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Dr.T.M 1
Absorption
and
Stripping
Dr.T.M 2
The objective is:
To understand the absorption( Gas-Liquid) process
To make the material balance for a absorption system
To understand the concept of equilibrium stages and their
estimation
To understand the stripping process
To make the material balance for a stripping system
To understand the concept of equilibrium stages and their
estimation
Dr.T.M 3
Introduction
A mass transfer operation – same category as distillation
Exclusive to gas-liquid separation
Distillation uses the VLE, i.e. difference in boiling temperatures
Absorption uses the GLE, i.e. solubilitygas is absorbed into liquid
liquid solvent or absorbent
gas absorbed solute or absorbate
Stripping is reverse of absorptionliquid absorbed into gas
act of regenerating the absorbent
Dr.T.M 4
Introduction
Absorption in the industry
Air pollution control – scrubbing of SO2 , NO2 , from combustion exhaust (power plant flue gas)
Absorption of ammonia from air with water
Hydrogenation of edible oils – H2 is absorbed in oil and reacts with the oil in the presence of catalyst
Dr.T.M 5
How does it work?
Solvent
Solute with
inert gas
Good
product
unwanted gas
solution to
disposal or
recovery
This section can be
trayed or packed
Dr.T.M 6
How does it work?
Tray
tower Packed
tower
Dr.T.M 7
How does it work?Tray tower:
Absorption on each tray
Dr.T.M 8
How does it work?
Tray tower:
Types of traySieve Valve
Bubble
Cap
A full tray
Dr.T.M 9
How does it work?
Packed tower:
1. Structured packing
2. Random packing
Dr.T.M 10
How does it work?Packed tower:
Structured packing
Dr.T.M 11
How does it work?Packed tower:
Structured packing
Dr.T.M 12
How does it work?Packed tower:
Random packing
Dr.T.M 13
How does it work?Spray tower
Dr.T.M 14
How does it work?Bubble Column
Liquid solvent “bed”
Dr.T.M 15
• Entering gas (liquid) flow rate, composition,
temperature and pressure
• Desired degree of recovery of one or more
solutes
• Choice of absorbent (stripping agent)
• Operating pressure and temperature, and
allowable gas pressure drop
General Design Considerations
Dr.T.M 16
• Minimum absorbent (stripping agent) flow rate
and actual absorbent (stripping agent) flow rate
as a multiple of the minimum flow rate
• Number of equilibrium stages
• Heat effects and need for cooling (heating)
• Type of absorber (stripper) equipment
• Diameter of absorber (stripper)
General Design Considerations
Dr.T.M 17
The ideal absorbent should:
• have a high solubility for the solute
• have a low volatility
• be stable
• be noncorrosive
• have a low viscosity
• be nonfoaming
• be nontoxic and nonflammable
• be available, if possible, within the
process
Dr.T.M 18
The most widely used absorbents are:
• water
• hydrocarbon oil
• aqueous solution of acids and bases
The most widely used stripping agents are:
• water vapor
• air
• inert gases
• hydrocarbon gases
Dr.T.M 19
Equilibrium Contact Stages
Single
multiple
Dr.T.M 20
Single Equilibrium Stage
Single equilibrium stage system above
Mass balance:
L0 + V2 = L1 + V1
V1
L0
V2
L1
Dr.T.M 21
Single Equilibrium Stage
Mass balance: L0 + V2 = L1 + V1
Gas-liquid absorption – usually 3 components
involved. Let A, B and C be the components, then
L0xA0 + V2yA2 = L1xA1 + V1yA1
L0xC0 + V2yC2 = L1xC1 + V1yC1
and xA + xB + xC = 1.0
V1
L0
V2
L1
Dr.T.M 22
Single Equilibrium Stage
L0xA0 + V2yA2 = L1xA1 + V1yA1
L0xC0 + V2yC2 = L1xC1 + V1yC1
xA + xB + xC = 1.0
To solve these 3 equations – their
equilibrium relations will be required
V1
L0
V2
L1
Dr.T.M 23
Single Equilibrium Stage
Gas phase – V
Components – A (solute) and B (inert)
Liquid phase – L
Components – C
In gas phase you have binary A-B
In liquid phase you have binary A-C
V1
L0
V2
L1
Dr.T.M 24
Single Equilibrium Stage
Only A redistributes between both phases.
Take mole balance of A:
where L’ moles of C and V’ moles of B
V1
L0
V2
L1
1
1'
1
1'
2
2'
0
0'
1111 A
A
A
A
A
A
A
A
y
yV
x
xL
y
yV
x
xL
Tutorial: Derive/Proof this
Dr.T.M 25
Single Equilibrium Stage
V1
L0
V2
L1
1
1'
1
1'
2
2'
0
0'
1111 A
A
A
A
A
A
A
A
y
yV
x
xL
y
yV
x
xL
To solve this, equilibrium relationship
between yA1 and xA1 is needed.
Use Henry’s Law: yA1= H’ xA1
H’ – Henry’s law constant (obtainable in
Handbooks eg Perry’s)
Dr.T.M 26
Countercurrent Multiple-Contact Stages
V1 V2 Vn+1V3 VnVN+1VN
L0 L1L2 Ln-1 Ln LN-1 LN
1 2 n N
Total overall balance:
L0 + VN + 1 = LN + V1 = M where M is the total flow
Overall Component Balance:
L0xo + VN + 1 yN +1 = LNxN + V1 y1 = Mxm
Dr.T.M 27
Making a total balance over the first n stages,
L0 + Vn + 1 = Ln + V1
Making a component balance over the first n stages,
L0xo + Vn + 1 yn +1 = Lnxn + V1 y1
Solving for yn +1,
1
0011
1
1
nn
nnn
V
xLyV
V
xLy Operating Line
Dr.T.M 28
Countercurrent contact with immiscible streams
• An important case in which the solute A is being transferred occurs when
the solvent stream V contains components A and B with no C and solvent
stream L contains A and C with no B.
x0 x1 x3x2 x4
1
2
3
4
Operating line
Equilibrium line
y1
y2
y3
y4
yN + 1
yN + 1
y4
y3
y2
y1x0
x1
x2
x5
xN
N = 4
3
2
1
Note: If the streams L and V are dilute in key
species, the operating line is a straight line
Dr.T.M 29
Analytical Equations for Countercurrent Stage
Contact (Kremser Equation)
• When the flow rates V and L in a countercurrent
process are essentially constant, the operating line
equation becomes straight
• If the equilibrium line is also a straight line over the
concentration range, simplified analytical expressions
can be derived for number of equilibrium stages in a
countercurrent stage process
Overall component balance on component A:
L0xo + Vn + 1 yn +1 = Lnxn + V1 y1
Dr.T.M 30
Rearranging,
LNxN - VN + 1 y N + 1 = Loxo - V1y1
Component balance for A on the first n stages,
Rearranging,
Loxo - V1y1 = Lnxn - Vn + 1y n+1
L0xo + Vn + 1 yn +1 = Lnxn + V1 y1
Thus,
LNxN - VN + 1 y N + 1 = Lnxn - Vn + 1y n+1
Dr.T.M 31
Since the molar flows are constant, Ln = LN =
constant = L and Vn+1= VN+1 = constant = V.
L (xn - xN) = V(yn+1 - yN+1)
Since yn + 1 and xn + 1 are in equilibrium and the equilibrium
line is straight, yn + 1 = mxn + 1. Also, yN + 1 = mxN + 1
Substituting mxn + 1 for yn + 1 and calling A = L/mV,
NN
nn Axm
yAxx 1
1
(A)
(B)
Dr.T.M 32
For transfer of solute A from phase V to L (absorption)
11
1
01
11
N
N
N
N
A
AA
mxy
yy
A
AAmxy
mxy
N
N
ln
111ln
01
01
01
11
mxy
yyN N
When A = 1,
Dr.T.M 33
Solving (B),
For transfer of solute A from phase L to V (stripping),
1)/1(
)/1()/1(
)/( 1
1
1
N
N
No
No
A
AA
myx
xx
)/1ln(
)1(/
/ln
1
10
A
AAmyx
myx
N NN
N
When A =1,
myx
xxN
NN
N
/1
0
Dr.T.M 34
If equilibrium line is not straight,
11
01
1NN
NN1N
Vm
L Aand
Vm
L A whereAAA
Dr.T.M 35
Example: Number of stages by analytical equation
It is desired to absorb 90% of the acetone in a gas containing 1.0 mol%
acetone in air in a countercurrent stage tower. The total inlet gas flow to the
tower is 30.0 kg mol/h, and the total inlet pure water flow to be used to
absorb the acetone is 90 kg mol H2O/h. The process is to operate
isothermally at 300K and a total pressure of 101.3 kPa. The equilibrium
relation for the acetone in the gas-liquid is yA = 1.5 xA. Determine the
number of theoretical stages required for this separation by graphical
method and compare it with Kremser equation.
Estimate the minimum solvent ratio for the process.
If 2 times of minimum solvent is used estimate the number of theoretical
stages required.
Dr.T.M 36
A gas mixture of air and CO2 is contacted in a multi stage mixer
with pure water at atmospheric conditions. The exit gas and liquid
streams are in equilibrium. The inlet gas and liquid flow rate are
100 kg/h and 300 kg/h respectively. The entering gas contains 0.2
mole fraction of CO2. If 90 % of CO2 is observed, Calculate the
composition of the leaving liquid and suggest the coordinates of
the operating line. Estimate the number of stages required for the
absorption.
Assume the equilibrium relation is y = 2.52x. Estimate the number
of stages required if, 1.5 times of minimum solvent is required..
Problem - 1&
Dr.T.M 37
An adsorption oil containing 0.12 moles of benzene per mole of
benzene free oil is to be stripped by using a superheated steam
at 121.1oC and at 1 atmospheric pressure. For every 200 kgmol
of benzene free oil, 100 kgmol of pure steam was used. The
outlet concentration of oil should not exceed 0.005 mol of
benzene per mole of benzene free oil. Estimate the number of
theoretical stages required for the stripping process. The
equilibrium data are given below:
X’ 0.02 0.04 0.06 0.08 0.10 0.12 0.13
Y’ 0.07 0.13 0.22 0.30 0.40 0.51 0.58
Dr.T.M 38
Solution:
V1 = 29.73 kg mol/h, yA1 = 0.00101, L0 = 90.0 and xA0 = 0.
Thus,
A1 = L/mV = L0/mV1 = 90.0 / (2.53 x 29.73) = 1.20
At stage N, VN + 1 = 30.0, yAN +1 = 0.01, LN = 90.27, and xAN = 0.00300
Thus,
AN = LN/mVN + 1 = 90.27/(2.53 x 30.0) = 1.19
The geometric average, A = (A1AN)1/2 = (1.20x1.19)1/2 = 1.195
For absorption and by using kremser equation,
stages 5.04(1.195) ln
1.195
1
1.195
11
2.53(0)0.00101
2.53(0)0.01ln
N
Dr.T.M 39
Graphical Equilibrium-Stage Method for Trayed
Towers
• Consider the countercurrent-flow, trayed tower for
absorption (or stripping) operating under isobaric,
isothermal, continuous, steady-state flow conditions
• Phase equilibrium is assumed to be achieved at
each tray between the vapor and liquid streams
leaving the tray. ====> equilibrium stage
• Assume that the only component transferred from
one phase to the other is the solute,
• For application to an absorber, let:
Dr.T.M 40
L’ = molar flow rate of solute-free absorbent
G’ = molar flow rate of solute-free gas (carrier gas)
X = mole fraction of solute to solute-free absorbent in the
liquid
Y = mole ratio of solute to solute-free gas in the vapor
Note that with these definitions, values of L’ and G’ remain constant
through the tower, assuming no vaporization of absorbent into carrier gas
or absorption of carrier gas by liquid. For the solute at any equilibrium
stage, n,
nn
nn
n
nn
XX
YY
x
yK
1/
1/
Dr.T.M 41
X0,L’ Y1,G’
YN+1,G’ XN,L’
1
n
N
(bottom)
(top)
Operating line
Equilibrium curve
XN + 1,L’ YN,G’
Y0,G’ X1,L’
1
n
N
top
bottom
Operating line
Equilibrium curve
absorber Stripper
O.P: YN + 1 = Xn(L’/G’)+ Y1 - X0(L’/G’) Yn = Xn + 1(L’/G’) + Y0 - X1(L’/G’)
Dr.T.M 42
Minimum Absorbent Flow Rate
1
N
X0 Y1
YN + 1 XN
Moles solute/mole solute-free liquid, X
Mole
s s
olu
te/m
ole
solu
te-f
ree g
as, Y
YN + 1 (gas in)
XN
(for Lmin)
X0
Y1
(gas out)
Op
era
tin
g lin
e 1
Dr.T.M 43
Consider, for n = N
X0L’ + YN + 1G’ = XNL’ + Y1G’
or
(C)0
11''
XX
YYGL
N
N
For stage N, for the minimum absorbent rate,
NN
NNN
XX
YYK
1/
1/ 11(D)
Solving for XN in (D) and substituting it into (C) gives
Dr.T.M 44
011
11'
min1/
'
XKKYY
YYGL
NNNN
N
For dilute solution, where Y y and X x, (E) becomes
(E)
01
11min ''
xK
y
yyGL
N
N
N
If the entering liquid contains no solute, that is, X0 0
L’min = G’KN(fraction of solute absorbed)
For Stripper,
stripped solute of fraction'
'min
NK
LG
Dr.T.M 45
Number of Equilibrium StagesX0’ Y1’
YN+1 XN
1
N
XN + 1, YN,
Y0, X1,
1
N
Stage 1
(top)
Stage 1
(bottom)
x0
Y1
YN + 1
xN
Y0
YN
x1xN + 1
Dr.T.M 46
Example:
When molasses is fermented to produce a liqour
containing ethyl alcohol, a CO2-rich vapour containing
a small amount of ethyl alcohol is evolved. The alcohol
can be recovered by absorption with water in a sieve-
tray tower. For the following conditions, determine the
number of equilibrium stages required countercurrent
flow of liquid and gas.K-value=0.57. Given
L/V=1.5(L/V)min . State your assumption.
Entering gas: 180 kmol/h; 98% CO2, 2% ethyl alcohol,
Entering liquid absorbent:100% water.
Required recovery of ethyl alcohol:97%
Dr.T.M 47
Packed absorption tower design
Packed-tower performance is often analysed on the
basis of equivalent equilibrium stages using packing
Height Equivalent to a Theoretical (equilibrium ) Plate
(staged),
OGOG
tt
NHz
N
z
NstagesmequilibriuequivalentofNumber
zheightPackedHETP
)(
)(
where HOG is the overall Height Transfer Unit (HTU) and
NOG is the overall Number of Transfer Unit (NTU)
Dr.T.M 48
SaK
VH
y
OG'
HOG , Height Transfer Unit (HTU)
V; average liquid flow rate
Ky’; Overall transfer coefficient
a: area for mass trasfer per unit volume of packed bed,
S; cross sectional area of the tower
NOG , Number of Transfer Unit (NTU)
A
A
AKxy
Kxy
A
A
Ninout
inin
OG 1
11ln
K ; equilibrium ratio
A, absorption factor = L/KV
Dr.T.M 49
Example:
When molasses is fermented to produce a liqour
containing ethyl alcohol, a CO2-rich vapour containing
a small amount of ethyl alcohol is evolved. The alcohol
can be recovered by absorption with water in a packed
tower. The tower is packed with 1.5in metal Pall rings.
K-value=0.57. Given L/V=1.5(L/V)min . If HOG = 2.0 ft,
determine the required packed height.
Entering gas: 180 kmol/h; 98% CO2, 2% ethyl alcohol,
Entering liquid absorbent:100% water.
Required recovery of ethyl alcohol:97%
Dr.T.M 50
b. A tray tower is used to absorb SO2 from an air stream by using pure water at 25oC.
The entering gas contains 20 mole percent of SO2. The tower is designed to absorb
90% of SO2. The flow rate of pure air is 150 kg/h.m2. The entering water flow
rate is 6000 kg water/h.m2. The equilibrium data are on solute free basis are given below
Equilibrium data for SO2 - water
Mole fraction of SO2
in water, X
Mole fraction of SO2
in vapour, Y
0.00000 0.00000
0.00150 0.03420
0.00200 0.05140
0.00280 0.07750
0.00420 0.12140
0.00700 0.21200
Dr.T.M 51
i. Estimate the concentration of SO2 in the exit water leaving the
tower.
ii.i. Estimate the number of theoretical stages required for the
desired absorption.
iii. If the overall efficiency of the tower is 40%, how many number
of actual stages are required.
I
Dr.T.M 52
Since the equilibrium data are given in molar units, calculate the molar flow rates
V’ = 150/29 = 5.18 Kg mol inert air/m2 h
L’ = 6000/18.0 = 333 Kg mol inert water/ m2 h
Y N+1 = 0.20; Y1 = 0.02
X0 = 0; XN = ????
substituting into the material balance equation
02.01
02.018.5
1333
20.01
20.018.5
01
0333
N
N
x
x
XN = ----------- ????
yN + 1
y4
y3
y2
y1x0
x1
x2
x5
xN
N = 4
3
2
1
Dr.T.M 53
Number of theoretical trays = 2.4
x0 x1 x3x2 xN
1
2
3
4
Operating line
Equilibrium line
y1
y2
y3
y4
yN + 1
Dr.T.M 54
Acetone is being absorbed by water in a packed tower having a cross-sectional area of 0.186m2
at 293 K and 101.32 kPa. The inlet air contains 2.6 mol% acetone and outlet 0.5 mol%. The gas flow is 13.65 kgmol air/h. The pure water flow is 45.36 kgmol/h. Film coefficients for the given flows in the tower are k’ya = 3.78x10-2
kgmol/s.m3.mol frac and k’xa = 6.16x10-2. Calculate packing height, z. The equilibrium relation is given by y = 1.186x.
Packed Tower Design
Dr.T.M 55
Solution: First calc HOG
Vav = (V1 + V2)/2 = 3.852 x 10-3 kg mol/s
Packed Tower Design
SaK
VH
y
OG'
kgmol/s 10 x 3.892 13.65/3600 3-
026011 1
1.y
'VV
kgmol/s 10 x 3.811 13.65/3600 3-
005011 2
2.y
'VV
Dr.T.M 56
Solution: First calc HOG
m is from y = mx = 1.186x relation established
K’ya = 2.19 x 10-2 kgmol/s.m3.mol frac
So,
HOG = 3.852 x 10-3/(2.19 x 10-2 x 0.186) = 0.947 m
Packed Tower Design
SaK
VH
y
OG'
2-2- 10 x 6.16
1.186
10 x 3.78
111
a'k
m
a'ka'K xyy
45.7
Dr.T.M 57
Solution: Next calc NOG :
A = L/mV = (45.36/3600)/(1.186)(3.852x10-3)
= 2.758
NOG = 1.28 transfer units
Packed Tower Design
Amxy
mxy
AAN
inout
ininOG
111ln
ln
1
758.2
1
x0186.1005.0
x0186.1026.0
758.2
11ln
758.2ln
1OGN
Dr.T.M 58
Solution:
NOG = 1.28 transfer units
HOG = 0.947 m
So,
z = 0.947 x 2.043 = 1.935 m
Packed Tower Design
OGOG NHz
Dr.T.M 59
A tray tower is used to absorb the ethanol vapors from an inert gas
stream using pure water at 30oC and atmospheric pressure. The
concentration of ethanol vapors in the gas stream is 2.2 mol %.
The gas stream flow rate is 100.Kg mol/h. It is desired to recover
90% of the alcohol. The equilibrium relation is Y* = 0.68 X. If
1.5 times of the minimum water flow rate is used, estimate the
number of theoretical stages required for the absorption. Calculate
the number of stages using kremser’s equation.
Problem - 1 (Assignment)
Dr.T.M 60
A hydrocarbon oil containing 4.0 mol % propane is being
stripped by direct superheated steam in a tower to reduce the
propane content to 0.2 mol %. A total of 11.42 kg mol of direct
steam is used for 300 kgmol of entering liquid. An equilibrium
relation of Y = 25x may be assumed. Steam can be considered
as an inert gas and will not condense. Find out the number of
theoretical stages needed for the operation
Problem - 2 (Assignment)
Dr.T.M 61
Problem- 3
When molasses is fermented to produce a liqour
containing ethyl alcohol, a CO2-rich vapour containing
a small amount of ethyl alcohol is evolved. The alcohol
can be recovered by absorption with water in a sieve-
tray tower. For the following conditions, determine the
number of equilibrium stages required countercurrent
flow of liquid and gas.K-value=0.57. Given
L/V=1.5(L/V)min . State your assumption.
Entering gas: 180 kmol/h; 98% CO2, 2% ethyl alcohol,
Entering liquid absorbent:100% water.
Required recovery of ethyl alcohol:97%