AZEOTROPIC DISTILLATION

• Azeotrope mixtures
• Minimum boiling point
• Maximum boiling point
• Azeotropic Distillation
 AZEOTROPE MIXTURES
• Liquid and vapor are exactly the same at a certain temperature
• It is a special class of liquid mixture that boils at a constant
temperature at a certain composition
 Cannot be separated by a simple/conventional distillation
 An introduction of a new component called entrainer is added to
the original mixture to form an azeotrope with one or more of
feed component
 The azeotrope is then removed as either the distillate or bottoms
 The purpose of the introduction of entrainer is to break an
azeotrope from being formed by the original feed mixture
 AZEOTROPIC DISTILLATION
 Function of entrainer:
◦ To separate one component of a closely boiling point
◦ To separate one component of an azeotrope
 Azeotropic distillation is a widely practiced process for the dehydration of a
wide range of materials including acetic acid, chloroform, ethanol, and many
higher alcohols.
 The technique involves separating close boiling components by adding a third
component, called an entrainer, to form a minimum boiling.
 Normally ternary azeotrope which carries the water overhead and leaves dry
product in the bottom.
 AZEOTROPIC DISTILLATION
• The overhead is condensed to two liquid phases; the organic, "entrainer rich"
phase being refluxed while the aqueous phase is decanted.
• A common example of distillation with an azeotrope is the distillation of ethanol
and water.
• Using normal distillation techniques, ethanol can only be purified to
approximately 89.4%
• Further conventional distillation is ineffective.
• Other separation methods may be used are azeotropic distillation or solvent
extraction
 AZEOTROPIC DISTILLATION
• The concentration in the vapor phase is the same as the
concentration in the liquid phase (y=x)
• At this point, the mixture boils at constant temperature and
doesn’t change in composition
• This is called as minimum boiling point (positive deviation)
 AZEOTROPIC DISTILLATION
• The characteristic of such mixture is boiling point
curve goes through maximum phase diagram
• The most common examples:
– Ethanol-water (89.4 mole%, 78.25 oC, 1 atm)
– Carbon Disulfide-acetone (61 mol% CS2,
39.25oC, 1 atm)
Example : Acetone-chloroform
– Benzene-water (29.6 mol% water, 69.25 oC, 1
atm)
 AZEOTROPIC DISTILLATION
• Let say binary mixture: A-B formed an azeotrope mixture
• Entrainer C is added to form a new azeotrope with the original
components, often in the LVC, say A
• The new azeotrope (A-C) is separated from the other original component
B
• This new azeotrope is then separated into entrainer C and original
component A.
• Hence the separation of A and B can be achieved
 AZEOTROPIC DISTILLATION
Example: Acetic acid-water using entrainer n-butyl acetate
 Boiling point of acetic acid is 118.1 oC, water is 100 oC & n-butyl
acetate is 125 oC
 The addition of the entrainer results in the formation of a minimum
boiling point azeotrope with water with a boiling point = 90.2 oC.
 The azeotropic mixture therefore be distilled over as a vapor
product & acetic acid as a bottom product
 The distillate is condensed and collected in a decanter where it
forms 2 insoluble layers
 AZEOTROPIC DISTILLATION
Example: Acetic acid-water using entrainer n-butyl acetate
 Top layer consist of nearly pure n-butyl acetate in water, whereas
bottom layer of nearly pure water saturated with butyl acetate
 The liquid from top layer is returned to column as reflux and
entrainer
 The liquid from bottom layer is sent to another column to
recover the entrainer (by stream stripping)
MULTICOMPONENT
DISTILL ATION
INTRODUCTION TO MULTICOMPONENT
DISTILLATION (MCD)
• INTRODUCTION OF MULTI – COMP. DISTILLATION
• LIGHT AND HEAVY KEY COMPONENT
• BOILING POINT, DEW POINT AND FLASH DISTILLATION
• EQUILIBRIUM DATA IN MULTI. DISTILLATION
FRACTIONATION OF MULTICOMPONENT MIXTURES
• Three or more components in the products
• The designer generally choose two components whose
concentrations or fractional recoveries in the distillate and bottom
products are a good index of the separation achieved
• These two components are called key components
• They must differ in volatility
• Light key (L) is more volatile (vapor)
• The heavy key (H) is less volatile (liquid)
• Normally the key components are adjacent in the rank order of
volatility. Such choice is called a sharp separation
• In sharp separations the keys are only components that appear in
both products in appreciable concentrations
 LIGHT AND HEAVY KEY COMPONENT

The essential requirement in Multi – comp. Distillation is to separate the two components.

These components are called key component.

Suppose a four component mixture A-B-C-D where:

A is MVC
where the mixture is to be
B is LVC
separated as below:
Feed Top Product Bottom Product
A A
B B B
C C C
D D

Then; B – is the lightest comp. in distillate and called Light Key Component (LKC)

C – is the heaviest comp. in top / bottom and called Heavy Key Component (HKC)

The main purpose of fractionating is to separate B from C

INTRODUCTION TO MULTICOMPONENT
DISTILLATION (MCD)
• For binary mixture at constant P – a vapor liquid equilibrium curve rather
unique.

• With a ternary mixture, the conditions of equilibrium curve are more

complex for constant P where mol fraction of the two components in liq.
phase must be given before composition of vapor can be fixed/determine.

• Therefore, the mol fraction yA not only depends on xA but also on the
proportions of the other two components.
INTRODUCTION TO MULTICOMPONENT
DISTILLATION (MCD)
• Graphical method are not generally suitable.

• The possible method used – plate to plate analysis.

• For many systems consisting particularly similar chemically substances, the

relative volatilities of the components remain constant over a wide range of
temperature and composition.
MULTICOMPONENT DISTILLATION
This is illustrated by the table below:
Temperature (K)
Component 353 393 453
Phenol 1.25 1.25 1.25
o – Cresol 1 1 1
M – Cresol 0.57 0.62 0.70
Xylenols 0.30 0.38 0.42

An alternative method is to use the simple relation.

𝑦 =K𝑥 where K = equilibrium constant

(or Equilibrium Vaporization ratio)
Particularly useful for hydrocarbon processing in petroleum industry

The K values can be measured .

MULTICOMPONENT DISTILLATION
Consider as ideal mixture, Raoult’s Law can be applied to determine the composition of
vapor in equilibrium with liquid.

Consider a four – component (A,B,C,D) mixture:

p A  x A P Ao p B  x B P Bo p C  x C PCo p D  x D P Do
pA P Ao P Bo PCo P Do
yA   xA yB  xB yC  xC yD  xD
P P P P P

In hydrocarbon system, because non-idealities, the equilibrium data are often represented
by: yA  K xA yB  K xB yC  K C xC yD  K xD
A B D

KA is the vapor – liquid equilibrium constant or distribution coefficient for component A.

Pi o for component i individual component.

Ki 
P
 MULTICOMPONENT DISTILLATION
K value is easily determined (refer K-chart) for hydrocarbons system but need to be
calculated for other system.

The relative volatility, i for each individual component in a multi component mixture can
be defined in a similar manner to that of a binary mixture.

Need to select a base component, say component C:

Ki KA KB KC KD
αi  , hence αA  αB  αC   1.0 αD 
KC KC KC KC KC

The values of Ki depend strongly on T.

MULTICOMPONENT DISTILLATION
Boiling Point.
At a specified pressure, the boiling point or bubble point of a given multi –
component mixture satisfy the relation  yi = 1.0. For a mixture A, B, C, and D, with
C as the base component,
 y i   K i x i  K C  α i x i  1.0

The calculation is a trial – and – error process where:

T is assumed.
Values of i are then calculated using K values at the assume T.
Then calculate value of KC where
1.0
KC 
 α i xi

Find the T that corresponds to the calculated value of KC

Compare with T value read from table that corresponds to the KC (calculated)
MULTICOMPONENT DISTILLATION
Boiling Point.
If values is differ, the calculated T is used for next iteration.
After the final T is known, the vapor composition is calculated from:
α i xi
yi 
 αi xi
Dew Point.
Also based on the trial and error method:
yi  1   yi 
 xi  
Ki
 
 KC


 
 α i
  1.0


The values of KC is calculated from:

1.0
KC 
 α i xi

After the final T is known, the liquid composition is calculated from:

yi αi
xi 
 yi αi
BUBBLE POINT@BOILING POINT
Bubble point@boiling point
=temperature at which liquid begins to vaporize

Dew point
=temperature at which liquid begins to condense out of the
vapor
 Discuss
example
11.7-1
BOILING POINT OF MC LIQUID
Example 11.7.1
A liquid feed to a distillation tower at 405.3 kPa abs is fed to a distillation tower. The
composition in mol fractions is as follows:
n – butane (xA = 0.40) n –pentane (xB = 0.25)
n – hexane (xC = 0.20) n – heptane (xD = 0.15)
Calculate the boiling point and the vapor in equilibrium with liquid.

Solution:
Assume first trial and error temperature, let say T=65OC. Find the values of K from Figure
11.7.2

Using component C (n – hexane) as base component, the following values are calculated for
the first trial:
Figure 11.7.2: Equilibrium K values for light hydrocarbon system at 405.3 kPa absolute.
Assume first trial and error temperature, let say T=65OC.

Set n-hexane as base component. Initial Kc = Ki (hexane) = 0.245

Ki
αi 
KC
Ki 1.68
For n - butane, α    6.857
KC 0.245
Do same calculation for n-pentane, n-hexane and n-heptane

For the first trial (T=65OC),

Component xi Ki (from chart) i=Ki/KC ixi
n-butane (xA) 0.40 1.68 6.857 2.743
n-pentane (xB) 0.25 0.63 2.571 0.643
n-hexane (xC) 0.20 0.245 1.000 0.200
n-heptane (xD) 0.15 0.093 0.380 0.057
1.00 ixi = 3.643
BOILING POINT, DEW POINT AND FLASH DISTILLATION

1 1
From table above, K C    0.2745
  i xi 3.643

From graph, for n-hexane at K = 0.275, T = 69OC

Using T = 69OC for trial 2, a temperature of 70OC is obtained.

Using T = 70OC for trial 3, the calculation shown final value of 70OC, which is bubble point.

Values y is calculated from: yi=ixi/ixi

From trial 3,

Component xi Ki (from chart) i=Ki/KC ixi yi

n-butane (xA) 0.40 1.86 6.607 2.643 0.748
n-pentane (xB) 0.25 0.710 2.522 0.631 0.178
n-hexane (xC) 0.20 0.2815 1.000 0.200 0.057
n-heptane (xD) 0.15 0.110 0.391 0.059 0.017
1.00 ixi = 3.643 1.000
DETERMINATION OF BOILING POINT TEMPERATURE
IN MULTI COMPONENT DISTILL ATION
The calculation is a trial and error process where :
1. T is assumed
2. Value of relative volatility of each component are then calculated using K
values at the assume T.
3. Then calculate value of Kc where
Kc= 1/(∑relative volatility x liq mole fraction)
4. Find the T that corresponds to the calculated value of Kc
5. Compare with T value read from table that corresponds to the Kc.
6. If value is differ, the calculated T is used for the next iteration.
7. After the final T is known, the vapor composition is calculated from
Yi= (relative volatility x liq mole fraction)/∑(relative volatility x liq mole
fraction)