5.Absorption and Stripping L1

Lecture 1: Absorption & Stripping - Introduction

• A mass transfer operation – same category as distillation• Exclusive to gas-liquid separation • Distillation uses the VLE, i.e. difference in boiling

temperatures• Absorption uses the GLE, i.e. solubility

– gas is absorbed into liquid– liquid solvent or absorbent – gas absorbed solute or absorbate

• Stripping is reverse of absorption– liquid absorbed into gas– act of regenerating the absorbent

Absorption in the industry• Air pollution control – scrubbing of SO2 , NO2 ,

from combustion exhaust (power plant flue gas)• Absorption of ammonia from air with water

• Hydrogenation of edible oils – H2 is absorbed in oil and reacts with the oil in the presence of catalyst

Continued…

How does it work?

Solvent

Solute with inert gas

Good product

unwanted gas solution to disposal or recovery

This section can be trayed or packed

How does it work?

Tray tower

Packed tower

How does it work?

Tray tower:

Absorption on each tray

How does it work?

Tray tower:

Types of traySieve Valve

Bubble Cap

A full tray

How does it work?

Packed tower:

1. Structured packing

2. Random packing

How does it work?

Packed tower:

Structured packing

How does it work?

Packed tower:

Structured packing

How does it work?

Packed tower:

Random packing

How does it work?

Spray tower

How does it work?

Bubble Column

Liquid solvent “bed”

• Entering gas (liquid) flow rate, composition, temperature and pressure

• Desired degree of recovery of one or more solutes

• Choice of absorbent (stripping agent)

• Operating pressure and temperature, and allowable gas pressure drop

General Design Considerations

• Minimum absorbent (stripping agent) flow rate and actual absorbent (stripping agent) flow rate as a multiple of the minimum flow rate

• Number of equilibrium stages

• Heat effects and need for cooling (heating)

• Type of absorber (stripper) equipment

• Diameter of absorber (stripper)

General Design Considerations

The ideal absorbent should:

• have a high solubility for the solute• have a low volatility• be stable• be noncorrosive• have a low viscosity• be nonfoaming• be nontoxic and nonflammable• be available, if possible, within the

process

The most widely used absorbents are:• water• hydrocarbon oil• aqueous solution of acids and bases

The most widely used stripping agents are:• water vapor• air• inert gases• hydrocarbon gases

Equilibrium Contact Stages: Single Equilibrium Stage

• Single equilibrium stage system above

• Mass balance:

L0 + V2 = L1 + V1 = M

V1

L0

V2

L1

Single Equilibrium Stage

Mass balance: L0 + V2 = L1 + V1 = M

Gas-liquid absorption – usually 3 components

involved. Let A, B and C be the components, then

L0xA0 + V2yA2 = L1xA1 + V1yA1 = MxAM

L0xC0 + V2yC2 = L1xC1 + V1yC1 = MxCM

and xA + xB + xC = 1.0

V1

L0

V2

L1


L0xA0 + V2yA2 = L1xA1 + V1yA1 = MxAM

L0xC0 + V2yC2 = L1xC1 + V1yC1 = MxCM

xA + xB + xC = 1.0

To solve these 3 equations – their

equilibrium relations will be required

V1

L0

V2

L1


• Gas phase – V

Components – A (solute) and B (inert)

• Liquid phase – L

Components – C

• In gas phase you have binary A-B

• In liquid phase you have binary A-C

V1

L0

V2

L1


• Only A redistributes between both phases.

• Take mole balance of A:

where L’ moles of C and V’ moles of B

V1

L0

V2

L1

1

1'

1

1'

2

2'

0

0'

1111 A

A

A

A

A

A

A

A

y

yV

x

xL

y

yV

x

xL

Tutorial: Derive/Proof this

Single Equilibrium StageV1

L0

V2

L1

1

1'

1

1'

2

2'

0

0'

1111 A

A

A

A

A

A

A

A

y

yV

x

xL

y

yV

x

xL

To solve this, equilibrium relationship between yA1 and xA1 is needed.

Use Henry’s Law: yA1= H’ xA1

H’ – Henry’s law constant (obtainable in Handbooks eg Perry’s)

Lecture 3: Single Equilibrium Stage: Example

A gas mixture at 1.0 atm abs containing air and CO2 is contacted in a single-stage mixer continuously with water at 293 K. The two exit gas and liquid streams reach equilibrium. The inlet gas flow is 100 kgmol/h, with a mole fraction of CO2 of yA2 = 0.20. The liquid flow rate is 300 kgmol/h water. Calculate the amounts and compositions of the two outlet phases. Assume that water does not vapourise to gas.

Solution V1

L0

V2

L1


1

1'

1

1'

2

2'

0

0'

1111 A

A

A

A

A

A

A

A

y

yV

x

xL

y

yV

x

xL

Flow of water 300kgmol/h is L0 = L’

Flow of air, V’ must exclude CO2 thus V’ = V2(1-yA2) = 100(1-0.20) = 80 kgmol/h

Substitute in equation:

1

1

1

1

180

1300

20.01

20.080

01

0300

A

A

A

A

y

y

x

x

Solution V1

L0

V2

L1


1

1

1

1

180

1300

20.01

20.080

01

0300

A

A

A

A

y

y

x

x

1 eqn, 2 unknowns unsolvable

Use Henry’s law, yA1= H’ xA1 to eliminate an unknown.

yA1= 0.142 x 104/1.0 xA1 is substituted into the eqn and solve for xA1:

xA1= 1.41x10-4 and yA1= 0.20

Solution V1

L0

V2

L1


xA1= 1.41x10-4 and yA1= 0.20

We have V’ = V1(1-yA1) and L’ = L1(1-xA1) , so

41

'

1 1041.11

300

1 Ax

LL

20.01

80

1 1

'

1Ay

VV

300 kgmol/h

100 kgmol/h

V1 V2 Vn+1V3 VnVN+1VN

L0 L1L2 Ln-1 Ln LN-1 LN

1 2 n N

Total overall balance:

L0 + VN + 1 = LN + V1 = M where M is the total flow

Overall Component Mole Balance:

L0x0 + VN + 1 yN +1 = LNxN + V1 y1 = Mxm compare to single stage

Equilibrium Contact Stages: (multiple)Countercurrent Multiple-Contact Stages

Making a total balance over the first n stages,

L0 + Vn + 1 = Ln + V1

Making a component balance over the first n stages,

L0x0 + Vn + 1 yn +1 = Lnxn + V1 y1

Solving for yn +1,

1

0011

11

nn

nnn V

xLyV

V

xLy Operating Line

Countercurrent Multiple-Contact Stages

An important case in which the solute A is being transferred occurs when the solvent stream V contains components A and B with no C and solvent stream L contains A and C with no B. (Immiscible contact)


yN + 1

y4

y3

y2

y1x0

x1

x2

x5

xN

N = 4

3

2

1


Consider a 4-stage contact

operating line equation

1

0011

11

nn

nnn V

xLyV

V

xLy

x0 x1 x3x2 x4

1

2

3

4

Operating line

Equilibrium line

y1

y2

y3

y4

yN + 1

yN + 1

y4

y3

y2

y1x0

x1

x2

x5

xN

N = 4

3

2

1

Note: If the streams L and V are dilute in key species, the operating line is a straight line


Example: Absorption of acetone in countercurrent stage tower: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 toal 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 operate isothermally at 300K and a total pressure of 101.3 kPa. The equilibrium relation for the acetone (a) in the gas-liquid is yA = 2.53 xA. Determine the number of theoretical stages required for this separation


Solution:

Given values are: yAN+1 = 0.01, xA0 = 0, VN+1 = 30.0 kg/h and L0 = 90.0 kg mol/h

Acetone material balance,

Amount of entering acetone = yAn + 1V N + 1 = 0.01(30.0) = 0.30 kg mol/h

Entering air = 30.0 - 0.3 = 29.7 kg mol/h

Acetone leaving in V1 = 0.10 (0.30) = 0.030 kg mol/h

Acetone leaving in LN = 0.90(0.30) = 0.27 kg mol/h


Solution:

V1 = 29.7 + 0.0 = 29.73 kg mol air + acetone/h

yA1 = 0.030/29.73 = 0.00101

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

xAN = 0.27/90.27 = 0.00300

Since the flow of liquid varies only slightly from L0 = 90.0 at the inlet to LN = 90.27 at the outlet and V from 30.0 to 29.73, the slope of the operating line is essentially constant (O.L. straight line)


Plot the x-y relations and draw the operating line calculated to find the number of stages

1

2

3

4

5

No. of theoretical stages are required = 5.2 stages


Analytical Equations (Kremser Equation)

Premise:

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

• If the equilibrium line is also a straight line over the concentration range, simplified analytical expressions can be derived for number of equilibrium stages in a countercurrent stage process


Overall balance: L0x0 + VN + 1 yN +1 = LNxN + V1 y1

Rearranging,LNxN - VN +1 yN +1 = Loxo - V1y1

Component balance on the first n stages,

L0x0 + Vn +1 yn +1 = Lnxn + V1 y1

Rearranging,

Loxo - V1 y1 = Lnxn - Vn +1 y n+1

Thus,

LNxN - VN + 1 y N + 1 = Lnxn - Vn +1 y n+1

Countercurrent Multiple-Contact StagesV1

V2 Vn+1V3 Vn VN+1VN

L0 L1 L2 Ln1 Ln LN-1 LN

1 2 n N

Since the molar flows are constant, Ln = LN = constant = L and Vn+1= VN+1 = constant = V

Then, LNxN VN + 1 y N + 1 = Lnxn Vn +1 y n+1

simplifies to L(xN xn) = V(yN+1 yn+1) (A)

Since yn+1 and xn+1 are in equilibrium and the equilibrium line is

straight, yn+1= mxn+1. Similarly, yN+1= mxN+1

Substituting mxn+1 f or yn+1 and calling A = L/mV,

Eqn (A) becomes: NN

nn Axm

yAxx

1

1 (B)


Solving (B),

For transfer of solute A from phase L to V (stripping),

1)/1(

)/1()/1(

)/( 1

1

1

N

N

No

No

A

AA

myx

xx

)/1ln(

)1(//

ln1

10

A

AAmyxmyx

N NN

N

When A =1,

myx

xxN

NN

N

/1

0


For transfer of solute A from phase V to L (absorption)

11

1

01

11

N

N

N

N

A

AA

mxy

yy

A

AAmxy

mxy

N

N

ln

111ln

01

01

01

11

mxy

yyN N

When A = 1,


If equilibrium line is not straight,

11

01

1

1

Vm

L A

Vm

L A

AAA

NN

NN

N

andwhere


Example: Number of stages by analytical equation

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 operate isothermally at 300K and a total pressure of 101.3 kPa. The equilibrium relation for the acetone (a) in the gas-liquid is yA = 2.53xA. Determine the number of theoretical stages required for this separation by using Kremser equation


Solution:

V1 = 29.73 kg mol/h, yA1 = 0.00101, L0 = 90.0 and xA0 = 0

Thus,

A1 = L / mV = L0 / mV1 = 90.0 / (2.53 x 29.73) = 1.20

At stage N,

VN + 1 = 30.0, yAN +1 = 0.01, LN = 90.27, and xAN = 0.0030

Thus, AN = LN/mVN + 1 = 90.27/(2.53 x 30.0) = 1.19

The geometric average,

A = (A1 AN)1/2 = (1.20 x 1.19)1/2 = 1.195


For absorption and by using Kremser equation,

stages)(

)(

.). (ln

......

ln

N 0451951

19511

19511

10532001010

0532010


Graphical Equilibrium-Stage Method for Trayed Towers

• Consider the countercurrent flow, trayed tower for absorption (or stripping) operating under isobaric, isothermal, continuous, steady-state flow conditions

• Phase equilibrium is assumed to be achieved at each tray between the vapour and liquid streams leaving the tray equilibrium stage

• Assume that the only component transferred from one phase to the other is the solute


For application to an absorber, let:

L’ = molar flow rate of solute-free absorbent

G’ = molar flow rate of solute-free gas (carrier gas)

X = mole fraction of solute to solute-free absorbent in the

liquid

Y = mole ratio of solute to solute-free gas in the vapor


Note that with these definitions, values of L’ and G’ remain constant through the tower, assuming no vaporization of absorbent into carrier gas or absorption of carrier gas by liquid. For the solute at any equilibrium stage, n,

nn

nn

n

nn XX

YY

x

yK

1/

1/

Lecture 4: Graphical Equilibrium-Stage Method for Trayed Towers

• Consider the countercurrent flow, trayed tower for absorption (or stripping) operating under isobaric, isothermal, continuous, steady-state flow conditions

• Phase equilibrium is assumed to be achieved at each tray between the vapour and liquid streams leaving the tray equilibrium stage

• Assume that the only component transferred from one phase to the other is the solute


For application to an absorber, let:

L’ = molar flow rate of solute-free absorbent

G’ = molar flow rate of solute-free gas (carrier gas)

X = mole fraction of solute to solute-free absorbent in the liquid

Y = mole ratio of solute to solute-free gas in the vapor


Note that with these definitions, values of L’ and G’ remain constant through the tower, assuming no vaporization of absorbent into carrier gas or absorption of carrier gas by liquid. For the solute at any equilibrium stage, n,

nn

nn

n

nn XX

YY

x

yK

1/

1/


• You can obtain equilibrium data from this expression• How?• By assuming Henry’s and Dalton’s law are applicable

pi=Hxi yi=pi /P

• Use to get y/x • Then rearrange above expression to get Y=f (X)• Calculate Y and X, then make the X-Y plot

nn

nn

n

nn XX

YY

x

yK

1/

1/

Graphical Equilibrium-Stage Method for Trayed TowersX0,L’ Y1,G’

YN+1,G’ XN,L’

1

n

N

(bottom)

(top)

Operating line

Equilibrium curve

XN + 1,L’ YN,G’

Y0,G’ X1,L’

1

n

N

top

bottom

Operating line

Equilibrium curve

absorberStripper

How would you obtain the

expressions for these?

Do mass balances

around unit


• Operating line for absorber:

Yn + 1 = Xn(L’/G’)+ Y1 - X0(L’/G’)

• For stripper:

Yn = Xn + 1(L’/G’) + Y0 - X1(L’/G’)

• It’s a straight line with slope L’/G’• Basically the terminal points (X,Y) for top and bottom can be used and a

straight line drawn between them

Graphical Equilibrium-Stage Method for Trayed Towers - Min Absorbent Flow

Consider, for n = N

X0L’ + YN + 1G’ = XNL’ + Y1G’

or

0

11

XX

YY'G'L

N

N

For stage N, for the minimum absorbent rate,

NN

NNN X/X

Y/YK

1

1 11 (2)

Solving for XN in (2) and substituting it into (1)

(1)

1

N

X0 Y1

YN + 1 XN


011

11

1 XKKY/Y

YY'GL

NNNN

N'min

For dilute solution, where Y y and X x, (3) becomes

(3)

01

11

xK

yyy

'G'L

N

N

Nmin

We obtain:


If the entering liquid contains no solute, that is, X0 0

L’min = G’KN (fraction of solute absorbed)

For Stripper,

stripped solute of fractionN

min K

'L'G

1

N

X0 Y1

YN + 1 XN

Moles solute/mole solute-free liquid, X

Mol

es s

olut

e/m

ole

sol

ute-

free

gas

, Y

YN + 1 (gas in)

XN

(for Lmin)X0

Y1

(gas out)

Ope

ratin

g lin

e 1

Ope

ratin

g lin

e 2

2x m

in a

bsor

bent

rat

eO

pera

ting

line

3

1.5x

min

abs

orbe

nt ra

te

Opera

ting

line

4

Min ab

sorb

ent r

ate


Number of Equilibrium Stages

Similar to Distillation, except start

step from bottom of graph

X0’ Y1’

YN+1 XN

1

N

XN + 1, YN,

Y0, X1,

1

N

Equilib

rium

cur

ve

Equilib

rium

cur

ve

Operatin

g line

Ope

ratin

g lin

e

Stage 1(top)

Stage 1(bottom)

x0

Y1

YN + 1

xN

Y0

YN

x1xN + 1

Packed-tower performance is often analysed on the basis of equivalent equilibrium stages using packing Height Equivalent to a Theoretical (equilibrium ) Plate (staged),

OGOG

tt

NHz

N

z

)N(stagesmequilibriuequivalentofNumber

)z(heightPackedHETP

where

HOG is the overall Height Transfer Unit (HTU) and

NOG is the overall Number of Transfer Unit (NTU)

Packed Tower Design

SaK

VH

yOG '

HOG , Height Transfer Unit (HTU)

V average liquid flow rateKy’ Overall transfer coefficient

a area for mass transfer per unit volume of packed bed,

S cross sectional area of the tower

NOG , Number of Transfer Unit (NTU)

Amxy

mxy

Aln

A

Ninout

ininOG

111

11

1

Packed Tower Design

Some issues in packed tower design:

1. Pressure drop across packing

2. Flooding – gas velocity < flooding velocity

3. Types of packing – usually structured, proprietary design

Packed Tower Design

Acetone is being absorbed by water in a packed tower having a cross-sectional area of 0.186m2 at 293 K and 101.32 kPa. The inlet air contains 2.6 mol% acetone and outlet 0.5 mol%. The gas flow is 13.65 kgmol/h. The pure water flow is 45.36 kgmol/h. Film coefficients for the given flows in the tower are k’ya = 3.78x10-2 kgmol/s.m3.mol frac and k’xa = 6.16x10-2. Calculate packing height, z.

Packed Tower Design

Solution: First calc HOG

y is obtained from x-y plot

Vav = (V1 + V2)/2 = 3.852 x 10-3 kg mol/s

Packed Tower Design

SaK

VH

yOG '

kgmol/s 10 x 3.892 13.65/3600 3-

026011 11 .y

'VV

kgmol/s 10 x 3.811 13.65/3600 3-

005011 22 .y

'VV

Solution: First calc HOG

m is from y = mx = 1.186x relation established

K’ya = 2.19 x 10-2 kgmol/s.m2.mol frac

So,

HOG = 3.852 x 10-3/(2.19 x 10-2 x 0.186) = 0.947 m

Packed Tower Design

SaK

VH

yOG '

2-2- 10 x 6.16

1.186

10 x 3.78

111

a'k

m

a'ka'K xyy

45.7

Solution: Next calc NOG :

A = L/mV = (45.36/3600)/(1.186)(3.852x10-3) = 2.758

NOG = 2.043 transfer units

Packed Tower Design

Amxy

mxy

Aln

A

Ninout

ininOG

111

11

1

7582

1

18610050

18610260

7582

11

75821

1

1

...

..

.ln

.

NOG x0

x0

Solution:

NOG = 2.043 transfer units

HOG = 0.947 m

So,

z = 0.947 x 2.043 = 1.935 m

Packed Tower Design

OGOG NHz

5.Absorption and Stripping L1

Documents

gasliquid absorption

absorbent gas

gas liquid flow rate

gas mixture

exit gas

solubility gas

gas act

inlet gas flow