Distillation I
Mass Transfer for 4th Year
Chemical Engineering Department
Faculty of Engineering
Cairo University
THINGS YOU HAVE TO REMEMBER
•
•
•
•
Phase Equilibria
Dalton’s Law
Raoult’s law
Antoine Equation
LET’S SAY THAT
For a binary system:
A = More Volatile Component
B = Less Volatile Component
PoA>PoB
a=Relative volatility (no more recovery)
PAo
a o
PB
AND OF COURSE
xi= composition of component i in liquid phase
yi= composition of component i in vapour phase
Pi=yiPT
Dalton’s law
Pi=xiPoi
Raoult’s law
B
ln P  A 
C T
o
Antoine Equation
Where T in oK and Po in mmHg
B
A

Or
o
P  e C T
Binary System Phase Diagram
V
T
L+V
y
L
0% A
100%B
x,y
100% A
0%B
Temperature-Composition Diagram
x
x-y Diagram
Binary System Phase Diagram
For a pure component
evaporation occurs at a
constant temperature, so it’s
called “BOILING POINT”
For a mixture of two or more
components
evaporation
occurs
at
increasing
temperature so it’s called
“BOILING RANGE” starting by
Bubble point and ending at
Dew point
T
Tdew
Tbubble
0% A
100%B
x,y
100% A
0%B
WHAT IS THIS??????
It is PTxy diagram for phase equilibrium
Effect of Pressure on Equilibrium
AZEOTROPIC MIXTURE
Some systems has what is so
called AZEOTROPE like CS2Acetone and MethanolWater systems.
An azeotropic mixture has
a=1 so can’t be separated by
distillation.
This figure represents a
“minimum azeotrope”
AZEOTROPIC MIXTURE
This figure represents a
“maximum azeotrope” in
the system acetonechloroform
CALCULATING LIQUID VAPOUR
EQUILIBRIA FOR BINARY SYSTEMS
RELATIONS USED
x-y diagram
1- I f we have TEMPERATURECOMPOSITION DIAGRAM
T
1- Draw tie line
2- Project x and y of that
line on the x-y diagram and
get a point on the
equilibrium curve
3- Repeat the last step
several times
4- Connect the points to
get the equilibrium curve
y
x
2- If vapour pressures are known as a function
of temperature:
T
PAo
PBo
PT  PA  PB
xB 1  xA
PT  xA P  xB P
o
o
o
o
PT  xA P  1  x A  PB  xA PA  PB   PB
o
A
o
A
o
B
PT  P
xA  o
PA  P
o
B
o
B
And by the general equilibrium relation for ideal
binary systems:
At equilibrium
PGas phase  PLiquidinterface
y A PT  x P
o
A A
o
A
P
yA 
xA  K A x A
PT
The most used form of equilibrium relation for liquid-vapour systems
(Either ideal or non-ideal, binary or multicomponent)
3- Another form of equilibrium relation ( Only for
binary systems if a is known):
PAo
a o
PB
yA
a
yB
xA
xB
o
A
P
yA 
xA
PT
yA

1 - y A 
xA
1 - x A 
y A 1 - x A 

x A 1 - y A 
a  x A 1- yA   yA 1- x A 
α xA
yA 
1  a - 1x A
yA
P 
PT
xA
o
A
A NEW PHASE DIAGRAM
ENTHALPY COMPOSITION DIAGRAM
H= specific enthalpy
of vapour.
h= specific enthalpy
of liquid.
H
BTU/Lb
H
Vapourous Curve
h
Liquidous Curve
0% A
100%B
x,y
100% A
0%B
Tie line is used for getting points
on equilibrium curve as what was
done in case of Temperature
composition diagram.
H
This diagram is more difficult to
draw, however it provides more
information about enthalpy of
streams.
y
x
THEORY OF DISTILLATION
For any system, once
it’s in the wet region, it
splits into two phases
in equilibrium each of
different compositions.
The compositions of
the two phases can be
determined using “TIE
LINE”
y
T
x
0% A
100%B
x,y
100% A
0%B
Distillation
operations
Single Stage
Simple
Differential
Distillation
Steam
Distillation
Multistage
Flash
vaporization
Distillation
Binary
system
Multicomponent
systems
Simple Differential Distillation
Sometimes
called
“ASTM
distillation” and Used in labs
Distillation occurs so that the
feed is slowly evaporated and
then condensed “DISTILLATE (D)”
and
remains
un-evaporated
portion “RESIDUES (W)”.
Now to get xw and ῩD
Residues
(W)
By Differential MB:
F
ln 
W
xf
dx


 
 xw y *  x
Distillate
(D)
Simple Differential Distillation
F
ln 
W
xf
dx

 
 xw y *  x
Solved by graphical integration
OR
Algebraically using the relation: (use when α given)
 1  xW
xF 1  xW 
1
F
ln   
ln
 ln 
 W  a  1 xW 1  xF 
 1  xF



Then calculate ῩD using C.M.B equation
4- 100 kmoles of a mixture of A and B is fed to a simple
still. The feed contains 50 mole % A and a product
contains 5 mole % A is required. Calculate the
quantity of product obtained. The equilibrium data is
presented as follows:
Mole
fraction of
A in liquid
0
0.2
0.4
0.6
0.8
1
Mole
fraction of
A in vapor
0
0.35
0.58
0.75
0.9
1
Givens:
F=100kmole
xw=0.05
xf=0.5
Here we will solve using graphical integration:
From xw to xf
F
ln 
W
xf
dx


 
 xw y  x
x
1/(y-x)
0.05
26.67
0.2
6.67
0.4
5.56
0.5
6.06
The additional points are calculated by inter polation :
dx
1
F
ln   
 [[(0.2  0.05) * (6.67  26.67)] 
 W  0.05 y  x 2
[0.2 * (6.67  5.56)]  [0.1 * (5.56  6.06)]]  4.3045
F
 exp(4.3045)
W
W  1.35kmoles
0.5
5- A liquid containing 50% n-heptane and 50% noctane is differentially distilled at 1 atmosphere
to vaporize 60 mole% of feed. Find the
composition of the distillate and the residue.
Find the boiling range during distillation. (The
average volatility α=2.15).
F
ln
W
 1  xW
x F 1  xW 
1

ln
 ln

 a  1 xW 1  x F 
 1  xF
By trial and error : X w=0.3285
From M.B: ῩD =0.614
Boiling range =109.78-114.586oC



Steam Distillation
Experimental method used for thermally sensible
materials.
For insoluble mixtures (hydrocarbon-water
mixtures) each component exerts pressure equals
the vapour pressure (x=1). If the summation of the
vapour pressures equals the total pressure then the
mixture will start boiling at a temperature lower
than boiling point of the pure hydrocarbon.
Steam is used to perform evaporation at a reduced
temperature.
Steam Distillation
Steam functions:
1- Heating the batch (material to be distilled) to the
bubble point.
2- Giving the batch the latent heat needed to
vaporize.
3- Carrying the vapours.
Steam Distillation
To calculate the necessary amount of steam needed
we will need to know:
1- Q1=heat required to heat the batch to Tb
2- Q2=Latent heat gained by the batch
3- amount of steam carrying the vapours
Steam Distillation
Q1=mbatch*CP ) batch*(Tb-TF)
Q2=mbatch*lbatch
Amount of steam carrying the vapours:
Vapours leaving=water vapour+hydrocarbon vapour
PT=Pwater+PHC
mw
nw  Mw
Pw M w



m HC n HC  M HC PHC M HC
Steam Distillation
Q1=mw1*l w
Q2=mw2*l w
Pw M w
m w3 

 m batch
PHC M HC
mw1=Q1/l w
mw2=Q2/l w
Problem 8:
10 Kg batch of ethylaniline is to be steam distilled from small amount of
non-volatile impurity. Saturated steam at 25 Psia is used. Initial
temperature of ethylaniline is 40 oC and the distillation takes place at
atmospheric pressure.
a- At what temperature will the distillation proceeds?
b- Determine the composition of the vapour phase.
c- How much steam is used?
Data:
• Heat capacity of ethylaniline is 0.4 KCal/Kg.C
• Heat capacity of steam is 0.35 KCal/Kg.C
• Latent heat of vaporization of ethylaniline is 72 Kcal/Kg
• Vapour pressures of water and ethylaniline are given in the table
below:
T (oC)
Pw (mmHg)
PEA (mmHg)
38.5
51.1
1
64.4
199.7
5
80.6
363.9
10
96
657.6
20
99.15
737.2
22.8
113.2
1225
40
T
Pw
PEA
PT
38.5
51.1
1
52.1
64.4
199.7
5
204.7
80.6
363.9
10
373.9
96
657.6
20
677.6
99.15
737.2
22.8
760
113.2
1225
40
1265
Operation occurs at 1 atm=760 mmHg
SO Temperature at which Ptotal=760 mmHg is
99.15oC
Composition of vapour phase:
At 99.15oC:
Pw=737.2 mmHg
PEA=22.8 mmHg
yw=737.2/760=0.97
yEA=22.8/760=0.03
Amount of steam used:
CP)EA= 0.4 KCal/Kg.C
CP)Steam= 0.35 KCal/Kg.C
lEA= 72 Kcal/Kg
lSteam= 540 Kcal/Kg
Q1=10*0.4*(99.15-40)=236.5 Kcal
m1=Q1/lSteam= 236.5/540=0.438 Kg
Q2=10*72=720 Kcal
m2=Q2/lSteam= 720/540=1.333 Kg
Pw M w
737.2 18
m w3 

 m batch 

10  48.1 Kg
PHC M HC
22.8 121
mw=0.438+1.333+48.1=49.87 Kg
Flash Vaporization Distillation
Flash vaporization or equilibrium distillation is a
single stage operation where a liquid mixture is
partially vaporized, the vapour allowed to come
to equilibrium with the residual liquid and the
resulting vapour and liquid phases are separated
and removed.
It’s easy but not efficient, and no packing or trays
are needed.
Generally used for easy separation (very high
relative volatility) or as a primary separation step.
How to Reach Flashing Conditions
Changing Temperature
T
T
0% A
100%B
Changing Pressure
x,y
100% A
0%B
0% A
100%B
x,y
100% A
0%B
How to Reach Flashing Conditions
Changing Temperature
Changing Pressure
V
y
F
xf
V
y
F
xf
L
x
L
x
Calculating Tbubble and Tdew
Flashing occurs at a temperature between
bubble and dew points. So calculating bubble
and dew points is necessary to specify suitable
flashing conditions.
T bubble > T flashing
> T dew
Calculating Tbubble
Bubble point is the
temperature at which
JUST ONE bubble of
gas evaporates and is
in equilibrium with
the liquid whose
composition will not
be affected.
T
y
Tbubble
x
0% A
100%B
xf
100% A
0%B
Calculating Tbubble
Then xi=xFi
And yi=KixFi
Where Ki=Poi/PT
TO GET Tbubble:
Tbubble
1. Assume Tbubble
2. Calculate yi’s
3. Calculate Syi. if equals 1
OK, else reassume Tbubble
0% A
4. Interpolate
100%B
T
y
x
xf
100% A
0%B
(Tbubble)1
(Tbubble)TRUE
(Tbubble)2
Sy1
Sy=1
Sy2
Sy
OR
1
Tb 1  Tb TRUE
Σy1  1

Tb 1  Tb 2
Σy1  Σy 2
(Tbubble)TRUE
Tbubble
Calculating Tdew
Dew point is the
temperature at which
JUST ONE point of
liquid condenses and
is in equilibrium with
the vapour whose
composition will not
be affected.
T
y
Tdew
x
0% A
100%B
yf
100% A
0%B
Calculating Tdew
Then yi=xFi
And xi=yFi/Ki
Where Ki=Poi/PT
Tdew
TO GET Tdew:
1. Assume Tdew
2. Calculate xi’s
3. Calculate Sxi. if equals 1
OK, else reassume Tdew
0% A
4. Interpolate
100%B
T
y
x
xf
100% A
0%B
Sx1
Sx=1
Sx2
(Tdew)1
(Tdew)TRUE
(Tdew)2
Sx
OR
1
Td 1  Td TRUE
Σx 1  1

Td 1  Td 2
Σx 1  Σx 2
(Tdew)TRUE
Tdew
Flash Distillation Calculations
Overall Material Balance:
F=L+V
Component Material Balance:
F
F xFi=L xi+V yi
xF
Equilibrium Relation:
yi=ki xi
V
y
L
x
Flash Distillation Calculations
(L+V) xFi=L xi+V yi
(L+V) xFi=L xi+V ki xi
(L+V) xFi=(L+V ki) xi
V
y
F
LV
xF
xi 
x Fi
L  Vk i
It’s also solved by trial and error
on liquid and vapour flow rates.
L
x
Flash Distillation Calculations
V
y
For the easiness of trials:
LV
xi 
xi
L  Vk i
L
1
V
xfi
L  ki
V
L 1
V
yi 
xi
L
1
V ki
xi 
(÷V)
F
xF
L
x
Flash Distillation Calculations
Finally:
xi 
L
1
V
x Fi
L  ki
V
yi 
L
L
V
1
V ki
1
x Fi
Now trials are done by changing ONLY L/V till Sx
and Sy are BOTH equal one
Flash Distillation Calculations
Once L/V is calculated, all needed variables can
be calculated:
F=L+V
F
L 
V
F    1V
L 
V

  1
V

L  F V
REMEMBER .....
The case we always deal with throughout our
course is the ideal case, that is why we can say:
Ki=Poi/PT
HOWEVER .....
In general, the systems are non-ideal. K values
then can be calculated from the relation
 i fio
Ki 
i p
Solving Flashing Problems IN EXAM
Due to time limitations:
1- Assume L/V (say L/V=1)
2- Calculate Sx and Sy.
3- If Sx=1 AND Sy=1 L/V is correct
4- If Sx ≠1 OR Sy≠1 assume another L/V
5- Do the last step twice then interpolate
(L/V)1
(L/V)operating
(L/V)2
Sx1
Sx=1
Sx2
Sx
OR
1
Σx1  1
Σx 1  Σx 2
(L/V)operating
L/V
LV   LV 

LV   LV 
1
operting
1
2
Solving Flashing Problems Practically
More iterations must be done to get more accurate
results.
Computer softwares such as Microsoft Excel can be
used to do such iterations.
Sometimes system is non-ideal, so other softwares
like Hysys or Aspen can be used.
We will now see how to do such calculations using
Microsoft Excel.
Microsoft Excel
How about non-ideal systems
All calculation steps will not be changed, only
the method of calculating “K” will change.
K can be got thermodynamically using a proper
equation of state (somehow difficult).
K can be got from practical data, i.e. charts.
For systems of oil fractionation it’s so common
to use charts that are got from experiments
instead of calculating them.
For More Information
For more information please check:
1- Robert E. Treybal, “Mass Transfer Operations”, Third
ed., Ch. 9, PP. 342
2- McCabe and Smith, “Unit Operations for chemical
engineers”, Fifth ed., Ch. 19, pp. 588
3- Perry, R. H. and D. Green, eds., “Perry’s Chemical
Engineer’s Handbook”, Seventh ed., Section 13,
McGraw-Hill, New York, 1997.