ABSORPTION



Bringing the dirty effluent gas into contact with the
scrubbing liquid and subsequently separating the
cleaned gas from the contaminated liquid
Absorption is a basic chemical enginnering unit
operation which in the APC field is reffered as
scrubbing
They have wide use in controlling SO2, H2S, and
light hydrocarbons
ABSORPTION
_Wet scrubbers can be categorized into 3 groups:
1.
Packed-bed counterflow scrubbers
2.
Cross-flow scrubbers
3.
Bubble plate and tray scrubbers
SCRUBBER TYPES
Packed Tower
Spray tower
Venturi Absorber
CONCEPT OF ABSORPTION
Gas absorption is the removal of one or
more pollutants from a contaminated gas
stream by allowing the gas to come into
intimate contact with a liquid that
enables the pollutatants to become
dissolved by the liquid. The principal
factor dictating performance is the
solubility of the pollutants in the
absorbing liquid.
The rate of transfer in the liquid is dictated
by the diffusion processes occurring on
each side of the gas liquid interface.
LIQUID WASTE
Pollutants removed from the gas stream transferred
into liquid phase whose disposal is another issue
to deal with.
Therefore scrubber needs other units such as
storage vessels, additives to treat the scrubbing
liquid according to required discharge standards.
ABSORPTION


Absorption units must provide large surface area of
liquid-gas interface
Therefore the units are designed to provide large
liquid surface area with a minimum of gas pressure
drop
PACKED TOWER
PACKING MATERIAL AND SHAPES


Packing material (must be inert) is designed to
increase the liquid-film surface
Many geometric shapes are available : Raschig ring,
pall ring, berl saddle, tellerete etc.
PACKING MATERIAL AND SHAPES
PACKING
MATERIAL
PROPERTIES
ABSORPTION THEORY


Physically, the absorption of a pollutant gas from a
moving gas stream into an appropriate liquid
stream is quite complex
Basically the transfer process into each fluid stream
is accomplished by 2 mechanisms:



The pollutant species is transferred from the bulk of
the gas stream toward the gas-liquid interface by
turbulent eddy motions
Very close to the interface laminant flow is valid and
transfer is accomplished by molecular diffusion
On the liquid side of the interface process is
reversed
ABSORPTION THEORY
ABSORPTION THEORY

On the basis of Fick's Law, the diffusion of one gas
(A) through a second stagnant gas B, NA, the molar
rate of transfer of A per unit cross-sectional area is
given by;
NA = -DAB (dcA/dz)/(1-(cA/c)
DAB: molecula diffusion coef. (m2/t)
cA: molar concentration of species A (mol/L)
c: molar concentration of the gas mixture (mol/L)
z: the direction of mass transfer (m)


DAB tables are available for a number of binary gas
mixtures
ABSORPTION THEORY

Mass transfer rate per unit area for molecular
diffusion of A through a second liquid is given by:
NA = -DL/z (cA2-cA1)
DL: liquid phase molecular diffusion coef. (m2/t)
cA2-cA1: concentration difference of A over the distance z


Typical values of DL for binary mixtures are tabulated
in the literature.
THE EQUILIBRIUM DISTRIBUTION CURVE

Before entering into details of mass transfer, let's
summarize the method of presenting equilibrium
data for a pollutant A distributed between liquid
and gas phase
P=c
Inject
solute A
Inert carrier
gas
Inert Liquid
Solvent
Mole
fraction in
gas, yA
Exp. Equilib.
Distribution
curve
Mole fraction in lqiuid, xA
THE EQUILIBRIUM DISTRIBUTION CURVE

After sufficient time, no further change in the
concentration of A in two phases. These
concentration can be measured and converted into
mole fraction xA in the liquid phase and yA in the
gas phase
P=c
Inject
solute A
Inert carrier
gas
Inert Liquid
Solvent
Mole
fraction in
gas, yA
Exp. Equilib.
Distribution
curve
Mole fraction in lqiuid, xA
MASS TRANSFER COEFFICIENTS BASED ON
INTERFACIAL CONCENTRATIONS


When mass transfer occurs in moving liquid and
gaseous streams, it is difficult to evaluate the
separate effects of molecular and turbulent diffusion
An alternative to this is to express NA for each
phase in terms of mass transfer coefficient k and a
driving force based on the bulk and interfacial
concentrations for that phase
MASS TRANSFER COEFFICIENTS BASED ON
INTERFACIAL CONCENTRATIONS

For the liquid phase:



NA = kL(cAi-cAL) = kx(xAi-xAL)
kL(is the liquid mass transfer coeff. Based on
concentration, in length per unit of time, cAiis the
concentration of A in the liquid phase at the interface,
cALis the concentration of A in the bulk of the phase, in
moles per unit volume.
kx is the liquid mass transfer coefficient based on mole
fractions, in moles per units of time and length
squared, xA is the mole fraction of A in the liquid
interface, and xAL is the mole fraction of A in the bulk of
the liquid phase
MASS TRANSFER COEFFICIENTS BASED ON
INTERFACIAL CONCENTRATIONS

For the gas phase:



NA = kG(pAG-cAi) = ky(yAG-yAi)
kGis the gas phase mass transfer coeff. based on partial
pressures, in moles/length2 time, pAGis the partial
pressure of A in the bulk of gas phase pAi is the partial
pressure of A in the gas interface
ky is the gas phase mass transfer coefficient based on mole
fractions, in moles per units of time and length squared,
yAG is the mole fraction of A in the bulk of the gas phase,
and yAi is the mole fraction of A in the gas phase interface
MASS TRANSFER COEFFICIENTS BASED ON
INTERFACIAL CONCENTRATIONS

However this approach to determining NA is not practical
since kx and ky are difficult to obtain and no way to
measure the values of yAi and xAi experimentally since
any attempt to do it will perturb the equilibrium
between the two streams
OVERALL MASS TRANSFER COEFFICIENTS

When mass transfer rates are reasonably low, NA can be
expressed as:


NA = KG(pAG-pA*) = Ky(yAG-yA*)
KG and Ky are local overall mass transfer coefficients


pA* : equilibrium partial pressure of solute
A in a gas phase which is in contact with a
liquid having the composition of cAL of the
main body of the absorption liquid
yA*: defined similarly in terms of a liquid
with mole fraction xAL of the bulk liquid
OVERALL MASS TRANSFER COEFFICIENTS
Point P represents the state of the
bulk phase of the 2 fluid streams,
yAG and xAL.
The point M represents the state
(yAi and xAi) associated with
equilibrium at the interface
slope=m'
The distance between P and C is a
measure of the driving force.
OVERALL MASS TRANSFER COEFFICIENTS





NA = KG(pAG-pA*) = Ky(yAG-yA*) This equation is
usually restricted the resistance to mass transfer is
primarily in the gas phase, which characterizes the
majority of absorption problems in air pollution work
The solubility of the polutant gas normally determines
the liquid that is chosen
The major physical problem is getting the pollutant to
diffuse through the gas phase to the interface,
consequently gas phase controls the process.
If the liquid phase controls:
NA = KL(cA*-cAL) = Kx(xA*-xAL)
OVERALLItM
ASS
T
RANSFER
C
OEFFICIENTS
is important to note that the quantities





pA*,yA*,cA*,xA* do not represent any actual
condition in the absorption process but are related
in each case to a real concentration in one of the
bulk fluids through the equilibrium data for the twophase system. From the geometry of the previous
figure:
yAG-yA*= yAG-yAi+(yAi-yA*)
yAi-yA*=m'(xAi-xAL)
yAG-yA*= yAG-yAi+m'(xAi-xAL)
1/Ky=1/ky+m'/kx
MASS BALANCES AND THE OPERATING LINE
FOR PACKED TOWERS
Lm,2
Ls
x2
X2
Gm,2
Gc
y2
Y2
T = const
P = const
Cross-sectional
area, A
dz
Lm,1
Ls
x1
X1
Gm,1
Gc
y1
Y1
Gm molar total gas flow
rate (carrier gas +
pollutant)
Gc molar inert carrier
gas flow rate
Lm molar total solvent
flow rate (solvent +
absorbed pollutant)
Ls molar solvent flow
rate
x is the liquid mole
fraction of pollutant, y
is the gas phase mole
fraction of the
pollutants, X is the
liquid phase mole ratio
and Y is the gas phase
mole ratio
MASS BALANCES AND THE OPERATING LINE
FOR PACKED TOWERS
Mole fraction and mole ratio:

X = x/(1-x)
Y = y/(1-y)

Subscript m denotes that rates are in the units of
mole basis

The conservation of mass principle applied to the
pollutant species in terms of total mass flow rates at
top and bottom yields:
Gm,1y1+ Lm,2x2 = Gm,2y2+ Lm,1x1

or Gm,1y1 -Gm,2y2 = Lm,1x1 -Lm,2x2

MASS BALANCES AND THE OPERATING LINE
FOR PACKED TOWERS




In Gm,1y1 -Gm,2y2 = Lm,1x1 -Lm,2x2 total gas and liquid
flow rates are not equal at the top and the bottom of
the column, therefore we cannot further simplify
this equation.
When we write the equation in terms of the carrier
gas and liquid solvent rates then:
GC,m(Y2-Y1) = LS,m(X2-X1)
These two equations above gives a straight line on
Y-X coordinates with a slope of Lsm/GCm and called
operating lines.


Dirty air
Clean air
Clean water
Dirty water
The operating line lies
above the equilibrium line
for absorption
For a stripping (removal of
gas from liquid stream)
the operating line must lie
below the equilibrium line
in order for the drving
force to act from the liquid
phase toward the gas
phase
THE MINIMUM AND DESIGN LIQUID- GAS
RATIO





At the bottom and top of the absorber, parameters
Gm,1, Gc, y1, Gm2, y2, and x2 are known.
We need to determine Ls, and x1
So we have one equation with 2 unknowns...
However selection of one of these values, obviously
fixes the other.
How to select a value?
THE MINIMUM AND DESIGN LIQUID- GAS
The minimum rate
RATIO
is highly
undesirable. At
this point driving
force is almost 0.
Hence it would
take an infinetely
tall absorber to
accomplish the
desired separation
As a general
operating principle
an absorber is
typically designed
to operate at liquid
rates which are 30
to 70 % greater
than minimum
rate.
TOWER DIAMETER AND PRESSURE DROP PER
UNIT TOWER HEIGHT



For a given packing and liquid flow rate in an
absorption tower variation in the gas velocity has a
significant effect on the pressure drop
As the gas velocity is increased, the liquid tends to
be retarded in its downward flow, giving rise to
term liquid holdup (LH)
A LH increases, the free cross-sectional area for gas
flow decreases and pressure drop per unit height
increases.
PROBLEMS WITH HIGH GAS VELOCITY




Channeling: the gas or liquid flow is much greater at
some points than at others
Loading: the liquid flow is reduced due to the increased
gas flow; liquid is held in the void space between
packing
•Flooding: the liquid stops flowing altogether and
collects in the top of the column due to very high gas
flow
TO AVOID this condition experience dictates operating
at gas velocities which are 40 to 70 % of those which
causing flooding
FLOOD POINT



The relationship between DP/Z and other important
tower variables-liquid and gas rates, liquid and gas
stream densities and viscosities, and type of
packing has been extensively studied on an
experimental basis.
A widely accepted correlation among these
parameters can be seen in below figure
Where G' and L': superficial gas and liquid mass
flow rate defined as actual flow rates divided by the
empty cross-sectional area of the tower.
(G ' ) 2 Fm L0.1
g c ( r L  rG ) rG
G’:gas mass flux (lb/sft2)
F:packing factor (ft2/ft3)
mL:liquid viscosity, cp
gc: proportıonality
constant, 32.17 ftlb/s2-lbf
rL:liquid density, lb/ft3
rG:gas density,
lb/ft3
L’/G’√(rG/rL-rG)
L’: liquid mass flux (lb/s-ft2)
In Cooper and Alley’s book, Figure 13.6 can be used. Note that in Figure 13.6 Gx and Gy are
liquid and gas flux (lg/s-ft2), respectively. In our notation G’ and L’ correspond to Gx and Gy
PACKING FACTOR F


The top line in the figure represents the general
flooding condition for many packings. The flooding
condition however has been found to vary as a
function of the packing factor F (dimensionless
packing factor tabulated below)
Recent studies showed that when F is in the range
of 10 to 60, the pressure drop can be expressed by:
DPflood = 0.115F0.7
PACKING DATA
DETERMINING TOWER DIAMETER

First abscissa value is calculated
(L'/G')(pG/(pL-pG))0.5

Where this value intercepts the flooding line on
Figure A, move horizontally to the left and read
the value of the ordinate:
(G')2F(mL)0.1/gc(pL-pG)pG

Calculate the G’ and take 30 to 70% of it to
prevent flooding

Tower crossectional area: A = G/G‘

Evaluate the tower diameter
DETERMINING EXPECTED PRESSURE DROP PER
UNIT HEIGHT OF TOWER


First calculate actual G’ and L’ and then
calculate the abscissa and the ordinate for use in
Figure 13.6
From those values the intersection on the figure
defines the pressure drop per foot of packed
height
Another emprical correlation found in the litrature
for the DP in packing when operating below the
load point is
DP/Z = 10-8m[10nL’/rL](G’2/rG) m and n are packing
constants see Table 6.2
DETERMINING TOWER DIAMETER AND EXPECTED
PRESSURE DROP PER UNIT HEIGHT OF TOWER
EXAMPLE
A packed tower is to be designed to remove 95% of the
ammonia from a gaseous mixture of 8 percent ammonia and
92% air, by volume. The flow rate of the gas mixture
entering the tower at 68 F and 1 atm is 80 lb-moles/hr.
Water containing no ammonia is to be the solvent, and 1-in.
Ceramic raschig rings will be used as the packing. The
tower is to operated at 60% of theflood point and the liquid
water rate is to be 30% greater than the minimum rate.
Determine





1. The gas-phase flow rates, in lb-moles/hr, for the
solute and carrier gas
2. The mole ratios of the gas and liquid phases at inlet
and outlet and the required water rate in lbmoles/hr.
3.The gas and liquid rates (lb/hr) for carrier gas, solute
gas, total gas, liquid solvent, solute in liquid, and total
liquid
4. The tower area and diameter
5. The pressure drop based on the two methods given in
the lecture notes.
EXAMPLE
Removal efficiency: 95%
Effluent Stream Composition: 8% ammonia and 92% air
Gas T and P: 68F and 1 atm
Flowrate: 80 lb-moles/hr
Liquid phase: Containing no ammonia
EXAMPLE
Determine composition of the liquid at the exit (X1)
(Inlet liquid concentration since pure water is used is x2=X2=0)
Use equilibrium data for ammonia-air-water mixtures which are given
below for 68 F and 14,7 psia. :
X
Y
0.0206
0.0158
0.0310 0.0407 0.0502 0.0735 0.0962
0.024 0.0329 0.0418 0.0660 0.0920
In order to determine composition of liquid at the exit, we need to
calculate the minimum solvent flow rate first.
By plotting X vrs Y at the equilibrium, we can evaluate the minimum
solvent and then operating solvent rate.
In Cooper and Alley’s book, use Table B4 in the Appendix.
y, moles solute per mole
carrier gas
EXAMPLE
0,1
Y1=0.087
0,08
0,06
(Lm,S/Gm,C)min= 0,90
(0,087-0,00435)/0,092
0,04
0,02
X2,Y2
0
Y2=0.00435
0
0.092
0,02
0,04
0,06
0,08
X, moles solute per mole solvent
Since the liquid rate is to be 30% greater than the miniumu rate
(Lm,S)/Gm,C)design = 1,30(0.90) = 1.17 mole/mole
Lm,S = Gm,C*1,17 = 1.17*73.6 = 86.1 lb moles/hr
0,1
y, moles solute per mole
carrier gas
EXAMPLE
0,1
Y1=0.087
0,08
0,06
(Lm,S/Gm,C)=1.17
0,90
0,04
0,02
X2,Y2
0.00435
0
X1: 0.0707
0,02
0,04
0,06
0,08
0,1
X, moles solute per mole solvent
Now, X1 can now be found.
1. Graphically by drawing operating line with a slope of 1.17 with starting
point of (0, 0.00435) and the point crosses Y1=0.087 can be read. OR
2. From Lms/Gm,C = Y2-Y1/(X2-X1)=0.00435-0.087/(0-X1) = 1.17
X1 = 0.0707 lm mole A/lm mole water or x1 = 0.066 lb mole A /lb moles
solution
FLOW RATES
The gas and liquid rates:
Lm,2
GC = 73.6*29 = 2134 lb/hr
Ls
GA,1 = 6.4*17 = 109 lb/hr
x2
X2
GA,2 = 0.32*(17) = 5.4 lb/hr
LS = 86.1*18=1550 lb/hr
LA,1 = DGA= 109*0.95=104 lb/hr
Therefore:
G1 = 2134 +109=2243 lb/hr bottom
L1 = 1550 +104 = 1654 lb/hr
G2 = 21345+5=2139 lb/hr top Lm,1
L
L2=1550 + 0 = 1550 lb/hr top s
x1
X1
Gm,2
Gc
y2
Y2
T = const
P = const
Cross-sectional
area, A
dz
Gm,1
Gc
y1
Y1
TOWER AREA
To determine the tower area, we need to use Figure flooding
correlation plot.
Therefore we need to calculate gas and liquid phase densities
at the top and bottom of the tower. Since the ammonia
content is very low in liquid phase, use the density of pure
water, 62.3 lb/ft3 as the solution density through the tower.
For the gas phase assume ideal gas behavior:
r= P/RT = MwP/RT
At the top: Mw= SyiMi = 0.00435*17 + 0.9957*29 =
28.95
r= 28.95*14.7/(10.73*528) = 0.075 lb/ft3
At the bottom Mw= 0.08*17 + 0.92*29 = 28.04
r= 28.04*14.7/(10.73*528) = 0.0728 lb/ft3
Now calculate the abscissa of Flooding Figure
TOWER AREA
TOWER AREA
PRESSURE DROP
Pressure drop can be determined from the flooding figure
or from an emprical equation
DETERMINATION OF AN ABSORPTION TOWER
HEIGHT
Height of a packed tower = f(the overall resistance to mass
transfer between the gas and liquid phases, the average
driving force and interfacial area)
Consider a differential height of the absorber dZ. In height
dZ, the rate of mass transfer of species A
N A A(adZ )  d (Gm y )  Ad (G' y )
a: interfacial area available to mass transfer per unit
volume of the packing
A: cross-sectional area of the tower
TOWER HEIGHT
NA A(adz) = d(Gmy) = Ad(G' y)
K y a ( y AG  y *A ) dz  d (Gm ' y )
Gm ' dy
Gm ' y
dZ 

*
K y a( y  y )
K G aP ( y  y * )
The equation can be also written for liquid resistance
part.
TOWER HEIGHT
Gm ' y
Gm ' y
dZ =
=
*
K y a(y - y )
KG aP(y - y* )
To solve the above equation we can determine the overall
value of Kya (Kga) based on experimental “pilot plant”
operated with a certain packing and gas/liquid rate.
The right side of the equation can be integrated from
the knowledge of the operating line and equilibrium
line chracteristics.
This method can be modified to deal with the “height of a
transfer unit” and “the number of transfer units” by
modifiying the equation somewhat
TOWER HEIGHT
Gm ' dy
Gm ' dy
dZ 

*
K y a( y  y )
K G aP( y  y * )
To solve the above equation we can determine the overall
value of Kya (Kga) based on experimental “pilot plant”
operated with a certain packing and gas/liquid rate.
The right side of the equation can be integrated from
the knowledge of the operating line and equilibrium
line chracteristics.
This method can be modified to deal with the “height of a
transfer unit” and “the number of transfer units” by
modifiying the equation somewhat
Tower Height
The equation can be expressed in terms of height of
transfer unit (HTU ) and number of transfer units :
y1
G 'm
dy
Z
K OG aP yZ  y  y *
HTU or
Hoy
NTU or
Noy
HTU is reaonably constant through the absorber and has
55
unit of length. NTU is dimensionless.
x1, y1
2020/4/12
For dilute gas
streams, transfer
unit equation can
be simplified:
y1
G 'm
dy
Z
K OG aP yZ  y  y *
xZ, yZ
xZ, yZ*
y1 - yz
G'm
Z=
´
;
KOG aP DyLM
DyLM
y - y ) - (y - y )
(
=
1
*
1
z
´y - y* ´
ln ´ 1 1* ´
´yz - yz ´
*
z
Aerosol & Particulate Research Lab
x1, y1*
56
0.04% CO2
Pure amine
Lm = 0.46 gmole/s
2020/4/12
Aerosol & Particulate Research Lab
Q: A Packed tower using organic amine at 14 oC to absorb CO2.
The entering gas contains 1.27% CO2 and is in equilibrium with
a solution of amine containing 7.3% mole CO2. The gas leaves
containing 0.04% CO2. The amine, flowing counter-currently,
enters pure. Gas flow rate is 2.31 gmole/s and liquid flow rate is
0.46 gmole/s. The tower’s cross-sectional area is 0.84 m2.
KOGa = 9.34×10-6 s-1atm-1cm-3. The pressure is 1 atm.
Determine the tower height that can achieve this goal.
1.27% CO2
Gm = 2.31
gmole/s
C* = 7.3%
CO2 in
amine
57
Absorption of concentrated vapor
Mole balance on the controlled volume
Gas flux
 1 

G 'm  G 'm 0 
1 y 
x1, y1*
Liquid flux
 1 
L 'm  L 'm 0 

1 x 
 y1  L'm 0  x
x1 

 



1  y1  G 'm 0  1  x 1  x1 

y
 y1  L'm 0  x
x1 
 


1  

 1  y1  G 'm 0  1  x 1  x1 
Aerosol & Particulate Research Lab
d
d
0   (G 'm y )  ( L'm x)
dz
dz
2020/4/12
x1, y1
xZ, yZ
xZ, yZ*
58
EXAMPLE
A 1‐ft diameter packed column is used to scrub a soluble
gas (MW = 22) from an air‐gas mixture. Pure water enters
the top of the column at 1000 lbm/hr. The entering gas
stream contains 5% soluble gas and 95% air. Ninety‐five
percent of the soluble gas is removed. Both the operating
line and equilibrium curve may be assumed to be straight.
The equation for the equilibrium curve is y = 1.2x, where x,
y = mole fractions. The entering gas mixture flow rate is
800 lbm/hr. The column operates at 30 °C and 1 atm, and
 Kya = 4.29 lbmol/hr‐ft3‐Δy 

EXAMPLE
 Calculate or find:
 a) Concentration of the soluble gas in the effluent liquid if the column is operated at
minimum liquid flow rate
 b) Concentration of soluble gas in the liquid at a point in column where y = 0.02
 c) Height of packed section, ZT
 d) Hoy
 e) Whether columnis in danger of flooding if it is packed with 1⁄2‐in. ceramic
Raschig rings
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution
Solution