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Lecture #25 begins here
Chapter 13. Gas Absorption/Stripping
gas absorption -- 2 components of a gas are separated by contact
witha liquid (in which one component is preferentially soluble)
stripping -- 2 components of a liquid are separated by contact
with agas
An example of gas absorption is the removal ofammonia from air
by contact with liquid water.Ammonia is very soluble in water
whereas air is onlyslightly soluble.
Both gas absorption and stripping involve at leastthree
components. Usually only one of thesecomponents crosses the phase
boundary. In theexample of ammonia and air, ammonia is thecomponent
whose molar flowrate changes by thelargest percentage of the inlet
value. Although someair will also dissolve in water, and some water
willevaporate into the air, the molar flowrates of air andwater
change by negligible fractions: their flows canusually be
considered constant.
These are the two main differences between them anddistillation:
1) at least 3 components, and 2) oftenonly "one transferrable
component". By constrast, indistillation, all of the components are
present in bothphases.
EQUIPMENT FOR ABSORPTION/STRIPPING
Liquid and gas streams for absorption or strippingcould be
contacted using a tray column like that usedin distillation.
Instead of tray towers, we are going tolook at the design of packed
towers. Packed towersare a reasonable alternative to tray towers
insituations in which the tray efficiency is low (perhapsonly 5%
instead of 50%). Because of this very lowefficiency, very large
numbers of trays would berequired -- perhaps 100's or 1000's.
Fabrication costsjust become prohibitively expensive. Fortunately,
aviable alternative exists: the packed tower.
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A packed tower is simply a tube or pipe, which is filled with
some sort of "packing." Thepacking typically consists of particles
around an inch in diameter. In commercial packed towers,the usual
choice are particles with one of three different shapes:
raschig ring (which is just a piece of pipe which has been cut
into segments, whose lengthand diameter are about the same)
L D 12
to 1 inches12
Berl saddle
Pall ring
Although, in a pinch, almost anything you have laying around
would do -- ping-pong balls, golfballs, etc. The purpose of the
packing is to promote good contact between the liquid and
vaporstreams which are being brought together to permit interfacial
mass transfer. In particular, whatis desired is a large interfacial
area per unit volume.
The liquid stream is usually fed into the top of the tower while
the vapor is fed into the bottom.Thus we have countercurrent flow
of the two streams, which has the same advantages for masstransfer
as it did for heat transfer.
The packing promotes good contact between the phases by dividing
the two feed streams intomany parallel interconnected paths.
Ideally, you would like the liquid to flow downward as athin film
over the surface of the packing. This would give the maximum
surface area of contactbetween the gas and liquid.
If you just pour the liquid from the end of the pipe onto the
top of the packing in tower havingmuch larger diameter than the
pipe, most of the packing will not even be wet. Only some of
thechannels will be carrying flow. This is called:
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channeling -- maldistribution of liquid flow
So some sort of device to distribute the flow over the entire
cross section of the tower is needed.This device is called a
distributor.
Even if the flow is evenly distributed at the top of tower,
channeling might still develop as thefluid trickles down. When two
thin films converge they tend to form a thick film and a drypatch,
which results in a reduction in contact area. So redistributors are
placed every 10-15 feetalong the length of the tower.
A TYPICAL ABSORBER DESIGN PROBLEM
To motivate the next few lectures, lets pose a typical design
problem. For this example, I'mgoing to take Prob. 22-1 from McCabe,
Smith & Harriott.
Problem: treat 500 SCFM of air containing14 mol% acetone to
remove 95% of theacetone by absorption in liquid water in apacked
bed operating at 80F and 1 atm,with 1-inch raschig rings. The feed
watercontains 0.02% acetone and the flowrate is1.08 times the
minimum.
The partial pressure of acetone over anaqueous solution at 80F
can be calculatedfrom
2where ln 1.95 1oA A A A A Ap P x x
where PAo = 0.33 atm is the vapor pressure of acetone at 80F. As
the designer, you must select
the following:
500 SCFMair containing
14 mol% acetone
water with0.02% acetone
= 1.08L Lmin80 F1 atm1 rasching rings
recover 95%of acetone
yb
ya
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flowrate of water diameter of tower height of packing
Solution Overview:
Choice of liquid flowrate: L L 1.1 to 1.5 times min
Just as with distillation, there is a minimum L/V required to
achieve the desired separation. Intypical operation of an absorber,
the liquid rate is chosen to be just above the minimum.
Tower diameter: GG = 0.5 to 0.7 times GG,flood
The diameter of the tower is usually chosen on the basis of gas
mass velocity GG. Generally,for a particular L and V, the smaller
the column diameter, the larger the mass velocities will be,and the
larger the pressure drop. Generally, large mass velocities are
desirable because they givehigh mass transfer coefficients, but too
large mass velocities cause flooding which severelydecreases mass
transfer rates. The gas mass velocity above is about as close to
flooding as youdare get.
Tower height:* ZV S
K a
dy
y yT
y
Hy
y
NOya
b
Oy
z; *
The height of the tower is determined by mass transfer rates.
Basically, the gas and liquid phasesneed to be in contact for a
certain time for the acetone to have time to diffuse from the gas
phaseinto the liquid. The equation above is called the "design
equation". The integral
*
b
a
yb a
Oy
y
y ydyN
y y y
is called the number of transfer units, where y is the mole
fraction of the transferable
component, y y* is the local driving force for mass transfer and
y is some average driving
force (perhaps the log-mean of the driving force at the top and
bottom). The number of transferunits is loosely analogous to the
number of ideal trays required: both can be thought of asmeasures
of the difficulty of the separation.
Since NOy is dimensionless, to get units of height, the
remaining factor in the design equation(defined to be HOy) must
have units of height and is called the height of one transfer
unit:
* This particular formula is for dilute solutions, which is
probably not really applicable to the current example, but itis
applicable in the homework problems.
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HV S
K aOy
y
where Kya is the overall mass transfer coefficient times the
interfacial area per unit volume ofpacking.
Detailed Solution:
First of all, what is SCFM? This stands for standard cubic feet
per minute
SCFM = standard (0C, 1 atm) cubic feet per minute
where standard means that the volume is evaluated at standard
conditons. We can convertSCFM into moles. Starting with the ideal
gas law,
PV nRT
orV
n
RT
P
R K
atm
273
122 4 359
b gb g .
liters
gmol
ft
lbmol
3
depending of which set of units are used to express R. Recall
that a gmol is the quantity ofmaterial whose mass equals the
molecular weight in grams. Similarly, a lbmol is the quantity
ofmaterial whose mass equals the molecular weight in pounds. For
example, air has a molecularweight of 29:
MW of airg
gmol
lb
lbmol 29 29
Thus we can convert 500 SCFM into a molar flowrate:
500
359139
ft min
ft lbmol
lbmol
min
3
3 . V
Now lets move on to determine the liquid flowrate required.
Recall that in distillation the slopeof the ROL is given by
slope of ROL
L
V
R
R 1
You should also recall that there exists a minimum value of the
reflux ratio for any givencombination of product and feed
specifications. This minimum value of R translates into aminimum
allowable L/V. There is a similar minimum L/V for gas absorption.
To find theminimum reflux ratio in distillation, we needed to plot
an operating line and an equilibriumcurve.
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Lecture #26 begins here
EQUILIBRIUM CURVE
In Prob. 22-1, McCabe gives us equilibrium data in the form of
partial pressure of acetone inthe gas phase, pA, as a function of
the mole fraction of acetone in the liquid, x:
p P xA Ao
A (132)
PAo = vapor pressure of acetone at 80F = 0.33 atm
A = activity coefficient for liquid mixture
Activity coefficient is a measure of nonideality of the liquid
phase. For ideal solutions theactivity coefficient is unity:
A = 1 (ideal solution)
For ideal solutions and A = 1, (132) reduces to Raoults law. Our
solution of acetone in water isnot ideal, but McCabe kindly gives
us a model
ln . A x 195 12b g
Note that A1 (i.e. lnA 0) as x 1. This is ageneral rule: pure
components always behaveideally since the partial pressure pA
exerted by a
pure component is its vapor pressure PAo [for (132)
to predict this for x=1 we must require that A=1 fora pure
component]. We can obtain an expressionfor the mole fraction of
acetone in the gas just bydividing the partial pressure by the
total pressure:
21.95 10.33 atm
1 atm
xApy xeP
where we use P = 1 atm. Repeating this fordifferent xs to obtain
ys up to 0.14 (the feed concentration), we obtain the curve at
right, where
x = mole fraction of acetone in the liquid (2nd component is
water)y = mole fraction of acetone in the gas (2nd component is
air)
x
y
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0 0.02 0.04 0.06 0.08 0.1 0.12
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2010
OPERATING LINE
Suppose we are trying to absorb acetone from a mixture of air
and acetone by contacting theair mixture with water. Let
L,V = total molar flowrates
x,y = mol.frac. of transf. comp.
where the transferrable component is acetone in thisexample.
Performing a component balance on thetransferrable component about
the top section of thetower yields:
Laxa + Vy = Lx + Vaya (133)
This is identical to (78) to its counterpart indistillation (see
page 102). Solving for y:
yL
Vx
V y L x
Va a a a
This relationship y(x) between the mole fractions in the liquid
and gas streams is again known asthe operating line, since the
relationship is imposed by a component mole balance.
Unlike distillation (where the equimolal overflow assumption
makes the ratio L/V constantwithin any cascade), L and V are not
constant over the height of the absorber: as acetone istransferred
from the gas to liquid, L and V change. In particular, their ratio
changes. Thus theoperating line is not straight. This makes it
difficult to determine the minimum water flowrate.
There is one limiting case where we can easily predict the
change in L and V:
Only One Transferable Component
It is often possible to assume that only one component is
undergoing transfer between theliquid and gas streams. For example,
in our problem acetone is being transferred from the air tothe
water:
transferable: acetone
non-transferable: air, water
Although some of the water will evaporate when it contacts the
air and some of the air willdissolve in the water, the molar rates
of transfer of these components can often (but not always)be
neglected compared to rate of acetone transfer. When you can
neglect the molar transfer rateof all but one component, then
significant simplification can be made. If we can neglect
Laweak liquor
Lbstrong liquor
Valean gas
Vbrich gas
LV
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evaporation of water and dissolution of air, then the moles of
water (the nontransferablecomponent) in the liquid stream must be
the same at all elevations:
(1-x)L L' = const so LL
x
1(134)
In the above equation and what follows, a prime (' ) is used to
denote a flowrate or mole fractionof the nontransferable component.
Similarly, the moles of air in the gas stream must be the sameat
all elevations:
(1-y)V V' = const so VV
y
1(135)
(134) and (135) into (133):
L x
x
V y
y
L x
x
V y
ya
a
a
a1 1 1 1
where L',V' are constants. Solving thisequation for y(x):
1 1 1
1 1 1
a a a a
a a a
y x x R y x xy x
y x x R x x
where'
'
LR
V
Figure 12 shows several operating lines corres-ponding to
different values of R (equallyspaced from R = 3.82 for the upper
most redcurve to R = 1.91 for the lowest). Theequilibrium curve is
shown in black. All
curves were drawn with the help of Mathcad.While it is hard to
tell from the figure, none ofthe curves in Figure 12 is a straight
line. Thenit becomes a trial-and-error process to determine Rmin.
There is an easier way, which avoids thistrial-and-error.
If we now express concentrations in terms of mole ratios instead
of mole fraction:
Xx
x
1
moles of A in liquid
moles of non - A in liquid
Yy
y
1
moles of A in gas
moles of non - A in gas
0 0.01 0.02 0.03 0.04 0.05 0.06 0.070
0.02
0.04
0.06
0.08
0.1
0.12
0.14
ya
yb
Figure 12
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then this equation becomes very simple:
L'Xa + V'Y = L'X + V'Ya
or YL
VX Y
L
VXa a
FHG
IKJ (136)
Note that mole ratios need not be smaller than unity:
0 x 1
but 0 X
although they are in this example. SinceL'/V' = const, this is
the equation of astraight line. Thus for the special case
ofone-transferable component, the operatinglines are straight on
mole ratiocoordinates.
One point on the operating line is(Xa,Ya). We are told that the
inlet watercontains a small amount of acetone:
xa = 0.0002 Xx
xa
a
a
10 0002.
The concentration of acetone in the gasphase is determined from
the specificationthat we want to remove 95% of theacetone from the
feed:
y V y Va a b b 0 05.
yV
yy
V
ya
ab
b
10 05
1.
Since V' (the molar air flowrate) is the sameat either end of
the column, we can cancel itout, leaving
Y Ya b
0 05 0 00814
014
1 01401628
. .
.
..
500 SCFMair containing
14 mol% acetone
water= 0.0002xa
80 F1 atm
recover 95%of acetone
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or 0.008071
aa
a
Yy
Y
The second point on the operating linemust lie somewhere along
the line
Y = Yb = 0.1628
When you plot this up, you find that theequilibrium curve is
below the operatingline. This is generally true for gasabsorption
and makes sense when yourealize that the gas must be richer in
theammonia than at equilibrium otherwisethe acetone would not
spontaneouslyabsorb into the liquid.
Drawing a straight line through (Xa,Ya)= (0.0002, 0.0081), we
decrease the slope(starting at vertical) until the line touchesthe
equilibrium curve at any point in theinterval Ya Y Yb. While the
touch or pinch usually occurs at Y = Yb, this is not necessary.In
this particular case, the pinch occurs at a point in the middle of
the interval Ya Y Yb.
Extending this straight line from the pinch up to Y = Yb, we
locate a second point on the line:(Xb,max,Yb) = (0.081, 0.1628).
Using the two points (Xa,Ya) and (Xb,max,Yb), we can calculatethe
slope correspoinding to the minimum water flowrate:
FHG
IKJ
L
V min
. .
. ..
01628 0 00814
0 081 0 0002191
min lbmol lbmol gal1.91 1.91 1.39 1 0.14 2.29 4.94min min minL V
Multiplying this by the molecular weight of water (18 lb/lbmol) and
dividing by the density ofwater (8.33 lb/gal) yields a minimum
water flowrate of 4.94 gal/min.
Lecture #27 begins here
INTERFACIAL MASS TRANSFER: REVIEW
An important design parameter is the depth of packing ZT
required, which we will soon seecan be calculated from a design
equation like the following:
X
Y
0
0.04
0.08
0.12
0.16
0 0.02 0.04 0.06 0.08 0.1
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*
b
a
y
Ty y
V S dyZ
K a y y
where V is the molar flowrate of gas, S is the cross-sectional
area of the tower and Kya is a masstransfer coefficient. The
denominator of the integrand, y-y*, is the local driving force for
masstransfer; on an xy diagram, it is also the vertical distance
between the operating line y(x) and theequilibrium curve y*(x).
In distillation, the tower height was determined by the number
of plates required times theplate separation (which is usually 1-2
feet). The number of plates required for a given separationis
determined by the operating and equilibrium lines. For packed
towers, the height also dependson the operating and equilibrium
lines, but in addition it is inversely proportional to the
masstransfer coefficient, which in distillation plays only a minor
role in determining plate efficiency.
Now that we have established the importance of interfacial mass
transfer in packed towers,let's talk about modelling mass transfer
across a phase boundary. There are two main differencesbetween
interfacial heat transfer (as employed in Hxer design) and
interfacial mass transfer:
1. overall driving force2. reference frame for flux
Let's start by recalling the driving force for heattransfer
across an interface. Suppose I contact a hot gaswith a cold liquid.
The temperature profile near theinterface will look something like
that shown at right.There are two characteristics of this sketch
which areimportant:
1. heat flows from high to low temperature
2. temperature is continuous across the interface
Th
Ti
Tc
gas liquid
heattransport
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Interfacial mass transfer is similar to interfacial
heattransfer, but it is also different.
1. mass transfer occurs in the direction fromhigh to low
chemical potential (notnecessarily from high to low
concentration*)
2. chemical potential is continuous acrossinterface
(concentration is generally notcontinuous)
The main difference is evident in the sketch of theconcentration
profile near the interface. Note thediscontinuity in the mole
fraction at the interface.
yi xi
Tyi = Txi = Ti
The reason for this discontinuity in concentration across the
interface has to do withthermodynamic criteria for phase
equilibrium. Recall:
phase equilibrium: jV = jL for j=1,...,Nc
thermal equilibrium: TV = TL
where j is called the chemical potential which playsthe role of
temperature in mass transfer. Unfortunately,there exists no
thermometer for measuring chemicalpotential. Instead, we are forced
to measure chemicalconcentration. While chemical potential
usuallyincreases with concentration within any given phase(this is
why diffusion of a solute occurs from high tolow concentration),
when comparing chemical potentials between two phases, there is no
generalcorrelation between chemical potential and
concentration.
To illustrate this, recall the simplest case of VLE: an ideal
gas mixture in equilibrium with anideal solution. This leads to
Raoult's law:
oj j j jp y P x P
* Within a single phase, transport is usually from high to low
concentration.
Instead of component i, the superscript i will be used to denote
quantities evaluated at the interface between twophases. In place
of i, we will use j to denote components.
y
xi
yi
gas
liquid
masstransport
y*
mV
mi
mL
gas liquid
masstransport
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Note that:y
x
P
P
j
j
jo
1
This ratio is never unity (except for pure components).
Thus yj xj for VLE
even in the simplest case of vapor-liquid equilibrium, the mole
fractions are not equal, except inthe trivial case when you have
only one component and the total pressure is the vapor
pressure.
Definitions of Transfer Coefficients
Recall from the first part of this course, the local heat flux
through the interface can berelated to the local temperatures using
any one of three types of local heat transfer coefficients:
rate of heat transfer
interfacial areay h i x i c h ch T T h T T U T T (137)
where hx,hy = local, one-phase heat transfer coefficients
U = local, overall coefficient.
Basically these equations say that the rate of heat transfer is
proportional to the driving force(which is the departure from
equilibrium) and the proportionality constant is the heat
transfercoefficient.
Similar definitions of transp ort coefficients can be made for
mass transfer:
molar rate of transfer
interfacial area k x x k y y K x x K y yx i y i x y( ) ( ) ( * )
( *) (138)
where kx,ky = local, one-phase mass transfer coefficients
Kx,Ky = local, overall mass transfer coefficients
Comparing the mass transport expressions in (138) with their
heat-transfer analogues in (137),there is a good deal of similarity
especially in the one-phase relations. For a single phase,
thedriving force is the difference between the concentration or
temperature in the bulk and in theconcentration or temperature at
the interface.
But the driving force for the overall coefficients look a little
different. The overall driving forcefor heat transfer is just the
difference between the temperature of the hot and cold fluid
Th - Tc
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By simple analogy, you might expect the overall driving force
for mass transfer to be thedifference in concentration of the two
phases:
y - x overall driving force
Instead y - y* and x* - x
appear in (138) as the overall driving force. The *'s aredefined
as shown on the figure at right. Basically y - y* andx* - x are two
different measures of the distance between theoperating line and
the equilibrium curve at that elevation inthe packed bed where the
liquid and gas concentrations are(x,y).
x-y does not represent the overall driving force for
masstransfer, because the transport rate does not go to zero
whenx-y = 0. The transport rate is zero only at equilibrium, and
x-y 0 at equilibrium.
Lecture #28 begins here
Determining the Interfacial Concentrations: (xi,yi)
Although overall mass-transfer coefficients are the mosteasy to
use in design, correlations are usually more availablein the form
of single-phase coefficients. Then we need tocalculate overall
coefficients Kx and Ky from the single-phase coefficients kx and
ky. As the firststep, we will need to evaluate the interfacial
concentrations xi and yi, which appear in thedefinitions above.
Example #1
Given: x, y, kx, ky and the equilibrium curve
Find: xi and yi
Solution: Consider the rate of interfacial transport at some
arbitrary elevationin the absorber, where the local concentrations
in the liquid and gas are x and yand (x,y) is a point on the
operating line. At steady state, the flux of acetonethrough the gas
film must equal the flux of acetone through the liquid film:
NA = kx(xi-x) = ky(y-yi) (139)
If the fluxes were not equal, we would have acetone building up
at theinterface. We will denote this interfacial flux by NA.
Now the bulk compositions x,y are known, together with the
two
y
xi
yi
gas
liquid
masstransport
y*
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single-phase mass transfer coefficients. Think of the two
interfacial concentrations as twounknowns. We need two equations.
One equation is provided by the requirement that (xi,yi)must lie on
the equilibrium curve. The second relation is (139), which can be
re-written as alinear relationship between yi and xi:
yk
kx y
k
kxi
x
yi
x
y
slope intercept;
(140)
One point on this line is (xi, yi) = (x, y) and theslope is
kx/ky. The intersection of (140) and theequilibrium curve gives the
interfacialconcentrations.
Example #2: Next, lets determine the value of theoverall
coefficient which leads to the same flux for these interfacial
concentrations.
Given: kx and ky
Find: Ky
Solution: Using the definitions of the k's and of Ky,
NA = kx(xi-x) = ky(y-yi) = Ky(y-y*) (141)
The relationship among the concentrations is shown in the figure
at right. Adding andsubtracting yi from the overall driving force
y-y*:
* *
A A
y y
i i
N N
K k
y y y y y y (142)
Using (141) we can assign meaning to two of the three
differences appear above, leaving:
*
i
A Ai
y yx x m
N Ny y
K k
(143)
The remaining difference yi-y* can be related to xi-x if we
multiply and divide the last term byxi-x:
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*
*
A
x
i Ai i
i xN
mk
y y Ny y x x m
x x k
(144)
In the second equality above, xi-x is expressed interm of the
remaining single-phase mass transfercoefficient using (139).
Substituting (144) into(143) and dividing by NA:
1 1
K k
m
ky y x (145)
where*i
i
y ym
x x
= avg. slope of E.C.
is the average slope of the equilibrium curve in the region of
the xy diagram between x and xi.Of course, if the equilibrium curve
is straight (as it will be in dilute solutions), then m is its
trueslope over the entire range.
Comment: This is similar to the expression for overall heat
transfer coefficients for a double-pipeheat exchanger:
Di = Do:1 1 1
U h hx y (146)
We said that 1/U represented the total resistance to heat
transfer through the two phases, which isjust the sum of the
resistances of each phase. The main difference between (145) and
(146) is theappearance of m in (145). For heat transfer, the slope
of the equilibrium line is unity (m = 1)because at thermal
equilibrium Ty = m Tx = Tx.
In short, (145) states that the total resistance to mass
transfer equals the sum of the resistances ofthe gas and liquid
phases. We could also have showed, in a similar fashion, that
1 1 1
x y xK m k k
(147)
In our earlier treatment of heat transfer, we also distinguished
between inner and outer coefficients (i.e. Ui andUo). This was
because the inside area and outside area of a pipe were somewhat
different. We will drop thisdistinction here because the inside
area and outside area are equal for an interface. In other words,
the pipe wallthickness is zero.
slope k
kx
y
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where now:*
i
i
y ym
x x
is the average slope of the equilibrium curve in the regionof
the xy diagram between xi and x*. If the operating iscurved, this
slope might be slightly different from theaverage between x and xi;
thus we add the prime to thelabel. On the other hand, if the
equilibrium curve isstraight, these two slopes would be the same (
m m ) andwe could relate the two overall mass transfer
coefficient:
straight E.C.:1
y x
m
K K
Equimolar Counter-Diffusion vs. Diffusion through Stagnant
Fluid
We have just seen that the overall driving force for interfacial
mass transfer is y-y* or x*-x (itis not y-x). This is one of two
major differences between interfacial mass transfer and
interfacialheat transfer. The second major difference arises
because diffusion of mass usually inducesconvection of the fluid
(i.e. motion of the center of mass). Recall Fouriers law of
heatconduction:
energy
heat fluxarea time
zdT
q kdz
If there is also bulk fluid motion, an additional term is added
to the heat flux to account forconvection of heat:
convectionconduction
z p z refdT
q k c v T z Tdz
where vz is the local z-component of velocity of the fluid and
cp(TTref) is the enthalpy per unitvolume. Usually heat conduction
does not induce flow.*
By contrast, diffusion of mass usually always induces flow.
Ficks law of diffusion must bemodified by adding a contribution
from convection of mass:
* An important exception is natural convection which arises
because the nonuniform temperature of the fluidcauses nonuniform
density of fluid which in a gravitational field can cause flow.
*
m'
m
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diffusion mol average velocity
convection
A Az BzAz AB A
v
dc N NN D c
dz c
(148)
where c = cA+cB is the total molar concentration. The the second
term represents convection ofmatter which arises because the center
of mass move with time.
Case I: If we use a capillary tube to connect two gasreservoirs
having the same total pressure butdifferent partial pressures of
gases A and B (see Fig.6.2-1), we will obtain equimolar
counter-diffusionof A and B (i.e. NAz = NBz).* In this way,
themole-average velocity v = 0 and the total pressurein both
reservoirs remains constant. For NAz = -NBz,(148) gives
AAz AB
dcN D
dz (149)
See Example 6.2-1 in Geankoplis for details leading to a linear
profile of partial pressure or ofmolar concentration of either gas.
For 1-D steady equimolar counter-diffusion (from a reservoirhaving
cA1 to a second reservoir having cA2), taking the flux NAz in (149)
to be independent of z,(149) integrates to yield cA varying
linearly with z. Substituting this result into (149) yields
1 2 1 22 1 2 1
1 2
AB ABAz A A A A
y A A
D cDN c c y y
z z z z
k y y
(150)
where z2-z1 is the length of the capillary through which
diffusion is occuring. The last equationis the definition of k'y,
the single-phase (gas) mass transfer coefficient to be used with
molefractions and equimolar-counter diffusion.
* If the gases obey the ideal-gas law, then equal temperatures
and equal total pressures in the two reservoirs implyequal total
molar concentrations. If the molar fluxes of the two gases were not
equal in magnitude but opposite indirection, then the total molar
concentrations in the two reservoirs could not remain the same: and
the total pressurescould not remain the same. Any difference in
total pressure would generate a pressure-driven flow of gas in
adirection so as to eliminate the difference in pressure. Thus the
total pressure remains uniform thoughout thecapillary and the two
reservoirs.
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Lecture #29 begins here
Case II: If instead, we have diffusion of benzene vapor above
its liquid through air (which isvirtually insoluble in the liquid),
the air must remain stagnant (i.e. NBz = 0) since it cannot
enterthe liquid (see Fig. 6.2-2a). This is diffusion of A through
stagnant B. For NBz = 0, (148) gives
A
A AAz AB Az
y
dc cN D N
dz c or 1 A AA Az AB AB
dc dyy N D cD
dz dz
Solving this differential equation for the partial
pressureprofile yields a nonlinear profile (see Example 6.2-2)
asshown in the figure at right. The flux of of benzene iseventually
calculated as
1 22 1
1 2
1AB
Az A AA M
y A A
cDN y y
z z y
k y y
(151)
where
1 2
1
2
1 11
1ln
1
A AA M
A
A
y yy
y
y
represents the logmean of 1-yA evaluated at either end of the
diffusion path. Once again, thesecond equation in (151) is the
definition of ky, the single-phase (gas) mass transfer coefficient
tobe used with mole fractions and diffusion through stagnant film
of B.
Comparing (150) and (151), we see that
P
pA
pB
z1 z2
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1y
y yA M
kk k
y
(152)
For the same driving force (yA1 yA2), the fluxes NAz are
different. Since 1-yA is always lessthan one, we see that equimolar
counter-diffusion (150) is slower than diffusion through astagnant
fluid (151). This can be qualitatively understood as follows.
Suppose that to get toclass, you need to walk down a corridor
that's crowded with other students. If everyone else wasstanding
almost still (stagnant fluid), it would be easier to walk around
them than if everyone iswalking toward you (counter diffusion).
Athough (152) was derived for one particularly simple case (no
pressure-driven or gravity-driven convection), it applies much more
generally. Similar relations exist for diffusion in theliquid
phase:
1x
xA M
kk
x
where
1 2
1
2
1 11
1ln
1
A AA M
A
A
x xx
x
x
and between the overall mass-transfer coefficients for equimolar
counter-diffusion and diffusionthrough a stagnant film:
*1y
yA M
KK
y
and
*1x
xA M
KK
x
where
*
* *
1 11
1ln
1
A AGA M
A
AG
y yy
y
y
and
*
* *
1 11
1ln
1
A ALA M
A
AL
x xx
x
x
and where yAG is the mole fraction in the bulk of the gas,*Ay is
the mole fraction which would
be in equilibrium with the bulk liquid having a mole fraction of
xAL, and*Ax is mole fraction in
the bulk of the liquid which would be in equilibrium with the
bulk gas having a mole fraction ofyAG.
HEIGHT OF A PACKED TOWER
The analysis which follows has as its goal the determination
ofthe height of packing required. The approach is similar to that
usedin the design of double-pipe heat exchangers in which the goal
isthe determination of the area of heat-exchange surface
required.Because the driving force y-y* varies over the height of
the
V
ya
a
V
yb
b
L
xa
a
L
xb
b
ZT
z
Dz
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column, we have to chop up the tower into pieces which are small
enough so that the drivingforce is virtually uniform throughout
each piece. Since the compositions change only with z, wechop up
the tower in such a way that we produce pieces which have zconst,
which is a thinhorizontal slice.
Now let's take a closer look at what happens inside aparticular
slice of the tower whose lower surface is z=z andwhose upper
surface is z=z+z. The slice contains solidpacking as well as liquid
and gas streams. In what follows, wewill ignore the solid and treat
the liquid and gas phases as ifthey were completely separated,
rather than interspersed in eachother. Let's do a mass balance on
the acetone in the gas phaseonly contained within our slice of
column. Besides the liquidstreams entering and leaving the slice,
we have acetonecrossing the interface between the gas and liquid
phases. Therate of absorption per unit area can be expressed in
terms of the local overall mass transfercoefficient:
molar rate of transfer( *)
interfacial areayK y y (153)
We could also have used any one of the other expressions in
(138). Just like the overall heattransfer coefficient Uo depends on
the flowrates of both fluids being contacted in the heatexchanger,
the overall mass transfer coefficient Ky depends on both
flowrates
K K L Vy y ,a f
To obtain the rate of transfer across the interface in our slice
of column, we need to multiply(153) by the interfacial area in that
slice. In a heat exchanger, the heat transfer surface is fixedby
the geometry of the equipment selected: it is just the area of the
pipe wall or the tubes. Inparticular, the heat transfer area does
not depend on the flowrates of the hot and cold streams.On the
other hand, the boundary between liquid and gas in a packed bed is
very complex andvery hard to measure directly. Most importantly,
the area also depends on the flowrates of thegas and liquid streams
(L,V). The interfacial area is usually expressed as a, the
interfacial areaper unit volume of packing:
interfacial area
volume of packing a a L V( , ) (154)
One empirical correlation relating area to flowrates is the
Schulman Correlation (Table 6.3 ofTreybal):
a mG Gx y
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where Gx and Gy are the mass flowrates of the liquid or gas
stream divided by the cross-sectional area of the tower
GD
mass time
mass velocity 2
4
and where m, are constants which depend on the type and size of
packing used.
Multiplying (153) by (154):
rate of transfer
interf area
interf area
vol. of packing K y y ay ( *)
r K a y yy molar rate of transfer
volume of packing( *) (155)
Since Ky and a usually cannot be measured independently, what is
usually found in correlationsof mass transfer rates is their
product Kya.
If we now multiply by the volume of this slice of the
packedcolumn, we will obtain the rate at which acetone crosses
theinterface in this slice:
NA = rate of transfer = r S z
is the rate of transfer in a thin slice of column, where
S = empty tower cross-section =DT
2
4
S z = volume of slice
At steady state, the rate of transport of acetone into the vapor
must equal the rate out:
in = out
(Vy)z = (Vy)z+z + rSz (156)
Dividing by -z and letting z 0, (156) becomes:
d Vy
dzrS
b g (157)
Now the total molar flowrate (V) will change as the gas phase
loses acetone, but the flowrate ofacetone-free air (V' ) does not
change with z:
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VV
y
1 Vy
V y
y
1
21 1 11
d d V y d y dy dz V dyVy V V
dz dz y dz y y dzy
(157) becomes:V
y
dy
dzrS K a y y Sy
1 *b g (158)
The second equation is obtained by substituting our expression
for the rate of absorption r from(155). We can now solve this
equation for the height dz of the slice needed for a certain
changedy in concentration:
dzVdy
y y y K a Sy
1b gb gd i*
So if we know the local mole fractions y and y* and the change
in the mole fraction dy of the gasphase which occurred between the
top and bottom of this slice, we can calculate the height of
theslice. The total height of packing is the sum of the height of
each slice:
Z dzVdy
y y y K a ST
Z
yy
yT
b
a
z z0 1b gb gd i*Factoring out the S, which is constant along the
entire heightof the packing, and changing the order of integration
(sinceyb>ya):
ZS
Vdy
y y y K aT
yy
y
a
b
z1 1b gb gd i*
The equation above represents the most general form of thedesign
equation. Unfortunately, it hard to give any meaningto the integral
in this general form. So lets examine somelimiting cases. For the
case of a dilute gas stream, we knowthat
y y V V K ay 1 1 1 const const
After making these approximations, we have
OL EC
xa xb
yb
ya
y*y
x
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avg*
b
a
y
Ty y
V S dyZ
K a y y
(159)
which is a little simpler. We have added the subscript avg to
suggest that, in case V and Kyaare not quite constant, we might
still evaluate this quotient at each end of the packed height
(i.e.at y=ya and at y=yb) and average the two values.
Lecture #30 begins here
EVALUATION OF NTU'S
To better understand how toevaluate this integral, we will
continuethe example we started a couple oflectures back (see page
165). Theproblem is summarized in the figureat right. The water
flowrate given inthe figure is 1.08 times the minimumwhich was
determined on page 172:
min
1.08 2.06
L L
V V
We want to evaluate NOy from (163). The y in this expression is
the mole fraction of acetone inthe absorber at some particular
elevation. If the corresponding mole fraction of acetone in
thewater is x, then two are related by the operating line:
y(x):
FHG
IKJ
FHG
IKJL
x
x
x
xV
y
y
y
ya
a
a
a1 1 1 1(160)
To obtain y = yb = 0.14 with xa = 0.0002 and ya = 0.00807 (5% of
acetone remains with gas), wefind that x = xb = 0.07. Given a value
of y in the interval ya y yb, we can use the operatingline to solve
for x(y). On the other hand, y*(x) is the mole fraction of acetone
in the gas, whichwould be in equilibrium with a liquid having mole
fraction x.
2* 0.33 exp 1.95 1ace acep P
y x x x xP P
(161)
water= 0.0002= 2.87 lbmol/min
xL'
a
acetone in air= 0.14
= 1.39 lbmol/miny
V'b
xb = 0.07
ya = 0.00807
(95% recoveredin liquid)
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A plot of y(x) and y*(x) is shown below at left.Neither curve is
a straight line on thesecoordinates, although the operating line is
close.
To evaluate the integral, we need to compute theintegrand at
various points in the domain:
0 0.02 0.04 0.06 0.080
0.05
0.1
0.15
0
y b
yi
yei
x a x b
xi
y x from OL y* from EC1
y y *
ya = 8.07E-03 2.00E-04 4.64E-04 131.4010.022 7.18E-03 0.016
168.6350.036 0.014 0.031 204.7470.05 0.021 0.045 225.474
0.063 0.028 0.059 219.9020.076 0.035 0.071 192.770.089 0.042
0.083 158.1470.102 0.049 0.094 126.1930.115 0.056 0.105 100.380.128
0.063 0.115 80.583
yb = 0.14 0.07 0.125 65.615
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A plot of the integrand might look somethinglike that shown
above at right. The area underthis curve represents.
21.0*
b
a
y
y
dy
y y
To evaluate this integral in Mathcad, itconvenient to pose y and
y* as functions of x:y(x) is the operating line given by (160)
andy*(x) is the equilibrium curve given by (161).Then
21.0
* *
b b
a a
dydy dx y x dx
y xdx
y x
dy y xdx
y y y x y x
The maximum value of the integrand occurred at the point in the
column where the operating linecomes closest to the equilibrium
curve. The denominator y-y* is a measure of this distance.Because
the equilibrium curve turns downward in this case, the "pinch"
point occurs in themiddle of the range. In other cases (e.g. NH3 in
water), the equilibrium curve turns upwardinstead.
Special Case: ya
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Thus we plot y/(y-y*) versus lny. Since boththe numerator and
the denominator becomesmall at the lower limit, the quotient tends
tobe more near constant with y, allowing theintegration to be done
with fewer points and auniformly spaced grid.
TRANSFER UNIT
Even if the integrand varies significantlyover the domain of
integration, we can still pullit outside the integral, provided we
replace itwith some appropriate constant, which can bethought of as
the mean value for this range.Recall the Mean Value Theorem of
Calculus:
dy
y y y ydy
y y
y yy
y
y
yb a
a
b
a
b
z z* * *1b g b gavg avg (162)Of course, it is not usually known
what type of average to use, but for the present purpose,
theprecise value is not important. The integral can be thought of
as the total change in gas molefraction divided by the average
driving force.
(Gas Phase) Transfer Unit -- a slice (not necessarily thin) of
an absorber in which the gasundergoes a change in y equal to y y *b
gavg which represents the average driving forceover the entire
absorber.
No. of TU's -- the number of these it takes to make the entire
absorber (not necessarily aninteger)
Height of TU -- height of a slice of an absorber corresponding
to one TU
From (162), we see that the integral clearly represents the
number of transfer units:
Ndy
y yOy
y
y
a
b
z * (163)This is called the number of overall gas phase transfer
units since it is based on the overalldriving force expressed as
the gas-phase mole fraction. Since the product in (159) represents
thetotal height of packing, the coefficient of this integral must
represent the height of a transferunit:
6 5 4 3 2 10
1
2
3
0
lny0
lnyNy
y yeq
ln( )y
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HV S
K aOy
yFHG
IKJ avg
So that our design equation can be rewritten as:
ZT = HOyNOy
which is called the HTU method for sizing an absorber.
Throughout the derivation above wemade use of one type of mass
transfer coefficient:
molar rate of mass transfer
packed volume= r = Kya(y-y*)
Of course, other types of mass transfer coefficient are commonly
encountered and can also beused to compute the height of packing
required:
r = Kxa(x*-x) = kxa(xi-x) = kya(y-yi)
to mention a few. Of course, there are still more definitions
based on driving forces expressed asdifferences in molar
concentration or partial pressure. Any one of these mass
transfercoefficients can be used in the HTU method:
ZT = HOyNOy = HOxNOx = HxNx= HyNy
where
Ndx
x xH
L S
K a
Ndx
x xH
L S
k a
Ndy
y yH
V S
k a
Oxx
x
Oxx
xix
x
xx
yiy
y
yy
b
a
b
a
a
b
FHG
IKJ
FHG
IKJ
FHG
IKJ
zzz
*avg
avg
avg
Note that the N's are not equal to each other; thus the number
of transfer units depends on whichdriving force you use.
Ny NOy
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However, its still not apparent what average to use to compute
avg*y y . There is one
case in which we can provide a simple answer: For very dilute
solutions,* this integral in (162)can be evaluated analytically
(i.e. numerical integration is not required). If both x
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Recall the design equation: Z H NV S
K a
y y
y yT Oy Oy
y
b a
L
*b g (166)
This contains the product that appears in (165). Solving (166)
for V(yb-ya)
V y y K a SZ y yb a y T L b g : b gtotal volume
of packing
*
where SZT is the total volume of packing in the absorber and a
is the interfacial area per unitvolume of packing. Thus
aSZT = total interfacial area in entire absorber AT
and substituting in (165) yields:
qT = KyAT y y L *b g
which is analogous to: qT = UATTL
Thus, for dilute solutions, the design equation for an absorber
is identical in form to that for heatexchangers: with the overall
mass transfer coefficient Ky analogous to the overall heat
transfercoefficient U and the log-mean of y-y* at the two ends of
the absorber analogous to the log-meanT.
RELATIONSHIPS AMONG HOX, HOY, HX AND HY
In (145) on page 178, we obtained a relationship between the
overall mass-transfercoefficient and the single-phase mass-transfer
coefficients. This relationship can be easilymodified to relate the
corresponding heights of transfer units, which are often used in
theliterature to report correlations of mass transfer rates.
Dividing both sides of (145) by a (the interfacial area per
volume of packing), we obtain:
1 1
K a k a
m
k ay y x (167)
where m = (yi-y*)/(xi-x) = avg. slope of E.C.
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Of course, if the equilibrium curve is straight (as it will be
in dilute solutions), then m is its slope.If we now multiply (167)
by V/S we have the height on an overall gas transfer unit on the
left-hand side:
V S
K a
V S
k a
mV S
k a
V S
k am
V
L
L S
k ay
H
y x y
H
x
HOy y x
; : :
Recalling the definitions for various heights of a transfer unit
we end up with
H H mV
LHOy y x
Had instead we started with (147) and multiplied by L/S we would
obtained
Ox y xL
H H Hm V
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Lecture #31 begins here
Fig. 7. Pressure drop across a packed bed of 1-inch ceramic
Intalox saddles as a function of air and waterflowrates. Taken from
Fig. 22-4 of MS&H5.
PRESSURE DROP IN PACK BEDS
Usually, flow in a packed-bed absorber is countercurrent.Gravity
causes the liquid to flow down through the packing,whereas a small
pressure drop drives the gas flow upward. Thispressure drop plays
an important role in packed beds. Withoutliquid present, the
pressure-drop across dry packing increasesapproximately with the
square of the gas flowrate (see Fig. 7).
GG = mass velocity of gas =lb/hr of gas
cross-section of towerGM V
S
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where MG is the average molecular weight of thegas. Once liquid
is flowing down the column asgas is flowing up, some of the space
between thepacking particles is taken up by liquid, leavingless
space for the gas to flow. Forcing the samegas flow through a
smaller opening will increasethe frictional contribution to the
pressure drop.Thus the curves for increasing liquid flow areabove
that for dry packing although the curvestend to be parallel.
Note that the curves with liquid flow curl up at high gas flow
rates. This can be explained asfollows:
packingparticle
packingparticle
stagnantair
flowingair
packingparticle
fastflowingair
The upward flow of the gas exerts a shear force on the liquid,
retarding its downward motion. Toget the same flowrate with a lower
velocity, you need a thicker film. Thus liquid holdupincreases with
increased gas flowrate. As the gas flowrate increases so does this
shear force.When the shear force becomes comparable to gravity, the
liquid flow might slow to zero(rightmost figure above) and flooding
occurs in the tower.
The amount of liquid residing or accumulating in the tower is
called:
(liquid) holdup -- fraction of interstitial volume occupied by
liquid
intersticial volume -- "empty" space between and inside packing
particles; that volumeoccupied by liquid or gas.
loading -- an increase in liquid holdup caused by an increase in
gas flowrate
For a given liquid flowrate, there is a maximum flowrate of gas
which can be forced through thecolumn. If you exceed this maximum
the shear forces on the liquid are so high that they exceedthe
weight of the liquid. Then the net force on the liquid is upward
and you blow the liquid backout the top of the column. This is
called:
flooding -- liquid downflow is essentially stopped by high gas
upflow
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Since you can't get the liquid through the column, flooding is
clearly a condition to be avoided.Generally, to get a high area of
contact between liquid and gas, you want to operate well belowthe
flooding velocity:
Flooding According to McCabe, Smith & Harriott, 5th Ed.
at flooding: Pflood 2 inch H2O/ft of packing
loading: P 0.5 inch H2O/ft
normal oper.: P 0.25 to 0.5 inch H2O/ft
Normal operating pressure drops are just below those for which
loading begins.
Flooding According to Geankoplis
Flooding occurs at a pressure drop depending on the packing
factor:
0.7 -1
flood -1
0.115 for 60 ft
2 for 60 ft
p p
p
F FP
F
(168)
where this formula gives Pflood in inches of water per foot of
packed height. Once GG atflooding is determined then
loading: GG/GG,flood
normal oper.: GG/GG,flood 0.5 0.7
with the high ratios used for high flowparameters (see x-axis of
Fig. 9).
Notice (from Fig. 7) that when we increasethe liquid flowrate,
flooding occurs at lowergas flows. If we increase the size of
thepacking particle, we can tolerate higher gasflows. While higher
gas flows can beobtained with larger packing particles, theamount
of interfacial area is generally less (i.e.a is decreased). These
effects are summarizedby Fig. 8 at right.
Fig. 8. Air flowrate at which flooding occurs for ceramicIntalox
saddles at different water flowrates. Taken from
Fig. 22.5 in MS&H5.
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Fig. 7 and Fig. 8 are for a particular packing (1 Intalox
saddles) and for a particulartemperature and pressure (20 C and 1
atm). A more general correlation of pressure drop isprovided by
Fig. 9 where
GG
mG
S = mass velocity of gas [=] kg-s-1-m-2
GG
G
Gv
= superficial velocity of gas [=] m/s
L = dynamic viscosity of liquid [=] kg-m-1-s-1
L
L
= kinematic viscosity of liquid [=] m2/s*
The y-coordinate of this graph in Fig. 9 is not dimensionless,
soyou need to use the units summarized in the table at right.
1 poise = g-cm-1-s-1 and 1 centipoise = 1 cp = 10-2 g-cm-1-s-1
(approximate viscosity of water at roomtemperature).
* 1 stokes = 1 cm2/s and 1 centistokes = 10-2 cm2/s. Yes, the s
belongs at the end here: it does not denote theplural but is part
of a mans name (George Gabriel Stokes, who gave us Stokes law).
Fig. 9. General correlation of pressure drop across packed beds
for random packings.Taken from Fig. 10.6-5 of Geankoplis.
Symbol UnitsGL, GG lb/ft2-sL cp
L, G lb/ft3
GG
G
Gv
ft/s
Fp ft-1
LL
centistokes
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The size and type of packing is accounted for in this
correlation through the parameter
Fp = packing factor
Values of the packing factor for various types of packing are
given in the following table (Fig.10).
Fig. 10. Characteristics of common packings. Taken from Table
10.6-1 of Geankoplis.
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TOWER DIAMETER
Once the total molar liquid and gas flowrates (L and V) are
known, we can choose thediameter of tower we need. The diameter of
the tower is usually chosen such that the pressuredrop is some
prescribed value below flooding:
choose DT such that: GG/GG,flood 0.5 0.7
where the flooding velocity GG,flood is found from Fig. 9 such
that the pressure drop is given by(168).
where2
4
mass flowrate of liquid
cross-sectional area of column T
L LL
D
M L M LG
S
and where ML is the average molecular weight of the liquid.
Similarly, GGM V
GS
So to get a certain pressure drop for a certain set of
flowrates, we must choose a particular valuefor the tower diameter.
Although it looks like you have to guess the diameter DT in order
tocalculate GL and GG to get p, a trial-and-error can be avoided by
noting that the abscissa(x-coordinate) does not depend on the
diameter:
LL L
GG G
M LG M LS
M VG M VS
Note that the unknown cross-sectional area S conveniently
cancels out. What remains is known.So the procedure is as
follows:
Given: L, V
Find: DT
Solution:
Step 1) calculate the flow parameter (abscissa) of Fig. 9
Step 2) calculate Pflood from (168).
Step 3) locate the point on the curve of Fig. 9 corresponding to
Pflood which also has thecalculated value of the flow parameter
(abscissa)
Step 4) read the corresponding capacity parameter (ordinate of
this point) from Fig. 9
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06-361 page 200 Fall, 2008
Copyright 2010 by Dennis C. Prieve PDF generated September 25,
2010
Step 5) calculate GG which gives this value for the capacity
parameter. This givesGG,flood
Step 6) calculate the operating value of GG using GG/GG,flood
0.5 0.7
Step 7) calculate the desired tower diameter from
2
4GT
G
M VDS
G