1

MULTIPLE STAGE EQUILIBRIUM PROCESSES

1. VAPOUR LIQUID EQUILIBRIA AND DISTILLATION

Let us consider a simple problem:

A process produces a product B from reactant A with only limited conversion, so that the reactor

outlet contains 60 mol% A and 40 mol% B. It is desirable to separate the reactant (A) from the

product (B) in order to recycle the reactants and to obtain a high purity product. We will choose a

basis of 100 kmol of reactor product:

Suppose that A has a lower boiling point than B. It is proposed that 60% (on a mole basis) of the

mixture is boiled off to produce a vapour rich in A and a liquid rich in B.

We thus have two unknowns: xA and yA. We have already ensured that the total mole balance is

satisfied [input = output; 100 kmol = (60 + 40) kmol] so we can develop only one equation by

performing a material balance for one of the components (A or B).

Balance for A: 60 = 60 yA + 40 xA

Balance for B: 40 = 60 yB + 40 xB

∴ 40 = 60 (1 – yA) + 40 (1 – xA)

∴ 40 = 60 + 40 – 60 yA – 40 xA

∴ 60 = 60 yA + 40 xA

100 kmol mixture

0.6 mol A/mol mixture

0.4 mol B/mol mixture

A rich stream

B rich stream

100 kmol

0.6 kmol A/kmol

0.4 kmol B/kmol

60 kmol

yA kmol A/kmol

yB = (1-yA) kmol B/kmol

40 kmol

xA kmol A/kmol

xB = (1-xA) kmol B/kmol Heat

Boiling

solution

at TºC

2

Thus the problem cannot be solved without further information. In order to determine the

composition of the streams we need to know how the composition of the vapour produced from a

boiling mixture varies with the composition of the mixture.

At its boiling point the vapour pressure of a liquid is equal to the ambient pressure. The vapour

pressure of the mixture is the sum of the vapour pressure of each component. Dalton’s law gives

the concentration of a component in terms of its vapour pressure (yA = PA/P). Thus, we need to

know how the vapour pressure of a component varies with the composition of the mixture and

temperature.

For an ‘ideal solution’ (where equal forces act between molecules of {A and A}, {A and B} and {B

and B}) we can use Raoult’s law. Raoult’s law states that the vapour pressure exerted by a

component above an ideal solution is proportional to its mole fraction. The constant of

proportionality is the vapour pressure of the pure component.

If the vapour pressure of pure A at temperature T is PA

0 then the vapour pressure of A solution with

a mole fraction xA mols A per mol of solution is

Raoult’s law Dalton’s law

PA = xA PA

0 and yA =

PA

P =

xAPA

0

P

where P is the ambient pressure. Similarily for B:

Raoult’s law Dalton’s law

PB = xB PB

0 and yB =

PB

P =

xBPB

0

P

Raoult’s law: vapour pressures of ideal solutions

(diagram shows behaviour at constant temperature)

xA 0 1

PA

PB

PA

0

PB

0

3

Thus we can solve the problem above if the solution is ideal and we know the ambient pressure and

the saturated vapour pressure of A or B.

e.g. If A is hexane, B is octane, P = 760 mm Hg and the solution is boiling at 94°C at which

temperature PB

0 = 288 mm Hg, determine xA and yA for the problem on page 1 above.

[Hint: use yB = xBPB

0

P and the material balance equation]

yB = (1 – yA) = xBPB

0

P =

(1 − xA)PB

0

P

∴ yA = 1 − PB

0

P +

xAPB

0

P

Substituting into the material balance equation at the top of the page with P = 760 mm Hg and PB

0 =

246 mm Hg

60 = 60

1 − 288

760 +

xA × 288

760 + 40 xA

60 . 288/760 = xA (60 . 288/760 + 40)

∴ xA = 0.362 kmol A/kmol

yA = 1 – 40 xA/60 = 0.758 kmol A/kmol

If the data available is PA

0 and PB

0 (but not P) then we can still solve the problem:

e.g. Solve the problem above if PA

0 = 1578 mm Hg and PB

0 = 288 mm Hg and P is unknown.

[Hint: divide yA = xAPA

0

P by yB =

xBPB

0

P to eliminate P]

Dividing yA = xAPA

0

P by yB =

xBPB

0

P to eliminate P:

yA

yB

= xA

xB

PA

0

PB

0 substituting for yB = (1 – yA) and xB = (1 – xA):

yA

1 − yA

= xA PA

0

(1 − xA) PB

0

ASIDE: The term

B

A

B

A

x

xy

y

is sometimes called the relative volatility α. For ideal systems α = P

A

0

PB

0

Substituting for yA = 1 – 40 xA/60 from the material balance

4

60 − 40 xA

40 xA

= xA PA

0

(1 − xA) PB

0

(1.5 – xA)(1 – xA)PB

0 = xA

2 PA

0

xA2(PA

0 – PB

0) + 2.5 PB

0 xA – 1.5 PB

0 = 0

xA = −2.5 PB

0 ± (2.5 PB

0)2 + 6 PB

0(PA

0 − PB

0)

2(PA

0 − PB

0)

xA = −2.5 . 288 + (2.5 . 288)

2 + 6 . 288 (1578 − 288)

2(1578 − 288) = 0.363 as before (nearly!).

In general if we select the operating pressure, we do not know the boiling point of the solution as

we do not know its composition. We thus need to consider how the boiling point of the solution

varies with composition, and thus determine the composition it generates as above.

For ideal solutions, the total vapour pressure above the solution P is thus

P = PA + PB = xA PA

0 + xB PB

0

= xA PA

0 + (1 − xA)PB

0

Rearranging:

xA = P − PB

0

PA

0 − PB

0

Thus if vapour pressure data (as a function of temperature) is available for the two components,

then this equation can be used to determine the boiling point of any liquid mixture at given pressure

P or the composition of any mixture boiling at a given temperature.

5

e.g. Hexane and Octane

P = 760 mm Hg

Mixture boiling point = 94°C

Vapour pressure of pure hexane at 94°C PA

0 = 1578 mm Hg

Vapour pressure of pure octane at 94°C PB

0 = 288 mm Hg

xA = 760 − 288

1578 − 288 = 0.366

yA = xAPA

0

P =

0.327 × 1819

760 = 0.760

Data is often calculated for a range mixtures between pure A and pure B which are plotted on a

T – x – y diagram, or if temperatures are not important an x-y diagram. Note that these diagrams

are normally drawn for the more volatile component, so on the T-x-y diagram the vapour (y) line is

above the liquid (x) line and on the x-y diagram the equilibrium line is above the diagonal.

6

T - x - y plot for hexane and octane at 1 atmosphere total pressure

65

70

75

80

85

90

95

100

105

110

115

120

125

130

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x or y , mol fraction of hexane in liquid/vapour

Tem

pera

ture

(d

eg

rees C

)

x

(mol hexane per mol liquid)

y (mol hexane per mol vapour)

7

Vapour-liquid equilibrium for hexane/octane at P=1 atm.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x (mol fraction hexane in liquid)

y (

mo

l fr

acti

on

hexan

e in

vap

ou

r)

8

For a non-ideal system we will need to use the graphical data with the material balance, requiring a

graphical solution. The values of yA and xA must

(i) satisfy the material balance equation

60 = 60 yA + 40 xA

AND

(ii) must lie on the equilibrium line on the x-y diagram.

Thus if we plot the material balance line: 60 = 60 yA + 40 xA on the x-y diagram, the

solution will be the point which crosses the equilibrium x-y line.

Although this simple process can only achieve a crude separation, it is often used in industry. The

process is sometimes called ‘equilibrium distillation’ or ‘Rayleigh distillation’. Heating is rarely

used however, as the same effect can be achieved by dropping the pressure so that the boiling point

of the solution falls to below its actual temperature.

P>1 atm

0.6 mol A/mol mixture

0.4 mol B/mol mixture

A rich stream

B rich stream

P = 1 atm

9

MULTI-STAGE DISTILLATION

The separation achieved in the example above was rather poor. In practice we want a high purity

product and very little of the product in the recycle. First consider the more volatile component A.

We could condense the vapour produced and boil it once again to obtain a higher concentration of

A, and this could be repeated any number of times to increase the concentration of A.

100 kmol mixture

0.6 mol A/mol mixture

0.4 mol B/mol mixture

Heat

Boiling

solution

Heat

Boiling

solution

cooling

water

Heat

Boiling

solution

cooling

water

This will certainly produce a stream concentrated in A,

but what are the practical disadvantages of this approach?

Expensive in both cooling water and heat.

Many side streams need handling with varying composition.

Very little concentrated A generated, most A lost in liquid streams.

High capital cost (many process stages).

10

Solution - multistage distillation:

Feed Boiling

solution

Boiling

solution

Boiling

solution

Boiling

solution

cooling

water

Boiling

solution

Heat

Boiling

solution

Boiling

solution

Feed

steam

Distillation trays or plates

Distillation

column

total

condenser

reboiler

reflux

Distillate or

top product

Bottom

product

11

By selecting the number of trays in a distillation column we can design the column to achieve a

desired separation. Thus a general distillation problem is where the feed flowrate and composition

is known and a column must be designed to achieve desired top and bottom product compositions.

E.g. 100 kmol/h of a mixture of 60 mol% hexane and 40 mol% octane must be separated to

achieve a top product containing 95 mol% hexane and a bottom product containing 95% octane.

The first step in solving this problem is to carry out an overall material balance for the column to

determine the top and bottom product flow rates.

Total balance F = D + B

100 = D + B

Balance on hexane F xF = D xD + B xB

60 = 0.95 D + 0.05 B

Substituting B = 100 – D from the total balance

60 = 0.95 D + 5 – 0.05 D

D = 55/0.9 = 61.11 kmol/h

B = 100 – 61.11 = 38.89 kmol/h

These overall balances should always be carried out first when tackling distillation problems.

F

100 kmol/h

D

B

12

Now we must consider what is happening on each tray, in order to determine the vapour and liquid

flows between the trays.

Liquid and Vapour Flow Rates

Consider how much vapour will be produced from a tray in the section of the column above the

feed (called the enrichment section). This amount of vapour produced is related to the amount of

heat supplied to the tray. Heat is supplied in the form latent heat from condensing vapour flowing

from the tray below.

The nomenclature is defined in the figure above:

The flowrate of liquid and vapour leaving tray n are called Vn and Ln respectively.

Assuming:

(i) There are no heat losses from the column

(ii) The difference in temperature between adjacent trays is negligible

(iii) The latent heat of the two components are equal

If this latent heat is λ, we can write a heat balance as:

λ Vn-1 = λ Vn

∴ Vn-1 = Vn

In words, for every mole of vapour condensed, one mole of vapour boils off.

A total material balance for the tray gives:

Vn-1 + Ln+1 = Vn + Ln

∴ Ln+1 = Ln

For every mole of liquid flowing to tray n, one mole of liquid flows from tray n to tray n-1.

Vn-1

Vn

Ln

Ln+1

13

Now consider the condenser and reflux. The vapour from the condenser is completely condensed

so that the resulting liquid has the same composition. This liquid is split to produce the top product

and to provide liquid for the top tray.

Thus these three streams, the vapour leaving the top tray, the top product and the reflux all

have the same composition, xD.

Notice that we have used Vn and Ln to denote the vapour and liquid flowrates since we have shown

above that these flowrates are constant throughout the enrichment section. We have also extended

the nomenclature to include the composition: the mole fraction of the more volatile component

(MVC) in the liquid leaving tray n is xn, while the mole fraction of the MVC in the vapour

generated on tray n is yn.

Initially we will assume that these two streams (the vapour and liquid leaving the tray) are in

equilibrium, i.e. the vapour leaving a tray is assumed to have the composition that would be

produced by boiling the liquid which is leaving the same tray. This means that (xn,yn) lies on the

equilibrium (x,y) line on the x-y diagram.

As the condenser at the top of the column is a total condenser the vapour flowrate Vn is equal to the

sum of the reflux and the distillate flowrate:

Vn = D + Ln

Thus in the enrichment section the vapour flow rate is greater than the liquid flowrate.

Ln

xN+1

= xD

Vn , y

N = x

D

D

xD

tray N

Vn

yN-1

Ln

xN

14

Now consider the feed tray:

Here we denote the vapour and liquid flow rates in the section of the column below the feed (called

the stripping section) as Vm and Lm. If the feed is a liquid at its boiling point, the heat balance

gives

Vm = Vn

and the material balance gives:

Lm = Ln + F

For each tray below the feed tray, by the same argument used above in the enrichment section, the

vapour and liquid flow rates are constant.

Thus the liquid flow rate in the stripping section is greater than that in the enrichment section, while

the vapour flow rates in the two sections are equal.

If the feed is a vapour at its boiling point, then the heat balance gives:

F + Vm = Vn

and the material balance gives:

Lm = Ln

In this case the liquid flowrates in the two sections are equal, while the vapour flow rate in the

enrichment section is greater than in the stripping section.

Reflux Ratio

The ratio of the reflux stream flow rate (Ln) to the top product flow rate (D) can be varied during

the operation of the column. This ratio is a very important parameter and is called the reflux ratio.

Reflux ratio R = LN+1

D =

Ln

D

Vn Ln

F

Vm

Lm

15

Thus

Ln = R D

and

Vn = (1 + R)D

Material Balances for Each Tray

Let us consider the condenser and the top two trays of the column:

e.g. 100 kmol/h of a mixture of 60 mol% hexane and 40 mol% octane must be separated to achieve

a top product containing 95 mol% hexane and a bottom product containing 95% octane. We now

need to specify the reflux ratio R. We will choose R=2 (the choice of the value of R will be

discussed later).

Carry out a material balance for tray N, and using the x-y diagram, determine xN and yN-1.

[HINT: remember (xN, yN) is on the equilibrium line.]

Material balance equation Vn yN-1 + Ln xD = Vn xD + Ln xN

yN-1 = Ln

Vn xN +

Vn − Ln

Vn xD

substituting for Vn and Ln

Ln

xN+1

= xD

Vn , yN = x

D

D

xD

tray N

Vn , y N-1

LN

xN

tray (N – 1)

Vn

yN-2

LN

xN-1

Ln

xN

16

yN-1 = R

1 + R xN +

xD

1 + R

yN-1 = 2

3 xN –

0.95

3 = 0.667 xN + 0.317

Using the x-y diagram, (xN,yN) = (xN,xD) so when y = 0.95 we can read off xN as 0.751.

Thus yN-1 = 0.667 × 0.75 + 0.317 = 0.818

Mark the point (xN,yN-1) on the x-y diagram.

Now carry out the material balance for tray N–1 and using the x-y diagram determine xN-1 and yN-2.

Material balance equation Vn yN-2 + Ln xN = Vn yN-1 + Ln xN-1

yN-2 = Ln

Vn xN-1 +

Vn xN-1 − Ln xN

Vn

From the material balance on tray N we can substitute for Vn xN-1 − Ln xN

yN-2 = Ln

Vn xN-1 +

Vn xN-1 − Ln xN

Vn

yN-2 = Ln

Vn xN-1 +

Vn − Ln

Vn xD

[ASIDE we could obtain this equation directly by considering an overall material balance for the

top two trays.]

Substituting for Ln and Vn as before:

yN-2 = R

1 + R xN-1 +

xD

1 + R

yN-2 = 0.667 xN-1 + 0.317

As before, using the x-y diagram, (xN-1,yN-1) = (xN-1,0.818) so when y = 0.818 we can read off xN as

0.442.

yN-2 = 0.667 × 0.457 + 0.317 = 0.612

Now plot the point (yN-2, xN-1) on the x-y diagram.

17

Notice that the equation between yN-2 and xN-1 is the same equation as that between yN-1 and xN.

This equation can be derived by considering a material balance for any tray n in the enrichment

section and the top section of the column:

Vn yn-1 = Ln xn + D xD

yn-1 = Ln

Vn xn +

D

Vn xD

yn-1 = R

1 + R xn +

xD

1 + R

This is the Enrichment Section Operating Line (ESOL). The composition on each tray can be

constructed by stepping between the ESOL and the equilibrium line.

Below the feed tray, the the ESOL no longer applies and we use the Stripping Section Operating

Line (SSOL).

Material balance Lm xm+1 = Vm ym + B xB

ym = Lm

Vm xm+1 –

B

Vm xB SSOL

For a liquid feed Vm = Vn and Lm= Ln + F

ym = R D + F

(1 + R)D xm+1 –

B

(1 + R)D xB

D

xD

Ln

xn

N

N-1

.

.

.

.

n+1

n

Vn

yn-1

B

xB

m

m-1

.

.

.

.

2

1

Vm

ym

Lm

xm+1

y

0 x 1

1

x

D

1 + R

xD

18

Consider the reboiler and bottom tray (tray 1). Unlike the condenser, the solution is only partially

boiled, so that the vapour flowing to the first tray has a different compostion to the liquid leaving

the first tray and going to the reboiler. In fact the reboiler acts like an equilibrium tray in the

column.

The vapour leaving the reboiler is assumed to in equilibrium with the bottom product, so (xB,yR)

lies on the equilibrium line, xB = 0.05 gives yR = 0.195 from the x-y diagram. The material balance

for the reboiler gives

Lm x1 = Vm yR + B xB

x1 = (1 + R)D

R D + F yR +

B

R D + F xB

x1 = 3 × 61.11

2 × 61.11 + 100 0.195 +

38.89

2 × 61.11 + 100 0.05

x1 = 0.170

The vapour leaving tray 1 is in equilibrium with x1 = 0.170, so from the equilibrium line on the x-y

diagram, y1 = 0.504.

Now we are stepping between the SSOL and the equilibrium line.

Lm , x2

Vm

y1

tray 1

Vm , yR

LN

xN

heat

B

xB

Lm

x1

reboiler

19

The intersection of the SSOL and ESOL occurs at the ‘feed line’. For a liquid feed this is at x = xF,

while for a vapour feed this is at y = xF.

PROOF:

ESOL Vn yn-1 = Ln xn + D xD

SSOL Lm xm+1 = Vm ym + B xB

Intersection at (xi,yi), so (xi,yi) lies on both the ESOL and SSOL

Thus xm+1 = xn = xi and ym = yn-1 = yi

ESOL Vn yi = Ln xi + D xD

SSOL Lm xi = Vm yi + B xB

Adding these two equations:

(Vn – Vm)yi = (Ln – Lm)xi + D xD + B xB

Substitute F xF = D xD + B xB

and for a liquid feed Vm = Vn and Lm = Ln + F

0 yi = – F x + F xF xi = xF

for a vapour feed Vm + F = Vn and Lm = Ln

F yi = 0 xi + F xF yi = xF

Liquid feed Vapour feed

Construction of the operating lines

y

0 x 1

1

x

D

1 + R

xD x

F x

B

y

0 x 1

1

x

D

1 + R

xD x

F x

B

20

Vapour-liquid equilibrium for hexane/octane at P=1 atm.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x (mol fraction hexane in liquid)

y (

mo

l fr

acti

on

hexan

e in

vap

ou

r)

N

N – 1

N – 2

R

1

21

We are now in a position to determine the number of plates required in a distillation column to

achieve a given separation. This involves a graphical construction which must be the best known

graphical procedure in Chemical Engineering and is called the McCabe-Thiele Method. A

detailed procedure is shown on the separate sheet.

The construction for the problem above is shown below, giving 4 equilibrium stages plus the

reboiler.

Vapour-liquid equilibrium for hexane/octane at P=1 atm.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x (mol fraction hexane in liquid)

y (

mo

l fr

acti

on

hexan

e in

vap

ou

r)

R

1

x

D

1 + R = 0.317

2

3

4

22

Choice of the Reflux ratio.

Decreasing the reflux ratio means decreasing the liquid flow rate to the top tray from the condenser

(Ln) and hence the liquid flow rates in throughout the column are also decreased. Since Vn = Ln +

D, this also implies decreasing the vapour flow throughout the column. Thus if the reflux ratio is

decreased, the amount of vapour generated in the reboiler and condensed in the condenser are

decreased. By implication the operating costs associated with steam and cooling water are reduced

when the reflux ratio is reduced. But what happens to the number of trays required to achieve a

given separation when the reflux ratio is reduced?

Consider the example above, but with the reflux ratio R reduced to a value of 0.5. The construction

is shown on the next page, with the SSOL intercept now at 0.633. The number of trays required is

increased from 4 to 5 trays. This means that the capital cost of the column will increase. The

choice of the reflux ratio is thus a balance between operating and capital costs.

Minimum Reflux Ratio

If the reflux ratio is reduced too much, it will no longer be possible to achieve the separation

required. The minimum reflux ratio is the value of R at which the number of trays approaches

infinity. This occurs when the intersection of the two operating lines occurs on the equilibrium

line.

Liquid feed Vapour feed

For the example above Rmin can be obtained from the x-y diagram:

xD

1 + Rmin = 0.81

Rmin = 0.173

The optimum reflux ratio is often expressed as a function of Rmin. Typically the optimum reflux

ratio is around 1.3 Rmin.

y

0 x 1

1

x

D

1 + Rmin

xD x

F x

B

y

0 x 1

1

x

D

1 + Rmin

xD x

F x

B

23

Vapour-liquid equilibrium for hexane/octane at P=1 atm.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x (mol fraction hexane in liquid)

y (

mo

l fr

acti

on

hexan

e in

vap

ou

r)

R

1

x

D

1 + R = 0.633

2

3

4

5