Chapter 4 - Gas Absorption

CHAPTER 4: GAS ABSORPTION

CHAPTER / CONTENT

Definition, Application and Notation Used in Gas Absorption

Packed tower Description and Design

Plate tower Description and Design

Conditions of Equilibrium Between Liquid and Gas According to Raoults Law

The Mechanism of Absorption

Definition, Application and Notation Used in Gas Absorption

A type of mass transfer operation (separation) between gas and liquid system.

The removal of one or more selected components from a mixture of gases.

Or A soluble vapor is absorbed from its mixture with an inert gas by meansof a liquid the soluble gas is more soluble.

Common example of gas absorption:

Ammonia can be absorbed by passing the gases (NH3 air) intowater where ammonia will be dissolved in water.

The ammonia (solute) can then be recovered by distillation and theabsorbing liquid can be either discarded or reused.

Ammonia Air System

Acetone Air mixture


The two phases (gas and liquid) when brought into contact tend to reachequilibrium.

Consider air water system, the water in contact with air evaporates untilthe air is saturated with water vapor, and the air is absorbed by the wateruntil it becomes saturated with the individual gases.

In any mixture of gases, the degree to which each gas is absorbed isdetermined by its partial pressure at a given temperature and pressure.

When a single gas (solute) and a liquid (solvent) are brought into contact(until equilibrium), the resulting concentration of dissolved gas (solvent) inliquid is called gas solubility (at T and P).

At fixed temperature, solubility concentration increased when pressureincreased .


Partial pressure of solute in gas phase

(kN/m2)

Concentration of solute in water kg/1000 kg water

Ammonia Sulfur dioxide Oxygen

1.3

6.7

13.3

26.7

66.7

11

50

93

160

315

1.9

6.8

12

24.4

56

-

-

0.08

0.13

0.33

Ammonia Very soluble

Sulfur Dioxide A moderate soluble

Oxygen Slightly soluble


In any mixture of gases the solubility of each gas depends on partialpressure

Solubility also depends on Temperature (Solubility as T ).

Recall Raoults Law for ideal solution:

o

AAA Pxp

etemperatur that at A pressure vapor

phase liquid in A of fraction mol

phase gas in A of pressure partial

o

A

A

A

P

x

p

In G.A.;

pressure partial mEquilibriu where ** PxPP o

In G.A, feed is a gas and enters at bottom of column and the solvent isfed at top column, the absorbed gas and solvent leave out the bottom andunabsorbed components leave as gas from the top.

Liquid / solvent is well below its boiling point and gas molecules arediffusing into liquid.

Equilibrium Distribution (Solubility Curve)


y

x

(y,x)

Equilibrium curve

Figure 1 Equilibrium curve

Due to this concept:

Film concept / Theory in mass transfer


Gas molecules must diffuse from main body of the gas phase to the gas liquid interphase, then

Cross this interface into the liquid side and finally diffuse from theinterface into the main body of the liquid.

For dilute concentration of most gases, and over a wide range for somegases, equilibrium relationship is given by Henrys Law.

AA CHp

constant sHenry'

liquid in component of ionconcentrat

phase gas in A of pressure partial

H

C

p

A

A

THE TWO FILM THEORY

RATE OF ABSORPTION

EVALUATION OF MASS TRANSFER COEFFICIENT

The Mechanism of Absorption

The Two Film Theory

Developed by Whitman.

According to this theory, material is transferred in the bulk of the phase byconvection currents, and concentration differences are considered asnegligible.

On either side of this interface it suppose that the currents die out and thatthere exists a thin film of fluid flowing through which the transfer occursonly to molecular diffusion.

GA

S F

ILM

BO

UN

DA

RY

LIQ

UID

FIL

M

BO

UN

DA

RY MAIN BULK

OF LIQUID

MAIN BULK OF GAS

GAS FILM

LIQUID FILM

PA

RT

IAL

PR

ES

SU

RE

(P

) O

F S

OL

UB

LE

GA

S

PG

Pi

A

B

D

E

Ci

CL

MO

LA

R C

ON

CE

NT

RA

TIO

N

OF

SO

LU

TE

IN

A L

IQU

ID

Figure 2 Concentration profile for absorbed component A

The Two Film Theory

According to Ficks Law, the rate transfer by diffusion is proportional to theconcentration gradient and to the area of interface over which the diffusionis occurring.

The direction of transfer of material across the interface, is, however, notdependent on the concentration difference, but on the equilibriumrelationship.

The controlling factor the rate of diffusion through the two film.

From Figure 2:

PG - Partial pressure in the bulk of the gas phase

Pi - Partial pressure at interface

CL - Concentration in the bulk of the liquid phase

Ci - Concentration at interface

The Two Film Theory

According to the two film theory:

The concentration at the interface are in equilibrium

The resistance to transfer is centred in the thin films on either side ofthe wall

Assumptions of two film theory:

Steady state concentration at any position do not change with time

Interface between the gas and liquid phase is a sharp boundary

Laminar film exist at the interface on both sides of the interface

Equation exist at the interface

No chemical reaction(rate of diff. across gas phase = rate of diff. across liquid phase)

The Two Film Theory

GA

S F

ILM

BO

UN

DA

RY

LIQ

UID

FIL

M

BO

UN

DA

RY MAIN BULK

OF LIQUID

MAIN BULK OF GAS

GAS FILM

LIQUID FILM

PA

RT

IAL

PR

ES

SU

RE

(P

) O

F S

OL

UB

LE

GA

S

PG

Pi

A

B

D

E

Ci

CL

MO

LA

R C

ON

CE

NT

RA

TIO

N

OF

SO

LU

TE

IN

A L

IQU

ID

Figure 2 Concentration profile for absorbed component A

Rate of Absorption

The process of absorption may be regarded as the diffusion of a soluble gasA into a liquid.

The molecules of A have to diffuse through a stagnant gas film and thenthrough a stagnant liquid film before entering the main bulk of liquid.

The rate of absorption of A per unit time over unit area is given:

tcoefficien transfer film Gas - GAAGA kPPkN 21'

The rate of diffusion in liquids is much slower than in gases, and mixturesof liquids may take a long time to reach equilibrium.

For the rate of absorption of A into liquid:

tcoefficien transfer film liquid - LAALA kCCkN 21'

Rate of Absorption

In steady state process of absorption, the rate of transfer of materialthrough the gas film will be the same as that through the liquid film.

The general equation for mass transfer can be written as:

LiLiGGA CCkPPkN '

PG - Partial pressure in the bulk of the gas phase

Pi - Partial pressure at interface

CL - Concentration in the bulk of the liquid phase

Ci - Concentration at interface

NA - Overall rate of mass transfer (mol/unit area.time)

iG

Li

L

G

PP

CC

k

k

Therefore:

Rate of Absorption

This condition can be shown graphically as below where ABF isequilibrium curve.

Figure 3 Driving forces in the gas and liquid

phase

Point D ( CL , PG ) represents conditions in bulk of gas and liquid

Point A ( Ce, PG ) represents concentration Ce in the liquid in equilibrium with PG in thegas.

Point B ( Ci, Pi ) represents concentration Ci in the liquid in equilibrium with Pi in thegas, and gives conditions at the interface.

Point F (CL, Pe ) represents the partial pressure Pe in the gas in equilibrium with CL in theliquid.

CL CeCi

PG

Pi

PB

A

B

D

F

E

PG

P

iCi CL

Rate of Absorption

Then the driving force causing transfer in the gas phase:

DEPP iG

Then,

The driving force causing transfer in the liquid phase:

BECC Li

G

L

Li

iG

k

k

CC

PP

The concentration at the interface (point B) are found by drawing a linethrough D of slope kL/kG to cut the equilibrium curve in B.

Rate of Absorption

Overall Coefficients

LeLeGGA CCKPPKN '

To obtain a direct measurement of the values kL and kG, we require themeasurement of the concentration at the interface.

Need to define two overall coefficient, KL and KG

tcoefficien phase liquid Overall -

tcoefficien phase gas Overall -

L

G

K

K

1Eq. LeLeGGLiLiGGA CCKPPKCCkPPkN'

The rate of transfer A can now be written as:

Rate of Absorption

2Eq.

iG

ei

GiG

iG

GG

iG

eG

GG

eGGiGG

PP

PP

kPP

PP

kK

PP

PP

kK

PPKPPk

111

11

From Eq. (1):

Li

iG

LG

LiLiGG

CC

PP

kk

CCkPPk

11

Insert equation above into Eq. (2):

3Eq. 1

1

Li

ei

LGG

iG

ei

Li

iG

LGG

iG

ei

GiG

iG

GG

CC

PP

kkK

PP

PP

CC

PP

kkK

PP

PP

kPP

PP

kK

11

11

111

Rate of Absorption

where term = average slope of equilibrium curve and,

Li

ei

CC

PP

if the solution obeys Henrys Law (H):

Li

ei

CC

PP

dC

dPH

Therefore Eq. (3) becomes:

LGG

Li

ei

LGG

k

H

kK

CC

PP

kkK

11

113Eq.

1

Similarly:

LGGLL K

H

KHkkK

1111 and

Rate of Absorption

Factors Influencing the Mass Transfer coefficient

Very soluble gas (e.g NH3 in H2O).- resistance is so small where H negligible. So kG KG

Low solubility gas (e.g O2 in H2O)- resistance is in the liquid H is large. So kL KL

Moderately soluble gas- both film offer resistance, so kG KG and kL KL

Example:

Show that from rate of transfer of A:

GLL

LGG

HkkKb

k

H

kKa

111)

11)

Evaluation of Mass Transfer Coefficient

Mass transfer coefficient Diffusivity (D) 1/film thickness (zg) for gas

g

g

Gz

Dk

Problem arises when measuring kG, kL since we do not know the value offilm thickness.

However, one piece of equipment where the surface area is known as the:Wetted Wall column

For gas:L

LL

z

Dk For liquid:

GAS

Slow flowing film of water- laminar flow condition

H

Glass tubeSometimes

turbulent flow could occur

D


HD

Method:

Surface area of film

*PPKN G Flux,

N unit in mol/(m2.s) P* = Equilibrium pressure

Measure gas concentration entering

Measure gas concentration leaving

Surface area of film is its area of the tube

Calculate the concentration driving force at inlet and outlet andtake log mean difference

*PPN

K AG


Example

The data given below were obtained from a wetted wall columnwith a constant liquid flow rate.

Molar Gas Flow rate, G (kmol/s) Overall Mass Transfer Coefficient, KG(kmol/s.m2 (kN/m2)

0.03

0.06

0.12

0.18

157.8

210.6

261.0

285.6

kG also related to the gas flow rate by:

constant is A where82.0GAkG

For molar gas flow rate of G = 0.1 kmol/s, evaluate the individualmass transfer coefficient (kG and kL) and overall mass transfercoefficient (KG) if H = 20 (kN/m2)/kmol


Solution

From Rate of Absorption:

LGG k

H

kK

11

Assuming kL constant and given,

constant is A where82.0GAkG

aEq.

LG k

H

GAK82.0

11

Equation (a) is straight line equation (y=mx+c) where:

y

x

GK1

82.01 G

m

c

A1

LkH


Solution

Construct graph:

y axis

x axis

GK1

82.01 G

G (kmol/s)

KG x106

(kmol/s.m2 (kN/m2))

0.03 157.8 17.732 6.337 x 10-3

0.06 210.6 10.044 4.748 x 10-3

0.12 261.0 5.689 3.831 x 10-3

0.18 285.6 4.080 3.501 x 10-3

82.01 GGK1


Solution

y = 2.079E-04x + 2.653E-03

0.0E+00

1.0E-03

2.0E-03

3.0E-03

4.0E-03

5.0E-03

6.0E-03

7.0E-03

0 2 4 6 8 10 12 14 16 18 20

1/G0.82

1/K

G x

10

-6

From the graph, straight line obtained: y = 2.079x10-4x + 2.653 x 10-3

m = = 2.079x10-4

c = = 2.653 x 10-3

A1

LkH


Solution

From equation obtained: y = 2.079x10-4x + 2.653 x 10-3

For molar gas flow rate of G = 0.1 kmol/s, evaluate the individualmass transfer coefficient (kG) and overall mass transfer coefficient (KG)if H = 20 (kN/m2)/kmol

G (kmol/s)

KG x106

(kmol/s.m2 (kN/m2)) x 10-6

0.10 248.3 6.607 4.027 x 10-3

82.01 G GK1

3.248

10027.41

10027.4

10653.2)607.6(10079.2

10653.210079.2

3

3

34

34

G

G

K

K

y

y

xy


Solution

From equation, slope:

481010079.2

1

10079.21

4

4

A

Am

0.728

10.0481082.0

82.0

G

G

G

k

k

GAk

Individual mass transfer coefficient for gas, kG:


Solution

From equation, y intercept

310653.2 Lk

Hc

Individual mass transfer coefficient for liquid, kL:

6.7538

10653.2 3

L

L

k

c

Hk

H

20

/kmolkN/m 20 For 2

INTRODUCTION TO PACKED TOWER

PRESSURE DROP AND FLOODING IN PACKED TOWER DETERMINATION OF TOWER DIAMETER

DETERMINATION OF HEIGHT OF TOWER

Packed tower Description and Design

Introduction to Packed Tower

Packed towers are used for continuous counter current in absorption.

The tower in Figure 4 consists of a cylindrical column containing:

A gas inlet and distributing space atthe bottom

A liquid inlet and distributing deviceat the top

A gas outlet at the top

A liquid outlet at the bottom

A packing filling in the tower.

A large intimate contact between theliquid and gas is provided by thepacking

Figure 4 Packed tower flows

Introduction to Packed Tower

Common types of packing which are dumped at random in the tower areshown in Figure 5.

Packing are available in size of 3 mm to about 75 mm and mostly are madeof materials such as clay, porcelain, metal or plastic.

High void spaces of 65 95% are characteristics of good packings.

The packings permit relatively large volumes of liquid to passcountercurrent to the gas flow through the openings with relatively lowpressure drops for the gas.

Figure 5 Typical random or dumped tower packings

(a) Rashig Ring (b) Berl Saddle (c) Pall Ring (d) Intalox Metal (e) Jaeger Metal Tri - Pack

Pressure Drop and Flooding in Packed Towers

In a given packed tower with a given type and size of packing and with adefinite flow of liquid, there is an upper limit to the rate of gas flow, calledthe flooding velocity.

The tower cannot be operated at gas flow velocity above flooding velocity.

At a low gas velocities, the liquid flows downward through the packing,essentially uninfluenced by the upward gas flow.

As the gas flow rate is increased at low gas velocities, the pressure drop isproportional to the flow rate to the 1.8 power.

At a gas flow rate called the loading point, the gas starts to hinder theliquid down flow, and local accumulations or pools of liquid start to appearin the packing (liquid holdup).

The pressure drop of the gas starts to rise at a faster rate.

As the flow rate of gas increased, the liquid holdup or accumulationincreases.


At the flooding point, the liquid can no longer flow down through thepacking and is blown out with the gas.

In actual operating tower, the gas velocity is well below flooding velocity.

The optimum economic gas velocity is about one half or more of floodingvelocity.

Flooding velocity depends on:

Type of packing / packing factor

Size of packing

Liquid mass velocity

Limiting pressure drop,

Factor Packing :

flooding at drop Pressure : where

p

flood

pflood

F

P

FP

7.0115.0

Example 1

Ammonia is being absorbed in a tower using pure water at 25OCand 1 atm abs pressure.

The feed rate is 1440 lbm/h (653.2 kg/h) and contains 3.0 mol% ofammonia in air.

The process design specifies a liquid to gas mass flow rate ratioGL / GG of 2/1 and use 1-in. metal Pall rings.


Calculate the pressure drop in the packing and the gas mass velocityat flooding. Using 50% of the flooding velocity, calculate the pressuredrop, gas and liquid flows, and tower diameter.

Repeat (a) above by use Mellapak 250Y structured packing.

Solution 1

Find the required data to be used in Figure 10.6-5 (Geankoplis pp.660)

Given mol fraction for ammonia = 0.03 Mwt ammonia = 17Given mol fraction for air = 0.97 Mwt air = 29

Average molecular weight of entering gas:


64282997017030 . x . x . x MwtyM iiav

333

307300101711

152980682

64281

lbm/ft. @ cmg.

K. mol.K

atmcm.

mol

g . atm

RT

PMg

Solution 1

From Appendix A.2-4, the water viscosity = 0.8937 cP. From A.2-3,the water density is 0.99708 g/cm3

1 Centistokes = 10-2 cm2/s


33

33

3

36

38561

59237453

1

1

30480

1

10990780 /ft lb.

g.

lbx

ft

m.x

m

cmx

cm

g. m

mL

es centistok.scm x .cm.s

g

cPx

g

cm

.cP x .

-

-

90201090201

10

990780

189370 22

23

From Table 10.6-1, for 1-in, Pall rings, Fp=56 ft-1. Using equation

10.6-1,

height packing ft / OH in. 2925156115011507070 ..F.P..

pflood

Solution 1

x-axis for Figure 10.6-5:


From y axis:

0687108561

07300

1

25050

..

.

G

G..

L

G

G

L

For flow parameter of 0.06871 (abscissa) and pressure drop 1.925in/ft at flooding, a capacity parameter (ordinate) of 1.7 is read offthe plot.

sft6.6381

G

G

pGLGG

F

2561.0902.05607310.085.6107310.07.1

7.1

05.05.05.0

05.05.05.0

Solution 1

at flooding


Using 50% of the flooding velocity for design,

sft

lbm.

s

ft.

ft

lbm.G GGG

234852063816073100

sft

lbm.

sft

lbm..GG

22242604852050

Liquid flow rate,sft

lbm

sft

lbm22

4852.02426.02LG

To calculate the pressure drop at 50% flooding,sft

lbm2

24260.GG

and . The new capacity parameter is 0.5 x 1.7 = 0.85.sft

lbm2

48520.GL

By using value of 0.85 and flow parameter of 0.06871 (abscissa), avalue of 18 in H2O/ft is obtained.

Solution 1


The tower cross sectional area, At

2

m

2m ft

s 3600

hr 1x

lb 0.2426

sft

hr

1440lbrate Feed6488.1

G

tG

A

4

2

tt

DA

22 099.2142.3

6488.144ft

AD tt

ft `1.448 5.02099.2 ftDt

Example 2

Ammonia (NH3) is being removed from air by scrubbing withwater in a packed tower with 6 mm ceramic Berl Saddles (Cf = 900.

The gas entering at 1.2 m3/s contains 15% NH3.

The water enters at a rate of 4 kg/s and has a specific gravity of 1,viscosity of 2.5 x 10-3 kg/m.s.

The gas mixture enters at 27OC and 1 bar (0.987 atm). GivenMolecular weight for ammonia = 17, water = 18.

Calculate the diameter of the packed tower when 80% of NH3 isremoved and the pressure drop is 400 N/m2 per m packing.


Solution 2

Find the required data to be used.

Given mol fraction for ammonia = 0.15 Mwt ammonia = 17Given mol fraction for air = 0.85 Mwt air = 29

Average molecular weight of entering gas:


27OC

0.987 atm

80% NH3 is

removed

Lin = 4 kg/s

xA = 0

Lout = LLwater = 4 kg/s

xA = ?

Gin = 1.2 m3/s

yA = 0.15

Gout = ?

yA = ?

2272985017150 . x . x . x MwtyM iiav

Solution 2


33

3gcmg101.090

K 300.15 mol.K

atmcm82.06

mol

g 27.2 atm0.987

RT

PM

Given Gin = 1.2 m3/s

kg/s 1.308 @ g/s 1308m 1

cm10

s

m 1.2

cm

g101.090 G of Mass

3

363

3

3gin

V

kmol/s 0.04809 @ mol/s 48.0871

K 300.15 mol.K

atmcm82.06

s

cm 1.2x10 atm0.987

,G of Mol3

36

in

RT

PVn

kg/s 0.1226 kmol

kgx17

s

kmolx0.04809 0.15in G of Mass

3NH

Solution 2


kg/s 0.0981 kg/s 0.1226 x 0.80 waterin absorbed NH of Mass 3

kg/s 1.2099 0.0981 1.308 absorbed NH of Mass in gas of MassG of Mass 3out

kg 4.09814 0.0981 absorbed NH of Mass waterof MassL of Mass 3out

From information, the water viscosity = 2.5 x 10-3 kg/m.s. Waterdensity is 1000 kg/m3. Since the larger flow quantities are at the bottomof absorber, the diameter will be chosen to accommodate the bottomcondition. From Treybal:

abscissa (x-coordinate) = 50

50

.

GL

.

G

G'

L'

(Refer to bottom condition)

10.0

0.53-

0.53-

10 x 1.0901s

kg1.308

10 x 1.090s

kg4.0981


CG

LG

Lf

gJC

G

1.0

2'

Solution 2


At pressure drop of and x-coordinate = 0.10,

y-coordinate = 0.066

m

mN400

2

0660

102

.g

JCG'

cGLG

.

Lf

sm

kg

0.145g

2

cGLG

38.0

1105.2900

109.1100009.1066.0066.0

''

1.031.0

2''

G

JCG

Lf

Cross sectional area, 2

2

4423

380

3081

m.

.sm

kg.

s

kg .

G''

G'A

m . D

.

A D

DA 092

442344

4

2

Exercise

A packed tower is to be designed for a counter current contact of an NH3- air mixture with water to wash out NH3 from the gas.

The conditions are:

Gas in: Gas out: All NH3 is removed.

Flow rate = 1.5 m3/s

Temperature = 27 C

Pressure = 1 bar (0.987 atm)

Contains 8 mol % NH3

Liquid in:

Flow rate = 4.8 kg/s

Density = 996 kg/m3

Viscosity = 2.5 x 10-3 kg/m.s

Packing used is 38mm Raschig ring (Cf = 95, gc = 1)

(a) Calculate the flow rate of liquid out.

(b) If the pressure drop of the packed tower is 400 N/m2, by using the diagram, calculate the required diameter for the tower.

Component denoted in Figure 6:

Determination of Height of Tower

Gm = Mols of inert gas / (unit time) (unit cross section of tower)

Lm = Mols of inert liquid / (unit time) (unit cross section of tower)

Y = Mols of soluble gas A / mol of inert gas Bin gas phase

X = Mols of soute A / mol of inert solvent C inliquid phase

dZ

x

x + dx

y

y + dy

y2

Gm

x2

Lm

y1

Gm

x1

Lm

Figure 6 Countercurrent absorption towerMass balance over differential / smallsection of column:

tower of section

small of Vol. ere wh

AdzCCakAdZN

AdzAdzPPakAdZN

LiLA

iGGA

'

' NA = mol / (m2.s)

a = Interfacial area / Volume ofcolumn


From the tower,

AdyGdZAPPak

AdyGAdZN

miGG

mA

'

From Dalton and Raoults Law,

iTi

T

ii

GT

T

G

PPyP

Py

PyPP

Py

and

Therefore,

iTG

m

iTGm

yy

dy

aPk

GdZ

dZyyaPkdyG

Integrating both sides; get:

1

2

y

y iTG

m

yy

dy

aPk

GZ Z = height of tower


Similarly for concentration in liquid phase:

xx

dx

ak

LZ

iL

m

Normally, Z is written in terms of KG.a & KL.a & in terms of mol fraction:

OGOG NHZ

Z = height of tower

HOG = height of transfer unit constant.

NOG = number of transfer unit constant

*yy

dy

aPK

GZ

TG

m


And for concentration in liquid phase:

Simplifying:

xx

dx

aCK

LZ

TL

m

*

OGOG NHZ

OLOL NHZ

For gas phase based on equilibrium concentration

For liquid phase based on equilibrium concentration

The number of transfer unit (NOG) can be calculated using several methodwhich will be discussed later.

Graphical Method

Log Mean Driving Force Method

Colburns Method

Methods for Evaluation of NOG

Based on interface concentration:

Based on Area, A:

Graphical Methods

LLGG NHNHZ

xx

yy

G

Lxx

G

Lyy

xxLyyG

m

m

m

m

mm

1

111

11

Operating line equation relates concentration in gas phase to that in liquidphase. From operating line above:

Graphical Methods

y

x

Equilibrium line normally linear at dilute concentration

Operating line

axis- y withonintersecti

slope @ gradient :where

line operating

11

11

11

yxG

L

G

L

yxG

Lx

G

Ly

yxxG

Ly

Example

Graphical Methods

An acetone air mixture containing 0.015 mol fraction of acetone willbe reduced to 1% of this value by countercurrent absorption in freshpure water in packed tower.

The gas flow rate is 1.0 kg/m2.s and the liquid flow rate is 1.6 kg/m2.s.

For the system, Henrys law holds and y* = 1.75 x where y* is the molfraction of acetone in the vapor in equilibrium with x mol fraction inliquid.

Calculate the height of the tower / absorber if HOG = 0.3 m.

Data given:

Cross sectional A of column = 1 m2

Mwt air = 29Mwt acetone = 58Mwt H2O = 18

Solution

Graphical Methods

Given:

Cross sectional Aof column = 1 m2

Mwt air = 29Mwt acetone = 58Mwt H2O = 18HOG = 0.3 my* = 1.75 x

Reduced to 1%

1

2

Gm=1.0 kg/m2.s

Gm=1.0 kg/m2.s

Lm=1.6 kg/m2.s

Lm=1.6 kg/m2.s

y2 = 0.00015

y1 = 0.015 x1 = ?

x2 = 0

Solution

Graphical Methods

The given mass flow rate in kg/m2.s. Need to convert into kmol /s.

s

kmolm

kmol

kgsm

kg.

G

AreaM

itin mass un GG

m

wt

mm

03448.0129

012

2

s

kmolm

kmol

kgsm

kg.

L

AreaM

itin mass un LL

m

wt

mm

08889.0118

612

2

slope

line operating

578.2/03448.0

/08889.0

11

skmol

skmol

G

L

yxG

Lx

G

Ly

Find slope of the graph.

Solution

Graphical Methods

Calculate the mol ratio for each location.

?

0152.0015.01

015.0

1

1

m

m

L

Ax

G

Ay

Location 1

0

01

0

00015.000015.01

00015.0

2

2

m

m

L

Ax

G

Ay

Location 2

Need to find x1 in order to plot operating line.

Cxy

yxG

Lx

G

Ly

578.2

11

Solution

Graphical Methods

Use coordinate at location 2 in order to find C value.

00015.0

0578.200015.0

578.2

C

C

Cxy

Operating line, 00015.0578.2 xy

Insert value at location 1 in order to find x1 value

00576.0

578.2

00015.0015.0

00015.0578.2015.0

00015.0578.2

1

1

11

x

x

xy

Construct operating line by plotting coordinate at location 1 and 2

Location 2

00015.0,0, 22 yx

Location 1

015.0,00576.0, 11 yx

Graphical Methods

No. of stages = 11 NOG = 11

Equilibrium line

Operating line

Solution

Graphical Methods

Height of absorber, Z

mZ

mZ

NHZ OGOG

3.3

113.0

To evaluate NOG, we can take an average driving force in log mean.


xx

dx

yy

dy

** or

2

*

21

*

1

*

21

2

*

1

*

2

*

1

*

21

ln

ln

HxyHxy

yy

yyN

yy

yy

yyyy

yyN

OG

OG

y - line

y* - line

y1

(y - y*)

(y - y*)1

(y - y*)2

y2

Example





Data given:




Solution

Given:



Reduced to 1%

1

2

Gm=1.0 kg/m2.s

Gm=1.0 kg/m2.s

Lm=1.6 kg/m2.s

Lm=1.6 kg/m2.s

y2 = 0.00015

y1 = 0.015 x1 = ?

x2 = 0


Solution


s

kmolm

kmol

kgsm

kg.

G

AreaM

itin mass un GG

m

wt

mm

03448.0129

012

2

s

kmolm

kmol

kgsm

kg.

L

AreaM

itin mass un LL

m

wt

mm

08889.0118

612

2

slope

line operating

578.2/03448.0

/08889.0

11

skmol

skmol

G

L

yxG

Lx

G

Ly

Find slope of the graph.


Solution

Calculate the mol ratio for each location.

?

0152.0015.01

015.0

1

1

m

m

L

Ax

G

Ay

Location 1

0

01

0

00015.000015.01

00015.0

2

2

m

m

L

Ax

G

Ay

Location 2

Need to find x2 in order to plot operating line.

Cxy

yxG

Lx

G

Ly

578.2

11


Solution

Use coordinate at location 2 in order to find C value.

00015.0

0578.200015.0

578.2

C

C

Cxy

Operating line, 00015.0578.2 xy

Insert value at location 1 in order to find x1 value

00576.0

578.2

00015.0015.0

00015.0578.2015.0

00015.0578.2

1

1

11

x

x

xy


Solution


Need to find y1* and y2* using equilibrium line given in question.

0

075.175.1

010.0

00576.075.175.1

75.1

*

2

2

*

2

*

1

1

*

1

*

1

*

1

y

xy

y

xy

xyHxy

Solution


Calculate NOG.

*

21

ln yy

yyNOG

33

3*

2

*

1

*

2

*

1

**

104242.1

00015.0

102.5ln

1005.5ln

000015.0

010.00152.0ln

000015.0010.00152.0

ln

ln

yy

yy

yy

yyyyyy

Solution


Calculate NOG.

stages 1157.10

104242.1

00015.00152.0

ln 3*21

yy

yyNOG


mZ

mZ

NHZ OGOG

3.3

113.0

Objective to evaluate:

Assume mol fraction are so small fraction; mol fraction = mol ratio

Colburns Method

2121 xxLyyG mm

Also, y*=Hx, but instead of H, use m. So y*=mx

If we substitute, we get

1

2

*

y

yyy

dy

1

22

y

y

m

m yyL

mGy

dy

2* yy

L

mGy

m

m

Solution as log function and represented as a nomogram (graphical solution)

Example





Data given:



Colburns Method

Solution

Given:



Reduced to 1%

1

2

Gm=1.0 kg/m2.s

Gm=1.0 kg/m2.s

Lm=1.6 kg/m2.s

Lm=1.6 kg/m2.s

y2 = 0.00015

y1 = 0.015 x1 = ?

x2 = 0

Colburns Method

Solution


s

kmolm

kmol

kgsm

kg.

G

AreaM

itin mass un GG

m

wt

mm

03448.0129

012

2

s

kmolm

kmol

kgsm

kg.

L

AreaM

itin mass un LL

m

wt

mm

08889.0118

612

2

10000015.0

015.0

6788.008889.0

03448.075.1

2

1

y

y

L

Gm

m

m

Find mGm/Lm and y1/y2

Colburns Method

Colburns Method

Solution

From nomogram, NOG = 10.6 = 11

Colburns Method


mZ

mZ

NHZ OGOG

3.3

113.0

INTRODUCTION TO PLATE TOWER

DETERMINATION STAGES OF TOWER

Plate tower Description and Design

MINIMUM LIQUID FLOW RATE TO OBTAIN SPECIFIC SEPARATION

Introduction to Plate tower

Bubble cap columns or sieve trays, are sometimes used for gas absorption.

Application of plate tower particularly when the load is more than can behandled in a packed tower about 1 m diameter; and

- when there is any likelihood of deposition of solids which would quicklychoke a packing.

Plate tower are particularly useful when the liquid rate is sufficient to flooda packed tower.

Determination stages of tower

It will be assumed that dilute solutions are used so that mole fraction andmole ratio approximately equal.

A material balance for the absorbed component from the bottom to a planeabove plate n will give:

line Operating

s

m

msn

m

mn

nmsmsmnm

xG

Lyx

G

Ly

xLyGxLyG

11

11

Operating line also describes such a line passes through two point, 1 (top oftower) and 2 (bottom of tower)


Example 1

A bubble cap absorption column is to be used to absorb ammonia,NH3 by using water.

A gaseous mixture containing 20.5 mol% NH3 and 79.5 mol% air entersthe bottom of the absorption tower.

60.5 kmol of gaseous NH3 enters the tower per hour while 5500 kgaqueous NH3 solution containing 0.1% by mass NH3 enters the top ofthe tower per hour.

The column operates at the atmospheric pressure and at a constanttemperature of 30OC.

It is desired to absorb 95% of the entering gas NH3. Assume that theeffect of water vapor in the gases is negligible.


Example

Determine:

The equilibrium data given as follows:

The molar flow rate of entering gaseous mixture

The molar flow rate of the raffinate

The molar flow rate of the extract

The mol ratio of NH3 in the raffinate and extract

The number of ideal stages required

Mol NH3mol water

0.000 0.053 0.111 0.177 0.250

Mol NH3mol air

0.000 0.044 0.089 0.159 0.280


Solution

95% removal T=30OCP=1 atm

1

2

Gm

Gm

Lm

Lm

y2 = 0.00015

yA1 = 0.205 kmol A/kmol

x1 = ?

xA2 = 0.001 kg A/kg

yB1 = 0.795 kmol B/kmol

Feed of NH3 in= 60.5 kmol/h

Feed of NH3 aq in= 5500 kg/h

xB2 = 0.999 kg B/kg

A = NH3B = AirC = H2O


Solution

Determine:

The molar flow rate of entering gaseous mixture, G1

From Gaseous mixture inlet (at point 1),Feed NH3 in = 60.5 kmol/h

h

A kmol

A kmol

kmol

h

A kmol

h

A kmol

kmol

A kmol

12.295

205.0

15.60

5.60205.0

1

1

1

G

G

G


Solution

Determine:

The molar flow rate of the raffinate, G2

From information, 95% of the entering NH3 is being absorbed.

h

kmol

h

kmol 025.35.6005.0 Unabsorbed NH3 =

The molar flow rate of the raffinate, G2 = Gm + unabsorbed NH3

h

kmol

h

kmol NH unabsorbed

h

Bkmol

h

kmol

kmol

Bkmol

kmol

Bkmol

3

645.237

025.362.234

62.23412.295795.0795.0

2

2

1

G

GG

GG

m

m


Solution

Determine:

The molar flow rate of the extract, L1

h

kmol

h

kmol 475.575.6095.0 Absorbed NH3 =

From information at L2, feed NH3 aq. in = 5500 kg/h

h

A kg

h

kg

kg

A kg

kg

A kg

h

C kg

h

kg

kg

C kg

kg

C kg

5.55500001.0001.0

5.54945500999.0999.0

22

2

LL

LL

A

m

Convert Lm and LA2 in kmol/h unit. Given:

Mwt NH3 = 17 Mwt H2O = 18


Solution

h

A kmol

A kg

A kmol

h

A kg

h

C kmol

C kg

C kmol

h

C kg

324.017

15.5

25.30518

15.5494

2

A

m

L

L

The molar flow rate of the extract, L1

h

kmolh

kmol5

NH Absorbed 3

049.363

475.7324.025.305

1

1

21

L

L

LLL Am


Solution

Determine: The mol ratio of NH3 in all stream

Location 1

Gas Feed stream, G1

Mol of NH3 = 60.5 kmol/h

Mol of Air, Gm = 234.62 kmol/h

258.0'1

'

1

'

1

Y

Y

Y

kmol 234.62

kmol 60.5

Air of mol

NH of mol 3

Extract stream, L1

Mol of NH3 = 0.324 + 57.475 kmol/h= 57.799 kmol/h

Mol of Water, Lm = 305.25 kmol/h

189.0'1

'

1

'

1

X

X

X

kmol 305.25

kmol 57.799

waterof mol

NH of mol 3

Coordinate for Location 1 = (X1, Y1) = (0.189,0.258)


Solution

Location 2

Raffinate stream, G2


Mol of Air, Gm = 234.62 kmol/h

013.0'2

'

2

'

2

Y

Y

Y

kmol 234.62

kmol 3.025

Air of mol

NH of mol 3

Liquid feed stream, L2


Mol of Water, Lm = 305.25 kmol/h

001.0'2

'

2

'

2

X

X

X

kmol 305.25

kmol 0.324

waterof mol

NH of mol 3

Coordinate for Location 2 = (X2, Y2) = (0.001,0.013)


Solution

Plot the equilibrium line and the operating line in order tocalculate number of stages

No of theoretical stages = 5 theoretical stages


0

0.025

0.05

0.075

0.1

0.125

0.15

0.175

0.2

0.225

0.25

0.275

0.3

0 0.025 0.05 0.075 0.1 0.125 0.15 0.175 0.2 0.225 0.25 0.275 0.3

x

y

A bubble cap absorption column is to be used to absorb ammonia,NH3 by using water.

A gaseous mixture containing 15 mol% NH3 and 85 mol% air entersthe bottom of the absorption tower.

57 mol of gaseous NH3 enters the tower per hour, while 3.7 kg of purewater enters the top of the tower per hour.

The column operates at the atmospheric pressure and at a constanttemperature of 30OC.

It is desired to absorb NH3 such as only 2.5% NH3 leaves the towerwith air. Determine the number of theoretical plates required for theabove process.

Equilibrium data is given as in Example 1. Assume that the effect ofwater vapor in the gases is negligible.

Example 2


Minimum Liquid Flow rate

Min. flow rate = Infinite number of stages = Minimum reflux in distillation.

y

x

as L decreases, line approaching equilibrium line

equilibrium line

Minimum value of flow rate is where:

Exit liquid [ ] = Equilibrium [ ]or

Inlet gas [ ] = Equilibrium [ ]

Documents

Chapter 4 - Gas Absorption