123chem

1. ABSORPTION

2. CONTENT
 Principles, mechanism
 Gas Liquid Equlibrium
 Single stage and multi stage
 Henry Law
 Design of packed and plate tower

3. ABSORPTION PRINCIPLES

4. PRINCIPLES
 2 contacting phases are gas and liquid
 Solute A (or more) are absorbed from the gas phase into the
liquid phase
 This process involves molecular and turbulent diffusion or mass
transfer of solute A through a stagnant, non-diffusing gas B into
stagnant liq C
 Examples of absorption
 Absorption of ammonia from air by liquid water
 Absorption of SO2 from flue gases by alkaline solution

5. DESORPTION
 Reverse of absorption is stripping
 Stripping = Desorption, from liq into gas
 Example
 Steam stripping of non-volatile oils in which the steam contacts the
oil and small amount of volatile components of the oil pass out with
the steam

6. EQUILIBRIUM RELATIONS
 Equilibrium relation between pA and xA can be expressed by Henry’s law at low
concentrations:
pA=HxA
 H: Henry’s law constant in atm/mole fraction
 If both sides are devided by total pressure (P) in atm:
yA=H’xA
 H’: Henry’s law constant in mole frac gas/mole frac liquid = H/P
 Data for some common gases with water can be found in Appendix A.3-18

7. GAS-LIQUID EQUILIBRIUM DATA
 Example: SO2-air-water
 An amount of gaseous SO2, air
and water are put in closed
container
 Shaken repeatedly at a given
temperature until equilibrium is
reached
 Samples of the gas and liquid
are analyzed to determine the
partial pressure pA in atm of SO2
in the gas and mole fraction xA
in the liquid
 Fig 10.2.1 shows a plot of data
from appendix A3 of partial
pressure of SO2 in vapor in eqm
with mole fraction of SO2 in liq

9. HENRY’S LAWVALIDITY
 Henry’s law only valid at low concentration
 The SO2-water system follow Henry’s law up to a concentration
xA of about 0.005 where H=29.6 atm/mol frac
 In general up to a total pressure of about 5 x 105
Pa (5 atm) the
value of H is independent of P.

10. EXAMPLE 10.2.1 P 628
Find the concentration of oxygen (A) in water at 298K, 1
atm.The air and water is in equilibrium. Henry’s law
constant is 4.38 x 104
atm/mol fraction.
pA in air is 0.21 atm.Why?
0.21=HxA=4.38 X 104
xA
xA=4.80 X10-6
mol O2 is dissolved in 1.0 mol water plus
oxygen or 0.000853 part O2/100 parts water

11. EXAMPLE 2
The partial pressure of CO2 in air is 1.333 x 104 Pa and the total
pressure is 1.133 x 105 Pa. the gas phase is in equilibrium with a
water solution at 303K. What is the value of xA for CO2 in
equilibrium in the solution?

12. EXAMPLE 3
At 303K, the concentration of CO2 in water is 0.90 x 10-4
kg
CO2/kg water. Using the Henry’s Law constant, what partial
pressure of CO2 must be kept in the gas to prevent the CO2 from
vaporizing from the aqueous solution?

13. SINGLE-STAGE EQUILIBRIUM
CONTACT
 2 different phase are brought into intimate contact with each
other and then are separated.
 Intimate mixing occurs during the time of contact.
 Various components diffuse and redistribute themselves.
 If mixing time is long enough, the components are essentially at
equilibrium in the two phases after separation.

14. Overall balance:
Lo +V2 = L1 +V1 = M
L andV in kg and M is total kg
Component balance:
If three components (A, B, C) are present:
LoxAo +V2yA2 = L1xA1 +V1yA1 = MxAm
LoxCo +V2yC2= L1xC1 +V1yC1 = MxCm
How about balance for component B?
 Not needed because xA+xB+xC = 1.0

15. SINGLE STAGE
 Gas phase (V) : soluteA, inert air B
 Liquid phase (L) : inert water C
 Assumptions:
 Air is insoluble in the water phase
 Water does not vaporize to the gas phase
 Gas phase is a binary A-B
 Liquid phase is a binary A-C
 Balance onA – Operating Line:
 L’: moles inert water C
 V’: moles inert air B
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A
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A
A
A
y
y
V
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x
L
y
y
V
x
x
L

16. SINGLE STAGE
 Need Equilibrium data to solve. Given by Henry’s Law
 If solution is not dilute, equilibrium data is in the form of plot pA
or ya versus xa is available (Like Fig 10.2.1)
1
1 ' A
A x
H
y 

17. EXAMPLE 10.3.1 P 630
A gas mixture at 1.0 atm containing air and CO2 is contacted in a
single-stage mixer continuously with pure water at 293K. The
two exit gas and liquid streams reach equilibrium. The inlet gas
flow rate is 100 kgmol/h, with a mole fraction of COS 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 not vaporize to the gas phase.

19. EXAMPLE 5
A gas mixture at 2.026 x 105
Pa total pressure containing air and
SO2 is brought into contact in a single stage equilibrium mixer
with pure water at 293K. The partial pressure of SO2 in the
original gas is 1.52 x 104
Pa. the inlet gas contains 5.70 total
kgmol and the inlet water 2.20 total kgmol. The exit gas and
liquid leaving are in equilibrium. Calculate the amounts and
compositions of the outlet phases.

20. COUNTERCURRENT MULTIPLE
CONTACT STAGE
Overall balance
L0 +VN+1 =LN +V1 = M
Overall component balance
L0x0+VN+1yN+1 =LNxN +V1y1 = MxM

21. COUNTERCURRENT MULTIPLE
CONTACT STAGE
Total balance over the first n stage:
L0 +Vn+1 = Ln +V1
Component balance over the first n stages
L0x0 +Vn+1yn+1 = Lnxn +V1y1
Solving yn+1:
21
1
0
0
1
1
1
1
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n
n
n
n
n
V
x
L
y
V
V
x
L
y

22. COUNTERCURRENT CONTACT
WITH IMMISCIBLE STREAMS

23. EXAMPLE 10.3.2 P 633
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 kgmol/h, and the total inlet
pure water flow to be used to absorb the acetone is 90.0 kgmol
H2O/h. The process is to operate isothermally at 300 K and a
total pressure of 101.3 kPa. The equilibrium relation for the
acetone (A) in the gas-liquid us yA=2.53xA. Determine the
number of theoretical stages required for this separation.

24. 24
Example
1
N=?
yN+1
1 mol%
Acetone in air
xN
y1
x0
VN+1=
30 kgmol/h
L0=90.0 kgmol/h
300K
P=101.3 kPa
yA=2.53xA
V1=29.7+ 0.1(0.3)
=29.73 kgmol/h
Amount of
Acetone =
yAN+1VN+1
=0.01(30.0)
=0.30 kgmol/h
Amount of air
=30-0.3=29.7 kg mol/h
LN=90+ 0.9(0.3)
=90.27 kgmol/h
yA1=0.030/29.73
=0.00101
xAN=0.27/90.27
=0.0030

26. 26
Since L andV are essentially constant throughout the
system, the slope of the operating line Ln/Vn+1
is constant
5.2 theoretical stages
yA=2.53xA

27. EXAMPLE 7
 Repeat example 6 using the same conditions but with the
following change: use a pure water flow to the tower of 108 kg
mol H2O/h. Determine the number of stages required.

28. ABSORPTION IN PLATE AND PACKEDTOWERS

29. DESIGN OF PACKEDTOWER

30. PRESSURE DROP AND FLOODING
 Flooding velocity is an upper limit to the rate of gas flow given type
and size of packing and liquid flowrate for possible operation of the
tower
 Pressure drop α (gas flow rate)1.8
 Loading point is the intermediate point where the gas start to
hinder liquid flow, and local accumulation of liquid start to appear in
the packing. The pressure drop of the gas starts to rise at a faster
rate.
 At the flooding point, liquid can no longer flow down through the
packing and is blown out with the gas
 In actual operating tower, the optimum economic gas velocity is
about one-half or more of the flooding velocity.

31. PRESSURE DROP IN RANDOM
PACKING

32. FLOODING PRESSURE DROP
• 1 in.H2O/ft = 83.33 mm H2O/m height of packing
• Fp is packing factor in ft-1
and is given inTable
10.6-1
This equation can be used for 9< FP <60
For Fp > 60, Δpflood= 2 in.H2O/ft
packing
of
height
/
in.H
115
.
0 2
7
.
0
ft
O
F
P p
flood 


33. PROCEDURESTO CALCULATE
LIMITING FLOW RATES ANDTOWER
DIAMETER
 Select a suitable random/structured packing, find the FP value
 A suitable liquid-to-gas ratio GL/GG is selected along with the total gas flow
rate.
 Calculate pressure drop
 Calculate flow parameter, capacity parameter is read off from Fig 10.6-5 or
10.6-6
 Using capacity parameter, value of GG is obtained
 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 Fig 10.6-5 or
10.6-6
 Knowing the total gas flow rate and GG, the tower cross-sectional area and
ID can be calculated.

34. EXAMPLE 8
Ammonia is being absorbed in a tower using pure water at
25 C and 1.0 atm. The feed rate is 1440 lb
◦ m/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.
(a) 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.
(b) repeat using Mellapak 250Y structured packing

35. DESIGN OF PLATETOWER

36.  Similar to countercurrent multiple stage process

37. MATERIAL BALANCE EQUATIONS
Graphical determination of the number of trays: the plot y vs x
of the material balance equation will give a curve line unless:
• (1-x) and (1-y) ~ 1 (dilute solution)
• The line will be straight with slope=L’/V’
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stage
n
to
up
Balance
1
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balance
Overall
y
y
V
x
x
L
y
y
V
x
x
L
y
y
V
x
x
L
y
y
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L
n
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N
N
N
N

38. EXAMPLE 9
A tray tower is to be designed to absorb SO2 from an air stream
by using pure water at 293K. The entering gas contains 20 mol%
SO2 and that leaving 2 mol % at a 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 operates at 293K.