Chapter 1 distillation

CHAPTER 1: DISTILLATION 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 1.1: Definition & process descriptiondescription

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 vodka2)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 1.1: Definition & process descriptiondescription

Other definitionDistillation 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 / TestingLaboratory / Testing

1.2: Physical Concept of 1.2: Physical Concept of distillationdistillation

Carried out by either 2 principal methodsFirst 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 1.3: Vapor – liquid equilibrium 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 a) VAPOR – LIQUID EQUILIBRIUM OF AN ORDINARY BINARY LIQUID AN ORDINARY BINARY LIQUID MIXTUREMIXTUREEquilibrium 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 VAPOR – LIQUID EQUILIBRIUM CURVECURVE

b) Prediction of vapor-liquid b) Prediction of vapor-liquid equilibrium compositions for equilibrium compositions for ordinary binary mixturesordinary 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 Prediction of vapor-liquid equilibrium compositions for equilibrium compositions for ordinary binary mixturesordinary binary mixturesFor 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 phaseyA = mole fraction of A in vapor phasePT = total pressure of the system

Phase Rule

Example: 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

SolutionSolution

1.4: Relative1.4: Relative volatility ( volatility (αα) ) of a of a mixturemixture

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 (Relative volatility (αα) of a ) of a mixturemixtureThe 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 (Relative volatility (αα) of a ) of a mixturemixture

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

xAAB xy

xy

y

y

B

A

Relative volatility (Relative volatility (αα) of a ) of a mixturemixtureSeparation is possible for > 1.0For 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.

Relative volatility (Relative volatility (αα) of a ) of a mixturemixture

Example:Example:

Using the data from table below, determine

the relative volatility for the benzene-

toluene system at 85°C and 105°C

Exercise 1Exercise 1A 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 temperatureii. Plot a boiling point diagram for n-hexaneiii. Plot an equilibrium diagram for the mixtureiv. 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.

SolutionSolutionVapor 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 2PART 2

Flash & batch distillation Flash & batch distillation

Flash (equilibrium) distillationSimple batch distillation

Flash (Equilibrium) DistillationFlash (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) DistillationFlash (Equilibrium) DistillationAs 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) DistillationFlash (Equilibrium) Distillation

ExampleExampleA 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 1374 0.825 0.92377 0.647 0.784380 0.504 0.669383 0.387 0.558386 0.288 0.449389 0.204 0.342392 0.132 0.236395 0.068 0.132398.2 0 0

Solution Solution Basis = 100 moles of liquid feed (F)Given, xF = 0.7V = 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* = 0.62equilibrium mol fraction of n-heptane in vapor, y* = 0.76the temperature of the separator at equilibrium ≈ 378oC

Determination of vapor-liquid equilibrium Determination of vapor-liquid equilibrium composition for a flash distillation of n-composition for a flash distillation of n-heptane/n-octane mixtureheptane/n-octane mixture

Figure: Equilibrium curve and operating line

Determination of equilibrium temperature Determination of equilibrium temperature for a flash distillation of n-heptane-n-octane for a flash distillation of n-heptane-n-octane mixturemixture

x

y

Exercise 11.2-1 (page Exercise 11.2-1 (page 752)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 distillationSimple batch distillationSimple 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.

Simple batch distillationSimple batch distillation


1. Raleigh equation for ideal and non-ideal mixtures

Consider a typical differential distillation at an instant time, t1 as shown below:

Simple batch distillationSimple batch distillationNow 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 + dBy

Simple batch distillationSimple batch distillationWe know from definitions,

Since is constant for an ideal mixture,

After simplifying,

Rearranging,

BA

Bx

BA

Ax

dBdA

dBy

dBdA

dAy

BA

BA

AB

BAA

dBdAdB

BAB

dBdAdA

xy

xy

BB

AAAB

/

/

dBA

dABAB

A

dA

B

dBAB

Simple batch distillationSimple batch distillationIntegrating 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

AB A

dA

B

dB

2

1

2

1

2

1

2

1

lnln AA

BBAB

A

A

B

B

AB ABA

dA

B

dB

AB

)5.(lnln1

2

1

2 EqA

A

B

BAB

Simple batch distillationSimple batch distillationExample 1A 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

SolutionSolution

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

molFrom 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

SolutionSolutionBy 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

835.040

60277.01005.022112211

av

avavav

y

yV

LxLxyVyLxLx

PART 3Continuous distillation

Continuous / retrification Continuous / retrification distillation distillation

Continuous / retrification Continuous / retrification distillation 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.

Continuous / retrification Continuous / retrification distillation distillation 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.

Continuous / retrification Continuous / retrification distillation distillation 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 MethodMc-Cabe Thiele Method

The intersection of operating The intersection of operating lines, lines, qqFeed 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.

The intersection of operating The intersection of operating lines, lines, qq

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

The intersection of operating The intersection of operating lines, lines, qqSubstituting: , 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

The intersection of operating The intersection of operating lines, lines, qq

•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 Animation of the construction of enriching, stripping & q of enriching, stripping & q operating linesoperating lines

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

ExerciseExercise

1) 11.4-12) 11.4-2

Steps of McCabe Thiele Steps of McCabe Thiele MethodMethod1. 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 line5. Plot stripping line6. Determine the stages7. Feed tray location

TutorialTutorial

11.4-5

Exercise 11.4-6Exercise 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 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 Example

Solution Solution Feed, xF = 0.4Product, xD = 0.9Bottom, xw = 0.10Taking 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 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

SolutionSolutionSince 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 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 Solution xF = 0.5xD = 0.94Xw = 0.02From 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

SolutionSolutionFind q-line. Feed enters at 75%

vapor.

lineforqandPlot

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

SolutionSolutionFrom 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

1 mmm

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

28.31

1

1

28.3)64.1(22

11

1

1

1

min

SolutionSolutionThe number of theoretical stages

required = 10.5 stages including boiler

Feed plate location: 5 from top.

Q1 Final Exam Jan 2012Q1 Final Exam Jan 2012A 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 compositionof 2.0 mole% ethanol are desired. The externalreflux 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.molEnthalpy 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 trayii)The total number of equilibrium trayiii)The feed location

Q2, Final Exam Jan 2013Q2, Final Exam Jan 2013A 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.

Determinea)The flowrates of distillate and

bottom product in kg/hrb)The minimum reflux ratioc)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 AZEOTROPIC DISTILLATION

Azeotrope mixturesMinimum boiling pointMaximum boiling pointAzeotropic Distillation

Azeotrope mixturesAzeotrope mixturesLiquid and vapor are exactly the same

at a certain temperatureIt 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

Azeotropic DistillationAzeotropic DistillationAn 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 DistillationAzeotropic DistillationAzeotropic 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.

Azeotropic DistillationAzeotropic DistillationA 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 DistillationAzeotropic DistillationThe 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 DistillationAzeotropic DistillationThe characteristic of such mixture is

boiling point curve goes through maximum phase diagram

Example: Acetone-chloroform

Azeotropic DistillationAzeotropic DistillationThe 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)

Azeotropic DistillationAzeotropic DistillationLet say binary mixture: A-B formed an

azeotrope mixtureEntrainer 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 DistillationAzeotropic DistillationExample: Acetic acid-water using entrainer n-butyl acetateBoiling point of acetic acid is 118.1 oC, water is 100 oC & n-butyl acetate is 125 oCThe 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 productThe distillate is condensed and collected in a decanter where it forms 2 insoluble layers

Azeotropic DistillationAzeotropic DistillationExample: Acetic acid-water using entrainer n-butyl acetateTop layer consist of nearly pure n-butyl acetate in water, whereas bottom layer of nearly pure water saturated with butyl acetateThe liquid from top layer is returned to column as reflux and entrainerThe liquid from bottom layer is sent to another column to recover the entrainer (by stream stripping)

Determination of Boiling Point Determination of Boiling Point Temperature in multi component Temperature in multi component distillationdistillationThe calculation is a trial and error

process where1.T is assumed2.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 11.7-1Example 11.7-1

Bubble point@boiling Bubble point@boiling pointpointBubble point@boiling point=temperature at which liquid

begins to vaporizeDew point=temperature at which liquid

begins to condense out of the vapor

Chapter 1 distillation

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Transcript of Chapter 1 distillation