Unit-2: Distillation: Boiling point diagram, Azeotropes-

types, Partial vaporization and condensation and Relative

volatility, flash distillation and differential distillation, Steam

distillation, Azeotropic distillation and Extractive distillation.

 What is distillation?
Separation of components of solution
Distribution of substance between gas and vapor phase

Distillation is used to separate liquids from nonvolatiles, as in the

separation of alcoholic liquors from fermented materials,
separation of two or more liquids having different boiling points, as in
the separation of gasoline, kerosene, and lubricating oil from crude
oil.
Evaporation- complete solvent removal
Distillation – volatiles separation
Absorption – Dissolution in gas phase, not in pure form
Volatile
component to
Volatile+non- condensation
Heat
volatile
 How distillation differs?

Absorption – New phase

Distillation - New phase differs only in heat content, chemically similar

to the original phase

Disadvantage ! – Sometimes the chemical similarity gets closer make

separation impossible
Vapor liquid equilibrium

P-T-C phase diagram

Assumption:
Homogenous solution
No azeotrope formation

Binary mixture

A is more volatile

VPA>VPB at any temperature

 T-x-y diagram

Ease of separation, difference in

vapor and liquid composition
 Relative volatility

α should be higher for easier separation

α =1, no separation possible
Effect of
pressure
 Ideal solutions-Raoult’s Law

Vapor pressure of
pure component

Equilibrium partial
pressure
PA*

Pt
For A

For B
Deviations from ideality Positive

< Pt Positive deviation

Ideal Mixture pressure

Raoult’s law valid at higher

concentrations
 Min. boiling azeotrope – Positive
deviation
Larger positive deviation from ideality
Closely spaced vapor pressures
Total pressure curves present a maximum
 V

y =x

Vapor y<x
y>x
 Boiling at D / H

Changes with pressure ( T and C)

heteroazeotrope
Completely insoluble components
Vapor pressure of one not influenced by other
True vapor pressure is exerted at any temperature by the components

High boiling compounds can be separated at low

temperatures

Heat requirement
 *
t

solution (VP)
pure
For A

For B
 Deviations from ideality Positive

< Pt Positive deviation

Ideal Mixture pressure

Raoult’s law valid at higher

concentrations
 Min. boiling azeotrope – Positive deviation

Larger positive deviation from ideality

Closely spaced vapor pressures
Total pressure curves present a maximum
 V

y =x

Vapor y<x
y>x
 Boiling at D / H

Changes with pressure ( T and C)

heteroazeotrope
Completely insoluble components
Vapor pressure of one not influenced by other
True vapor pressure is exerted at any temperature by the components

High boiling compounds can be separated at low

temperatures

Heat requirement
Negative deviations from
ideality
Acetone-Chloroform, 1 atm
HCl-H2O, 11.1 mole% HCl 110
C, 1 atm
 Heat of solution

Specific heat

For sat. liquids, tL is bubble point

Sat. vapor enthalpy, at dew point

HM,ZM HN,ZN
N
M P
 A is immiscible with water, Feed contains A and traces of non-volatiles

when m A moles of A is volatilized by m B moles of steam

mA Ex A PAv

mB P  Ex A PAv

Feed is a mixture of A and C, which form ideal solution , insoluble in water, C is non-
volatile

If mAi is initial moles of A and mAf is final number of moles

when m A moles of A is volatilized by m B moles of steam

mA Ex A PAv

mB P  Ex A PAv
 dmA EmA PAv EmA PAv EmA PAv
  mB  mB  mB
dt P(mA  mc )  EmA PAv PmA  Pmc  EmA PAv mA ( P  EPAv )  Pmc

mA ( P  EPAv )  Pmc P Pmc

v
 v
1
EmA PA EPA EmA PAv
Pmc
 P  EPAv 
mA
Y*=xD
  ( x(1  y*))  y *(1  x)
 x   xy*  y *(1  x)
 x  y *[(1  x)   x]
x x
y*  
(1  x)   x (  1) x  1

x y
0 0
0.2 0.384615
0.4 0.625
0.6 0.789474
0.8 0.909091
1 1
 1

0.8

0.6
yD*=0.5
0.4

Xf=0.4
0.2
Xw=0.3
0
0 0.2 0.4 0.6 0.8 1
 1

0.8
Slope=-W/D= -0.8

0.6
yD*=0.5
0.4

Xf=0.4
0.2
Xw=0.3
0
0 0.2 0.4 0.6 0.8 1
F=50 kmol; Acetone in feed= 50 * 0.35=17.5 kmole acetone
Amount vaporized from feed liquid = 50 * 35= 17.5 kmoles=D
F=D+W
50=17.5+W
W=32.5
-W/D=32.5/17.5 = 1.86
 12000

Hg vs y
10000

8000

6000
Enthalpy
4000

2000 HL vs x
x and y
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
1

0.9

0.8

0.7

y 0.6 Slope = -1.86

0.5

0.4

0.3

0.2

0.1
x
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
 12000

10000

8000

6000

4000

2000

0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
1

0.9

0.8
0.77=xD
0.7

0.6

0.5

0.4

0.3

0.2

0.1
Xw=0.1 Xf=0.35
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
 12000

10000

8000

4
6000

6 3
4000

2000
5
0
10 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
1

0.9

0.8 2
0.77=xD
0.7

0.6

0.5

0.4

0.3

0.2

0.1
Xw=0.1 Xf=0.35
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Binary mixtures Vapor yD

AT equilibrium
Liquid xw

Equilibrium

Rayleigh
equation
Equilibrium data, The BP of A and B are 98.4 and 125.6 C respectively. Choose any 5
temp in between this range, and find the VP for pre components at these temperature,
then find x and y* using Raoult's law at 1 atm.

T PA PB x y* α
F=100 mols
98.4 760 333 1 1 2.282282
zf= 0.5
105 940 417 0.655832 0.81116 2.254197 D=60 mol
110 1050 484 0.487633 0.673703 2.169421 W=100-60 = 40 mol
115 1200 561 0.311424 0.491722 2.139037 -(W/D) = (-40/60) = -0.667
120 1350 650 0.157143 0.279135 2.076923
125.6 1540 760 0 0 2.026316
 1
dy/dx=-0.667
0.9
dx=0.1
dy= 0.0667
0.8

0.7

0.6 yD = 0.59
0.0667
0.5 Feed

0.4 From Table T is in

between 110 and 115
0.3 C, interpolation
For 5 C T difference,
0.2 diff. in y is 0.181
For diff. in y of 0.8,
0.1 diff in T is + 2.22 C
Therefore T is 112.22
xW =0.38
0
0 0.2 0.4 0.6 0.8 1
 Generate equilibrium data, The BP of A and B are 98.4 and 125.6 C respectively. Choose
any 5 temp in between this range, and find the VP for pre components at these
temperature, then find x and y* using Raoult's law at 1 atm.

T PA PB x y* α
F=100 mols
98.4 760 333 1 1 2.282282
zf= 0.5
105 940 417 0.655832 0.81116 2.254197 D=60 mol
110 1050 484 0.487633 0.673703 2.169421 W=100-60 = 40 mol
115 1200 561 0.311424 0.491722 2.139037 -(W/D) = (-40/60) = -0.667
120 1350 650 0.157143 0.279135 2.076923
125.6 1540 760 0 0 2.026316
 1
dy/dx=-0.667
0.9
dx=0.1
dy= 0.0667
0.8

0.7

0.6 yD = 0.59
0.0667
0.5 Feed

0.4 From Table T is in

between 110 and 115
0.3 C, interpolation
For 5 C T difference,
0.2 diff. in y is 0.181
For diff. in y of 0.8,
0.1 diff in T is + 2.22 C
Therefore T is 112.22
xW =0.38
0
0 0.2 0.4 0.6 0.8 1
 Azeotropic distillation

Add third component

Change system pressure

Decantation (2
Benzene layers)
Benzene,
C alcohol,water New
Entrainer
azeotrope
boils at low T
A+B
(Azeotrope)

Ethanol, water (96%)

Pure alcohol
Should form low boiling azeotrope with minority component
 Similar molecular weight
compounds
Isooctane 99. 3 C
Unit-3: Distillation: Continuous distillation with rectification, Sieve plate column, Bubble cap

column: Packed column; calculation of number of plates, Lewis sorel Method, McCabe Thiele

Method, Economic reflux ratio, Minimum reflux Ratio, Fenske Underwood Equation. Plate

efficiency, Ponchon Savarit method, Packed Column; HETP – method, Transfer unit and Batch

distillation.
https://www.sulzer.com/en/shared/about-us/predict-distillation-tray-efficiency-
white-paper-4-2018
 Fractionator-Continuous
Enriching/absorb
ing/rectifying
section

Stripping/exhausting
There are four streams (2 vapour and 2 liquid) associated with this plate, each with its own
flow rate and concentration:

Composition (Mole
Stream Total Molar Flowrate
Fraction MVC)

Vapour leaving plate Vn yn

Liquid leaving plate Ln xn

Vapour entering plate Vn+1 yn+1

Liquid entering plate Ln-1 xn-1

An ideal plate is one where the vapour leaving the plate is in equilibrium with the liquid
leaving the same plate (see the equilibrium diagram : xn and yn are in equilibrium).

On each plate, the vapour rising to it and the liquid flowing down to it are not in equilibrium.
Thus there is a concentration difference (driving force) for mass transfer between the two
phases.

The system tends to reach equilibrium on each tray because

some of the less volatile component condenses from the rising vapour into the liquid, thus
increasing the concentration of the more volatile component in the vapour as it moves
upwards, e.g. yn > yn+1.

some of the more volatile component is vaporized from the liquid on the tray, thus
decreasing the concentration of the more volatile component in the liquid as it moves
downwards, e.g. xn-1 > xn.

The combined effect is a gradual increase in concentration of the more volatile component
in the vapor as it moves up the column, and a gradual increase in concentration of the less
volatile component in the liquid as it moves down the column.
 Constant molar overflow
The vapour and liquid streams inside the column are assumed to be saturated at
their respective dew points and bubble points corresponding to the position in the
column.
The heat released by one mole of vapour upon condensation is approximately
equal to the heat required to vaporize one mole of the liquid: the number of
molecules passing from the vapour phase to the liquid phase and vice versa will be
approximately the same: equimolar counter-diffusion.
If the molar heats of vaporization are approximately constant, we can assume that
the flows of liquid and vapour are nearly constant in each section of the column,
i.e.

L1 = L2 = L3 = ......... = Ln = constant
V1 = V2 = V3 = ......... = Vn = constant

This is the important concept of constant molal overflow.

Separation is achieved with the vapour rich in the more volatile component leaving
the top of the column, and the liquid rich in the less volatile component leaving the
bottom of the column.
The temperature decreases as one moves up the column, i.e. Tn+1 > Tn > Tn-1.
http://www.separationprocesses.com/Distillation/DT_Chp04g.htm
 Enthalpy balance

Heat economy-Heat
exchange
 Ratio of moles of reflux to withdrawn distillate , reflux ratio or external reflux
ratio

Envelope I

Heat load on condenser

Envelope II
Heat load on reboiler
 Negligible heat loss

Envelope III 1

Constant
Difference in component A flow up and down i.e net flow
upward

Net heat flow upward

Ratio of moles of reflux to withdrawn distillate , reflux ratio or external reflux
ratio

Envelope I

Heat load on condenser

Qc  DRH G1  DH G1  DRH L 0  DH D
Qc  DH G1  DH D  R[ DH G1  DH L 0 ]
[Qc / D]  H G1  H D  R[ H G1  H L 0 ]
Locate Q’ point on Hxy plot, Join zF=0.36 and Q’point. That should give delD. Extend
this to cut the vertical line from xw, which will give delW.
Interpolation to find
HG1
0 to 1 -9000
Difefrence -9000

For 0.915 -8235

HG1 39765
140000
120000
100000
80000
60000
40000
20000
0
0 0.2 0.4 0.6 0.8 1
 1

0.8
0.69

0.6

0.4

0.2
0.36
0
0 0.2 0.4 0.6 0.8 1
 140000 ΔD Qact=136432.5
120000
100000
80000
60000
Qmin=66000

40000
G1

20000
0
0 0.2 0.4 0.6 0.8 1
1

0.8

ΔW 0.6

0.4

0.2

Xw=0.01 0 0.36 Xd=0.915

0 0.2 0.4 0.6 0.8 1
 140000 ΔD Qact=136432.5
120000
100000
80000
60000
Qmin=66000

40000 G1.HG1
20000
0
0 0.2 0.4 0.6 0.8 1
1 L1

0.8

ΔW 0.6

0.4

0.2

Xw=0.01 0 Xd=0.915
0 0.2 0.4 0.6 0.8 1
 140000 ΔD Qact=136432.5
120000
100000
80000
60000
Qmin=66000

40000 G1.HG1
20000
0
0 0.2 0.4 0.6 0.8 1
1 L1

0.8
G2
ΔW 0.6

0.4

0.2

Xw=0.01 0 Xd=0.915
0 0.2 0.4 0.6 0.8 1
 140000 ΔD Qact=136432.5
120000
100000
80000
60000
Qmin=66000

40000 G1.HG1
20000
0
0 0.2 0.4 0.6 0.8 1
1 L1

0.8
G2
ΔW 0.6

0.4

0.2

Xw=0.01 0 Xd=0.915
0 0.2 0.4 0.6 0.8 1
 140000 ΔD Qact=136432.5
120000
100000
80000
60000
Qmin=66000

40000
20000
0
0 0.2 0.4 0.6 0.8 1
1

0.8

ΔW 0.6

0.4

0.2

Xw=0.01 0 Xd=0.915
0 0.2 0.4 0.6 0.8 1
 140000 ΔD Qact=136432.5
120000
100000
80000
60000
Qmin=66000
Total plates= 6-1 =5 40000
20000
0
0 0.2 0.4 0.6 0.8 1
1

0.8

ΔW 0.6

0.4

0.2

Xw=0.01 0 Xd=0.915
0 0.2 0.4 0.6 0.8 1
 1

0.8

0.6

0.4

0.2

0
0 0.2 0.4 0.6 0.8 1
Enthalpy balance

Heat economy-Heat
exchange
L=DR; G=D(R+1)
Inside the tower

V and L are saturated

G G  L  L F
[ L  L  F ]H G  FH F
[ L  L ]H G  F [ H F  H G ]
L L  G G  F
G G  F
q
F
G  G  qF  F  F (q  1)
 1

0.8

0.6

0.4

0.2

0
0 0.2 0.4 0.6 0.8 1
 1

0.8

0.6

0.4

0.2

0
0 0.2 0.4 0.6 0.8 1
 1

0.8

y
0.6

0.4

0.2
xd
xw xf
0
0 0.2 0.4 0.6 0.8 1
 1

0.8

y 0.6

0.4

0.2

0
0 0.2 0.4 0.6 0.8 1
Total plates =20-1=19

Considering efficiency

Number of plates =19/0.6 = 31.66= 32 plates

Feed is mixture of liquid and vapor
 1
0.8
0.6
y

0.4
0.2
xf

0 xw
xd
0 0.2 0.4 0.6 0.8 1
x
50% = LF

q/(q-1) = Lf/(Lf-F)

Slope = - 0.5/0.5 = -1
 1
0.8
0.6
y

0.4
0.2
0
0 0.2 0.4 0.6 0.8 1
x
 1
0.8
0.6
Rmin=0.35
y

0.4
0.2
0
0 0.2 0.4 0.6 0.8 1
x
 1
0.8
0.6
y

0.4
0.2
0
0 0.2 0.4 0.6 0.8 1
x
0.35

0.35
2.74

2.74 =5.5

0.15
5.5+1
 1
0.8
0.6
y

0.4
0.2
0
0 0.2 0.4 0.6 0.8 1
x
 1
0.8
0.6
y

0.4
0.2
0
0 0.2 0.4 0.6 0.8 1
x
N=9-1 = 8
A equimolar feed mixture containing A and B is differentially distilled such that 70% of the

feed is distilled out. Estimate the composition of the distillate and residue.

Equilibrium data

Rayleigh equation
 60

50

40

1/(y-x) 30

20

10 xW xF
0.23 0.5

0
0 0.2 0.4 0.6 0.8 1 1.2

x
 60

50

40

1/(y-x) 30

20 1.2 area

10 xW xF
0.23 0.5

0
0 0.2 0.4 0.6 0.8 1 1.2

x
 Limitations of McCabe Thiele Method

1.3<α<5
Trays upto 25
R>1.1 Rmin

Lewis Sorel Method

G1, y1
1

L1, x1
G2,y2 2

L2,x2
Overall balance
 Tray n=1

For tray 1, liquid

source is reflux

y
For feed tray

Y1=0.9

0.8
For tray 1, n=1
y2  0.75( x1 )  0.225
y2  0.75(0.8)  0.225  0.818
 y2  0.818 Corresponding x2 from chart is 0.644 Tray 2

n=2
y3  0.75( x2 )  0.225
Tray 3
y3  0.75(0.644)  0.225  0.708

n=3
Equilibrium x3, corresponding to y3 is 0.49
y4  0.75( x3 )  0.225
Tray 4
y4  0.75(0.49)  0.225  0.59
Equilibrium x4 corresponding to y4 is 0.39 –equal to feed
composition.
Stripping section

yn 1  1.415 xn  0.042
n=4
y5  1.415 x4  0.042
Tray 5
y5  1.415(0.39)  0.042  0.49
Equilibrium x5 for y5 is 0.3

n=5 y6  1.415(0.29)  0.042  0.38

Tray 6
y7  1.415(0.21)  0.042  0.25
y8  1.415(0.11)  0.042  0.12
Corresponding equilibrium x6 is 0.21

n=6
y7  1.415(0.21)  0.042  0.25 Tray 7

Corresponding equilibrium x7 is 0.11

y8  1.415(0.11)  0.042  0.12

Corresponding equilibrium x8 is 0.05
Xw is attained, total number of trays = 8 including reboiler

8-1 = 7, Theoretical number of trays

 1

0.8

0.6

0.4

0.2

0
0 0.2 0.4 0.6 0.8 1
A binary mixture containing 40 mole% light component is to be separated

into 5 mole% bottom product and 90 mole% distillate product. The feed flow

rate is 100 kmol/h. Feed is 50% vapor and 50% liquid . R= 2.625. Relative

volatility is 2.5. Assume theoretical trays are partial reboiler, total condenser

and saturated liquid reflux. Find optimum feed tray location and theoretical

number of trays required. If overall plate efficiency is 60% find actual

number of plates.
 x y
0 0
0.1 0.217391
0.2 0.384615
x
y 0.3 0.517241
1  (  1) x 0.4 0.625
0.5 0.714286
0.6 0.789474
0.7 0.853659
0.8 0.909091
0.9 0.957447
1 1
Xf= 0.4

Xw= 0.05

XD = 0.9
Slope of feed line
For a mix of vapor and liquid feed
LF
q=  0.5 / 0.5  1
LF  F

Intercept of enriching section op. line

xD / R  1  0.9 / (2.625  1)  0.248

 1

0.8

0.6

0.4

0.248
0.2

0
0 0.2 0.4 0.6 0.8 1
 1

0.8

0.6

0.4

0.248
0.2

0
0 0.2 0.4 0.6 0.8 1
 1

0.8

0.6

0.4

0.248
0.2

0
0 0.2 0.4 0.6 0.8 1
 1

0.8

0.6

0.4

0.248
0.2

0
0 0.2 0.4 0.6 0.8 1
Total theoretical trays 10
Actual = 10/0.65= 17
Feed plate location=6
Unit-3: Distillation: Continuous distillation with rectification, Sieve plate column, Bubble cap

column: Packed column; calculation of number of plates, Lewis sorel Method, McCabe Thiele

Method, Economic reflux ratio, Minimum reflux Ratio, Fenske Underwood Equation. Plate

efficiency, Ponchon Savarit method, Packed Column; HETP – method, Transfer unit and Batch

distillation.
 Tray types

Gas-liquid contactors
 Trays

https://www.wermac.org/equipment/distillation_part2.html
BUBBLE CAP TRAY Although many bubble cap columns still are in operation, bubble cap

trays rarely are specified today because of high cost factors and the excellent performance

of the modern valve type tray. The bubble cap, however, does have a good turndown ratio

and is good for low liquid loads.

 Sieve plate

SIEVE TRAY The sieve tray is a low cost device which consists of a perforated plate
that usually has holes of 3/16 inch to one inch diameter, a downcomer, and an outlet
weir. Although inexpensive, a correctly designed sieve tray can be comparable to
other styles in vapor and liquid capacities, pressure drop, and efficiency. For
flexibility, however, it is inferior to valve and bubble cap trays and sometimes is
unacceptable for low liquid loads when weeping has to be minimized. Depending
upon process conditions and allowable pressure drop, the turndown ratio of a sieve
tray can vary from 1.5 to 3 and occasionally higher. Ratios of 5 as sometimes claimed
can be achieved only when the tray spacing is large, available pressure drop is very
high, liquid loadings are high, and the system is non-foaming. For many applications,
a turndown of 1.5 is quite acceptable. It also is possible to increase the flexibility of a
sieve tray for occasional low throughput operation by maintaining a high reboil and
increasing the reflux ratio. This may be economically desirable when the low
throughput occurs for a small fraction of the operating time.
http://www.separationprocesses.com/Operations/Fig022b.htm
https://www.wermac.org/equipment/distillation_part2.html
Throughputs and efficiencies can be as high as sieve or bubble cap trays

Very high flexibility can be achieved and turndown ratios of 4 to 1 are easily

obtained without having to resort to large pressure drops at the high end of the

operating range

Special valve designs with venturi shaped orifices are available for duties

involving low pressure drops

Although slightly more expensive than sieve trays, the valve tray is very

economical in view of its numerous advantages

 Packed column

Heat sensitive materials

Low pressure drop
 1

0.8

0.6xD/(Rmin+1)
0.53
0.4

0.2

0
0 0.2 0.4 0.6 0.8 1
 1

0.8

0.6
xD/(R+1)
0.4

0.2

0
0 0.2 0.4 0.6 0.8 1
 1

0.8

0.6
xD/(R+1)
0.4

0.2

0
0 0.2 0.4 0.6 0.8 1
 1

0.8

0.6

0.4

0.2

0
0 0.2 0.4 0.6 0.8 1
 1

0.8

0.6

0.4

0.2

0
0 0.2 0.4 0.6 0.8 1
y yi y 1/yi-y
33.3333
0.96 0.99 0.96 3
0.84 0.89 0.84 20
0.76 0.84 0.76 12.5
8.33333
0.6 0.72 0.6 3
7.14285
0.4 0.54 0.4 7
11.1111
0.2 0.29 0.2 1
0.04 0.08 0.04 25
30 Area = 16

25
20
15
10
5
0
0 0.2 0.4 0.6 0.8 1
(0.49x 7)+ (0.42 x 9) = 7.21 m
Absorption: Design of packed tower, Principles of absorption
and rate of absorption,
Flooding point, Loading point. Mass transfer – Liquid phase &
Gas phase, Overall MTC,
H.T.U. method- absorption with chemical reaction, Adsorption-
types and applications.
 Absorption

Liquid

Washing of by product coke oven gas with

Absorption water to remove ammonia
Gas Liquid H2S removal using alkaline solutions

Benzene and toluene removal using steam

Stripping

Gas
 Equilibrium solubility

Concentration of gas in non

volatile liquid solvent

Used in ore
leaching
Ideal Solutions
 Non ideal solutions

Henry’s law

If not valid, then there is chemical interaction or electrolytic dissociation

Valid upto 5 atm
Vapor upto 50% of saturation
 Solvent

Solubility Volatility

Higher value preferred Low vapor pressure

Easily recoverable to avoid loss
Decreased volume
Chemical similarity
No chemical reactions

H2S+amine
H2S+NaOH
 Removal of one component- Counter
flow

H2S with other gases

H2S is transferrable
Others- constant
Counterflow
With mole fractions
Minimum flow conditions
Cocurrent flow
• Pure substance
absorption
• Very tall tower
• Chemical reaction
Multistage Unit
Dilute solutions-Kremser’s equation
Continuous contact
Packed tower
For equilibrium relations in straight line form, overall mass transfer
coefficients can be applied

Mass transfer coefficients become constant

Operating line straight- Dilute
solutions
 m3

98%NH3

1200 kg/h
Y1=5%
m2
Air+NH3
 0.07
Y
0.06
Y1
0.05

0.04

0.03

0.02

0.01
0.045
Y2 X1min
0 X
X2 0 0.01 0.02 0.03 0.04 0.05 0.06
The given mass transfer coefficient is interms of pressure, we have mole fraction, so we
need MTC interms of mole fraction, use the relation given below
 Adsorption
Adsorbent

Adsorbate

Solid-Fluid
 Requirements

Large surface area 10^ 6 m2/g

Porous
Fullers earth, silica gel, activated carbon etc

Decreases with temperature-exothermic