Absorption and

Stripping

The objective is:

To understand the absorption( Gas-Liquid) process

To make the material balance for a absorption system

To understand the concept of equilibrium stages and their estimation

To understand the stripping process

To make the material balance for a stripping system

To understand the concept of equilibrium stages and their estimation

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

Introduction

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

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

• multiple

Single Equilibrium Stage

• Single equilibrium stage system above

• Mass balance:

L0 + V2 = L1 + V1

V1

L0

V2

L1

Single Equilibrium Stage

Mass balance: L0 + V2 = L1 + V1

Gas-liquid absorption – usually 3 components

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

L0xA0 + V2yA2 = L1xA1 + V1yA1

L0xC0 + V2yC2 = L1xC1 + V1yC1

and xA + xB + xC = 1.0

V1

L0

V2

L1

Single Equilibrium Stage

L0xA0 + V2yA2 = L1xA1 + V1yA1

L0xC0 + V2yC2 = L1xC1 + V1yC1

xA + xB + xC = 1.0

To solve these 3 equations – their

equilibrium relations will be required

V1

L0

V2

L1

Single Equilibrium Stage

• 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

Single Equilibrium Stage

• 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)

Countercurrent Multiple-Contact Stages

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 Balance:

L0xo + VN + 1 yN +1 = LNxN + V1 y1 = Mxm

Making a total balance over the first n stages,

L0 + Vn + 1 = Ln + V1

Making a component balance over the first n stages,

L0xo + Vn + 1 yn +1 = Lnxn + V1 y1

Solving for yn +1,

1

0011

11

nn

nnn V

xLyV

V

xLy Operating Line

Countercurrent contact with immiscible streams

• 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.

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

Analytical Equations for Countercurrent Stage Contact (Kremser Equation)

• 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 component balance on component A:

L0xo + Vn + 1 yn +1 = Lnxn + V1 y1

Rearranging,

LNxN - VN + 1 y N + 1 = Loxo - V1y1

Component balance for A on the first n stages,

Rearranging, Loxo - V1y1 = Lnxn - Vn + 1y n+1

L0xo + Vn + 1 yn +1 = Lnxn + V1 y1

Thus,

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

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

L (xn - xN) = V(yn+1 - yN+1)

Since yn + 1 and xn + 1 are in equilibrium and the equilibrium line is straight, yn + 1 = mxn + 1. Also, yN + 1 = mxN + 1

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

NN

nn Axm

yAxx

1

1

(A)

(B)

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

11

1

01

11

N

N

N

N

A

AA

mxy

yy

A

AAmxymxy

N

N

ln

111ln

01

01

01

11

mxy

yyN N

When A = 1,

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

If equilibrium line is not straight,

11

01

1NN

NN1N Vm

L Aand

Vm

L A whereAAA

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 in the gas-liquid is yA = 1.5 xA. Determine the number of theoretical stages required for this separation by graphical method and compare it with Kremser equation.

Estimate the minimum solvent ratio for the process.

If 2 times of minimum solvent is used estimate the number of theoretical stages required.

A gas mixture of air and CO2 is contacted in a single stage mixer with pure water at atmospheric conditions. The exit gas and liquid streams are in equilibrium. The inlet gas and liquid flow rate are 100 kg/h and 300 kg/h respectively. The entering gas contains 0.2 mole fraction of CO2. If 90 % of CO2 is observed, Calculate the composition of the leaving liquid and suggest the coordinates of the operating line. Estimate the number of stages required for the absorption.

Assume the equilibrium relation is y = 2.52x. Estimate the number of stages required if, 1.5 times of minimum solvent is required..

Problem - 1&

An adsorption oil containing 0.12 moles of benzene per mole of benzene free oil is to be stripped by using a superheated steam at 121.1oC and at 1 atmospheric pressure. For every 200 kgmol of benzene free oil, 100 kgmol of pure steam was used. The outlet concentration of oil should not exceed 0.005 mol of benzene per mole of benzene free oil. Estimate the number of theoretical stages required for the stripping process. The equilibrium data are given below:

X’ 0.02 0.04 0.06 0.08 0.10 0.12 0.13Y’ 0.07 0.13 0.22 0.30 0.40 0.51 0.58

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 = 0.00300

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

The geometric average, A = (A1AN)1/2 = (1.20x1.19)1/2 = 1.195

For absorption and by using kremser equation,

stages 5.04(1.195) ln

1.1951

1.1951

12.53(0)0.00101

2.53(0)0.01ln

N

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 vapor 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/

X0,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

absorber Stripper

O.P: YN + 1 = Xn(L’/G’)+ Y1 - X0(L’/G’) Yn = Xn + 1(L’/G’) + Y0 - X1(L’/G’)

Minimum Absorbent Flow Rate

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

Ope

ratin

g lin

e 3

Operatin

g line 4

Consider, for n = N

X0L’ + YN + 1G’ = XNL’ + Y1G’

or

(C)

0

11''

XX

YYGL

N

N

For stage N, for the minimum absorbent rate,

NN

NNN XX

YYK

1/

1/ 11 (D)

Solving for XN in (D) and substituting it into (C) gives

011

11'min 1/

'

XKKYY

YYGL

NNNN

N

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

(E)

01

11min ''

xKy

yyGL

N

N

N

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

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

For Stripper,

stripped solute of fraction'

'minNK

LG

Number of Equilibrium StagesX0’ 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

Example:

When molasses is fermented to produce a liqour containing ethyl alcohol, a CO2-rich vapour containing a small amount of ethyl alcohol is evolved. The alcohol can be recovered by absorption with water in a sieve-tray tower. For the following conditions, determine the number of equilibrium stages required countercurrent flow of liquid and gas.K-value=0.57. Given L/V=1.5(L/V)min . State your assumption.

Entering gas: 180 kmol/h; 98% CO2, 2% ethyl alcohol,

Entering liquid absorbent:100% water.

Required recovery of ethyl alcohol:97%

Packed absorption tower design

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

NstagesmequilibriuequivalentofNumber

zheightPackedHETP

)(

)(

where HOG is the overall Height Transfer Unit (HTU) and

NOG is the overall Number of Transfer Unit (NTU)

SaK

VH

yOG '

HOG , Height Transfer Unit (HTU)

V; average liquid flow rateKy’; Overall transfer coefficienta: area for mass trasfer per unit volume of packed bed,S; cross sectional area of the tower

NOG , Number of Transfer Unit (NTU)

AA

AKxyKxy

AA

N inout

inin

OG 1

11ln

K ; equilibrium ratioA, absorption factor = L/KV

Example:

When molasses is fermented to produce a liqour containing ethyl alcohol, a CO2-rich vapour containing a small amount of ethyl alcohol is evolved. The alcohol can be recovered by absorption with water in a packed tower. The tower is packed with 1.5in metal Pall rings. K-value=0.57. Given L/V=1.5(L/V)min . If HOG = 2.0 ft, determine the required packed height.

Entering gas: 180 kmol/h; 98% CO2, 2% ethyl alcohol,

Entering liquid absorbent:100% water.

Required recovery of ethyl alcohol:97%

b.         A tray tower is used to absorb SO2 from an air stream by using pure water at 25oC.

The entering gas contains 20 mole percent of SO2. The tower is designed to absorb

90% of SO2. The flow rate of pure air is 150 moles/h.m2. The entering water flow

rate is 333moles of water/h.m2. The equilibrium data are on solute free basis are given below

Equilibrium data for SO2 - water

Mole fraction of SO2

in water, X

Mole fraction of SO2 in

vapour, Y

0.00000 0.00000

0.00150 0.03420

0.00200 0.05140

0.00280 0.07750

0.00420 0.12140

0.00700 0.21200

i.    ii.i.   Estimate the number of theoretical stages required for the desired absorption.

 

Since the equilibrium data are given in molar units, calculate the molar flow rates

V’ = 150/29 = 5.18 Kg mol inert air/m2 h

L’ = 6000/18.0 = 333 Kg mol inert water/ m2 h

Y N+1 = 0.20; Y1 = 0.02

X0 = 0; XN = ????

substituting into the material balance equation

02.01

02.018.5

1333

20.01

20.018.5

01

0333

N

N

x

x

XN = ----------- ????

yN + 1

y4

y3

y2

y1x0

x1

x2

x5

xN

N = 4

3

2

1

Number of theoretical trays = 2.4

x0 x1 x3x2 xN

1

2

3

4

Operating line

Equilibrium line

y1

y2

y3

y4

yN + 1

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 air/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. The equilibrium relation is given by y = 1.186x.

Packed Tower Design

Solution: First calc HOG

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.m3.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 = 1.28 transfer units

Packed Tower Design

Amxy

mxy

AAN

inout

ininOG

111ln

ln

1

758.2

1

x0186.1005.0

x0186.1026.0

758.2

11ln

758.2ln

1OGN

Solution:

NOG = 1.28 transfer units

HOG = 0.947 m

So,

z = 0.947 x 2.043 = 1.935 m

Packed Tower Design

OGOG NHz