Chapter 1 distillation

CHAPTER 1: DISTILLATION  Part

1

 Definition  Physical

& process description

concept of distillation

 Vapor-liquid

equilibrium

relationship  Relative  Batch

volatility

distillation –Part 2

 Continuous  Azeotropic

distillation –Part 3

distillation -Part 4

 Multicomponent

5

distillation –Part

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)

2)

3)

distillation of crude fermentation broths into alcoholic spirits such as gin and vodka fractionation of crude oil into useful products such as gasoline and heating oil. In the organic lab, distillation is used for purifying solvents and liquid

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: p A = y A · 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 benzenetoluene using the vapor pressure from Table 11.1-1 at 101.32 kPa. Table 11.1-1

Solution

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  AB 

y A / xA y B / xB

y A / xA  (1  y A ) /(1  x A )

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

If the system obeys Raoult‟s law for an ideal system: PA x A yA  PT



PB xB yB  PT

 AB

PA  PB

x A yA  1  (  1) x A

Separation is possible for > 1.0

Relative volatility (α) of a mixture 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. 

Relative volatility (α) of a mixture

Example: Using the data from table below, determine the relative volatility for the benzenetoluene 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

Use the following list of vapor pressure for pure n-heptane & n-octane at various temperature.

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

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

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

Flash (Equilibrium) Distillation

Example A liquid mixture containing 70 mol% nheptane (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 nheptane 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) 371.6 374 377 380 383 386 389 392 395 398.2

xA 1 0.825 0.647 0.504 0.387 0.288 0.204 0.132 0.068 0

yA 1 0.92 0.784 0.669 0.558 0.449 0.342 0.236 0.132 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

Determination of vapor-liquid equilibrium composition for a flash distillation of n-heptane/noctane mixture y, mol fraction of n-heptane in vapor

1.2

1

0.8

0.6

0.4

0.2

0 0

0.2

0.4

0.6

0.8

1

x, mol fraction of n-heptane in liquid

Figure: Equilibrium curve and operating line

1.2

Temperature (K)

Determination of equilibrium temperature for a flash distillation of n-heptane-n-octane mixture 400

400

395

395

390

390

385

385

380

380

375

375

370

370 0

x

0.2

0.4

0.6

0.8

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

1

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

The equilibrium data are as follows, where xx and y areofmole fraction of n-pentane. (mol fraction ny (mol fraction of npentane in liquid)

pentane in vapor)

1.000 0.867 0.594 0.398 0.254 0.145 0.059 0

1.000 0.984 0.925 0.836 0.701 0.521 0.271 0

Answer: xA = 0.430 yA = 0.854

Simple batch distillation 

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.

Simple batch distillation

Simple 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 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:

Simple batch distillation

Applicable for ideal and non-ideal solution

Simple batch distillation The average composition of total material distilled yav can be obtained by material balance: (integrate from Rayleigh Lequation) 1 x1  L2 x2  ( L1  L2 ) y av

L1 x1  L2 x2  Vy av

Simple batch distillation 2. Simplified Raleigh equation for ideal mixture Consider a simple batch distillation process at an initial time, t1, as shown below

Simple batch distillation

dL= dA + dB y

L1= A1 + B1 X1

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

Simple batch distillation 

We know from definitions,  dA  yA    dA  dB    A  xA    A  B  



Since AB is constant for an ideal mixture,  AB 



 dB  yB    dA  dB    B  xB    A  B  

y A / xA y B / xB

After simplifying,  AB 



dA B   dA  dB A  B dB A  dA  dB A  B

Rearranging,

B  dA A  dB

dB dA   AB  B A

Simple batch distillation 

Integrating within the limits of t1 and t2, B2





B1

dB   AB  B

A2



A1

dA A

Since AB is constant, B2

 AB



B1

dB  B

A2



A1

dA A

 B2   A2     AB ln   ln   B1   A1 



 AB ln B BB2  ln A AA2 1

1

 Eq .(5)

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

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 xA yA fractions of n-pentane: 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: 

L1  L2  V



L2  L1  V  100  40

L2  60 mol

Substiting into Eq. (4) x1

 L1   1    dx ln     L2  x 2 y  x 



 100  ln   60 

0.5

 1   dx y x x2 



0.5

0.510 



 1   dx y x x2 



The unknown, x2 is the composition n of the liquid L2 at the end of batch distillation.

Solution 





 1  graph y  x 

0.5

 1   dx y x x2 

 By plotting 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:

x1L1  x 2L2  y av V

y av 

x1L1  x 2L2 V

y av 

0.5  100  0.277  60 40

y av  0.835



Average composition of total vapor distilled, yav = 0.835

PART 3 Continuous distillation

Continuous / retrification 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

Continuous / retrification distillation 

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 distillation

Continuous / retrification 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: Ln 1  Ln  Ln 1  ....etc. Vn 1  Vn  Vn 1  ....etc. 

Conditions for constant molal overflow:

◦ Heat loses negligible (achieved more easily in industrial column) ◦ Negligible heat of mixing ◦ Equal or close heats of vaporization

Continuous / retrification distillation

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 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

Continuous / retrification distillation

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. 

The intersection of operating lines, q heat needed to vaporize 1 mol of feed at entering conditions q

molar latent heat of vaporizati on of feed q





Hv  H F Hv  H L

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 Lm  Ln  qF (1) entrance of feed: Vn  Vm  (1  q )F (2)



Rewrite the equations Vn y of Ln enriching x  Dx D (3& ) stripping without the tray subscripts: Vm y  Lm x  Wxw ( 4)



Subtracting from (4) (Vm  V(3) n )y  (Lm  Ln )x  (DxD  Wxw ) (5)

The intersection of operating lines, q 

Substituting:FxF  DxD  Wxw will produce: xF q y x q 1 q 1





 

, Eq. (1) & (2) into (5)

(q  line equation )

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.

The intersection of operating lines, q •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_ Animation/McCabeThiele.html

Exercise 1) 11.4-1 2) 11.4-2

Steps of McCabe Thiele Method 1.

2. 3. 4.

5. 6. 7.

Plot equilibrium mole fraction for component that more volatile. [ y(mole fraction A in vapor) vs x(mole fraction A in liquid)] Make 45° line (x=y) Plot enriching line; Plot q line Plot stripping line Determine the stages 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



A balance on MVC (benzene); F  x F  D  x D  W  xw 40  0.9D  0.1W



hence; W = 100 – D (Eq. 1) 100(0.4)  D(0.9)  W (0.1) (Eq.2)

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

Solution 

Using notation from reflux: Ln Ln  RD D Ln  112 .5 R



Ln  3(37.5)

From material balance at the top stage; Vn 1  Ln  D



Ln  3D

Vn 1  150kmol

Thus, the operating line equation: y n 1 

Ln D xn  xD Vn 1 Vn 1

y n 1  0.75 x n  0.225

y n 1 

112 .5 D xn  xD 150 Vn 1

Solution 

Since the feed is all liquid at its boiling point, it will all run down as increased reflux to the plate below: Lm  Ln  F



Lm  212 .5kmol

The material balance at the bottom:

Lm  Vm1  W



Lm  112.5  100

Vm  212 .5  62.5

Vm  150 kmol  Vn

Bottom operating line equation: y m 1 

Lm W xm  xw Vm 1 Vm 1

y m 1  1.417 x m  0.0417

y m 1 

212 .5 62.5 xm  (0.1) 150 150

Example 



11,200 kg/h of equal parts (in wt) of BenzeneToluene 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. q  0.75(qvapor )  0.25(q liquid ) q  0.75(0)  0.25(1.0)

q  0.25

q  line equation  q   x   x q   f  y q    q  1  q  1  0.25   0.54  yq    xq     0.25  1   0.25  1  Let x  0.3, y q  0.333 (0.3)  0.72

y q  0.333 x q  0.72 y q  0.62

Plot (0.54, 0.54 ) and (0.3, 0.62 ) forq  line

Solution 

From the graph, y intercept for q-line = x 0.95 0.36  0.36  0.36 R 1.64 D

Rm  1



Rm  1

m

The number of theoretical stages required if the reflux ratio used is 2 times R  2Rmin  2(reflux 1.64 ) Rratio  3.28 the minimum R 1 xn  xD R 1 R 1 3.28 1  xn  (0.95 ) 3.28  1 3.28  1  0.766 x n  0.222

y n 1  y n 1 y n 1

At x  0.5,

y n 1  0.766 (0.5)  0.222

y n 1  0.605

Plot (0.95, 0.95 ) and (0.5, 0.605 ) for enriching OL

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 compositio of 80 mole% ethanol and bottom composition of 2.0 mole% ethanol are desired. The extern reflux ratio is 5/3 and it is returned as a saturated liquid. It is assumed the condition i 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 =

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  

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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

Azeotropic Distillation 

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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 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. 

Azeotropic Distillation 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  Example: Acetone-chloroform 

Azeotropic Distillation 

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)

Azeotropic Distillation 

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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  

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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 

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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