absorption prie ve

06-361 page 163 Fall, 2008

Copyright 2010 by Dennis C. Prieve PDF generated September 25, 2010

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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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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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

06-361 page 170 Fall, 2008


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

06-361 page 171 Fall, 2008


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

06-361 page 172 Fall, 2008


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.


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.


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

06-361 page 180 Fall, 2008


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.

06-361 page 181 Fall, 2008



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

06-361 page 182 Fall, 2008


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

06-361 page 183 Fall, 2008


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

06-361 page 184 Fall, 2008


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:

06-361 page 185 Fall, 2008


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

06-361 page 186 Fall, 2008


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.


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)

06-361 page 187 Fall, 2008


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

06-361 page 188 Fall, 2008


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

06-361 page 189 Fall, 2008


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

06-361 page 190 Fall, 2008


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

06-361 page 191 Fall, 2008


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

06-361 page 192 Fall, 2008


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.

06-361 page 193 Fall, 2008


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

06-361 page 194 Fall, 2008



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

06-361 page 195 Fall, 2008


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

06-361 page 196 Fall, 2008


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.

06-361 page 197 Fall, 2008


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

06-361 page 198 Fall, 2008


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.

06-361 page 199 Fall, 2008


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

06-361 page 200 Fall, 2008


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

absorption prie ve

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