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BITS Pilani Pilani Campus Separation Processes - I (CHE F244) Dr. Suresh Gupta Associate Professor & Head, Deptt. of Chemical Engg.
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Page 1: First part of distillation slides

BITS PilaniPilani Campus

Separation Processes - I (CHE F244)

Dr. Suresh GuptaAssociate Professor & Head, Deptt. of Chemical Engg.

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BITS PilaniPilani Campus

Distillation of Binary Mixtures

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• A feed mixture of two or more components is separated

into two or more products

• often limited to, an overhead distillate and

• a bottoms product

• Most often, the feed is a liquid or a vapor–liquid mixture

• The bottoms product is almost always a liquid

• The distillate may be a liquid, a vapor, or both

• The separation requires that

• a second phase be formed so that both liquid and vapor are

present and can make contact while flowing counter currently to

each other in a trayed or packed column

Distillation (Fractionation)

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• components have different volatilities so that they partition

between phases to different extents, and

• the two phases are separable by gravity or mechanical means

• Distillation differs from absorption and stripping

• the second fluid phase is usually created by thermal means

(vaporization and condensation)

• rather than by the introduction of a second phase that may contain

an additional component or components not present in the feed

mixture

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Distillation of a binary mixture of benzene and toluene

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• Feed flow rate, composition, temperature, pressure, and phase

condition

• Desired degree of component separation

• Operating pressure (which must be below the critical pressure of

the mixture)

• Pressure drop, particularly for vacuum operation

• Minimum reflux ratio and actual reflux ratio

• Minimum number of equilibrium stages and actual number of

equilibrium stages (stage efficiency)

• Type of condenser (total, partial, or mixed)

• Degrees of liquid reflux subcooling

• Type of reboiler (partial or total)

Design and Analysis Factors

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• Type of trays or packing

• Column height

• Feed-entry stage

• Column diameter

• Column internals, and materials of construction

• Heat lability and chemical reactivity of feed components

Design and Analysis Factors

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• Temperature and phase of the feed are determined at

the feed-tray pressure by an adiabatic-flash calculation

across the feed valve

• As the feed vapor fraction increases

• the required reflux ratio (L/D) increases

• but the boilup ratio (V/B) decreases

• The column operating pressure in the reflux drum should correspond to a distillate temperature

• Some what greater than the supply temperature of the cooling

water to the overhead condenser

Some Initial Considerations

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• However, if this pressure approaches the critical

pressure of the more volatile component

• then a lower pressure must be used and a refrigerant is required

as coolant

• If the estimated pressure is less than atmospheric,

• the operating pressure at the top is often set just above

atmospheric to avoid vacuum operation

• unless the temperature at the bottom of the column is limited by

decomposition, polymerization, excessive corrosion, or other

chemical reactions

• In that case, vacuum operation is necessary

Some Initial Considerations

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• For given (1) feed, (2) desired degree of separation, and

(3) operating pressure

• A minimum reflux ratio exists

• that corresponds to an infinite number of theoretical stages

• A minimum number of theoretical stages exists

• that corresponds to an infinite reflux ratio

• The design trade-off is between the number of stages and the

reflux ratio

Some Initial Considerations

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• Successful applications of distillation methods

• Depends greatly upon an understanding of the equilibria existing

between vapor and liquid phases of the mixture

Vapor-Liquid Equilibrium Relations

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• An ideal law, can be defined for vapor-liquid phases in

equilibrium (only ideal solution e.g. benzene-toluene,

hexane-heptane etc.)

• Composition in liquid:

• Composition in vapor:

Phase Rule and Raoult’s Law

AAA xPp =

BA xx +=1

BA yy +=1

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• Boiling-point diagram for system benzene (A)-toluene (B)

at a total pressure of 101.32 kPa.

Constant Pressure Equilibria

Dew point is the temperature at which the

saturated vapour starts to condense.

Bubble-point is the temperature at which

the liquid starts to boil.

The difference between liquid and vapour

compositions is the basis for distillation operations.

If we start with a cold liquid composition is

xA1 = 0.318 (xB1 = 0.682) and heat the

mixture, it will start to boil at 98ºC.

The first vapor composition in equilibrium

is yA1 = 0.532 (yB1 = 0.468).

Continue boiling, the composition xA will

move to the left since yA is richer in A.

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• The boiling point diagram can be calculated from

• (1) the pure vapor-pressure data in the table below and

• (2) the following equations:

Constant Pressure Equilibria

Ppp BA =+

PxPxP ABAA =−+ )1(

P

xP

P

py AAA

A ==

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Constant Pressure Equilibria

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• Calculate the vapor and liquid compositions in

equilibrium at 95ºC (368.2K) for benzene-toluene using

the vapor pressure from the table at 101.32 kPa.

• Solution: At 95ºC from Table for benzene, PA = 155.7 kPa and PB = 63.3 kPa. Substituting into Eq.(5) and

solving,

155.7(xA) + 63.3(1-xA) = 101.32 kPa (760 mmHg)

Hence, xA= 0.411 and xB= 1 – xA = 1 - 0.411 = 0.589.

Substituting into eqn.(6),

Constant Pressure Equilibria

PxPxP ABAA =−+ )1(

632.032.101

)411.0(7.155====

P

xP

P

py AAA

A

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• A common method of plotting the equilibrium data is

• yA is plotted versus xA for the benzene-toluene system

• The 45º line is given to show that yA is richer in component A

than is xA.

Constant Pressure Equilibria

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• An azeotrope is a mixture of two or more liquids in such

a ratio that its composition cannot be changed by simple

distillation.

• The maximum temperature Tmax corresponds to a

concentration xAZ and xAZ = yAZ

• The plot of yA versus xA would show the curve crossing the 45o

line at this point

• Acetone-chloroform is an example

• A minimum boiling azeotrope with yAZ = xAZ at Tmin

• Ethanol-water is such a system

Nonideal System

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Nonideal System

Maximum-boiling azeotropeMinimum-boiling azeotrope

A mixture whose total pressure is

greater than that computed from

ideality

A mixture whose total pressure is

less than that computed from

ideality

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Nonideal System

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• It is a measure of the differences in volatility between 2

components

• hence their boiling points

• It indicates how easy or difficult a particular separation will be

• Where αAB is the relative volatility of A with respect to B in the

binary system.

• Raoult’s Law:

• when αAB is above 1.0, a separation is possible.

Relative Volatility of Vapor-Liquid Systems

)1)(1(

/

/

/

AA

AA

BB

AAAB

xy

xy

xy

xy

−−==α

AAB

AAB

Ax

xy

)1(1 −+=

α

α

P

xPy AA

A =

B

AAB

P

P=α

P

xPy BB

B =

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• A single equilibrium stage is

• the two different phases are brought into intimate contact with

each other

• The mixing time is long enough and the components are

essentially at equilibrium in the two phases after separation

• Total mass balance:

• Component balance:

Single-Stage Equilibrium Contact for Vapor-Liquid System

V1 V2

L0 L1

Where

V1, V2 is a vapor

L0, L1is a liquid

MVLVL =+=+1120

AMAAAA MxyVxLyVxL =+=+ 11112200

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• Distillation has two main methods in practice:

• Production of vapor by boiling the liquid mixture to be separated

in a single stage and recovering and condensing the vapors

• No liquid is allowed to return to the single-stage still to contact

the rising vapors

• Returning of a portion of the condensate to the still

• The vapors rise through a series of stages or trays, and part of the

condensate flows downward through the series of stages or trays

countercurrently to the vapors (“fractional distillation, distillation with

reflux, or rectification”)

Simple Distillation Methods

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• There are 3 important types of distillation that occur in a

single stage or still

• Equilibrium or Flash Distillation

• Simple batch or differential distillation

• Simple steam distillation

Simple Distillation Methods

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• Single stage separation technique

• A liquid mixture is pumped through a heater to raise the

temperature and enthalpy of the mixture

• It then flows through a valve and the pressure is reduced,

causing the liquid to partially vaporize

• Once the mixture enters a big enough volume (the “flash drum”),

the liquid and vapor separate

• Because the vapor and liquid are in such close contact up until

the “flash” occurs, the product liquid and vapor phases approach

equilibrium

Equilibrium or Flash Distillation

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• Total mass balance:

• Component balance:

• Material balance for more volatile component :

• Where, f = V/F = molal fraction of the feed that is vaporized and

withdrawn continuously as vapor

• 1-f = one as liquid

heater

SeparatorLVF +=

AAF LxyVFx +=

AAF xfyfx )1( −+=AAF xF

V

F

Fy

F

Vx )()( −+=

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• A mixture of 50% mole normal heptane and 50% normal

octane at 30ºC is continuously flash distilled at 1

standard atmosphere so that 60 mol% of the feed is

vaporized. What will be the composition of the vapor and

liquid products?

Problem

xA 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

yA 0.247 0.453 0.607 0.717 0.796 0.853 0.898 0.935 0.968

Solution: Given: xF = 0.5, f = 0.6; Find: xA, yA

Basis: F = 100 mols

Applying the mass balance yields:

Since

LVF +=

60)100(6.0 === fFV

4060100 =−=−= VFL

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Material balance for more volatile component,

Subtituting value of f =0.6 and xF =0.5 we get,

Assuming that xA = 0.5 and solving yA = 0.5.

Next, assuming that xA=0 and solving, yA = 0.83.

(These point are plotted on the graph.)

At the intersection of this line with the equilibrium curve,

yA = 0.58 and xA = 0.39.

AAF xfyfx )1( −+=

AAF xF

V

F

Fy

F

Vx )()( −+=

AA xy )6.01(6.05.0 −+=

AA xy 4.06.05.0 +=

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yA

xA

xF =0.5

yF = 0.5

x=0

y-intercept= 0.834

xA =0.39

yA = 0.581st

2nd

3th

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Simple Batch or Differential Distillation

• The pot is filled with liquid

mixture and heated.

• Vapour flows upwards though

the column and condenses at

the top.

• Part of the iquid is returned to

the column as reflux, and the

remainder withdrawn as

distillate.

• Nothing is added or withdrawn

from the still until the run is

completed.

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Simple Batch or Differential Distillation

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• The total moles of component A left in the still nA will be

nA = xn

• Where,

• n is the moles of liquid left in the still at a given time

• y and x is the vapor and liquid compositions

• If a small amount of liquid dn is vaporized

• the change in the moles of component A is ydn, or dnA.

• Differentiating equation gives

Simple Batch or Differential Distillation

ydnxdnndx

xdnndxxnddnA

=+

+== )(xy

dx

n

dn

−=

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• dx/(y-x) can be integrated graphically or numerically

using tabulated equilibrium data or an equilibrium curve

• For ideal mixture:

Simple Batch or Differential Distillation

∫∫ =−

=

1

0 0

1

1

0

ln

x

x

n

nn

n

xy

dx

n

dn

xy

dx

n

dn

−= Rayleigh equation

B

A

AB

B

A

x

x

y

yα=

B

B

AB

A

A

B

AAB

B

A

B

A

n

dn

n

dn

n

n

dn

dn

dndn

dndn

α

α

=

==AB

A

A

B

B

n

n

n

nα/1

00

=

B

B

AB

A

A

n

n

n

n

00

lnln α=

Integrating

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• A batch of crude pentane contains 15 mole percent n-butane and 85 percent n-pentane. If a simple batch distillation at atmospheric pressure is used to remove 90 percent of butane, how much pentane will be removed? What will be the composition of the remaining liquid?

Solution: An average value of 3.5 is used for αAB.

Basis: 1 mol feed

(butane) (pentane)

From equation:

nB = total mole of B left in still, nA = total mole A left in still.

n0B = total initial mole of B in still, n0A = total initial mole A in still.

Problem

15.0=OAn 015.0=An 85.0=OBn

AB

A

A

B

B

n

n

n

nα/1

00

=

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• Total mole of liquid left in still:

• Mole fraction of butane in liquid left:

( ) 518.01.085.0

5.3/1==Bn

440.0)85.0(518.0 ==Bn

moln 455.0015.044.0 =+=

033.0455.0

015.0==Ax

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• At atmospheric pressure high boiling liquids cannot be

purified by distillation

• Since the components of the liquid may decompose at high

temperature required

Simple Steam Distillation

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• Often the high temperature substances are essentially

insoluble in water

• So a separation at lower temperatures can be obtained by

simple steam distillation

• This method is often used to separate a high boiling

component from small amount of nonvolatile impurities

• If the total pressure is fixed

• Since there are two liquid phases, each will exert its own

vapor pressure at the prevailing temperature and cannot

be influenced by the presence of the other

Simple Steam Distillation

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• When the sum of the separate vapor pressures equals

the total pressure, the mixture boils and

• Where

• PA is vapor pressure of pure water A

• PB is vapor pressure of pure B

• Then the vapor composition is

• The ratio moles of B distilled to moles of A distilled is

Simple Steam Distillation

PPP BA =+

P

Py A

A =P

Py B

B =

A

B

A

B

P

P

n

n=

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• This method has the disadvantage that

• Large amount of heat must be used to evaporate the water

simultaneously with the high boiling compounds

Simple Steam Distillation

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• A mixture contains 100 kg of H2O and 100 kg of ethyaniline (mol wt = 121.1

kg/kg mol), which is immiscible with water. A very slight amount of

nonvolatile impurity is dissolved in the organic. To purify the ethyaniline it is

steam-distilled by bubbling saturated steam into the mixture at a total

pressure of 101.32 kPa (1 atm). Determine the boiling point of the mixture

and the composition of the vapor. The vapor pressure of each of the pure

compounds is as follows (T1):

Problem

Temperature PA(water)

(kPa)

PB(ethylaniline)

(kPa)K ºC

353.8 80.6 48.5 1.33

369.2 96.0 87.7 2.67

372.3 99.15 98.3 3.04

386.4 113.2 163.3 5.33

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Solution

PPP BA =+

Temperature PA

(water)

(kPa)

PB

(ethylaniline)

(kPa)

P=PA+PB

(kPa)K ºC

353.8 80.6 48.5 1.33 49.83

369.2 96.0 87.7 2.67 90.37

372.3 99.15 98.3 3.04 101.34

386.4 113.2 163.3 5.33 169.23

The boiling temperature = 99.15ºC since total pressure in this temperature is

equal to atmospheric pressure.

The vapor composition are:

97.032.101

3.98===

kPa

kPa

P

Py A

A03.0

32.101

04.3===

P

Py B

B

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• A column containing the equivalent of N theoretical

stages

• a total condenser in which the overhead vapor leaving the top

stage is totally condensed to a bubble point

• liquid distillate and a liquid reflux that is returned to the top stage

• a partial reboiler in which liquid from the bottom stage is partially

vaporized to give a liquid bottoms product

• vapor boilup that is returned to the bottom stage and

• An intermediate feed stage.

Distillation with Reflux and McCabe- Thiele method

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Distillation with Reflux and McCabe- Thiele method

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• For components with close boiling points

• the temperature change over the column is small and relative

volatility is almost constant

Distillation with Reflux and McCabe- Thiele method

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• In 1925, McCabe and Thiele [5] published a graphical

method for combining the equilibrium curve with material

balance operating lines to obtain

• for a binary-feed mixture and selected column pressure

• the number of equilibrium stages and

• reflux required for a desired separation of feed components

Distillation with Reflux and McCabe- Thiele method

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• Typical input specifications and results (outputs) from the

McCabe–Thiele construction for a single-feed, two-

product distillation are

Distillation with Reflux and McCabe- Thiele method

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Distillation with Reflux and McCabe- Thiele method

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• From the specification of xD and xB for the LK, distillate

and bottoms rates, D and B, are fixed by material

balance, since

• But, B = F – D and therefore,

• Besides the equilibrium curve, the McCabe–Thiele method includes

• A 45o reference line, operating lines for the upper rectifying

section and the lower stripping section of the column, and a fifth

line (the q-line or feed line) for the phase or thermal condition of

the feed.

Distillation with Reflux and McCabe- Thiele method

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• The rectifying section of equilibrium stages extends from

the top stage, 1, to just above the feed stage, f

• Consider a top portion of the rectifying stages, including

the total condenser

Rectifying-Section Operating Line

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• A material balance for the LK over the envelope for the

total condenser and stages 1 to n is as follows:

• Solving for yn+1 gives the equation for the rectifying section operating line:

• This equation relates LK compositions yn+1 and xn of passing streams Vn+1 and Ln, respectively

Rectifying-Section Operating Line

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• This straight line equation is the locus of compositions of

all passing streams in the rectifying section

• L and V must not vary from stage to stage in the rectifying

section

• This is the case if:

• The two components have equal and constant molar enthalpies

of vaporization (latent heats)

• Component sensible-enthalpy changes (CPDT) and heat of

mixing are negligible compared to latent heat changes

• The column is insulated, so heat loss is negligible

• Column pressure is uniform (thus, no pressure drop).

Rectifying-Section Operating Line

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• These assumptions leading to the condition of constant

molar overflow in the rectifying section

• Since a total material balance for the rectifying-section

envelope gives Vn+1 = Ln + D

• if L is constant, then V is also constant for a fixed D

• Rewriting equation:

• Thus, the slope of the operating line in the rectifying

section is a constant L/V, with V > L and L/V < 1

Rectifying-Section Operating Line

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Rectifying-Section Operating Line

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• For constant molar overflow in either the rectifying or the

stripping section, only material balances and an

equilibrium curve are required

• Energy balances are needed only to determine

condenser and reboiler duties

• Liquid entering stage 1 at the top is the external reflux

rate, L0, and its ratio to the distillate rate, L0=D, is reflux ratio R

• Because of constant molar overflow, R = L/D is a

constant in the rectifying section

Rectifying-Section Operating Line

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• Since V = L+ D, the slope of the operating line is readily

related to the reflux ratio:

• Similarly,

• Combining equations produces the most useful form of

the operating line for the rectifying section:

• If R and xD are specified, plots as a straight line

Rectifying-Section Operating Line

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• The stripping section extends from the feed to the

bottom stage

Stripping-Section Operating Line

Consider a bottom portion of stripping

stages including the partial re-boiler and

extending up from stage N to stage m+1,

below the feed entry

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• A material balance for the LK over the envelope results

in

• Solving for ym+1:

• where L and V are total molar flows (which may be different from

L and V in the rectifying section because of feed addition)

• Vapor leaving the partial reboiler is assumed to be in

equilibrium with the liquid bottoms product, B, making

the partial reboiler an equilibrium stage

Stripping-Section Operating Line

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• With the constant-molar overflow assumption, VB =

is constant in the stripping section

• Since

Stripping-Section Operating Line

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Stripping-Section Operating Line