Chapter 1 Dc

CHAPTER 1: DISTILLATION

Part 1

Definition & process description

Physical concept of distillation

Vapor-liquid equilibrium relationship

Relative volatility

Batch distillation –Part 2

Continuous distillation –Part 3

Azeotropic distillation -Part 4

Multicomponent distillation –Part 5

1.1: Definition & process description

Distillation is a process of separating various

components of a liquid solution by heating the liquid

to forms its vapors and then condensing the vapors to

form the liquid.

It is use to separate 2 or more substances present in

the liquid OR for purification purpose.

Distillation is a commonly used method for purifying

liquids and separating mixtures of liquids into their

individual components

All components presents in both phases

Familiar examples include

1) distillation of crude fermentation broths into alcoholic spirits such as gin and vodka

2) fractionation of crude oil into useful products such as gasoline and heating oil.

3) In the organic lab, distillation is used for purifying solvents and liquid reaction products.

1.1: Definition & process description

Other definition

Distillation is done by vaporizing a definite fraction

of a liquid mixture in a such way that the evolved

vapor is in equilibrium with the residual liquid

The equilibrium vapor is then separated from the

equilibrium residual liquid by condensing the vapor

Laboratory / Testing

1.2: Physical Concept of distillation

Carried out by either 2 principal methods

First method: based on the production of a vapor

by boiling the liquid mixture to be separated and

condensing the vapors without allowing any liquid

to return to the still - NO REFLUX (E.g. Flash,

simple distillation)

Second method: based on the return part of the

condensate to the still under such condition that

this returning liquid is brought into intimate

contact with the vapors on their way to the

condenser – conducted as continuous / batch

process (E.g. continuous distillation)

1.3: Vapor – liquid equilibrium

DEFINITION:

EVAPORATION: The phase transformation

processes from liquid to gas/vapor phase

VOLATILITY: The tendency of liquid to change

form to gas/vapor phase

a) VAPOR – LIQUID EQUILIBRIUM OF AN

ORDINARY BINARY LIQUID MIXTURE

b) PREDICTION OF VAPOR – LIQUID

EQUILIBRIUM COMPOSITIONS FOR

ORDINARY BINARY MIXTURES

a) VAPOR – LIQUID EQUILIBRIUM OF AN

ORDINARY BINARY LIQUID MIXTURE

Equilibrium curve: shows the relationship

between composition of residual liquid and

vapor that are in dynamic phase equilibrium.

The curve will be very useful in calculations to

predict the number of stages required for a

specified distillation process.

VAPOR – LIQUID EQUILIBRIUM CURVE

b) Prediction of vapor-liquid equilibrium

compositions for ordinary binary mixtures

Raoult’s Law for ideal solution & Dalton’s Law of partial pressure can be manipulated in order to calculate compostions of liquid and vapor, which are in equilibrium.

Raoult’s Law – the partial pressure of a component in the vapor phase is equal to the mole fraction of the component in the liquid multiplied by its pure vapor pressure at the temperature:

pA = xA · PAo

pA = partial pressure of A in a vapor phase

xA = mole fraction of A in liquid phase

PAo = vapor pressure of A at the temperature

Prediction of vapor-liquid equilibrium

compositions for ordinary binary mixtures

For a mixture of the different gases inside a close container, Dalton’s law stated that the resultant total pressure of the container is the summation of partial pressures of each of all gases that make up the gas mixture:

PT = pA + pB

Dalton also state that the partial pressure of gas (pA) is:

pA = yA · PT

pA = partial pressure of A in vapor phase

yA = mole fraction of A in vapor phase

PT = total pressure of the system

Phase Rule

Example: Calculate the vapor and liquid compositions

in equilibrium at 95oC (368.2K) for

benzene-toluene using the vapor pressure

from Table 11.1-1 at 101.32 kPa. Table 11.1-1

1.4: Relative volatility (α) of a mixture

Separations of components by distillation

process depends on the differences in

volatilities of components that make up the

solution to be distilled.

The greater difference in their volatility, the

better is separation by heating (distillation).

Conversely if their volatility differ only slightly,

the separation by heating becomes difficult.

Relative volatility (α) of a mixture The greater the distance between the equilibrium

line & 45o line, the greater the difference the vapor composition and a liquid composition. Separation is more easily made.

A numerical measure of ‘how easy’ separation – relative volatility, αAB

αAB – relative volatility of A with respect to B in the binary system

Relative volatility – ratio of the concentration of A in the

vapor to the concentration of A in liquid divided by the ratio of the concentration B in the vapor to the concentration of B in the liquid:

Relative volatility (α) of a mixture

αAB – relative volatility of A with respect to B in the

binary system

If the system obeys Raoult’s law for an ideal

system:

Separation is possible for > 1.0

A

AA

B

AAB

T

BBB

T

AAA

x

xy

P

P

P

xPy

P

xPy

)1(1

)1/()1(

/

/

/

AA

AA

xB

xA

ABxy

xy

y

y

B

A


Separation is possible for > 1.0

For non-ideal solution, the values of change with

temperature.

For ideal solution, the values of doesn’t change with

temperature.

For solution that approaches ideal solution, its

would fairly constant.

Example:

Using the data from table below, determine

the relative volatility for the benzene-

toluene system at 85°C and 105°C

Exercise 1

A liquid mixture is formed by mixing n-hexane (A) & n-

octane (B) in a closed container at constant pressure of 1

atm (101.3kPa).

i. Calculate the equilibrium vapor and liquid composition of

the mixture at each temperature

ii. Plot a boiling point diagram for n-hexane

iii. Plot an equilibrium diagram for the mixture

iv. Calculate the αAB at 100 °C

Vapor Pressure

Temperature n-Hexane n-Octane

(°C) kPa mm Hg kPa mm Hg

68.7 101.3 760 16.1 121

79.4 136.7 1025 23.1 173

93.3 197.3 1480 37.1 278

107.2 284.0 2130 57.9 434

125.7 456.0 3420 101.3 760

Use the following list of vapor pressure for pure n-heptane & n-octane

at various temperature.

Solution

Vapor Pressure

Temperature n-Hexane (A) n-Octane (B)

(°C) kPa XA YA kPa XB YB

68.7 101.3 1 1 16.1 0 0

79.4 136.7 0.6884 0.9290 23.1 0.3116 0.071

93.3 197.3 0.4007 0.7804 37.1 0.5993 0.2196

107.2 284.0 0.1920 0.5383 57.9 0.8080 0.4617

125.7 456.0 0 0 101.3 1 1

PART 2 - DISTILLATION COLUMN

Flash & batch distillation

Flash (equilibrium) distillation

Simple batch distillation

Flash (Equilibrium) Distillation Flash distillation – a single stage process because it has only

one vaporization stage (means one liquid phase is expected

to one vapor phase)

The vapor is allowed to come to equilibrium with the liquid

The equilibrium vapor is then separated from the equilibrium

residual liquid by condensing the vapor

Flash distillation can be either by batch or continuous

Flash (Equilibrium) Distillation

As illustrated in Figure 3, a liquid mixture feed, with initial mole fraction of A at XF, is pre-heated by a heater and its pressure is then reduced by an expansion valve.

Because of the large drop in pressure, part of liquid vaporizes.

The vapor is taken off overhead, while the liquid drains to the bottom of the drum

The system is called “flash” distillation because the vaporization is extremely rapid after the feed enters the drum.

Now, we interested to predict the composition (x and y) of these vapor and liquid that are in equilibrium with each other.

Flash (Equilibrium) Distillation

Example A liquid mixture containing 70 mol% n-heptane (A)

and 30 mol % n-octane (B) at 30oC is to be

continuously flash at the standard atmospheric

pressure vaporized 60 mol% of the feed.

Determine

1) the compositions of vapor and liquid for n-

heptane

2) temperature of the separator for an equilibrium

stage?

The equilibrium data for n-heptane – n-octane

mixture at 1 atm and 30°C is given as follows:

T (K) xA yA

371.6 1 1

374 0.825 0.92

377 0.647 0.784

380 0.504 0.669

383 0.387 0.558

386 0.288 0.449

389 0.204 0.342

392 0.132 0.236

395 0.068 0.132

398.2 0 0

Solution

Basis = 100 moles of liquid feed (F)

Given, xF = 0.7

V = 0.6(100) = 60 moles f = V/F = 60/100 = 0.6

We want to fine the equilibrium composition of liquid and liquid;

y* & x*

The operating line: y* = (0.6-1)x* + 0.7

0.6 0.6

= -0.667x* + 1.167

From the intersection of the operating line & the equilibrium

curve as shown in the graph:

equilibrium mol fraction of n-heptane in liquid, x* = 0.62

equilibrium mol fraction of n-heptane in vapor, y* = 0.76

the temperature of the separator at equilibrium ≈ 378oC

0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2

y, m

ol fr

acti

on

of

n-h

ep

tan

e in

vap

or

x, mol fraction of n-heptane in liquid

Determination of vapor-liquid equilibrium composition

for a flash distillation of n-heptane/n-octane mixture

Figure: Equilibrium curve and operating line

370

375

380

385

390

395

400

370

375

380

385

390

395

400

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Tem

pera

ture

(K

)

Mol fraction of n-heptane in vapor (y) and liquid (x)

Determination of equilibrium temperature for a flash

distillation of n-heptane-n-octane mixture

x

y

Exercise 11.2-1 (page 752) A mixture of 100 mol containing 60 mol%

n-pentane (A) and 40 mol% n-heptane (B) is

vaporized at 101.32 kPa abs pressure until 40

mol of vapor and 60 mol of liquid in

equilibrium with each other are produced.

This occur in a single stage system, and the

vapor and liquid are kept in contact with each

other until vaporization is complete.

Determine the composition of the vapor and

liquid.

The equilibrium data are as follows, where

x and y are mole fraction of n-pentane.

x (mol fraction of n-

pentane in liquid)

y (mol fraction of n-

pentane in vapor)

1.000 1.000

0.867 0.984

0.594 0.925

0.398 0.836

0.254 0.701

0.145 0.521

0.059 0.271

0 0

Answer:

xA = 0.430

yA = 0.854


Simple batch distillation which is also known as

differential distillation refer to a batch distillation in

which only one vaporization stage (or one exposed

liquid surface) is involved.

Simple batch distillation is done by boiling a liquid

mixture in a stream-jacketed-kettle (pot) and the vapor

generated is withdrawn and condensed (distillate) as

fast as it forms.

The first portion of vapor condensed will be richest in

the more volatile component A. As the vaporization

proceeds, the vaporized product becomes leaner in A.


1. Raleigh equation for ideal and non-ideal mixtures

Consider a typical differential distillation at an

instant time, t1 as shown below:

Simple batch distillation Now consider that the differential distillation at certain infinitesimal time

lapse,(dt), at t2 where t2=t1 + dt, after an infinitesimal amount of liquid has

vaporized as shown below:


Applicable for ideal and

non-ideal solution


The average composition of total material

distilled yav can be obtained by material

balance: (integrate from Rayleigh equation)

avyLLxLxL )( 212211

avVyxLxL 2211


2. Simplified Raleigh equation for ideal mixture

Consider a simple batch distillation process at

an initial time, t1, as shown below


L1=no of moles of binary mixture

containing A and B at t1

A1= no of moles comp A in L1 at t1

B1= no of moles comp B in L1 at t1

x1= mol fraction of A in L1 at t1

dL= infinitesimal amount of liq that

has vaporized

dA=infinitesimal amount of A that

has vaporized

dB = infinitesimal of B that has

vaporized

L1= A1 + B1

X1

dL= dA + dB

y

Simple batch distillation We know from definitions,

Since is constant for an ideal mixture,

After simplifying,

Rearranging,

BA

Bx

BA

Ax

dBdA

dBy

dBdA

dAy

BA

BA

AB

BA

A

dBdA

dBBA

B

dBdA

dA

xy

xy

BB

AAAB

/

/

dBA

dABAB

A

dA

B

dBAB

Simple batch distillation Integrating within the limits of t1 and t2,

Since is constant,

Equation 5 known as simplified Raleigh equation for simple batch distillation which applicable for ideal solution.

2

1

2

1

B

B

A

A

ABA

dA

B

dB

2

1

2

1

2

1

2

1

lnlnA

A

B

BAB

A

A

B

B

AB ABA

dA

B

dB

AB

)5.(lnln1

2

1

2 EqA

A

B

BAB

Simple batch distillation Example 1

A mixture of 100 mol containing 50 mol% n-pentane and 50

mol% n-heptane is distilled under differential (batch) conditions at

101.3 kPa until 40 mol is distilled. What is the average

composition of the total vapor distilled and the composition of the

n-pentane in the liquid left. The equilibrium data as follows, where

x and y are mole fractions of n-pentane: xA yA

1.000 1.000

0.867 0.984

0.594 0.925

0.398 0.836

0.254 0.701

0.145 0.521

0.059 0.271

0 0

Solution

xA yA 1/(yA –xA)

1.000 1.000 -

0.867 0.984 8.5470

0.594 0.925 3.0211

0.398 0.836 2.2831

0.254 0.701 2.2371

0.145 0.521 2.6596

0.059 0.271 4.7170

0 0 -

Solution

Given L1 = 100 mol V (mol distilled) = 40 mol

From material balance:

Substiting into Eq. (4)

The unknown, x2 is the composition n of the liquid L2

at the end of batch distillation.

molLVLLVLL 6040100 21221

1

2

5.0

22

1 1

60

100ln

1ln

x

x x

dxxy

dxxyL

L

dxxy

x

5.0

2

1510.0

Solution By plotting graph versus x, the value is

referring to the value of area under the curve.

From the graph, area under curve = 0.510 at x2 = 0.277. Composition of the liquid L2, x2 = 0.277.

From material balance on more volatile component:

Average composition of total vapor distilled, yav = 0.835

xy

1dx

xyx

5.0

2

1

835.0

40

60277.01005.022112211

av

avavav

y

yV

LxLxyVyLxLx

PART 3

Continuous distillation

Continuous / retrification distillation


Retrification (fractionation) - or stage distillation with

reflux can be considered as a process in which a series of

flash – vaporization stages are arranged in a series in such

a manner that the vapor and liquid products from each

stage flow counter-current to each other.

Continuous distillation - the process is more suitable for

mixtures of about the same volatility and the condensed

vapor and residual liquid are more pure (since it is re-

distilled)

The fractionator consists of many trays which have holes to

permit the vapor, V which rises up from the lower tray to

bubble through and mixes with the liquid, L on the upper

tray and equilibrated, and V and L stream leaves in

equilibrium.


During the mixing, the vapor will pick up more of

component A from the liquid while the liquid will richer and

richer in component B. As the vapor rises, it becomes richer

and richer in component A but poorer with component B.

Conversely, as the liquid falls further down, it becomes

poorer with A but richer in B. Thus we obtain a bottom

product and an overhead product of higher purity in

comparison to those obtained by single-stage simple batch

or flash distillation.

NOTE: Fractionation refers to a process where a part or

whole of distillate is being recycled to the fractionator. The

recycled distillation (reflux) will supply the bulk of liquid

need to mix with vapor.


The feed stream is introduced on some intermediate tray where the liquid has approximately the same composition as the feed.

The system is kept steady-state: quantities (feed input rate, output stream rates, heating and cooling rates, reflux ratio, and temperatures, pressures, and compositions at every point) related to the process do not change as time passes during operation.

With constant molal overflow assumption:

Conditions for constant molal overflow:

◦ Heat loses negligible (achieved more easily in industrial column)

◦ Negligible heat of mixing

◦ Equal or close heats of vaporization

.......... 1111 etcVVVetcLLL nnnnnn


Number of plates required in a distillation column

Four streams are involved in the transfer of heat and material across a plate, as shown in figure above:

Plate n receives liquid Ln+1 from plate n+1 above, and

vapor, Vn-1 from plate n-1 below.

Plate n supplies liquid Ln to plate n-1, and vapor Vn to

plate n+1

Action of the plate is to bring about mixing so that the vapor Vn of composition yn reaches equilibrium with the liquid Ln of composition xn.


Design and operation of a distillation column

depends on the feed and desired products

A continuous distillation is often a fractional

distillation and can be a vacuum distillation or a

steam distillation.

Calculation for number of plates:

Mc-Cabe & Thiele

Lewis-Sorel Method

Mc-Cabe Thiele Method

The intersection of operating lines, q

Feed enters as liquid at its boiling point that the two operating lines intersect at point having an x-coordinate of xF.

The locus point of the intersection of the operating lines is considerable importance since it is dependent on the temperature and physical condition of feed.

The condition of the feed (F) determines the relation between the vapor (Vm) in the stripping section and (Vn) in the enriching section, as well as between Lm and Ln.


q also as the no. of moles of saturated liquid produced on the feed plate by each mole of feed added to tower.

The relationship between flows above & below entrance of feed:

Rewrite the equations of enriching & stripping without the tray subscripts:

Subtracting (3) from (4)

feedofonvaporizatiofheatlatentmolar

conditionsenteringatfeedofmolvaporizetoneededheatq

1

Lv

Fv

HH

HHq

)2()1(

)1(

FqVV

qFLL

mn

nm

)4(

)3(

wmm

Dnn

WxxLyV

DxxLyV

)5()()()( wDnmnm WxDxxLLyVV


Substituting: , Eq. (1) & (2) into (5) will produce:

The equation – locus of the intersection of the two operating lines

Setting y = x in the equation, the intersection of the q-line equation with the 45o line is y = x = xF, where xF is the overall composition of the feed.

Slope = q/(q-1).

A convenient way to locate a stripping line operating line is 1st to plot the enriching operating line and then q-line.

wDF WxDxFx

)(11

equationlineqq

xx

q

qy F


•Depending on the state of

the feed, the feed lines

will have different slopes:

q = 0 (saturated vapour)

q = 1 (saturated liquid)

0 < q < 1 (mix of liquid and

vapour)

q > 1 (subcooled liquid)

q < 0 (superheated vapour)

Animation of the construction of

enriching, stripping & q operating lines

http://www.separationprocesses.com/Distillation/DT_Anima

tion/McCabeThiele.html

http://www.separationprocesses.com/Distillation/DT_Animation/McCabeThiele.html

http://www.separationprocesses.com/Distillation/DT_Animation/McCabeThiele.html

Exercise

1) 11.4-1

2) 11.4-2

Steps of McCabe Thiele Method

1. Plot equilibrium mole fraction for component

that more volatile. [ y(mole fraction A in

vapor) vs x(mole fraction A in liquid)]

2. Make 45° line (x=y)

3. Plot enriching line;

4. Plot q line

5. Plot stripping line

6. Determine the stages

7. Feed tray location

Tutorial

11.4-5

Exercise 11.4-6

Repeat Problem 11.4-1 but use a feed that

is saturated vapor at dew point. Determine

(a) Minimum reflux ratio, Rm

(b) Minimum number of theoretical plates at total

reflux

(c) Theoretical number of trays at an operating

reflux ratio of 1.5Rm

Example

A mixture of benzene and toluene containing 40

mole% benzene is to be separated to give a product

of 90 mole% benzene at the top, and a bottom

product with not more than 10 mole% of benzene.

The feed is heated so that it enters the column at its

boiling point, and the vapor leaving the column is

condensed but not cooled, and provides reflux and

product.

It is proposed to operate the unit with a reflux ratio

of 3 kmol/kmol product. It is required to find the

number of theoretical stages needed and the

position of entry for the feed.

Example

Solution

Feed, xF = 0.4 Product, xD = 0.9 Bottom, xw = 0.10 Taking basis; 100 kmol of feed. A total mass balance: F = D + W hence; W = 100 – D (Eq. 1)

A balance on MVC (benzene);

From the calculations; D = 37.5 kmol, W = 62.5 kmol

)2.(1.09.040

)1.0()9.0()4.0(100

EqWD

WDxWxDxF wDF

Solution

Using notation from reflux:

From material balance at the top stage;

Thus, the operating line equation:

5.112

)5.37(33

n

nnnn

L

LDLRDLD

LR

kmolVDLV nnn 15011

225.075.0

150

5.112

1

11

111

nn

Dn

nnDn

nn

nn

xy

xV

Dxyx

V

Dx

V

Ly

Solution

Since the feed is all liquid at its boiling point, it will

all run down as increased reflux to the plate

below:

The material balance at the bottom:

Bottom operating line equation:

nmmmm VkmolVVWVL 1505.625.2121

kmolLLFLL mmnm 5.2121005.112

0417.0417.1

)1.0(150

5.62

150

5.212

1

111

1

mm

mmwm

mm

mm

xy

xyxV

Wx

V

Ly

Example 11,200 kg/h of equal parts (in wt) of Benzene-Toluene

solution is to be distilled in a fractionating tower at atmospheric pressure.

The liquid is fed as a liquid-vapor mixture in which the feed consist of 75% vapor. The distillate contains 94 wt% Benzene whereas the bottom products contains 98 wt% toluene. Determine;

◦ The flowrate of distillate and bottom product (kg/h)

◦ The minimum reflux ratio, Rm.

◦ The number of theoretical stages required if the reflux ratio used is 1.5 times the minimum reflux ratio

◦ The position of the feed tray The MW of Benzene = 78

The MW of Toluene = 92

Solution

xF = 0.5

xD = 0.94

Xw = 0.02

From the total & component material balance:

D = 5739.1 kg/h, W = 5260.9 kg/h

Convert mass fraction to mol fraction. (Basis of

calculation = 100kg)

Mol fraction: xF = 0.54, xD = 0.95, xw = 0.03

Solution Find q-line. Feed enters at 75% vapor.

l ineforqandPlot

yyxLet

xyxy

q

xx

q

qy

equationlineq

qq

qqq

qq

qqqq

fqq

liquidvapor

)62.0,3.0()54.0,54.0(

62.072.0)3.0(333.0,3.0

72.0333.0125.0

54.0

125.0

25.0

11

25.0)0.1(25.0)0(75.0

)(25.0)(75.0

Solution

From the graph, y intercept for q-line = 0.36

The number of theoretical stages required if the

reflux ratio used is 2 times the minimum reflux ratio

64.136.01

95.036.0

1m

mm

D RRR

x

OLenrichingforandPlot

yyxAt

xy

xy

xR

xR

Ry

RRR

nn

nn

nn

Dnn

)605.0,5.0()95.0,95.0(

605.0222.0)5.0(766.0,5.0

222.0766.0

)95.0(128.3

1

128.3

28.3

1

1

1

28.3)64.1(22

11

1

1

1

min

Solution

The number of theoretical stages required = 10.5

stages including boiler

Feed plate location: 5 from top.

Q1 Final Exam Jan 2012 A distillation column with a total condenser

and partial reboiler is used to separate an

ethanol-water mixture. The feed containing

20 mole% ethanol enters the column at feed

rate 1000 kg.moles/hr. A distillate composition

of 80 mole% ethanol and bottom composition

of 2.0 mole% ethanol are desired. The external

reflux ratio is 5/3 and it is returned as a

saturated liquid. It is assumed the condition is

at constant molal overflow.

Given;

Enthalpy of feed at dew point, Hv =485 kJ/kg.mol

Enthalpy of feed at boiling point, HL =70 kJ/kg.mol

Enthalpy of feed at entrance condition, HF =15 kJ/kg.mol

The equilibrium curve of ethanol-water is provided in

Appendix 1.

Determine;

i) The minimum number of tray

ii) The total number of equilibrium tray

iii) The feed location

Q2, Final Exam Jan 2013

A total feed of 500 kmol/hr having an

overall composition of 55 mol% heptane

and 45mol% ethyl benzene is to be

fractionated at 1.0 bar to give a distillate

containing 95 mol% heptane and bottom

containing 2 mol% heptane. The feed enters

the tower at equimolar vapor and liquid.

The molecular weight of heptane = 100.2

kg/kmol and ethyl benzene =106.6 kg/kmol.

Equilibrium data for heptane-ethylbenzene is

given in Table 1.

Determine

a) The flowrates of distillate and bottom product in

kg/hr

b) The minimum reflux ratio

c) The number of theoretical stages if the reflux

ratio used is 1.3 times the minimum reflux ratio.

Table 1: Equilibrium data for heptane

ethylbenzene

Temperature (°C) Mole fraction

XH YH

98.3 1.00 1.00

102.8 0.79 0.90

110.6 0.49 0.73

119.4 0.25 0.51

129.4 0.08 0.23

136.1 0.00 0.00

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

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.

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


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)


The characteristic of such mixture is boiling point

curve goes through maximum phase diagram

Example: Acetone-chloroform


The most common examples:

Ethanol-water (89.4 mole%, 78.25 oC, 1 atm)

Carbon Disulfide-acetone (61 mol% CS2, 39.25oC, 1

atm)

Benzene-water (29.6 mol% water, 69.25 oC, 1 atm)


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


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


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)

Determination of Boiling Point Temperature in multi

component distillation

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)

Example 11.7-1

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

Chapter 1 Dc

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