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
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 −
o
Antoine Equation
C +T
Where T in oK and Po in mmHg
B
Or A −
P = e C +T
o
 Binary System Phase Diagram
V

T
L+V

0% A 100% A
100%B
x,y 0%B x
Temperature-Composition Diagram x-y Diagram
 Binary System Phase Diagram
For a pure component
evaporation occurs at a T
constant temperature, so it’s
Tdew
called “BOILING POINT”
For a mixture of two or more Tbubble
components evaporation
occurs at increasing
temperature so it’s called
“BOILING RANGE” starting by
Bubble point and ending at
Dew point
0% A 100% A
100%B
x,y 0%B
Effect of Pressure on Equilibrium
AZEOTROPIC MIXTURE
Some systems has what is so
called AZEOTROPE like CS2-
Acetone and Methanol-
Water 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 acetone-
chloroform
 CALCULATING LIQUID VAPOUR
EQUILIBRIA FOR BINARY SYSTEMS
RELATIONS USED
x-y diagram
1- I f we have TEMPERATURE-
COMPOSITION 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
y
4- Connect the points to
get the equilibrium curve

x
2- If vapour pressures are known as a function
of temperature:
T
PAo
PBo

PT = PA + PB
PT = xA P + xB P
o
A
o
B
xB = 1 − x A
PT = xA P + (1 − x A ) PB = xA (PA − PB ) + PB
o
A
o o o o

PT − P o
xA = o B
PA − P o
B
 And by the general equilibrium relation for ideal
binary systems:
At equilibrium
PGas phase = PLiquid interface
y A PT = x P o
A A

o
P
yA = xA = K A xA
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 P o
yA
a= o yA = xA A
P =
o
A PT
PB PT xA
yA yA
xA xA y A (1 - x A )
a= = =
yB (1 - y A ) x A (1 - y A )
xB (1 - x A )
a  x A (1 - y A ) = y A (1 - x A )

α xA
yA =
1 + (a - 1)x A
 A NEW PHASE DIAGRAM
ENTHALPY COMPOSITION DIAGRAM
H
BTU/Lb
H= specific enthalpy
H Vapourous Curve
of vapour.
h= specific enthalpy
of liquid.
h Liquidous Curve

0% A 100% A
100%B x,y 0%B
 H
Tie line is used for getting points
on equilibrium curve as what was
done in case of Temperature
composition diagram.

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 y
T
splits into two phases
in equilibrium each of
x
different compositions.
The compositions of
the two phases can be
determined using “TIE
LINE” 0% A 100% A
100%B
x,y 0%B
 Distillation
operations

Single Stage Multistage

Simple Flash Binary

Differential
Steam
vaporization
Multicomponent
Distillation system systems
Distillation Distillation
 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
By Differential MB: (W)

xf
F dx
ln   =  Distillate
 W  xw y * − x (D)
 Simple Differential Distillation
xf
F dx
ln   = 
 W  xw y * − x
Solved by graphical integration
OR
Algebraically using the relation: (use when α given)
F 1 xF (1 − xW )  1 − xW 
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 0 0.2 0.4 0.6 0.8 1
A in liquid

Mole
fraction of 0 0.35 0.58 0.75 0.9 1
A in vapor
Givens:
F=100kmole xf=0.5
xw=0.05

Here we will solve using graphical integration:

From xw to xf
xf
F dx
ln   = 
 W  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 :

F
0 .5
dx 1
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
5- A liquid containing 50% n-heptane and 50% n-
octane 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  1 x F (1 − xW )  1 − xW 
ln = ln + ln 
W  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) =mw1*l w
mw1=Q1/l w

Q2= =mbatch*lbatch =mw2*l w

mw2=Q2/l w

Pw M w
m w3 =   m batch
PHC M HC
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) 38.5 64.4 80.6 96 99.15 113.2
Pw (mmHg) 51.1 199.7 363.9 657.6 737.2 1225
PEA (mmHg) 1 5 10 20 22.8 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
Thanks