1

Chapter 4 Distillation Design

Subject: 1304 332 Unit Operation in Heat transfer

Instructor: Chakkrit UmpuchDepartment of Chemical Engineering

Faculty of EngineeringUbon Ratchathani University

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What are you going to learn in this chapter?

4.1 Vapor-Liquid Equilibrium Relations

4.2 Single-Stage Equilibrium Contact for Vapor-Liquid

System

4.3 Simple Distillation Methods

4.4 Distillation with reflux and McCabe-Thiele Method

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4.1 Vapor-Liquid Equilibrium Relations4.1.1 Raoult’s Law

An ideal law, Raoult’s law, can be defined for vapor-liquid phases in equilibrium (only ideal solution e.g. benzene-toluene, hexane-heptane etc.

Where

pA is the partial pressure of component A in the vapor in Pa (atm)

PA is the vapor pressure of pure A in Pa (atm)

xA is the mole fraction of A in the liquid.

AAA xPp

BA xx 1

BA yy 1Composition in liquid:

Composition in vapor:

(1)

(2)(3)

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Boiling-point diagram for system benzene (A)-toluene (B) at a total pressure of 101.32 kPa.

4.1.2 Boiling-Point Diagrams and xy Plots

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.

5

Boiling-point diagram for system benzene (A)-toluene (B) at a total pressure of 101.32 kPa.

4.1.2 Boiling-Point Diagrams and xy Plots

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.

6

4.1.2 Boiling-Point Diagrams and xy Plots

Ppp BA

PxPxP ABAA )1(

PxP

Pp

y AAAA

The boiling point diagram can be calculated from (1) the pure vapor-pressure data in the table below and (2) the following equations:

(4)

(5)

(6)

Where

pA, pB are the partial pressure of component A and B in the vapor in Pa (atm)

PA , PB are the vapor pressure of pure A and pure B in Pa (atm)

P is total pressure in Pa (atm)

xA is the mole fraction of A in the liquid.

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4.1.2 Boiling-Point Diagrams and xy Plots

The boiling point diagram can be calculated from (1) the pure vapor-pressure data in the table below and (2) the following equations:

1

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Ex 4.1 Use of Raoult’s Law for Boiling-Point Diagram

Calculate the vapor and liquid compositions in equilibrium at 95ºC (368.2K) for benzene-toluene using the vapor pressure from the

table 1 at 101.32 kPa.

Solution: At 95ºC from Table 1 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),

PxPxP ABAA )1(

632.032.101

)411.0(7.155

PxP

Ppy AAA

A

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The boiling point diagram can be calculated from the pure vapor-pressure data in the table below and the following equations:

1

10

A common method of plotting the equilibrium data is shown in Fig. 2 where 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.

Fig. 2 Equilibrium diagram for system benzene(A) – toluene(B) at 101.32 kPa (1atm).

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4.1 Vapor-Liquid Equilibrium Relations4.1.2 Boiling-Point Diagrams and xy Plots

Maximum-boiling azeotropeMinimum-boiling azeotropeIdeal boiling point diagram

An azeotrope is a mixture of two or more liquids in such a ratio that its composition cannot be changed by simple distillation.

This occurs because, when an azeotrope is boiled, the resulting vapor has the same ratio of constituents as the original mixture.

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4.1.2 Boiling-Point Diagrams and xy Plots

4.1 Vapor-Liquid Equilibrium Relations

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4.2 Single-Stage Equilibrium Contact for Vapor-Liquid System

V1 V2

L0 L1

MVLVL 1120Total mass balance:

AMAAAA MxyVxLyVxL 11112200Mass A balance:

Where

V1, V2 is a vapor

L0, L1is a liquid

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.

In case of constant molal overflow : V1 = V2 and L0 = L1

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Ex 4.2 Equilibrium Contact of Vapor-Liquid Mixture

A vapor at the dew point and 101.32 kPa containing a mole fraction of 0.40 benzene (A) and 0.60 toluene (B) and 100 kg mol total is

contacted with 110 kg mol of a liquid at the boiling point containing a mole fraction of 0.30 benzene and 0.70 toluene. The two streams are

contacted in a single stage, and the outlet streams leave in equilibrium with each other. Assume constant molal overflow. Calculate the amounts and compositions of the exit streams.

Solution: The given values are V2 = 100 kg mol, yA2 = 0.40, L0=110 kg mol , and xA0 = 0.30.

V1 V2

L0 L1

For constant molal overflow,

V2 = V1 and L0 = L1.

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Material balance on component A,

To solve equation above, the equilibrium relation between yA1 and xA1 in figure below must be used.

First, we assume that xA1 = 0.20 and substitute into equation above to solve for yA1.

Assuming that xA1 = 0.20 and solving yA1 = 0.51.

Next, assuming that xA1=0.40 and solving, yA1 = 0.29.

Next, assuming that xA1=0.40 and solving, yA1 = 0.29.

(These point are plotted on the graph.)

At the intersection of this line with the equilibrium curve,

yA1 = 0.455 and xA1 = 0.25. Answer

AMAAAA MxyVxLyVxL 11112200

1100)2.0(110)40.0(100)30.0(110 Ay

11 100110)40.0(100)30.0(110 AA yx

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4.3 Simple Distillation Methods

4.3.1 Introduction

Distillation is a method used to separate the components of liquid solution, which depends upon the distribution of these various components between a vapor and a liquid phase.

The vapor phase is created from the liquid phase by vaporization at the boiling point.

Distillation is concerned with solution where all components are appreciably volatile such as in ammonia-water or ethanol-water solutions, where both components will be in the vapor phase.

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4.3.2 Relative Volatility of Vapor-Liquid Systems

)1)(1(/

//

AA

AA

BB

AAAB xy

xyxyxy

AAB

AABA x

xy

)1(1

PxPy AA

A

B

AAB P

P

PxPy BB

B

Relative volatility

It is a measure of the differences in volatility between 2 components, and 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.

when αAB is above 1.0, a separation is possible.

Raoult’s law:

)( AB

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Ex 4.3 Using data from table 1 calculate the relative volatility for the benzene-toluene system at 85ºC (358.2K) and 105ºC (378.2K)

Solution: At 85ºC, substituting into equation below for a system following Raoutl’s law,

Similarly at 105ºC,

The variation in α is about 7%. Answer

54.20.469.116

B

AAB P

P

38.20.862.204

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4.3.3 Equilibrium or Flash Distillation

Distillation has two main methods in practice.

4.3.3.1 Introduction to distillation methods

1. 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.

2. 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”).

There are 3 important types of distillation that occur in a single stage or still: Equilibrium or flash distillation, Simple batch or differential distillation and simple steam distillation

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4.3.3.2 Equilibrium or Flash Distillation

heater

Separator

xA

yA

Flash distillation is a single stage separation technique.

1. A liquid mixture is pumped through a heater to raise the temperature

and enthalpy of the mixture.

2. It then flows through a valve and the pressure is reduced, causing

the liquid to partially vaporize.

3. Once the mixture enters a big enough volume (the “flash drum”), the liquid and

vapor separate.

4. Because the vapor and liquid are in such close contact up until the “flash” occurs,

the product liquid and vapor phases approach equilibrium.

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4.3.3.2 Equilibrium or Flush Distillation

heater

Separator

Total mass balance:

Component A balance:

where

F, V and L are flow rate of feed, vapor and liquid phases.

xF, yA and xA are mole fraction of component A in feed, vapor and liquid.

AAF xfyfx )1( Where

f = V/F = molal fraction of the feed that is vaporized and withdrawn continuously as vapor.

1-f = one as liquid

Material balance for more volatile component :

xA

yA

AAF xFV

FFy

FVx )()(

AAF LxyVFx

LVF

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Ex 4.4 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?

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

FVf /

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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. Answer

AAF xfyfx )1(

AAF xFV

FFy

FVx )()(

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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4.3.4 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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4.3.4 Simple Batch or Differential Distillation

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

ydnxdnndxxdnndxxnddnA

)(

xydx

ndn

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B

AAB

B

A

xx

yy

dx/(y-x) can be integrated graphically or numerically using tabulated equilibrium data

or an equilibrium curve.

For ideal mixture:

1

0 0

11

0

lnx

x

n

n nn

xydx

ndn

xydx

ndn

B

BAB

A

A

B

AAB

B

A

B

A

ndn

ndn

nn

dndn

dndndndn

AB

A

A

B

B

nn

nn

/1

00

B

BAB

A

A

nn

nn

00

lnln

4.3.4 Simple Batch or Differential Distillation

Integrating

Rayleigh equation

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Ex 4.5 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.

15.0OAn 015.0An 85.0OBn

AB

A

A

B

B

nn

nn

/1

00

30

518.01.085.0

5.3/1 Bn

440.0)85.0(518.0 Bn

moln 455.0015.044.0

033.0455.0015.0

Ax

Total mole of liquid left in still:

Mole fraction of butane in liquid left:

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4.3.5 Simple Steam Distillation

Note that by steam distillation, as long as water is present, the high-boiling component B vaporizes at a temperature well below its normal

boiling point without using a vacuum. The A and B are usually condensed in condenser and the resulting two immiscible liquid

phases separated. Disadvantage: large amounts of heat must be used to simultaneously

evaporate the water with high-boiling compound.

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4.3.5 Simple Steam Distillation

When the sum of the separate vapor pressures equals the total pressure, the mixture boils and

Where is vapor pressure of pure water A

is vapor pressure of pure B

Then the vapor composition is

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

PPP BA

BP

PPy A

A

AP

PPy B

B

A

B

A

B

PP

nn

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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

Ex 4.6 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):

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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 3.04 101.34386.4 113.2 163.3 5.33 169.23

The boiling temperature = 99.15ºC since total pressure in this temperature is equal to atmospheric pressure.

The vapor composition are:

97.032.1013.98

kPakPa

PPy A

A03.0

32.10104.3

PPy B

B

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4.4 Distillation with Reflux and McCabe-Thiele Method4.4.1 Introduction to Distillation with Reflux

Rectificaition (fractionation) or stage distillation with reflux is

a series of flash-vaporization stages are arranged in a series which the vapor and liquid products from each stage flow countercurrently to each other.

V2

L1

V1 V2

L0 L1

V3

L2

Vn Vn+1

Ln-1 Ln

The liquid in a stage is conducted or flows to the stage below and the vapor from a stage flow upward to the stage above.

nnnn LVLV 11

nnnnnnnn xLyVxLyV 1111

A total material balance:

A component balance on A:

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4.4.1 Introduction to Distillation with Reflux

1. Feed enters the column somewhere in the middle of the column.

5. The vapor continues up to the next tray or stage, where it is again contacted with

a downflowing liquid.

2. Feed is liquid, it flows down to a sieve tray or stage.

4. The vapor and liquid leaving the tray are essentially in equilibrium.

In a distillation column the stages (referred to as sieve plates or trays) in a distillation tower are arranged vertically, as shown schematically in figure below.

3. Vapor enters the tray and bubbles through the liquid on this tray as the

entering liquid flows across.

6. The concentration of the more volatile component is being increased in the

vapor form each stage going upward and decreased in the liquid from each stage

going donwards.

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4.4.1 Introduction to Distillation with Reflux

7. The final vapor product coming overhead is condensed in a condenser

and a portion of the liquid product (distillate) is removed, which contains a

high concentration of A.

9. The liquid leaving the bottom tray enters a reboilier, where it partially

vaporized, and the remaining liquid, which is lean in A or rich in B, is withdrawn as

liquid product.

In a distillation column the stages (referred to as sieve plates or trays) in a distillation tower are arranged vertically, as shown schematically in figure below.

8. The remaining liquid from the condenser is returned (refluxed) as a

liquid to the top tray.

10. The vapor from the reboiler is sent back to the bottom stage or trays is much

greater.

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4.4.3 McCabe-Thiele Method of Calculation for Number of Theoretical Stages

A mathematical – graphical method for determining the number of theoretical trays or stages needed for a given separation of a binary mixture of A and B has been developed by McCabe and Thiele.

The method uses material balances around certain parts of the tower, which give operating lines and the xy equilibrium curve for the system.

Main assumption

1) Equimolar overflow through the tower between the feed inlet and the top tray and the feed inlet and bottom tray.

2) Liquid and vapor streams enter a tray, are equilibrated, and leave.

4.4.3A) Introduction and assumptions

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A total material balance:

nnnn LVLV 11

A component A balance:

nnnnnnnn xLyVxLyV 1111

Where

Vn+1 is mol/h of vapor from tray n+1

Ln is mol/h liquid from tray n

yn+1 is mole fraction of A in Vn+1 and so on.

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A total material balance:

A component A balance:

Where

F is the entering feed (mol/h)

D is the distillate (mol/h)

W is the bottoms (mol/h)

WDF

wDF WxDxFx

4.4.3B) Equation for enriching section

(1)

(2)

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Material balance over dashed-line section: DLV nn 1

A balance on component A: DDxxLyV nnnn 11 (4)

(3)

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Solving for yn+1, the enriching-section operating line is

)1/(/, 11 RRVLDLV nnnn

DLR n /

111

R

xx

RRy D

nn

where = reflux ratio = constant.

The eqn. (1) is a straight line on a plot of vapor composition versus liquid composition.

Since and equation becomes

111

n

Dn

n

nn V

DxxVLy

(6)

(5)

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The slope is or . It intersects the y=x line (45º diagonal

line) at . The intercept of the operating line at x = 0 is .1/ nn VL )1/( RR

)1/( Rxy DDxx

The theoretical stages are determined by starting at xD and stepping off the first

plate to x1. Then y2 is the composition of the vapor passing the liquid x1.

In a similar manner, the other theoretical trays are stepped off down the tower in

the enriching section to the feed tray.

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4.4.3C Equation for stripping section

Material balance over dashed-line section: WLV mm 1

A component A balance: wWxxLyV mmmm 11 (8)

(7)

45

Solving for ym+1, the enriching-section operating line is

111

m

Wm

m

mm V

WxxVLy (9)

1 mm VL

Again, since equimolal flow is assumed, = constant and

= constant, eqn. (2) is a straight line when plotted as y versus x,

with a slope of . It intersects the y = x line at x = xw.

The intercept at x = 0 is .

Nm LL

Nm VV 1

1/ mW VWxy

46

The theoretical stages for the stripping section are determined by starting at

xW, going up to yW, and then across to the operating line, etc.

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4.4.3D Effect of feed conditions

The condition of feed stream is represented by the quantity q, which is the mole fraction of liquid in feed.

qFLL nm

FqVV mn )1(

Dnn DxxLyV

wmm WxxLyV

)()()( wDnmnm WxDxxLLyVV

The enriching and striping operating-line equations on an xy diagram can be derived as follows:

Where the y and x values are the point of intersection of the two operating lines. Subtracting eqn.(12) from eqn.(13),

(12)

(13)

(14)

(10)

(11)

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4.4.3D Effect of feed conditions

Substituting eqn.(2), (10), and (11) into eqn.(14) and rearranging,

(15)

11

qx

xqqy F

)(

1 FbpL TTcq

)(

1 dFpV TTcq

Cold-liquid feed Superheated vapor

where

CpL, CpV = specific heats of liquid and vapor, respectively

TF = temperature of feed

Tb, Td = bubble point and dew point of feed respectively

λ = heat of vaporization

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4.4.3E Location of the feed tray in a tower and number of trays.

q = 0 (saturated vapor)

q = 1 (saturated liquid)

q > 1(subcooled liquid)

q < 0 (superheated vapor)

0 < q < 1 (mix of liquid and vapor)

From eqn.(15), the q-line equation and is the locus of the intersection of the two operating lines. Setting y = x in eqn(15), the intersection of the q-line equation with the 45º line is y=x=xF, where xF is the overall composition of the feed.

In given below the figure, the q line is plotted for various feed conditions. The slope of the q line is q/(q-1).

1st point

2nd point

3rd pointn = 7 =number of tray + reboiler

Number of tray = 6

4.4.3D Number of stages and trays

Slope = R/(R+1)

Slope = q/(1-q)

Slope = L/ V

4.4.3F Using Operating Lines and the Feed Line in McCabe-Thiele Design

Ex 4.7 A continuous fractioning column is to be designed to separate 30,000 kg/h of a mixture of 40 percent benzene and 60 percent toluene into an overhead product containing 97 percent benzene and a bottom

product containing 98 percent toluene. These percentages are by weight. A reflux ratio of 3.5 mol to 1 mol of product is to be used. The molal latent heats of benzene and toluene are 7,360 and 7,960 cal/g mol, respectively. Benzene and toluene from a nearly ideal system with a

relative volatility of about 2.5. The feed has a boiling point of 95ºC at a pressure of 1 atm.

a) Calculate the moles of overhead product and bottom product per hour.

b) Determine the number of ideal plates and the position of the feed plate (i) if the feed is liquid and at its boiling point; (ii) if the feed is liquid and at 20ºC (specific heat 0.44 cal/g.ºC); (iii) if the feed is a mixture of two-thirds vapor and one-third liquid.

Solution (a)

The average molecular weight of the feed is

The average of heat vaporization is

The feed rate F is 30,000/85.8 = 350 kg mol/h. By an overall benzene balance, using Eq. below

440.0

9260

78407840

Fx 0235.0

9298

782782

Bx974.0

923

78977897

Dx

8.85

9260

7840100

gmolcal /696,7)960,7(56.0)360,7(44.0

hkgmolD /4.1530235.0974.00235.0440.0350

hkgmolB /6.1964.153350

Solution (b) (i),

We determine the number of ideal plates and position of the feed plate.

1) Plot the equilibrium diagram, erect verticals at xD, xF, and xB.

2) Draw the feed line. Here q=1, and the feed line is vertical.

3) Plot the operating lines. The intercept of the rectifying lie on the y axis is, xD/(R+1) = 0.974/(3.5+1) = 0.216 (eqn (6)). From the

intersection of the rectifying operating line and the feed line, the stripping line is drawn.

4) Draw the rectangular steps between the two operating lines and the equilibrium curve. The stripping line is at the seventh step. By counting steps it is found that, besides the reboiler, 11 ideal plates are

needed and feed should be introduced on the seventh plate from the top.

Solution (b) (ii),

The latent heat of vaporization of the feed λ is 7,696/85.8 = 89.7 cal/g.

The slope of the feed line is 1.37/(1.37-1) = 3.70. When steps are drawn for this case, as shown in Fig. below, it is found that a reboiler and 10 ideal plates are needed and that the feed should be introduced on the

sixth plate.

37.17.89

)2095(44.01

q

11

qx

xqqy F

Solution (b) (iii),

From the definition of q it follows that for this case q = 1/3 and the slope of the feed line is -0.5. The solution is shown in Fig. below. It calls for a

reboiler and 12 plates, with the feed entering on the seventh plate.

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4.4.4 Total and Minimum Reflux Ratio for McCabe-Thiele Method

4.4.4A Total flux

One limiting values of reflux ratio is that of total reflux, or R = ∞. Since R = Ln/D and, by eqn.(16).

DLV nn 1

Then Ln is very large, as is the vapor flow Vn. This means that the slope R/(R+1) of the enriching operating line becomes 1.0 and the operating lines of both sections of the column coincide with the 45º diagonol line, as shown in Fig below.

Minimum number of trays can be obtained by returning all the overhead condensed vapor V1 from the top of the tower back to the tower as reflux, i.e., total reflux. Also, the liquid in the bottoms is reboiled.

(16)

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Minimum number of theoretical steps Nm

when a total condenser is used (α is constant).av

w

w

D

D

m

xx

xx

Nlog

11

log

2/11 wav For small variations in α,

where α1 is the relative volatility of the overhead vapor

αw is the relative volatility of the bottoms liquid.

(17)

(18)

62

4.4.4B Minimum reflux ratio

The minimum reflux ratio (Rm) will require an infinite number of trays for the given separation desired of xD and xW.

If R is decreased, the slope of the enriching operating line R/(R+1) is decreased, and the intersection of this line and the stripping line with the q line moves farther from the 45º line and closer to the equilibrium line.

xxyx

RR

D

D

m

m

1

Two operating lines touch the equilibrium line (“pinch point”) at y’ and x’ (number of steps required becomes infinite).

The line passes through the points x’, y’ and xD (y=xD):

(19)

63

4.4.4C Operating and optimum reflux ratio

Total reflux = number of plates is a minimum, but the tower diameter is infinite.

This corresponds to an infinite cost of tower and steam and cooling water. This is the limit in the tower operation.

Minimum reflux = number of trays is infinite, which again gives an infinite cost. These are the two limits in operation of the tower.

Actual operating reflux ratio to use is in between these two limits. The optimum reflux ratio to use for lowest total cost per year is between the minimum Rm and total reflux (1.2Rm to 1.5Rm).

64

Ex 4.8 What are (a) the minimum reflux ratio and (b) the minimum number of plates for cases (b)(i), (b)(ii), and (b)(iii) of EX 4.7 if αavis given as 2.47.

Solution (a)

For minimum reflux ratio use eqn. (18). Here xD = 0.974. The results are given in Table below.

Case x’ y’ RDm

(b)(i) 0.440 0.658 1.45

(b)(ii) 0.521 0.730 1.17

(b)(iii) 0.300 0.513 2.16

65

Ex 4.8 What are (a) the minimum reflux ratio and (b) the minimum number of plates for cases (b)(i), (b)(ii), and (b)(iii) of EX 4.7 if αavis given as 2.47.

Solution (b)

For minimum number of plates, the reflux ratio is infinite, the operating lines coincide with the diagonal, and there are no differences among

the three cases. Use the αav = 2.47 and equation below we get,

The minimum number of ideal plates is 7 plus a reboiler.

71105.8

1026.0024.0976.0974.0ln

log

11

log

av

w

w

D

D

m

xx

xx

N

Homework No.6

1. The vapor-pressure data are given below for the system hexane-octane.

(a) Using Raoult’s law, calculate and plot the xy data at a total pressure of 101.32 kPa.

(b) Plot the boiling-point diagram.

T(ºF) T(ºC) Vapor Pressure

n-Hexane n-Octane

kPa mmHg kPa mmHg

155.6 68.7 101.3 760 16.1 121

175 79.4 136.7 1025 23.1 173

200 93.3 197.3 1480 37.1 278

225 107.2 284.0 2130 57.9 434

258.2 125.7 456.0 3420 101.3 760

Homework No.6 (continue)

2. A mixture of 100 kg mol which contains 60 mol% n-pentane (A) and 40 mol% n-heptane (B) is vaporized at 101.32 kPa pressure under differential conditions until 40 kg mol are distilled. Use vapor-liquid equilibrium data below. If flash distillation is done and 40 kg mol are distilled, what is the composition of the vapor distilled and of the liquid left?

equilibrium data for n-pentane-n-heptane, x and y are mole fraction of n-pentane

x 0.0 0.059 0.145 0.254 0.398 0.594 0.867 1.000y 0.0 0.271 0.521 0.701 0.836 0.925 0.984 1.000

Homework No.7

1. A mixture containing 50 g mol of benzene and 50 g mol of chlorobenzene is distilled by simple distillation without reflux until 40 percent of the initial charge is taken off as overhead. The system benzene-chlorobenzene may be considered ideal, with an averge relative volatility of 5.3

(a) What are the compositions of overhead and residue after distillation is complete?

(b) The overhead from the first distillation is subjected to a second simple distillation. Again 40 percent of the charge is taken overhead. What is the composition of the second over head product? What is its mass in grams? How many grams of chlorine does it contain?

Homework No.7 (continue)

2. A mixture of 50 g mol of liquid benzene and 50 g mole water. Determine the boiling at 101.32 kPa pressure. Liquid benzene is immiscible in water. Determine the boiling point of the mixture and the composition of the vapor. Which component will first be removed completely from the still? Vapor pressure data of the pure components are as follows:

Temperature Pwater

(mm Hg)Pbenzene(mm Hg)K ºC

308.5 35.3 43 150

325.9 52.7 106 300

345.8 72.6 261 600

353.3 80.1 356 760

Homework No.8

1. A saturated liquid feed of 200 mol/h at the boiling point containing 42mol% heptane and 58% ethyl benzene is to be fractionated at 101.32 kPa abs to give a distillate containing 97 mol% heptane and a bottoms containing 1.1 mol% heptane. The reflux ratio used is 2.5:1. Calculate the mol/h distillate, mol/h bottoms, theoretical number of trays, and the feed tray number. Equilibrium data are given below at 101.32 kPa abs pressure for the mole fraction n-heptane xH and yH.

Temperature Temperature

K ºC xH yH K ºC xH yH

409.3 136.1 0 0 383.8 110.6 0.485 0.730

402.6 129.4 0.08 0.23 376.0 102.8 0.790 0.904

392.6 119.4 0.25 0.514 371.5 98.3 1.000 1.000

Homework No.8 (continue)

2. Determine the minimum reflux ration Rm and the minimum number of theoretical plates at total reflux for the rectification of a mixture of heptane and ethyl benzene as given in problem 1. Do this by the graphical mehtods of McCabe-Thiele.