cab2054 l1 absorption and stripping

Dr.T.M 1

Absorption

and

Stripping

Dr.T.M 2

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

Dr.T.M 3

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. solubilitygas is absorbed into liquid

liquid solvent or absorbent

gas absorbed solute or absorbate

Stripping is reverse of absorptionliquid absorbed into gas

act of regenerating the absorbent

Dr.T.M 4

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

Dr.T.M 5

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

Dr.T.M 6

How does it work?

Tray

tower Packed

tower

Dr.T.M 7

How does it work?Tray tower:

Absorption on each tray

Dr.T.M 8

How does it work?

Tray tower:

Types of traySieve Valve

Bubble

Cap

A full tray

Dr.T.M 9

How does it work?

Packed tower:

1. Structured packing

2. Random packing

Dr.T.M 10

How does it work?Packed tower:

Structured packing

Dr.T.M 11

How does it work?Packed tower:

Structured packing

Dr.T.M 12

How does it work?Packed tower:

Random packing

Dr.T.M 13

How does it work?Spray tower

Dr.T.M 14

How does it work?Bubble Column

Liquid solvent “bed”

Dr.T.M 15

• 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

Dr.T.M 16

• 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

Dr.T.M 17

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

Dr.T.M 18

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

Dr.T.M 19

Equilibrium Contact Stages

Single

multiple

Dr.T.M 20

Single Equilibrium Stage

Single equilibrium stage system above

Mass balance:

L0 + V2 = L1 + V1

V1

L0

V2

L1

Dr.T.M 21

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

Dr.T.M 22

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

Dr.T.M 23

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

Dr.T.M 24

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

Dr.T.M 25

Single Equilibrium Stage

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

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)

Dr.T.M 26

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

Dr.T.M 27

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

1

1

nn

nnn

V

xLyV

V

xLy Operating Line

Dr.T.M 28

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

Dr.T.M 29

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

Dr.T.M 30

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

Dr.T.M 31

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)

Dr.T.M 32

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,

Dr.T.M 33

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(/

/ln

1

10

A

AAmyx

myx

N NN

N

When A =1,

myx

xxN

NN

N

/1

0

Dr.T.M 34

If equilibrium line is not straight,

11

01

1NN

NN1N

Vm

L Aand

Vm

L A whereAAA

Dr.T.M 35

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.

Dr.T.M 36

A gas mixture of air and CO2 is contacted in a multi 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&

Dr.T.M 37

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 0.13

Y’ 0.07 0.13 0.22 0.30 0.40 0.51 0.58

Dr.T.M 38

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

1.195

1

1.195

11

2.53(0)0.00101

2.53(0)0.01ln

N

Dr.T.M 39

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:

Dr.T.M 40

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/

Dr.T.M 41

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’)

Dr.T.M 42

Minimum Absorbent Flow Rate

1

N

X0 Y1

YN + 1 XN

Moles solute/mole solute-free liquid, X

Mole

s s

olu

te/m

ole

solu

te-f

ree g

as, Y

YN + 1 (gas in)

XN

(for Lmin)

X0

Y1

(gas out)

Op

era

tin

g lin

e 1

Dr.T.M 43

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

Dr.T.M 44

011

11'

min1/

'

XKKYY

YYGL

NNNN

N

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

(E)

01

11min ''

xK

y

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'

'min

NK

LG

Dr.T.M 45

Number of Equilibrium StagesX0’ Y1’

YN+1 XN

1

N

XN + 1, YN,

Y0, X1,

1

N

Stage 1

(top)

Stage 1

(bottom)

x0

Y1

YN + 1

xN

Y0

YN

x1xN + 1

Dr.T.M 46

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%

Dr.T.M 47

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)

Dr.T.M 48

SaK

VH

y

OG'

HOG , Height Transfer Unit (HTU)

V; average liquid flow rate

Ky’; Overall transfer coefficient

a: area for mass trasfer per unit volume of packed bed,

S; cross sectional area of the tower

NOG , Number of Transfer Unit (NTU)

A

A

AKxy

Kxy

A

A

Ninout

inin

OG 1

11ln

K ; equilibrium ratio

A, absorption factor = L/KV

Dr.T.M 49

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%

Dr.T.M 50

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 kg/h.m2. The entering water flow

rate is 6000 kg 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

Dr.T.M 51

i. Estimate the concentration of SO2 in the exit water leaving the

tower.

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

desired absorption.

iii. If the overall efficiency of the tower is 40%, how many number

of actual stages are required.

I

Dr.T.M 52

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

Dr.T.M 53

Number of theoretical trays = 2.4

x0 x1 x3x2 xN

1

2

3

4

Operating line

Equilibrium line

y1

y2

y3

y4

yN + 1

Dr.T.M 54

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

Dr.T.M 55

Solution: First calc HOG

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

Packed Tower Design

SaK

VH

y

OG'

kgmol/s 10 x 3.892 13.65/3600 3-

026011 1

1.y

'VV

kgmol/s 10 x 3.811 13.65/3600 3-

005011 2

2.y

'VV

Dr.T.M 56

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

y

OG'

2-2- 10 x 6.16

1.186

10 x 3.78

111

a'k

m

a'ka'K xyy

45.7

Dr.T.M 57

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

Dr.T.M 58

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

Dr.T.M 59

A tray tower is used to absorb the ethanol vapors from an inert gas

stream using pure water at 30oC and atmospheric pressure. The

concentration of ethanol vapors in the gas stream is 2.2 mol %.

The gas stream flow rate is 100.Kg mol/h. It is desired to recover

90% of the alcohol. The equilibrium relation is Y* = 0.68 X. If

1.5 times of the minimum water flow rate is used, estimate the

number of theoretical stages required for the absorption. Calculate

the number of stages using kremser’s equation.

Problem - 1 (Assignment)

Dr.T.M 60

A hydrocarbon oil containing 4.0 mol % propane is being

stripped by direct superheated steam in a tower to reduce the

propane content to 0.2 mol %. A total of 11.42 kg mol of direct

steam is used for 300 kgmol of entering liquid. An equilibrium

relation of Y = 25x may be assumed. Steam can be considered

as an inert gas and will not condense. Find out the number of

theoretical stages needed for the operation

Problem - 2 (Assignment)

Dr.T.M 61

Problem- 3

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%