Distillation

Separation of a liquid mixture by partial vaporization (on introducing heat)

Distillation can be done in one stage:

- flash distillation (continuous, usually assuming an equilibrium stage)

- batch distillation (unsteady process)

Or in a column containing a number of stages (equilibrium or not) with counter-current flow due to a

reflux – rectification

Rectification can be also operated in a packed bed column. Equilibrium is described by Raoult’s law or

empirically (tables graph)

Notation

phase x … liquid

phase y … vapour

A … more volatile component

B … less volatile component

Raoult’s and Dalton’s laws imply the equilibrium relation

𝑦𝑘 = 𝜓𝑘𝑥𝑘 =𝑝𝑘0

𝑝𝑥𝑘; 𝑘 = A,B; 𝑥, 𝑦 are molar fractions

𝑝 … pressure

𝑝𝑘0 … vapour pressure of k (from Antoine eq.)

The two equilibrium relations can be combined

𝑦𝐴𝑦𝐵=𝑝𝐴0

𝑝𝐵0

𝑥𝐴𝑥𝐵

1) In two-component mixtures: 𝑦𝐵 = 1 − 𝑦𝐴 , 𝑥𝐵 = 1 − 𝑥𝐴

2) We introduce relative volatility 𝛼𝐴𝐵:

𝛼𝐴𝐵 =𝑝𝐴0(𝑇)

𝑝𝐵0(𝑇)

For similar compounds the ration of vapour pressures is approximately constant → 𝛼𝐴𝐵 = 𝑐𝑜𝑛𝑠𝑡

Therefore the equilibrium reads:

𝑦𝐴 =𝛼𝐴𝐵

1 + (𝛼𝐴𝐵 − 1)𝑥𝐴𝑥𝐴

Remark: 𝛼𝐴𝐵 can be determined as geometric mean value of 𝑝𝐴0(𝑇)

𝑝𝐵0 (𝑇)

for several temperatures between

boiling points of A and B or as geometric mean of 𝑦𝐴

𝑥𝐴⁄𝑦𝐵

𝑥𝐵⁄ taken from a table of liquid-vapour equilibrium.

Data from table or from an equilibrium relation

can be plotted

Flash distillation

Mixture is fed into a still, where it is heated, pressurized and led into a chamber where vapour separates

from the boiling liquid. The vapour is saturated (= equilibrium) and later may be condensed.

The balance scheme:

ℎ … molar enthalpy

For example

ℎ𝐹 = 𝑥𝐴𝐹ℎ𝐴 + (1 − 𝑥𝐴𝐹)ℎ𝐵 + ∆ℎ𝑚𝑖𝑥⏟ 𝑐𝑎𝑛 𝑏𝑒

𝑛𝑒𝑔𝑙𝑒𝑐𝑡𝑒𝑑

where ℎ𝑘 = 𝑐𝑝𝑘(𝑡𝐹 − 𝑡𝑟𝑒𝑓), 𝑐𝑝𝑘 at ⟨𝑡⟩ =𝑡𝐹+𝑡𝑟𝑒𝑓

2; 𝑘 = 𝐴, 𝐵

For total condenser → 𝑥𝐴𝐷 = 𝑦𝐴; �̇�𝑉 = �̇�𝐷

Balance on moles at steady state (two independent equations because of 2 components)

- overall(i.e. A+B): �̇�𝐹 = �̇�𝐿 + �̇�𝑉

- component A: �̇�𝐹𝑥𝐴𝐹 = �̇�𝐿𝑥𝐴 + �̇�𝑉𝑦𝐴

Equilibrium conditions (L and V are in equilibrium)

𝑦𝑘 = 𝜓𝑘𝑥𝑘; 𝑘 = 𝐴, 𝐵

𝜓𝑘 could be 𝑝𝑘0

𝑝 or found in tables, etc.

Also pressure and temperature of L and V are the same 𝑝𝐿 = 𝑝𝑉; 𝑇𝐿 = 𝑇𝑉

Enthalpy balances

- still: �̇�𝐹ℎ𝐹 + �̇�𝑖 = �̇�𝐿ℎ𝐿 + �̇�𝑉ℎ𝑉 + �̇�𝑒

- condenser: �̇�𝐷 = (ℎ𝑉 − ℎ𝑉)�̇�𝑉

Molar balance + equilibrium relation serve to find output composition 𝑥𝐴, 𝑦𝐴 and

Enthalpy balance serves to find the heat input �̇�𝑖 in the still and heat �̇�𝐷 to be removed in the condenser

(by using a coolant → heat exchangers)

Graphical solution

Component A balance can be rewritten as:

𝑦𝐴 = −�̇�𝐿�̇�𝑉𝑥𝐴 +

�̇�𝐿𝑥𝐴𝐹�̇�𝑉

… operating line

Intersection of operating line and equilibrium curve → 𝑥𝐴, 𝑦𝐴

Remark: notice that operating line passes through the point [𝑥𝐴𝐹 , 𝑥𝐴𝐹] on the diagonal can be used instead

of the intercept

Solution on the assumption of 𝛼𝐴𝐵 = 𝑐𝑜𝑛𝑠𝑡

Equilibrium relation is

𝑦𝐴 =𝛼𝐴𝐵

1 + (𝛼𝐴𝐵 − 1)𝑥𝐴𝑥𝐴

By combining with the overall and component balances on moles we get quadratic equation for 𝑥𝐴:

(𝛼𝐴𝐵 − 1)𝑥𝐴2 + [1 − (𝛼𝐴𝐵 − 1)𝑥𝐴𝐹

�̇�𝐹�̇�𝐿+ 𝛼𝐴𝐵 (

�̇�𝐹�̇�𝐿− 1)] 𝑥𝐴 −

�̇�𝐹�̇�𝐿𝑥𝐴𝐹 = 0

Batch distillation (unsteady operation)

An initial batch of mixture is subject to heating vapours are leaving continually the still, condensed and

collected. We need to:

a) determine time until a required composition of vapours is reached

b) heat needed in the still and heat removed in the condenser

Liquid in the still

- at the beginning: 𝜏 = 𝜏1, 𝑛𝐿 = 𝑛𝐹 , 𝑥𝐴 = 𝑥𝐴𝐹 , ℎ𝐿 = ℎ𝐹

- at the end: 𝜏 = 𝜏2, 𝑛𝐿 = 𝑛𝑊, 𝑥𝐴 = 𝑥𝐴𝑊, ℎ𝐿 = ℎ𝑊

Balance on moles at any time 𝜏1 ≤ 𝜏 ≤ 𝜏2

- overall:

0⏟𝑖𝑛𝑝𝑢𝑡

= �̇�𝑉⏟𝑜𝑢𝑡𝑝𝑢𝑡

+d𝑛𝐿d𝜏⏟

𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛

- component A:

0 = �̇�𝑉𝑦𝐴 +d(𝑛𝐿𝑥𝐴)

d𝜏

Equilibrium relation: we assume instantaneous equilibrium in the still: 𝑦𝑘 = 𝜓𝑘𝑥𝑘; 𝑘 = 𝐴, 𝐵

By combining the two balance equations and eliminating we obtain:

𝑦𝐴d𝑛𝐿 = d(𝑛𝐿𝑥𝐴) = 𝑛𝐿d𝑥𝐴 + 𝑥𝐴d𝑛𝐿

On rearranging and integrating within the values of 𝑛𝐿 and 𝑥𝐴 at times 𝜏1 and 𝜏2 we have:

∫d𝑛𝐿𝑛𝐿

𝑛𝐹

𝑛𝑊

= ∫d𝑥𝐴

𝑦𝐴 − 𝑥𝐴

𝑥𝐴𝐹

𝑥𝐴𝑊

Finally, assuming instantaneous equilibrium:

ln𝑛𝐹𝑛𝑊

= ∫d𝑥𝐴

𝜓𝐴𝑥𝐴 − 𝑥𝐴

𝑥𝐴𝐹

𝑥𝐴𝑊

This can be integrated provided we know 𝜓𝐴 as a function of 𝑥𝐴.

For example, from Raoult’s law 𝜓𝐴 =𝑝𝐴0(𝑇)

𝑝 depends on temperature 𝑇, which is expressed as a function of

𝑥𝐴 by using

𝑦𝐴 + 𝑦𝐵 = 1 =𝑝𝐴0(𝑇)

𝑝𝑥𝐴 +

𝑝𝐵0(𝑇)

𝑝𝑥𝐵

Or by assuming constant relative volatility we can take

𝜓𝐴 =𝛼𝐴𝐵

1 + (𝛼𝐴𝐵 − 1)𝑥𝐴

In the case of 𝛼𝐴𝐵 = 𝑐𝑜𝑛𝑠𝑡, the integral can be more conveniently evaluated as follows:

balance of A:

d(𝑛𝐿𝑥𝐴)

d𝜏= −�̇�𝑉𝑦𝐴

balance of B:

d(𝑛𝐿𝑥𝐵)

d𝜏= −�̇�𝑉𝑦𝐵

By dividing:

d(𝑛𝐿𝑥𝐴)

d(𝑛𝐿𝑥𝐵)=𝑦𝐴𝑦𝐵=

𝑦𝐴𝑥𝐴(𝑛𝐿𝑥𝐴)

𝑦𝐵𝑥𝐵(𝑛𝐿𝑥𝐵)

= 𝛼𝐴𝐵𝑛𝐿𝑥𝐴𝑛𝐿𝑥𝐵

On separation and integration from 𝑥𝐴𝑊, 𝑥𝐵𝑊 to 𝑥𝐴𝐹 , 𝑥𝐵𝐹:

ln𝑛𝐹𝑥𝐴𝐹𝑛𝑊𝑥𝐴𝑊

= 𝛼𝐴𝐵ln𝑛𝐹𝑥𝐵𝐹𝑛𝑊𝑥𝐵𝑊

Finally:

ln𝑛𝐹𝑛𝑊

=1

𝛼𝐴𝐵 − 1ln𝑥𝐴𝐹𝑥𝐴𝑊

1 − 𝑥𝐴𝑊1 − 𝑥𝐴𝐹

+ ln1 − 𝑥𝐴𝑊1 − 𝑥𝐴𝐹

This formula (or the earlier form with 𝜓𝐴) serves to calculate 𝑛𝑊 if 𝑥𝐴𝑊 is given or vice versa.

To calculate 𝑄𝑊 and 𝑄𝐷 one needs to write down mass and enthalpy balance within the whole time

interval from 𝜏1 to 𝜏2

Balance on moles

- overall: 𝑛𝐹 = 𝑛𝑊 + 𝑛𝐷

- component A: 𝑛𝐹𝑥𝐴𝐹 = 𝑛𝑊𝑥𝐴𝑊 + 𝑛𝐷⟨𝑥𝐴𝐷⟩

which serve to calculate 𝑛𝐷 and ⟨𝑥𝐴𝐷⟩

Enthalpy balance

- still:

𝑄𝑊 + 𝑛𝐹ℎ𝐹 = 𝑛𝑊ℎ𝑊 +∫ �̇�𝑉ℎ𝑉

𝜏2

𝜏1

d𝜏

- condenser:

∫ �̇�𝑉ℎ𝑉

𝜏2

𝜏1

d𝜏 = 𝑄𝐷 + 𝑛𝐷⟨ℎ𝐷⟩

Rectification in a column with stages

Feed is introduced to an internal plate, vapour rises to the top, then is condensed and partly removed as a

top product and partly returned as a reflux to the column, which generates a counter-current flow of

vapour and liquid. Liquid flows down the column to a reboiler, where it is heated and partly removed as a

bottom product and partly evaporated. The vapour then rises throughout the column.

- plates above the feed plate f form the rectifying section

- plates below the feed plate f form the stripping section

- condenser is assumed total

- reboiler is assumed as additional equilibrium stage

- outflowing vapour Vn and liquid Ln from a plate n (any in the column) are assumed to be in

equilibrium

Molar balances for the entire column at steady state:

- overall:�̇�𝐹 = �̇�𝐷 + �̇�𝑊

- component A: �̇�𝐹𝑧𝐴𝐹 = �̇�𝐷𝑥𝐴𝐷 + �̇�𝑊𝑥𝐴𝑊

Enthalpy balance of the column (assumed adiabatic)

�̇�𝑊 + �̇�𝐹ℎ𝐹 = �̇�𝐷ℎ𝐷 + �̇�𝑊ℎ𝑊 + �̇�𝐷

From molar balances → composition of top and bottom products.

Then ℎ𝐹 , ℎ𝐷 and ℎ𝑊 can be evaluated and from the enthalpy balance → �̇�𝑊 provided we know �̇�𝐷 which

is calculated from the balances of condenser as follows:

- molar balance: �̇�𝑉1 = �̇�𝐿0 + �̇�𝐷

upon introducing the reflux ratio 𝑅 =�̇�𝐿0

�̇�𝐷⁄

we can write : �̇�𝑉1 = (𝑅 + 1)�̇�𝐷 notice that: 𝑥𝐴0 = 𝑥𝐴𝐷 , ℎ𝐿0 = ℎ𝐷

- enthalpy balance:

�̇�𝑉1ℎ𝑉1 = �̇�𝐷 + �̇�𝐷ℎ𝐷 + �̇�𝐿0ℎ𝐿0⏟ �̇�𝑉1ℎ𝐷

and so

�̇�𝐷 = �̇�𝑉1(ℎ𝑉1 − ℎ𝐷) = (𝑅 + 1)�̇�𝐷(ℎ𝑉1 − ℎ𝐷)

From the balances of the entire column we can calculate energy requirements → �̇�𝑊 and �̇�𝐷

But of we ask for the number of plates N, we need to write down balances and separately and solve all

these equations together:

Balance on moles in plate n

overall: �̇�𝑉,𝑛+1 + �̇�𝐿,𝑛−1 = �̇�𝑉,𝑛 + �̇�𝐿,𝑛 𝑛 = 1,… ,𝑁

component A: �̇�𝑉,𝑛+1𝑦𝐴,𝑛+1 + �̇�𝐿,𝑛−1𝑥𝐴,𝑛−1 = �̇�𝑉,𝑛𝑦𝐴,𝑛 + �̇�𝐿,𝑛 𝑥𝐴,𝑛 𝑛 = 1,… ,𝑁

Equilibrium at plate n:

𝑦𝐴,𝑛 = 𝜓𝐴,𝑛𝑥𝐴,𝑛 𝑛 = 1,… ,𝑁

By solving these equations simultaneously (an a computer) N is found so that the composition at the feed

F and outputs D and W satisfy requirements on the entire column.

McCabe-Thiele diagram

A convenient graphical solution based on drawing operational lines and equilibrium curves in the x-y

diagram described below. The McCabe-Thiele method of graphical solution is based on the assumption

that the molar latent heat of evaporation of both A and B are the same. This in turn implies that the same

amount of moles is being evaporated on a given plate as there is condensed. And that implies a constant

molar flow of both the vapour and liquid in the rectifying section and in the stripping section.

Molar balance of the rectifying section

�̇�𝑉 = 𝑐𝑜𝑛𝑠𝑡 regardless of the plate number

�̇�𝐿 = 𝑐𝑜𝑛𝑠𝑡

then

𝑅 =�̇�𝐿�̇�𝐷

For a block of plates from 1 to n < f:

- overall: �̇�𝑉 = �̇�𝐿 + �̇�𝐷

- component A: �̇�𝑉𝑦𝐴,𝑛+1 = �̇�𝐿𝑥𝐴,𝑛 + �̇�𝐷𝑥𝐴𝐷

by using R: �̇�𝑉 = (𝑅 + 1)�̇�𝐷 and �̇�𝐿 = 𝑅�̇�𝐷

thus

𝑦𝐴,𝑛+1 =�̇�𝐿�̇�𝑉𝑥𝐴,𝑛 +

�̇�𝐷𝑥𝐴𝐷�̇�𝑉

=𝑅

𝑅 + 1𝑥𝐴,𝑛 +

𝑥𝐴𝐷𝑅 + 1

This relation is a straight line with slope =𝑅

𝑅+1 and intercept =

𝑥𝐴𝐷

𝑅+1. In addition, this line passes through

the point [𝑥𝐴𝐷 , 𝑥𝐴𝐷]. This line is called operational line (in

rectifying section).

A similar result can be obtained for the stripping section (i.e.

for stages from n to N+1, where n<f)

Here

�̇�𝑉′ = 𝑐𝑜𝑛𝑠𝑡

�̇�𝐿′ = 𝑐𝑜𝑛𝑠𝑡

These are different values than those in the rectifying section

(because of the feed).

- overall balance: �̇�𝐿′ = �̇�𝑉

′ + �̇�𝑊

- component A: �̇�𝐿′𝑥𝐴,𝑛−1 = �̇�𝑉

′𝑦𝐴,𝑛 + �̇�𝑊𝑥𝐴𝑊

- operational line:

𝑦𝐴,𝑛 =�̇�𝐿′

�̇�𝑉′ 𝑥𝐴,𝑛−1 −

�̇�𝑊𝑥𝐴𝑊

�̇�𝑉′

- slope: �̇�𝐿

′

�̇�𝑉′

- passes through the point [𝑥𝐴𝑊, 𝑥𝐴𝑊]

McCabe-Thiele diagram

The two operational line meet at a point, which corresponds to the feed plate. In addition, a third line –

called the feed line – passes through this intersection.

Practical construction is as follows:

a) draw the operational line in the rectifying section with the use of the point [𝑥𝐴𝐷 , 𝑥𝐴𝐷] on the

diagonal and the intercept 𝑥𝐴𝐷

𝑅+1

b) the operational line in the stripping section starts in the point [𝑥𝐴𝑊, 𝑥𝐴𝑊] but another point (or

slope) is inconvenient to construct (even though it would be possible). It is easier to draw the feed

line to find the intersection point and only then to draw the other operational line

How to construct the feed line

It will be explained a little later, at this moment, we consider the simplest situation: the feed line passes

through the point [𝑧𝐴𝐹 , 𝑧𝐴𝐹] and is parallel with the vertical axis (as shown in the Figure). This situation

corresponds to the feed being introduced at its boiling point.

Finding the number of plates

Each plate is represented by a

rectangular step between operational

lines and equilibrium curve →

equilibrium plate Point 1’ indicates

composition of streams L0 and V1

(output streams from plate 1) etc. see

scheme. The last plate is incomplete, its

fraction 𝑄𝑅̅̅ ̅̅

𝑃𝑅̅̅ ̅̅. The number of steps is N+1

(reboiler makes one stage), i.e. in the

Figure 𝑁 = 3 +𝑄𝑅̅̅ ̅̅

𝑃𝑅̅̅ ̅̅− 1 = 2 +

a little bit.

Limits of the operation of the column

The intersection of both operational lines can in general move along the feed line. Obviously, there are

two limiting situations:

a) the intersection moves down to touch the diagonal → then operational lines coincide with the

diagonal (and one another). N becomes minimal, 𝑥𝐴𝐷

𝑅+1 becomes zero → R → ∞. Infinite 𝑅 =

�̇�𝐿0�̇�𝐷

means total reflux (no top product is collected)

b) the intersection moves up adn touches the equilibrium line, N becomes infinite, 𝑥𝐴𝐷

𝑅+1 becomes

maximal → R becomes minimal → Rmin. This means that for R < Rmin we would NOT reach the

required composition of top product xAD even if N → ∞.

c)

In practice, total reflux means cheap columns (minimum plates) but no production, minimal reflux means

maximum production but infinitely expensive column. There is an economic optimum at 𝑅 = 𝛾𝑅𝑚𝑖𝑛

where 𝛾 is and empirical value (typically 1.3 < 𝛾 < 2)

Construction of the feed line revisited

To find the feed line, we simply combine component A balances in the rectifying a stripping sections

(here written in a difference form).

balance of A in rectifying section: �̇�𝑉𝑦𝐴,𝑛+1 − �̇�𝐿𝑥𝐴,𝑛 = �̇�𝐷𝑥𝐴𝐷

balance of A in stripping section: �̇�𝑉′𝑦𝐴,𝑛 − �̇�𝐿

′𝑥𝐴,𝑛−1 = −�̇�𝑊𝑥𝐴𝑊

By subtracting and assuming coordinates 𝑥𝐴+, 𝑦𝐴

+ for the intersection point we get:

(�̇�𝑉′ − �̇�𝑉)𝑦𝐴

+ − (�̇�𝐿′− �̇�𝐿)𝑥𝐴

+ = −�̇�𝑊𝑥𝐴𝑊 − �̇�𝐷𝑥𝐴𝐷⏟ −�̇�𝐹𝑧𝐴𝐹

Now we need to express the differences (�̇�𝑉′ − �̇�𝑉) and (�̇�𝐿

′ − �̇�𝐿) from the molar and enthalpy balances

of the feed plate. In the balances we assume that molar enthalpy in the vapour streams is the same at input

and output, and likewise for liquid streams.

Overall molar balance

�̇�𝐹 + �̇�𝐿 + �̇�𝑉′ = �̇�𝑉 + �̇�𝐿

′

or

�̇�𝐿′ − �̇�𝐿 = �̇�𝑉

′ − �̇�𝑉 + �̇�𝐹 ≝ 𝑞�̇�𝐹

Here 𝑞 is introduced so that

𝑞 ≝�̇�𝐿′ − �̇�𝐿�̇�𝐹

=�̇�𝑉

′ − �̇�𝑉�̇�𝐹

+ 1 (a relative flow of liquid)

However, importance of 𝑞 becomes clear from the enthalpy balance

�̇�𝐹ℎ𝐹 + �̇�𝐿ℎ𝐿𝑓 + �̇�𝑉′ℎ𝑉𝑓 = �̇�𝑉ℎ𝑉𝑓 + �̇�𝐿

′ℎ𝐿𝑓

or

ℎ𝐿𝑓 (�̇�𝐿′ − �̇�𝐿)⏟ 𝑞�̇�𝐹

= ℎ𝑉𝑓 (�̇�𝑉′ − �̇�𝑉)⏟

(𝑞−1)�̇�𝐹

+ �̇�𝐹ℎ𝐹

and finally

𝑞 =ℎ𝑉𝑓 − ℎ𝐹

ℎ𝑉𝑓 − ℎ𝐿𝑓

properties of the vapour a liquid streams may be approximately take at composition of the feed F:

𝑞 =ℎ𝑉𝐹 − ℎ𝐹ℎ𝑉𝐹 − ℎ𝐿𝐹

ℎ𝐹 … enthalpy of the feed as is

ℎ𝑉𝐹 … enthalpy of the feed as saturated vapour

ℎ𝐿𝐹 … enthalpy of the feed as saturated liquid

Interpretation

𝑞 tells us what is the enthalpic state of the feed, for example, if ℎ𝐹 = ℎ𝐿𝐹 then the feed is at its boiling

point and

𝑞 =ℎ𝑉𝐹 − ℎ𝐿𝐹ℎ𝑉𝐹 − ℎ𝐿𝐹

= 1

The feed line

Knowing the difference flows �̇�𝑉′ − �̇�𝑉 and �̇�𝐿

′ − �̇�𝐿 in terms of 𝑞 we can write the equation for the feed

line:

𝑦𝐴+ =

𝑞

𝑞 − 1𝑥𝐴+ −

𝑧𝐴𝐹𝑞 − 1

which has slope =𝑞

𝑞−1 and passes through the point [𝑧𝐴𝐹 , 𝑧𝐴𝐹].

The enthalpic states of the feed on terms of and are as follows:

𝑞 𝑞

𝑞 − 1

1 unsaturated liquid (below boiling point) > 1 > 1

2 saturated liquid (at b. p.) 1 ∞

3 mixture of liquid and vapour 0 < 𝑞 < 1 < 0

4 saturated vapour (at dew point) 0 0

5 oversaturated vapour (above d.p) < 0 (0,1)

Remark: notation of the feed composition reflects the flexibility of phase conditions.

Feed line orientation in the graph

Feed line in sector 1 corresponds to unsaturated liquid, in position 2 to saturated liquid, etc. → see the

Table.

Higher enthalpic state of the feed means preheating of the feed, which implies lower number of plates

below the feed plate and increased number above it.

Remark: In reality, plates do not operate as equilibrium plates → correction by efficiency. either overall

efficiency is used

𝐸𝑜𝑣𝑒𝑟𝑎𝑙𝑙 =𝑁

𝑁𝑟𝑒𝑎𝑙

𝑁𝑟𝑒𝑎𝑙 … number of real plates

or a Murphree efficiency of a plate is used → see Absorption.