectively. The solution density is 6.1lb/ft3. Ata given temperature (40 0C) and pressure (10 atm), the distribution of SO2 beetween air and water can be approximately described as pA = 25 xA, where is the partial pressure of SO2 in the gas phase in atm. The individual mass transfer coefficient are kx= 10 kmol/hm2(delta x) and ky = 8 kmol/hm2(delta y). Calculate the overall coefficient K
G in kmol/hm2(delta p in mm Hg) and xAi and yAi at the gas-liquid interface.
SOLUTION
17
LOCAL MASS TRANSFER COEFFICIENTS For general case
Diffusion of more than one species No equimolar counter diffusion The mass transfer rates are large
k-type diffusion coefficients cannot be used F-type diffusion coefficient has to be used.
General approach is same for finding expression
A ,G A ,i A ,L A ,LA A ,G G A ,L L
A ,G A ,G A ,L A ,i
y xN F ln F ln
y x
A ,L L
A ,G G
FF
A ,G A ,i A ,L A ,L
A ,G A ,G A ,L A ,i
y x
y x
Psi values are molal ratio of a component to the total moles, and F values are the mass transfer coefficients
18
EXAMPLEA wetted-wall absorption tower is fed with water as the wall liquid and an ammonia air mixture as the central-core gas. At a particular point in the lower, The ammonia concentration in the bulk gas is 0.60 mole fraction, that in the bulk liquid is 0.12 mole fraction. The temperature is 300 K and the pressure is 1 atm. Ignoring the vaporization of water, calculate the local ammonia mass-transfer flux. The rates of flow are such that FL = 0.0035 kmol/m2-s. and FG = 0.0020 kmol/m2-s. The equilibrium-distribution data for the system at 300° K and 1 atm is: A A A A A A Ay 10.51 x ; 0.156 0.622 x 5.765 x 1 ; x 0.3
19
SOLUTION Given
T=300 K; P=1 atmyA,G=0.60; xA,L=0.12FL = 0.0035 kmol/m2-s; . FG = 0.0020
kmol/m2-s
Although this is a diffusion of A through stagnant B, the ammonia concentration is too high to use k-type mass transfer coefficient, F-type coefficient must be used
A A A A A A Ay 10.51 x ; 0.156 0.622 x 5.765 x 1 ; x 0.3
A ,G A ,i A ,L A ,LA A ,G G A ,L L
A ,G A ,G A ,L A ,i
y xN F ln F ln
y x
A ,G A ,L 1
A ,i A ,LA G L
A ,G A ,i
1 y 1 xN F ln F ln
1 y 1 x
L
G
F
FA ,L
A ,i A ,GA ,i
1 xy 1 1 y
1 x
20
SOLUTION
Graphically or numerically yA,i=0.49; xAi=0.23
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0 0.05 0.1 0.15 0.2 0.25 0.3
xAy A
A A A A A A Ay 10.51 x ; 0.156 0.622 x 5.765 x 1 ; x 0.3
A ,iA G
A ,G
3
4 2
1 yN F ln
1 y
1 0.492 x10 ln
1 0.6
0.4.7 x10 kmol / m s
L
G
F1.75
FA ,L
A ,i A ,G A ,iA ,i A ,i
1 x 0.88y 1 1 y y 0.6
1 x 1 x
equilibriu
m
From flux
A A A A Ay 10.51 x 0.156 0.622 x 5.765 x 1 ; x 0.3
=
21
MATERIAL BALANCE Consider SS mass transfer
operation involves countercurrent contact of two immiscible phases V is total moles of phase V Vs is moles of A-free V L is total moles of phase L Ls is moles of A-free L y is mole fraction of component A
in V x is mole fraction of component A
in L Y is the moles of A per mole A-free
V X is mole of component A per mole
A-free L
L, x, X V, y, Y
L2, x2, X2 V2, y2, Y2
V1, y1, Y1L1, x1, X1
Z=z1
Z=z2
yY
1 y
x
X1 x
A = solute
22
MATERIAL BALANCE: COUNTER FLOW
Mole balance around the column
Using solute free basis
Mole balance around plane z
Using solute free basis
moles of A entering moles of A leaving
the column the column
1 1 1 1V y Lx Vy L x
1 1 2 2 2 2 1 1V y L x V y L x
S 1 S 2 S 2 S 1V Y L X V Y L X
L, x, X V, y, Y
L2, x2, X2 V2, y2, Y2
V1, y1, Y1L1, x1, X1
Z=z1
Z=z2
z
S 1 S S S 1V Y L X V Y L X
23
MATERIAL BALANCE: COUNTER FLOW The mole balance in terms of solute free
basis can be expressed as:
or
X1
Y
Slope=Ls/Vs
operating Line
Y2
Y1
X2
X
equilibrium curve
X1
YLs/Vs
operating LineY1
Y2
X2
X
equilibrium curve
L, x, X V, y, Y
L2, x2, X2 V2, y2, Y2
V1, y1, Y1L1, x1, X1
Z=z1
Z=z2
z
Transfer from phase V to phase L (Absorption)
Transfer from phase L to phase V (Stripping)
S 1
S 1
L Y Y
V X X
S 1 2
S 1 2
L Y Y
V X X
24
MATERIAL BALANCE: COCURRENT FLOW
Mole balance around the column
Using solute free basis:
Using the operating line approach:
moles of A entering moles of A leaving
the column the column
1 1 1 1 2 2 2 2V y L x Vy Lx V y L x
L, x, X V, y, Y
L2, x2, X2 V2, y2, Y2
V1, y1, Y1L1, x1, X1
Z=z1
Z=z2
zS 1 s 1 S S S 2 S 2V Y L X V Y L X V Y L X
S1 2 1
1 2 S 1
LY Y Y Y
X X V X X
25
MATERIAL BALANCE: COCURRENT FLOW
The mole balance in terms of solute free basis can be expressed as:
X1
YSlope=-Ls/Vs
operating LineY1
Y2
X2
X
equilibrium curve
L, x, X V, y, Y
L2, x2, X2 V2, y2, Y2
V1, y1, Y1L1, x1, X1
Z=z1
Z=z2
z
Transfer from phase V to phase L (Absorption)
Transfer from phase L to phase V (Stripping)
S1 2 1
1 2 S 1
LY Y Y Y
X X V X X
X1
Y
Slope=-Ls/Vs
operating Line
Y1
Y2
X2
X
equilibrium curve
26
MATERIAL BALANCE
Masses, mass fraction, and mass ratio can be substituted consistently for moles, mole fractions, and mole ratios in the mass balance equations
Counter flow
Cocurrent Flow
Where the prime (´) indicates mass based property
' ' 'S 1 2' ' 'S 1 2
L Y Y
V X X
' ' 'S 1 2' ' 'S 1 2
L Y Y
V X X
27
EXAMPLE: ADSORPTION OF NO2 ON SILICA GELNO2 produced by thermal process for fixation of nitrogen, is to be removed from dilute mixture with air by adsorption on silica gel in a continuous adsorber. The mass flow rate of the gas entering the adsorber is 0.5 kg/s. It contains 1.5% NO2, by volume, and 85% of NO2 is to be removed. Operation to be isothermal at 298° K and 1 atm. The entering gel will be free of NO2. the equilibrium adsorption data at this temperature are:
If twice the minimum gel rate is to be used, calculate the gel mass flow rate and the composition of the gel leaving the process for: i. Counter flow operationii. Cocurrent flow operation
pNO2, mmHg 0 2 4 6 8 10 12
Solid conc. (m), kg NO2/ 100 kg gel
0 0.4 0.9 1.65 2.60 3.65 4.85
28
SOLUTION:
1. Plot the equilibrium data Since the equilibrium data is given in
terms of mass ratios, it is easier to use mass based equation in this case
The partial pressure data have to be converted to mass ratio data (Y’)
Now the equilibrium data become:
2 2NO NO' ii i
A
i
ir i A
ii
i
ir
i
M kgp 46Y Y x
M 760 p 29
y pY
1 y P
kg
p
Gel+NO2 Air/NO2
GelAir/NO2
Cocurrent
Gel Air/NO2
Gel/NO2Air/NO2
Countercurrent
Xi’, kg NO2/ 100 kg gel 0 0.4 0.9 1.65 2.60 3.65 4.85
Yi’, kgNO2/ 100 kg Air
29
CONTD.
Minimum gel rate
When the operating line touches (reaches) the equilibrium line
X1´=0.037 (From graph)
2 2NO NO' 11 1
Air 1 Air
M kgy 46 0.015 46Y Y x * 0.0242
M 1 y 29 1 0.015 29 kg
0
0.005
0.01
0.015
0.02
0.025
0.03
0 0.01 0.02 0.03 0.04 0.05
Y' ,
kg
NO
2/kg
Air
(Ls/Vs) Min
operating Line
Y1'
Y2'
X2'
X' kg NO2/kg gel
equilibrium curve
X1', max
2NO' '2 1
Air
kgY 1 0.15 xY 0.15 * 2.42 0.0036
kg
' 22
kg NOX 0.00
kg gel
(1-0.85)*Y1’
*100
30
CONTD.
' ' 'S 1 2' ' '
1 2S'
' '' ' '1 2
minS S S' '1 2
' 's 1 B 1 '
1
'S
'S
' '1 2' ' '
1 2 S 'S
L Y Y
X XV
Y Y 0.0242 0.0036L xV xV
0.037 0X X
1 1V V 0.5 x 0.5 0.488 kgAir / s
1 0.02421 Y
L (min) 0.268 kg gel / s
L 2 x0.268 0.536 kg gel / s
Y Y 0.0X X V 0.00 0.488 x
L
2
24 0.00360.0186 kg NO / kg gel
0.536
31
0
0.005
0.01
0.015
0.02
0.025
0.03
0 0.01 0.02 0.03 0.04 0.05
Y' ,
kg
NO
2/kg
Air
(Ls/Vs) Min
operating Line
Y1'
Y2'
X2' X' kg NO2/kg gel
equilibrium curve
X1', max
32
COCURRENT FLOW Y1’=0.024
Y2’=0.0036
X1’=0.00
For (LS)min, (X2’)max can be found be drawing the operating line reaching
the equilibrium line (X2’)max=0.0034
Gel+NO2 Air/NO2
GelAir/NO2
Cocurrent
0
0.005
0.01
0.015
0.02
0.025
0.03
0 0.01 0.02 0.03 0.04 0.05
Y' ,
k
g N
O2
/kg
Air
Slope (-Ls/Vs) Min
operating Line
X1', Y1'
X' kg NO2/kg gel
equilibrium curve
X2', max,Y2'
33
SOLUTION, CONTINUED
' ' 'S 1 2' ' '
1 2S'
' '' '1 2
minS S' '2 ,max 1
'S
' '1 2' ' '
2 1 S 'S
2
L Y Y
X XV
Y Y 0.0242 0.0036L xV x0.488 2.957 kg gel / s
0.0034 0X X
L 2 x 2.957 5.92 kg gel / s
Y YX X V
L
0.024 0.00360 0.488 x 0.00168 kg NO / kg gel
5.92
To reach the same degree of removal of NO2, countercurrent flow is much more effective compared to cocurrent
The amount of gel needed for cocurrent flow (5.92 kg/s) is about 11 times of that needed if countercurrent flow is used (0.536 kg/s)
34
EQUILIBRIUM-STAGE OPERATION
In many instances mass transfer devices are assembled by interconnecting individual units (stages) The material passes through each one of these stages Two material streams moves countercurrently
(cascades) In each stage the two streams are contacted, mixed, and
then separated As the stream moves between stage they come close to
equilibrium conditions If the leaving streams from a certain stage is in
equilibrium, this stage is an ideal stage If the stage are connected cocurrently they represent a
single stage Batch mass-transfer operations are also a single stage
35
STAGE-OPERATIONS
The flow rate and composition of each stream are numbered corresponding to the effluent from a stage X2 is the mole ratio in stream leaving stage 2
YN is the mole ratio of stream leaving For ideal stages, the effluents from each stage are in
equilibrium Y2 is in equilibrium with X2 and so on The cascade has the characteristics of the
countercurrent process with operating line goes through points (X0, Y1) and (XN, YN+1)
The cas
Stage N-1
StageN
Stage 2
Stage 1
V1
VS
Y1
L0
LS
X0
V2
VS
Y2
L1
LS
X1
VN
VS
YN
LN-1
LS
XN-1
VN+1
VS
YN+1
LN
LS
XN
n
36
NUMBER OF IDEAL STAGES
Number of ideal stages can be determined graphically (for two component systems)
Y
XN, YN+1
X
equilibrium curve
X0, Y1
X0X1X3
YN+1
Y3
Y2
Y1
37
NUMBER OF STAGES For linear equilibrium line (Yi=mXi),
analytical solution is possible (Kremser):Define Absorption factor (ratio of the
slope of the operating line to the slope of the equilibrium line):S
S
LA
mV
For transfer from L to V (Stripping)
0 N 1
N N 1
0 N
N N 1
X Y / mln 1 A A
X Y / mN A 1
ln 1 / A
X XN A 1
X Y
For transfer from V to L (Absorption)
N 1 0
1 0
N 1 1
1 0
Y mX 1 1ln 1
Y mX A AN A 1
ln A
Y YN A 1
Y mX
38
EXAMPLE:
A flue gas flows at a rate of 10 kmol.s at 298 K and 1 atm with a SO2 content of 0.15 mole%. Ninety percent of sulfur dioxide is to be removed by absorption with pure water at 298° K. The design water flow rate will be 50% higher than the minimum. Under these conditions, the equilibrium line is:i iY 10 X
39
SOLUTION
Given Data T=298° K y1=yN+1=0.0015 x2=x0=0 m=10 V1=10 kmol/s LS=1.5 (LS) min
Single Stage
V1, VS, y1, Y1
L2, LS, x2, X2
L1, LS, x1, X1
V2, VS, y2, Y2 Casca
de
N
VN+1, VS, yN+1, YN+1
L0, LS, x0, X0
LN, LS, xN, XN
V1, VS, y1, Y1
2
1
3
N-1
40
SOLUTION y1=yN+1=0.0015Y1=(YN+1)cascade=
0.0015/(1-0.0015)=0.001502 Y2=(Y1)cascade=(1-0.9)*Y1=0.00015 x2=(x0)cascade=0 Vs=V1(1-y1)=10*(1-
0.0015)=9.985 kmol/s From the graph
X1, max=0.00015 (LS/VS)min=(Y1-Y2)/(X1, max-X2)
= 9 LS min=9*9.985=89.9 kmol/s LS =1.5 Ls min=134.8 kmol/s
=134.8*18=2426 kg/s LS/VS=13.5 X1=X2+(Y1-Y2)/(LS/VS)
=10-4 mol SO2/mol H2O = 10-4 *64/18=0.356x10-3
= 0.356 g SO2/kg H2O
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0.0014
0.0016
0.0018
0.002
0 0.00002 0.00004 0.00006 0.00008 0.0001 0.00012 0.00014 0.00016 0.00018 0.0002
X
Y
Y2
Y1
X2
X1 maxX1
(LS/VS)
min
LS/V
S
Equilibrium
41
SOLUTION: NUMBER OF STAGES
Graphically:N ≈ 4 stages
Analytically
N 1 0
1 0
Y mX 1 1ln 1
Y mX A AN A 1
ln A
S
S
L 13.5A 1.35
mV 10
0.001502 10 x0.0 1 1ln 1
0.0015 10 x0.0 1.35 1.35N 4.01 Stages
ln 1.35
0
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0.0014
0.0016
0.0018
0.002
0 0.00002 0.00004 0.00006 0.00008 0.0001 0.00012 0.00014 0.00016 0.00018 0.0002
X
Y
2
1
3
4
Y1
YN+1
X0
XN
Operating Line
Slope=LS/VS
Equilibrium lineSlope = m