Absorption Tray Towers

Absorption operationsAbsorption operations

Gas absorption is a unit operation in which one or more soluble components of a gas mixture are dissolved in a liquid

Stripping is an inverse operations performed when it is desired to transfer a volatile component from a liquid into a gas


Examples:

- NH3 is removed from oven gas by water

- CO2, H2S are removed from natural gas using water solutions of alkaline salts

- Benzene, toluene are removed from natural gas using hydrocarbon oil

Equilibrium relations for dilute solutions:Equilibrium relations for dilute solutions:solubilitysolubility

- The maximum amount of the gas that can be dissolved in a solvent at specific conditions (T,P) is called solubility

Equilibrium relations for dilute solutions:Equilibrium relations for dilute solutions:Henry’s lawHenry’s law

AAA xTHPyp )( Henry’s law is valid for dilute solutions, where A does not ionize, dissociate or react in the liquid phase

For water H [atm/mole fraction]

Equilibrium relations for dilute solutions: Equilibrium relations for dilute solutions: Solubility dataSolubility data

Absorption: Design considerationsAbsorption: Design considerationsVa, ya

La, xa

Vb, yb

Lb, xb

Plate 1

Plate 2

Plate n

Plate N

L1 x1V2 y2

L2 x2V3 y3

Plate 3

L(n-1) x(n-1)Vn yn

Ln xn

V(n+1) y(n+1)

Absorption: Design considerationsAbsorption: Design considerationsVa, ya

La, xa

Vb, yb

Lb, xb

Plate 1

Plate 2

Plate n

Plate N

L1 x1V2 y2

L2 x2V3 y3

Plate 3

L(n-1) x(n-1)Vn yn

Ln xn

V(n+1) y(n+1)

We create a contour incorporating a particular number of plates and consider the Elements inside as a steady state system:

- for this system two mass balance equations can be written

- Total mass balance

- Component mass balance

anna VLVL 1

aannnnaa yVxLyVxL 11

1n

aaaan

1n

n

VxLyVx

VL1ny

-

Operating line

Absorption: Design considerationsAbsorption: Design considerations

y

x(mole fraction of A in L)

(mole fraction of A in V)

xa xb

ya

yb

equilibrium line

x*b

1n

aaaan

1n

n

VxLyVx

VL1ny

-

Operating line

)(*AA xfy

PxTHy AA /)(*

Equilibrium line


y



xa xb

ya

yb

equilibrium line

x*b

1 ne yyThe driving force for mass transfer

xn


y



xa xb

ya

yb

equilibrium line

x*b

McCabe-Thiele graphical construction

xn


y



xa xb

ya

yb

equilibrium line

x*b

McCabe-Thiele graphical construction


Limiting conditions: gas-liquid ratio

1n

aaaan

1n

n

VxLyVx

VL1ny

-

In general this is not a straight line because mass is constantly transferred from phase V to phase L (so La<Lb and Va<Vb and Ln/Vn+1is not constant)

If we decrease L, yb does not change but overall concentration of A in L increases. We can continue this process until the operating line crosses the equilibriumline.

At this point (x*b, yb) the driving force of mass transfery-y* is equal to zero. This means we need infinite number of stages to reach this separation.

y



xa xb

ya

yb

equilibrium line

x*b


Limiting conditions: gas-liquid ratio

This condition is called the limiting (L/V)min ratio. This is the lowest flow of solvent possible in the system to achievethe desired separation, although in an infinite number of stages.

The actual solvent flow is usually calculated a multiple of (L/V)min (1.1-1.5 times)

y



xa xb

ya

yb

equilibrium line

x*bba

abab

b

aaaab

b

bb VL

VVLLV

xLyVxVLy

/)(

-**


Limiting conditions: Gas-liquid ratio; straight operating line

Condition: L, V constant -> L/V constant

This is possible for very dilute (<5% mole fraction) mixtures so change in total number of moles of each flow is insignificant

A) Limiting (L/V)min value:

y



xa xb

ya

yb

equilibrium line

x*b

B) Number of ideal stages: the actual L/V ratio is calculated as a multiple of the limiting value (f* (L/V)min) this gives a steeper slope of the operating line.

The number of ideal stages can be then constructed using McCabe-Thiele method.

ab

abaabb xx

yyVL

VLxVyx

VLy

*

min

* -


Limiting conditions: Gas-liquid ratio; straight operating and equilibrium lines

Condition: L, V constant -> L/V constant, ye=mxe

This is possible for very dilute (<5% mole fraction) mixtures so change in total number of moles of each flow is insignificant

and

the region of interest is in the Henry’s law regime

A) Limiting (L/V)min value:

y



xa xb

ya

yb

equilibrium line

x*bB) Number of ideal stages: can be calculated analytically

ab

ab

ab

ab

aabb

xmyyy

xxyy

VL

VLxVyx

VLy

/

-

*min

*

Design of absorbers: simplified casesDesign of absorbers: simplified cases

To calculate the number of ideal stages let’s consider the following:

nnee mxymxy mVLA /

aanaann yAyAyAmxyAyy *

1

)...()...1(

)...()...1(

)()1(

)(

)1(

2*2

2*21

2*2

**22*23

**2

Na

NabN

na

nan

aa

aaaaaa

aaaaa

AAAyAAAyyy

AAAyAAAyy

AAyAAy

yAyyAAAyyAyAyy

AyAyyAyAyy

Define adsorption factor

Then operating line equation becomes

The applying this equation stage by stage:


Let’s consider the following:

)...()...1(

)...()...1(2*2

2*21

Na

NabN

na

nan

AAAyAAAyyy

AAAyAAAyy

Inside the brackets we have geometric series: rrasn

n

1

)1(1

AAAy

AAyy

N

a

N

ab

1)1(

11 *

1

Therefore:

This is Kremser equation


Let’s consider the following:

)( **

***

abba

aabaaNb

yyAyy

yAyAyyAyAyy

baabaa

N

Na

Nab

N

a

N

ab

yyyyAyyA

AAyAyAyAAAy

AAyy

)()(

)(1)1(1

)1(11

**1

1*1

*1

From operating line:

VN,yN=Vb,yb

LN,xN=Lb,xb


From the previous slide:

)( **abba yyAyy

baabaaN yyyyAyyA )()( **1

;

)()()(

*

*

****1

aa

bbN

ababaaN

yyyyA

yyAyyAyyA

A

yyyyN aabb

ln)/()(ln **

Kremser equation: the number of ideal Kremser equation: the number of ideal stagesstages

A

yyyyN aabb

ln)/()(ln **

nnee mxymxy Straight equilibrium line

constmVLA / Adsorption factor, straight operating line

The Kremser equation (1930)

y


xa xb

equilibrium line

y*b

yb

y*b

yb

A stream of gas vented from a condenser in an aromatics plant has a flowrate of 200 kmol h -1, a temperature of 25C and a pressure of 5 bar. The composition (mole fractions) is

Hydrogen 0.900

Methane 0.07

Benzene 0.03

It is proposed to recover 98% of the benzene by absorption into an initially pure, non-volatile hydrocarbon oil using a plate column. Find:

[i] The minimum feed rate kmol h-1 of the oil for this separation [ii] The minimum number of theoretical stages required at an oil rate 30.25 kmol h-1

Additional data:

m is taken as the slope of the X-Y equilibrium line.Equilibrium data at 25C and 5 bar.

Substance Ki = yi/xi

Methane 43.0

Benzene 0.132

Absorption: ExampleAbsorption: Example

La, xa

Va, ya

Vb, yb

Lb, xb

Some preliminary considerations

The mole fraction of benzene in Vb yb=0.03.

The goal is to recover 98% of it.

The equilibrium line is a straight line ye=0.132xe. This follows from the relativevolatility table provided in the problemstatement. This table essentially says that benzene is will be essentially all dissolved in the oiled compared to methane (and even more so hydrogen). Therefore we we can consider hydrogen + methane as inert carrier. Hence the equilibrium line equation


y


xa xb

ya

yb

equilibrium line

x*b

Let’s solve the first part. In the figure on the right we schematically show the condition for the minimum feed rate.

We are interested in finding xb, yb and the liquid flow rate La corresponding to this condition.

Vb=200kmol/hye=0.132xeyb=0.03

therefore(xb)*=(yb)/0.132=0.03/0.132=0.227

Amount of benzene in Vb is equal toNVb=yb*Vb=0.03*200=6.0kmol/h

98% of this has to end in Lb (this is our recovery target). NLb98%*NVb/100=5.88kmol/h

xb=NLb/Lb from this Lb=NLb/xb=5.88/0.277=25.9kmol/h

La=Lb-NLb => La=25.9-5.88=20.02kmol/h


In the second part we need to find the minimum number of stages given La=30.25 kmol/h

Lb=La+NLb=30.25+5.88=36.13kmol/h

xb=NLb/Lb=0.1627

Va=Vb-NLb=200-5.88=194.12kmol/h

ya=NVa/Va=0.12/194.12=6.18*10-4

So we have two points on the operating line:

xa=0.0, ya=6.18*10-4

xb=0.1627 yb=0.03

In addition we can construct an additional point,(or any number of additional points) by looking at a mass balance within the column corresponding to some intermediate stage

La=30.25, xa=0

Va=194.12, ya=6.18*10-4

Vb=200, yb=0.03

Lb=36.13, xb=0.1627

NVb=6

NVa=0.12

NLb=5.88


La=32.25, xa=0

Va, ya

Vb=200, yb=0.03

Lb, xb

Vn+1,yn+1Ln,xn

Let’s for example consider xn=0.1

Then NLn/Ln=0.1 Ln=La+NLn

So NLn=3.36 and Ln=33.6kmol/h

The overall balance is:

La+Vn+1=Ln+Va30.25+Vn+1=33.6+194.12Vn+1=197.8

yn+1 can be found from the operating lineequation

1n

aaaan

1n

n

VxLyVx

VL1ny

-

yn+1=0.017


So we have three points of the operating line and an equilibrium line (shown on the right).

Graphical McCabe-Thiele construction gives11 theoretical stages.

An alternative solution could be obtained using the Kremser equation (based on straight line assumption for both the equilibrium and operatinglines).

yb*=0.132*(xb)=0.132*0.1627=0.021ya*=0.132*(xa)=0

So Nmin=11.2

A

yyyyN aabb

ln)/()(ln **

27.10.20013.36

12.19425.30

132.0111

b

b

a

a

Av VL

VL

mVL

mA

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0 0.05 0.1 0.15 0.2

x

y equilibriumoperating


Absorption Tray Towers

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