gas absorption in packed tower (s1 2015) (note)
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NotesTRANSCRIPT
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1CHE3165
Separation Processes
S1 2015
GAS ABSORPTION IN PACKED
TOWERS
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2Leaning Outcomes
By the end of the lecture you should be able to:
1. Define rate based method for packed columns
2. Define Height Equivalent to a Theoretical Plate (HETP) and explain
how it and the number of equilibrium stages differ with the height of a
transfer unit (HTU) and number of transfer units (NTU)
3. Explain the differences between loading point and flooding point in a
packed column
4. Estimate the packed height
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3Contents
1 Introduction
2 Mole balance
3 Absorption of very dilute mixtures
4 Transfer unit
5 Absorption of concentrated mixtures
6 Absorption of semi-dilute mixtures
7 Controlling resistance
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41. Introduction
Absorption:
Solute A absorbed from the gas phase into the liquid phase OR a
process involves molecular/mass transfer of solute A through a
stagnant, non-diffusing gas B into a stagnant liquid C
Gas-Liquid System: Solute transfer from Gas Liquid
Adsorption:
Components of a liquid or gas stream adsorbed on the surface or in the
pores of a solid adsorbent.
Gas/Liquid Solid System: Solute transfer from Gas/LiquidSolid
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5 Stripping:
Reverse of absorption
Liquid Gas System: Solute transfer from Liquid Gas
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6Tray and packed bed columns
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7yo, V, Vo
yi, V, Vi
xi, L, Li
xo, L, Lo
L = Liquid stream total flow rate
V = Gas stream total flow rate
L = Inert (carrier) liquid flow rateV = Inert gas flow rateyA = Mass/mole fraction of A in gas
stream
xA = Mass/mole fraction of A in liquid
stream
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8Example 1 (Geankoplis: Example 10.3-2)
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 90kg mol H2O/h.
yo, V, Vo
yi, V, Vi
xi, L, Li
xo, L, Lo
Identify all the unknowns: xi, L, Li, yi, V, Vi, yo, Vo, xo, Lo
Vi = 30 kgmol/hyi = 0.01V = 30 (1-0.01) = 29.7 kgmol/hL = 90 kgmol/hxi = 0 Li = 90 kgmol/hAc in V0 = 30 (0.01) (1-0.9) = 0.03 kgmol/hAc in L0 = 30 (0.01) (0.9) = 0.27 kgmol/hV0 = 29.7 + 0.03 kgmol/h = 29.73 kgmol/hL0 = 90 + 0.27 kgmol/h = 90.27 kgmol/hy0 = 0.03/29.73 = 0.00101x0 = 0.27/90.27 = 0.003
Solution:
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9Example 2 (Geankoplis: Example 10.3-1)
A gas mixture at 1.0 atm pressure abs containing air and CO2 is contacted in a single-stage mixer
continuously with pure water at 293 K. The two exit gas and liquid streams reach equilibrium. The inlet
gas flow rate is 100 kgmol/h, with a mole fraction of CO2 of yA2 = 0.20. The liquid flow rate entering is 300
kgmol water/h. Calculate the amounts and compositions of the two outlet phases. Assume that water
does not vaporize to the gas phase.
V1?yA1?
V=100yA2=0.2V =?
L=300xA2=0
L1?XA1?
Solution:
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10
Gas
V1, y1
V2, y2Gas
L2, x2Liquid
Liquid
L1, x1
z1=0
z2=Z
(1)
(2)
V and L = total molar flow rates
Packing
A total interfacial area (m2)
a interfacial area per unit volume of packed tower (m2/m3)
S cross sectional area of tower (m)
Z Bed height (m)
Function of packing is to generate largest
possible interfacial area for the smallest
possible gas pressure drop.
A = a x vol. of packed tower
= a x (S x Z)
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Packed Tower Operation
Flooding velocity:
Upper limit to the rate of gas
flow. Liquid can no longer flow
Optimum economic: V 0.5 flooding velocity = f(equipment
cost, P, processing variables)
Loading point:
Gas flow rate where liquid down
flow starts to be hindered by
gas
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Packing Materials
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Packing Materials
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Packing Materials
http://finepacstructures.tradeindia.com
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Packing Materials
Random Structured
Raschig rings
and saddles
Through flow
Relative cost Low Moderate High
Pressure drop Moderate Low Very low
Efficiency Moderate High Very high
Vapor capacity Fairly high High High
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Example 3 (Geankoplis: Example 10.6-2)
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 293 K. Lxo
Vyn+1
y1
xn ??
Solution
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2. Mole Balance
Mole balance for solute over differential volume of tower:
- We can choose either gas or liquid phase.
VY = V(Y+dY) + d(in at z) (out at z+dz) (transfer to liquid)
VdY = - d
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Write in terms of flux and area, and Y in terms of y:
nd- dY V '
N.dA- y-1
ydV '
dz NaS- y)-(1
dyV
2
'
dz NaS- y)-(1
dyV
Volume
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Separating variables
Integrating from (1) to (2)
This is the most general equation relating total packed height (z) to gas-
phase variables.
y)-(1
dy
NaS
Vdz
Zy)-(1
dy
NaS
Vdz
z2
z1
2
1
y
y
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If we balance the liquid phase alone, an analogous equation is obtained:
To evaluate the integrals, we need V(y), N(y) or L(x), N(x)
x2
x1x)-(1
dx
NaS
LZ
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For gas phase analysis:
Solute flux
Choose overall Ky as typical case
AiAG
*AAG
iy
*y
P - Pk N
or P-PK N
or yykN
or yyKN
**LM
'y*
BLM
'y
yyy-1
K yy
y
KN
Across gas film
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varies up the tower
f(y)
*y-1
y-1ln
*y-1-y-1y-1 *LM
*LMy-1
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yyKN *y
2
1
y
yy)-(1
dy
NaS
Vz
*
*LM
'y
yyy-1
KN
y2
y1
*
*LM
'y )y(y
dy
y)-(1
y-1
aSK
Vz
y2
y1
*2'y
*LM'
)y(yy)-aS(1K
dyy-1
S
Vz
y1V
V '
a may vary with V (i.e. with y)
Ky is a function of flowrate V
V solvent-only flow and S tower cross section area are the only true
constant
To further progress, need to make
simplifying assumptions!!!
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Assumptions:
1. Very dilute solutions (x, y less than 0.1 or 10%)
2. Linear equilibrium (i.e. Henry)
Start with
V varies only slightly from (1) to (2)
Use
y2
y1
*
*LM
'y )y(y
dy
y)-(1
y-1
aSK
Vz
y1V
V '
V = f(y)
2
VVV 21AVE
Kya will be nearly constant
because V ~ constant
3. Absorption of very dilute mixtures
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(1 y) ~ 1
(1 y)*LM ~ 1
1
y)-(1
y-1 *LM
y2
y1
*
*LM
'y )y(y
dy
y)-(1
y-1
aSK
Vz
1
y2
y1
*'y
AVE
)y(y
dy-
aSK
Vz Approximate
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This integral can be evaluated analytically if Henrys law holds
Mass balance (operating line):
straight line if V & L are constant
from (1) to (2)
y
y*)(y
y-y
)y-(y)y-(y
)y y*,y(y & )y y*,y(y put
dy y*)-d(y
y y*)-(ySay
line stright is f(y) y*)(y
12
*1
*22
*22
*1
1
1
y*)-d(ydy
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(2)-(1) LM
12
'y
AVE
*11
*22
*11
*22
12
'y
AVE
*11
*22
'y
AVE
y2
y1
'y
AVE
y*)-(y
)y(-1)(y
aSK
Vz
yy
y-y
)y-(y)y-(y
1))(y(y
aSK
Vz
yy
y-y
aSK
Vz
y*)(y
y*)-d(y-
aSK
Vz
ln
ln1 12
*1
*22
y-y
)y-(y)y-(y 1
These terms make up a log mean
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Similar expressions hold for the various forms of N
(2)-(1) LMi
12
'y
AVE
)y-(y
)y(-1)(y
aSk
Vz
(2)-(1) LMi
12
'x
AVE
x)-(x
)x(-1)(x
aSk
Lz
(2)-(1) LM*
12
'x
AVE
x)-(x
)x(-1)(x
aSK
Lz
We usually choose the phase which is controlling the rate
(i.e. the phase with the greatest mass transfer resistance)
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Re-arrange
M'x21
M'y21
Mi'x21
Mi'y21
-x)*az(xKx-xS
L
y*)az(yKy-yS
V
x)az(xkx-xS
L
)yaz(yky-yS
V
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How to find mole fraction at interface and equilibrium ???
y1,x1
y2,x2
y
x
y*1
y*2xi1
yi1
xi2
yi2
Dilute solution
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Example 4 (Geankoplis: Example 10.6-4)
Acetone is being absorbed by water in a packed tower having a cross-sectional
area of 0.186 m2 at 293K and 101.32 kPa. The inlet air contains 2.6 mol%
acetone and outlet 0.5%. The gas flow is 13.65 kgmol inert air/h. Pure water inlet
flow is 45.36 kg mol water/h. Film coefficients for the given flows in the tower at
kya = 3.78 x 10-2 kgmol/s.m3.mol frac and kxa = 6.16 x 10
-2 kgmol/s.m3.mol frac.
(a) Calculate the tower height using kya.(b) Repeat using kxa.(c) Calculate Kya and the tower height.
Solution
M'x21
M'y21
Mi'x21
Mi'y21
-x)*az(xKx-xS
L
y*)az(yKy-yS
V
x)az(xkx-xS
L
)yaz(yky-yS
V
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4. Transfer Units
y2
y1 i'y
AVE
)y(y
dy-
aSk
Vz
(m) Dimensionless
(m)
HEIGHT of a
gas film
transfer unit
Number of a
gas film
transfer unit
Z = Hy Ny
Z = Hx Nx
If OVERALL coefficients are
used to describe the flux:
Z = Hoy Noy
Z = Hox Nox
Depending on concentration
units and coefficients used
Z = HG NG = HL NL
Z = HOG NOG = HOL NOL
HG Hy; NG Ny; HOG Hoy; NOG Noy
Heights of transfer units have been
correlated with operating variables
(i.e. V, L, packing type and size)
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Sx)-a(1K
L
aSK
LH
Sy)-a(1K
V
aSK
VH
Sx)-a(1k
L
aSk
LH
Sy)-a(1k
V
aSk
VH
*Mx'x
OL
*My'y
OG
iMx'x
L
iMy'y
G
M
1OL
M
1OG
Mi
1L
Mi
1G
-y)*(x
x-xN
y*)-(y
y-yN
x)-(x
x-xN
)y-(y
y-yN
2
2
2
2
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Example 5 (Geankoplis: Example 10.6-5)
Repeat Example 5 using transfer units and height of a transfer unit as
follows:
(a) Use HG and NG to calculate tower height
(b) Use HOG and NOG to calculate tower height
Solution
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0.5
LL
0.5
G
ScS
LH
ScS
L
S
VH
Correlation form:
Constants , , , , for different packings and sizes(Geankoplis 10.8b)
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Slope ~ 0.43 on log-log graph
eg. correlation for inch Raschig
rings with NH3/H2O
yS
VH
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Absorption Column Size
Height Mass transfer (Hy Ny)
Diameter Hydraulics Two phase flow over packing; gas
pressure drop - Design (Flooding
and loading)
Difficult to locate precisely
(deviation from standard line)
Loading curve nearly vertical
Operate close to loading, ~1/2
flooding
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Analytical equation to calculate theoretical number of trays
For transfer of the solute from phase V to phase L (absorption)
For transfer of the solute from phase L to phase V (stripping)
Aln
1/A1/A)(1mx-y
mx-y ln
N22
21
(1/A) ln
AA)(1/my-x
/my-x ln
N11
12
2122221111 AA A and V/mL A,V/mLA
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Analytical equation to calculate packed bed height
Absorption
Stripping
Operating and/or equilibrium lines are slightly curved
A/my-x
/my-xA)(1 ln
A-1N
1/Amx-y
mx-y1/A)(1 ln
1/A-1N
11
12OL
22
21OG
1
1
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Analytical equation to calculate packed bed height
When the operating and equilibrium lines are straight
z = HOG x NOGz = N x HETP
See Example 10.6-5
1/A-1lnA
NOGN
The height of a theoretical tray or stage, HETP, in m is related to HOG by
/AA-1ln(1/A)
HHETP OG
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Example 6:
Experimental data have been obtained for air containing 1.6% of SO2being scrubbed by pure water in a packed bed column of 1.5m2
cross-sectional area and 3.5 m in packed height. Entering gas and
liquid flow rates are 0.062 and 2.2 kmol/s, respectively. If the outlet
mole fraction of SO2 in the gas is 0.004. m = 40. Calculate NOG.
Solutions
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5. Absorption of concentrated mixtures
y2
y1 i
iLM
'y
)y(y
dy
y)-(1
y-1
aSk
Vz
1
y2
y1
*'y
AVE
)y(y
dy-
aSK
Vz
Approximate
Dilute mixtures
y2
y1 i
iLM
'y
)y(y
dy
y)-(1
y-1
aSk
Vz
Concentrated mixtures
The only constant
Each term (or bracketed terms) needs to
be calculated for a range of values of y
between y1 and y2. Then, evaluate the
integral numerically (or graphically) in
order to find yi between y1 and y2.
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We need the whole operating curve from (1) to (2). This is obtained from a
series of mass balances between (1) and a range of points up to (2).
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Sherwood suggests that (1-y)*LM and (1-y*) can often replace the true interface
values if the liquid resistance is small. The * values are much easier to get
since ky need not be known.
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Example 6 (Geankoplis: Example 10.7-1)
A tower packed with 25.4 mm ceramic rings is to be designed to absorb SO2from air by using pure water at 293K and 1.013 x 105Pa abs pressure. The
entering gas contains 20 mol% SO2 and that leaving 2 mol%. The inert air
flow is 6.53 x 10-4 kg mol air/s and the inert water flow is 4.20 x 10-2 kg mol
water/s. The tower cross-sectional area is 0.0929 m2. For dilute SO2, the
film mass-transfer coefficients at 293 K are, for 25.4 mm rings:
kya = 0.0594Gy0.7Gx
0.25 kxa = 0.152Gx0.82
Gx and Gy are kg total liquid/gas per sec per m2 tower cross section.
Calculate the tower height.
Solution
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6. Absorption of semi-dilute mixtures
Dilute enough to use VAVE or LAVE and constant Kya or Kxa
y2
y1
*LM
'y
y*)(y
dy
y)-(1
y-1
aSK
Vz
y2
y1
*LM
'y
AVE
y*)(y
(-dy)
y)-(1
y-1
aSK
Vz
Always check this for ~ 1.0. If yes, need only evaluate
Which is relatively simple (numerical or graphical)
y2
y1*yy
dy
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7. Controlling resistance
i) Liquid film controlling (eg. gas almost insoluble O2/H2O)
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7. Controlling resistance
ii) Gas film controlling (eg. gas very soluble or react with liquid NH3/H2O)