MASS TRANSFER Coefficient and Inter Phase Mass Transfer

41

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Page 1: MASS TRANSFER Coefficient and Inter Phase Mass Transfer

LECTURE 3: INTERPHASE MASS TRANSFER AND DIFFUSION COEFFICIENT

Dr Aradhana Srivastava,

Associate Professor and Group Leader,

Chemical engineering Group,

BITS Pilani, Hyderabad Campus

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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

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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

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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

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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

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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

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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

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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

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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

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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)

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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

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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

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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)

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SOLUTION

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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.

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SOLUTION

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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

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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

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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

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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

=

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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'

Page 33: MASS TRANSFER Coefficient and Inter Phase Mass Transfer

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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)

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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

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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

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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

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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

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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

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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

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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

Page 41: MASS TRANSFER Coefficient and Inter Phase Mass Transfer

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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