chapter 4 - gas absorption
DESCRIPTION
note gas absorptionTRANSCRIPT
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CHAPTER 4: GAS ABSORPTION
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CHAPTER / CONTENT
Definition, Application and Notation Used in Gas Absorption
Packed tower Description and Design
Plate tower Description and Design
Conditions of Equilibrium Between Liquid and Gas According to Raoults Law
The Mechanism of Absorption
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Definition, Application and Notation Used in Gas Absorption
A type of mass transfer operation (separation) between gas and liquid system.
The removal of one or more selected components from a mixture of gases.
Or A soluble vapor is absorbed from its mixture with an inert gas by meansof a liquid the soluble gas is more soluble.
Common example of gas absorption:
Ammonia can be absorbed by passing the gases (NH3 air) intowater where ammonia will be dissolved in water.
The ammonia (solute) can then be recovered by distillation and theabsorbing liquid can be either discarded or reused.
Ammonia Air System
Acetone Air mixture
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Conditions of Equilibrium Between Liquid and Gas According to Raoults Law
The two phases (gas and liquid) when brought into contact tend to reachequilibrium.
Consider air water system, the water in contact with air evaporates untilthe air is saturated with water vapor, and the air is absorbed by the wateruntil it becomes saturated with the individual gases.
In any mixture of gases, the degree to which each gas is absorbed isdetermined by its partial pressure at a given temperature and pressure.
When a single gas (solute) and a liquid (solvent) are brought into contact(until equilibrium), the resulting concentration of dissolved gas (solvent) inliquid is called gas solubility (at T and P).
At fixed temperature, solubility concentration increased when pressureincreased .
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Conditions of Equilibrium Between Liquid and Gas According to Raoults Law
Partial pressure of solute in gas phase
(kN/m2)
Concentration of solute in water kg/1000 kg water
Ammonia Sulfur dioxide Oxygen
1.3
6.7
13.3
26.7
66.7
11
50
93
160
315
1.9
6.8
12
24.4
56
-
-
0.08
0.13
0.33
Ammonia Very soluble
Sulfur Dioxide A moderate soluble
Oxygen Slightly soluble
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Conditions of Equilibrium Between Liquid and Gas According to Raoults Law
In any mixture of gases the solubility of each gas depends on partialpressure
Solubility also depends on Temperature (Solubility as T ).
Recall Raoults Law for ideal solution:
o
AAA Pxp
etemperatur that at A pressure vapor
phase liquid in A of fraction mol
phase gas in A of pressure partial
o
A
A
A
P
x
p
In G.A.;
pressure partial mEquilibriu where ** PxPP o
-
In G.A, feed is a gas and enters at bottom of column and the solvent isfed at top column, the absorbed gas and solvent leave out the bottom andunabsorbed components leave as gas from the top.
Liquid / solvent is well below its boiling point and gas molecules arediffusing into liquid.
Equilibrium Distribution (Solubility Curve)
Conditions of Equilibrium Between Liquid and Gas According to Raoults Law
y
x
(y,x)
Equilibrium curve
Figure 1 Equilibrium curve
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Due to this concept:
Film concept / Theory in mass transfer
Conditions of Equilibrium Between Liquid and Gas According to Raoults Law
Gas molecules must diffuse from main body of the gas phase to the gas liquid interphase, then
Cross this interface into the liquid side and finally diffuse from theinterface into the main body of the liquid.
For dilute concentration of most gases, and over a wide range for somegases, equilibrium relationship is given by Henrys Law.
AA CHp
constant sHenry'
liquid in component of ionconcentrat
phase gas in A of pressure partial
H
C
p
A
A
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THE TWO FILM THEORY
RATE OF ABSORPTION
EVALUATION OF MASS TRANSFER COEFFICIENT
The Mechanism of Absorption
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The Two Film Theory
Developed by Whitman.
According to this theory, material is transferred in the bulk of the phase byconvection currents, and concentration differences are considered asnegligible.
On either side of this interface it suppose that the currents die out and thatthere exists a thin film of fluid flowing through which the transfer occursonly to molecular diffusion.
GA
S F
ILM
BO
UN
DA
RY
LIQ
UID
FIL
M
BO
UN
DA
RY MAIN BULK
OF LIQUID
MAIN BULK OF GAS
GAS FILM
LIQUID FILM
PA
RT
IAL
PR
ES
SU
RE
(P
) O
F S
OL
UB
LE
GA
S
PG
Pi
A
B
D
E
Ci
CL
MO
LA
R C
ON
CE
NT
RA
TIO
N
OF
SO
LU
TE
IN
A L
IQU
ID
Figure 2 Concentration profile for absorbed component A
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The Two Film Theory
According to Ficks Law, the rate transfer by diffusion is proportional to theconcentration gradient and to the area of interface over which the diffusionis occurring.
The direction of transfer of material across the interface, is, however, notdependent on the concentration difference, but on the equilibriumrelationship.
The controlling factor the rate of diffusion through the two film.
From Figure 2:
PG - Partial pressure in the bulk of the gas phase
Pi - Partial pressure at interface
CL - Concentration in the bulk of the liquid phase
Ci - Concentration at interface
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The Two Film Theory
According to the two film theory:
The concentration at the interface are in equilibrium
The resistance to transfer is centred in the thin films on either side ofthe wall
Assumptions of two film theory:
Steady state concentration at any position do not change with time
Interface between the gas and liquid phase is a sharp boundary
Laminar film exist at the interface on both sides of the interface
Equation exist at the interface
No chemical reaction(rate of diff. across gas phase = rate of diff. across liquid phase)
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The Two Film Theory
GA
S F
ILM
BO
UN
DA
RY
LIQ
UID
FIL
M
BO
UN
DA
RY MAIN BULK
OF LIQUID
MAIN BULK OF GAS
GAS FILM
LIQUID FILM
PA
RT
IAL
PR
ES
SU
RE
(P
) O
F S
OL
UB
LE
GA
S
PG
Pi
A
B
D
E
Ci
CL
MO
LA
R C
ON
CE
NT
RA
TIO
N
OF
SO
LU
TE
IN
A L
IQU
ID
Figure 2 Concentration profile for absorbed component A
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Rate of Absorption
The process of absorption may be regarded as the diffusion of a soluble gasA into a liquid.
The molecules of A have to diffuse through a stagnant gas film and thenthrough a stagnant liquid film before entering the main bulk of liquid.
The rate of absorption of A per unit time over unit area is given:
tcoefficien transfer film Gas - GAAGA kPPkN 21'
The rate of diffusion in liquids is much slower than in gases, and mixturesof liquids may take a long time to reach equilibrium.
For the rate of absorption of A into liquid:
tcoefficien transfer film liquid - LAALA kCCkN 21'
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Rate of Absorption
In steady state process of absorption, the rate of transfer of materialthrough the gas film will be the same as that through the liquid film.
The general equation for mass transfer can be written as:
LiLiGGA CCkPPkN '
PG - Partial pressure in the bulk of the gas phase
Pi - Partial pressure at interface
CL - Concentration in the bulk of the liquid phase
Ci - Concentration at interface
NA - Overall rate of mass transfer (mol/unit area.time)
iG
Li
L
G
PP
CC
k
k
Therefore:
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Rate of Absorption
This condition can be shown graphically as below where ABF isequilibrium curve.
Figure 3 Driving forces in the gas and liquid
phase
Point D ( CL , PG ) represents conditions in bulk of gas and liquid
Point A ( Ce, PG ) represents concentration Ce in the liquid in equilibrium with PG in thegas.
Point B ( Ci, Pi ) represents concentration Ci in the liquid in equilibrium with Pi in thegas, and gives conditions at the interface.
Point F (CL, Pe ) represents the partial pressure Pe in the gas in equilibrium with CL in theliquid.
CL CeCi
PG
Pi
PB
A
B
D
F
E
PG
P
iCi CL
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Rate of Absorption
Then the driving force causing transfer in the gas phase:
DEPP iG
Then,
The driving force causing transfer in the liquid phase:
BECC Li
G
L
Li
iG
k
k
CC
PP
The concentration at the interface (point B) are found by drawing a linethrough D of slope kL/kG to cut the equilibrium curve in B.
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Rate of Absorption
Overall Coefficients
LeLeGGA CCKPPKN '
To obtain a direct measurement of the values kL and kG, we require themeasurement of the concentration at the interface.
Need to define two overall coefficient, KL and KG
tcoefficien phase liquid Overall -
tcoefficien phase gas Overall -
L
G
K
K
1Eq. LeLeGGLiLiGGA CCKPPKCCkPPkN'
The rate of transfer A can now be written as:
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Rate of Absorption
2Eq.
iG
ei
GiG
iG
GG
iG
eG
GG
eGGiGG
PP
PP
kPP
PP
kK
PP
PP
kK
PPKPPk
111
11
From Eq. (1):
Li
iG
LG
LiLiGG
CC
PP
kk
CCkPPk
11
Insert equation above into Eq. (2):
3Eq. 1
1
Li
ei
LGG
iG
ei
Li
iG
LGG
iG
ei
GiG
iG
GG
CC
PP
kkK
PP
PP
CC
PP
kkK
PP
PP
kPP
PP
kK
11
11
111
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Rate of Absorption
where term = average slope of equilibrium curve and,
Li
ei
CC
PP
if the solution obeys Henrys Law (H):
Li
ei
CC
PP
dC
dPH
Therefore Eq. (3) becomes:
LGG
Li
ei
LGG
k
H
kK
CC
PP
kkK
11
113Eq.
1
Similarly:
LGGLL K
H
KHkkK
1111 and
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Rate of Absorption
Factors Influencing the Mass Transfer coefficient
Very soluble gas (e.g NH3 in H2O).- resistance is so small where H negligible. So kG KG
Low solubility gas (e.g O2 in H2O)- resistance is in the liquid H is large. So kL KL
Moderately soluble gas- both film offer resistance, so kG KG and kL KL
Example:
Show that from rate of transfer of A:
GLL
LGG
HkkKb
k
H
kKa
111)
11)
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Evaluation of Mass Transfer Coefficient
Mass transfer coefficient Diffusivity (D) 1/film thickness (zg) for gas
g
g
Gz
Dk
Problem arises when measuring kG, kL since we do not know the value offilm thickness.
However, one piece of equipment where the surface area is known as the:Wetted Wall column
For gas:L
LL
z
Dk For liquid:
GAS
Slow flowing film of water- laminar flow condition
H
Glass tubeSometimes
turbulent flow could occur
D
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Evaluation of Mass Transfer Coefficient
HD
Method:
Surface area of film
*PPKN G Flux,
N unit in mol/(m2.s) P* = Equilibrium pressure
Measure gas concentration entering
Measure gas concentration leaving
Surface area of film is its area of the tube
Calculate the concentration driving force at inlet and outlet andtake log mean difference
*PPN
K AG
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Evaluation of Mass Transfer Coefficient
Example
The data given below were obtained from a wetted wall columnwith a constant liquid flow rate.
Molar Gas Flow rate, G (kmol/s) Overall Mass Transfer Coefficient, KG(kmol/s.m2 (kN/m2)
0.03
0.06
0.12
0.18
157.8
210.6
261.0
285.6
kG also related to the gas flow rate by:
constant is A where82.0GAkG
For molar gas flow rate of G = 0.1 kmol/s, evaluate the individualmass transfer coefficient (kG and kL) and overall mass transfercoefficient (KG) if H = 20 (kN/m2)/kmol
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Evaluation of Mass Transfer Coefficient
Solution
From Rate of Absorption:
LGG k
H
kK
11
Assuming kL constant and given,
constant is A where82.0GAkG
aEq.
LG k
H
GAK82.0
11
Equation (a) is straight line equation (y=mx+c) where:
y
x
GK1
82.01 G
m
c
A1
LkH
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Evaluation of Mass Transfer Coefficient
Solution
Construct graph:
y axis
x axis
GK1
82.01 G
G (kmol/s)
KG x106
(kmol/s.m2 (kN/m2))
0.03 157.8 17.732 6.337 x 10-3
0.06 210.6 10.044 4.748 x 10-3
0.12 261.0 5.689 3.831 x 10-3
0.18 285.6 4.080 3.501 x 10-3
82.01 GGK1
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Evaluation of Mass Transfer Coefficient
Solution
y = 2.079E-04x + 2.653E-03
0.0E+00
1.0E-03
2.0E-03
3.0E-03
4.0E-03
5.0E-03
6.0E-03
7.0E-03
0 2 4 6 8 10 12 14 16 18 20
1/G0.82
1/K
G x
10
-6
From the graph, straight line obtained: y = 2.079x10-4x + 2.653 x 10-3
m = = 2.079x10-4
c = = 2.653 x 10-3
A1
LkH
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Evaluation of Mass Transfer Coefficient
Solution
From equation obtained: y = 2.079x10-4x + 2.653 x 10-3
For molar gas flow rate of G = 0.1 kmol/s, evaluate the individualmass transfer coefficient (kG) and overall mass transfer coefficient (KG)if H = 20 (kN/m2)/kmol
G (kmol/s)
KG x106
(kmol/s.m2 (kN/m2)) x 10-6
0.10 248.3 6.607 4.027 x 10-3
82.01 G GK1
3.248
10027.41
10027.4
10653.2)607.6(10079.2
10653.210079.2
3
3
34
34
G
G
K
K
y
y
xy
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Evaluation of Mass Transfer Coefficient
Solution
From equation, slope:
481010079.2
1
10079.21
4
4
A
Am
0.728
10.0481082.0
82.0
G
G
G
k
k
GAk
Individual mass transfer coefficient for gas, kG:
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Evaluation of Mass Transfer Coefficient
Solution
From equation, y intercept
310653.2 Lk
Hc
Individual mass transfer coefficient for liquid, kL:
6.7538
10653.2 3
L
L
k
c
Hk
H
20
/kmolkN/m 20 For 2
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INTRODUCTION TO PACKED TOWER
PRESSURE DROP AND FLOODING IN PACKED TOWER DETERMINATION OF TOWER DIAMETER
DETERMINATION OF HEIGHT OF TOWER
Packed tower Description and Design
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Introduction to Packed Tower
Packed towers are used for continuous counter current in absorption.
The tower in Figure 4 consists of a cylindrical column containing:
A gas inlet and distributing space atthe bottom
A liquid inlet and distributing deviceat the top
A gas outlet at the top
A liquid outlet at the bottom
A packing filling in the tower.
A large intimate contact between theliquid and gas is provided by thepacking
Figure 4 Packed tower flows
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Introduction to Packed Tower
Common types of packing which are dumped at random in the tower areshown in Figure 5.
Packing are available in size of 3 mm to about 75 mm and mostly are madeof materials such as clay, porcelain, metal or plastic.
High void spaces of 65 95% are characteristics of good packings.
The packings permit relatively large volumes of liquid to passcountercurrent to the gas flow through the openings with relatively lowpressure drops for the gas.
Figure 5 Typical random or dumped tower packings
(a) Rashig Ring (b) Berl Saddle (c) Pall Ring (d) Intalox Metal (e) Jaeger Metal Tri - Pack
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Pressure Drop and Flooding in Packed Towers
In a given packed tower with a given type and size of packing and with adefinite flow of liquid, there is an upper limit to the rate of gas flow, calledthe flooding velocity.
The tower cannot be operated at gas flow velocity above flooding velocity.
At a low gas velocities, the liquid flows downward through the packing,essentially uninfluenced by the upward gas flow.
As the gas flow rate is increased at low gas velocities, the pressure drop isproportional to the flow rate to the 1.8 power.
At a gas flow rate called the loading point, the gas starts to hinder theliquid down flow, and local accumulations or pools of liquid start to appearin the packing (liquid holdup).
The pressure drop of the gas starts to rise at a faster rate.
As the flow rate of gas increased, the liquid holdup or accumulationincreases.
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Pressure Drop and Flooding in Packed Towers
At the flooding point, the liquid can no longer flow down through thepacking and is blown out with the gas.
In actual operating tower, the gas velocity is well below flooding velocity.
The optimum economic gas velocity is about one half or more of floodingvelocity.
Flooding velocity depends on:
Type of packing / packing factor
Size of packing
Liquid mass velocity
Limiting pressure drop,
Factor Packing :
flooding at drop Pressure : where
p
flood
pflood
F
P
FP
7.0115.0
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Pressure Drop and Flooding in Packed Towers
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Example 1
Ammonia is being absorbed in a tower using pure water at 25OCand 1 atm abs pressure.
The feed rate is 1440 lbm/h (653.2 kg/h) and contains 3.0 mol% ofammonia in air.
The process design specifies a liquid to gas mass flow rate ratioGL / GG of 2/1 and use 1-in. metal Pall rings.
Pressure Drop and Flooding in Packed Towers
Calculate the pressure drop in the packing and the gas mass velocityat flooding. Using 50% of the flooding velocity, calculate the pressuredrop, gas and liquid flows, and tower diameter.
Repeat (a) above by use Mellapak 250Y structured packing.
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Solution 1
Find the required data to be used in Figure 10.6-5 (Geankoplis pp.660)
Given mol fraction for ammonia = 0.03 Mwt ammonia = 17Given mol fraction for air = 0.97 Mwt air = 29
Average molecular weight of entering gas:
Pressure Drop and Flooding in Packed Towers
64282997017030 . x . x . x MwtyM iiav
333
307300101711
152980682
64281
lbm/ft. @ cmg.
K. mol.K
atmcm.
mol
g . atm
RT
PMg
-
Solution 1
From Appendix A.2-4, the water viscosity = 0.8937 cP. From A.2-3,the water density is 0.99708 g/cm3
1 Centistokes = 10-2 cm2/s
Pressure Drop and Flooding in Packed Towers
33
33
3
36
38561
59237453
1
1
30480
1
10990780 /ft lb.
g.
lbx
ft
m.x
m
cmx
cm
g. m
mL
es centistok.scm x .cm.s
g
cPx
g
cm
.cP x .
-
-
90201090201
10
990780
189370 22
23
From Table 10.6-1, for 1-in, Pall rings, Fp=56 ft-1. Using equation
10.6-1,
height packing ft / OH in. 2925156115011507070 ..F.P..
pflood
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Solution 1
x-axis for Figure 10.6-5:
Pressure Drop and Flooding in Packed Towers
From y axis:
0687108561
07300
1
25050
..
.
G
G..
L
G
G
L
For flow parameter of 0.06871 (abscissa) and pressure drop 1.925in/ft at flooding, a capacity parameter (ordinate) of 1.7 is read offthe plot.
sft6.6381
G
G
pGLGG
F
2561.0902.05607310.085.6107310.07.1
7.1
05.05.05.0
05.05.05.0
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Pressure Drop and Flooding in Packed Towers
-
Solution 1
at flooding
Pressure Drop and Flooding in Packed Towers
Using 50% of the flooding velocity for design,
sft
lbm.
s
ft.
ft
lbm.G GGG
234852063816073100
sft
lbm.
sft
lbm..GG
22242604852050
Liquid flow rate,sft
lbm
sft
lbm22
4852.02426.02LG
To calculate the pressure drop at 50% flooding,sft
lbm2
24260.GG
and . The new capacity parameter is 0.5 x 1.7 = 0.85.sft
lbm2
48520.GL
By using value of 0.85 and flow parameter of 0.06871 (abscissa), avalue of 18 in H2O/ft is obtained.
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Solution 1
Pressure Drop and Flooding in Packed Towers
The tower cross sectional area, At
2
m
2m ft
s 3600
hr 1x
lb 0.2426
sft
hr
1440lbrate Feed6488.1
G
tG
A
4
2
tt
DA
22 099.2142.3
6488.144ft
AD tt
ft `1.448 5.02099.2 ftDt
-
Example 2
Ammonia (NH3) is being removed from air by scrubbing withwater in a packed tower with 6 mm ceramic Berl Saddles (Cf = 900.
The gas entering at 1.2 m3/s contains 15% NH3.
The water enters at a rate of 4 kg/s and has a specific gravity of 1,viscosity of 2.5 x 10-3 kg/m.s.
The gas mixture enters at 27OC and 1 bar (0.987 atm). GivenMolecular weight for ammonia = 17, water = 18.
Calculate the diameter of the packed tower when 80% of NH3 isremoved and the pressure drop is 400 N/m2 per m packing.
Pressure Drop and Flooding in Packed Towers
-
Solution 2
Find the required data to be used.
Given mol fraction for ammonia = 0.15 Mwt ammonia = 17Given mol fraction for air = 0.85 Mwt air = 29
Average molecular weight of entering gas:
Pressure Drop and Flooding in Packed Towers
27OC
0.987 atm
80% NH3 is
removed
Lin = 4 kg/s
xA = 0
Lout = LLwater = 4 kg/s
xA = ?
Gin = 1.2 m3/s
yA = 0.15
Gout = ?
yA = ?
2272985017150 . x . x . x MwtyM iiav
-
Solution 2
Pressure Drop and Flooding in Packed Towers
33
3gcmg101.090
K 300.15 mol.K
atmcm82.06
mol
g 27.2 atm0.987
RT
PM
Given Gin = 1.2 m3/s
kg/s 1.308 @ g/s 1308m 1
cm10
s
m 1.2
cm
g101.090 G of Mass
3
363
3
3gin
V
kmol/s 0.04809 @ mol/s 48.0871
K 300.15 mol.K
atmcm82.06
s
cm 1.2x10 atm0.987
,G of Mol3
36
in
RT
PVn
kg/s 0.1226 kmol
kgx17
s
kmolx0.04809 0.15in G of Mass
3NH
-
Solution 2
Pressure Drop and Flooding in Packed Towers
kg/s 0.0981 kg/s 0.1226 x 0.80 waterin absorbed NH of Mass 3
kg/s 1.2099 0.0981 1.308 absorbed NH of Mass in gas of MassG of Mass 3out
kg 4.09814 0.0981 absorbed NH of Mass waterof MassL of Mass 3out
From information, the water viscosity = 2.5 x 10-3 kg/m.s. Waterdensity is 1000 kg/m3. Since the larger flow quantities are at the bottomof absorber, the diameter will be chosen to accommodate the bottomcondition. From Treybal:
abscissa (x-coordinate) = 50
50
.
GL
.
G
G'
L'
(Refer to bottom condition)
10.0
0.53-
0.53-
10 x 1.0901s
kg1.308
10 x 1.090s
kg4.0981
-
Pressure Drop and Flooding in Packed Towers
CG
LG
Lf
gJC
G
1.0
2'
-
Solution 2
Pressure Drop and Flooding in Packed Towers
At pressure drop of and x-coordinate = 0.10,
y-coordinate = 0.066
m
mN400
2
0660
102
.g
JCG'
cGLG
.
Lf
sm
kg
0.145g
2
cGLG
38.0
1105.2900
109.1100009.1066.0066.0
''
1.031.0
2''
G
JCG
Lf
Cross sectional area, 2
2
4423
380
3081
m.
.sm
kg.
s
kg .
G''
G'A
m . D
.
A D
DA 092
442344
4
2
-
Exercise
A packed tower is to be designed for a counter current contact of an NH3- air mixture with water to wash out NH3 from the gas.
The conditions are:
Gas in: Gas out: All NH3 is removed.
Flow rate = 1.5 m3/s
Temperature = 27 C
Pressure = 1 bar (0.987 atm)
Contains 8 mol % NH3
Liquid in:
Flow rate = 4.8 kg/s
Density = 996 kg/m3
Viscosity = 2.5 x 10-3 kg/m.s
Packing used is 38mm Raschig ring (Cf = 95, gc = 1)
(a) Calculate the flow rate of liquid out.
(b) If the pressure drop of the packed tower is 400 N/m2, by using the diagram, calculate the required diameter for the tower.
-
Component denoted in Figure 6:
Determination of Height of Tower
Gm = Mols of inert gas / (unit time) (unit cross section of tower)
Lm = Mols of inert liquid / (unit time) (unit cross section of tower)
Y = Mols of soluble gas A / mol of inert gas Bin gas phase
X = Mols of soute A / mol of inert solvent C inliquid phase
dZ
x
x + dx
y
y + dy
y2
Gm
x2
Lm
y1
Gm
x1
Lm
Figure 6 Countercurrent absorption towerMass balance over differential / smallsection of column:
tower of section
small of Vol. ere wh
AdzCCakAdZN
AdzAdzPPakAdZN
LiLA
iGGA
'
' NA = mol / (m2.s)
a = Interfacial area / Volume ofcolumn
-
Determination of Height of Tower
From the tower,
AdyGdZAPPak
AdyGAdZN
miGG
mA
'
From Dalton and Raoults Law,
iTi
T
ii
GT
T
G
PPyP
Py
PyPP
Py
and
Therefore,
iTG
m
iTGm
yy
dy
aPk
GdZ
dZyyaPkdyG
Integrating both sides; get:
1
2
y
y iTG
m
yy
dy
aPk
GZ Z = height of tower
-
Determination of Height of Tower
Similarly for concentration in liquid phase:
xx
dx
ak
LZ
iL
m
Normally, Z is written in terms of KG.a & KL.a & in terms of mol fraction:
OGOG NHZ
Z = height of tower
HOG = height of transfer unit constant.
NOG = number of transfer unit constant
*yy
dy
aPK
GZ
TG
m
-
Determination of Height of Tower
And for concentration in liquid phase:
Simplifying:
xx
dx
aCK
LZ
TL
m
*
OGOG NHZ
OLOL NHZ
For gas phase based on equilibrium concentration
For liquid phase based on equilibrium concentration
The number of transfer unit (NOG) can be calculated using several methodwhich will be discussed later.
-
Graphical Method
Log Mean Driving Force Method
Colburns Method
Methods for Evaluation of NOG
-
Based on interface concentration:
Based on Area, A:
Graphical Methods
LLGG NHNHZ
xx
yy
G
Lxx
G
Lyy
xxLyyG
m
m
m
m
mm
1
111
11
Operating line equation relates concentration in gas phase to that in liquidphase. From operating line above:
-
Graphical Methods
y
x
Equilibrium line normally linear at dilute concentration
Operating line
axis- y withonintersecti
slope @ gradient :where
line operating
11
11
11
yxG
L
G
L
yxG
Lx
G
Ly
yxxG
Ly
-
Example
Graphical Methods
An acetone air mixture containing 0.015 mol fraction of acetone willbe reduced to 1% of this value by countercurrent absorption in freshpure water in packed tower.
The gas flow rate is 1.0 kg/m2.s and the liquid flow rate is 1.6 kg/m2.s.
For the system, Henrys law holds and y* = 1.75 x where y* is the molfraction of acetone in the vapor in equilibrium with x mol fraction inliquid.
Calculate the height of the tower / absorber if HOG = 0.3 m.
Data given:
Cross sectional A of column = 1 m2
Mwt air = 29Mwt acetone = 58Mwt H2O = 18
-
Solution
Graphical Methods
Given:
Cross sectional Aof column = 1 m2
Mwt air = 29Mwt acetone = 58Mwt H2O = 18HOG = 0.3 my* = 1.75 x
Reduced to 1%
1
2
Gm=1.0 kg/m2.s
Gm=1.0 kg/m2.s
Lm=1.6 kg/m2.s
Lm=1.6 kg/m2.s
y2 = 0.00015
y1 = 0.015 x1 = ?
x2 = 0
-
Solution
Graphical Methods
The given mass flow rate in kg/m2.s. Need to convert into kmol /s.
s
kmolm
kmol
kgsm
kg.
G
AreaM
itin mass un GG
m
wt
mm
03448.0129
012
2
s
kmolm
kmol
kgsm
kg.
L
AreaM
itin mass un LL
m
wt
mm
08889.0118
612
2
slope
line operating
578.2/03448.0
/08889.0
11
skmol
skmol
G
L
yxG
Lx
G
Ly
Find slope of the graph.
-
Solution
Graphical Methods
Calculate the mol ratio for each location.
?
0152.0015.01
015.0
1
1
m
m
L
Ax
G
Ay
Location 1
0
01
0
00015.000015.01
00015.0
2
2
m
m
L
Ax
G
Ay
Location 2
Need to find x1 in order to plot operating line.
Cxy
yxG
Lx
G
Ly
578.2
11
-
Solution
Graphical Methods
Use coordinate at location 2 in order to find C value.
00015.0
0578.200015.0
578.2
C
C
Cxy
Operating line, 00015.0578.2 xy
Insert value at location 1 in order to find x1 value
00576.0
578.2
00015.0015.0
00015.0578.2015.0
00015.0578.2
1
1
11
x
x
xy
Construct operating line by plotting coordinate at location 1 and 2
Location 2
00015.0,0, 22 yx
Location 1
015.0,00576.0, 11 yx
-
Graphical Methods
No. of stages = 11 NOG = 11
Equilibrium line
Operating line
-
Solution
Graphical Methods
Height of absorber, Z
mZ
mZ
NHZ OGOG
3.3
113.0
-
To evaluate NOG, we can take an average driving force in log mean.
Log Mean Driving Force Method
xx
dx
yy
dy
** or
2
*
21
*
1
*
21
2
*
1
*
2
*
1
*
21
ln
ln
HxyHxy
yy
yyN
yy
yy
yyyy
yyN
OG
OG
y - line
y* - line
y1
(y - y*)
(y - y*)1
(y - y*)2
y2
-
Example
An acetone air mixture containing 0.015 mol fraction of acetone willbe reduced to 1% of this value by countercurrent absorption in freshpure water in packed tower.
The gas flow rate is 1.0 kg/m2.s and the liquid flow rate is 1.6 kg/m2.s.
For the system, Henrys law holds and y* = 1.75 x where y* is the molfraction of acetone in the vapor in equilibrium with x mol fraction inliquid.
Calculate the height of the tower / absorber if HOG = 0.3 m.
Data given:
Cross sectional A of column = 1 m2
Mwt air = 29Mwt acetone = 58Mwt H2O = 18
Log Mean Driving Force Method
-
Solution
Given:
Cross sectional Aof column = 1 m2
Mwt air = 29Mwt acetone = 58Mwt H2O = 18HOG = 0.3 my* = 1.75 x
Reduced to 1%
1
2
Gm=1.0 kg/m2.s
Gm=1.0 kg/m2.s
Lm=1.6 kg/m2.s
Lm=1.6 kg/m2.s
y2 = 0.00015
y1 = 0.015 x1 = ?
x2 = 0
Log Mean Driving Force Method
-
Solution
The given mass flow rate in kg/m2.s. Need to convert into kmol /s.
s
kmolm
kmol
kgsm
kg.
G
AreaM
itin mass un GG
m
wt
mm
03448.0129
012
2
s
kmolm
kmol
kgsm
kg.
L
AreaM
itin mass un LL
m
wt
mm
08889.0118
612
2
slope
line operating
578.2/03448.0
/08889.0
11
skmol
skmol
G
L
yxG
Lx
G
Ly
Find slope of the graph.
Log Mean Driving Force Method
-
Solution
Calculate the mol ratio for each location.
?
0152.0015.01
015.0
1
1
m
m
L
Ax
G
Ay
Location 1
0
01
0
00015.000015.01
00015.0
2
2
m
m
L
Ax
G
Ay
Location 2
Need to find x2 in order to plot operating line.
Cxy
yxG
Lx
G
Ly
578.2
11
Log Mean Driving Force Method
-
Solution
Use coordinate at location 2 in order to find C value.
00015.0
0578.200015.0
578.2
C
C
Cxy
Operating line, 00015.0578.2 xy
Insert value at location 1 in order to find x1 value
00576.0
578.2
00015.0015.0
00015.0578.2015.0
00015.0578.2
1
1
11
x
x
xy
Log Mean Driving Force Method
-
Solution
Log Mean Driving Force Method
Need to find y1* and y2* using equilibrium line given in question.
0
075.175.1
010.0
00576.075.175.1
75.1
*
2
2
*
2
*
1
1
*
1
*
1
*
1
y
xy
y
xy
xyHxy
-
Solution
Log Mean Driving Force Method
Calculate NOG.
*
21
ln yy
yyNOG
33
3*
2
*
1
*
2
*
1
**
104242.1
00015.0
102.5ln
1005.5ln
000015.0
010.00152.0ln
000015.0010.00152.0
ln
ln
yy
yy
yy
yyyyyy
-
Solution
Log Mean Driving Force Method
Calculate NOG.
stages 1157.10
104242.1
00015.00152.0
ln 3*21
yy
yyNOG
Height of absorber, Z
mZ
mZ
NHZ OGOG
3.3
113.0
-
Objective to evaluate:
Assume mol fraction are so small fraction; mol fraction = mol ratio
Colburns Method
2121 xxLyyG mm
Also, y*=Hx, but instead of H, use m. So y*=mx
If we substitute, we get
1
2
*
y
yyy
dy
1
22
y
y
m
m yyL
mGy
dy
2* yy
L
mGy
m
m
Solution as log function and represented as a nomogram (graphical solution)
-
Example
An acetone air mixture containing 0.015 mol fraction of acetone willbe reduced to 1% of this value by countercurrent absorption in freshpure water in packed tower.
The gas flow rate is 1.0 kg/m2.s and the liquid flow rate is 1.6 kg/m2.s.
For the system, Henrys law holds and y* = 1.75 x where y* is the molfraction of acetone in the vapor in equilibrium with x mol fraction inliquid.
Calculate the height of the tower / absorber if HOG = 0.3 m.
Data given:
Cross sectional A of column = 1 m2
Mwt air = 29Mwt acetone = 58Mwt H2O = 18
Colburns Method
-
Solution
Given:
Cross sectional Aof column = 1 m2
Mwt air = 29Mwt acetone = 58Mwt H2O = 18HOG = 0.3 my* = 1.75 x
Reduced to 1%
1
2
Gm=1.0 kg/m2.s
Gm=1.0 kg/m2.s
Lm=1.6 kg/m2.s
Lm=1.6 kg/m2.s
y2 = 0.00015
y1 = 0.015 x1 = ?
x2 = 0
Colburns Method
-
Solution
The given mass flow rate in kg/m2.s. Need to convert into kmol /s.
s
kmolm
kmol
kgsm
kg.
G
AreaM
itin mass un GG
m
wt
mm
03448.0129
012
2
s
kmolm
kmol
kgsm
kg.
L
AreaM
itin mass un LL
m
wt
mm
08889.0118
612
2
10000015.0
015.0
6788.008889.0
03448.075.1
2
1
y
y
L
Gm
m
m
Find mGm/Lm and y1/y2
Colburns Method
-
Colburns Method
-
Solution
From nomogram, NOG = 10.6 = 11
Colburns Method
Height of absorber, Z
mZ
mZ
NHZ OGOG
3.3
113.0
-
INTRODUCTION TO PLATE TOWER
DETERMINATION STAGES OF TOWER
Plate tower Description and Design
MINIMUM LIQUID FLOW RATE TO OBTAIN SPECIFIC SEPARATION
-
Introduction to Plate tower
Bubble cap columns or sieve trays, are sometimes used for gas absorption.
Application of plate tower particularly when the load is more than can behandled in a packed tower about 1 m diameter; and
- when there is any likelihood of deposition of solids which would quicklychoke a packing.
Plate tower are particularly useful when the liquid rate is sufficient to flooda packed tower.
-
Determination stages of tower
It will be assumed that dilute solutions are used so that mole fraction andmole ratio approximately equal.
A material balance for the absorbed component from the bottom to a planeabove plate n will give:
line Operating
s
m
msn
m
mn
nmsmsmnm
xG
Lyx
G
Ly
xLyGxLyG
11
11
Operating line also describes such a line passes through two point, 1 (top oftower) and 2 (bottom of tower)
-
Determination stages of tower
Example 1
A bubble cap absorption column is to be used to absorb ammonia,NH3 by using water.
A gaseous mixture containing 20.5 mol% NH3 and 79.5 mol% air entersthe bottom of the absorption tower.
60.5 kmol of gaseous NH3 enters the tower per hour while 5500 kgaqueous NH3 solution containing 0.1% by mass NH3 enters the top ofthe tower per hour.
The column operates at the atmospheric pressure and at a constanttemperature of 30OC.
It is desired to absorb 95% of the entering gas NH3. Assume that theeffect of water vapor in the gases is negligible.
-
Determination stages of tower
Example
Determine:
The equilibrium data given as follows:
The molar flow rate of entering gaseous mixture
The molar flow rate of the raffinate
The molar flow rate of the extract
The mol ratio of NH3 in the raffinate and extract
The number of ideal stages required
Mol NH3mol water
0.000 0.053 0.111 0.177 0.250
Mol NH3mol air
0.000 0.044 0.089 0.159 0.280
-
Determination stages of tower
Solution
95% removal T=30OCP=1 atm
1
2
Gm
Gm
Lm
Lm
y2 = 0.00015
yA1 = 0.205 kmol A/kmol
x1 = ?
xA2 = 0.001 kg A/kg
yB1 = 0.795 kmol B/kmol
Feed of NH3 in= 60.5 kmol/h
Feed of NH3 aq in= 5500 kg/h
xB2 = 0.999 kg B/kg
A = NH3B = AirC = H2O
-
Determination stages of tower
Solution
Determine:
The molar flow rate of entering gaseous mixture, G1
From Gaseous mixture inlet (at point 1),Feed NH3 in = 60.5 kmol/h
h
A kmol
A kmol
kmol
h
A kmol
h
A kmol
kmol
A kmol
12.295
205.0
15.60
5.60205.0
1
1
1
G
G
G
-
Determination stages of tower
Solution
Determine:
The molar flow rate of the raffinate, G2
From information, 95% of the entering NH3 is being absorbed.
h
kmol
h
kmol 025.35.6005.0 Unabsorbed NH3 =
The molar flow rate of the raffinate, G2 = Gm + unabsorbed NH3
h
kmol
h
kmol NH unabsorbed
h
Bkmol
h
kmol
kmol
Bkmol
kmol
Bkmol
3
645.237
025.362.234
62.23412.295795.0795.0
2
2
1
G
GG
GG
m
m
-
Determination stages of tower
Solution
Determine:
The molar flow rate of the extract, L1
h
kmol
h
kmol 475.575.6095.0 Absorbed NH3 =
From information at L2, feed NH3 aq. in = 5500 kg/h
h
A kg
h
kg
kg
A kg
kg
A kg
h
C kg
h
kg
kg
C kg
kg
C kg
5.55500001.0001.0
5.54945500999.0999.0
22
2
LL
LL
A
m
Convert Lm and LA2 in kmol/h unit. Given:
Mwt NH3 = 17 Mwt H2O = 18
-
Determination stages of tower
Solution
h
A kmol
A kg
A kmol
h
A kg
h
C kmol
C kg
C kmol
h
C kg
324.017
15.5
25.30518
15.5494
2
A
m
L
L
The molar flow rate of the extract, L1
h
kmolh
kmol5
NH Absorbed 3
049.363
475.7324.025.305
1
1
21
L
L
LLL Am
-
Determination stages of tower
Solution
Determine: The mol ratio of NH3 in all stream
Location 1
Gas Feed stream, G1
Mol of NH3 = 60.5 kmol/h
Mol of Air, Gm = 234.62 kmol/h
258.0'1
'
1
'
1
Y
Y
Y
kmol 234.62
kmol 60.5
Air of mol
NH of mol 3
Extract stream, L1
Mol of NH3 = 0.324 + 57.475 kmol/h= 57.799 kmol/h
Mol of Water, Lm = 305.25 kmol/h
189.0'1
'
1
'
1
X
X
X
kmol 305.25
kmol 57.799
waterof mol
NH of mol 3
Coordinate for Location 1 = (X1, Y1) = (0.189,0.258)
-
Determination stages of tower
Solution
Location 2
Raffinate stream, G2
Mol of NH3 = 3.025 kmol/h
Mol of Air, Gm = 234.62 kmol/h
013.0'2
'
2
'
2
Y
Y
Y
kmol 234.62
kmol 3.025
Air of mol
NH of mol 3
Liquid feed stream, L2
Mol of NH3 = 0.324 kmol/h
Mol of Water, Lm = 305.25 kmol/h
001.0'2
'
2
'
2
X
X
X
kmol 305.25
kmol 0.324
waterof mol
NH of mol 3
Coordinate for Location 2 = (X2, Y2) = (0.001,0.013)
-
Determination stages of tower
Solution
Plot the equilibrium line and the operating line in order tocalculate number of stages
No of theoretical stages = 5 theoretical stages
-
Determination stages of tower
0
0.025
0.05
0.075
0.1
0.125
0.15
0.175
0.2
0.225
0.25
0.275
0.3
0 0.025 0.05 0.075 0.1 0.125 0.15 0.175 0.2 0.225 0.25 0.275 0.3
x
y
-
A bubble cap absorption column is to be used to absorb ammonia,NH3 by using water.
A gaseous mixture containing 15 mol% NH3 and 85 mol% air entersthe bottom of the absorption tower.
57 mol of gaseous NH3 enters the tower per hour, while 3.7 kg of purewater enters the top of the tower per hour.
The column operates at the atmospheric pressure and at a constanttemperature of 30OC.
It is desired to absorb NH3 such as only 2.5% NH3 leaves the towerwith air. Determine the number of theoretical plates required for theabove process.
Equilibrium data is given as in Example 1. Assume that the effect ofwater vapor in the gases is negligible.
Example 2
Determination stages of tower
-
Minimum Liquid Flow rate
Min. flow rate = Infinite number of stages = Minimum reflux in distillation.
y
x
as L decreases, line approaching equilibrium line
equilibrium line
Minimum value of flow rate is where:
Exit liquid [ ] = Equilibrium [ ]or
Inlet gas [ ] = Equilibrium [ ]