5.Absorption and Stripping L1
Transcript of 5.Absorption and Stripping L1
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Lecture 1: Absorption & Stripping - 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. solubility
– gas is absorbed into liquid– liquid solvent or absorbent – gas absorbed solute or absorbate
• Stripping is reverse of absorption– liquid absorbed into gas– act of regenerating the absorbent
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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
Continued…
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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
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How does it work?
Tray tower
Packed tower
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How does it work?
Tray tower:
Absorption on each tray
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How does it work?
Tray tower:
Types of traySieve Valve
Bubble Cap
A full tray
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How does it work?
Packed tower:
1. Structured packing
2. Random packing
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How does it work?
Packed tower:
Structured packing
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How does it work?
Packed tower:
Structured packing
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How does it work?
Packed tower:
Random packing
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How does it work?
Spray tower
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How does it work?
Bubble Column
Liquid solvent “bed”
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• 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
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• 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
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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
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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
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Equilibrium Contact Stages: Single Equilibrium Stage
• Single equilibrium stage system above
• Mass balance:
L0 + V2 = L1 + V1 = M
V1
L0
V2
L1
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Single Equilibrium Stage
Mass balance: L0 + V2 = L1 + V1 = M
Gas-liquid absorption – usually 3 components
involved. Let A, B and C be the components, then
L0xA0 + V2yA2 = L1xA1 + V1yA1 = MxAM
L0xC0 + V2yC2 = L1xC1 + V1yC1 = MxCM
and xA + xB + xC = 1.0
V1
L0
V2
L1
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Single Equilibrium Stage
L0xA0 + V2yA2 = L1xA1 + V1yA1 = MxAM
L0xC0 + V2yC2 = L1xC1 + V1yC1 = MxCM
xA + xB + xC = 1.0
To solve these 3 equations – their
equilibrium relations will be required
V1
L0
V2
L1
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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
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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
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Single Equilibrium StageV1
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)
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Lecture 3: Single Equilibrium Stage: Example
A gas mixture at 1.0 atm abs containing air and CO2 is contacted in a single-stage mixer continuously with water at 293 K. The two exit gas and liquid streams reach equilibrium. The inlet gas flow is 100 kgmol/h, with a mole fraction of CO2 of yA2 = 0.20. The liquid flow rate is 300 kgmol/h water. Calculate the amounts and compositions of the two outlet phases. Assume that water does not vapourise to gas.
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Solution V1
L0
V2
L1
A gas mixture at 1.0 atm abs containing air and CO2 is contacted in a single-stage mixer continuously with water at 293 K. The two exit gas and liquid streams reach equilibrium. The inlet gas flow is 100 kgmol/h, with a mole fraction of CO2 of yA2 = 0.20. The liquid flow rate is 300 kgmol/h water. Calculate the amounts and compositions of the two outlet phases. Assume that water does not vapourise to gas.
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
Flow of water 300kgmol/h is L0 = L’
Flow of air, V’ must exclude CO2 thus V’ = V2(1-yA2) = 100(1-0.20) = 80 kgmol/h
Substitute in equation:
1
1
1
1
180
1300
20.01
20.080
01
0300
A
A
A
A
y
y
x
x
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Solution V1
L0
V2
L1
A gas mixture at 1.0 atm abs containing air and CO2 is contacted in a single-stage mixer continuously with water at 293 K. The two exit gas and liquid streams reach equilibrium. The inlet gas flow is 100 kgmol/h, with a mole fraction of CO2 of yA2 = 0.20. The liquid flow rate is 300 kgmol/h water. Calculate the amounts and compositions of the two outlet phases. Assume that water does not vapourise to gas.
1
1
1
1
180
1300
20.01
20.080
01
0300
A
A
A
A
y
y
x
x
1 eqn, 2 unknowns unsolvable
Use Henry’s law, yA1= H’ xA1 to eliminate an unknown.
yA1= 0.142 x 104/1.0 xA1 is substituted into the eqn and solve for xA1:
xA1= 1.41x10-4 and yA1= 0.20
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Solution V1
L0
V2
L1
A gas mixture at 1.0 atm abs containing air and CO2 is contacted in a single-stage mixer continuously with water at 293 K. The two exit gas and liquid streams reach equilibrium. The inlet gas flow is 100 kgmol/h, with a mole fraction of CO2 of yA2 = 0.20. The liquid flow rate is 300 kgmol/h water. Calculate the amounts and compositions of the two outlet phases. Assume that water does not vapourise to gas.
xA1= 1.41x10-4 and yA1= 0.20
We have V’ = V1(1-yA1) and L’ = L1(1-xA1) , so
41
'
1 1041.11
300
1 Ax
LL
20.01
80
1 1
'
1Ay
VV
300 kgmol/h
100 kgmol/h
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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 Mole Balance:
L0x0 + VN + 1 yN +1 = LNxN + V1 y1 = Mxm compare to single stage
Equilibrium Contact Stages: (multiple)Countercurrent Multiple-Contact Stages
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Making a total balance over the first n stages,
L0 + Vn + 1 = Ln + V1
Making a component balance over the first n stages,
L0x0 + Vn + 1 yn +1 = Lnxn + V1 y1
Solving for yn +1,
1
0011
11
nn
nnn V
xLyV
V
xLy Operating Line
Countercurrent Multiple-Contact Stages
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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. (Immiscible contact)
Countercurrent Multiple-Contact Stages
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yN + 1
y4
y3
y2
y1x0
x1
x2
x5
xN
N = 4
3
2
1
Countercurrent Multiple-Contact Stages
Consider a 4-stage contact
operating line equation
1
0011
11
nn
nnn V
xLyV
V
xLy
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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
Countercurrent Multiple-Contact Stages
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Example: Absorption of acetone in countercurrent stage tower: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 toal 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 (a) in the gas-liquid is yA = 2.53 xA. Determine the number of theoretical stages required for this separation
Countercurrent Multiple-Contact Stages
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Solution:
Given values are: yAN+1 = 0.01, xA0 = 0, VN+1 = 30.0 kg/h and L0 = 90.0 kg mol/h
Acetone material balance,
Amount of entering acetone = yAn + 1V N + 1 = 0.01(30.0) = 0.30 kg mol/h
Entering air = 30.0 - 0.3 = 29.7 kg mol/h
Acetone leaving in V1 = 0.10 (0.30) = 0.030 kg mol/h
Acetone leaving in LN = 0.90(0.30) = 0.27 kg mol/h
Countercurrent Multiple-Contact Stages
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Solution:
V1 = 29.7 + 0.0 = 29.73 kg mol air + acetone/h
yA1 = 0.030/29.73 = 0.00101
LN = 90.0 + 0.27 = 90.27 kg mol water + acetone/h
xAN = 0.27/90.27 = 0.00300
Since the flow of liquid varies only slightly from L0 = 90.0 at the inlet to LN = 90.27 at the outlet and V from 30.0 to 29.73, the slope of the operating line is essentially constant (O.L. straight line)
Countercurrent Multiple-Contact Stages
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Plot the x-y relations and draw the operating line calculated to find the number of stages
1
2
3
4
5
No. of theoretical stages are required = 5.2 stages
Countercurrent Multiple-Contact Stages
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Analytical Equations (Kremser Equation)
Premise:
• 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
Countercurrent Multiple-Contact Stages
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Overall balance: L0x0 + VN + 1 yN +1 = LNxN + V1 y1
Rearranging,LNxN - VN +1 yN +1 = Loxo - V1y1
Component balance on the first n stages,
L0x0 + Vn +1 yn +1 = Lnxn + V1 y1
Rearranging,
Loxo - V1 y1 = Lnxn - Vn +1 y n+1
Thus,
LNxN - VN + 1 y N + 1 = Lnxn - Vn +1 y n+1
Countercurrent Multiple-Contact StagesV1
V2 Vn+1V3 Vn VN+1VN
L0 L1 L2 Ln1 Ln LN-1 LN
1 2 n N
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Since the molar flows are constant, Ln = LN = constant = L and Vn+1= VN+1 = constant = V
Then, LNxN VN + 1 y N + 1 = Lnxn Vn +1 y n+1
simplifies to L(xN xn) = V(yN+1 yn+1) (A)
Since yn+1 and xn+1 are in equilibrium and the equilibrium line is
straight, yn+1= mxn+1. Similarly, yN+1= mxN+1
Substituting mxn+1 f or yn+1 and calling A = L/mV,
Eqn (A) becomes: NN
nn Axm
yAxx
1
1 (B)
Countercurrent Multiple-Contact Stages
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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(//
ln1
10
A
AAmyxmyx
N NN
N
When A =1,
myx
xxN
NN
N
/1
0
Countercurrent Multiple-Contact Stages
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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,
Countercurrent Multiple-Contact Stages
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If equilibrium line is not straight,
11
01
1
1
Vm
L A
Vm
L A
AAA
NN
NN
N
andwhere
Countercurrent Multiple-Contact Stages
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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 (a) in the gas-liquid is yA = 2.53xA. Determine the number of theoretical stages required for this separation by using Kremser equation
Countercurrent Multiple-Contact Stages
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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 = 0.0030
Thus, AN = LN/mVN + 1 = 90.27/(2.53 x 30.0) = 1.19
The geometric average,
A = (A1 AN)1/2 = (1.20 x 1.19)1/2 = 1.195
Countercurrent Multiple-Contact Stages
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For absorption and by using Kremser equation,
stages)(
)(
.). (ln
......
ln
N 0451951
19511
19511
10532001010
0532010
Countercurrent Multiple-Contact Stages
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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 vapour and liquid streams leaving the tray equilibrium stage
• Assume that the only component transferred from one phase to the other is the solute
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Graphical Equilibrium-Stage Method for Trayed Towers
For application to an absorber, let:
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
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Graphical Equilibrium-Stage Method for Trayed Towers
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/
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Lecture 4: 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 vapour and liquid streams leaving the tray equilibrium stage
• Assume that the only component transferred from one phase to the other is the solute
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Graphical Equilibrium-Stage Method for Trayed Towers
For application to an absorber, let:
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
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Graphical Equilibrium-Stage Method for Trayed Towers
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/
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Graphical Equilibrium-Stage Method for Trayed Towers
• You can obtain equilibrium data from this expression• How?• By assuming Henry’s and Dalton’s law are applicable
pi=Hxi yi=pi /P
• Use to get y/x • Then rearrange above expression to get Y=f (X)• Calculate Y and X, then make the X-Y plot
nn
nn
n
nn XX
YY
x
yK
1/
1/
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Graphical Equilibrium-Stage Method for Trayed TowersX0,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
absorberStripper
How would you obtain the
expressions for these?
Do mass balances
around unit
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Graphical Equilibrium-Stage Method for Trayed Towers
• Operating line for absorber:
Yn + 1 = Xn(L’/G’)+ Y1 - X0(L’/G’)
• For stripper:
Yn = Xn + 1(L’/G’) + Y0 - X1(L’/G’)
• It’s a straight line with slope L’/G’• Basically the terminal points (X,Y) for top and bottom can be used and a
straight line drawn between them
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Graphical Equilibrium-Stage Method for Trayed Towers - Min Absorbent Flow
Consider, for n = N
X0L’ + YN + 1G’ = XNL’ + Y1G’
or
0
11
XX
YY'G'L
N
N
For stage N, for the minimum absorbent rate,
NN
NNN X/X
Y/YK
1
1 11 (2)
Solving for XN in (2) and substituting it into (1)
(1)
1
N
X0 Y1
YN + 1 XN
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Graphical Equilibrium-Stage Method for Trayed Towers - Min Absorbent Flow
011
11
1 XKKY/Y
YY'GL
NNNN
N'min
For dilute solution, where Y y and X x, (3) becomes
(3)
01
11
xK
yyy
'G'L
N
N
Nmin
We obtain:
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Graphical Equilibrium-Stage Method for Trayed Towers - Min Absorbent Flow
If the entering liquid contains no solute, that is, X0 0
L’min = G’KN (fraction of solute absorbed)
For Stripper,
stripped solute of fractionN
min K
'L'G
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1
N
X0 Y1
YN + 1 XN
Moles solute/mole solute-free liquid, X
Mol
es s
olut
e/m
ole
sol
ute-
free
gas
, Y
YN + 1 (gas in)
XN
(for Lmin)X0
Y1
(gas out)
Ope
ratin
g lin
e 1
Ope
ratin
g lin
e 2
2x m
in a
bsor
bent
rat
eO
pera
ting
line
3
1.5x
min
abs
orbe
nt ra
te
Opera
ting
line
4
Min ab
sorb
ent r
ate
Graphical Equilibrium-Stage Method for Trayed Towers - Min Absorbent Flow
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Number of Equilibrium Stages
Similar to Distillation, except start
step from bottom of graph
X0’ Y1’
YN+1 XN
1
N
XN + 1, YN,
Y0, X1,
1
N
Equilib
rium
cur
ve
Equilib
rium
cur
ve
Operatin
g line
Ope
ratin
g lin
e
Stage 1(top)
Stage 1(bottom)
x0
Y1
YN + 1
xN
Y0
YN
x1xN + 1
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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
)N(stagesmequilibriuequivalentofNumber
)z(heightPackedHETP
where
HOG is the overall Height Transfer Unit (HTU) and
NOG is the overall Number of Transfer Unit (NTU)
Packed Tower Design
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SaK
VH
yOG '
HOG , Height Transfer Unit (HTU)
V average liquid flow rateKy’ Overall transfer coefficient
a area for mass transfer per unit volume of packed bed,
S cross sectional area of the tower
NOG , Number of Transfer Unit (NTU)
Amxy
mxy
Aln
A
Ninout
ininOG
111
11
1
Packed Tower Design
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Some issues in packed tower design:
1. Pressure drop across packing
2. Flooding – gas velocity < flooding velocity
3. Types of packing – usually structured, proprietary design
Packed Tower Design
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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/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.
Packed Tower Design
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Solution: First calc HOG
y is obtained from x-y plot
Vav = (V1 + V2)/2 = 3.852 x 10-3 kg mol/s
Packed Tower Design
SaK
VH
yOG '
kgmol/s 10 x 3.892 13.65/3600 3-
026011 11 .y
'VV
kgmol/s 10 x 3.811 13.65/3600 3-
005011 22 .y
'VV
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Solution: First calc HOG
m is from y = mx = 1.186x relation established
K’ya = 2.19 x 10-2 kgmol/s.m2.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
yOG '
2-2- 10 x 6.16
1.186
10 x 3.78
111
a'k
m
a'ka'K xyy
45.7
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Solution: Next calc NOG :
A = L/mV = (45.36/3600)/(1.186)(3.852x10-3) = 2.758
NOG = 2.043 transfer units
Packed Tower Design
Amxy
mxy
Aln
A
Ninout
ininOG
111
11
1
7582
1
18610050
18610260
7582
11
75821
1
1
...
..
.ln
.
NOG x0
x0
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Solution:
NOG = 2.043 transfer units
HOG = 0.947 m
So,
z = 0.947 x 2.043 = 1.935 m
Packed Tower Design
OGOG NHz