CHAPTER 3:ABSORPTION

PROCESS

LEARNING OUTCOMES

Explain gas absorption and (desorption), and the method of contacting between gas-liquid phase.

Explain the choice of solvents in gas absorption

Derive the operating line equation for counter-current operation and understand the graphical representation.

Use graphical method to determine the number of theoretical trays required for separation

INTRODUCTION

Absorption, or gas absorption, is a unit operation used in the chemical industry to separate gases by washing or scrubbing a gas mixture with a suitable liquid.

DEFINITION:

Mass transfer of the component of the gas from the gas phase to the liquid phase.

Gas mixture is contacted with a liquid for the purposes of preferentially dissolving one/more components of gas and to provide a solution of them in the liquid.

Gas mixture = inert gas + solute.Solute = transferred component.

To obtain or remove the DESIRED component of Vapor Mixture with the usage of a LIQUID SOLVENT based on

SOLUBILITY characteristic of the component towards the

Liquid Solvent.

Purpose Of Absorption

A

Gas

solventB

Liquid

BA

Gas

Solvent

Liquid

Nonvolatileeg: water Carrier & insoluble

e.g: air+ NH3

Component to be removed e.g: NH3

EXAMPLE: NH3 that is diffusing from an gaseous air-NH3 mixture into liquid

phase water.

Ammonia - desired componentAir – unwanted componentResult : To absorb ammonia, water is used as a solvent & air free of ammonia

Removal of ammonia, benzene and toluene from coke gas

Removal of SO2 from stack gas Removal of CO2 from synthesis gas Air pollution control, the various oxides of

nitrogen can be removed by absorption with water, sulphuric acid, and organic solutions

Example of applications

Table below showed representative commercial absorption applications:

GAS ABSORPTION OPERATION

Counter-current

Co-current

Counter-current operation Co-current operation

Two methods of contacting the gas and liquid :

Application of Packed Absorption Tower

Most common type of absorption towerHigh flowrate of liquid solventBased on diagram:Air-mixture enter at the bottomLiquid solvent enter via liquid distributor(3)Packed section (2) : air-mixture contact liquidAir will rise upLiquid+gas at the bottomLiquid solvent will recycle to fully utilizeSome of the liquid solvent will be remove for further treatment to obtain desired comp.Make-up stream : to add some more liq solvent if some amount is lost by vapourization or not much left at the bottom since some has already being withdrawn.

PLATE COLUMN ABSORPTION TOWER

Suitable for low flowrate of liquid solvent

Based on diagram :Air-mixture enter at the bottom.Liquid solvent enter at the top and flow downward by gravitySolvent will flow across each tray and flows down the tray belowType of column : bubble tray or sieve trayType of flow : counter-currrent most popular

SOLVENT

Gas solubility

VolatilityMiscellaneous

Corrosiveness

Cost

Viscosity

SOLVENT PROPERTIES

Gas solubilitySolute gas should be high solubility into solvent

thus increase the rate of absorption and decrease the quantity of solvent required.

VolatilitySolvent non volatile.Have low vapor pressure to reduce loss of solvent

in the gas leaving an absorption column.

CorrosivenessMaterial of construction should not be unusual or

expensive.

CostSolvent should be inexpensive, losses are not

costly and should be readily available.

ViscosityLow viscosity to increase absorption rate, improved

flooding characteristics in absorption tower, low pressure drop on pumping and good heat transfer characteristics.

OthersSolvent should be nontoxic, nonflammable and

chemically stable and have low freezing point.

►gas-phase pollutant is brought into close contact with the liquid to facilitate diffusion across the gas-liquid interface.►gaseous pollutant diffuse through a thin boundary layer on the gas side of the interface (gas film) and a thin boundary layer on the liquid side of the interface (liquid film.). ►Once the pollutant enters the liquid phase, it can simply dissolve, or it can react with other chemicals also in the liquid. ►Once the pollutant in the liquid phase has reached its solubility limit, there is no net transfer of pollutant across the gas-liquid interface. ►At this point, equilibrium has been reached whereby the amount of gaseous pollutant that continues to dissolve equals the amount coming out of solution and reentering the gas phase.►Henry's law can be used to determine the solubility limit of absorption.

GAS-LIQUID EQUILIBRIUM

Where,

pA = partial pressure of component A .H = Henry’s law constant (pressure/mol fraction).H’ = Henry’s law constant (mol frac.gas/mol frac.liquid)

= H/P(dimensionless).

xA = mole fraction of component A in liquid.

yA = mole fraction of component A in gas = pA/P.P = total pressure (atm).

AA

AA

xHy

xHp

'

.

AA

AA

xHy

XHP

'

.

When the solution is non-ideal, Raoult's Law cannot be applied. For non-ideal solution, use Henry's Law which states that:

In Figure 10.2-1, the data follow Henry’s Law up to a concentration xA of about 0.005, where H=29.6atm/mol frac. In general, up to a total pressure of about 5 atm, the value of H is independent of P.

EXAMPLE 1

• What will be the concentration of oxygen dissolved in

water at 298K when the solution is in equilibrium

with air at 1 atm total pressure? Given partial

pressure of oxygen in air is 0.21 atm and the Henry’s

Law constant is 4.38 X 104 atm/mol fraction.

SOLUTION…

• The partial pressure pA of oxygen (A) in air is 0.21atm.

• Apply Henry’s Law;

• pA=HxA

• 0.21atm=(4.38 X 104 atm/mol frac )(xA)

• xA=4.795 X 10-6 mol frac

• It means that 4.795 X 10-6 mol oxygen is dissolved in 1.0 mol water plus oxygen.

• Assuming 3 components A,B and C are present in the streams, making balance on A & C

• No need to do balance for B because:• xA + xB + xC = 1.0

SINGLE-STAGE EQUILIBRIUM CONTACT FOR GAS-LIQUID SYSTEM

• The mass fraction of A in L stream is xA and V stream is yA.• The mass fraction of A in M stream is xAM.

• Since the component A is the only component that redistribute between the two phases, a balance on A can be written as:

Where L’ = moles inert water C V’ = moles inert air B

Constant and usually known

EXAMPLE 2

• A gas mixture at 1.0 atm pressure abs containing air and

CO2 is contacted in a single stage mixer continuously with

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

reach equilibrium. The inlet gas flow rate is 100 kg mole/hr,

with a mole fraction of CO2 of yA2=0.20. The liquid flow rate

entering is 300 kg mole water/hr. Make a balance on CO2(A)

and calculate the amounts and compositions of the two

outlet phases. Assume that the water does not vaporize to

the gas phase.

SOLUTION:

The inert water flow is L’=L0=300 kg mole/hrThe inert air flow V’=V(1-yA)Hence, V’=V2(1-yA2)=100(1-0.20)=80 kg mole/hrThen, substitute into this equation:

Eq 2Balance on CO2

CONT…

• From properties table, Henry’s law constant, • H=0.142 X 104 atm/mole.frac.

• Then, H’=H/P=(0.142 X 104 atm/mole.frac)/1atm

• =0.142 X 104 mole frac gas/mole frac liq

• Substituting in Henry’s law equation,

• yA1=0.142 X 104 xA1 ------------Eq 2• Solve simultaneously

• xA1=1.41 X 10-4

• yA1=0.20

CONT…

*Since the liquid solution is so dilute, L0≈L1

COUNTER-CURRENT MULTIPLE CONTACT STAGES

Overall balance on all stages;

• VN+1 is mole/hr entering• LN is mole/hr leaving• M is total flow

Overall component balance on A,B or C;

• x and y are mole fraction• The flows are in mass flowrate or

mole flowrate

Total balance over the first n stages;

Component balance over the first n stages;

Rearrange the equation to get yn+1;

• Equilibrium line relates the compositions of two

streams leaving a stage in equilibrium which each

other.

• To determine the number of ideal stages required to

bring about a given separation or reduction of the

concentration of A from yN+1 to y1, the calculation is

often done graphically.

• Starting at stage 1, y1 and x0 are on the operating line.

• The vapor y1 leaving is in equilibrium with the leaving

x1 and both composition are on the equilibrium line.

• Then, y2 and x1 are on the operating line and y2 is

equilibrium in x2, and so on.

• Each stage is represented by a step drawn in the figure.

• The steps are continued on the graph until yN+1 is

reached.

• Alternatively, we can start at yN+1 and draw the steps

going to y1.

EXAMPLE 3• It is desired to absorb 90% of the acetone in the gas

containing 1.0 mol% acetone in air in counter-current

stage tower. The total inlet gas flow to the tower is 30

kg mol/hr, and total inlet pure water flow to be used to

absorb the acetone is 90 kg mol H20/hr. The process is

to operate isothermally at 300 K and a total pressure of

101.3 kPa. The equilibrium relation for the acetone (A) in

gas-liquid is yA=2.53xA. Determine number of theoretical

stages required for this separation.

SOLUTION…

• Given:

• yAN+1=0.01

• xA0=0

• VN+1=30 kg mol/hr

• L0=90 kg mol/hr

• Making acetone material balance;

• Amout of entering acetone: yAN+1VN+1=0.01(30)=0.30 kg mol/hr

• Entering air:

(1-yAN+1)VN+1=(1-0.01)(30)=29.7 kg mol air/hr

• Acetone leaving in V1= 0.1(0.30) =0.03 kg mol/hr

• Acetone leaving in LN = 0.9(0.30) =0.27 kg mol/hr

• V1 = 29.7+0.03= 29.73 kg mol (air+acetone)/hr

• yA1 =0.03/29.73 =0.00101

• LN = 90+0.27=90.27 kg mol (water+acetone)/hr

• xAN =0.27/90.27=0.003

• *Since the flow of liquid varies only slightly from L0=90 and LN=90.27 at the outlet and V from 30 to 29.73, the slope Ln/Vn+1 of operating line is constant.

• Apply this equation;

Slope of the graph

5.2 theoretical stages

ANALYTICAL EQUATIONS FOR COUNTERCURRENT STAGE CONTACT

• Can be applied if satisfied these 2 conditions:

• When the flow rates V and L in a countercurrent process are essentially constant, operating line becomes straight.

• If the equilibrium line is also a straight line.

*A refers to absorption factor

L0

x0

LN

xN

V1

y1

VN+1

yN+1

CONCENTRATION PROFILES IN INTERPHASE MASS TRANSFER

MASS TRANSFER BETWEEN PHASES

Solute A diffuse from bulk gas phase G to the liquid

phase L, must pass through (phase G, interphase,

phase L) in series.

Must have concentration gradient.

The average or bulk conc. of A in gas phase in mole

fraction unit is yAG where yAG=pA/P and in the bulk

liquid phase in mole fraction units is xAL.

Conc. in bulk gas phase yAG decreases to yAi at

interphase.

Liquid conc. start at xAi falls to xAL.

Since no resistance in the interphase, yAi and xAi are in

equilibrium.

MASS TRANSFER USING FILM MASS-TRANSFER

COEFFICIENTS AND INTERPHASE CONCENTRATION

2 main cases:

Equimolar counterdiffusio

n

Diffusion of A through

stagnant B

CASE 1: EQUIMOLAR COUNTERDIFFUSION

• Point P represents the bulk phase composition yAG and xAL of two phases.

• Point M is the interface concentrations yAi and xAi.

• For A diffusing from gas to liquid and B is equimolarcounterdiffusion;

• NA=k’y(yAG-yAi)=k’x(xAi-xAL)

• Where k’y=gas phase mass transfer coefficient in kgmol/s.m2.mol frac

• k’x=liquid phase mass transfer coefficient in kgmol/s.m2.mol frac

• =

• Where (yAG-yAi) is gas phase driving force

• (xAL-xAi) is liquid phase driving force

• is slope of PM line

CASE 2: DIFFUSION OF A THROUGH STAGNANT B

Point P represents the bulk phase composition yAG and xAL of two phases.

Point M is the interface concentrations yAi and xAi. NA=ky(yAG-yAi)=kx(xAi-xAL)

Rearrange Eq 1 &Eq 2; = , = is slope of PM line Where (1-yA)iM =

(1-xA)iM=

Then, NA==

= When A is diffusing through stagnant B and the

solutions are dilute, (1-yA)iM and (1-xA)iM ≈ 1

To get the slope (Eq 4), a trial and error method is needed.

1st trial: (1-yA)iM and (1-xA)iMare assumed to be 1.0

Then, get slope using Eq 4 and yAi and xAi

2nd trial: These values of yAi and xAiare used to calculate new slope to get new yAi and xAi

This is repeated until the interface compositions do not change. Three trials are usually sufficient.

OVERALL MASS TRANSFER COEFFICIENTS AND DRIVING FORCES

• Film or single phase mass transfer coefficient k’y and k’x or ky and kx are often difficult to measure experimentally.

• Thus, overall mass transfer coefficients K’y and K’x are measured based on the gas or liquid phase.

• NA=K’y(yAG-y*A) and NA=K’x(x*

A-xAL)

• Where;

• K’y is overall gas phase driving force in kgmol/s.m2.mol frac

• y*A is value that would be in equilibrium with xAL

• K’x is overall liquid phase driving force in kgmol/s.m2.mol frac

• x*A is value that would be in equilibrium with yAG

EQUIMOLAR COUNTER DIFFUSION AND/OR DIFFUSION IN DILUTE

SOLUTIONS

• NA=k’y(yAG-yAi)=k’x(xAi-xAL)

• yAG-y*A=(yAG-yAi)+(yAi-y*

A)

• Between point E and M, the slope m’ can be given as:• m’=• Rearrange equation,• yAG-y*

A=(yAG-yAi)+m’(xAi-xAL)• Rearrange equation to canceling out NA,

Total resistanceGas film resistance

Liquid film resistance

• In similar manner,

• x*A-xAL=(x*

A-x*Ai)+(xAi-xAL)

• Between point D and M, the slope m’’ can be given as:

• m’’=• Rearrange equation to canceling out NA,

• When m’ value is quite small, m’/k’xis very small.

• So, ≈ and yAG-y*A≈yAG-yAi

• When m’’ is very large, 1/(m’’k’y) becomes small.

• So,≈ and xAi≈x*A

DIFFUSION OF A THROUGH STAGNANT OR NONDIFFUSING B

• NA== yAG-y*

A=(yAG-yAi)+m’(xAi-xAL) We must define the equations for the flux

using overall coefficients as follows: NA=[=[ Ky=[ , Kx=[

Rearrange equation to canceling out NA

Where; (1-yA)*M=

Similarly for K’x,

Where; (1-xA)*M=

DESIGN OF PLATE ABSORPTION TOWERS

• A plate (tray) absorption tower has the same process flow diagram as counter current multiple stage process.

Overall material balance on component A

A balance around the dashed-line box

Material balance in absorption tray or plate tower

EXAMPLE: ABSORPTION OF SO2

IN A TRAY TOWERA tray tower is to be designed to absorb SO2 from an air stream by

using pure water at 293 K (680F). The entering gas contains 20 mol%

SO2 and that leaving 2 mol% at a total pressure of 101.3 kPa. The

inert air flow rate is 150 kg air/hr.m2, and the entering water flow rate

is 6000 kg water/hr.m2. Assuming an overall tray efficiency of 25%,

how many theoretical trays and actual trays are needed? Assume that

the tower operates at 293K (200C). Molecular weight of air and water

is 29 kg/kmol and 18 kg/kmol respectively. The equilibrium curve is

plotted using data for SO2-Water System in Appendix.

SOLUTION…

L0=6000X0=0 (pure water)

LN=?xN=?

V1=?y1=0.02

VN+1=150yN+1=0.2

T=293K (200C)P=1 atm

• Calculate the molar flow rates

• V’ =

=5.18 kmol inert air/hr.m2

• L’ =

=333 kmol inert water/hr.m2

Substituting in material balance,333()+5.18()=333()+5.18()=0.00355

From graph, 2.4 traysEfficiency=25%=0.25The actual no of tray is=2.4/0.25=9.6 trays

SIMPLIFIED DESIGN METHODS FOR ABSORPTION OF DILUTE

GAS MIXTURES IN PACKED TOWERS

• The concentration can be considered dilute when mole fraction gas (y) and liquid (x) are less than 10% which is 0.1

• Thus, , , , 1.0

• Since the solution dilute, operating line will be straight.

LENGTH OR HEIGHT OF THE ABSOPRTION COLUMN

Because , , , 1, substitute back to z, length or height equation will get:

Where S is the cross sectional area of the tower in m2

The left side is the kg mol absorbed/s.m2

The right side is rate equation for mass transfer Vav=(V1+V2)/2 and Lav=(L1+L2)/2

V1= and V2= , L’≈L1≈L2≈Lav

Material balance in absorption packed tower

STEPS TO CALCULATE HEIGHT,Z FOR PACKED ABSORPTION TOWER

• Plot the operating line (just need two coordinates [x1,y1] as

P1 and [x2,y2] as P2.------straight line

• Calculate Vav and Lav

• Plot equilibrium line (data refer to Appendix)

• k’ya and k’xa will be given

• If use K’ya is used, have to find y1* and y2* while if K’xa is

used, have to find x1* and x2*. Usually, you have to find

the K’ya or K’xa using the equation.

Find slope, PM1Trial 1 to get the value of yi1 and xi1

• 1st trial, use

• 2nd trial, get yi1 and xi1 from 1st trial and insert into this equation to get the value of (1-y)iM and (1-x)iM

• Then, find PM1Trial 2 using this equation

Stop the iteration when value for last trial of slope

is approximately same with the previous slope.

Then will get the final value of yi1 and xi1.

Next, find the PM2Trial 1 and get value of yi2 and xi2

using this equation

If the PM2Trial 1 value is almost same with the PM1Trial

2(or last trial), stop the iteration and get the value yi2

and xi2 from graph.

Calculate the log mean driving force (y-yi)M using this equation if k’ya is used

If K’ya is used, find the log mean driving force using this equation

If using the liquid coefficient which is k’xa and K’xa, use the appropriate driving force equation.

Lastly, calculate height, z using this equation relate to which gas coefficient used.

SIMPLIFIED METHOD:DESIGN OF PACKED TOWERS USING TRANSFER

UNITS• Useful design method of packed towers is the use of

the transfer unit concept.

• For the case of A diffusing through stagnant and non-diffusing B:

In term of Overall film coefficient

The height of the packed tower is then:

Gas phase

Liq phase

In term of film coefficientH= height of transfer unitN – number of transfer unit

In term of film coefficient

In term of Overall film coefficient

The height of the packed tower :

Where:

In term of film coefficient

In term of overall film coefficient

• When the solution are dilute with concentration below 10%:

• the term (1-y)iM/(1-y),(1-x)iM/(1-x),(1-y)*M/(1-y) and (1-x)*M/(1-x) can be taken outside the integral & average values used.

• Often they are quite close to 1 & can be dropped out.

Operating line and interphase compositions in a packed tower for absorption of dilute gases

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