CHAPTER THREE
Stage & Continuous Gas–Liquid Separation
Processes
1
Content
10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.8
Types of Separation Processes and Methods
Equilibrium Relation Between Phases
Single and Multiple Equilibrium Contact Stages
Mass Transfer Between Phases
Continuous Humidification Processes
Absorption on plate and Packed Tower
Absorption and Concentrated Mixtures in Packed Tower
Estimation of Mass Transfer Coefficient in Packed Tower
2
10.1 Types of Separation Processes
and Methods
 Absorption (When the two contacting phases are a gas and a liquid)
 Distillation (A volatile vapor phase and a liquid that vaporizes are
involved)
 Liquid-liquid Extraction (When the two phases are liquids)
 Leaching (Extract a solute from a solid, sometimes also called
extraction)
 Membrane Processing (Separation of molecules by the use of
membrane)
 Crystallization
 Adsorption
 Ion-Exchange
3
Definitions
Absorption
The removal of one or more selected components from a mixture
of gases by absorption into a suitable liquid is the second major
operation of chemical engineering that is based on interphase mass
transfer controlled largely by rates of diffusion.
Stripping/desorption
Reverse of absorption.
Humidification
Transfer of water vapor from liquid water into pure air.
Dehumidification
4
Removal of water vapor from air.
Absorption Unit
Pure Gas Out
A= 5 mole
Pure Liquid In
A= 0 mole
Gas In
A= 20mole
Liquid Out
A= 15 mole
5
Main design1/2
Pure Solvent
Gas
6
Main design2/2
7
10.2 Equilibrium Relation Between Phases
 Gas- Liquid Equilibrium
For dilute concentrations of most gases, and over a wide range for some gases, the
equilibrium relationship is given by Henry’s law. This law, can be written as:
PA  H.X A
YA  H ' X A
…………………………. (10.2.2)
…………………………. (10.2.3)
where,
PA = partial pressure of component A (atm).
H = Henry’s law constant (atm/mol fraction).
H’ = Henry’s law constant (mol frac.gas/mol frac.liquid). = H/P.
XA = mole fraction of component A in liquid. ( dimensionless )
YA = mole fraction of component A in gas = PA/P. ( dimensionless )
P = total pressure (atm).
8
Example 10.2-1
Dissolved Oxygen Concentration in Water.
What will to be the concentration of oxygen dissolved in water at
298 K when the solution is in equilibrium with air at 1 atm total
pressure ? The Henry’s law constant is 4.38 x 104 atm/mol fraction.
Solution,
PA  H . X A
The partial pressure pA of oxygen (A) in air is 0.21 atm. Using
Eq.(10.2-2)..
0.21=H xA
= 4.38 x104 xA
Solving , xA = 4.80 x10-6 mol fraction.
9
10.3 Single and Multiple Equilibrium
Contact Stages
10.3.A Single - stage Equilibrium contact
 It can be defined as one in which two different phases (liquid & gas for
absorption) are brought into contact & then separated.
 During the time of contact intimate mixing occurs and the various components
diffuse and redistribute themselves between the two phases.
 If mixing time is enough, the components are essentially at equilibrium in the two
phase after separation and the process is considered to be single equilibrium
stage
10.3.B Single-stage Equilibrium contact for Gas-Liquid System (Absorption)
Gas phase contains solute A & inert gas B.
Liquid phase contains solute A & inert liquid (solvent) C.
V = gas flowrate & L = liquid flowrate.
L’ = inert liquid (C) flowrate & V’ = inert gas (B) flowrate
x,y = mole frac. in liquid phase & gas phase, respectively.
10
Single-Stage Equilibrium Contact for
Gas-Liquid System 1/2
Gas phase
outlet
V1
yA1
V2
yA2
Gas phase
inlet
Single-stage
Liquid phase
inlet
L0
xA0
L1
xA1
Liquid phase
outlet
11
Single-Stage Equilibrium Contact for Gas-Liquid
System 2/2
In
=
Out
Gas phase
outlet
Liquid phase
inlet
Total material balance: L0 + V2 = L1 + V1
Component A balance: L0xA0 + V2yA2 = L1xA1 + V1yA1
Component C balance: L0xC0 + V2yC2 = L1xC1 + V1yC1
An equation for B is not needed since
In term of inert flow rate:
V1
yA1
L0
xA0
Singlestage
V2
yA2
Gas phase
inlet
L1
xA1
Liquid
phase
outlet
xA + xB + xC = 1.0
L’ = L(1-xA)  L = L’/(1-xA)
V’ = V(1-yA)  V = V’/(1-yA)
Operating line:
 x A0 
y A2 
x A1 
y A1 
'
'
'
  V 
  L 
  V 

L 
 1  y A2 
 1  x A1 
 1  y A1 
 1  x A0 
'
unknowns :-
xA1 & yA1
12
Example 10.3-1 1/5
Equilibrium Stage Contact for CO2–Air-water
A gas mixture at 1.0 atm pressure abs containing air and
CO2 is contacted in a single-stage mixer continuously
with pure water at 293 K. The two exit gas and liquid
streams reach equilibrium. The inlet gas flow rate is 100
kg mol/h, with a mole fraction of CO2 of yA2 = 0.20. The
liquid flow rate entering is 300 kg mol water/h. calculate
the amounts and compositions of the two outlet phases.
Assume that water does nor vaporize to the gas phase.
13
Example 10.3-1 2/5
Solution
The flow diagram,
(1) The inert water flow is L’ = L0 = 300 kg mol/h.
(2) The inert air flow V’ is obtained from,
Hence, the inert air flow is
V’ = V2 (1-yA2 )
= 100 (1-0.20)
= 80 kg mol/h .
V1
yA1
L0 = 300 kg mol/h
xA0
1 atm
293 K
V’ = V(1-yA ).
V2 = 100 kg mol/h
yA2 = 0.20
L1
xA1
14
Example 10.3-1 3/5
(3) Substituting into equation to make a balance on CO2 (A).
 x A0 
y A2 
x A1 
y A1 
'
'
'
  V 
  L 
  V 

L 
 1  y A2 
 1  x A1 
 1  y A1 
 1  x A0 
'
…(1)
 x A1 
 y A1 
 0 
 0.20 
  80

300
  80
  300
1 0 
 1  0.20 
 1  x A1 
 1  y A1 
Know …. The Resulting equation has two unknowns (XA1 and YA1 ) .. So it
is necessary to calculate one of these unknowns to solve the equation
above..!!!
It is possible to calculate the YA1 by the use of Henry’s law because the gas
and liquid are in equilibrium as mentioned in the question before
15
Example 10.3-1 4/5
YA  H ' X A
H` needed to calculated
At 293 K, the Henry’s law constant from Appendix A.3 is
H = 0.142 x 104 atm/mol frac.
Then
H’ = H/P = 0.142 x 104 /1.0 = 0.142 x 104 mol frac. gas/mol frac. Liquid.
Substituting into,
yA1 = 0.142 x 104 xA1
………(2)
Know it possible to use the value of YA1 in equation (1) to calculate the value of XA1
16
Example 10.3-1 5/5
Solving equation (1) and (2) simultaneously,
get xA1 = 1.41 x 104 and yA1 = 0.20. To calculate the total flow rates
leaving,
L'
300
L1 

 300kgmol / h
4
1  x A1
1  1.4110
V'
80
V1 

 100kgmol / h
1  y A1
1  0.20
In this case, since the liquid solution is so dilute, L0  L1 .
17
10.3.C Countercurrent Multiple-Contact
Stages 1/4.
V1
V2
Vn
2
1
L0
V3
L1
Vn+1
n
L2
Ln-1
VN+1
VN
N
Ln
LN-1
LN
18
10.3.C Countercurrent Multiple-Contact
Stages 2/4.
 Countercurrent multiple-contact stages.
more concentrated product.
total number of ideal stages = N.
B&C may or may be not be somewhat miscible in each other.
two streams leaving a stage in equilibrium with each other.
 Total material balance:
 Component A overall balance:
 For the first n stages balance :
L0 + VN+1 = LN + V1
L0 x0A + VN+1 yN+1A = LN xNA + V1 y1A
L0 x0A + Vn+1 yn+1A = Ln xnA + V1 y1A
This is why we write n instead of N
 Operating line:-
19
10.3.C Countercurrent Multiple-Contact
Stages 3/4.
 An operating line is an important material-balance equation because
it relates the concentration yn+1 in the V stream with xn in the L
stream passing it.
Ln
slope 
Vn 1
slope
streams L & V are immiscible in each other with only A being
transferred:-
 x A0 
'  y An1 
'  x An 
'  y A1 
  V 
  L 
  V 

L 
 1  y A1 
 1  x A0 
 1  y An1 
 1  x An 
'
20
10.3.C Countercurrent Multiple-Contact
Stages 4/4.
 Graphical calculation for determining N:
1.
Plot yA vs xA.
2.
Draw operating line.
3.
Draw equilibrium line (Henry’s law).
4.
Stepping upward (or downward) until yN+1 (or y1) is reached.
5.
N = number of steps/trays
 Dilute system (<10%):1. slope (flowrates V & L)  constant  operating line = straight
2. L’ & V’ = constant, dilute system (<10%)  operating line = straight.
21
Example 10.3-2 1/4
Absorption of Acetone in a Countercurrent Stage Tower
It is desired to absorb 90% of the acetone in a gas containing
1.0 mol % acetone in air in a countercurrent stage tower. The
total inlet gas flow to the tower is 30.0 kg mol/h and the total
inlet pure water flow to be used to absorb the acetone is 90
kg mol H2O/h.
The process is to be operate isothermally at 300 K and a total
pressure of 101.3 kPa. The equilibrium relation for the
acetone (A) in the gas-liquid is yA = 2.53xA.
Determine the number of theoretical stages required for this
separation.
22
Example 10.3-2 2/4
Solution
the first step is to have good identification of the data given in the question
Given values are
yAN+1= 0.01
(1 mole % of acetone in air entering)
xA0 = 0
(Pure water )
VN+1 = 30.0 kg mol/h,
(total inlet gas flow to the tower )
L0 = 90.0 kg mol/h.
(total inlet pure water )
 making an acetone material balance,
(1) amount of entering acetone = yAN+1 VN+1
= 0.01(30.0)
= 0.3 kg mol/h.
(2) entering air = (1- yAN+1 )VN+1
= (1-0.01)(30.0)
= 29.7 kg mol air/h
23
Example 10.3-2 3/4
(3) acetone leaving in V1= 0.10(0.30) = 0.030 kg mol/h.
(4) acetone leaving in L1 = 0.90(0.30) = 0.27 kg mol/h.
 from the four steps above, V1 , yA1 , LN ,and xAN can be calculated
V1 = 29.7 + 0.03 = 29.73 kg mol air + acetone/h.
yA1 = (0.030/29.73) = 0.00101
LN = 90.0 + 0.27 = 90.27 kg mol water + acetone /h.
xAN = (0.27/90.27) = 0.0030.
Since the flow of liquid varies only slightly from L0 = 90.0 at the inlet
to LN = 90.27 at the the outlet and V from 30.0 to 29.73, the slope Ln
/Vn+1 of the operating line is essentially constant.
 This line is plotted, and the equilibrium relation of Henry
yA = 2.53xA is also plotted.
 Starting at point yA1 ,xA0 the stages are drawn. About 5.2 theoretical
24
stages are required.
Example 10.3-2 4/4
Mole fraction acetone in air, yA
0.012
yAN+1
Operating line
0.008
5
4
0.004
Equilibrium line
3
2
yA1
0
0
xA0
1
0.003
xAN
Mole fraction acetone in water, xA
0.001
0.002
0.004
Figure 10.3-4: Theoretical stages for countercurrent absorption in example 10.3-2.
25
10.3.D: Analytical Equations For Countercurrent
Stage Contact 1/3
 This technique is used to calculate the number of theoretical
stages analytically without the need for the graphical method
 Its also used to make comparison and justifications in the results
of both methods
 Kremser equations
 This method is used to calculate the number of ideal stages.
 This method is valid only when operating & equilibrium lines are
straight.
 In case of Absorption system the equation used is :
 y N 1  mx0 
1  1
log 
1    
y1  mx0 
A  A

N 
log A
26
10.3.D Analytical Equations For Countercurrent
Stage Contact 2/3
where,
m = slope of equilibrium line.
A = absorption factor = constant = L/(mV).
L.V = molar flow rates.
y N 1  y1
N 
when A = 1
y1  mx0
Stripping:
y N 1


x

 0

m

log 
1  A  A
y
 x N  N 1

m


N 
1
log
A
27
10.3.D Analytical Equations For Countercurrent
Stage Contact 3/3
When A = 1,
x0  x N
N 
y N 1
xN 
m
 Procedure (for varying A):
1.
Calculate A1 at L0 & V1.
2.
Calculate AN at LN &VN+1.
3.
Calculate Aave. =
4.
Calculate N.
A1 AN
28
Example 10.3-3 1/2
Number of Stages by Analytical Equation.
Repeat Example 10.3-2 but use the Kremser analytical equation
for countercurrent stage processes.
Solution,
At Stages 1
V1 = 29.73 kg mol/h,
yA1 = 0.001001,
L0 = 90.0, and
xA0 = 0.
Also, the equilibrium relation is yA = 2.53xA where m = 2.53.
Then,
L
L
90.0
A1 
 0 
 1.20
mV
mV1
2.53  29.73
29
Example 10.3-3 2/2
At Stage N
VN+1 =30.0,
yAN+1 = 0.01,
LN = 90.27, and
xAN = 0.0030.
The geometric average,
Then,
AN 
A
LN
90.27

 1.19
mVN 1 2.53  30.0
A1 AN 
0.20 1.19  1.195
 0.01  2.53(0) 
1 
1 
log 
1





0
.
00101

2
.
53
(
0
)
1
.
195

 1.195 

N
 5.04
log( 1.195)
This compares closely with 5.2 stages using graphical method
30
10.4 Mass Transfer Between Phases 1/2
mole fraction
For absorption, the solute may diffuse through a gas phase and
then diffuse through and be absorbed in an adjacent and
immiscible liquid phase.
 The two phases are in direct contact with each other, and the
interfacial area between the phases is usually not well defined.
 A concentration gradient must exist to cause this mass transfer
through the resistances in each phase.
Liquid
Gas
y
xi
x
yi
Mass transfer of A
31
10.4 Mass Transfer Between Phases 2/2
liquid-phase solution
of A in liquid L.
gas-phase mixture
of A in gas G
yAG
yAi
xAL
xAi
NA
interface
Where
distance from interface
yAG = concentration of A in the bulk gas phase
yAi = concentration of gas A at the interface
xAi =concentration of liquid A at the interface
xAL = concentration of A in the bulk liquid phase
32
Film-transfer Coefficient and Interface
Concentration.
 Equimolar counterdiffusion.
NA = k’y(yAG – yAi) = k’x(xAi – xAL)
Where
k’y = Gas phase mass transfer coefficient (kg mol/s.m2.mol frac)
k’x = Liquid phase mass transfer coefficient (kg mol/s.m2.mol frac)
 Diffusion of A through stagnant or nondiffusing B.
NA = ky(yAG – yAi) = kx(xAi – xAL)
where,
ky 
k y'
(1  y A ) iM
k x'
kx 
(1  x A )iM
33
NOTE (for equimolar counter-diffusion)
The interface composition (xAi and yAi ) can be determined by drawing the
line PM with the slope (-kx’ /ky’) intersecting the equilibrium line
k x'
y  y Ai
slope   '  AG
ky
x AL  x Ai
yAG
D
P
equilibrium
line
slope = m”
yAi
y*A
M
slope = m’
E
xAL
xAi
x*A
34
NOTE (for diffusion of A through stagnant
or nondiffusing B)
The interface composition (xAi and yAi ) can be determined by drawing the
line PM with the slope (
) intersecting the equilibrium line
k x'
y  y Ai
slope   '  AG
ky
x AL  x Ai
yAG
D
P
equilibrium
line
slope = m”
yAi
y*A
M
slope = m’
E
xAL
xAi
x*A
35
Example 10.4-1 1/8
Interface Composition in Interphase Mass Transfer
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 mol fraction and the bulk liquid concentration is
xAL = 0.1. The tower is operating at 298 K and 1.013 x 105 Pa and the
equilibrium data are as follows:
xA
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
yA
0
0.022
0.052
0.087
0.131
0.187
0.265
0.385
36
Example 10.4-1 2/8
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.
37
Example 10.4-1 3/8
Solution
First we plot the data
0.4
0.3
yA
0.2
0.1
0
0
0.1
0.2
0.3
xA
0.4
38
yAG
Example 10.4-1 4/8
0.4
P
D
0.3
yAi
0.2
M
M1
0.1
y
Now we need to find the point P on the graph
E
0
0
0.1
0.2
So :
x
x
 Since the correlations are for dilute solutions, (1-yA)iM and (1-xA)iM are
approximately 1.0 and the coefficients are the same as k’y and k’x .
*
A
AL
0.3
Ai
0.4
x*A
 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 and the slope of line
PM is, from Eq.(10.4-9).
k x' /(1  x A ) iM
1.967 10 3 / 1.0
slope   '

 1.342
3
k y /(1  y A ) iM
1.465 10 / 1.0
 A line through point P with a slope of –1.342 is plotted in the figure intersecting
the equilibrium line at M1, where
yAi = 0.183 and xAi = 0.247.
39
yAG
Example 10.4-1 5/8
0.4
P
D
0.3
yAi
0.2
M
M1
0.1
y*A
0
E
0
0.1
xAL
0.2
0.3
xAi
0.4
x*A
 For the second trial we use yAi and xAi from the first trial to
calculate the new slope. Substituting into Eqs.(10.4-6) and (10.4-7),
(1  y A ) iM
(1  y Ai )  (1  y AG )

ln[( 1  y Ai ) /(1  y AG )]
(1  0.183)  (1  0.380)

 0.715
ln[( 1  0.183) /(1  0.38)]
(1  x AL )  (1  x Ai )
(1  x A ) iM 
ln[( 1  x AL ) /(1  x Ai )]
(1  0.1)  (1  0.247)

 0.825
ln[( 1  0.1) /(1  0.247)]
40
yAG
Example 10.4-1 6/8
0.4
P
D
0.3
yAi
0.2
M
M1
0.1
y*A
 Substituting into Eq. (10.4-9) to obtain the new slope,
0
E
0
0.1
xAL
0.2
0.3
xAi
0.4
x*A
k x' /(1  x A ) iM
1.967 10 3 / 0.825
slope   '

 1.163
3
k y /(1  y A )iM
1.465 10 / 0.715
 A line through point P with a slope of –1.163 is plotted and intersects
the equilibrium line at M, where yAi = 0.197 and xAi = 0.257. Using
these new values for the third trial, the following values are
calculated:
(1  y A )iM
(1  0.197)  (1  0.380)

 0.709
ln[( 1  0.197) /(1  0.38)]
(1  x A )iM 
(1  0.1)  (1  0.257)
 0.820
ln[( 1  0.1) /(1  0.257)]
Please notice the values of xAi
and yAi didn’t change much
from the first trial and that
means we are on the right way
of solving the problem
41
(refining the answer)
yAG
Example 10.4-1 7/8
0.4
P
D
0.3
yAi
0.2
M
M1
0.1
y*A
0
3
E
0
0.1
xAL
0.2
k /(1  x A ) iM
1.967 10 / 0.820
slope  

 1.160
3
k /(1  y A )iM
1.465 10 / 0.709
'
x
'
y
0.3
xAi
0.4
x*A
 This slope of –1.160 is essentially the same as the slope of –1.163 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,
NA 
k y'
(1  y A )iM
1.967 10 3
( y AG  y Ai ) 
(0.380  0.197)
0.709
 3.78 10 4 kgmol / s.m 2
Note that the flux NA through each phase is the same as in other phase, which
should be the case at steady state.
42
Example 10.4-1 8/8
yAG
0.4
P
D
0.3
yAi
0.2
M
M1
0.1
y*A
0
0
E
0.1
xAL
0.2
0.3
0.4
xAi
x*A
Fig.10.4-4: Location of interface concentrations for example 10.4-1.
43
Overall Mass-transfer Coefficients
and Driving Force.
 For equimolar counterdiffusion and/or diffusion in dilute
solutions,
NA = k’y (yAG – yAi ) = k’x (xAi – xAL)
 K’y(yAG – y*A )  K’x (x*A – xAL)
x*A is the value that would be in equilibrium with yAG
y*A is the value that would be in equilibrium with xAL
 For diffusion of A through stagnant or nondiffusing B.

K y'
NA  
 (1  y A )M
'


K

x
(
y

y
)

 AG
A


 (1  x A )M
 
 ( x A  x AL )

44
10.5 Continuous Humidification Processes
Natural draft wet cooling hyperboloid towers at Didcot
Power Station, UK
45
Water-Cooling Tower 1/3
46
Water-Cooling Tower 2/3
Warm water flows counter-currently to an air stream. The warm water enters the top of
a packed tower and cascades down through the packing, leaving at the bottom.
Air enters at the bottom of the tower and flows upward through the descending water
by the natural draft or by the action of a fan.
The water is distributed by troughs and overflows to cascade over slat gratings or
packing that provide large interfacial areas of contact between the water and air in the
form of droplets and film of water.
The tower packing often consists of slats of wood or plastic or of a packed bed.
47
Water-Cooling Tower 3/3
48
Theory and Calculation of Water-Cooling
Towers
interface
liquid water
Hi
HG humidity
water vapor
TL
sensible heat
in liquid
Ti
air
TG temperature
latent heat in gas
sensible heat in gas
Figure10.5-1: Temperature and concentration profile in upper
part of cooling tower.
49
Vapor Pressure of Water & Humidity
 Introduction
calculation involve properties and concentration of
mixtures of water vapor and air.
 Humidification
transfer of water from the liquid phase into a gaseous
mixture of air and water vapor.
 Dehumidification
reverse transfer where the water vapor is transferred
from the vapor state to the liquid state.
50
Humidity & Humidity Chart 1/4
(1) Humidity, H : the kg of water vapor contained in 1 kg of dry air.
18.02
pA
H 
28.97 P  p A
where,
pA = partial pressure of water vapor in the air.
Saturated air – water vapor in equilibrium with liquid water.
pA = pAS
where,
pAS = saturated vapor pressure.
51
Humidity & Humidity Chart 2/4
(2) Humid volume, vH
It can be defined as total volume (m3) of 1 kg of dry air plus the vapor it contains at
1 atm abs pressure and the given gas temp.
vH (m3/kg dry air) = (2.83 x 10-3 + 4.56 x 10-3 H) T (K).
(3) Total enthalpy of an air-water mixture, HY
- the total enthalpy of 1 kg of air plus its water vapor.
- sensible heat of the air-water vapor mixture plus the latent heat.
HY (kJ/kg dry air) = (1.005 + 1.88 H) (T ºC-0) + 2501.4H
where,
Tref for both components = 0 ºC
52
Humidity & Humidity Chart 3/4
(4) Wet bulb temperature TW,
 It’s the steady-state nonequilibrium temperature reached when a small
amount of water is contacted under adiabatic conditions by a
continuous stream of gas.
Steady state temp. attained by a wet-bulb thermometer under
standardized condition
 The temperature and humidity of the gas are not changed.
 At TW = TS, the convective heat transfer and wet bulb lines.
q  M B k y W H W  H  A
or,
h M Bk y
H  HW

T  TW
W
53
Humidity & Humidity Chart 4/4
h M Bk y
H  HW

T  TW
W
MB = molecular weight of Air
ky= Mass transfer coefficient
w = latent heat of vaporization at Tw
A = surface area
h = Heat transfer coefficient
q  h. A.(T  Tw)
54
Humidity Chart
Please insert figure 9.3-2 (Humidity
chart) page 529 (Geankoplis)
55
The operating line
G(Hy – Hy1) = LcL (TL – TL1)
where,
G = dry air flow, kg/s.m2.
L = water flow, kg water/s.m2
cL = heat capacity of water, assumed constant at 4187 kJ/kg.K.
TL = temperature of water, ºC or K.
Hy = enthalpy of air-water vapor mixture, J/kg air.
= cs (T-T0) + H0 = (1.005 + 1.88H)103 (T-0) + 2.501x106H
H = humidity of air, kg water/kg dry air.
56
Figure 10.5-3: Temperature enthalpy diagram
and operating line for water-cooling.
Hy*2
equilibrium line
Hy2
Hy*
Enthalpy of airHyi
vapor mixture,
Hy (J/kg dry gas)
Hy
Hy*1
operating line,
slope = LcL/G
R
M
S
P
slope = -hLa
kGaMBP
Hy1
TL1
Ti TL
TL2
Liquid temperature ºC
57
Design of Water-Cooling Tower Using
Film Mass-Transfer Coef.
 Please follow the steps mentioned in section 10.5C in book
 To calculate the tower height

z
0
G
dz  z 
M B kG aP

H y2
H y1
dH y
H yi  H y
where,
z = tower height
P = atm pressure.
MB = molecular weight of air
kGa = volumetric mass transfer coeff. in gas, kg mol/s.m3
58
Design Using Overall Mass-Transfer
Coefficients.

z
0
H y2
dH y
G
dz  z 
M B K G aP H y1 H y  H y
where,
KGa = overall mass transfer coefficient.
 If experimental cooling data in an actual run in a cooling tower
with known height z are available, then the value of KGa can be
obtained.
59
Table 10.5-1
60
Example 10.5-1 1/6
Design of Water-Cooling Tower Using Film Coefficients.
A packed countercurrent water-cooling tower using a gas flow
rate of G = 1.356 kg dry air/s. m2 and a water flow rate of
L = 1.356 kg water/s. m2 to cool the water from TL2 = 43.3 ºC
to TL1 = 29.4 ºC.
The entering air at 29.4 ºC has a wet bulb temperature of
23.9 ºC . The mass-transfer coefficient kG a is estimated as
1.207 x 10-7 kg mol/s.m3.Pa and hL a / kGaMBP as 4.187 x 104
J/kg.K. .
Calculate the height of packed tower z. The tower operates at a
pressure of 1.013 x 105 Pa.
61
Example 10.5-1
Solution
200
2/6
Slope =
-41.87 x103
180
equilibrium line
160
140
Enthalpy Hy
[(J/kg)10-3]
120
100
operating line
80
60
Following the step outlined
The enthalpies from the saturated air-water vapor mixtures from
Table 10.5-1 are plotted in Fig. 10.5-4.
The inlet air at TG1 = 29.4 ºC has a wet bulb temperature of 23.9 ºC
.
The humidity from the humidity chart is H1 = 0.0165 kg H2O/kg
dry air.
Substituting into Eq.(9.3-8),
28
30
TL1
1.
Data
preparation
to be used in
equation
(9.3.8)
2.
3.
4.
32
34
36
Liquid Temperature (ºC)
38
40
42
44
46
TL2
Hy1 = cs (T-T0) + H0 = (1.005 + 1.88H) 103 (T-T0) + H0
Hy1 = (1.005 + 1.88 x 0.0165)103 (29.4-0) + 2.501 x 106(0.0165).
= 71.7 x 103 J/kg.
The point Hy1 = 71.7 x 103 and TL1 = 29.4 ºC is plotted. Then
substituting into Eq. (10.5-2) and solving,
62
Example 10.5-1 3/6
G(Hy2 –Hy1) = LcL (TL2 – TL1)
1.356 (Hy2 – 71.7 x 103) = 1.356 (4.187 x 103) (43.3 – 29.4).
Hy2 = 129.9 x 103 J/kg dry air.
Now both Hy1 and Hy2 are calculated
Then
(1) The point Hy2 = 129.9 x 103 and TL2 = 43.3 ºC is plotted, giving the
operating line.
(2) Lines with slope - hL a / kG aMBP = -41.87 x 103 J/kg.K are plotted
giving Hyi and Hy values, which are tabulated in Table 10.5-2 along
with derived values as shown.
(3) Values of 1/(Hyi – Hy) are plotted versus Hy and the area under the
curve from Hy1 = 71.7 x 103 to Hy2 = 129.9 x 103 is
H y2
dH y
H y1
H yi  H y

 1.82
63
Example 10.5-1
200
4/6
Slope =
-41.87 x103
180
equilibrium line
160
140
Enthalpy Hy
[(J/kg)10-3]

H y2
H y1
dH y
H yi  H y
120
100
 1.82
operating line
80
60
28
30
TL1
32
34
36
38
Liquid Temperature (ºC)
40
42
44
46
TL2
Substituting into Eq. (10.5-13),
dH y
G
1.356
z

(1.82)
7
5

M B kG aP H yi  H y
29(1.207  10 )(1.013  10 )
 z  6.98m
64
Example 10.5-1 5/6
Hyi
94.4 x 103
108.4 x 103
124.4 x 103
141.8 x 103
162.1 x 103
184.7 x 103
Hy
Hyi – Hy
1/(Hyi – Hy)
71.7 x 103
83.5 x 103
94.9 x 103
106.5 x 103
118.4 x 103
129.9 x 103
22.7 x 103
24.9 x 103
29.5 x 103
35.3 x 103
43.7 x 103
54.8 x 103
4.41 x 10-5
4.02 x 10-5
3.39 x 10-5
2.83 x 10-5
2.29 x 10-5
1.85 x 10-5
Table 10.5-2: Enthalpy Values for Solution to Example 10.5-1
(enthalpy in J/kg dry air).
65
Example 10.5-1 6/6
200
Slope =
-41.87 x103
180
equilibrium line
160
140
Enthalpy Hy
120
[(J/kg)10-3]
100
operating line
80
60
28
30
TL1
32
34
36
38
Liquid Temperature (ºC)
40
42
44
46
TL2
Figure 10.5-4: Graphical solution of Example 10.5-1
66
Minimum Value of Air Flow 1/2
N
Hy2
Hy1
equilibrium
line
M
TL1
P
operating line for
Gmin,
slope
operating
line,= LcL/Gmin
slope = LcL/G
TL2
 The air flow G is not fixed but must be set for the design of the
cooling tower.
 For a minimum value of G, the operating line MN is drawn through
the point Hy1 and TL1 with a slope that touches the equilibrium line
at TL2, point N.
 If the equilibrium line is quite curved, line MN could become
tangent to the equilibrium line at a point farther down the
equilibrium line than point N.
 For the actual tower, a value of G greater than Gmin must be used.
Often, a value of G equal to 1.3 to 1.5 times Gmin is used.
67
Minimum Value of Air Flow 2/2
N
equilibrium line
Hy2
P
operating line for Gmin,
slope = LcL/Gmin
Hy1
M
TL1
operating line,
slope = LcL/G
TL2
Figure 10.5-5: Operating-line construction for
minimum gas flow.
68
Design Using Height of a Transfer Unit
dH y
dH y
H y2
H y2
G
z
 H OG 


H y1 H   H
M B K G aP H y1 H y  H y
y
y
Where :
G= air flow rate
KGa= overall mass transfer coefficient (kg.mole/s.m3)
HOG= height of overall gas enthalpy transfer unit (m)
Temperature and Humidity of Air Stream in Tower
dH y
dTG

H yi  H y
Ti  TG
69
10.6 Absorption on plate and Packed Tower
10.6.1 Absorption on plate Tower
V1, y1
1
2
Vn+1, yn+1
n
L0, x0
Ln, xn
n+1
N-1
N
VN+1, yN+1
LN, xN
Fig.10.6-4: Material balance in an absorption tray tower.
70
10.6.1 Absorption on plate Tower
 The operating line
A plate (tray) absorption tower has the same process flow diagram
as the countercurrent multiple-stage process (as shown in the
figure)
 x0
L 
 1  x0
'

y N 1 
x
'
'
  V 
  L  N

 1  y N 1 
 1  xN

y1 
'
  V 

 1  y1 

 A balance around the dashed-line box gives
 x0
L 
 1  x0
'
Where

yn 1 
x
'
'
  V 
  L  n

 1  yn 1 
 1  xn

 y 
  V '  1 
 1  y1 

x= mole fraction A in the liquid
y= mole fraction A in the gas
Ln= total moles of liquid/s
Vn+1= total moles of gas/s
71
Example 10.6-2 1/4
Absorption of SO2 in a Tray Tower.
A tray tower is to be designed to absorb SO2 from an air
stream by using pure water at 293 K.
The entering gas contains 20 mol % SO2 and that leaving
2 mol % at the total pressure of 101.3 kPa .
The inert air flow rate is 150 kg air/h.m2 and the entering
water flow rate is 6000 kg water/h.m2. Assuming an
overall tray efficiency of 25 %.
How many theoretical trays and actual trays are
needed? Assume that the tower operate at 293 K.
72
Example 10.6-2 2/4
Solution,
(1) calculating the molar flow rates,
V’ = (150/29) = 5.18 kg mol inert air/h.m2
L’ = (6000/18.0) = 333 kg mol inert water/h.m2.
(2) Referring to figure 10.6.4
yN+1 = 0.20 (Entering Gas from the bottom)
y1 = 0.02, (2% leaving in the gas )
x0 = 0. (pure liquid )
(3)Substituting into operating line equation and solving for xN.
 xN
 0 
 0.2 
333
  5.18
  333
1 0 
 1  0.20 
 1  xN
 xN  0.00355

0.02 
  5.18

1

0
.
02



(4) Substituting into dashed-line eq., using V’ and L’ as kg mol/h.m2
instead of kg mol/s.m2.
73
Example 10.6-2 3/4
 yn 1 
 xn
 0 


333
 333
  5.18

1 0 
 1  yn 1 
 1  xn

0.02 
  5.18

 1  0.02 

(5)In order to plot the operating line, several intermediate points will be
calculated. Setting yn+1 =0.07 and substituting into the operating
equation,
 xn
 0.07 
0  5.18
  333
 1  0.07 
 1  xn

0.02 
  5.18

1

0
.
02



Hence, xn = 0.000855. To calculate another intermediate point, we ser
yn+1 = 0.13, and xn is calculated as 0.00201.
(6)The two end points and the two intermediate points on the operating
line are plotted in Fig.10.6-5, as are the equilibrium data from
Appendix A.3.
(7) The operating line is somewhat curved. The number of theoretical
trays is determined by stepping off the steps to give 2.4. The actual
number of trays is 2.4/0.25 = 9.6 trays.
74
Example 10.6-2 4/4
YN+1
Mole
fraction,
y
y1
0.20
0.18
0.16
0.14
0.12
0.10
0.08
0.06
0.04
0.02
0
operating
line
equilibrium
line
2
1
0
x0
0.002
0.004
0.006
xN
Mole fraction, x
0.008
Fig.10.6-4 : Theoretical number of trays for absorption of SO2 in example 10.6-2.
75
10.6.2 Absorption on Packed Tower
Design of Packed Tower for Absorption
V2,y2
L2,x2
dz
V,y
V1,y1
z
L,x
L1,x1
76
Structured Packing
Figure 10.6-6. Pressure-drop correlation
for structured packings
where ΔPflood : is in in. H2O/ft height of packing
Fp :is the packing factor in ft-1 given in
77
Table 10.6-1 for random or structured packing
Table 10.6-1. Packing Factors for Random
and Structured Packing 1/2
78
Random Packing
Figure 10.6-5. Pressure-drop correlation
for random packing
79
procedure used to determine the limiting flow
rates and the tower diameter.
1
First, a suitable random packing or structured packing is selected, giving an
Fp value.
2
A suitable liquid-to-gas ratio GL/GG is selected along with the total gas flow
rate.
3
The pressure drop at flooding is calculated using Eq. (10.6-1), or if Fp is 60 or
over, the ΔPflooding is taken as 2.0 in./ft packing height.
4
Then the flow parameter is calculated, and using the pressure drop at flooding
and either Fig. 10.6-5 or 10.6-6, the capacity parameter is read off the plot.
5
Using the capacity parameter, the value of GG is obtained, which is the
maximum value at flooding.
6
Using a suitable % of the flooding value of GG for design, a new GG and GL
are obtained. The pressure drop can also be obtained from Figure 10.6-5 or
10.6-6.
7
Knowing the total gas flow rate and GG, the tower cross-sectional area and ID
80
can be calculated.
Table 10.6-1. Packing Factors for Random
and Structured Packing 2/2
81
EXAMPLE 10.6-1 1/5
Pressure Drop and Tower Diameter for
Ammonia Absorption
Ammonia is being absorbed in a tower using pure
water at 25°C and 1.0 atm abs pressure. The feed
rate is 1440 lbm/h (653.2 kg/h) and contains 3.0 mol
% ammonia in air. The process design specifies a
liquid-to-gas mass flow rate ratio GL/GG of 2/1 and
the use of 1-in. metal Pall rings.
Calculate the pressure drop in the packing and gas
mass velocity at flooding. Using 50% of the
flooding velocity, calculate the pressure drop, gas
and liquid flows, and tower diameter.
82
EXAMPLE 10.6-1 2/5
Solution:
The gas and liquid flows in the bottom of the tower are the largest, so
the tower will be sized for these flows. Assume that approximately all
of the ammonia is absorbed.
The gas average mol wt = 28.97(0.97) + 17.0(0.03) = 28.61.
The weight fraction of ammonia = 0.03(17)/(28.61) = 0.01783.
Assuming the water is dilute, from Appendix A.2-4, the water viscosity
μ = 0.8937 cp. From A.2-3, the water density is 0.99708 gm/cm3.
Then, ρL = 0.99708(62.43) = 62.25 lbm/ft3.
Also, v = μ/ρ = 0.8937/0.99708 = 0.8963 centistokes.
83
EXAMPLE 10.6-1 3/5
 From Table 10.6-1, for 1-in. Pall rings, Fp = 56 ft-1. Using
Eq. 10.6-1, ΔPflood = 0.115 = 0.115(56)0.7 = 1.925 in. H2O/ft
packing height. The flow parameter for Fig. 10.6-5 is :
 Using Fig. 10.6-5, for a flow parameter of 0.06853
(abscissa) and a pressure drop of 1.925 in./ft at flooding, a
capacity parameter (ordinate) of 1.7 is read off the plot.
Then, substituting into the capacity parameter equation and
solving for vG,
84
EXAMPLE 10.6-1 4/5
 vG = 6.663 ft/s.
 Then GG = vGρG = 6.663(0.07309) = 0.4870 lbm/(s · ft2) at
flooding. Using 50% of the flooding velocity for design,
GG = 0.5(0.4870) = 0.2435 lbm/(s · ft2) [1.189 kg/(s · m2)].
 Also, the liquid flow rate
GL = 2.0(0.2435) = 0.4870 lbm/(s · ft2) [2.378 kg/(s · m2)].
 To calculate the tower pressure drop at 50% of flooding, GG =
0.2435 and GL = 0.4870, the new capacity parameter is 0.5(1.7)
= 0.85. Using this value of 0.85 and the same flow parameter,
0.06853, a value of 0.18 in. water/ft is obtained from Fig. 10.6-5.
85
EXAMPLE 10.6-1 5/5
 The tower cross-sectional area =
(1440/3600 lbm/s)(1/0.2435 lbm/(s · ft2))
= 1.6427 ft2 = (π/4)D2.
Solving, D = 1.446 ft (0.441 m).
 The amount of ammonia in the outlet water assuming all
of the ammonia is absorbed is 0.01783(1440) = 25.68 lb.
 Since the liquid flow rate is 2 times the gas flow rate,
 the total liquid flow rate = 2.0(1440) = 2880 lbm/hr.
 Hence, the flow rate of the pure inlet water=
2880 - 25.68 = 2858.3 lbm/s.
86
Design of Packed Tower for Absorption
 Overall material balance on component A:
 x2
L 
 1  x2
'

y1 
x1 
y2
'
'
'
  V 
  L 
  V 

 1  y1 
 1  x1 
 1  y2



 Operating line:y1 
x1 
y 
 x 
'
'
'






L' 

V

L

V






1 x 
1 y 
 1  y1 
 1  x1 
 Very dilute system,
L' x  V ' y1  L' x1  V ' y
 with a slope of L’/V’
 minimum liquid flow rate L’min at x1max.
87
Design of Packed Tower for Absorption
To Solve for L’min:
1. Plot yA vs xA.
2. Draw the equilibrium line.
3. Draw a straight line from (x2,y2) to intersect the equilibrium line
at (x1 max ,y1)
In some cases, if the equilibrium line is curved concavely downward,
the minimum value of L is reached by the operating line becoming
tangent to the equilibrium line instead of intersecting it to give x1 max.
4. Calculate L’min from,
 x1max
 x2 
'  y1 
'
  V 
  Lmin 
L 
 1  x2 
 1  y1 
 1  x1max
'
min

'  y2 
  V 

 1  y2 

88
Example 10.6.3 1/4
Minimum Liquid Flow Rate and Analytical Determination of
Number of Trays
A tray tower is absorbing ethyl alcohol from an inert gas using
pure water at 303 K and 101.3 kPa . The inlet gas stream flow
rate is 100 kg.mol/h and it contains 2.2 mol% alcohol .
It is desired to recover 90% of the alcohol . The equilibrium
relationship is y=m.x = 0.68x for this dilute stream.
Using 1.5 times the minimum liquid flow rate, determine the
number of trays needed graphically
89
Example 10.6.3 2/4

Solution
The given data are :
y1= 0.022, x2= 0, V1= 100 kg mol/ h, m=0.68
(1)V`=V1(1-y1) = 100(1-0.022) = 97.8 kg mol/h
(2) moles of alcohol/h in V1 are 100-97.8=2.2
(3) Removing 90%, moles/h in outlet gas V2is 0.1(2.2) = 0.22
(4) V2=V`+0.22=97.8+0.22=98.02
(5) y2= 0.22/98.02= 0.00244
(6) the equilibrium line is plotted in figure (10.6.12) along with x2,y2and y1
(7) the operating line for minimum liquid flow rate Lmin is drawn from y2,x2
to point P, touching the equilibrium line where x1max = y1/m =
0.022/0.68=0.03235
90
(8) substituting into operation – line equation (10.6.4) and solving for Lmin,
Example 10.6.3
3/4
 x 
 y 
 x 
 y 
L'  2   V '  1   L'  1   V '  2 
 1  x2 
 1  y1 
 1  x1 
 1  y2 
 0 
 0.022 
 0.002244 
'  0.03235 
'
L min 
  97.8
  Lmin 
  97.8

1

0
1

0
.
022
1

0
.
03235
1

0
.
002244








Lmin  59.24kg.mol / h
(9)
Using the relation mentioned in the question
L`=1.5Lmin=1.5(59.24)=88.86
(10) Using L` in equation (10.6.4) and solving for the outlet concentration , x1 = 0.0218
(11) The top operating line now is plotted as a straight line through the points y2,x2 and
y1,x1 in figure (10.6.12)
(12) An intermediate point is calculated by setting y=0.012 in equation (10.6.5) and solving
for x= 0.01078. plotting this point shows that the operating line is very linear . This occurs
because the solutions are dilute
91
(13) The number of theoretical trays obtained by stepping them off is 4.0 trays
Example 10.6.3 4/4
92
Simplified Methods For Dilute Gas Mixtures
(Height Of Packed Towers) 1/3


Since a considerable percentage of the absorption processes include absorption
of a dilute gas A , these cases will be considered using a simplified design
procedure
These cases are taken under consideration under the followings :
1.
2.
3.
4.
Dilute (<10%)
Operating line = straight
Equilibrium line = curve
Height of tower z:-
 V 1  y iM
z '
 k y aS 1  y
 1  y iM 
 y1 dy
 y
 av 2 y  yi
1  yi   1  y 
ln 1  yi  1  y 
93
Simplified Methods For Dilute Gas Mixtures
(Height Of Packed Towers) 2/3
x1
 L 1  x iM 
dx
z '
 x2
k
aS
1

x
xi  x
 x
 av
 1  x iM

1  x   1  xi 

ln 1  x  1  xi 
 V

1  y M
z '

 K y aS 1  y

y1
dy
 y

2 y  y

 av

1  y   1  y 
 1  y 

ln 1  y  1  y 

M

94
Simplified Methods For Dilute Gas Mixtures
(Height Of Packed Towers) 3/3

1  x   1  x  
 1  x M 
ln 1  x  1  x  
x
 L 1  x M 
dx
z
 '
 K x aS
 x2 
x x
 av
1
1 x
Where,
k’x ,k’y = film mass-transfer coefficient for gas phase & liquid phase,
respectively.
K’x ,K’y = overall mass-transfer coefficient for gas phase & liquid phase,
respectively.
a = interfacial area per volume of packed section.
S = cross-sectional area of tower.
yi ,xi = gas & liquid conc. at the interface, respectively.
y* ,x* = gas & liquid conc. that would be in equilibrium with y,x.
95
Solving Procedure 1/9
There are six major steps in the calculations:
1. On xy plot, draw operating line & equilibrium line.
2. By trial-and-error, determine yi ,xi or y*,x* using
k x' a /(1  x) iM
Slope ( PM )   '
k y a /(1  y ) iM
Between the top & bottom of the tower.
Point 1:- 1st trial:
Let (1-y)iM  (1-y1)
(1-x)iM  (1-x1)
&
k x' a /(1  x1 )
k x' a /(1  x) iM
Slope   '
 '
k y a /(1  y1 )
k y a /(1  y ) iM
Plot this slope on xy plot & get yi1 , xi1.
96
Solving Procedure 2/9
2nd trial:
Calculate (1-y)iM & (1-x)iM
Get new slope using,
k x' a /(1  x) iM
Slope   '
k y a /(1  y ) iM
Get new values of yi1 , xi1
3rd trial:
Calculate new slope using values from of yi1 , xi1 2nd trial
Get yi1 , xi1
Stop when slopen1 = slope(n+1)1
Do the same for point 2 and any points between the two.
97
Solving Procedure 3/9
3. Plot y vs 1/(y-yi), x vs 1/(xi – x), 1/(y-y*) or 1/(x* - x).
4. Calculate area under the curve.
5. Calculate the term in the brackets at point 1 & 2. (Get the average.)
6. Calculate z from the appropriate equation.
To calculate the overall mass transfer coefficients:
K’ya:
Get K’yaave = (K’ya1 + K’ya2)/2
Use
1
1
m'
 '
 '
'
K y a /(1  y) M
k y a /(1  y) iM k x a /(1  x) iM
at point 1 & point 2.
98
Solving Procedure 4/9
K’Xa:
Get K’Xaave = (K’Xa1 + K’Xa2)/2
Use,
1
1
1
 '' '
 '
'
K x a /(1  x)M m k y a /(1  y)iM k x a /(1  x)iM
at point 1 & point 2.
99
Solving Procedure 5/9
If solution = dilute (straight operating line) & a straight equilibrium line
Height of tower z :1.
V
( y1  y2 )  k y' az ( y  yi ) M
S
where,
( y  yi ) M
( y1  yi1 )  ( y 2  yi 2 )

 ( y1  yi1 ) 
ln 

 y 2  yi 2 ) 
100
Solving Procedure 6/9
2.
3.
L
( x1  x2 )  k x' az ( xi  x) M
S
( xi1  x1 )  ( xi 2  x2 )
 ( xi  x) M 
 ( xi1  x1 ) 
ln 

 xi 2  x2 ) 
V
( y1  y2 )  K y' az ( y  y  ) M
S


(
y

y
)

(
y

y

1
1
2
2)
 ( y  y )M 
 ( y1  y1 ) 
ln 


 y2  y2 ) 
101
Solving Procedure 7/9
4.
L
( x1  x2 )  K x' az ( x   x) M
S


(
x

x
)

(
x

1
1
2  x2 )
 ( x  x) M 
 ( x1  x1 ) 
ln  

 x2  x2 ) 
Use Vave = (V1 + V2)/2
and
Lave = (L1 + L2)/2
in the above equations.
102
Solving Procedure 8/9
 These
equations may be used in different ways .the general steps are
shown in the following figure:
103
Solving Procedure 9/9
(1) Calculate Vav
Vav=(V1+V2)/2
(2) the interface composition yi1 and xi1 at point y1,x1 in the tower must be determined by
plotting line P1M1 , whose slope is calculated by (please see P672 for more details)
k x` .a /(1  x) iM
k .a
slope   `
 x
k y .a
k y .a /(1  y ) iM
slope  
k x` .a /(1  x1 )
k y` .a /(1  y1 )
(3) If the overall coefficient K`ya is being used , y*1and y*2 are determined as shown in
the past figure . If K`ya is used x1*and x2*are obtained
(4) Calculate the log mean driving force (y-yi)M from equation (10.6.27) if k`ya is used .
For K`ya, (y-y*)M is calculated by equation (10.6.28)
(5) Calculate the column height z by substituting into the appropriate equation
104
Example 10.6-4 1/12
Absorption of Acetone in a Packed Tower.
Acetone is being absorbed by water in a packed tower having a
cross sectional area of 0.186 m2 at 293 K and 101.32 (1 atm).
The inlet air contains 2.6 mol % acetone and outlet 0.5%. The gas
flow is 13.65 kg mol inert air/h. The pure water inlet flow is 45.36
kg mol water/h. Film coefficients for the given flows in the tower
are k’y a = 3.78 x 10-2 kg mol/s.m3.mol frac. And k’x a = 6.16 x10-2
kg mol/s.m3.mol frac. Equilibrium data are given in Appendix A.3.
(a) Calculate the tower height using k’y a.
(b) Repeat using k’x a.
(c) Calculate K’y a, and the tower height.
105
Example 10.6-4 2/12
y1
0.028
slope = -1.62
0.024
From Appendix A.3 for acetone-water
and xA = 0.0333 mol frac.
pA = 30/760 = 0.0395 atm or
yA = 0.0395 mol frac.
Hence, the equilibrium line is
yA = mxA or 0.0395 = m(0.0333).
Then, y=1.186x. This equilibrium line is plotted in Fg. 10.6-10. The given
data are L’ = 45.36 kg mol/h, V’=13.65 kg mol/h, y1= 0.026, y2 = 0.005,
and x2 = 0.
Substituting into Eq.10.6-3 for an overall material balance using flow
rates as kg mol/h instead of kg mol/s.
operating line
0.020
yi
1
y*
1y2
yi
2y*
0.016
0.012
0.008
equilibrium line
0.004
0
0
0.002
x2
xi2
0.004
0.006
0.008
0.010
0.012
0.014
2
x1
xi1
Fig. 10.6-10: Location of interface composition for example 10.6-2.
x1 = 0.00648
 x1 
 0 
 0.026 
 0.005 
  13.65
45.36
  13.65
  45.36

1 0 
 1  0.026 
 1  0.005 
 1  x1 
106
y1
Example 10.6-4
0.028
slope = -1.62
0.024
3/12
operating line
0.020
yi
1
y*
1y2
yi
2y*
0.016
0.012
0.008
equilibrium line
0.004
0
0
0.002
x2
xi2
0.004
0.006
0.008
0.010
0.012
0.014
2
x1
xi1
Fig. 10.6-10: Location of interface composition for example 10.6-2.
The points y1 , x1 and y2 , x2 are plotted in Fig. 10.6-10 and a
straight line is drawn for the operating line. Using Eq.(10.6-31)
the approximate slope at y1 , x1 is,
k x' a /(1  x1 )
6.16 10 2 /(1  0.00648)
slope   '

 1.60
2
k y a /(1  y1 )
3.78 10 /(1  0.026)
Plotting this line through y1 ,x1 , the line intersects the
equilibrium line at yi1 = 0.0154 and xi1 = 0.0130. Also,
y*1 = 0.0077. Using Eq. (10.6-30) to calculate a more accurate
slope, the preliminary values of yi1 and xi1 will be used in the
trial-and-error solution. Substituting into Eq. (10.4-6),
107
y1
Example 10.6-4 4/12
(1  y ) iM
slope = -1.62
0.024
operating line
0.020
yi
1
(1  yi1 )  (1  y1 )

ln[( 1  yi1 ) /(1  y1 )]
0.028
y*
1y2
yi
y2 *
0.016
0.012
0.008
equilibrium line
0.004
0
0
0.002
x2
xi2
0.004
0.006
0.008
0.010
0.012
0.014
2
x1
xi1
Fig. 10.6-10: Location of interface composition for example 10.6-2.
(1  0.0154)  (1  0.026)

 0.979
ln[(1  0.0154) /(1  0.026)]
Using Eq.(10.4-7),
(1  x) iM
(1  x1 )  (1  xi1 )

ln[( 1  x1 ) /(1  xi1 )]
(1  0.00648)  (1  0.0130)

 0.993
ln[( 1  0.00648) /(1  0.0130)]
108
y1
Example 10.6-4
5/12
slope = -1.62
operating line
0.020
yi
1
y*
1y2
yi
2y*
substituting into Eq. (10.6-30),
0.028
0.024
0.016
0.012
0.008
equilibrium line
0.004
0
0
0.002
x2
xi2
0.004
0.006
0.008
0.010
0.012
0.014
2
x1
xi1
Fig. 10.6-10: Location of interface composition for example 10.6-2.
k x' a /(1  x1 )iM
6.16 10 2 / 0.993
slope   '

 1.61
2
k y a /(1  y1 )iM
3.78 10 / 0.929
Hence, the approximate slope and interface values are accurate enough. For the
slope at point y2, x2.
k x' a /(1  x2 )
6.16 10 2 /(1  0)
slope   '

 1.62
2
k y a /(1  y2 )
3.78 10 /(1  0.005)
The slope changes little in the tower. Plotting this line, yi2 = 0.0020, xi2 = 0.0018,
and y*2 = 0. Substituting into Eq. (10.6-24).
109
y1
Example 10.6-4
6/12
( y  yi ) M
slope = -1.62
operating line
0.020
yi
1
( y1  yi1 )  ( y2  yi 2 )

ln[( y1  yi1 ) /( y2  yi 2 )]
0.028
0.024
y*
y2
1
yi
2
y*
0.016
0.012
0.008
equilibrium line
0.004
0
0
0.002
x2
xi2
0.004
0.006
0.008
0.010
0.012
0.014
2
xi1
x1
Fig. 10.6-10: Location of interface composition for example 10.6-2.
(0.026  0.0154)  (0.005  0.0020)

 0.00602
ln[( 0.026  0.0154) /( 0.005  0.0020)]
To calculate the total molar flow rates in kg mol/s,
V'
13.65 / 3600
V1 

 3.893 103 kgmol / s
1  y1
1  0.026
V'
13.65 / 3600
V2 

 3.81110 3 kgmol / s
1  y2
1  0.005
V1  V2 3.893 103  3.811103
Vav 

 3.852 103 kgmol / s
2
2
110
y1
Example 10.6-4
0.028
slope = -1.62
0.024
7/12
operating line
0.020
yi
1
y*
y2
1
yi
2
y*
0.016
0.012
0.008
equilibrium line
0.004
0
0
0.002
x2
xi2
0.004
0.006
0.008
0.010
0.012
0.014
2
xi1
x1
Fig. 10.6-10: Location of interface composition for example 10.6-2.
45.36
L  L1  L2  Lav 
 1.260 10 2 kgmol / s
3600
'
For part (a), substituting into Eq.(10.6-26) and solving.
Vav
( y1  y2 )  k y' az ( y  yi ) M
S
3.852 10 3
(0.026  0.005)  (3.78 10  2 ) z (0.00602)
0.186
 z  1.911m
111
y1
0.028
slope = -1.62
0.024
Example 10.6-4
8/12
operating line
0.020
yi
1
y*
y2
1
yi
For part (b), using an equation similar to Eq.(10.6-27),
2
y*
0.016
0.012
0.008
equilibrium line
0.004
0
0
0.002
x2
xi2
0.004
0.006
0.008
0.010
0.012
0.014
2
( xi  x) M 
( xi1  x1 )  ( xi 2  x2 )
ln[( xi1  x1 ) /( xi 2  x2 )]
xi1
x1
Fig. 10.6-10: Location of interface composition for example 10.6-2.
(0.013  0.00648)  (0.0018  0)

 0.00368
ln[( 0.013  0.00648) /( 0.0018  0)]
substituting into Eq. (10.6-30) and solving,
1.260 10 2
2
(00.00648
.026  0-.005
0 )  (6.16 10 ) z (0.00368)
0.186
 z  1.936m
This checks part (a) quite closely.
112
y1
0.028
slope = -1.62
0.024
Example 10.6-4 9/12
operating line
0.020
yi
1
y*
y2
1
yi
2
y*
0.016
0.012
0.008
equilibrium line
0.004
0
0
0.002
x2
xi2
0.004
0.006
0.008
0.010
0.012
0.014
2
xi1
x1
Fig. 10.6-10: Location of interface composition for example 10.6-2.
For part (c), substituting into Eq. (10.1-25) for point y1 ,x1.
(1  y )M
(1  y1 )  (1  y1 )

ln[( 1  y1 ) /(1  y1 )]
(1  0.0077)  (1  0.026)

 0.983
ln[( 1  0.0077) /(1  0.026)]
The overall mass-transfer coefficient K’y a at point y1 ,x1 is calculated by
substituting into Eq. (10.4-24).
1
1
m'
 '
 '
'
K y a /(1  y )M
k y a /(1  y ) iM k x a /(1  x) iM
113
Example 10.6-4 10/12
1
1
m'
 '
 '
'
K y a /(1  y )M
k y a /(1  y ) iM k x a /(1  x) iM
1
1
1.186


'
2
K y a / 0.983 3.78 10 / 0.979 6.16 10  2 / 0.993
 K y' a  2.183 10  2 kgmol / s.m3 .molfrac.
Substituting into Eq. (10.6-25),
( y  y )M
( y1  y1 )  ( y2  y2 )

ln[( y1  y1 ) /( y2  y2 )]
(0.026  0.0077)  (0.005  0)

 0.01025
ln[( 0.026  0.0077) /( 0.005  0)]
114
Example 10.6-4 11/12
Finally substituting into Eq. (10.6-28),
V
( y1  y2 )  K y' az ( y  y  ) M
S
3.852 10 3
(0.026  0.005)  (2.183 10  2 ) z (0.01025)
0.186
z  1.944m
115
Example 10.6-4 12/12
y1
0.028
slope = -1.62
0.024
operating line
0.020
yi1 0.016
0.012
y*1 0.008
y2
0.004
yi2
0
y*2
equilibrium line
0
0.002
x2
xi2
0.004
0.006
0.008
0.010
0.012
x1
Fig. 10.6-10: Location of interface composition for example 10.6-2.
0.014
xi1
116