Liquid Gas Absorption Process

8/19/2019 Liquid Gas Absorption Process

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+

1

Gas-Liquid Separation

Absorption Process

Part 01


2/89

+ 2Content

Types of Separation Processes and Methods Equilibrium Relation Between Phases

Single and Multiple Equilibrium Contact Stages

Mass Transfer Between Phases

Continuous Humidification Processes

Absorption on plate and Packed Tower

Absorption and Concentrated Mixtures in Packed

Tower

Estimation of Mass Transfer Coefficient in Packed Tower


3/89

+ 3Types of Separation Processes and Methods

In order to separate or remove one or more of the component

from its original mixture, it must be contact with anotherphase to allow solute to diffuse from one phase to the others

phase.

During the contact of the two phase, the component of the

original mixture redistribute themselves between the two

phases. The phases are then separated by simple physical

methods.

By choosing the proper conditions and phases, one phase is

enriched while the other phase is depleted in one or more

components.

The two phase pair can be:

1. Gas (vapor)-liquid :absorption, distillation, etc

2. Gas –solid :adsorption, etc

3. Liquid-liquid :liquid-liquid extraction, etc

4. Liquid- solid :leaching, crystallization, etc


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+ 4Gas-Liquid Separation

Absorption

A solute A or several solutes from a gas phase are absorbed into aliquid phase.

involves molecular and turbulent diffussion or mass transfer of solute A through a stagnant, nondifussing gas B into a stagnantliquid C

e.g: absorbing NH3 (A) from air (B) using liquid water (C).

Stripping/desorption

reverse of absorption: separating a solute A or several solutesfrom a liquid phase by contacting the liquid with a gas phase.

E.g. removal of volatile component of the oils by passing out withthe steam.

Humidification

When the gas is pure air and the liquid is pure water

transfer of water vapor from liquid water into pure air. Dehumidification

removal of water vapor from air.


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+ 5Equilibrium Relation Between PhasesGas-Liquid Equilibrium – Henry’s Law

Henry’s Laws:

where,

p A = partial pressure of component A (atm).

H = Henry’s law constant (atm/mol

fraction).

H’ = Henry’s law constant (mol frac.gas/mol

frac.liquid). = H/P.x A = mole fraction of component A in

liquid.

y A = mole fraction of component A in gas =

p A /P.

P = total pressure (atm).

H’ depend on total pressure, whereas H does

not.

At low concentration – a straight line on plot

p A vs x A .

A A Hx p A A x H y '


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+ 6Single Stage Equilibrium Contact

Single-stage process

two different phases (liquid & gas for absorption) are brought into contact

& then separated.

various component diffuse and redistribute themselves between two

phase

long enough, equilibrium is reached.

Two exit stream L1 andV1 leave in equilibrium with each other.

Gas phase contains solute A & inert gas B

Liquid phase contains solute A & inert liquid (solvent) C.

Inert gas is insoluble in the solvent and the solvent does not vaporize to

the gas phase.

L1 ,xA1Lo ,xAo

V2, yA2V1 , yA1

SingleStageLiquid phase

inletLiquid phase

outlet

Gas phaseinlet

Gas phaseoutlet


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+ Single Stage Equilibrium Contact for Absorption

Total material balance

Component A balance on single stage equilibrium contact for

absorption

Both L’ and V’ are constant and usually is known

To solve the value x A1 and y A1, we use a Henry’s Law relationship

Remember the unit H’ is mol frac.gas/mol frac.liquid

7

Lo x Ao +V 2 y A2 = L1 x A1 +V 1 y A1 = Mx AM

1

1

1

1

2

2

1'1'1'1' A

A

A

A

A

A

Ao

Ao

y

y

V x

x

L y

y

V x

x

L

A A x H y '

Lo +V 2 = L1 +V 1 = M

L1 ,xA1Lo ,xAo

V2, yA2V1 , yA1

SingleStageLiquid phase

inletLiquid phase

outlet

Gas phaseinlet

Gas phaseoutlet

V = gas flow rate

V’= inert/pure gas (B) flow rate = V(1-y A )

L = liquid flow rate

L’= inert/pure liquid (C) flow rate = L(1-x A )

x= mole fraction in liquid phase

y = mole fraction in gas phase


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+ 8

A gas mixture at 1.0 atm pressure abs containing air and CO2is contacted in a single-stage mixer continuously with pure

water at 293 K. The two exit gas and liquid streams reach

equilibrium. The inlet gas flow rate is 100 kg mol/h, with a

mole fraction of CO2 of y A2 = 0.20. The liquid flow rate

entering is 300 kg mol water/h. calculate the amounts and

compositions of the two outlet phases. Assume that waterdoes not vaporize to the gas phase.

Example 10.3-1

L1= ?xA1 =?

L0= 300 kg/h

xAo =0

V2= 100 kg/hyA2 =0.20

Single-stageV1= ?

yA1 =?


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+ 9Solution Example 10.3-1

The inert water flow is L’ = L0 = 300 kg mol/h.

The inert air flow V’ is obtained from, V’ = V(1-y A ). Hence, the inert air flowis

V’ = V2 (1-y A2 )

= 100 (1-0.20)

= 80 kg mol/h .

Substituting into equation to make a balance on CO2 (A).

……….(1)

1

1

1

1

180

1300

20.01

20.080

01

0300

A

A

A

A

y

y

x

x


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+ 10

At 293 K, the Henry’s law constant from Appendix A.3 is

H = 0.142 x 104 atm/mol frac.

Then

H’ = H/P = 0.142 x 104 /1.0 = 0.142 x 104 mol frac. gas/mol frac.liquid.

Substituting into,

y A1 = 0.142 x 104 x A1 ………(2)

Solve simultaneous equation.

x A1 = 1.41 x 104 and y A1 = 0.20

To calculate the total flow rates leaving,

hkgmol x

L

L A

/ 3001041.11

300

1 41

'

1

hkgmol y

V V

A

/ 10020.01

80

1 1

'

1

Solution Example 10.3-1


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+ 11

Countercurrent multiple-contact stages.

more concentrated product.

total number of ideal stages = N.

Component B&C may or may be not be somewhat miscible in each other.

two streams leaving a stage in equilibrium with each other. Total overall material balance: L 0 + V N+1 = L N + V 1 = M

Component A balance: L 0 x 0 + V N+1 y N+1 = L N x N + V 1 y 1 = Mx M

For the first n stages: L 0 x 0 + V n+1 y n+1 = L n x n + V 1 y 1

Operating line:

An operating line is an important material-balance equation because it relatesthe concentration yn+1 in the V stream with xn in the L stream passing it.

Slope= Ln/Vn+1 in operating line is varies if the L and V streams vary from stage tostage

1 2 n N

L0

L1

L2

Ln - 1 L n L N - 1 L N

V N + 1

V N

V n + 1

V n

V 3

V 2

V 1

Countercurrent Multiple Contact Stage

yn+1 = LnV n+1

xn +V

1 y

1 − L0 x0V n+1


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+ 12Countercurrent Multiple Contact Stage – Number of Ideal Stages


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+ 13

Graphical calculation for determining number of ideal stage,

N:

1) Plot y A vs x A .

2) Draw operating line.

3) Draw equilibrium line (Henry’s law).

4) Stepping upward (or downward) until yN+1 (or y1) is reached5) N = number of steps/trays

Dilute system (


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+ 14Example 10.3-2

It is desired to absorb 90% of the acetone in a gas containing

1.0 mol % acetone in air in a countercurrent stage tower. The

total inlet gas flow to the tower is 30.0 kg mol/h and the total

inlet pure water flow to be used to absorb the acetone is 90

kg mol H2O/h. The process is to be operate isothermally at

300 K and a total pressure of 101.3 kPa. The equilibrium

relation for the acetone (A) in the gas-liquid is y A = 2.53x A .Determine the number of theoretical stages required for this

separation.


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Given values are y AN+1 = 0.01, x A0 = 0,VN+1 = 30.0 kg mol/h, and L0 =L’= 90.0 kg mol/h.

Amount of entering acetone = y AN+1 VN+1= 0.01(30.0)

= 0.3 kg mol/h.

Entering air,V’ = (1- y AN+1 )VN+1= (1-0.01)(30.0)

= 29.7 kg mol air/h

Acetone leaving in V1 = 0.10(0.30) = 0.03 kg mol/h.

Acetone leaving in L1

= 0.90(0.30) = 0.27 kg mol/h. (10% avetone enter is absorbed)

V1 = 29.7 + 0.03 = 29.73 kg mol air + acetone/h.

y1 = (0.030/29.73) = 0.00101

LN = 90.0 + 0.27 = 90.27 kg mol water + acetone /h.

xN = (0.27/90.27) = 0.0030.

Since the flow of liquid varies only slightly from L0 = 90.0 at the inlet to LN = 90.27 at the the outlet and V from 30.0 to 29.73, the slope Ln /Vn+1 of the operating line is essentially constant. This line is plotted,and the equilibrium relation y A = 2.53x A is also plotted.

Starting at point y A1 ,x A0 the stages are drawn. About 5.2 theoretical stages are required.


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0 0 0.001 0.002 0.003 0.004

0.004

0.008

0.012

xANxA0

yA1

yAN+1

Equilibrium line

Operating line

3

2

1

4

5

Mole fraction acetone in water, xA

Mole fraction acetone in air, yA


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+ 17 Analytical Equations For Countercurrent StageContact : Kremser equations

Kremser equations

calculate the number of ideal stages.

valid only when operating & equilibrium lines are straight.

v, L have a constant value

Absorption Stripping

When A = 1

N = y N +1 − y1 y

1 −mx

0

N = In

x0 − ( y N +1 / m) x N − ( y N +1 / m) 1− A( ) + A

In(1 / A) N =

In y N +1 −mx0 y1 −mx01− 1 A +

1 A

InA

N = x0 − x N x N − y N +1 / m

Procedure (for varying A):

1.Calculate A 1 = L0/mV1 at L0 & V1.

2.Calculate A N = LN/mVN+1 at LN &VN+1.3.Calculate geometric average area, A ave. =

4. Calculate N.

m = slope of equilibrium line.

A = absorption factor = constant = L/(mV).

L, V = molar flow rates

N A A

1


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+ 18Example 10.3-3

Repeat Example 10.3-2 but use the Kremser analytical equation

for countercurrent stage processes.

Solution

At one end of the process at stages 1,V 1 = 29.73 kg mol/h y A1 = 0.001001

L 0 = 90.0, and x A0 = 0.

Also, the equilibrium relation is y A = 2.53x A where m = 2.53.

Then,

20.173.2953.2

0.90

1

0

1 mV

L

mV

L

A


19/89


At stage N,

VN+1 =30.0, y AN+1 = 0.01,LN = 90.27, and x AN = 0.0030.

The geometric average,

Then,

This compares closely with 5.2 stages using graphical method

19.10.3053.2

27.90

1

N

N N

mV

L A

195.119.120.01 N A A A

04.5)195.1log(

195.1

1

195.1

11

)0(53.200101.0

)0(53.201.0log

N


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+Next Part

Types of Separation Processes and Methods

Equilibrium Relation Between Phases




Absorption on plate and Packed Tower

Absorption and Concentrated Mixtures in Packed Tower

Estimation of Mass Transfer Coefficient in Packed Tower

20


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+

21

Gas-Liquid Separation

Part 02


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+Next Part

Types of Separation Processes and Methods Equilibrium Relation Between Phases



Continuous Humidification Processes Absorption on plate and Packed Tower

Absorption and Concentrated Mixtures in Packed

Tower

Estimation of Mass Transfer Coefficient in PackedTower

22


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+Subtopics:


Introduction

Film Mass Transfer Coefficient and Interface Concentration :

Equimolar Counterdiffusion

Film Mass Transfer Coefficient and Interface Concentration :

Diffusion of A through Stagnant & Nondiffusing B

Overall Mass-Transfer Coefficients:

Equimolar Counterdiffusion/Diffusion in Dilute Solutions

Overall Mass-Transfer Coefficients:

Diffusion of A Through Stagnant & Nondiffusing B

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25/89

+Mass Transfer Between Phases

y AG - average or bulk mole fraction of A in gas phase

x AL - average or bulk mole fraction of A in liquid phase

y Ai & x Ai - equilibrium mole fraction at interface

25

Distance from interface

NA

yAGyAi

interface

gas-phase mixture

of A in gas G

liquid-phase solution

of A in liquid L.

xAixAL

Concentration profile of solute A diffusing through two phases.


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+Film Mass Transfer Coefficient and InterfaceConcentration : Equimolar Counterdiffusion

k’y -gas phase mass transfer coefficient (kgmol/s.m2.mol frac.)

k’x -liquid phase mass transfer coefficient (kgmol/s.m2.mol frac.)

Point P is bulk phase compositions y AG and x AL

Point M is interface concentration x Ai and y Ai

26

)(')(' AL Ai x Ai AG y A x xk y yk N

)(

)(

'

'

Ai AL

Ai AG

y

x

x x

y y

k

k

If two film coefficient k’x and k’yare known, the interface

composition can be

determined by drawing line

PM with slope –k’x/k’yintersecting the equilibrium

line


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+ Film Mass Transfer Coefficient and InterfaceConcentration : Diffusion of A through Stagnant &

Nondiffusing B

Trial and error need to get the slope. For first trial assume (1-y A )iMand (1-x A )iM equal to 1.

27

)]1 /()1[(

)1()1()1(

)]1 /()1[(

)1()1()1(

)1(

'

)1(

')()(

Ai AL

Ai AL

iM A

AG Ai

AG Ai

iM A

iM A

x

x

iM A

y

y

AL Ai x Ai AG y A

x x In

x x x

y y In

y y y

x

k k

y

k k

x xk y yk N

)(

)(

)1 /('

)1 /('

)()1(

')()1(

'

Ai AL

Ai AG

iM A y

iM A x

AL Ai

iM A

x Ai AG

iM A

y A

x x

y y

yk

xk

x x xk y y

yk N


28/89

+ Example 10.4-1The solute A is being absorbed from a gas mixture of A and B in a wetted-wall

tower with the liquid flowing as a film downward along the wall. At certain point

in the tower the bulk gas concentration y AG = 0.380 mol fraction and the bulk

liquid concentration is x AL = 0.1. The tower is operating at 298 K and 1.013 x 105

Pa and the equilibrium data are as follows:

x A y A x A y A

0 0 0.20 0.131

0.05 0.022 0.25 0.187

0.10 0.052 0.30 0.265

0.15 0.087 0.35 0.385

The solute A diffuse through stagnant B in the gas phase and then through a

nondiffusing liquid. Using correlations for dilute solutions in wetted-wall

towers, the film mass-transfer coefficient for A in the gas phase is predicted as

ky = 1.465 x 10-3 kg mol A/s.m2 mol frac. And for the liquid phase as kx = 1.967

x 10-3 kg mol A/s.m2 mol frac. Calculate the interface concentrations y Ai and x Ai

and the flux N A .

28


29/89

+Overall Mass-Transfer Coefficients andDriving Force

Film or single phase mass transfer coefficient k’y and k’xor ky and kx are often difficult to measure

As a result, overall mass transfer coefficient K’y and Kx’ are

measured based on gas phase or liquid phase.

K’y - overall gas phase driving force (kgmol/s.m2.mol frac.)

K’x - overall liquid phase driving force (kgmol/s.m2.mol frac.)

y* A - mole fraction that equilibrium with x AL

X* A - mole fraction that equilibrium with y AG

29

N A = K ' y ( y AG − y A* ) = K ' x ( x A* − x AL )


30/89

+ Overall Mass-Transfer Coefficients:Equimolar Counterdiffusion/Diffusion in Dilute Solutions

30

Gas phase controlling (i.e major resistance in gas

phase)- m’ is small, so that the equilibrium curve is

almost horizontal, a small value of yA in the gas willgive a large value of xA in equilibrium in liquid. A is

very soluble in liquid phase

Liquid phase controlling (i.e major resistance in

liquid phase) - m” is large and A is very insoluble

in liquid phase

N A = K ' y ( y AG − y A*

) = K ' x ( x A* − x AL )

1

K ' y = 1

k y ' + m '

k x 'm ' =

y Ai − y A*

x Ai − x AL1

K ' x= 1m"k y '

+ 1k x '

m" = y AG − y Ai x A

* − x Ai

'

1

'

1

y y k K

'

1

'

1

x x k K


31/89

+ Overall Mass-Transfer Coefficients:Diffusion of A Through Stagnant & Nondiffusing B

31

)()1(

)()1(

''

AL A

M A

x A AG

M A

y

A x x x

K y y

y

K N

)]1 /()1[()1()1()1(

)]1 /()1[()1()1()1(

)1(

'

)1(

'

*

*

**

*

*

**

A AL

A AL M A

AG A

AG A M A

M A

x

x

M A

y

y

x x In x x x

y y In y y y

x

K K

y

K K

iM A xiM A y M A y xk

m

yk yK )1 /('

'

)1 /('

1

)1 /('

1

*

iM A xiM A y M A x xk yk m xK )1 /('

1

)1 /('"

1

)1 /('

1

*


32/89

+Example 10.4-2

Using the same data as in Example 10.4-1, calculate the overall

mass transfer coefficient K’y and the percent resistance in the

gas and liquid films. Do this for the case of A diffusing through

stagnant B.

32


33/89

+Subtopics:


Water-Gas Interface for Water Cooling Tower

Common Term in Humidification

Operating Line for Water Cooling Tower

Tower Height For Water-Cooling

Design of Water Cooling Tower Using Film Mass Transfer

Coefficient

Design of Water Cooling Tower Using Overall Mass Transfer

Coefficient

Minimum Value of Air Flow

Design of Water Cooling using Height of Transfer Unit

33


34/89

+Continuous Humidification Process

When a relatively warm liquid is brought into direct contact with a gas that is

unsaturated, some of the liquid is vaporized to the gas phase.

This done for the following purpose

Humidifying of air for control the moisture content of air in drying or air conditioning

Dehumidifying air where cold water condenses some water vapor from air.

Water cooling where water is evaporated to air to cool the warm water.

Typical water cooling tower

Warm water flows counter currently to an air stream. The warm water enters the top of a packed tower and cascades down through the

packing,leaving at the bottom.

Air enters at the bottom of the tower and flows upward through the descending water

by the natural draft or by the action of a fan.

The water is distributed by troughs and overflows to cascade over slat gratings or

packing that provide large interfacial areas of contact between the water air air in

the form of droplets and film of water. The tower packing often consists of slats of wood or plastic or of a packed bed

In humidification and dehumidification, the gas phase resistance controls the

rate of the mass transfer.

34


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+ Water-Gas Interface for Water CoolingTower

Sensible heat flow from liquid to the interface= sensible heat flow in

the gas phase + latent heat (vaporization) flow in the gas.

35


36/89

+ Common Term in HumidificationChapter 9.3

Humidity (H) of an air-water vapor is the kg of water vapor

contained in 1 kg of dry air.

Saturated humidity (Hs)- air in which the water vapor is equilibrium

with liquid water at given P & T. Partial pressure of water vapor inair-water mixture is equal to the vapor pressure p As of pure water at

given T

Percentage humidity (Hp)

Percentage relative humidity (HR)

36


37/89

+Common Term in Humidification

Dew point

the temperature at which a given mixture of air and water vapor would be saturated.

or temperature at which vapor begins to condense when the gas

phase is cooled at constant pressure.

Humid heat cS amount of heat required to raise the temperature of 1 kg of dry air

plus the water vapor present by 1 K.

cS (kJ/kg dry air.K) = 1.005 + 1.88H (SI)

where, cP water(v) = 1.88 kJ/kg water vapor. K and cP air = 1.005

kJ/kg dry air. K

37


38/89

+Common Term in Humidification

Humid volume, vH - total volume (m3) of 1 kg of dry air plus

the vapor it contains at 1 atm abs pressure and the given gastemp.

vH (m3/kg dry air) = (2.83 x 10-3 + 4.56 x 10-3 H) T (K).

Total enthalpy of an air-water mixture, H Y - the total enthalpyof 1 kg of air plus its water vapor. Sensible heat of the air-

water vapor mixture plus the latent heat.

H Y (kJ/kg dry air) = (1.005 + 1.88 H) (T ºC-0) + 2501.4

where,Tref for both components = 0 ºC

38


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+ Humidity Chart39


40/89

+ Operating Line for Water CoolingTower

Assumption:

1. Process carried out adiabatic

2. L is constant

3. cL is constant at 4.187 x 103 J/kg. K

Slope= LcL/G

G = dry air flow, kg/s.m2.

L = water flow, kg water/s.

cL = heat capacity of water, assumed constant at 4187 kJ/kg.K.

TL= temperature of water, ºC or K.

Hy= enthalpy of air-water vapor mixture, J/kg air.

= cs (T-T0) + H0 = (1.005 + 1.88H)103 (T-0) + 2.501x106H

H = humidity of air, kg water/kg dry air – Humidity chart in Fig. 9.3-2pg. 568.

40

)()( 11 L L L y y T T Lc H H G


41/89

+ 41Tower Height For Water-Cooling

Tower height for water-cooling

z = column height

MB = molecular weight of air

a = m2 of interfacial area per m3 volume of packed section

kG = gas phase mass transfer coefficient ( kgmol/s. m2)

kGa = volumetric coefficient in kgmol/s.m3 volume

P = pressure in atm

Slope of line from interface composition to point in equilibrium line

2

1

Hy

Hy y yi

y

G B H H

dH

aPk M

G z

Li

y yi

BG

L

T T

H H

PaM k

ah


42/89

+ Temperature Enthalpy Diagram AndOperating Line For Water- Cooling

42


43/89

+ Enthalpies of Saturated Air-Water VaporMixture (0C Base Temperature)

43


44/89

+ Design of Water Cooling Tower UsingFilm Mass Transfer Coefficient

1. Draw equilibrium line ( Data from table 10.5-1) in H-versus-T plot

2. Draw operating line with slope LcL/G.

3. Determine Hyi-Hy by plot a line from P (TL, Hy) to M (Ti, Hyi). The slope of thisline –hLa/kGaMBP. Do for various P point between (TL1, Hy1) to (TL2, Hy2).

4. Use numerical or graphical integration to obtain the value of integration in

height tower equation. For example using graphical method, plot 1/(Hyi-Hy)

versus Hy.

44

2

1

Hy

Hy y yi

y

G B H H

dH

aPk M

G z

Li

y yi

BG

L

T T

H H

PaM k

ah


45/89

+ Design of Water Cooling Tower UsingOverall Mass Transfer Coefficient

Often overall mass transfer coefficient KGa (kgmol/s.m3.Pa) is available. Then

to determine tower height, z

1. Draw equilibrium line ( Data from table 10.5-1) in H-versus-T plot

2. Draw operating line with slope LcL/G.

3. Determine H*y-Hy by plot a line from P (TL, Hy) to R (TL, H*y). The slope of thisline –hLa/kGaMBP. Do for various P point between (TL1, Hy1) to (TL2, Hy2).

4. Use numerical or graphical integration to obtain the value of integration inheight tower equation. For example using graphical method, plot 1/(Hyi-Hy)

versus Hy.

45

Li

y yi

BG

L

T T

H H

PaM k

ah

2

1 *

Hy

Hy y y

y

G B H H

dH

aPK M

G z

46


46/89

+Example 10.5-1

A packed countercurrent water-cooling tower using a gas flow

rate of G = 1.356 kg dry air/s. m2 and a water flow rate of L =1.356 kg water/s. m2 to cool the water from TL2 = 43.3 ºC to TL1= 29.4 ºC. The entering air at 29.4 ºC has a wet bulb

temperature of 23.9 ºC . The mass-transfer coefficient kG a is

estimated as 1.207 x 10-7 kg mol/s.m3.Pa and hL a / kG aMBP as

4.187 x 104

J/kg.K . Calculate the height of packed tower z. Thetower operates at a pressure of 1.013 x 105 Pa.

46

47


47/89


Following the step outlined, the enthalpies from the saturated air-water vapor

mixtures from Table 10.5-1 are plotted in Fig. 10.5-4.

The inlet air at TG1 = 29.4 ºC has a wet bulb temperature of 23.9 ºC . The

humidity from the humidity chart is H1 = 0.0165 kg H2O/kg dry air.

Substituting into Eq.(9.3-8),

Hy1 = cs (T-T0) + H0 = (1.005 + 1.88H) (T-T0) + H0

Hy1 = (1.005 + 1.88 x 0.0165)103 (29.4-0) + 2.501 x 106 (0.0165).

= 71.7 x 103 J/kg.

The point Hy1 = 71.7 x 103 and TL1 = 29.4 ºC is plotted.

Then substituting into Eq. (10.5-2) and solving,

G(Hy2 –Hy1) = LcL (TL2 – TL1)

1.356 (Hy2 – 71.7 x 103) = 1.356 (4.187 x 103) (43.3 – 29.4).

Hy2 = 129.9 x 103 J/kg dry air.

+ 48


48/89


+ 49


49/89

+ 49

+ 50


50/89

+ 50

+ Mi i V l f Ai Fl 51


51/89

+ Minimum Value of Air Flow The air flow G is not fixed but must be set for the design of the cooling

tower.

For a minimum value of G, the operating line MN is drawn through the

point Hy1 and TL1 with a slope that touches the equilibrium line at TL2,point N.

If the equilibrium line is quite curved, line MN could become tangent tothe equilibrium line at a point farther down the equilibrium line thanpoint N.

For the actual tower, a value of G greater than Gmin must be used. Often,a value of G equal to 1.3 to 1.5 times Gmin is used.

51

+ D i f W t C li i H i ht f 52


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+ Design of Water Cooling using Height of Transfer Unit

Using Film Mass Transfer Coefficient

Using Overall Mass Transfer Coefficient

52

2

1

2

1

Hy

Hy y yi

y

G

G B

G

Hy

Hy y yi

y

G B

H H

dH N

aPk M

G H

H H

dH

aPk M

G z

2

1 *

2

1 *

Hy

Hy y y

y

OG

G B

OG

Hy

Hy

y y

y

G B

H H

dH N

aPK M

G H

H H

dH

aPK M

G z

z= height of transfer unit

X number of transfer unit

+


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+

53

Gas-Liquid SeparationPart 03

+ 54


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+Next Part

Types of Separation Processes and Methods

Equilibrium Relation Between Phases




Absorption in Plate and Packed Tower Pressure Drop & Flooding in Packed Towers

Absorption of Dilute Gas Mixtures in Packed Tower

Absorption of Concentrated Gas Mixtures in

Packed Tower Design of Packed Towers Using Transfer Units

54

+ 55


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+Equipment for Absorption & Distillation

1. Plate (tray) tower

Sieve tray

Valve tray

Bubble cap tray

2. Structured packing

3. Packed tower

Raschig ring

Lessing ring

Berl saddle

Pall ring

etc

55

+ 56


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+Tray/Plate Tower

56

+ 57


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+Tray (Plate) Tower

Sieve tray

Valve tray

57

+ 58


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+Tray (Plate) Tower

Bubble cap tray

58

+ S d P ki59


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+ Structured Packing59

+ S d P ki60


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+ Structured Packing

Corrugated structured packing

60

+ 61


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+Packed Tower

+ Packed Tower 62


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+ Packed Tower

+ P D & Fl di i P k d T 63


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+ Pressure Drop & Flooding in Packed Towers

Flooding velocity – upper limit of gas flow rate

At low gas velocities the liquid flow downward through the

the packing, essentially uninfluenced by the upward gas flow

As the gas flow rate increased at low gas velocities, the

pressure drop is proportional to the flow rate to the 1.8

power.

At a gas flow rate called the loading point, the gas start to

hinder the liquid downflow, and local accumulations or pools

of liquid start to appear in the packing. The pressure drop of

the gas starts to rise at a faster rate

As the gas flow arte is increased, the liquid holdup or

accumulation increase. At the flooding point, the liquid no

longer flow down through the packing and is blown out withthe gas.

The optimum economic gas velocity is about 0ne-half or

more of the flooding velocity

+ P D i R d P ki64


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+ Pressure Drop in Random Packing

Note: this capacity parameters is not dimensionless and that onlythese unit (english) should be used

+ 65


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+ Pressure Drop in Structured Packing

+ Packing Factors (F ) for Random and 66


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+ Packing Factors (Fp) for Random andStructured packing

+ Flooding Pressure Drops in Packed and 67


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+ Flooding Pressure Drops in Packed andStructured Packing

Can be used for packing factors from 9 up to 60. It predicts all the

data for flooding within +- 15% and most for +-10% for packing factor of 60 or higher, this equation is not valid and the

pressure drop can be taken as 2 in H2O/ft( 166.7 mm H2O/m)

∆ P flood = 0.115 F 0.7

+ Procedure to Determine Limiting Flow Rate 68


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+ Procedure to Determine Limiting Flow Rateand Tower Diameter

1. A suitable random packing or structured packing is selected, giving an Fp value

2. A suitable GL/GG is selected along with the total gas flow rate

3. The pressure drop at flooding is calculated

4. The flow parameter is calculated, and using the pressure drop at flooding and either fig 10.6-5 or10.6-6, the capacity parameter is read off the plot

5. Using the capacity parameters, the GG is obtained, which is the maximum at flooding

6. Using a suitable % of the flooding value of GG for design, a new GG and GL are obtained. Thepressure drop can also be obtained from the plot.

7. Knowing the total gas flow rate and GG, the tower cross sectional are and ID can be calculated.

+ 69


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+Example 10.6-1

Ammonia is being absorbed in a tower using pure water at 25C

and 1 atm abs pressure. The feed rate is 1440 lbm/h (653 kg/h)and contains 3 mol% ammonia in air. The process design

specifies a liquid to gas mass flaw rate ration GL/GG of 2/1 and

the use of 1-in metal Pall rings.

a) calculate the pressure drop n the packing and gas mass

velocity at flooding. Using 50% of the flooding velocity,calculate the pressure drop, gas and liquid flows and tower

diameter.

b) Repeat(a) above but use Mellapak 250Y structured

packing.

+ Design of Plate Absorption Tower70


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+ Design of Plate Absorption Tower

Operating line assumption

solute A diffusing through a stagnant gas B

the moles of pure air and pure water remain constant throughout the

entire tower

similar equation to countercurrent stages process

Overall material balance

Balance around the dashed-lined box

If x and y are very dilute, (1-x) and (1-y)can be taken as 1.0 and the operating lineis straight with a slope L’/V’

+ N b f T71


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+ Number of Tray

Graphical determination of the number of trays

+ 72


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+Example 10.6-2

A tray tower is to be designed to absorb SO2 from an air stream

by using pure water at 293 K. the entering gas contains 20 mol% SO2 and that leaving 2 mol % at the total pressure of 101.3

kPa . The inert air flow rate is 150 kg air/h.m2 and the entering

water flow rate is 6000 kg water/h.m2. Assuming an overall tray

efficiency of 25 %, how many theoretical trays and actual trays

are needed? Assume that the tower operate at 293 K.

+Design of Packed Tower for Absorption73


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+Design of Packed Tower for Absorption

solute A diffusing through a

stagnant gas B Overall material balance

Balance around the dashed line box

give a curve line on yx plot

If x and y are very dilute, (1-x) and (1-y)can be taken as 1.0 and the operating lineis straight with a slope L’/V’

+ L ti f O ti Li74


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+ Location of Operating Line

+ 75


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+Limiting and Optimum L’/V’ Ratios

In absorption, V1, y1, y2, x2 is always known.

Only amount of the entering liquid flow (L2

or

L) is open to choice.

L’min – when operating line has a minimum

slope and touch (becoming tangent) to the

equilibrium line.The value of x1 is x1max.

To solve L’min, insert the value y1 and x1max in

operating line equation.

The choice of optimum L’

/V’

ratio dependon an economic balance

too high ratio value require a large liquid flow

and a large diameter tower.

too small value give small liquid flow result in a

high tower,which is also costly.

An optimum liquid flow rate can be taken as

1.2-1.5 times L’ min.

For stripping, optimum gas flow rate is takenat about 1.5 timesV’ min.

+ Analytical Equations for Theoretical76


76/89

+ Analytical Equations for TheoreticalNumber of Steps or Trays

Absorption


1. Calculate m1 and m2

2. Calculate A1 and A2

3. Calculate geometric average area, A ave.= /A1A2

4. Used m2 in equation

Stripping


1. Calculate m1 and m2

2. Calculate A1 and A2

3. Calculate geometric average area, A ave.= /A1A2

4. Used m1 in equation

N = In

y1 −mx2 y2 − mx2

1− 1 A

+ 1

A

InA N =

In ( x2 − y1 / m x1 − y1 / m

) 1− A( ) + A

In(1 / A)

+ 77


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+Example 10.6-3

A tray tower is absorbing ethyl alcohol from an inert gas stream

using pure water at 303K and 101.3 kPa. The inlet gas streamflow rate is 100 kmol/h and it contains 2.2 mol% alcohol. It is

desired to recover 90% of the alcohol. The equilibrium

relationship is y=0.68x for this dilute stream. Using 1.5 times

the minimum liquid flow rate. determine the number of trays

needed. Do this graphically and also using the analytical

equations.

+ Film and Overall Mass Transfer 78


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+ Film and Overall Mass TransferCoefficients in Packed Towers

Defining a as interfacial area in m2 per m3 volume of packed

section, the volume of packing in a height dz m is S dz, where S ism2 cross sectional area of tower. Also

dA = aS dz

The volumetric film and overall mass transfer coefficient are then

defined as

+Design Method for Packed Tower using Mass 79


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+Design Method for Packed Tower using MassTransfer Coefficient

Absorption of A from stagnant B and the moles of A leaving V equal

to the moles entering L (i.e d( Vy)= d(Lx)

Using film coefficient

Using overall coefficient

The equilibrium and operating

lines are usually curved, and k’xa,

k’ya, K’ya and K’xa vary somewhat with total gas and liquid flows

Then, these equations must be

integrated numerically or

graphically

+ Simplified Design Method for Absorption of 80


80/89

+ Simplified Design Method for Absorption of Dilute Gas Mixture in Packed Towers

Assumptions

Concentration can be considered dilute when mole fraction y or xare less than about 0.10 (10%).

Operating line is straight

L = (L1+L2)/2

V = (V1+V2)/2

M x M y

M i x M i y

x xazK x xS

L y yazK y y

S

V

x xazk x xS

L y yazk y y

S

V

)(')()(')(

)(')()(')(

*

21

*

21

2121

+ Simplified Design Method for Absorption of 81


81/89

+ Simplified Design Method for Absorption of Dilute Gas Mixture in Packed Towers

1. Plot operating line and equilibrium line

2. Determine interface composition yi1and xi1 at point y1, x1. To get this , findintersection point by draw a line toequilibrium line with slope

Then determine interface compositionyi2, xi2 at point y2,x2 using the samemethod. Do trial and error if necessary.

If overall mass transfer coefficient use,find appropriate y1

*, y2*, x1

* or x2*.

3. Calculate related (y-yi)M or (y-y*)M for

gas coefficient and (xi-x)M or (x*-x)M.

4. Calculate the column height byappropriate simplified equation.

)1 /('

)1 /('

1

1

yak

xak slope

y

x

+E l 10 6 4

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Example 10.6-4

Acetone is being absorbed by water in a packed tower having

a cross sectional area of 0.186 m2 at 293 K and 101.32 (1 atm).The inlet air contains 2.6 mol % acetone and outlet 0.5%. The

gas flow is 13.65 kg mol inert air/h. The pure water inlet flow is

45.36 kg mol water/h. Film coefficients for the given flows in

the tower are k’y a = 3.78 x 10-2 kg mol/s.m3.mol frac. And k’x a

= 6.16 x10-2 kg mol/s.m3.mol frac. Equilibrium data are given in

Appendix A.3.

(a) Calculate the tower height using k’y a.

(b) Repeat using k’x a.

(c) Calculate K’y a, and the tower height.

+ Design Method for Absorption of 83


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g p

Concentrated Mixtures in Packed Tower

The design equation must be integrated graphically or

numerically.

1

2*

*

1

2*

*

1

2

1

2

))(1()1(

'))(1(

)1(

'

))(1()1(

'))(1(

)1(

'

x

x

M

x

y

y

M

y

x

x

i

iM

x

y

y

i

iM

y

x x x x

aS K

Ldx z

y y y y

aS K

Vdy z

x x x x

aS k

Ldx z

y y y y

aS k

Vdy z

+ Design Method for Absorption of 84


84/89

Concentrated Mixtures in Packed Tower

1. Plot operating line and equilibrium line

2. Calculate value k’ya and k’xa from empirical equation given

3. Use trial and error to determine interface composition at several

point between P1(y1,x1) and P2(y2,x2) by draw a line with following

slope from the point to intersect to equilibrium line.

4. Using the value yi and xi, calculate the value f(y) as follows. Then

plot f(y) vs. y. Calculate numerically or graphically area under

curve to insert into design equation for height.

iM y

iM x

yak

xak slope

)1 /('

)1 /('

))(1()1(

'

)(

i

iM

y y y y yaS k

V y f

+E l 10 7 1

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Example 10.7-1

A tower packed with 25.4 mm ceramic rings is to be designed

to absorb SO2 from air by using pure water at 293 K and 1.013 x105 Pa abs pressure. The entering gas contains 20 mol % SO2and that leaving 2mol %. The inert air flow is 6.53 x10-4 kg mol

air /s. the tower cross-sectional area is 0.0929 m2. For dilute SO2,the film mass transfer coefficients at 293 K are for 25.4 mm

rings,

k’ya = 0.0594 Gy0.7 Gx

0.25 and k’xa = 0.152GX0.82.

where k’ya & k’xa is kg mol/s.m3. mol frac., and Gx ,Gy are kg

total liquid or gas, respectively, per sec per m2

tower crosssection. Calculate the tower height.


86/89

+ Design Of Packed Towers Using Transfer Units: 87


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Design for Dilute Solutions

The terms (1-y)iM/(1-y), (1-x)iM/(1-x), (1-y)*M/(1-y) and (1-x)*M/(1-x)

are close to one, and if the operating lines and equilibrium line arestraight and dilute,

z = H G N G = H Gdy

y− yi y2 y1∫ = H G × y1 − y2

( y − yi ) M

z = H OG N OG = H OG dy y− y* y2 y1

∫ = H G × y1 − y2( y − y* ) M

z = H L N L = H Ldx

xi − x x2 x1∫ = H L × x1 − x2

( xi − x) M

z = H OL N OL = H OLdx

x* − x x2 x1

∫ = H OL ×

x1 − x2( x* − x) M

+ Design Of Packed Towers Using Transfer Units: 88


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Design for Dilute Solutions

Using the equilibrium line equation y = mx and letting A = L/mV, the number

of transfer unit for absorption and stripping are

When the operating and equilibrium lines are straight and not parallel, NOGis related to the number of theoretical trays N by equation:

The height of theoretical trays (HETP)

N OG = 1

(1−1 / A) In[(1−1 / A)( y1 −mx2

y2 −mx2) +1 / A]

N OL = 1

(1− A) In[(1− A)( x2 − y1 / m

x1 − y1 / m) + A]

HETP = H OG In(1 / A)

(1− A) / AThen, z = N x HETP

N OG = N In A

(1−1 / A)

+E l 10 6 5

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Example 10.6-5

Acetone is being absorbed by water in a packed tower having

a cross sectional area of 0.186 m2 at 293 K and 101.32 (1 atm).The inlet air contains 2.6 mol % acetone and outlet 0.5%. The

gas flow is 13.65 kg mol inert air/h. The pure water inlet flow is

45.36 kg mol water/h. Film coefficients for the given flows in

the tower are k’y a = 3.78 x 10-2 kg mol/s.m3.mol frac. And k’x a

= 6.16 x10-2 kg mol/s.m3.mol frac. Equilibrium data are given in

Appendix A.3.

a) Use HG and NG to calculate tower height

b) Use HOG and NOG to calculate tower height

c) Use E 10.6-52 to calculate NOG and tower height

d) Using analytical equations, calculate HETP, theoreticalsteps N and tower height.

Documents

Liquid Gas Absorption Process