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GAS ABSORPTION
MASS TRANSFER
OPERATIONS -І
WHAT IS GAS ABSORPTION????
 Absorption, or gas absorption, is a unit operation used in the chemical
industry to separate gases by washing or scrubbing a gas mixture with a
suitable liquid .
 The fundamental physical principles underlying the process of gas
absorption are the solubility of the absorbed gas and the rate of mass
transfer. One or more of the constituents of the gas mixture dissolves or
is absorbed in the liquid and can thus be removed from the mixture. In
some systems, this gaseous constituent forms a physical solution with the
liquid or the solvent, and in other cases , it reacts with the liquid
chemically.
 The purpose of such scrubbing operations may be any of the
following :
Gas purification (eg , removal of air pollutants from exhausts
gases or contaminants from gases that will be further processed) ,
Product Recovery , or production of solutions of gases for various
purposes.
 The absorber may be a packed column , plate column , spray
column , venturi scrubbers , bubble column , falling films , wet
scrubbers ,stirred tanks
 Gas absorption is usually carried out in vertical
counter current columns.
 The solvent is fed at the top of the absorber ,
whereas the gas mixture enters from the bottom
.The absorbed substence is washed out by the
solvent and leaves the absorber at the bottom as
a liquid solution .
 The solvent is often recovered in a subsequent
stripping or desorption operation . This second
step is essentially the reverse of absorption and
involves counter current contacting of the liquid
loaded with solute using and inert gas or water
vapor .
Choice Of Solvent for Absorption
 If the principal purpose of the absorption operation is to
produce a specific solution, as in the manufacture of
hydrochloric acid, for example, the solvent is specified by
the nature of the product, i.e. water is to be the solvent. If
the principal purpose is to remove some components (e.g.
impurities) from the gas, some choice is frequently possible.
 The factors to be considered are:
 GAS SOLUBILITY :
The gas solubility should be high, thus increasing the rate of
absorption and decreasing the quantity of solvent required.
Solvent with a chemical nature similar to the solute to be
absorbed will provide good solubility.
 VOLATALITY :
The solvent should have a low vapour pressure to reduce loss
of solvent in the gas leaving an absorption column.
 CORROSIVENESS :
The materials of construction required for the equipment
should not be unusual or expensive
 COST :
The materials of construction required for the
equipment should not be unusual or expensive.
 VISCOSITY :
Low viscosity is preferred for reasons of rapid
absorption rates, improved flooding characteristics in packed
column, low pressure drops on pumping, and good heat
transfer characteristics.
 The solvent should be non-toxic, non-flammable and
chemically stable.
ONE COMPONENT TRANSFERRED
MATERIAL BALANCES
COUNTERCURRENT FLOW
• Countercurrent tower which may be either a packed
or spray tower, filled with bubble cape trays or of
any internal construction to bring about liquid gas
contact.
• Mole ratio may be define as
Y = y/(1-y)
X = x/(1-x)
• Here y for gas stream and x for liquid stream.
Material balance
Gs = G(1-y) = G/(1+Y)
Ls = L(1-x) = L/(1+X)
NOW
Gs(Y1 - Y) = Ls(Xs - X)
AND
Gs(y1/(1-y1) – y/(1-y)) = Gs(p1/(1-p1) – p/(1-p)) = Gs(x1/(1-x1) –
x/(1-yx))
MINIMUM LIQUID-GAS RATIO
• In the design of absorbers, the quantity of gas to be treated G or Gs,
the terminal concentrations Y1 and Y2,and the composition of the
entering liquid X2 are ordinarily fixed by process requirements.
• The operating line must be pass through point D and must end at the
ordinate Y.
• Liquid gas ratio is define by
L/G
• Graph of minimum liquid-gas ratio can be shown as behind slide.
CO-CURRENT FLOW
• When gas and liquid flow cocurrently, the operating line
has a negative slope.
-L/G
• There is no limit of this ratio, but an infinitely tall tower
would produce an exit liquid and gas in equilibrium
• In cocurrent fow requirred height of tower should be
higher then in counter current flow.
Counter-current multi-stage absorption
(Tray absorber):
 In tray absorption tower, multi-stage contact between gas and liquid takes
place.
 In each tray, the liquid is brought into intimate contact of gas and
equilibrium is reached thus making an ideal stage.
 In ideal stage, average composition of liquid leaving the tray is in
equilibrium with liquid leaving that tray.
 The most important step in design of tray absorber is the
determination of number of trays.
 The schematic of tray tower is presented in figure.
 The liquid enters from top of the column whereas gas is added from
the bottom.
 The efficiency of the stages can be calculated as:
𝑆𝑡𝑎𝑔𝑒 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑑𝑒𝑎𝑙 𝑠𝑡𝑎
𝑔𝑒𝑠/𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑎𝑙 𝑠𝑡𝑎𝑔𝑒𝑠
…..Eq.(1)
Gs,Y1
Gs, YN+1
Liquid out
Ls, XN
Gas in
Gas outGs, Y1
Liquid in
Ls, X0
Ls, X1
Gs,YN
Ls, X1
N
N-1
1
2
 The following parameters should be known for the
determination of “number of stages”:
(1) Gas feed rate
(2) Concentration of gas at inlet and outlet of
the tower
(3) Minimum liquid rate; actual liquid rate is
1.2 to 2 times the minimum liquid rate.
(4) Equilibrium data for construction of
equilibrium curve
 Now, the number of theoretic stages can be obtained
graphically:
(A) Graphical Method for the Determination of Number of Ideal
Stages:
 Overall material balance on tray tower:
Gs(YN+1 -Y1) = Ls(XN -X0)
This is the operating line for tray tower
 If the stage (plate) is ideal, (Xn, Yn) must lie on the equilibrium line, Y*=f(X).
 Top plate is located at P(X0, Y1) and bottom plate is marked as Q(XN,
YN+1) in X-Y plane.
 A vertical line is drawn from Q point to D point in equilibrium line at (XN, YN).
 From point D in equilibrium line, a horizontal line is extended up to operating
line at E (XN-1, YN).
 The region QDE stands for N-th plate. We may get fraction of plates.
 In that situation, the next whole number will be the actual number of ideal
plates.
 If the overall stage efficiency is known, the number of real plates can be
obtained from Equation A.
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
E ( X
N-1
,Y
N
)
X
P ( X
0
,Y
1
)
Q ( X
N
,Y
N+1
)
Equilibrium line
Operating line
N
N-1
N-2
D ( X
N
,Y
N
)
Example:
It is desired to absorb 95% of acetone by water from a mixture of
acetone and nitrogen containing 1.5% of the component in a
countercurrent tray tower. Total gas input is 30 kmol/hr and water enters
the tower at a rate of 90 kmol/hr. The tower operates at 27ºC and 1 atm.
The equilibrium relation is Y=2.53X. Determine the number of ideal
stages necessary for the separation using graphical method.
Solution:
 Basis: 1 hour
 G(N+1)=30 kmol
 Y(N+1)=0.015
 Lo=90 kmol
 Moles acetone in = 30×0.015 moles=0.45 moles
 Moles nitrogen in = (30-0.45) moles=29.55 moles
 Moles acetone leaving (95% absorbed) = 0.45×(1-0.95) moles=0.0225
moles
 Gs=29.55 moles
 Ls=90 moles
 α=2.53 [as, Y=2.53X]
 𝑌1 = 0.0225/29.55 = 7.61 × 10^−4
 𝑌(𝑁+1)= 0.015
 Equation.. 𝐺𝑠𝑌(𝑁+1)− 𝑌1= 𝐿𝑠(𝑋𝑁− 𝑋0)
29.55 × 0.015 − 7.61 × 10^−4= 90(𝑋𝑁− 0)
 XN=4.68×10^-3
 Solution by graphical method,Construction of operating line PQ:
P(X0,Y1)=P(0, 7.61×10^-4)
Q(XN, YN+1)=Q(4.68×10^-3, 0.015)
Construction of equilibrium line (Y=2.53X):
X 0 0.001 0.002 0.003 0.004 0.005
Y 0 0.00253 0.00506 0.00759 0.01012 0.01265
From graphical construction in Figure , the number of triangles obtained is
more than 7. Hence number of ideal stages is 8.
23
 Rate of absorption
•Volumetric mass transfer coefficients (Kya, etc.) are used for most
calculations, because it is more difficult to determine the coefficients per
unit area and because the purpose of the design calculation is generally to
determine the total absorber volume.
•Kya=overall volumetric mass-transfer coefficient, kmol/(m3·h·unit mole
fraction).
•a=effective area of interface per unit packed volume, m2/m3
24
•Simplicity Treatment
•The following treatment applies to lean gases (up to 10% solute):
•(a) Correction factors for one-way diffusion are omitted for simplicity.
•(b) Changes in gas and liquid flow rates (V and L) are neglected.
•(c) kxa, kya, Kya, Kxa can be considered as constants.
25
•the rate of mass transfer:
r=NA [kgmol/(m2·h·unit mole fraction)]
)()(
)(
AAixAiAyA xxkyykN
r


)(
)(



AAyA yyKN
r
)(
)(
AAxA xxKN
r



26
•Let r =rate of absorption per unit volume, kgmol/(m3·h)
)(
)(
)(
)(
xxaKr
yyaKr
xxakr
yyakr
x
y
ix
iy






•It is hard to measure or to predict a, but in most cases it is not necessary
to know its actual value since design calculations can be based on the
volumetric coefficients.
27
•Determining the interface composition (yi, xi)
•(yi, xi) is also hard to measure, but it can be obtained from the
operating-line diagram
ak
ak
xx
yy
y
x
i
i



)(
)(
•Thus a line drawn from the operating line with a slope –kxa/kya will
intersect the equilibrium line at (yi, xi).
28
Operating line
Equilibrium curve

y
ak
ak
Slope
y
x


y
x
y
x
y
ak
ak
Slope
i
i
y
x

x
y
ak
ak
Slope
y
x


y
x
y
x
y
ak
ak
Slope
i
i
y
x


y
x
y
x
y
ak
ak
Slope
i
i
y
x



y
x
y
x
y
ak
ak
Slope
i
i
y
x



x
y
x
y
x
y
ak
ak
Slope
i
i
y
x
ak
ak
xx
yy
y
x
i
i



)(
)(
)(
)(
)(
)(
:
xx
yy
xx
yy
forceDriving
i
i






29
xyy
AAAiAi
AAixAiAyy
k
m
kK
mxymxy
xxkyykK




11
,,
)()(

•Determining the overall coefficients: Using the local slope of the
equilibrium curve m, we have
Therefore,
ak
m
akaK
mxymxy
xxk
yy
yyk
yy
K
xyy
AAAiAi
AAix
AAi
AiAy
AiA
y










11
,,
)56.17(
)()(
1

amkakaK
mxymxy
xxk
yy
yyk
yy
K
yxx
AAAiAi
AAix
AAi
AiAy
AiA
y
111
,,
)()(
1










Similarly,
30
=overall resistance to mass transfer
= resistance to mass transfer in the gas film
=resistance to mass transfer in the liquid film
ak
m
ak
aK
x
y
y
1
1
ak
m
ak
aK
x
y
y
1
1
ak
m
ak
aK
x
y
y
1
1
ak
m
akaK
mxymxy
xyy
AAAiAi


11
,,
31
•Gas film “controls” and Liquid film “controls”
When ""
11
,
1
controlsfilmGas
akaKak
m
ak yyxy

ak
m
akaK
mxymxy
xyy
AAAiAi
AAixAiAyy

 
11
,,
•Or, When the coefficients kya and kxa are of the same order of
magnitude, and m is very much greater than 1.0, the liquid film
resistance is said to be controlling. That is,
""
1
,
1
controlsfilmLiquid
ak
m
aKak
m
ak xyxy

32
•Liquid film controlling means that any change in kxa has a nearly
proportional effect on both Kxa and Kya and on the rate of absorption,
whereas a change in kya has little effect.
•Examples of Liquid film controls: Absorption of CO2, H2,O2,Cl2 in
water;
•When the solubility of the gas is very high, m is very small and the gas-
film resistance controls the rate of absorption.
•Examples of gas-film controls: Absorption of HCl, NH3 in water; NH3
in acid solution; SO2, H2S in basic solvent.
33
•With gases of intermediate solubility both resistances are important,
but the term controlling resistance is sometimes used for the larger
resistance.
•The absorption of NH3 in water is often cited as an example of gas-film
control, since the gas film has about 80 to 90 percent of the total
resistance.
34
b
b
T
a
a
xL
yV
Z
x
y
dZ
Z
yV
xL
,
,
,
,
b
b
T
a
a
xL
yV
Z
x
y
dZ
Z
yV
xL
,
,
,
,
b
b
T
a
a
xL
yV
Z
x
y
dZ
Z
yV
xL
,
,
,
,
b
b
T
a
a
xL
yV
Z
x
y
dZ
Z
yV
xL
,
,
,
,
b
b
T
a
a
xL
yV
Z
x
y
dZ
Z
yV
xL
,
,
,
,
b
b
T
a
a
xL
yV
Z
x
y
dZ
Z
yV
xL
,
,
,
,
b
b
T
a
xL
yV
Z
x
y
dZ
Z
yV
,
,
,
b
b
T
a
a
xL
yV
Z
x
y
dZ
Z
yV
xL
,
,
,
,
b
b
T
a
a
xL
yV
Z
x
y
dZ
Z
yV
xL
,
,
,
,
b
T
a
a
xL
dyy
Z
x
y
dZ
Z
yV
,
,

dxx
yV
Z
x
y
dZ
Z
yV
b
T
a

,
,
•Fig. Diagram of packed
absorption tower
S=cross sectional area;
SdZ=differential volume in height dZ.
•The amount absorbed in section dZ
is Vdy.
 





b
a
T y
yy
T
Z
y
yy
dy
aSK
V
ZdZ
SdZyyaKVdy
rSdZVdy
)(
)(
0
35
•Number of transfer units
 


b
a
y
yy
T
yy
dy
aK
SV
Z
)(
/
•The equation for column height can be written as follows:
 


b
a
y
y
oy
yy
dy
N
)(
=overall number of transfer units [NTU],
based on gas phase.
aK
SV
H
y
oy
/
 =overall height of a transfer unit [HTU], based
on gas phase.
36
oyoyT NHZ 
(1)If the operating line and equilibrium line are
straight and parallel,
Equilibrium
line
Operating line
a
a
b
x
x
yy
y
y
y
 1
2
3
b
a
a
b
xx
x
x
x
yy
y
y
y


3
2
1
1
2
3
a
a
b
x
yy
y
y
y
 1
2
3
a
a
b
x
x
x
yy
y
y
y

1
1
2
3
a
b
x
yy
y
y
y

1
2
3
b
a
b
b
xx
x
x
x
yy
y
y
y


3
2
1
1
2
3














 
bb
ab
oy
aa
ab
oy
ab
oy
y
y
oy
yy
yy
N
yy
yy
N
yy
yy
N
yy
dy
N
b
a
.18(
)(














 
bb
ab
oy
aa
ab
oy
ab
oy
y
y
oy
yy
yy
N
yy
yy
N
yy
yy
N
yy
dy
N
b
a
1(
)(














 
bb
ab
oy
aa
ab
oy
ab
oy
y
y
oy
yy
yy
N
yy
yy
N
yy
yy
N
yy
dy
N
b
a
)(
37















 
bb
ab
oy
aa
ab
oy
ab
oy
y
y
oy
yy
yy
N
yy
yy
N
NTP
yy
yy
N
yy
dy
N
b
a
)(
•If the operating line and equilibrium line are straight and parallel,
NTP=Number of theoretical plates








 
ab
oy
L
ab
oy
y
y
oy
yy
N
y
yy
N
yy
dy
N
b
a
)(
•(2) For straight operating and equilibrium line (not parallel),







aa
bb
aabb
L
yy
yy
yyyy
y
ln
)()(
Similarly,















 
bb
ab
oy
aa
ab
oy
ab
ox
y
y
oy
yy
yy
N
yy
yy
N
NTP
xx
xx
N
yy
dy
N
b
a
)(











 
bb
ab
oy
aa
ab
oy
oyox
y
y
oy
yy
yy
N
yy
yy
N
NTPNN
yy
dy
N
b
a
)(
38
•(3) When the operating line is straight but steeper than the
equilibrium line,











 
bb
ab
oy
aa
ab
oy
oy
y
y
oy
yy
yy
N
yy
yy
N
NTPN
yy
dy
N
b
a
)(
NTP=Number of theoretical plates
NTP
y
yy
NTUN
yyy
yyyyyy
yy
y
x
y
L
ab
oy
Lab
aabbab
ba
a
b
b










1
Equilibrium
line
Operating line
yyyyyy
yy
y
x
y
aabbab
ba
a
b
b





yy
y
x
y
ba
a
b
b




x
y
b
b
y
x
y
a
b
b

x
y
a
b
39
•Similarly, for straight operating and equilibrium line (not parallel),













 
bb
ab
oy
aa
ab
oy
L
ab
ox
y
y
oy
yy
xx
N
yy
yy
N
x
xx
N
yy
dy
N
b
a
)(
aa
bb
aabb
L
xx
xx
xxxx
x







ln
)()(
oxoxT NHZ 


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 
x
ox
y
y
oy
m
aK
SL
H
yy
dy
N
b
a
][
/
)(
40
•Common equations for calculations of height of packed section:
height of packed section=height of a transfer unit  number of transfer
units
•There are four kinds of transfer units:
•The overall height of a transfer unit [Hoy OR Hox] can be defined as
the height of a packed section required to accomplish a change in
concentration equal to the average driving force in that section.












 
bb
ab
oy
aa
ab
oy
oyTmab
y
y
oy
yy
yy
N
yy
yy
N
HZyyyy
yy
dy
N
b
a
)(
)(
41
•Four kinds of transfer units:
Gas film:
Liquid film:
Overall gas:
Overall liquid: 














b
a
b
a
b
a
b
a
x
x
ox
y
ox
y
y
oy
y
oy
x
x i
x
x
x
y
y i
y
y
y
oxoxoyoyxxyyT
xx
dx
N
aK
SL
H
yy
dy
N
aK
SV
H
xx
dx
N
ak
SL
H
yy
dy
N
ak
SV
H
NHNHNHNHZ
/
/
/
/
42
• Alternate forms of transfer coefficients
The gas-film coefficients reported in the literature are often based on a
partial –pressure driving force instead of a mole-fraction difference and
are written as kga or Kga.
•Similarly liquid-film coefficients may be given as kLa or KLa, where the
driving force is a volumetric concentration difference. [kL=kc defined by
Eq.(17.36)]
)20.17()(
,
0
AAi
T
v
AA
y
g
y
g
y
y
AMv
B
A
A
MvAA
cc
B
D
JN
P
aK
aK
P
ak
ak
dyDdbN
db
dy
DNJ
A
Ai
T




 

43
aK
G
Hand
ak
G
H
aPK
G
Hand
aPk
G
H
M
L
S
L
M
G
S
V
M
G
G
L
xx
ox
L
xx
x
g
M
oy
g
M
y
xM
M
xy
M


//
,





Let
=molal mass velocity, kgmol/m2 h
=mass velocity of gas stream based on total tower cross
section, kg/m2 h
Where,
=mass velocity of liquid stream based on total tower
cross section, kg/m2 h
//
)25.18(
,
GG
aPK
G
Hand
aPk
G
H
M
S
L
M
G
S
V
M
G
G
xxxx
g
M
oy
g
M
y
xM
xy
M






)25.18(
,
G
Hand
G
H
M
S
L
M
G
S
V
M
G
G
M
oy
M
y
xM
xy
M




 ,
M
S
L
M
G
S
V
M
G
xM
xy
M
 

44
•The terms HG, HL, NG AND NL often appear in the literature instead of
Hy, Hx, Ny AND Nx, as well as the corresponding terms for overall values,
but here the different subscripts do not signify any difference in either
units or magnitude.
•Relationships among different kinds of height of a transfer unit:
ak
m
akaK
mxymxy
xxk
yy
yyk
yy
K
xyy
AAAiAi
AAix
AAi
AiAy
AiA
y










11
,,
)()(
1

M
M
x
M
y
M
y
M
AAAiAi
AAix
AAi
AiAy
AiA
y
L
L
ak
mG
ak
G
aK
G
mxymxy
xxk
yy
yyk
yy
K










,,
)()(
1

45
)20.17()(
,
0
AAi
T
v
AA
y
g
y
g
y
y
AMv
B
A
cc
B
D
JN
P
aK
aK
P
ak
ak
dyDdbN
db
A
Ai
T


  
aK
G
Hand
ak
G
H
aPK
G
Hand
aPk
G
H
M
S
L
M
G
S
V
M
G
G
L
xx
ox
L
xx
x
g
M
oy
g
M
y
xM
xy
M


//
,





x
M
M
yoy
AAAiAi
AAix
AAi
AiAy
AiA
y
H
L
mG
HH
mxymxy
xxk
yy
yyk
yy
K










,,
)()(
1

)26.18(
/
)25.18(
,
aK
G
Hand
ak
L
H
aPK
G
Hand
aPk
G
H
M
S
L
M
G
S
V
M
G
G
L
xx
ox
x
M
x
g
M
oy
g
M
y
xM
xy
M







M
M
x
M
y
M
y
M
AAAiAi
L
L
ak
mG
ak
G
aK
G

46
amkakaK
mxymxy
yxx
AAAiAi
111
,,

 
Similarly,
M
M
y
M
x
M
x
M
AAAiAi
AAix
AAi
AiAy
AiA
y
G
G
amk
L
ak
L
aK
L
mxymxy
xxk
yy
yyk
yy
K










,,
)56.17(
)()(
1

y
M
M
xox
AAAiAi
AAix
AAi
AiAy
AiA
y
H
mG
L
HH
mxymxy
xxk
yy
yyk
yy
K










,,
)()(
1

47

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gas absorption

  • 2. WHAT IS GAS ABSORPTION????  Absorption, or gas absorption, is a unit operation used in the chemical industry to separate gases by washing or scrubbing a gas mixture with a suitable liquid .  The fundamental physical principles underlying the process of gas absorption are the solubility of the absorbed gas and the rate of mass transfer. One or more of the constituents of the gas mixture dissolves or is absorbed in the liquid and can thus be removed from the mixture. In some systems, this gaseous constituent forms a physical solution with the liquid or the solvent, and in other cases , it reacts with the liquid chemically.
  • 3.  The purpose of such scrubbing operations may be any of the following : Gas purification (eg , removal of air pollutants from exhausts gases or contaminants from gases that will be further processed) , Product Recovery , or production of solutions of gases for various purposes.  The absorber may be a packed column , plate column , spray column , venturi scrubbers , bubble column , falling films , wet scrubbers ,stirred tanks
  • 4.  Gas absorption is usually carried out in vertical counter current columns.  The solvent is fed at the top of the absorber , whereas the gas mixture enters from the bottom .The absorbed substence is washed out by the solvent and leaves the absorber at the bottom as a liquid solution .  The solvent is often recovered in a subsequent stripping or desorption operation . This second step is essentially the reverse of absorption and involves counter current contacting of the liquid loaded with solute using and inert gas or water vapor .
  • 5. Choice Of Solvent for Absorption  If the principal purpose of the absorption operation is to produce a specific solution, as in the manufacture of hydrochloric acid, for example, the solvent is specified by the nature of the product, i.e. water is to be the solvent. If the principal purpose is to remove some components (e.g. impurities) from the gas, some choice is frequently possible.  The factors to be considered are:
  • 6.  GAS SOLUBILITY : The gas solubility should be high, thus increasing the rate of absorption and decreasing the quantity of solvent required. Solvent with a chemical nature similar to the solute to be absorbed will provide good solubility.  VOLATALITY : The solvent should have a low vapour pressure to reduce loss of solvent in the gas leaving an absorption column.  CORROSIVENESS : The materials of construction required for the equipment should not be unusual or expensive
  • 7.  COST : The materials of construction required for the equipment should not be unusual or expensive.  VISCOSITY : Low viscosity is preferred for reasons of rapid absorption rates, improved flooding characteristics in packed column, low pressure drops on pumping, and good heat transfer characteristics.  The solvent should be non-toxic, non-flammable and chemically stable.
  • 9. COUNTERCURRENT FLOW • Countercurrent tower which may be either a packed or spray tower, filled with bubble cape trays or of any internal construction to bring about liquid gas contact. • Mole ratio may be define as Y = y/(1-y) X = x/(1-x) • Here y for gas stream and x for liquid stream.
  • 10. Material balance Gs = G(1-y) = G/(1+Y) Ls = L(1-x) = L/(1+X) NOW Gs(Y1 - Y) = Ls(Xs - X) AND Gs(y1/(1-y1) – y/(1-y)) = Gs(p1/(1-p1) – p/(1-p)) = Gs(x1/(1-x1) – x/(1-yx))
  • 11. MINIMUM LIQUID-GAS RATIO • In the design of absorbers, the quantity of gas to be treated G or Gs, the terminal concentrations Y1 and Y2,and the composition of the entering liquid X2 are ordinarily fixed by process requirements. • The operating line must be pass through point D and must end at the ordinate Y. • Liquid gas ratio is define by L/G • Graph of minimum liquid-gas ratio can be shown as behind slide.
  • 12.
  • 13. CO-CURRENT FLOW • When gas and liquid flow cocurrently, the operating line has a negative slope. -L/G • There is no limit of this ratio, but an infinitely tall tower would produce an exit liquid and gas in equilibrium • In cocurrent fow requirred height of tower should be higher then in counter current flow.
  • 14. Counter-current multi-stage absorption (Tray absorber):  In tray absorption tower, multi-stage contact between gas and liquid takes place.  In each tray, the liquid is brought into intimate contact of gas and equilibrium is reached thus making an ideal stage.  In ideal stage, average composition of liquid leaving the tray is in equilibrium with liquid leaving that tray.  The most important step in design of tray absorber is the determination of number of trays.  The schematic of tray tower is presented in figure.  The liquid enters from top of the column whereas gas is added from the bottom.
  • 15.  The efficiency of the stages can be calculated as: 𝑆𝑡𝑎𝑔𝑒 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑑𝑒𝑎𝑙 𝑠𝑡𝑎 𝑔𝑒𝑠/𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑎𝑙 𝑠𝑡𝑎𝑔𝑒𝑠 …..Eq.(1) Gs,Y1 Gs, YN+1 Liquid out Ls, XN Gas in Gas outGs, Y1 Liquid in Ls, X0 Ls, X1 Gs,YN Ls, X1 N N-1 1 2
  • 16.  The following parameters should be known for the determination of “number of stages”: (1) Gas feed rate (2) Concentration of gas at inlet and outlet of the tower (3) Minimum liquid rate; actual liquid rate is 1.2 to 2 times the minimum liquid rate. (4) Equilibrium data for construction of equilibrium curve  Now, the number of theoretic stages can be obtained graphically:
  • 17. (A) Graphical Method for the Determination of Number of Ideal Stages:  Overall material balance on tray tower: Gs(YN+1 -Y1) = Ls(XN -X0) This is the operating line for tray tower  If the stage (plate) is ideal, (Xn, Yn) must lie on the equilibrium line, Y*=f(X).  Top plate is located at P(X0, Y1) and bottom plate is marked as Q(XN, YN+1) in X-Y plane.  A vertical line is drawn from Q point to D point in equilibrium line at (XN, YN).  From point D in equilibrium line, a horizontal line is extended up to operating line at E (XN-1, YN).
  • 18.  The region QDE stands for N-th plate. We may get fraction of plates.  In that situation, the next whole number will be the actual number of ideal plates.  If the overall stage efficiency is known, the number of real plates can be obtained from Equation A. 0.0 0.2 0.4 0.6 0.8 1.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 E ( X N-1 ,Y N ) X P ( X 0 ,Y 1 ) Q ( X N ,Y N+1 ) Equilibrium line Operating line N N-1 N-2 D ( X N ,Y N )
  • 19. Example: It is desired to absorb 95% of acetone by water from a mixture of acetone and nitrogen containing 1.5% of the component in a countercurrent tray tower. Total gas input is 30 kmol/hr and water enters the tower at a rate of 90 kmol/hr. The tower operates at 27ºC and 1 atm. The equilibrium relation is Y=2.53X. Determine the number of ideal stages necessary for the separation using graphical method.
  • 20. Solution:  Basis: 1 hour  G(N+1)=30 kmol  Y(N+1)=0.015  Lo=90 kmol  Moles acetone in = 30×0.015 moles=0.45 moles  Moles nitrogen in = (30-0.45) moles=29.55 moles  Moles acetone leaving (95% absorbed) = 0.45×(1-0.95) moles=0.0225 moles  Gs=29.55 moles  Ls=90 moles  α=2.53 [as, Y=2.53X]
  • 21.  𝑌1 = 0.0225/29.55 = 7.61 × 10^−4  𝑌(𝑁+1)= 0.015  Equation.. 𝐺𝑠𝑌(𝑁+1)− 𝑌1= 𝐿𝑠(𝑋𝑁− 𝑋0) 29.55 × 0.015 − 7.61 × 10^−4= 90(𝑋𝑁− 0)  XN=4.68×10^-3  Solution by graphical method,Construction of operating line PQ: P(X0,Y1)=P(0, 7.61×10^-4) Q(XN, YN+1)=Q(4.68×10^-3, 0.015) Construction of equilibrium line (Y=2.53X): X 0 0.001 0.002 0.003 0.004 0.005 Y 0 0.00253 0.00506 0.00759 0.01012 0.01265 From graphical construction in Figure , the number of triangles obtained is more than 7. Hence number of ideal stages is 8.
  • 22.
  • 23. 23  Rate of absorption •Volumetric mass transfer coefficients (Kya, etc.) are used for most calculations, because it is more difficult to determine the coefficients per unit area and because the purpose of the design calculation is generally to determine the total absorber volume. •Kya=overall volumetric mass-transfer coefficient, kmol/(m3·h·unit mole fraction). •a=effective area of interface per unit packed volume, m2/m3
  • 24. 24 •Simplicity Treatment •The following treatment applies to lean gases (up to 10% solute): •(a) Correction factors for one-way diffusion are omitted for simplicity. •(b) Changes in gas and liquid flow rates (V and L) are neglected. •(c) kxa, kya, Kya, Kxa can be considered as constants.
  • 25. 25 •the rate of mass transfer: r=NA [kgmol/(m2·h·unit mole fraction)] )()( )( AAixAiAyA xxkyykN r   )( )(    AAyA yyKN r )( )( AAxA xxKN r   
  • 26. 26 •Let r =rate of absorption per unit volume, kgmol/(m3·h) )( )( )( )( xxaKr yyaKr xxakr yyakr x y ix iy       •It is hard to measure or to predict a, but in most cases it is not necessary to know its actual value since design calculations can be based on the volumetric coefficients.
  • 27. 27 •Determining the interface composition (yi, xi) •(yi, xi) is also hard to measure, but it can be obtained from the operating-line diagram ak ak xx yy y x i i    )( )( •Thus a line drawn from the operating line with a slope –kxa/kya will intersect the equilibrium line at (yi, xi).
  • 29. 29 xyy AAAiAi AAixAiAyy k m kK mxymxy xxkyykK     11 ,, )()(  •Determining the overall coefficients: Using the local slope of the equilibrium curve m, we have Therefore, ak m akaK mxymxy xxk yy yyk yy K xyy AAAiAi AAix AAi AiAy AiA y           11 ,, )56.17( )()( 1  amkakaK mxymxy xxk yy yyk yy K yxx AAAiAi AAix AAi AiAy AiA y 111 ,, )()( 1           Similarly,
  • 30. 30 =overall resistance to mass transfer = resistance to mass transfer in the gas film =resistance to mass transfer in the liquid film ak m ak aK x y y 1 1 ak m ak aK x y y 1 1 ak m ak aK x y y 1 1 ak m akaK mxymxy xyy AAAiAi   11 ,,
  • 31. 31 •Gas film “controls” and Liquid film “controls” When "" 11 , 1 controlsfilmGas akaKak m ak yyxy  ak m akaK mxymxy xyy AAAiAi AAixAiAyy    11 ,, •Or, When the coefficients kya and kxa are of the same order of magnitude, and m is very much greater than 1.0, the liquid film resistance is said to be controlling. That is, "" 1 , 1 controlsfilmLiquid ak m aKak m ak xyxy 
  • 32. 32 •Liquid film controlling means that any change in kxa has a nearly proportional effect on both Kxa and Kya and on the rate of absorption, whereas a change in kya has little effect. •Examples of Liquid film controls: Absorption of CO2, H2,O2,Cl2 in water; •When the solubility of the gas is very high, m is very small and the gas- film resistance controls the rate of absorption. •Examples of gas-film controls: Absorption of HCl, NH3 in water; NH3 in acid solution; SO2, H2S in basic solvent.
  • 33. 33 •With gases of intermediate solubility both resistances are important, but the term controlling resistance is sometimes used for the larger resistance. •The absorption of NH3 in water is often cited as an example of gas-film control, since the gas film has about 80 to 90 percent of the total resistance.
  • 35. 35 •Number of transfer units     b a y yy T yy dy aK SV Z )( / •The equation for column height can be written as follows:     b a y y oy yy dy N )( =overall number of transfer units [NTU], based on gas phase. aK SV H y oy /  =overall height of a transfer unit [HTU], based on gas phase.
  • 36. 36 oyoyT NHZ  (1)If the operating line and equilibrium line are straight and parallel, Equilibrium line Operating line a a b x x yy y y y  1 2 3 b a a b xx x x x yy y y y   3 2 1 1 2 3 a a b x yy y y y  1 2 3 a a b x x x yy y y y  1 1 2 3 a b x yy y y y  1 2 3 b a b b xx x x x yy y y y   3 2 1 1 2 3                 bb ab oy aa ab oy ab oy y y oy yy yy N yy yy N yy yy N yy dy N b a .18( )(                 bb ab oy aa ab oy ab oy y y oy yy yy N yy yy N yy yy N yy dy N b a 1( )(                 bb ab oy aa ab oy ab oy y y oy yy yy N yy yy N yy yy N yy dy N b a )(
  • 37. 37                  bb ab oy aa ab oy ab oy y y oy yy yy N yy yy N NTP yy yy N yy dy N b a )( •If the operating line and equilibrium line are straight and parallel, NTP=Number of theoretical plates           ab oy L ab oy y y oy yy N y yy N yy dy N b a )( •(2) For straight operating and equilibrium line (not parallel),        aa bb aabb L yy yy yyyy y ln )()( Similarly,                  bb ab oy aa ab oy ab ox y y oy yy yy N yy yy N NTP xx xx N yy dy N b a )(              bb ab oy aa ab oy oyox y y oy yy yy N yy yy N NTPNN yy dy N b a )(
  • 38. 38 •(3) When the operating line is straight but steeper than the equilibrium line,              bb ab oy aa ab oy oy y y oy yy yy N yy yy N NTPN yy dy N b a )( NTP=Number of theoretical plates NTP y yy NTUN yyy yyyyyy yy y x y L ab oy Lab aabbab ba a b b           1 Equilibrium line Operating line yyyyyy yy y x y aabbab ba a b b      yy y x y ba a b b     x y b b y x y a b b  x y a b
  • 39. 39 •Similarly, for straight operating and equilibrium line (not parallel),                bb ab oy aa ab oy L ab ox y y oy yy xx N yy yy N x xx N yy dy N b a )( aa bb aabb L xx xx xxxx x        ln )()( oxoxT NHZ       x ox y y oy m aK SL H yy dy N b a ][ / )(
  • 40. 40 •Common equations for calculations of height of packed section: height of packed section=height of a transfer unit  number of transfer units •There are four kinds of transfer units: •The overall height of a transfer unit [Hoy OR Hox] can be defined as the height of a packed section required to accomplish a change in concentration equal to the average driving force in that section.               bb ab oy aa ab oy oyTmab y y oy yy yy N yy yy N HZyyyy yy dy N b a )( )(
  • 41. 41 •Four kinds of transfer units: Gas film: Liquid film: Overall gas: Overall liquid:                b a b a b a b a x x ox y ox y y oy y oy x x i x x x y y i y y y oxoxoyoyxxyyT xx dx N aK SL H yy dy N aK SV H xx dx N ak SL H yy dy N ak SV H NHNHNHNHZ / / / /
  • 42. 42 • Alternate forms of transfer coefficients The gas-film coefficients reported in the literature are often based on a partial –pressure driving force instead of a mole-fraction difference and are written as kga or Kga. •Similarly liquid-film coefficients may be given as kLa or KLa, where the driving force is a volumetric concentration difference. [kL=kc defined by Eq.(17.36)] )20.17()( , 0 AAi T v AA y g y g y y AMv B A A MvAA cc B D JN P aK aK P ak ak dyDdbN db dy DNJ A Ai T       
  • 43. 43 aK G Hand ak G H aPK G Hand aPk G H M L S L M G S V M G G L xx ox L xx x g M oy g M y xM M xy M   // ,      Let =molal mass velocity, kgmol/m2 h =mass velocity of gas stream based on total tower cross section, kg/m2 h Where, =mass velocity of liquid stream based on total tower cross section, kg/m2 h // )25.18( , GG aPK G Hand aPk G H M S L M G S V M G G xxxx g M oy g M y xM xy M       )25.18( , G Hand G H M S L M G S V M G G M oy M y xM xy M      , M S L M G S V M G xM xy M   
  • 44. 44 •The terms HG, HL, NG AND NL often appear in the literature instead of Hy, Hx, Ny AND Nx, as well as the corresponding terms for overall values, but here the different subscripts do not signify any difference in either units or magnitude. •Relationships among different kinds of height of a transfer unit: ak m akaK mxymxy xxk yy yyk yy K xyy AAAiAi AAix AAi AiAy AiA y           11 ,, )()( 1  M M x M y M y M AAAiAi AAix AAi AiAy AiA y L L ak mG ak G aK G mxymxy xxk yy yyk yy K           ,, )()( 1 
  • 47. 47