1

PART – A

1. Define Gas absorption?

Gas absorption is an operation in which a gas mixture is contacted with a liquid for

the purposes of dissolving one or more components of the gas and to provide a solution of

them in the liquid.

2. Give one-example for gas absorption?

Gas from by product coke ovens is washed with H2O to remove NH3 and again with

an oil to remove benzene and toluene vapours.

3. Define Raoult’s law?

When the gas mixture in equilibrium with an ideal solution also follows ideal gas law,

the partial pressure p * of a solute gas A equals the product of its vapor pressure P at the

same temperature and its mole fraction in the solution x.

*p px=

4. Define Absorption Factor?

Absorption factor A = slope of operating line

of equilibrium curve

L

mG slope=

5. What happens when pressure drop is high in absorber?

For absorber, high pressure drops results in high operating cost.

6. Write down the expression for the height of packing?

Z = NLOG HLOG

Where Z = height of packing, NLOG = no. of overall gas transfer unit

HLOG = height of an overall gas transfer unit.

7. When will be the operating line and Equilibrium curve will be straight for an

absorber?

In an absorber, both equilibrium curve and operating line will be straight for dilute

solution and non isothermal operation.

8. What is the relation between the individual film coefficients and the overall mass

transfer coefficient?

L

1 1 &

1 1 1

K

G g l

l Kg

H

K K K

K H

= +

= +

where

KG = overall mass transfer coefficient based on gas phase

KL = overall mass transfer coefficient based on liquid phase.

Kg = gas film mass transfer coefficient

Kl = liquid film mass transfer coefficient, H = Henry’s law constant.

9. What condition HTU and HETP will be equal?

HETP is numerically equal to HTU, only when the operating line and equilibrium

lines are parallel.

2

10. What is the uses of co-current absorbers?

Co-current absorbers are usually used when the gas to be dissolved in the liquid is a

pure substance.

11. What is the valve of absorption factor?

Desirable value of absorption factors in an absorber is > 1.

12. When will be the tower height in absorber will increase?

For the same system if the same liquid used in an absorber is decreased, the tower

height in absorber will increase.

13. What is the range of Absorption factor?

The most economical absorption factors, A will be in the range from 1.25 to 2.

14. Define stripping factor?

It is defined as the reciprocal of absorption factor.

1

SA

=

15. What will be the temperature in the absorption?

For absorption, an exothermic r x n, the temperature tends to rise and for stripping, an

endothermic r x n, the temperature tends to fall.

16. What is the Gas velocity in Absorber?

The design velocity is normally specified at about 60% of the flooding value.

17. What will be the operating pressure in absorber?

In general an absorber should be operated at the highest pressure consistent with the

requirement of other steps in the process.

18. Write down the formula for NTU?

Number of Transfer unit (NTU or HLOG) is a measure of difficulty of the separation. If

the NTU’s are large, the separation is hard.

1

2

2

1

11ln

* 2 1

y

LOG

y

ydyN

y y y

−= +

− −

19. Write down the formula for HTU?

Height of Transfer unit (HTU or HLOG) give an ideal of efficiency of the equipment.

HTU = ( )1y

G

k a y −

A small HTU is a good tower, implying a large surface area per unit volume.

20. What is the practical range of HTU?

It tends to be between 0.3 meter to 1 meter.

21. What is the Relation between overall and individual transfer units?

1 1 1

LOG LG LL

mG

N N L N= +

3

LOG LG LL

mGH H H

L= +

22. What are the common packings used in the absorption?

(i) Berl saddle and Raschig rings are older types of packing that are not used now adays.

(ii) Intalox saddles are somewhat like Berl saddles where bed porosity increases.

(iii) Super Intalox saddles made up of plastic or ceramic form.

(iv) Pall rings are made from thin metal and have 90% void fraction and low pr drop.

(v) Hypak metal and flexi rings are similar in shape.

23. What is the diameter of rasching rings used in packed tower?

Diameter of rasching rings used in packed lower in industry is normally around 2

inches.

24. When will the exothermic reaction take place in absorption?

1. pr = sc; % change in heat & mass transfer flux will be the same for a given change in

the degree of turbulence

2. pr = sc = 1, total mass, momentum and thermal diffusivity will be the same.

25. What happens when absorption with evolution of heat as compared to isothermal

absorption?

1. Decreased solute solubility.

2. Large minimum liquid to gas ( )LG

ratio.

3. Large number of trays.

26. In the absorption of NH3 in H20, the main resistance to absorption is by the gas

phase.

27. What is value of rasching ring?

Diameter

ring = 1Raschingheight

=

28. Define gas solubility?

The gas solubility should be high, increasing the rate of absorption and decreasing the

quantity of solvent.

29. Give some Example used in gas solubility?

(i) Hydrocarbon oil is used to remove benzene from coke-oven gas.

(ii) Hydrogen sulfide can be removed from gas mixtures using ethanolamine solutions.

30. What is condition used in volatility?

The solvent should have low vapor pressure since the gas leaving an absorption

operation is saturate with the solvent and much may be lost.

31. When will be the operating line straight?

The operating line is straight only when plotted in terms of the mole ratio unit.

32. When will be the operating line is curve?

4

The operating line is cure when plotted in terms of mole fraction and partial pressure.

33. What is the Equation of line in an absorber?

1 1 1

1 111 1 1 1

s s

t

y P xy xG Gs L

y y x xP P

− = = −

− − − −−

34. Where will the operating line lie in an stripper and absorber?

The operating line lies above the equilibrium curve in an absorber where as the

operating line lies below the equilibrium curve.

35. What is the principle applied to stripper?

The operating line which anywhere touches the equilibrium curve represents a

maximum ratio of liquid to gas and a maximum exit gas concentration.

36. When will be the operating will be negative?

When gas and liquid flow concurrently, the operating line has a negative

slope s

s

LG

−.

37. Define ideal tray or theoretical tray?

It is defined as one where the average composition of all the gas leaving the tray is in

equilibrium with the average composition of all the liquid leaving the tray.

38. What is the formula for the number of equilibrium trays in terms of mole fraction of

an absorber?

1 0

1 0

1 1 of 1

Equilibriumlog A

trays

Np

P

y mxNumber log

y mx A AN

+ − − +

− =

39. What is the formula for Np of an stripper?

1

1

1 11

log S

Npo

NpNp

P

yx

mlogy S S

xm

N

+

+

−

− + − =

40. Define number of real tray D?

1log 1 1

Equilibrium trays

1 trayslog

MGE

o

EA

Ereal

A

+ −

= =

41. Expand HETP?

Height equivalents to an equilibrium stage, Height equivalent to an Theoretical plate.

42. CO2 can be absorbed by heated charcoal

5

43. The process employing desorption of the absorbed solute by a solvent is called

elution

44. What happens when absorption factor increases?

To increase absorption factor, increase ‘s’ and decrease ‘G’

Where S – solvent flow rate, G – Gas flow rate

45. Absorption factor method is used to calculate the number of ideal stages when the

operating and equilibrium lines are parallel.

6

PART – B

1. 5000 kg/hr of a SO2 – air mixture containing 5% by volume SO2 is to be scrubbed

with 200,000 kg/hr of water in a packed tower. The exit concentration of SO2 is reduced

to 0.15%. The tower operates at 1 atmosphere. The equilibrium relationship is given by

y = 30 x

Where 2 2

2

SO Mole SO; x =

air H

molesy

mole Mole O=

If the packed height of the tower is 420 cm, Estimate the height of transfer unit

(HTU).

Solution

Average Molecular of SO2 – air mixture = xiMi

= 64 x 0.05 + 29 x 0.95 = 30.75

Total gas rate = rate 5000

162.6.. Molecular wt 30.75

flow molkghrAve

= =

Inert gas rate = Total gas rate x gas in SO2 air mixture

= 162.6 x 0.95 = 154.5 kg mol/hr.

11

1

22

2

5 50.0526

1 1 5 95

0.15 0.150.015

1 1 0.15 99.85

yy

y

yy

y

= = = =− −

= = = =− −

Amount of water = 200,000 kg/hr

Amount of H2O = 2

200,000 200,000

. of H 18M wt O= = 11,111 kg mol/hr

formula : GM (y1 – y2) = L (x1 – x2)

where; x2 = 0

So ; 154.5 (0.0526 – 0.0015) = 11,111 x1, x1 = 0.0007

ye = 30 x

Where, ye is equilibrium value, Xe1 = 30 x 0.0007 = 0.021

Formula ( ) ( )

2

1

2

1 2

1 1 2

1

2

ln

e e

e

e

y y

y y y yNTU

y y

y y

−

− − −=

−

−

Since, X2 = 0, ye2 = 0

7

( ) ( )0.0526 0.0015

0.0526 0.021 0.0015

0.0526 0.021ln

0.0015

5.17

Height of the tower = 420 cm

NTU

NTU

−

− −=

−

=

Formula: Height = HTU x NTU

of the tower

420

5.17

HeightHTU

NTU=

=

of transfer 81.24 cm

unit

Height

2. An air-ammonia mixture containing 5% ammonia by volume is absorbed in H2O in a

packed column operated at 20oC and 1 atm pressure so as to recover 98% of NH3. If the

inert gas flow rate in the column is 1200 kg/hr m2, calculate.

a) The minimum mass velocity of water for this column.

b) The number of transfer unit in the column taking the operating liquid rate to be 1.25

times the minimum.

c) The height of the packed lower taking the overall transfer coefficient, KG a to be

128.0 kg mole hr (m3) (atm). The relationship for equilibrium in the column is y = 1.154

x1 where y and x are in mole fraction units.

Solution:-

Insert flow rate = 1200 kg/hr m2

11

1

5 50.0526

1 1 5 95

yy

y= = = =

− −

( ) ( )22 1

2

1 0.98 1 0.98 0.05261

yy y

y= = − = −

−= 0.001

Equilibrium relation is y = 1.154 x which can be rewritten as formula

y = 1.154 x

1.1541 1

e

e

y y

y x=

+ +

When y = 0.0526, value of x from the above is 0.0453. This corresponds to minimum liquid

rate.

a) Inert rate = 1200 kg/hrm2

21200

41.38 kg mol/hr m29

= =

Formula: GM (y1 – y2) = L(x1 – x2)

41.38 (0.0526 – 0.001) = L (0.0453)

Lm = 47.13 kg mol / hr m2.

8

b) Actual liquid rate = 1.25 Lm = 1.25 x 47.13 = 58.9 kg mol/hr m2

Let x1’ be the liquid composition corresponding to this condition,

Hence, 41.38 (0.0526 – 0.001) 58.9 x’1

X’1 = 0.0363

Putting this value in equilibrium relation of

1

1

1

'

11 '

1

1.1541 1

0.0421

e

e

e

y xy

y x

y

= =− +

=

( ) ( )1 2

1

2

1 2

1 2

1

2

ln

e e

e

e

y y

y y y yNTU

y y

y y

−

− − −=

−

−

( ) ( )0.0526 0.001

0.0526 0.0421 0.001 0

0.0526 0.0421ln

0.001

12.76

NTU

NTU

−

− − −=

−

=

c) Height of the tower = HTU x NTU

M

G T

GHTU

K aP=

Where

GM = inert gas rate kg mol/hr m2

41.38

0.323128 1

HTU m= =

Height = 0.323 x 12.76 Height = 4.12 m

3. A mixture containing 10% solute and rest insert is fed to a packed tower in which

90% if the solute is absorbed solute free H2O used for absorption contains 5% solute

when it leaves the tower at the bottom. If the equilibrium relationship is ye = 0.05 Xe and

Hy = 0.5 m and Hx = 0.4 m, what is the height of the packed section.

Solution:

1

1

2

2

1

100.1111

90

100.0111

90

0

50.0526

95

0.05e

y

y y

x

x

y e

= =

= =

=

= =

=

= 0.05 x 0.0526

= 0.0026

Mass velocity = 848.3 kg/hr m2 of H2O

9

( ) ( )0.1111 0.0111

0.1111 0.0026 0.0111 0

0.1111 0.0026ln

0.1111

NTU−

=− − −

−

NTU = 2.34

Formula:

(HTU)OG = (HTU)g + mG

L (HTU)l

which is also identical to

(HTU)G = (HTU)g + mG

L (HTU)l

(HTU)G = Hg + mG

L Hx

m = 0.05; Hg = 0.5m; Hx = 0.4 m

G(y1 – y2) = L(x1 – x2) = LX1

1

1 2

0.05260.526

0.1111 0.0111

XG

L y y= = =

− −

(HTU)G = 0.5 + 0.05 x 0.526 x 0.4 = 0.51

Height of packed section = HTU x NTU = 0.51 x 2.34 = 1.19 metre

4. An ammonia – air mixture containing 2% ammonia at 25oC and 1 atm is to be

scrubbed with water in a tower packed with 2.54 cm stoneware Rasching rings. The

water and gas rate will be 1200 kg/hr m2 each water and gas are fed at the top and

bottom respectively. Assume that the tower operates isothermally at 25oC. At this

temperature, the partial pressures of ammonia (pg) over aqueous solution of ammonia

are as follows:

Pg mm of Hg 12 18.2 31.7 50 69.5 166

Conc kg NH3/100 kg H2O 2.0 3 5 7.5 10 20

For the above packing, KG a = 62.39 kg mol/hr m3 atm.

Using the logarithmic mean driving force, estimate the required height for the

absorption of 98% of the ammonia in the entering gas.

Solution:

760

g

T g

g

g

pY

p p

p

p

=−

=−

Pg – partial pressure of NH3.

PT – total pressure

Liquid concentration (c) is to be converted to x as under

10

17 0.0106100

18

c

x c= = where C is 3

2

NH

100 kg H

kg

O

The equilibrium relation in terms of y and x are calculated from Pg and c and tabulated as

Y 0.016 0.0245 0.0435 0.0704 01007 0.2795

X 0.0212 0.0318 0.0530 0.0795 0.1060 0.2128

1

20.0204

98y = =

y2 = 0.02y1 = 0.0004

Average Molecular wt = 0.02 x 17 + 0.98 x 29 = 28.76

Total gas rate = 1200

28.76 = 41.72 kg mol/hr m2

Insert gas rate = 41.72 x 0.98 = 40.89 kg mol/hr m2.

Liquid rate = 1200

18 = 66.67 kg mol/hr m2

40.89 (0.0204 – 0.0004) = 66.67 x1

y1 = 0.0123

It is an equilibrium plot between y and x. value of ye1 = 0.0095 corresponding to x1 = 0.0123

( ) ( )0.0204 0.0004

0.0204 0.0095 0.0004 0

0.0204 0.0095ln

0.0004

NTU

−

− − −=

− = 6.3

HTU = 40.89

0.655 metre62.39 1

=

Height for absorption = HTU x NTU = 0.655 x 6.3 = 4.13 meter.

5. It is desired to absorb 96% of the acetone in a 2% mixture of acetone in air in a

0 .2 .4 .6 .8 .1 .12

.02

.04

.06

.08

.10

.12

y

x Equilibrium Relation

11

continuous counter current absorption tower using 20% more than the minimum liquid

rate pure water, used as solvent is introduced at the top of the tower and the gas

mixture is blown into the bottom of the tower at 450 kg/hr. Find the height and

diameter of the tower packed with 2.45 cm wet packet stone ware Rasching rings and

operating at 50% of the flooding velocity. The tower may be assumed to be operating at

1 atmosphere pressure and 300oK. The equilibrium relation is y = 2.5 x where y and x

are mole fractions of acetone in air-acetone mixture and acetone H2O solution reply.

The following relationship and data are available.

OG g g

mGH H H

L= +

where

Hg = 0.54 m; Hl = 0.32 m

Density of acetone air mixture = 1.181 kg/m3 (0.0737 lb/ft3)

Density of the solution of water = 998.4 kg/m3 (62.3 lb/ft3)

Viscosity of the solution = 0.86 cp. Characteristic factor of the packing = 160.

Solution:

Average Molecular wt of gas mixture = 0.02 x 58 + 0.98 x 29 = 29.58

Gas rate = 450

29.58 = 15.21 kg mol/hr

Inert gas rate = 15.21 x 0.98 = 14.91 kg mol/hr

1

20.0204

98y = =

y2 = 0.04 y1 = 0.0008

Equilibrium relation is y = 2.5 x

Which can be written in terms of mole rated as

2.51 1

y x

y x=

+ +

y = 0.0204; x = 0.0081

14.91 x (0.0204 – 0.0008) = Lmin x 0.0081

Lm = 36.08 kg mol/hr

Actual liquid rate = 1.2 Lmin = 1.2 x 36.08 = 43.3 kg mol/hr

Actual composition of outgoing liquor

( )14.91 0.0204 0.0008

43.3

−= = 0.0067

From Equilibrium relation

1

1

1

0.00672.5

1 1 0.0067

0.0173

e

e

e

y

y

y

=+ +

=

( ) ( )0.0204 0.0008

0.0204 0.0173 0.0008

0.0204 0.0173ln

0.0008

NTU−

=− −

−

= 11.5

Calculation of height: Height = HTU x NTU

12

HTU = Hg + mG

LHl

1

1 2

0.0067

0.0204 0.0008

0.34

XG

L y y

G

L

= =− −

=

m = 2.5

HTU = 0.54 + 2.5 x 0.34 x 0.32 = 0.812 metre

Height = 0.812 x 11.5 = 9.34 metre

Calculation of diameter:

Gas rate = 15.21 kg mol/hr

Ideal gas behaviour is assumed

PV = nRT

15.21 0.082 300

1

nRTV

P

= = = 374 m3/hr

Flooding velocity is calculated from the flooding correlation which is plot between 2

0.2

3&G G

L L

PL U a

G P gc

The quantities in the above correlation are in F.P.S unit

43.3 18 2.204 0.0737

450 2.204 62.3

G

L

L

G

=

From the above mentioned figure.

2

0.2

30.16G

L

U a

gc

=

3

160, =0.86a

=

Substituting the values, U = 5.3 ft/sec

Flooding velocity = 5.3 ft / sec

Operating velocity = 0.5 x 5.3 = 2.65 ft/sec = 0.81 m/sec

Gas rate = 374 m3/hr

Cross – sectional area of the column = A 374

0.81 3600=

= 0.128 m2

D = column diameter

2 0.1284

D

=

D = 0.40 m

Tower diameter = 40 cm

6. The acetone from an acetone – air mixture is scrubbed with water in a packed

absorption tower using 1” Rasching rings. The entering gas mixture has 2 mole % of

acetone and the gas leaving the tower is expected to have acetone only to the extend of

0.2 mole %. The gas rate and liquid rate as 160 and 290 kg/hr m2 respectively. The

temperature is 25oC and pressure is 1 atm. The equilibrium relationship is given by y =

2.53 x. Expressions are available to calculate height of transfer units as under (in ft).

HG = 6.41 G0.32 L-0.51 (Nsc)g0.5

13

HL = 0.01( )

( )

0.22

0.5

2.42 lSC

LN

Where

G = lb/hrft2 (gas) = lb/hr ft2 (liquid)

= viscosity = 0.82 cp; (NSC)g = Schmidt number for gas phase = 1.6 (NSC)l =

Schmidt number for liquid phase = 690. Calculate the height of packed tower.

Solution:

1

2

20.0204

98

0.20.002

99.8

y

y

= =

= =

y = 2.53 x is the equilibrium relation

G = 160 kg/hr m2 = 32.77 lb/hr ft2

L = 290 kg/hr m2 = 59.40 lb/hr ft2

HG = 6.41 G0.32 L-0.51 (NSC)g0.50 = 6.41 (32.77)0.32 (59.4)-0.51 (1.6)0.5 = 3.08

HL = 0.01 ( )

( )

0.22

0.5

2.42SC l

LN

( )0.22

0.5059.40.01 690

0.82 2.42

=

= 0.555

Average Molecular wt of gas = 0.02 x 58 + 0.98 x 29 = 29.58

Total gas rate = 160

29.58 = 5.409 kg mol/hr m2.

GH = 0.98 x 5.409 = 5.301 kg mol / hr m2

Liquid rate = 290 kg / hr m2 = 16.11 kg mol / hr m2.

From solute balance,

GM (y1 – y2) = Lx1

( )

( )

1 1 2

1

1 2

5.301 = 0.0204 0.002

16.11

0.329

M

H

Gx y y

L

G x

L y y

= −

−

= =−

m = 2.53

( ) HG LG

mGHTU H H

L= + = 3.08 + 2.53 x 0.329 x 0.555 = 3.54’ = 1.08 m

Ye1 = 2.53 x = 2.53 x 0.00605 = 0.0153

( ) ( )1 2

1

2

1 2

1 2

1

2

ln

e e

e

e

y yNTU

y y y y

y y

y y

−=

− − −

−

−

( ) ( )

0.0204 0.002

0.0204 0.0153 0.002 0

0.0204 0.0153ln

0.002

−=

− − −

−

14

NTU = 5.56

Height of packed Tower = HTU x NTU = 1.08 x 5.56 = 6 metre.

7. Determine the minimum liquid rate needed, the CO2 using out of a fermenter

contains 0.01 mole fraction of ethanol, which has to be reduced to 0.0001 mole fraction

by scrubbing with H2O in a countercurrent packed tower. The gas flow rate is 227.3

kmol/hr2 may be assumed constant throughout the tower. The equilibrium mole

fraction of ethanol in the gas phase y* is related that in the liquid x as y* = 1.07 x and

the number of overall gas-side transfer units needed at 1.5 times the minimum liquid

rate. The entering liquid may be assumed to be free of ethanol?

Solution:

Mole fraction x, y and mole ratio x, y are related as

&1 1

y xy x

y x= =

+ +

In reverse,

&1 1

y xy x

y x= =

− −

Equilibrium relation becomes

*

1.071 * 1

y x

y x=

+ +

Investing the above equation

1 * 1 1

* 1.07

1 1 11

* 1.07 1.07

0.93460.0654

0.9346 0.0654

y x

y x

y x

x

x

x

+ +=

= + −

= −

−=

Inverting again

*0.9346 0.0654

xy

x=

−

Nature of curve

X 0.1 0.2 0.3

y* 0.108 0.217 0.328

opera

tin

g lin

e

Equilibrium curve

X

y

y1 x1

y2

G2

x2

L2

15

These values indicates that, the equilibrium curve in terms of mole ratios y* vs X is concave

upward. The operating line y vs x is straight line. The minimum liquid gas ratio then

corresponds to an exit liquid concentration in equilibrium with the entering gas

Entering mole ratio of gas y1 = 0.01

0.01011 0.01

=−

Exit liquid concentration x2* in equilibrium with y, is obtained from equation

y* = 0.9846 0.0654

x

x−

*

2

*

2

0.01010.9346 0.0654

x

x=

−

0.00944 – 0.00066 x2* = x2

*

x2* = 0.00943

Minimum liquid rate Lmin

2

0.00010.0001

1 0.0001y = =

−

G(y1 – y2) = Lmin (x1 – y2*)

Lmin = 0.0101 0.0001

227.30.00943 0

−

−

= 241.04 kmol/hr

Operating liquid rate L = 1.5 Lmin = 1.5 (241.04) = 361.56 kmol/hr

Absorption factor A:

( )( )

Slope of operating line

of equilibrium curve

361.561.4455

1.07 227.3

1 0.692

Aslope

LA

mG

A

=

=

= =

=

For absorption of dilute solutions, overall number of transfer units NLOG is gn by

16

1 2

2 2

1 1ln 1

11

LOG

y mx

y mx A AN

A

− − +

− =

−

Substituting for the values

( )0.01

ln 1 0692 0.6920.0001

1 0.692LOGN

− +

=−

= 11.2

Number of overall gas – side transfer unit = 11.2

8. Equilibrium relationship for the system heptane- oil – air is gn by y = 2x (y and x are

kg heptane / kg – air and kg – heptane/ kg oil respectively). Oil containing 0.005 kg

heptane / kg oil is being used as solvent for reducing the heptane content of air from

0.10 to 0.02 kg – heptane / kg air in a continuous counter current packed bed absorber.

What column height is required to treat 1400 kg / hrm2 of empty lower cross section) of

pure air containing heptane if the overall gas mass transfer coefficient is 320 kg/hr

m3y. The oil rate employed is 3100 kg/hr m2.

Solution:

Material balance for absorber:

Ls (X1 – X2) = Gs (Y1-Y2)

3100 (X1- 0.005) = 1400 (0.1 – 0.02)

X1- 0.005 = 0.03613

X1 = 0.04113

Equation of operating line

17

( )2 2

LsY Y x x

Gs

3100Y 0.02 (x 0.005)

1400

Y = 2.2143 x +0.0089---------(1)

− = −

− = −

Number of overall gas transfer units

2

y

2LOG *

1y

1 ydy 1N ln (2)

y y 2 1 y

+= + − − − −

− +

Where

Y* = Equilibrium gas composition in mole fraction,

Y* = 2x

Sub/: for Y* in eqn (2) & from eqn (1) Eqn (2)

becomes

1

2

Y

2

toG 1Y

becomes 1 ydy 1ln (3)

N 6.2143x 0.0089 2 1 y

From eqn (1)

y = 2.2143x + 0.0089

+= + − − − −

+ +

diff/: = dy = 2.2143dx

Using this & changing the integration limits items of x Eqn (3) becomes

1

2

X

2toG

1X

1 x2.2143dx 1N ln

0.2143 0.0089 2 1 x

+= +

+

1 2toG

2 1

using

0.2143x 0.0089 1 x2.2143 1N ln ln (4)

0.2143 0.2143x 0.0089 2 1 x

+= + − − −

+

sub/: for X1 = 0.04113 & x2 = 0.005 in eqn (4)

toG

2.2143 (0.2143)(0.04113) 0.0089 1 1 0.005N ln ln

0.2143 (0.2143)(0.005) 0.0089 2 1 0.04113

= 5.9198

+ += +

+ +

Height of overall gas transfer unit HtoG is given by

toG

y

GsH (5)

K a(1 y) M= − − − − −

−

Where

Kya = overall gas phase mass transfer coefficient (1-y) *M = mean of logarithmic

average of (1-y) & (1-y*) between terminals of lower, gn as

18

1 1 2 2 2

1 0 2

2

(1 y*) (1 y*) (1 y ) (1 y*)1(1 y) * M (6)

2 (1 y) (1 y )ln ln

(1 y*) (1 y*)

− − − − − − − = + − − − − − − −

from the diagram shown, it can be visualized that when

Y1 = 0.02 Y1* = 0

Y2 = 0.1 Y2* = 2 x 0.04113 = 0.08226

Sub/: these in eqn (6)

1 (1 0.02) 1 (1 01) (1 0.08226)(1 y) M

1 0.02 1 0.12ln ln

1 1 0.08226

− − − − − − = +

− − −

= (0.5)(0.99 0.9088) 0.9494+ =

Then from eqn (5)

toG

1400H 4.6081m

(320)(0.9494)= =

Height of column = NoG HoG = 5.9198 x 4.6081 = 27.29 m

9. In a mass transfer apparatus operating at atmosphere the individual mass transfer

coefficients have the following values

Kx = 22 kg mol/m2 h (x = 1) Ky = 1.07 kg mol/m2h (y = 1)

The equilibrium composition of the gaseous & liquid phases are characterized by

Henry’s Law P* = 0.08 x 106 x mm Hg

1. Determine the overall mass transfer coefficient Kx & Ky

2. How many times the diffusion resistance of he liquid phase differs from that of

the gaseous phase?

Solution:

Kx = 22 Kg mol/hr m2 (x=1) = kl

KY = 1.07 Kg mol/hr m2 (y=1) = kg

P* = HC = 0.08 x 106 mmHg = 105.26 x atm

H = 105.26

2

1 1 1i)

Kx Kx H Kg

1 1 = 0.0539

22 105.26 x 1.07

Kx = 18.55 kg mol/m h

= +

+ =

19

ly

2

1 1 HIII

Ky Ky Kx

1 105.26 =

1.07 22

1 = 5.715

kg

ky = 0.175 kg mol/m h

= +

+

Rx 1/Kx 0.0539ii) 0.0094

Ry 1/Ky 5.715= + =

10. A tower packed with 0.5 cm Ranching rings of 12 metre height is to be used for

absorption of hydrogen sulphide (H2S) from natural gas (may be treated as mellane), by

using mono ethanolamine as a solvent. The operation in carried out at 300C 1atm

pressure & counter currently. The entering gas contains 18% H2S by volume 90% of

this to be absorbed. The gas flow rate is 200 m3 /m2 hr

Equilibrium line is straight in the operating limits and is given by y = 1.1x

Operating line is also straight and parallel to equilibrium line.

• Find the liquid flow rate

• Find the number of stages & HETP

•

Solution:

H = Height of the absorber = 12 m

P = 1 atmosphere; T = 300C

Average molecular wt of islet gas = ximi = 0.18 x 34 + 0.82 x 16 = 19.24

Taking ideal gas law to be valid

2

Pv 1 2000n

RT 0.082(273 30)

= 80.5 kg mol/hr m

= =

+

2inert rate = 80.5 0.82 = 66.01 kg moles/hr m

1

2 1

18Y = 0.22

2

Y = 0.1,X 0.022

=

=

Equation for equilibrium line in

Ye 1.1 c (lineA)=

Operating line in parallel to the above line The equation is Y 1.1 c Where c=1/2= +

Equation for operating line in

2

2

Y 1.1 0.022 (line B)

X 0

Y 0.022

= +

=

=

For Y1 = 0.22

X1 = 0.18

a) Let l = Flow of liquid, kg moles/hr m2

20

1 2

1 2

1

2

G(Y Y ) LX,

G(Y Y ) L =

X

66.01 (0.22-0.022) =

0.18

= 72.61 kg mol/hr m

− =

−

b) from (X1, Y1) i.e, (o.18, 0.22) slips are constructed in fig

No: of stages =9

Height Equivalent to theoretical Height

= Plates number of stage

12 1.33 metre

9

= =

11. In an experiment 200 liters of air So2 mixture per minute in scrubbed continuously

by H20 in counter. Current fashion is a packed lower. The mixture contains 10%, So2

by volume and is admitted at 200C & 1000 mm Hg pressure into the lower, during

absorption the lower temperature is maintained constants at 200C by means of cooling

arrangement. Determine the lower diameter required for absorbing 95% of So2. It

may be assumed that the height of the lower may have no limitation. Vapour pressure

of So2 over aqueous So2 Solution at 200C is given below:

%conc of So2 in water 0.5 1.0 2.0 3.0 5.0 10

Partial pr of So2 mm Hg 26 59 123 191 336 698

Maximum allowable flow rate of H2O is 200 litres per/hr/m2.

Solution:

The equilibrium data are to be converted to its conventional form

2

2

2 2

0.5mole SO 64X 0.0014

99.5mole H O18

mole SO Partial pf of SOY

mole air Partial pr of air

260.0267

1000 26

= = =

= =

= =−

In the above manner, the values are converted & the equilibrium relationship is obtained as

21

under.

The values are plotted

( )

1

2

1

10Y 0.111

90

Y 0.05

Y 0.0055

P 1000mmHg; T=293 K

PU 1000 0.2n=

RT 760 0.082 293

0.011Kg mole/min

= =

=

=

=

=

=

In GT rate = 0.11 x 0.9 = 0.01 mole / min

Since there is no limitation for scrubber height, minimum liquid can be used when minimum

liquid is used, the liquid concentration is obtained from the equilibrium plot.

X1 = 0.004n from the graph (corresponding to Y10.111)

Over all transfer equation is

( )

( )

1 2

1

GM Y YLM

K

0.01 0.111 0.0055

0.0045

0.231kgmol /min

−=

−=

=

Maximum flow rate of H2O = 200 lit/hrm2 2200

0.185kg mole/min m60 18

= =

Cross – sectional area of the scrubber 20.231

1.249m0.185

= =

Scrubber diameter 1.249 9

1.25metre

= =

0.24 0.20 0.16 0.12 Y1 0.08 0.04

0 0.003 X1 0.006 0.009

22

12. Write down the relationship between the overall coefficients & Transfer units?

The equilibrium distribution curve is straight in some case & ratio of mass – transfer

coefficients is constant that the OU & convenient.

The expressions for the height of packing can be written

( )( )( )

( ) ( )

1

2

1

2

1

2

toG toG

y

toG

y

y

2toG

1y

y

2toG

1y

toG a

y G

Z N H

1 y * MdyN

1 y 1 y *

1 ydy 1N ln

y y * 2 1 y

1 ydy 1N ln

y y * 2 1 y

G G GH

FoG K a 1 y * M K aPt 1 y * M

=

−=

− −

−= +

− −

+= +

− +

= = =− −

Where y* (or y*) is the solute concentration in the gas corresponding to equilibrium with the

bulk liquid concentration x (or x), so that y – y* ( or y-y*) is the vertical distance between

operating line & equilibrium curve (1-y) * M is the logarithmic average of 1 – y & 1-y*

NtoG = number of overall gas transfer units

HtoG = light of an overall gas transfer unit

For cases where the principal mass – transfer resistance ties within the liquid, it is

more convenient to use.

23

( )( )( )

( )

( )

1

2

1

2

1

2

toL toL

x

toL

x

x

1toL

2x

x

1toL

2x

toL a a

L

Z N H

1 x * MdxN

1 x x * x

1 xdx 1N ln

x * x 2 1 x

1 xdx 1N ln

x * x 2 1 x

L LH

FOL Kx 1x * M

L

K ac 1 x * M

=

−=

− −

−= +

− −

+= +

− +

= =

=−

13. Explain about the graphical construction for transfer units?

Overall gas transfer units results when the change in gas composition equals the

average overall driving force causing the change. Consider now the operating diagram in the

graph, where line KB has been drawn so as to be everywhere vertically halfway between the

operating lines equilibrium curve. The step CFD, which corresponds to one transfer unit, has

been constructed by drawing the horizontal line CEF so that line CE = CF & continuing

vertically to D. YG – YH can be considered as the average driving force for the change in gas

composition YD – YF corresponding to this slip since GE = EH and if the operating line can

be considered straight DF = 2(GE) = GH, the step CFD corresponds to one transfer unit. In

similar manner, the other transfer units were stepped off (JK = KL). For computing NtoL, the

line KB would be drawn horizontally half way between equilibrium curve and operating line

& would bisect the vertical positions of the steps.

Overall height of Transfer units

When the overall numbers of transfer units are appropriate, the over all heights of

transfer units can be synthesized from those for the individual phase through the relationship

m1 = m11 = m111 = ….. = const

( )( )

( )( )a

G

G 1 y iM 1 x iMG mG L

FoG F a 1 y *M L FLa 1 y *M

− −= + − −

Heights of transfer units

( )( )

( )( )toG tG tL

1 y iM 1 x iMmGH H H

1 y *M L 1 y *M

− −= +

− −

If the mass transfer resistance in the gas yi y*

( )( )toG tG tL

1 x iMMGH H H

L 1 y *M

−= +

−

24

For dilute solution

( )( )

( )( )toL tL tG

1 x iM 1 y iMLH H H

1 x *M mG 1 x *M

− −= +

− −

For mass transfer resistance is essentially all in the liquid

( )( )toL tL tG

1 y iMLH H H

mG 1 x *M

−= +

−

14. Explain about the equilibrium solubility of gases in liquids?

The rate at which a gaseous constituent of a mixture will dissolve in an absorbent

liquid depends upon the departure from equilibrium which exists & consider the equilibrium

characteristic of gas liquid system.

Two Component systems:

If a quantity of a single gas & nonvolatile liquid are brought to equilibrium &

resulting concentration of dissolved gas in the liquid is said to be gas solubility at the

prevailing temp & pr. At fixed temp, the solubility concentration with increase with pre –

fore y curve which shares the solubility of NH3 in H2O at 30C.

Different gases and liquids separate solubility curves, which must be determined

experimentally for each system. If the equilibrium pressure of a gas at a given liquid

concentration is high as in the case of curves the gas is said to be relatively in soluble in the

liquid, while if it is low as for curve C, the solubility is said to be high.

The solubility of any gas in influenced by the temperature in a manner by van’t Hoff’s

law of mobile equilibrium. If the temperature of a system at equilibrium is raised, that

change will occur which will absorb heat. The solution of a gas results in an evolution of

heat & it follows that in most cases the solubility of a gas decreases with increasing temp.

Curve A for NH3 in H2O at 30C lies above the corresponding curve for 10C. At the boiling

pt of the solvent, its vapour pressure in less than that of the gas or vapor solute, the gas

solubility will be zero. The solubility of many of the low – molecular wt gases such as H2,

O2, N2, N2, CH4 & others in H2O increases with increases with increased temperature above

100C & pr above atmospheric.

Uses: ore – leaching operation where

25

O2 – bearing

solution s are required

15. Gas from a petroleum distillation column has its concentration of H2S reduced from

0.03 K mol H2S per kmol inert hydrocarbon gas to 1% of this value by scrubbing with a

trial than of amine H2O solvent in a counter current tower of height 7.79 m operating at

300 K & atmospheric pressure.

The equilibrium relation may be taken as Y = 2X where,

2 2Kmol H S Kmol H SY = , X =

Kmol inert gas Kmol solvent

Pure solvent enters the lower & leaves containing 0.013 K mol solvent.

If the flow of inert hydrocarbon gas in 0.015 K mol/m2s & gas phase resistance controls

the process, calculate the overall coefficients for absorption KGa

Solution:

Y1 = 0.03, Y2 = 0.01, Y1 = 0.0003, T = 300 K, P = 1 atm, X2 = 0

Ye2 = 0, X1 = 0.013

Inert rate = 0.015 k mol/m2s

Equilibrium relation is Y = 2X

X1 = 0.013, Ye1 = 0.026

26

( ) ( )

( )( ) ( )

1 2

1 e1 2 e2

1 e1

2 e2

Y YNTU

Y Y Y Y

Y Yln

Y Y

0.03 0.0003

0.03 0.026 0.0003 0

0.03 0.026ln

0.0003

NTU 20.79

−=

− − −

−

−

−=

− − −

−

=

Height of tower = 7.79

= HTU x NTU

7.79 7.79HTU

NTU 20.79

0.375

= =

=

G

3Ga

3

GMHTU

K aPT

GM 0.015

0.015 KmolK 0.04Sm atm0.375

K mol144hrm atm

=

=

= =

=

16. A packed tower in designed to recover 98% CO2 from a gas mixture containing 10%

CO2 & 90% air using H2O. A relation y = 14x, can be used for equilibrium conditions

where y is 2kg CO

kg dry air & x is 2

2

kg CO

kg H O.

The H2O to gas rate is kept 30% more than the minimum value. Calculate the height of

the tower if (HTU) OG is 1 metre.

Solution:

( )

1

2

2

10Y 0.1111

90

10Y 0.02 0.0022

90

X 0

Y 14x in mole fraction

= =

= =

=

=

27

( )

y x

44 44141 1

29 18

Y 8.69 x,y are mole ratio

=

=

For minimum liquid rate

( )

11

M 1 2 min 1

1 2

min 1

act min

YX 0.0128

8.69

G Y Y L X

Y YL 0.1111 0.00228.5

G X 0.0128

L L1.3

G G

1.3 8.5 11.05

= =

− =

− − = = =

=

= =

Hence liquid concentration is

1 2Y Y

11.05

i.e., 0.0098

−

=

e1 1Y 8.69X

8.69 0.0098

=

=

( ) ( )

( ) ( )

1 2

1 e1 2 e2

1 e1

2 e2

Y YNTU

Y Y Y Y

Y Yln

Y Y

0.1111 0.0022

0.1111 0.085 0.0022 0

0.1111 0.085ln

0.0022

NTU 11.29

−=

− − −

−

−

−=

− − −

−

=

Height = NTU HTU

=1.0 11.29=11.29m

28

17.A packed tower is to be designed to absorb sulfur dioxide from air by scrubbing the gas

with water. The entering gas is 20% SO2 by volume and the leaving gas is to contain 0.5%

SO2 by volume. The entering water is SO2 free. The water flow is to be twice the minimum.

The air flow rate (SO2 free basis) is 975 kg/hr m2. The temperature is 30oC and the total

pressure is 2 atm. The equilibrium data is governed by y = 21.8 x where y and x are in mole

fraction units. Compute the number of overall gas phase transfer units. (GK Roy)

Ans : L min = 890.7 kg moles /hr m2 Ye1 = 0.1136 N.T.U = 6.16

Solution :

Y1 = 20 / (100-20) = 0.25

Y2 = 0.5 / 99.5 = 0.005

X2 = 0

Gs = 975 kg/hr m2 = 975/29 = 33.81kgmoles/hr m2

Equilibrium Relation is y = 21.8x

Which can be written as Y / (1 + Y ) = 21.8 ( X / ( 1 + X ))

29