absorbtion pe

8/12/2019 Absorbtion Pe

1/70

ABSORPTION

Bringing the dirty effluent gas into contact with the

scrubbing liquid and subsequently separating the

cleaned gas from the contaminated liquid

Absorption is a basic chemical enginnering unit

operation which in the APC field is reffered as

scrubbing

They have wide use in controlling SO2, H2S, and light

hydrocarbons


2/70

ABSORPTION

_Wet scrubbers can be categorized into 3 groups:

1. Packed-bed counterflow scrubbers

2. Cross-flow scrubbers

3. Bubble plate and tray scrubbers


3/70

SCRUBBERTYPES

Packed TowerSpray tower Venturi Absorber


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CONCEPTOFABSORPTION

Gas absorption is the removal of one or morepollutants from a contaminated gas streamby allowing the gas to come into intimatecontact with a liquid that enables thepollutatants to become dissolved by theliquid. The principal factor dictatingperformance is the solubility of thepollutants in the absorbing liquid.

The rate of transfer in the liquid is dictated bythe diffusion processes occurring on eachside of the gas liquid interface.


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LIQUIDWASTE

Pollutants removed from the gas stream transferred

into liquid phase whose disposal is another issue to

deal with.

Therefore scrubber needs other units such as storage

vessels, additives to treat the scrubbing liquid

according to required discharge standards.


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ABSORPTION

Absorption units must provide large surface area of

liquid-gas interface

Therefore the units are designed to provide large

liquid surface area with a minimum of gas pressure

drop


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PACKEDTOWER


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PACKINGMATERIALANDSHAPES

Packing material (must be inert) is designed toincrease the liquid-film surface

Many geometric shapes are available : Raschig ring,

pall ring, berl saddle, tellerete etc.


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PACKINGMATERIALANDSHAPES


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PACKINGMATERIAL

PROPERTIES


11/70

ABSORPTIONTHEORY Physically, the absorption of a pollutant gas from a

moving gas stream into an appropriate liquid stream isquite complex

Basically the transfer process into each fluid stream is

accomplished by 2 mechanisms:

The pollutant species is transferred from the bulk of thegas stream toward the gas-liquid interface by turbulent

eddy motions

Very close to the interface laminant flow is valid and

transfer is accomplished by molecular diffusion

On the liquid side of the interface process is reversed


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ABSORPTIONTHEORY


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ABSORPTIONTHEORY

On the basis of Fick's Law, the diffusion of one gas

(A) through a second stagnant gas B, NA, the molarrate of transfer of A per unit cross-sectional area isgiven by;

NA= -DAB(dcA/dz)/(1-(cA/c)

DAB: molecula diffusion coef. (m2/t)

cA: molar concentration of species A (mol/L)

c: molar concentration of the gas mixture (mol/L)

z: the direction of mass transfer (m)

DABtables are available for a number of binary gasmixtures


14/70

ABSORPTIONTHEORY

Mass transfer rate per unit area for molecular

diffusion of A through a second liquid is given by: NA= -DL/z (cA2-cA1)

DL: liquid phase molecular diffusion coef. (m2/t)

cA2-cA1: concentration difference of A over the distance z

Typical values of DLfor binary mixtures are tabulatedin the literature.


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THEEQUILIBRIUMDISTRIBUTIONCURVE

Before entering into details of mass transfer, let's

summarize the method of presenting equilibrium data

for a pollutant A distributed between liquid and gas

phase

Inert LiquidSolvent

Inert carriergas

Injectsolute A

P=c

Molefraction ingas, yA

Mole fraction in l iuid, xA

Exp. Equilib.Distribution

curve


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THEEQUILIBRIUMDISTRIBUTIONCURVE

After sufficient time, no further change in the

concentration of A in two phases. These concentration

can be measured and converted into mole fraction xA

in the liquid phase and yAin the gas phase

Inert LiquidSolvent

Inert carriergas

Injectsolute A

P=c

Molefraction ingas, yA

Mole fraction in lqiuid, xA

Exp. Equilib.Distribution

curve


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MASSTRANSFERCOEFFICIENTSBASEDON

INTERFACIALCONCENTRATIONS

When mass transfer occurs in moving liquid and

gaseous streams, it is difficult to evaluate the separate

effects of molecular and turbulent diffusion

An alternative to this is to express NAfor each phase

in terms of mass transfer coefficient k and a drivingforce based on the bulk and interfacial concentrations

for that phase


18/70



For the liquid phase:

NA

= kL(c

Ai-c

AL) = k

x(x

Ai-x

AL)

kL(is the liquid mass transfer coeff. Based on

concentration, in length per unit of time, cAi

is the

concentration of A in the liquid phase at the interface, cAListhe concentration of A in the bulk of the phase, in moles

per unit volume.

kxis the liquid mass transfer coefficient based on mole

fractions, in moles per units of time and length squared, xA

is the mole fraction of A in the liquid interface, and xAListhe mole fraction of A in the bulk of the liquid phase


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INTERFACIALCONCENTRATIONS For the gas phase:

NA

= kG(p

AG-c

Ai) = k

y(y

AG-y

Ai)

kGis the gas phase mass transfer coeff. based on partial

pressures, in moles/length2 time, pAG

is the partial pressure of

A in the bulk of gas phase pAi

is the partial pressure of A in

the gas interface

kyis the gas phase mass transfer coefficient based on mole

fractions, in moles per units of time and length squared, yAG

is the mole fraction of A in the bulk of the gas phase, and yAi

is the mole fraction of A in the gas phase interface


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However this approach to determining NA

is not practical

since kxand kyare difficult to obtain and no way to measurethe values of yAiand xAiexperimentally since any attempt to

do it will perturb the equilibrium between the two streams


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OVERALLMASSTRANSFERCOEFFICIENTS When mass transfer rates are reasonably low, NAcan be

expressed as: N

A= K

G(p

AG-p

A*) = K

y(y

AG-y

A*)

KGand K

yare local overall mass transfer coefficients

pA

* : equilibrium partial pressure of solute A

in a gas phase which is in contact with a

liquid having the composition of cALof the

main body of the absorption liquid

yA

*: defined similarly in terms of a liquid with

mole fraction xAL

of the bulk liquid


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OVERALLMASSTRANSFERCOEFFICIENTS

slope=m'

Point P represents the state of thebulk phase of the 2 fluid streams,y

AGand x

AL.

The point M represents the state(y

Aiand x

Ai) associated with

equilibrium at the interface

The distance between P and C is ameasure of the driving force.


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OVERALLMASSTRANSFERCOEFFICIENTS

NA

= KG(p

AG-p

A*) = K

y(y

AG-y

A*) This equation is usually

restricted the resistance to mass transfer is primarily in the

gas phase, which characterizes the majority of absorption

problems in air pollution work

The solubility of the polutant gas normally determines the

liquid that is chosen

The major physical problem is getting the pollutant to

diffuse through the gas phase to the interface,

consequently gas phase controls the process.

If the liquid phase controls: N

A= K

L(c

A*-c

AL) = K

x(x

A*-x

AL)


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OVERALLMASSTRANSFERCOEFFICIENTS It is important to note that the quantities p

A*

,yA

*,cA

*,xA

*

do not represent any actual condition in the absorption

process but are related in each case to a real

concentration in one of the bulk fluids through the

equilibrium data for the two-phase system. From the

geometry of the previous figure:

yAG-yA*= yAG-yAi+(yAi-yA*) y

Ai-y

A*=m'(x

Ai-x

AL)

yAG

-yA

*= yAG

-yAi

+m'(xAi

-xAL

)

1/Ky=1/k

y+m'/k

x


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MASSBALANCESANDTHEOPERATINGLINEFOR

PACKEDTOWERS

T = constP = constCross-sectionalarea, A

Gm,2

Gc

y2

Y2

Lm,2

Ls

x2

X2

Gm,1

Gc

y1

Y1

Lm,1

Ls

x1

X1

dz

Gmmolar total gas flowrate (carrier gas +pollutant)Gc molar inert carriergas flow rateL

mmolar total solvent

flow rate (solvent +

absorbed pollutant)Ls molar solvent flowratex is the liquid molefraction of pollutant, yis the gas phase molefraction of the

pollutants, X is theliquid phase mole ratioand Y is the gas phasemole ratio


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PACKEDTOWERS

Mole fraction and mole ratio:

X = x/(1-x) Y = y/(1-y)

Subscript m denotes that rates are in the units of mole

basis

The conservation of mass principle applied to the

pollutant species in terms of total mass flow rates at

top and bottom yields:

Gm,1

y1

+ Lm,2

x2

= Gm,2

y2

+ Lm,1

x1

or Gm,1

y1-G

m,2y

2= L

m,1x

1-L

m,2x

2


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PACKEDTOWERS

In Gm,1

y1-G

m,2y

2= L

m,1x

1-L

m,2x

2total gas and liquid

flow rates are not equal at the top and the bottom of

the column, therefore we cannot further simplify this

equation.

When we write the equation in terms of the carrier gasand liquid solvent rates then:

GC,m

(Y2-Y

1) = L

S,m(X

2-X

1)

These two equations above gives a straight line on Y-

X coordinates with a slope of Lsm/GCmand calledoperating lines.


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The operating line lies

above the equilibrium line

for absorption

For a stripping (removal of

gas from liquid stream) theoperating line must lie

below the equilibrium line in

order for the drving force to

act from the liquid phasetoward the gas phase

Dirty air

Dirty waterClean water

Clean air


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THEMINIMUMANDDESIGNLIQUID- GASRATIO

At the bottom and top of the absorber, parameters

Gm,1

, Gc, y

1, G

m2, y

2, and x

2are known.

We need to determine Ls, and x

1

So we have one equation with 2 unknowns...

However selection of one of these values, obviously

fixes the other.

How to select a value?


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THEMINIMUMANDDESIGNLIQUID- GASRATIOThe minimum rateis highly

undesirable. Atthis point drivingforce is almost 0.Hence it wouldtake an infinetelytall absorber to

accomplish thedesired separation

As a generaloperating principlean absorber istypically designed

to operate at liquidrates which are 30to 70 % greaterthan minimumrate.


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TOWERDIAMETERANDPRESSUREDROPPER

UNITTOWERHEIGHT

For a given packing and liquid flow rate in an

absorption tower variation in the gas velocity has a

significant effect on the pressure drop

As the gas velocity is increased, the liquid tends to be

retarded in its downward flow, giving rise to term liquidholdup (LH)

A LH increases, the free cross-sectional area for gas

flow decreases and pressure drop per unit height

increases.


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PROBLEMSWITHHIGHGASVELOCITY Channeling: the gas or liquid flow is much greater at some

points than at others Loading: the liquid flow is reduced due to the increased

gas flow; liquid is held in the void space between packing

Flooding: the liquid stops flowing altogether and collects

in the top of the column due to very high gas flow TO AVOID this condition experience dictates operating at

gas velocities which are 40 to 70 % of those which

causing flooding


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FLOODPOINT

The relationship between DP/Z and other importanttower variables-liquid and gas rates, liquid and gas

stream densities and viscosities, and type of packing

has been extensively studied on an experimental

basis. A widely accepted correlation among these

parameters can be seen in below figure

Where G' and L': superficial gas and liquid mass flow

rate defined as actual flow rates divided by the emptycross-sectional area of the tower.


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L/G(rG/rL-rG)

L:liquid mass flux (lb/s-ft2)

G:gas mass flux (lb/s-ft2)

F:packing factor (ft2/ft3)

mL:liquid viscosity, cpgc: proportonality

constant, 32.17 ft-lb/s2-lbf

rL:liquid density, lb/ft3

rG:gas density, lb/ft3

GGLc

L

g

FG

rrr

m

)(

)'( 1.02

In Cooper and Alleys book, Figure 13.6 can be used. Note that in Figure 13.6 Gx and Gy are

liquid and gas flux (lg/s-ft2), respectively. In our notation G and L correspond to Gx and Gy


36/70

PACKINGFACTORF

The top line in the figure represents the generalflooding condition for many packings. The flooding

condition however has been found to vary as a

function of the packing factor F (dimensionless

packing factor tabulated below) Recent studies showed that when F is in the range of

10 to 60, the pressure drop can be expressed by:

DPflood

= 0.115F0.7


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PACKINGDATA

DETERMININGTOWERDIAMETER


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First abscissa value is calculated

(L'/G')(pG/(pL-pG))0.5

Where this value intercepts the flooding line onFigure A, move horizontally to the left and read thevalue of the ordinate:

(G')2F(mL)0.1/g

c(p

L-p

G)p

G

Calculate the G and take 30 to 70% of it to preventflooding

Tower crossectional area: A = G/G

Evaluate the tower diameter

DETERMINING EXPECTED PRESSURE DROP PER


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DETERMININGEXPECTEDPRESSUREDROPPER

UNITHEIGHTOFTOWER

First calculate actual G and L and then calculatethe abscissa and the ordinate for use in Figure13.6

From those values the intersection on the figuredefines the pressure drop per foot of packed

height

Another emprical correlation found in the litrature forthe DP in packing when operating below the loadpoint is

DP/Z = 10-8

m[10nL/rL

](G2

/rG) m and n are packingconstants see Table 6.2

DETERMININGTOWERDIAMETERANDEXPECTED


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PRESSUREDROPPERUNITHEIGHTOFTOWER

EXAMPLE


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A packed tower is to be designed to remove 95% of theammonia from a gaseous mixture of 8 percent ammonia and92% air, by volume. The flow rate of the gas mixture entering

the tower at 68 F and 1 atm is 80 lb-moles/hr. Watercontaining no ammonia is to be the solvent, and 1-in. Ceramicraschig rings will be used as the packing. The tower is tooperated at 60% of theflood point and the liquid water rate isto be 30% greater than the minimum rate. Determine

1. The gas-phase flow rates, in lb-moles/hr, for the soluteand carrier gas

2. The mole ratios of the gas and liquid phases at inlet andoutlet and the required water rate in lbmoles/hr.

3.The gas and liquid rates (lb/hr) for carrier gas, solutegas, total gas, liquid solvent, solute in liquid, and total

liquid 4. The tower area and diameter 5. The pressure drop based on the two methods given in

the lecture notes.

EXAMPLE


42/70

Removal efficiency: 95%

Effluent Stream Composition: 8% ammonia and 92% air

Gas T and P: 68F and 1 atm

Flowrate: 80 lb-moles/hr

Liquid phase: Containing no ammonia

EXAMPLE


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X 0.0206 0.0310 0.0407 0.0502 0.0735 0.0962

Y 0.0158 0.024 0.0329 0.0418 0.0660 0.0920

Determine composition of the liquid at the exit (X1)

(Inlet liquid concentration since pure water is used is x2=X2=0)

Use equilibrium data for ammonia-air-water mixtures which are givenbelow for 68 F and 14,7 psia. :

In order to determine composition of liquid at the exit, we need tocalculate the minimum solvent flow rate first.

By plotting X vrs Y at the equilibrium, we can evaluate the minimumsolvent and then operating solvent rate.

In Cooper and Alleys book, use Table B4 in the Appendix.

EXAMPLEe


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0

0,02

0,04

0,06

0,08

0,1

0 0,02 0,04 0,06 0,08 0,1

X, moles solute per mole solvent

y,molessolute

permole

carriergas

X2,Y2

(Lm,S/Gm,C)min=

(0,087-0,00435)/0,092

Y1=0.087

Since the liquid rate is to be 30% greater than the miniumu rate

(Lm,S)/Gm,C)design= 1,30(0.90) = 1.17 mole/mole

Lm,S= Gm,C*1,17 = 1.17*73.6 = 86.1 lb moles/hr

0,90

0.092Y2=0.00435

EXAMPLE

0,1ole


45/70

Now, X1 can now be found.

1. Graphically by drawing operating line with a slope of 1.17 with startingpoint of (0, 0.00435) and the point crosses Y1=0.087 can be read. OR

2. From Lms/Gm,C= Y2-Y1/(X2-X1)=0.00435-0.087/(0-X1) = 1.17

X1= 0.0707 lm mole A/lm mole water or x1= 0.066 lb mole A /lb moles

solution

0,90

0.00435

0,02

0,04

0,06

0,08

0 0,02 0,04 0,06 0,08 0,1

X, moles solute per mole solvent

y,molessolutepermo

carrier

gas

X2,Y2

(Lm,S/Gm,C)=1.17

Y1=0.087

X1: 0.0707

F R


46/70

FLOWRATES

The gas and liquid rates:GC= 73.6*29 = 2134 lb/hrGA,1= 6.4*17 = 109 lb/hrGA,2= 0.32*(17) = 5.4 lb/hrLS= 86.1*18=1550 lb/hr

LA,1= DGA= 109*0.95=104 lb/hr

Therefore:G1= 2134 +109=2243 lb/hr bottomL1= 1550 +104 = 1654 lb/hr

G2= 21345+5=2139 lb/hr topL2=1550 + 0 = 1550 lb/hr top

T = constP = const

Cross-sectionalarea, A

Gm,2G

c

y2

Y2

Lm,2L

s

x2

X2

Gm,1Gc

y1

Y1

Lm,1Ls

x1

X1

dz

TOWERAREA


47/70

To determine the tower area, we need to use Figure floodingcorrelation plot.

Therefore we need to calculate gas and liquid phase densitiesat the top and bottom of the tower. Since the ammoniacontent is very low in liquid phase, use the density of purewater, 62.3 lb/ft3 as the solution density through the tower.

For the gas phase assume ideal gas behavior:r= P/RT = MwP/RTAt the top: Mw= SyiMi= 0.00435*17 + 0.9957*29 = 28.95

r= 28.95*14.7/(10.73*528) = 0.075 lb/ft3At the bottom Mw= 0.08*17 + 0.92*29 = 28.04

r= 28.04*14.7/(10.73*528) = 0.0728 lb/ft3

Now calculate the abscissa of Flooding Figure

TOWERAREA


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TOWERAREA


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PRESSUREDROPP d b d t i d f th fl di fi


50/70

Pressure drop can be determined from the flooding figure or

from an emprical equation

DETERMINATIONOFANABSORPTIONTOWERHEIGHT


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HEIGHT

Height of a packed tower = f(the overall resistance to masstransfer between the gas and liquid phases, the averagedriving force and interfacial area)

Consider a differential height of the absorber dZ. In height dZ,the rate of mass transfer of species A

a: interfacial area available to mass transfer per unit volume of

the packingA: cross-sectional area of the tower

)'()()( yGAdyGdadZAN mA


52/70

TOWERHEIGHT

The equation can be also written for liquid resistance part.

)(

'

)(

'

)'()(

**

*

yyaPK

yG

yyaK

dyG

dZ

yGddzyyaK

G

m

y

m

mAAGy

NAA(adz)= d(Gmy)= Ad(G 'y)

TOWER HEIGHT


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TOWERHEIGHT

To solve the above equation we can determine the overall

value of Kya (Kga) based on experimental pilot plantoperated with a certain packing and gas/liquid rate. The

right side of the equation can be integrated from theknowledge of the operating line and equilibrium line

chracteristics.

This method can be modified to deal with the height of a

transfer unitand the number of transfer units bymodifiying the equation somewhat

dZ=Gm'y

Kya(y y*)

=

Gm'y

KGaP(y y*)

TOWERHEIGHT


54/70

To solve the above equation we can determine the overall

value of Kya (Kga) based on experimental pilot plantoperated with a certain packing and gas/liquid rate. The

right side of the equation can be integrated from theknowledge of the operating line and equilibrium line

chracteristics.

This method can be modified to deal with the height of a

transfer unitand the number of transfer units bymodifiying the equation somewhat

)(

'

)(

'** yyaPK

dyG

yyaK

dyGdZ

G

m

y

m

The equation can be expressed in terms of height of

Tower Height


55/70

The equation can be expressed in terms of height oftransfer unit (HTU ) and number of transfer units :

HTU is reaonably constant through the absorber and hasunit of length. NTU is dimensionless.

55

1

*

' y

yOG

m

Z yy

dy

aPK

GZ

HTU orHoy

NTU or

Noy

x1, y1For dilute gas


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2014/6/11

Aerosol&

Par

ticulateResearchLab

56

1

*

' y

yOG

m

Z yy

dy

aPK

GZ

x1, y1*

xZ, yZ*

xZ, yZ

Z = G 'm

KOGaPy1 yz yLM ; yLM = y

1 y1*

( ) yz yz*

( )ln y1 y1*yz yz*

streams, transferunit equation can

be simplified:

Pure amineLm= 0.46 gmole/s

0.04% CO2


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2014/6/11

Aerosol&

Par

ticulateResearchLab

57

1.27% CO2Gm= 2.31

gmole/s

C* = 7.3%CO2inamine

Q: A Packed tower using organic amine at 14 oC to absorb CO2.The entering gas contains 1.27% CO

2

and is in equilibrium witha solution of amine containing 7.3% mole CO2. The gas leavescontaining 0.04% CO2. The amine, flowing counter-currently,enters pure. Gas flow rate is 2.31 gmole/s and liquid flow rate is0.46 gmole/s. The towers cross-sectional area is 0.84 m2.KOGa = 9.3410

-6s-1atm-1cm-3. The pressure is 1 atm.Determine the tower height that can achieve this goal.

Absorption of concentrated vapor


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2014/6/11

Aerosol&

Par

ticulateResearchLab

58

Mole balance on the controlled volume

)'()'(0 xLdzdyG

dzd

mm

Gas flux

yGG mm

11'' 0

Liquid flux

xLL mm

11'' 0

1

1

0

0

1

1

1

1

0

0

1

1

11'

'

11

11'

'

1

x

x

x

x

G

L

y

y

x

x

x

x

G

L

y

y

y

m

m

m

m

x1, y1

x1, y1*

xZ, yZ*

xZ, yZ

EXAMPLE


59/70

EXAMPLEA 1ft diameter packed column is used to scrub a soluble gas

(MW = 22) from an airgas mixture. Pure water enters the top

of the column at 1000 lbm/hr. The entering gas streamcontains 5% soluble gas and 95% air. Ninetyfive percent of

the soluble gas is removed. Both the operating line and

equilibrium curve may be assumed to be straight. The

equation for the equilibrium curve is y = 1.2x, wherex, y =mole fractions. The entering gas mixture flow rate is 800

lbm/hr. The column operates at 30C and 1 atm, and

Kya = 4.29 lbmol/hrft3y

EXAMPLE


60/70

EXAMPLE

Calculate or find:

a) Concentration of the soluble gas in the effluent liquid if the column is operated at

minimum liquid flow rate

b) Concentration of soluble gas in the liquid at a point in column where y = 0.02

c) Height of packed section, ZT

d) Hoy

e) Whether columnis in danger of flooding if it is packed with 12in. ceramic Raschig

rings


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Solution


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Solution


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Solution


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Solution


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Solution


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Solution


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Solution


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Solution


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Solution


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absorbtion pe

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