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Transcript Distillation

Prepared by Dr.Nagwa El Mansi
Chemical Engineering Department
Cairo University
Fourth year
References:1-Coluson and Richerdson, Chemical Engineering vol ,
vol II , vol III.
2- Geancoplis, Principles of Unit Operation.
3- Mc-Cabe and Smith, Unit operations for Chemical
4- Traybal, Mass Transfer Operations.
5- Sherwood, Mass Transfer.
6-Perry’s , Chemical Engineering.
7- “Separation Process Principles”, 2nd ed, Seader et’al .
8- Site on Google search, Separation Processes.
Distillation:Is a method of separating homogenous mixture
based on differences in boiling points.
Distillation is done by:a-partial vaporization ( liq → vap)
b-partial condensation (vap → liq)
c- changing in pressure ( gas → liq)
Distillation has a number of applications:1- It is used to separate crude oil into more
fractions for specific uses such as transport,
power generation and heating.
2-Water is distilled to remove impurities, such as
salt from seawater.
3- Air is distilled to separate its components
oxygen, nitrogen, and argon for industrial use.
Separation in distillation depend on:Relative volatility ( α ) :α = α AB
= o
Where :-
A= more volatile component (mvc).
B = less volatile component (lvc).
PAo = vapor pressure of more volatile component (mvc).
PBo = vapor pressure of less volatile component (lvc).
When α > 1 or α < 1
separation takes place.
But when α = 1 there is no separation.
Vapor-liquid Equilibrium:[ When the liquid phase behaves as an ideal solution all
molecules have the same size; all intermolecular forces are
equal; the properties of the mixture depend only on the
properties of the pure components of the mixture].
Phase Diagram (Binary Mixture)
In our analysis of phase diagram, we shall consider only Two
component mixture, e.g. A (more volatile) and B (less volatile).
There are 2 types of phase diagram:
constant pressure or constant temperature.
Constant Pressure Phase Diagram
Figure(1) shows a constant pressure phase diagram for an ideal
solution (one that obey ) Raoult's Law
A typical equilibrium curve
for a binary mixture on x-y
diagram. It contains less
information than the phase
diagram (i.e. temperature is
not included), but it is most
commonly used. It is useful
for graphical design in
determining the number of
theoretical stages required
for a distillation column.
Constant Temperature (Isothermal) Phase Diagram:Figure(2)shows the constant temperature phase diagram.
The constant pressure phase diagram is more commonly
used in the analysis of VLE, but the constant temperature
Phase diagram is also useful in the analysis of solution
behavior that deviates from Raoult's Law.
[ From this Figure (constant temperature phase diagram)
we see that the more volatile liquid will have a higher
vapor pressure (i.e. pA at xA = 1.0). Note also the regions
of vapor-only, liquid-only and vapor-liquid mixture. ]
Effects of Increased Pressures:Although most distillations are carried out at
atmospheric or near atmospheric pressure, it is not
uncommon to distill at other pressures. High pressure
distillation (typically 3 - 20 atm). At elevated pressures,
the vapor phase deviates from ideal gas behavior, and
modifications to the VLE data is required. After pressure
P3, the critical pressure of the more volatile component
is exceeded, and there is no longer a distinction between
vapor and liquid. Distillation is no longer possible
beyond this point. The majority of distillations are
carried out at pressures below 70% of the critical
Abnormal mixtures:1- Azeotropic Mixture:Very large deviations from ideality lead to a special
class of mixtures known as azeotropes , azeotropic
mixtures, or constant-boiling mixtures. Azeotrope is
a special class of liquid mixture that boils at a constant
temperature at a certain composition. At this condition,
it behaves as if it was one component with one
constant boiling point. A boiling liquid mixture at the
azeotropic composition produces a vapor of exactly the
same composition, and the Liquid does not change its
composition as it evaporates.
Two types of azeotropes are known:minimum boiling and Maximum boiling
a- Minimum-boiling azeotropes:
a- One of the best known minimum-boiling azeotrope is the
ethanol-water system which at 1 atm occurs at 89.4 mole%
ethanol and 78.2 oC.
b- carbon-disulfide - acetone (61.0 mole% CS2, 39.25 oC,
1 atm)
c- benzene - water (29.6 mole% H2O, 69.25 oC, 1 atm)
b-Maximum-boiling azeotropes:a-Hydrochloric acid - water (11.1 mole% HCl, 110 oC, 1 atm)
b-Acetone - chloroform (65.5 mole% chloroform, 64.5 oC,
1 atm)
a-Minimum Azeotrope:Figure(5) show the constant pressure
phase diagram plus equilibrium curve
for a minimum-boiling azeotropic
mixture of carbon disulfide (CS2) and
acetone. At point L, the concentration in
the vapor phase is the same as the
concentration in the liquid phase ( y = x ),
and a = 1.0. This concentration is known
as the azeotropic composition (0.61 mole
fraction CS2). At this point , the mixture
boils at a constant temperature (39.25
oC under 1 atm) and without change in
composition. On the equilibrium diagram,
it can be seen that at this point,
the equilibrium curve crossed the diagonal
b-Maximum boiling Azeotrope:The Figure(6) shows the constant
pressure phase diagram plus
equilibrium curve for a maximum
boiling azeotrope mixture of
acetone and chloroform. The
Azeotropic composition is 0.345
mole fraction acetone.
Point L in the Figures is now a
minimum on the constant
temperature phase diagram, and a
maximum (64.5 oC, under 1 atm)
on the constant pressure phase
2- Partial liquid miscbility:Some substances exhibit such
large deviations from ideality
that they do not dissolve
completely in liquid state, e.g.,
3-Complete immisciblity Mixtures:The mutual solubility of some
Liquids is so small that they can be
Considered substantially insoluble
(Figure(8)) e.g., hydrocarbon and
water. The vapor pressure of either
component cannot be influenced by
the other and each one exerts its
vapor pressure at the prevailing
temperature. When the sum of the
separate vapor pressures equals the
total pressure, the mixture boils at a
temperature lower than the boiling
point of each component:PA + PB = PT , or P◦A + P◦B = PT
Calculation of Vapor/Liquid Equilibrium:1- By experimental methods:- (under certain P&T to get
y = f(x)).
2-Published data.
3-Prediction of K values.
For binary mixture:o
Pi = P . x i ------ (Raoult's law)
 PT . y i ------- ( Dalton's law)
 yi =
xi = Ki xi
where Pi :- partial pressure of component i
Pio :- vapor pressure of component i
Ki =
 K-value for component i
x i , yi :- mole fractions of liquid and vapor respectively
PT =
 Pi =
 i . xi
For binary mixture ( A & B ):PT = PA + PB
PT = PAo . x A + PBo . x B
 PT = P . x A + (1- x A ) P , where x A + x B = 1
PT = PAo . x A + PBo - PBo . x A
PT - PBo
 xA = o
PA - PBo
and y A =
x A ----(equilibrium relation)
Also, another form of equilibrium relation by using relative volatility( )
PAo y A /x A
y A /x A
y A (1-x A )
α= o =
PB y B /x B (1-y A )/(1-x A ) x A (1-y A )
 α x A - αx A y A = y A - y A x A
α x A = (αx A - x A +1) y A
α x A = [(α - 1) x A +1] y A
yA =
α xA
------ (equilibrium relation)
[(α - 1) x A +1]
For multi-component system:For ideal system:Pio
yi = k i . x i =
For non-ideal system:Pio
y i = k i . i . x i =
.  i . xi
where  i is the activation coefficient = measure of non-ideality
One Stage Distillation
1-Simple (differential) distillation (ASTM):We will consider a binary mixture of A (more volatile)
and B (less volatile). The set-up is as shown in the
Figure(9) . The system consist of a batch of liquid feed (F)
(fixed quantity) with a composition( xF) inside a kettle (or
still) fitted with heating element or steam jacket, and a
condenser to condense the vapor produced. The condensed
Vapor is known as the distillate (D) with a composition (yD) .
The distillate is collected In a condensate receiver. The
liquid remaining in the still is known as the residual (W)
With a composition ( xW).
At any time t, the amount of liquid in the still is L , with
mole fraction of A in the liquid being (x). After a small
differential time (t + dt) , a small amount of vapor dL is
produced, and the composition of A in the vapor is (y)
(mole fraction). The vapor is assumed to be in equilibrium
with the residue liquid . The amount of liquid in the still is
thus reduced from L to (L - dL), while the liquid
composition changed from x to (x - dx). See the following
Figure (10):-
Performing a material balance on A :
Initial amount in still  Amount left in still  Amount vaporized
We have,
L . x = ( L - dL) ( x - dx ) + dL .y
L . x = L . x - dL.x - L.dx - dL.dx + y . dL
Neglecting the term dx dL , the equation reduces to :
L . dx = y . dL - x dL = ( y - x ) dL
By separating variables then integrate:
dL W dx
L x F (y-x)
 Ln
W L = x (y-x)
x (y-x) --------(1)
There are two methods to solve equation (1):First method:- (by integration)
----- (equilibrium relation)
(α - 1) x+1
 Ln
x (y-x) -------- (1)
W (
- x)
(α - 1) x+1
x F (1 - x W )
1- x W
+ Ln
x W (1 - x F )
1- x F
After performing the integration assume xw,
(usually xw unknown) .
If left hand side=right hand side your assumption
is correct if not repeat your assumption.
Second method :- (graphical integration)
x (y-x) -------- (1)
 Ln
= Area under the curve (note x W is unknown)
Steps for calculation the correct xw:1-plot x-y diagram.
2- assume xw.
3-costruct the previous table between x and
1/(y*- x).
4 - plot x vs 1/(y*- x) as shown in the last figure.
5- calculate the area under the curve = Ln F/W.
6- if area = Ln F/W , your assumption of xw is correct
if not reassume xw.
7- finally calculate the correct xw.
Equation(1) can be used to
determine (xw) if (F, xF , D)
are known.
Also the average distillate
composition can be
Calculated ( yD ) by simple
material balance:F = D + W --------(2)
F xF = D yD + W xW -------(3)
By solving the three
equations all the quantities
and compositions can be
2-Flash (equilibrium) distillation:A single-stage continuous operation where a liquid mixture is
partially vaporized, then flows through a pressure reducing valve
to the separator. In the separator, separation between the vapor
and liquid takes place. How much of A is produced in the vapor
(and remained in the liquid) depends on the condition of the
Feed, (see Figure(13)).
Vapor amount (V) with composition (y), and a liquid amount (L)
with composition (x) are produced.
The two streams leaving the flash drum (y and x )are in
equilibrium with each others.
Flash distillation is done by:1-Flash vaporization (liquid →heating→throttling valve →
Separation ).
2-Flash condensation ( vapor →cooling →separation ).
3- Feed at high pressure → change to low pressure.
Some times flash distillation is used as a method for
changing conditions of a mixture.
A single-stage flash operation can rarely produce the
required purity or fractional recoveries. An obvious but
inefficient approach is to apply a series of flash
separators condensing part of the vapor and boiling part
of the liquid products from successive stages.
Its preferred to use systems with high relative volatility in
flash distillation to obtain considerable separation .
Operating line equation for flash distillation:For binary mixture(A &B)
Overall Material Balance:F=L+V
Component Material Balance:F. x F = L . x + V . y
 ( L + V ) xF = L . x + V . y
 L ( x F - x) = V ( y - x F )
(y - x F )
(y - x F )
 =
or =
-------(4) (operating line equation)
( x F - x)
( x - xF )
This line lies between two points (x,y) and (x F , x F )
and has a slope ( )
Figure(14) shows it's the graphical representation.
Multi-flash distillation:F=L+V
Component Material Balance:F. x Fi = L . x i + V . yi
( L + V ) x Fi = L . x i + V . yi
L ( x Fi - x i ) = V ( yi - x Fi )
(y - x Fi )
= i
-------- (5)
( x Fi - x i )
Equilibrium relation is yi = k i . x i or can be written as x i = yi / k i
substitute in equation (5) :(yi - x Fi )
( x Fi - yi / k i )
( x Fi - yi / k i ) = (yi - x Fi )
x Fi (1+
) = yi (1 +
V. k i
--------- (6)
(1 +
V. k i
( 1+
 yi  x Fi
( ki +
( 1+
or x i  x Fi
where :-
x  x
------- (7)
 1 and ,
( ki +
( 1+
y  x
(1 +
V. k i
( 1+
Stepes for calculation of multi-component distillation:Pio
1- Calculate k i values for each component ( k i 
at certain temperature.
2- Assume L/V.
3- Calculate
4- If
y  x
V  1
V ki
( 1+
 1 repeat your assumption of L/V .
 1  your assumption is correct then calculate x i  1
( solution is by trial and error)
Temperature calculations inside flash drum:-
Figure(16) shows an enthalpy balance for continuous
operation. The material and enthalpy balance are:-
F. x F = L . x + V. y
F . h F + Q = L . h L + V. H V -----(8)
where :h F = feed enthalpy
Q = enthalpy gained by the heater
h L = x [cp L,A (T-Tr )] + (1-x) [cp L,B (T-Tr )] + H mix -----(9)
H V = y[cp v,A (T-Tr ) +λ A ] + (1-y) [cp v,B (T-Tr ) + λ B ] -----(10)
T = operating temperature inside the drum
Tr = reference temperature
cp L = liquid heat capacity , energy/mol.degree
cp V = vapor heat capacity , energy/mol.degree
H mix  heat of mixing (neglected for ideal system)
λ A , λ B = latent heats of vaporization at Tr , energy/mol
Steps for calculation T :1- assume T.
2- calculate Ki.
3- assume L/V.
4- calculate yi .
5- check ∑ yi =1.
6- calculate xi .
7- check the enthalpy balance,
if (LHS = RHS )→ your assumption is correct, if not
reassume T .
Usually, T is assumed between the bubble and the dew
Points of the mixture.
Calculations of dew point and bubble point:A-Bubble point:-
yi = k i . x i
(for the first bubble x i = x F )
. x Fi
given, x Fi , P assume Tbubble and calculate k i
 
x Fi =
x FA +
x FB  1
if 
x Fi = 1  your assumption is correct
if not repeat your assumption
B- Dew point :yi
xi =
 xi = 1 =
(for the first liquid droplet x i = x Fi )
given, x Fi , P assume Tdew and calculate k i
  xi = 1 =
= 1  your assumption is correct
if not repeat your assumption
Trials to calculate bubble
Trials to calculate dew
3- Steam distillation:-
It is a method for distilling organic compounds which are heat
sensitive materials (contaminated by traces of non volatile
impurities).This process involves using bubbling steam (which
is completely insoluble with the organic compound) through a
heated mixture of the raw material to give maximum contact
between steam and the compound.
By Raoult's law, some of the target compound will
vaporize (in accordance with its partial pressure)leaving
Impurities and condensed steam in the still. The vapor mixture
(carried by steam) is cooled and condensed, usually yielding a
layer of organic compound(c) and a layer of water(w).
The vapor pressure of each component (c&w) cannot be
influenced by the presence of the other and each one exerts
its vapor pressure at the Prevailing temperature.
When the sum of the separate vapor pressures equal the total
pressure, the mixture boils at a lower temperature (lower than
boiling points of c & w). The use of steam reduces the partial
pressure of feed components and thus permits there
vaporization at temperature below there normal boiling point.
PW = P
, yW
PW = partial pressure of water
PWo = vapor pressure of water
Also for the component (c):o
PC = P
, yC =
PC = partial pressure of water
PCo = vapor pressure of water
 PT = PW + PC
(at boiling)
At boiling:PC
moles of materials n C
moles of water
y .P
moles of (c)
 C T 
PW PT  PC y W . PT moles of (w)
moles of (c) × M C
mass of (c)
moles of (w) × M W mass of (w)
mass of (w) = mass of steam used as a carrier.
Total amount of steam used = Steam used for heating  m cp (Tb mix -Tinital )
+ Steam used as a carrier
+ Steam used for vaporization  m
Advantage of steam distillation:1-Distillation takes place under low temperature.
2-Prevent decomposition of thermally sensitive
3- Cheap and economic.
4-Avalability of heating medium.
Fractional Distillation or Differential Counter-Current
Vapor/ Liquid Operations (Multi-stage Distillation):-
To obtain high purity use multi-stage arrangement
These stages can be combined in one column as
shown in Figure(23)
Steam Distillation:Steam distillation refers to a process
in which live steam is in direct contact
with the distilling system. The basis
of steam distillation rest on the fact
That water forms immiscible mixtures
with most organic substances(most
commonly employed in petroleum
refining operation) and these mixtures
will boil at a temperature below that
of either water or the other materials.
Vaporization takes place at a temp.
below its normal boiling point. Steam
is widely used because of its energy
level, cheapness, and availability.
Total and partial condenser arrangements:1-Total condenser:In total condenser all saturated vapors at the top of the
distillation column
condensed to saturated
 y1  x0  xD
QC = V1 λ mix
QC = ( L0 + D) λ mix
= ( +1) λ mix
= ( R +1 ) λ mix
Where:Q C = condenser duty
V1 = vapor flow rate at the top plate in the distillation column
y1 = vapor composition at the top plate in the distillation column
D = distillate flow rate
x D = distillate composition leaving condenser
L 0 = liquid flow rate returning back to the top plate
x 0 = liquid composition returning back to the top plate
= reflux ratio
2- Partial Condenser:Partial condenser
acts as one plate
with yD
in equilibrium with
top plate
condensate x0
yD = KD x0
and yD = xD
Types of Reflux:1- Cold Reflux:- Liquid stream returning back to the
Column is less than bubble point of the mixture.
2- Hot Reflux:-Liquid stream returning back to the
Column is more than bubble point of the mixture.
3- Recirculating Reflux or pumping around:A side stream is withdrawn from the column to be
cooled and return back to the column.
Reboilers:1- Internal Reboiler:The tubular heat exchanger built into the bottom of the
tower (Figure(33))
provides large surface area
(no shell), but Cleaning
requires a shut-down
of the distillation operation.
Liquid level is controlling
for small capacity towers
and for materials which
aren’t corrosive and
2-External Reboilers:A-Thermosiphon Reboiler:The vertical thermosiphon
reboiler as shown in
Figure(34) with the heating
medium outside the tubes,
can be operated so as to
vaporize all the liquid
entering it to produce a
vapor of the same composition as the residue product,
in which no enrichment is provided.
B- Kettle Reboiler:The kettle reboiler
shown in Figure (35),
with heating medium
inside the tubes,
provides a vapor to the
tower essentially in
equilibrium with the
residue product and
Thus it behaves like
a theoretical stage.
It’s large in size , better
to control, long residence
time , not used for thermally
sensitive materials
Pipe still heater:Sometimes we use a
Furnace or a pipe
still heater instead
of the reboiler.
Advantages:1-Used when no steam
is available(heating media is
hot oil or hot gases).
2- high temperatures is
2- Not suitable for
sensitive materials.
Distillation Calculations :Overall Material balance (O.M):F=D+W
Component material balance(CMB):F xF = D xD + W xW
Overall Heat Balance (OHB):F h F + QC = D h D + W h W + Q r
where:- QC = condenser duty
 QC = V λ
= V ( y λ + (1-y ) λ )
1 mix 1 1 1
1 2
V =L +D
 QC =(L + D) λ
water water
water out water in
and Qr = reboiler duty = m
steam steam
Methods for calculation number of stages in distillation
(1)Sorel plate to plate calculations:For upper section:OMB
Vn+1 = L n + D ----(11)
Vn+1 y n+1 = L n x n + D x D --(12)
Vn+1 H n+1 = L n h n + D h D  QC ---(13)
Where:H n+1 = f (y n+1 ) and h n = f (x n )
For bottom section:OMB
L'm+1 = V'm + W ---(14)
L'm+1 x'm+1 = V'm y'm + W x w ---(15)
L'm+1 h'm+1 + Q r = V'm H'm + W h W ----(16)
Where:h'm+1 = f (x'm+1 )
H'm = f (y'm )
Top or rectifying section ( total condenser conditions):First plate ( n = 1 )
V2 = L1 + D ---(OMB)
V2 y 2 = L1 x1 + D x D ---(CMB)
V2 H 2 = L1 h1 + D h D  Q C ---(HB)
where:- y1  k1x1 --- (Eq.relation)
Calculation steps :1-Assume V2 then get L1 from
(OMB) equation.
2- Calculate y2 after calculating
x1 ( from Eq.relation → y1 = k1 x1 )
3- If LHS = RHS (in HB equation) → your
assumption is correct if not repeat your assumption.
Second plate ( n = 2 )
V3 = L 2 + D ---(OMB)
V3 y3 = L 2 x 2 + D x D ---(CMB)
V3 H 3 = L 2 h 2 + D h D  QC ---(HB)
where:- y 2  k 2 x 2 --- (Eq.relation)
same previous steps for calculation V3 and L 2
then calculation of y3 , then checking your assumption
from heat balance equation.
Repeat your calculations till you reach
x n = x f or x n  x f
then calculate the number of stages in the upper section
Bottom section (Using partial vaporizer = Reboiler ):Reboiler acts as one
theoretical stage outside
the distillation column.
The two streams leaving it
are in equilibrium with
each other:yr = kr xw
For Reboiler (m = 0 )
L'1 = V'r + W ---(OMB)
L'1 x'1 = V'r y'r + W x w ---(CMB)
L'1 h'1 + Q r = V'r H'r + W h W ---(HB)
Steps for calculations:1-Assume Vr then get L’1 from(OMB) equation.
2- Calculate x’1 after calculating yr
( from Eq.relation → yr = kr xw )
3-Substituting by Vr , H’r , L’1 , h’1
in the (HB) equation.
(after calculating the correct
temperature inside the reboiler)
4- If LHS = RHS
→ your assumption is correct if not repeat
your assumption.
First plate 1' from the bottom (m = 1 )
L'2 = V'1 + W ---(OMB)
L'2 x'2 = V'1 y'1 + W x w ---(CMB)
L'2 h'2 + Q r = V'1 H'1 + W h W ---(HB)
Where :  y'2 = k'2 x 2
Same steps were followed till the vapor composition equal
or more than feed composition (y m  x F ).
All the above depends on the reflux ratio, in which QC and Qr
were calculated.
2- Ponchon Savarit method (Graphical representation
of Sorel method ) :-
F = D + W -----(OMB on the distillation column)
Fx F = D x D + W x W -----(CMB)
F h F + Q r = D h D + W h W + QC ----(HB)
F h F = D h D + W h W + QC - Qr
 F h F = D (h D +
) + W (h W )
put Q' = ( h D +
) & Q" = ( h W )
 F h F = D Q' + W Q" ---------(17)
Substitute in (CMB eq.) with (OMB eq.): (D + W) x F = D x D + W x W
 D ( x F - x D ) = W (x W - x F )
(x W - x F )
( xF - xD )
Substitute in eq. (17) with (OMB) eq.:(D + W) h F = D Q' + W Q"
D ( h F - Q') = W ( Q" - h F )
( Q" - h F )
( h F - Q' )
(x W - x F )
( Q" - h F )
( xF - xD )
( h F - Q' )
Equation  20  is an operating line equation for the hole column
between three points (x D , Q' ) , ( x F , h F ) , ( x W ,Q").
For top section:Vn+1 = L n + D ---(11)
Vn+1 y n+1 = L n x n + D x D ---(12)
Vn+1 H n+1 = L n h n + D h D  QC ---(13)
 Vn+1
 Vn+1
H n+1 = L n h n + D ( h D 
H n+1 = L n h n + D Q' -----(21)
By substituting eq.(11) in eq.(12)
 (L n + D) y n+1 = L n x n + D x D
 D ( y n+1  x D ) = L n ( x n  y n+1 )
( x n  y n+1 )
( y n+1  x D )
By substituting eq.(11) in eq.(21)
 ( L n + D) H n+1  L n h n + D Q'
 D ( H n+1  Q' ) = L n ( h n  H n+1 )
( h n  H n+1 )
(H n+1  Q' )
( x n  y n+1 )
( h n  H n+1 )
( y n+1  x D )
(H n+1  Q' )
Equation  24  is an operating line equation for the top section
between three points (x n , h n ) , (y n+1 , H n+1 ) , (x D , Q' ).
For bottom section:L'm+1 = V'm + W ---(14)
L'm+1 x'm+1 = V'm y'm + W x w ---(15)
L'm+1 h'm+1 + Q r = V'm H'm + W h W ----(16)
L'm+1 h'm+1 = V'm H'm + W ( h W )
 L'm+1 h'm+1 = V'm H'm + W Q" ----(25)
By substituting eq.(14) in eq.(15)
 (V'm + W ) x'm+1 = V'm y'm + W x w
W ( x'm+1  x w ) = V'm ( y'm  x'm+1 )
W ( y'm  x'm+1 )
V'm ( x'm+1  x w )
By substituting eq.(14) in eq.(25)
 ( V'm + W) h'm+1 = V'm H'm + W Q"
 W ( h'm+1  Q") = V'm ( H'm  h'm+1 )
W ( H'm  h'm+1 )
( h'm+1  Q'')
( H'm  h'm+1 )
W ( y'm  x'm+1 )
V'm ( x'm+1  x w )
( h'm+1  Q'')
Equation (28) is an operating line eq. for the bottom
between three points (y'm , H'm ) , ( x'm+1 , h'm+1 ), ( x w , Q'')
Enthalpy-composition diagram:- (Ponchon-Savarit
Method )
1-Takes into account latent heats, heats of mixing &
sensible heats.
2-No assumption of molal flow rates.
3-Graphical procedure combining enthalpy & material
4- Provides information on condenser & reboiler duties.
The diagram at a given constant pressure is based on
reference states of liquid and temperature such as 273K.
Use enthalpy concentration data:The data that require for enthalpy concentration
diagram are:
1-Heat capacity of the liquid, Cp
2-Boiling temperature, Tb
3-Latent heats of vaporization, λ
4-Heat of solution, ∆Hmixing
The saturated liquid line in enthalpy h kJ/kg can be
obtained from the following :-
hL = x [cpL,A (T-Tr )] + (1-x) [cpL,B (T-Tr )] + Hmix -----(9)
The saturated vapor line can be obtained from
the following equation below:HV = y[cpv,A (T-Tr ) +λA ] + (1-y) [cpv,B (T-Tr ) + λB ] -----(10)
x & y = mol or mass fractions of liquid & vapor.
T= boiling point of temperature
Tr = reference temperature
C p,L & C p,v = liquid & vapor heat capacity
∆Hmix = heat of solution at Tr ( neglected for ideal
Again the difference in enthalpy between the
streams passing each other is constant, with the
enthalpy of this stream being Q’ = hD + Qc / D ,
where Q c is the condenser duty in kJ/h.
Also, Q C  V1 λ = ( L 0 + D )
+ 1 ) = D ( R + 1)
 Q C = D ( R + 1) λ
= D(
= ( R + 1 ) λ ------(29)
If R is assumed or given, calculate Q C then Q' = ( h D +
which is the difference point ( or polar ) of the top section.
This difference Q’ is a common operating point
for all values of yn+1 and xn in enriching section
of the distillation tower that having an enthalpy
[Q’ = hD + Qc / D ] and a composition of xD . The
intersection of y1 is shown in the diagram having
the composition y1= xD for a total condenser. The
liquid x1 is in equilibrium with y1 and is located by
drawing a tie line through y1 intersecting the
saturated liquid line at x1. Next, a line is drawn as
x1Q’ , which is intersects saturated vapor line at y2.
following the operating line equation of the top
Section (eq.(24)).
Also , the difference point (or polar) in the stripping
section is called Q”, having an enthalpy Q”= ( hw -Qr/W)
and a composition xw . This point Q” is an operating
point just as Q’ for the enriching section. The point Q”
is plotted below.
Starting at point xw a tie line is drawn through this point
intersecting the saturated vapor line at yr, is in
equilibrium with xw . Next, the line yr Q” is drawn,
which intersects the saturated liquid line at x’1.
Hence, x’1 must be on the line between yr and Q”.
Next, the tie line x’1y’1 is drawn. This process is
continued in stepping off the theoretical plates.
Steps for calculation the theoretical number of stages:1-Draw the enthalpy-concentration plot & the x-y
equilibrium plot on the same graph.
2-Determine the points xF , xD , xW.
3-Use the following equation to calculate Q’= hD + Qc / D.
4-Locate point Q’ at (xD , Q’)
5-Locate point y1 at (xD , H1)
6-Locate point x0 at (xD , hD) → for total condenser.
7- use the following equation to calculate
Q” = hw - Qr/W.
8-Draw the locus of Q” which is a vertical line from xw,
{ point Q” at (xw , Q”)}.
9-Draw the line from Q’ to intersect hF (enthalpy of
saturated liquid at xF and the vertical line of xW). The
point of intersection shows Q” .
10-From point y1, draw a line down to 45o line. Then,
draw a horizontal tie line to cut the x-y equilibrium
curve. Draw a line up to the liquid enthalpy curve. This
point is x1.
11-Plot an operating line from x1 to Q’, intersecting the
vapor enthalpy curve at y2. From y2, draw a vertical line
down to the equilibrium curve. Draw yet another line,
go up to obtain x2; repeat the process until you exceed xF
12- Also, from point xw draw a tie line connecting
Between xw & yr.
11-Draw an operating line from Q”yr to intersect the
liquid enthalpy curve at x’1 .
12-To obtain y’2 repeat the process until you exceed xF
13-The tie lines drawn between x1 to y1, x2 to y2, x3 to y3
and so on = number of stages.
Minimum reflux ratio:Determine Q’min from the graph.
-From point, (xF ,hF), draw a line down to equilibrium
curve. Then, draw a horizontal line to 45o line. Next,
draw a line up to saturated vapor line. Let say the point
is y.
-Then, draw a line from point (xF ,hF) to point y till you
reach the vertical line of xD.
- Q’min is the point intersect between the line from
point (xF ,hF) to point y and vertical line of xD .
3-Lewis plate to plate calculations:Lewis assumes that:1- If the amount of molecules which evaporate and
which condensate are the same. This means that within
each section the liquid and the vapor flow rates remain
constant in the whole section.
This can be translated into
the following equations:
For Top or Rectifying Section:L1 = L2 = L3 = ......... = Ln = constant
V1 = V2 = V3 = ......... = Vn = constant
For Bottom or Stripping Section:L’1 = L’2 = L’3 = ......... = L’m+1 = constant
V’1 = V’2 = V’3 = ......... = V’m = constant
2- The heat of vaporization (λ) of the two components
of the feeding mixture must be the same ( λmvc = λLvc ) .
3- There is no losses from the column.
4-No heat of mixing ∆ H mix = 0 (Ideal system).
{ In this case the saturated liquid and vapor curves
becomes two straight parallel lines in the enthalpy
composition diagram}.
Based on the assumptions made and the identified
data and unknows, we can rewrite the mass balances
and equilibrium equations for both the sections as
For upper section:Vn+1 = Ln + D ----(OMB eq.)
 V = L + D -----(OMB eq. in Lewis method)
Vn+1 y n+1 = L n x n + D x D ----(CMB eq.)
 V y n+1 = L x n + D x D ----(CMB eq. in Lewis method)
 y n+1
xn +
x D -----( 30)
V = L + D  y n+1 =
xn +
xD 
L+ D
L+ D
 y n+1 =
xn +
xD 
L/D + D/D
L/D + D/D
 y n+1 =
xn +
x D -----( 31)
Equation (30) or (31) are operating line equations
having slop L/V or R/R+1 and there intersection with
y-axis are D/V or xD/ R+1 .
For bottom section:L'm+1 = V'm + W ---(OMB eq.)
 L' = V' + W -----(OMB eq. in Lewis method)
L'm+1 x'm+1 = V'm y'm + W x w ---(CMB eq.)
 L' x'm+1 = V' y'm + W x w ---(CMB eq.in Lewis method )
 y'm =
x'm+1 x w -----(32)
Equation (32) is an operating line equation has
a slope L’/V’ and intersection with y-axis is
- Wxw/V’ .
Plate to plate calculations using total condenser:Top section:-
First plate ( n = 1 )
V = L + D ---(OMB)
V y 2 = L x1 + D x D ---(CMB)
 y2 =
x1 +
y1  k1x1 --- (Eq.re. at certain T )
Calculation steps:1-For total condenser y1 = xD = x0 , calculate x1 from the
Eq. re. → y1 = K1 x1 at certain assumed temperature for
the top plate.
2-Substitute in equation (30) or (31) then calculate y2
(notice that the trails only on temperature not on the
amount as in Sorel method).
3-After calculating the correct y2 , calculate y3 from:-
 y3 =
x2 +
& y3 = k 3 x 3
4- Continue your calculation and so on till you reach
xn ≤ xF .
Bottom section:Reboiler ( plate out side the column , m = 0 )
y'r = k r x w
L' = V' + W
y'r =
x'1 
calculation steps:1- By assuming the temperature inside the reboiler ,the
composition yr is calculated.
2- Calculate x’1 from eq.(32) , for m = 0
First plate inside the column , m =1
y'1 = k1 x1
y'1 =
x'2 
3- calculate y’1 from the equilibrium relation, then
calculate x’2
4- Continue your calculation and so on till you reach
y’m ≥ xF .
4- Mc-Cabe Thiele Method (graphical representation of
Lewis method):To obtain the number of
theoretical trays using the
McCabe-Thiele method,
we shall divide the column
into 3 sections: rectifying,
feed and stripping sections.
As these sections are then
represented on the (x-y)
equilibrium curve for the
binary mixture in question
and re-combined to make a
complete design, as shown
in Figure (49).
1- Top section:V = L + D ---(OMB)
V y n+1 = L x n + D x D ---(CMB)
 ( L + D ) y n+1 = L x n + D x D
The point of intersection between this operating line equation
and 45 line (x = y) is:( L + D ) y = L y + D xD
L y + D y = L y + D xD
D y  D xD   y = x = x D
 the point of intersection between top operating line
and 45 line is (x D , x D )
The top operating line is drawn from its upper end at
y1 = x0 = xD (on the 45o line), with a slope L/V = R/R+1
and an
intercept =(xD /R+1)
with y-axis constant
for given R and
purity of
distillate xD .
2- Bottom section:V' = L' - W ---(OMB)
V' y'm = L' x'm1 + W x w ---(CMB)
 ( L'  W ) y'm = L' x'm+1 + W x w
The point of intersection between this operating line equation
and 45 line (x = y) is:( L'  W ) y' = L' y'  W x w
L' y'  W y' = L' y'  W x w
 W y' =  W x w   y' = x' = x w
 the point of intersection between top operating line
and 45 line is (x w ,x w )
The bottom operating line is
drawn from its lower end
at (xw , xw )
(on the 45o line)
with a slope L’/V’
and an intercept
= - W xw/ V’
with y-axis .
3-Feed stage consideration:By solving the two operating
line equations (top and
Bottom) at the point
of intersection (xi , yi)
  ( V yi = L x i + D x D )  (Top)
+  (V' yi = L' x i  W x w )  (Bottom)
 (V'  V) y i = (L'  L) x i  (D x D +Wx W )
(V'  V) yi = (L'  L) x i  F x F
(V'  V)
(L'  L)
yi =
x i  x F ----(33)
q- line definition:Material balance on feed entrance:F + L + V' = V + L'
F + ( V'  V) = ( L'  L)
( V'  V)
( L'  L)
( L'  L)
( V'  V)
q 
& q 1 =
By substitution in equation (33) with these values:xF
 yi =
xi 
q 1
q 1
It is the q line equation , it's slope
q 1
Heat balance on feed entrance:F h F + L h L + V' H'V = V H V + L' h'L
h L  h'L
& H V  H'V
 F h F + ( V' - V) H V = ( L'- L) h L
( V' - V)
( L'- L)
hF +
HV =
( L'- L)
( V' - V)
& q -1 =
 h F + ( q - 1) H V = q h L
 h F + q HV - HV = q h L
HV - h F
  q=
HV - h L
amount of heat necessary to vaporize one mole of the feed
q =
latent heat of vaporization
For different feed conditions, q has the following
numerical limits :L
1  Saturated liquid feed:F
 hF = hL
HV  h F
HV  h L
HV  h L 
 q=
= =1
HV  h L 
q  line:q
q 1
1 1
 q-line is a vertical line
bubble point
liquid feed
2Saturated vapor feed:-
 hF = hV
HV  h V
HV  h L
HV  h V
 q=
HV  h L
q  line:q
q 1 0 1
 q-line is a horizontal line
dew point
vapour feed
3  Partial vaporized feed:
0<q< 1
q = 0.7
q  line:q
=  ve
q 1
0.7  1  0.3
 q-line has a -ve slope
partially vaporized
4Subcooled feed: hF < hV
HV  h V
HV  h L
m cp ΔT + λ
q  line:- ( assume q = 1.5 )
q  1 1.5  1 +ve
 line in the first quarter
liquid feed
5  Superheated feed:-
 h F > HV
HV  h F
 ve
=  ve
HV  h L
q  line:q
q 1
ve  1  ve
 q-line in the third quarter
vapour feed
Number of theoretical stages using Mc-Cabe Thiele
The number of theoretical
stages required for a given
separation is then the
number of triangles that
can be drawn between
these operating lines and
the equilibrium curve.
The last triangle on the
diagram represents the reboiler.
Some cases in Mc-Cabe Thiele method:(1) Calculation of no. of stages in complex feed( more
than one feed):For first feed:L'  L
q1 
 get L'  assume sat.liq.
V = L + D  get R =
D is calculated from (OMB)
 calculate L & V .
q1  known , F1  known
 caculate L'.
V'  V
q1  1 
 get V'
For second feed:L"  L
q2 
 get L"
 assume vap.liq.
V"  V
q2 1 
 get V"
Draw operating line for top
Section, the draw operating
Line for bottom section, then
connect between the two
Lines as shown in the
opposite figure.
If the two feeds were mixed together the introduced to
The column, how many stages would be required to
perform this separation.
By mixing:F = F1 + F2
F x F = F1 x F1 + F2 x F2
F q mix = F1 q1 + F2 q 2
No. of stages in second
case > No. of stages in
the first case.
 It's prefered to introduce each feed
at it's corresponding composition
(2) Calculation of no. of stages in case of enriching
(3) Calculation of no. of stages in case of stripping
(4) Calculation of no. of stages in case of open
Bottom section :steam:L' + S = V' + W or V' - L' = S -W---(OMB)
L' x'm+1 + S ys = V' y'm+1 + W x W ----(CMB)
ys  0
 L' x'm+1 = V' y'm+1 + W x W ---(op.line eq.)
By solving the operating line equation
with y = x line : L' x' = V' x'+ W x W
 ( L' - V' ) = W x W &
 - ( S - W ) x = W xW
W xW
 y=x=
V' - L' = S - W
Open steam presentation Open steam presentation
in Mc-Cabe Thiele method in Ponchon Savarit method
(5) Calculation of no. of stages in case of side
steam:In some applications, a product other than top or
Bottom side products are also withdrawn from
the column, which is usually a saturated liquid
product . Assume the flow rate of intermediate
product is S with composition xs. The column is
divided into three sections:- top , middle and
Material balance on the top
Section:V = L + S + D ---(OMB)
 (V-L) (S+D)
V y n+1 = L x n + S x S + D x D (CMB)
The point of intersection between this
operating line and 45 line
 V y = L y + S xS + D x D
( V - L ) y = S xS + D x D
S xS + D x D S xS + D x D
 y=x=
Material balance on the bottom
section:L' = V' + S + W ---(OMB)
 ( L' - V' ) = ( S + W )
L' x'm+1 = V' y'm + S x S + W x w (CMB)
The point of intersection between this
operating line and 45 line
 L' y = V' y + S x S + W x w
( L' - V' ) y = S x S + W x w
S xS + W x w S xS + W x w
 y=x=
( L' - V' )
Packed tower for small capacities:-
Comparison between Mc-Cabe Thiele method
and Ponchon-Savarit method
Mc-Cabe Thiele method
Ponchon-Savarit method
1- Simple
1- Complex
2-Require (x-y )diagram
2-Require enthalpy comp.
Diagram & (x-y )diagram
3-Less data
Much more data
4-Equi-molar flow rates
4-Require more data on
each section( M.B, H.B ,
Operating conditions:(1) Reflux ratio , R :As R decreases, separation becomes difficult, driving
force(D.F) decreases, number of stages increase the
operating line for the rectification section moves
towards the equilibrium line.
The ‘pinch’ between operating and
equilibrium lines becomes more
pronounced and more and more
trays are required. In this case
R = Rmin and the number of stages
reaches infinite ( N=∞)→ Pinch point, and D.F = zero.
As R increases, separation is improved, D.F increases,
the two operating lines goes down till they coincide
With 45◦ line and L/V =1. At this condition R= ∞ = total
reflux and number of stages is minimum.
Relation between number of stages and reflux ratio:
The minimum reflux ratio and the infinite reflux ratio
place a constraint on the range of separation operation.
Any reflux ratio between
Rmin and Total R will
produce the desired
separation, with the
corresponding number
of theoretical stages
varying from infinity at
Rmin to the minimum number (Nmin at Total R).
Optimum reflux ratio:A trade off between operating cost and equipment cost
is needed.
As R increases →less stages ,
less equipment cost is needed.
As R decreases →more stages ,
more equipment cost is
needed, more boiling and
Equipment and operating costs
Combine to give total cost( Ropt = 1.1 to 1.5 Rmin)
(2) Choice of pressure:Some times high pressure (especially for gases) is used
in the distillation column, the equilibrium T/C diagram
goes up , temperature increases and the equilibrium
curve approaches diagonal (as shown in Figure (4)).
This results in decreasing the relative volatility (α) , and
The D.F decreases, thus distillation becomes difficult.
Number of stages increase(N), and the operating cost
Increase ,and (R) increases.
Vapor load (V) decreases, as the pressure increases
which leads to decrease in column diameter , but
column thickness will increase.
The high pressure column is tall and thin.
Explosion problems must be taken in to consideration.
As pressure decreases(e.g vacuum distillation) , α
increases, separation is more easier, number of stages
decreases, vapor load increases, which leads to big
column diameter. The temperature decreases ( which is
suitable for highly sensitive materials), but column
thickness will increase.
The low pressure column is short and fat .
Collapse problems must be taken into consideration.
The highest temperature and pressure usually at the
Bottom of the distillation column.
Note:- T bottom < T thermal decomposition
P bottom < P critical
The atmospheric distillation column operates at 1atm
and it is the most economical one.
(4) Feed temperature:When hF decreases , Qr increases
and Qc decreases. And when hF
increases Qr decreases and Qc
increases, and R increases. Also
the q-line moves from up to
down as feed temperature is
(5)Feed location:The state of the feed mixture and feed composition
affects the operating lines and hence the number of
stages required for separation (feed is introduced at the
point of intersection of the two operating lines).
During operation, if the deviations from design
specifications are excessive, then the column may no
longer be able handle the separation task. To overcome
the problems associated with the feed, some column are
designed to have multiple feed points ( two or three
nozzles)when the feed is expected to containing varying
amounts of components.
Batch Distillation:When the quantity of the solution to be separated is small
or products at different purities are required, then the
rectification operation is carried out batch wise. A plate
type batch rectification
column is shown in Figure( 56) .
Column operates first under total
reflux until the top product reaches
the desired composition,from there
on top product is withdrawn.
During the operation, no bottom
Product is withdrawn.
The purity of top product depends on the number of
the plates available in the column and the composition
of the liquid in the boiler. The liquid in the boiler
becomes poorer in the mvc as the distillation proceeds;
as no feed is given to the column and as the vapor
formed is always richer in the mvc. As a result of this,
top product becomes also poorer in the mvc as the
time passes. If the purity of the top product is to be
kept constant , then the reflux ratio must be increased
steadily . Batch distillation carried out in two different
ways:-either under constant reflux ratio or under
increases reflux ratio.
As the column consists of only enriching section, there
is only one operating line which is given as :R
y n+1 =
xn +
x D -----( 31)
While in operations under constant reflux ratio the slope
of this line remains constant but its intercept changes; in
operations under increasing reflux ratio, the slope of the
line changes but the intercept remains constant.
(a)Batch rectification under constant reflux ratio: In this case, xD continuously decreases as the reflux
ratio is kept constant. A mvc balance at any moment
during the rectification can be written as (ASTM):-
x (x D -x) -------- (1)
F = D + W ------(2)
Fx F = D x D + W x w -----(3)
With the help of equation (1), (2) and (3), the quantity of
top product (D) and it’s average composition (xD) at the
end of operation for a given F , xF , R , xw and N can be
calculated as follows:1-Draw the operating line from equation(31) starting at
point xF ,then locate the given number of stages.
2- By shifting this line up and down, the
2-Select an arbitrary xD value (more than xF) and draw
the operating line from this xD value and starting at this
point locate the given number of stages and then read
The corresponding
(x ) value. Repeat this
step by shifting this
line up and down until
reaching or exceeding
the given xw as shown in
the opposite Figure.
3- By plotting x values, 1/(xD-x) as shown in Figure( 57 )
the area under the curve between the limits of xD and
xw gives the value of the integral in equation(1) from
which W is then found .
4-Finally from equations
(2) and (3) D and xD are
The heat to be supplied
to provide the reflux is
Qc = λRD
(b)Batch Rectification at Constant Top Product :
If the composition of the top product is to be kept
constant, the reflux ratio must be increased steadily.
In this type of operation, generally N, F, xF , xD and xw are
1-First D and W are calculated from equations (1)and(2).
2-points xF , xD and xW are located on the xy diagram. A
Line from point xD is drawn and slope of this line is
changed until the given number of stages between point
xD and xW fit.
3-The reflux ratio, calculated from the slope of this line
gives the reflux ratio at which the operation will be stopped.
It is obvious that the
reflux ratio will be
increased continuously
from the initial value
to the final value .
The total amount of
distillate collected D
can be determined
from material balance
(before and after the
Distillation process):
F xF = D xD + W xw
 F x F = D x D + (F - D) x w
F xF = D xD + F xw - D xw
F ( xF - xw ) = D ( xD - xw )
F ( xF - xw )
D =
( xD - xw )
R assumed
R initial
R final
Xw ,assumed
x initial
D = F (XF –Xwass )/ (XD – Xwass)
D initial
x final
If the reflux ratio R is assumed to be adjusted
Continuously to keep the top product at constant value,
then at any moment the reflux ratio given by:R = dL/dD
R final
 dL = 
R initial
QR = λ  dL = λ
R final
R initial
By graphical integration will give the value of R  d D
Separation of Difficult Mixtures:Why mixtures are difficult to be separated by Distillation?
1-Low relative volatility of the mixture to be separated.
In such operation continuous distillation require high
reflux ratio , high reflux ratio, large cross-sectional
Diameter and high number of stages.
Ex:- separation of n-heptane from methyl cyclohexane
The relative volatility ≈ 1.08 , using large number of
Stages to achieve the separation.
2-when the relative volatility = 1 → azeotropic mixture ,
No straight forward distillation (equilibrium curve crosses
The diagonal indicating presence of azeotrope.
To over come this:1-Use another method for separation.
e.g:- Solvent extraction, adsorption , crystallization,…
2-Obvious change in pressure (high vacuum).
3-Addition of third or new material to change α of the
two component in which separation becomes easier.
Extractive distillation:Extractive distillation refers to those processes in which
a high-boiling solvent is added alter the relative
volatilities of components in the feed.
The boiling point of the solvent is generally much
higher than the boiling points of the feed mixture that
formation of new azeotropes is impossible. The high
boiling point will also ensure that the solvent is will not
vaporize in the distillation process .
consider the simplified system shown in the Figure (59)
for separation of toluene and iso-octane using phenol
as the solvent .
The separation of toluene (boiling point 110.8 oC) from
iso -octane (boiling point 99.3 oC) is difficult using
conventional distillation. Addition of phenol(boiling
point 181.4 oC) results in the formation of phenol
toluene mixture that leaves the extractive distillation
column as bottoms, while relatively iso- octane is
recovered as overhead product. The phenol-toluene
mixture is further separated in a second column
(solvent recovery column) whereby toluene appears as
distillate and the bottoms product, phenol, is recycled
back to the first column.
Azeotropic distillation:azeotrope is a special class of liquid mixture that boils
at a constant temperature at a certain composition. It
behaves as if it were one component with one constant
boiling point. Such mixture cannot be separated using
conventional distillation methods(α).
As an example the azeotrope in the ethanol water binary
system has a composition of 89 mole per cent of ethanol
The addition of benzene (entrainer) serves to volatilize
water to a greater extent than ethanol thus pure
ethanol may be obtained.
The first tower in Figure (60) gives the ternary azeotrope
as an overhead vapor, and nearly pure ethanol as
bottom product.
The ternary azeotrope is condensed and splits into liquid
phases in the decanter. The benzene-rich phase from
the decanter serves as reflux, while the water–ethanol
rich phase passes to two towers, one for benzene
recovery and the other for water removal.
The azeotropic overheads from these successive towers
are returned to appropriate points in the primary tower.