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![Page 1: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/1.jpg)
CHAPTER 1: DISTILLATION
Part 1
Definition & process description
Physical concept of distillation
Vapor-liquid equilibrium relationship
Relative volatility
Batch distillation –Part 2
Continuous distillation –Part 3
Azeotropic distillation -Part 4
Multicomponent distillation –Part 5
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1.1: Definition & process description
Distillation is a process of separating various
components of a liquid solution by heating the liquid
to forms its vapors and then condensing the vapors to
form the liquid.
It is use to separate 2 or more substances present in
the liquid OR for purification purpose.
Distillation is a commonly used method for purifying
liquids and separating mixtures of liquids into their
individual components
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All components presents in both phases
Familiar examples include
1) distillation of crude fermentation broths into alcoholic spirits such as gin and vodka
2) fractionation of crude oil into useful products such as gasoline and heating oil.
3) In the organic lab, distillation is used for purifying solvents and liquid reaction products.
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1.1: Definition & process description
Other definition
Distillation is done by vaporizing a definite fraction
of a liquid mixture in a such way that the evolved
vapor is in equilibrium with the residual liquid
The equilibrium vapor is then separated from the
equilibrium residual liquid by condensing the vapor
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Laboratory / Testing
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1.2: Physical Concept of distillation
Carried out by either 2 principal methods
First method: based on the production of a vapor
by boiling the liquid mixture to be separated and
condensing the vapors without allowing any liquid
to return to the still - NO REFLUX (E.g. Flash,
simple distillation)
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Second method: based on the return part of the
condensate to the still under such condition that
this returning liquid is brought into intimate
contact with the vapors on their way to the
condenser – conducted as continuous / batch
process (E.g. continuous distillation)
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1.3: Vapor – liquid equilibrium
DEFINITION:
EVAPORATION: The phase transformation
processes from liquid to gas/vapor phase
VOLATILITY: The tendency of liquid to change
form to gas/vapor phase
a) VAPOR – LIQUID EQUILIBRIUM OF AN
ORDINARY BINARY LIQUID MIXTURE
b) PREDICTION OF VAPOR – LIQUID
EQUILIBRIUM COMPOSITIONS FOR
ORDINARY BINARY MIXTURES
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a) VAPOR – LIQUID EQUILIBRIUM OF AN
ORDINARY BINARY LIQUID MIXTURE
Equilibrium curve: shows the relationship
between composition of residual liquid and
vapor that are in dynamic phase equilibrium.
The curve will be very useful in calculations to
predict the number of stages required for a
specified distillation process.
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VAPOR – LIQUID EQUILIBRIUM CURVE
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b) Prediction of vapor-liquid equilibrium
compositions for ordinary binary mixtures
Raoult’s Law for ideal solution & Dalton’s Law of partial pressure can be manipulated in order to calculate compostions of liquid and vapor, which are in equilibrium.
Raoult’s Law – the partial pressure of a component in the vapor phase is equal to the mole fraction of the component in the liquid multiplied by its pure vapor pressure at the temperature:
pA = xA · PAo
pA = partial pressure of A in a vapor phase
xA = mole fraction of A in liquid phase
PAo = vapor pressure of A at the temperature
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Prediction of vapor-liquid equilibrium
compositions for ordinary binary mixtures
For a mixture of the different gases inside a close container, Dalton’s law stated that the resultant total pressure of the container is the summation of partial pressures of each of all gases that make up the gas mixture:
PT = pA + pB
Dalton also state that the partial pressure of gas (pA) is:
pA = yA · PT
pA = partial pressure of A in vapor phase
yA = mole fraction of A in vapor phase
PT = total pressure of the system
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Phase Rule
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Example: Calculate the vapor and liquid compositions
in equilibrium at 95oC (368.2K) for
benzene-toluene using the vapor pressure
from Table 11.1-1 at 101.32 kPa. Table 11.1-1
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1.4: Relative volatility (α) of a mixture
Separations of components by distillation
process depends on the differences in
volatilities of components that make up the
solution to be distilled.
The greater difference in their volatility, the
better is separation by heating (distillation).
Conversely if their volatility differ only slightly,
the separation by heating becomes difficult.
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Relative volatility (α) of a mixture The greater the distance between the equilibrium
line & 45o line, the greater the difference the vapor composition and a liquid composition. Separation is more easily made.
A numerical measure of ‘how easy’ separation – relative volatility, αAB
αAB – relative volatility of A with respect to B in the binary system
Relative volatility – ratio of the concentration of A in the
vapor to the concentration of A in liquid divided by the ratio of the concentration B in the vapor to the concentration of B in the liquid:
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Relative volatility (α) of a mixture
αAB – relative volatility of A with respect to B in the
binary system
If the system obeys Raoult’s law for an ideal
system:
Separation is possible for > 1.0
A
AA
B
AAB
T
BBB
T
AAA
x
xy
P
P
P
xPy
P
xPy
)1(1
)1/()1(
/
/
/
AA
AA
xB
xA
ABxy
xy
y
y
B
A
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Relative volatility (α) of a mixture
Separation is possible for > 1.0
For non-ideal solution, the values of change with
temperature.
For ideal solution, the values of doesn’t change with
temperature.
For solution that approaches ideal solution, its
would fairly constant.
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Relative volatility (α) of a mixture
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Example:
Using the data from table below, determine
the relative volatility for the benzene-
toluene system at 85°C and 105°C
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Exercise 1
A liquid mixture is formed by mixing n-hexane (A) & n-
octane (B) in a closed container at constant pressure of 1
atm (101.3kPa).
i. Calculate the equilibrium vapor and liquid composition of
the mixture at each temperature
ii. Plot a boiling point diagram for n-hexane
iii. Plot an equilibrium diagram for the mixture
iv. Calculate the αAB at 100 °C
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Vapor Pressure
Temperature n-Hexane n-Octane
(°C) kPa mm Hg kPa mm Hg
68.7 101.3 760 16.1 121
79.4 136.7 1025 23.1 173
93.3 197.3 1480 37.1 278
107.2 284.0 2130 57.9 434
125.7 456.0 3420 101.3 760
Use the following list of vapor pressure for pure n-heptane & n-octane
at various temperature.
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Solution
Vapor Pressure
Temperature n-Hexane (A) n-Octane (B)
(°C) kPa XA YA kPa XB YB
68.7 101.3 1 1 16.1 0 0
79.4 136.7 0.6884 0.9290 23.1 0.3116 0.071
93.3 197.3 0.4007 0.7804 37.1 0.5993 0.2196
107.2 284.0 0.1920 0.5383 57.9 0.8080 0.4617
125.7 456.0 0 0 101.3 1 1
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PART 2 - DISTILLATION COLUMN
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Flash & batch distillation
Flash (equilibrium) distillation
Simple batch distillation
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Flash (Equilibrium) Distillation Flash distillation – a single stage process because it has only
one vaporization stage (means one liquid phase is expected
to one vapor phase)
The vapor is allowed to come to equilibrium with the liquid
The equilibrium vapor is then separated from the equilibrium
residual liquid by condensing the vapor
Flash distillation can be either by batch or continuous
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Flash (Equilibrium) Distillation
As illustrated in Figure 3, a liquid mixture feed, with initial mole fraction of A at XF, is pre-heated by a heater and its pressure is then reduced by an expansion valve.
Because of the large drop in pressure, part of liquid vaporizes.
The vapor is taken off overhead, while the liquid drains to the bottom of the drum
The system is called “flash” distillation because the vaporization is extremely rapid after the feed enters the drum.
Now, we interested to predict the composition (x and y) of these vapor and liquid that are in equilibrium with each other.
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Flash (Equilibrium) Distillation
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Flash (Equilibrium) Distillation
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Example A liquid mixture containing 70 mol% n-heptane (A)
and 30 mol % n-octane (B) at 30oC is to be
continuously flash at the standard atmospheric
pressure vaporized 60 mol% of the feed.
Determine
1) the compositions of vapor and liquid for n-
heptane
2) temperature of the separator for an equilibrium
stage?
The equilibrium data for n-heptane – n-octane
mixture at 1 atm and 30°C is given as follows:
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T (K) xA yA
371.6 1 1
374 0.825 0.92
377 0.647 0.784
380 0.504 0.669
383 0.387 0.558
386 0.288 0.449
389 0.204 0.342
392 0.132 0.236
395 0.068 0.132
398.2 0 0
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Solution
Basis = 100 moles of liquid feed (F)
Given, xF = 0.7
V = 0.6(100) = 60 moles f = V/F = 60/100 = 0.6
We want to fine the equilibrium composition of liquid and liquid;
y* & x*
The operating line: y* = (0.6-1)x* + 0.7
0.6 0.6
= -0.667x* + 1.167
From the intersection of the operating line & the equilibrium
curve as shown in the graph:
equilibrium mol fraction of n-heptane in liquid, x* = 0.62
equilibrium mol fraction of n-heptane in vapor, y* = 0.76
the temperature of the separator at equilibrium ≈ 378oC
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0
0.2
0.4
0.6
0.8
1
1.2
0 0.2 0.4 0.6 0.8 1 1.2
y, m
ol fr
acti
on
of
n-h
ep
tan
e in
vap
or
x, mol fraction of n-heptane in liquid
Determination of vapor-liquid equilibrium composition
for a flash distillation of n-heptane/n-octane mixture
Figure: Equilibrium curve and operating line
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370
375
380
385
390
395
400
370
375
380
385
390
395
400
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Tem
pera
ture
(K
)
Mol fraction of n-heptane in vapor (y) and liquid (x)
Determination of equilibrium temperature for a flash
distillation of n-heptane-n-octane mixture
x
y
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Exercise 11.2-1 (page 752) A mixture of 100 mol containing 60 mol%
n-pentane (A) and 40 mol% n-heptane (B) is
vaporized at 101.32 kPa abs pressure until 40
mol of vapor and 60 mol of liquid in
equilibrium with each other are produced.
This occur in a single stage system, and the
vapor and liquid are kept in contact with each
other until vaporization is complete.
Determine the composition of the vapor and
liquid.
![Page 37: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/37.jpg)
The equilibrium data are as follows, where
x and y are mole fraction of n-pentane.
x (mol fraction of n-
pentane in liquid)
y (mol fraction of n-
pentane in vapor)
1.000 1.000
0.867 0.984
0.594 0.925
0.398 0.836
0.254 0.701
0.145 0.521
0.059 0.271
0 0
![Page 38: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/38.jpg)
Answer:
xA = 0.430
yA = 0.854
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Simple batch distillation
Simple batch distillation which is also known as
differential distillation refer to a batch distillation in
which only one vaporization stage (or one exposed
liquid surface) is involved.
Simple batch distillation is done by boiling a liquid
mixture in a stream-jacketed-kettle (pot) and the vapor
generated is withdrawn and condensed (distillate) as
fast as it forms.
The first portion of vapor condensed will be richest in
the more volatile component A. As the vaporization
proceeds, the vaporized product becomes leaner in A.
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Simple batch distillation
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Simple batch distillation
1. Raleigh equation for ideal and non-ideal mixtures
Consider a typical differential distillation at an
instant time, t1 as shown below:
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Simple batch distillation Now consider that the differential distillation at certain infinitesimal time
lapse,(dt), at t2 where t2=t1 + dt, after an infinitesimal amount of liquid has
vaporized as shown below:
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Simple batch distillation
Applicable for ideal and
non-ideal solution
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Simple batch distillation
The average composition of total material
distilled yav can be obtained by material
balance: (integrate from Rayleigh equation)
avyLLxLxL )( 212211
avVyxLxL 2211
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Simple batch distillation
2. Simplified Raleigh equation for ideal mixture
Consider a simple batch distillation process at
an initial time, t1, as shown below
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Simple batch distillation
L1=no of moles of binary mixture
containing A and B at t1
A1= no of moles comp A in L1 at t1
B1= no of moles comp B in L1 at t1
x1= mol fraction of A in L1 at t1
dL= infinitesimal amount of liq that
has vaporized
dA=infinitesimal amount of A that
has vaporized
dB = infinitesimal of B that has
vaporized
L1= A1 + B1
X1
dL= dA + dB
y
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Simple batch distillation We know from definitions,
Since is constant for an ideal mixture,
After simplifying,
Rearranging,
BA
Bx
BA
Ax
dBdA
dBy
dBdA
dAy
BA
BA
AB
BA
A
dBdA
dBBA
B
dBdA
dA
xy
xy
BB
AAAB
/
/
dBA
dABAB
A
dA
B
dBAB
![Page 48: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/48.jpg)
Simple batch distillation Integrating within the limits of t1 and t2,
Since is constant,
Equation 5 known as simplified Raleigh equation for simple batch distillation which applicable for ideal solution.
2
1
2
1
B
B
A
A
ABA
dA
B
dB
2
1
2
1
2
1
2
1
lnlnA
A
B
BAB
A
A
B
B
AB ABA
dA
B
dB
AB
)5.(lnln1
2
1
2 EqA
A
B
BAB
![Page 49: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/49.jpg)
Simple batch distillation Example 1
A mixture of 100 mol containing 50 mol% n-pentane and 50
mol% n-heptane is distilled under differential (batch) conditions at
101.3 kPa until 40 mol is distilled. What is the average
composition of the total vapor distilled and the composition of the
n-pentane in the liquid left. The equilibrium data as follows, where
x and y are mole fractions of n-pentane: xA yA
1.000 1.000
0.867 0.984
0.594 0.925
0.398 0.836
0.254 0.701
0.145 0.521
0.059 0.271
0 0
![Page 50: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/50.jpg)
Solution
xA yA 1/(yA –xA)
1.000 1.000 -
0.867 0.984 8.5470
0.594 0.925 3.0211
0.398 0.836 2.2831
0.254 0.701 2.2371
0.145 0.521 2.6596
0.059 0.271 4.7170
0 0 -
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Solution
Given L1 = 100 mol V (mol distilled) = 40 mol
From material balance:
Substiting into Eq. (4)
The unknown, x2 is the composition n of the liquid L2
at the end of batch distillation.
molLVLLVLL 6040100 21221
1
2
5.0
22
1 1
60
100ln
1ln
x
x x
dxxy
dxxyL
L
dxxy
x
5.0
2
1510.0
![Page 52: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/52.jpg)
Solution By plotting graph versus x, the value is
referring to the value of area under the curve.
From the graph, area under curve = 0.510 at x2 = 0.277. Composition of the liquid L2, x2 = 0.277.
From material balance on more volatile component:
Average composition of total vapor distilled, yav = 0.835
xy
1dx
xyx
5.0
2
1
835.0
40
60277.01005.022112211
av
avavav
y
yV
LxLxyVyLxLx
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PART 3
Continuous distillation
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Continuous / retrification distillation
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Continuous / retrification distillation
Retrification (fractionation) - or stage distillation with
reflux can be considered as a process in which a series of
flash – vaporization stages are arranged in a series in such
a manner that the vapor and liquid products from each
stage flow counter-current to each other.
Continuous distillation - the process is more suitable for
mixtures of about the same volatility and the condensed
vapor and residual liquid are more pure (since it is re-
distilled)
The fractionator consists of many trays which have holes to
permit the vapor, V which rises up from the lower tray to
bubble through and mixes with the liquid, L on the upper
tray and equilibrated, and V and L stream leaves in
equilibrium.
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Continuous / retrification distillation
During the mixing, the vapor will pick up more of
component A from the liquid while the liquid will richer and
richer in component B. As the vapor rises, it becomes richer
and richer in component A but poorer with component B.
Conversely, as the liquid falls further down, it becomes
poorer with A but richer in B. Thus we obtain a bottom
product and an overhead product of higher purity in
comparison to those obtained by single-stage simple batch
or flash distillation.
NOTE: Fractionation refers to a process where a part or
whole of distillate is being recycled to the fractionator. The
recycled distillation (reflux) will supply the bulk of liquid
need to mix with vapor.
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Continuous / retrification distillation
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Continuous / retrification distillation
The feed stream is introduced on some intermediate tray where the liquid has approximately the same composition as the feed.
The system is kept steady-state: quantities (feed input rate, output stream rates, heating and cooling rates, reflux ratio, and temperatures, pressures, and compositions at every point) related to the process do not change as time passes during operation.
With constant molal overflow assumption:
Conditions for constant molal overflow:
◦ Heat loses negligible (achieved more easily in industrial column)
◦ Negligible heat of mixing
◦ Equal or close heats of vaporization
.......... 1111 etcVVVetcLLL nnnnnn
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Continuous / retrification distillation
Number of plates required in a distillation column
Four streams are involved in the transfer of heat and material across a plate, as shown in figure above:
Plate n receives liquid Ln+1 from plate n+1 above, and
vapor, Vn-1 from plate n-1 below.
Plate n supplies liquid Ln to plate n-1, and vapor Vn to
plate n+1
Action of the plate is to bring about mixing so that the vapor Vn of composition yn reaches equilibrium with the liquid Ln of composition xn.
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Continuous / retrification distillation
Design and operation of a distillation column
depends on the feed and desired products
A continuous distillation is often a fractional
distillation and can be a vacuum distillation or a
steam distillation.
Calculation for number of plates:
Mc-Cabe & Thiele
Lewis-Sorel Method
![Page 62: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/62.jpg)
Continuous / retrification distillation
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![Page 67: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/67.jpg)
Mc-Cabe Thiele Method
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The intersection of operating lines, q
Feed enters as liquid at its boiling point that the two operating lines intersect at point having an x-coordinate of xF.
The locus point of the intersection of the operating lines is considerable importance since it is dependent on the temperature and physical condition of feed.
The condition of the feed (F) determines the relation between the vapor (Vm) in the stripping section and (Vn) in the enriching section, as well as between Lm and Ln.
![Page 69: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/69.jpg)
The intersection of operating lines, q
q also as the no. of moles of saturated liquid produced on the feed plate by each mole of feed added to tower.
The relationship between flows above & below entrance of feed:
Rewrite the equations of enriching & stripping without the tray subscripts:
Subtracting (3) from (4)
feedofonvaporizatiofheatlatentmolar
conditionsenteringatfeedofmolvaporizetoneededheatq
1
Lv
Fv
HH
HHq
)2()1(
)1(
FqVV
qFLL
mn
nm
)4(
)3(
wmm
Dnn
WxxLyV
DxxLyV
)5()()()( wDnmnm WxDxxLLyVV
![Page 70: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/70.jpg)
The intersection of operating lines, q
Substituting: , Eq. (1) & (2) into (5) will produce:
The equation – locus of the intersection of the two operating lines
Setting y = x in the equation, the intersection of the q-line equation with the 45o line is y = x = xF, where xF is the overall composition of the feed.
Slope = q/(q-1).
A convenient way to locate a stripping line operating line is 1st to plot the enriching operating line and then q-line.
wDF WxDxFx
)(11
equationlineqq
xx
q
qy F
![Page 71: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/71.jpg)
The intersection of operating lines, q
•Depending on the state of
the feed, the feed lines
will have different slopes:
q = 0 (saturated vapour)
q = 1 (saturated liquid)
0 < q < 1 (mix of liquid and
vapour)
q > 1 (subcooled liquid)
q < 0 (superheated vapour)
![Page 72: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/72.jpg)
Animation of the construction of
enriching, stripping & q operating lines
http://www.separationprocesses.com/Distillation/DT_Anima
tion/McCabeThiele.html
![Page 73: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/73.jpg)
Exercise
1) 11.4-1
2) 11.4-2
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Steps of McCabe Thiele Method
1. Plot equilibrium mole fraction for component
that more volatile. [ y(mole fraction A in
vapor) vs x(mole fraction A in liquid)]
2. Make 45° line (x=y)
3. Plot enriching line;
4. Plot q line
5. Plot stripping line
6. Determine the stages
7. Feed tray location
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Tutorial
11.4-5
![Page 79: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/79.jpg)
Exercise 11.4-6
Repeat Problem 11.4-1 but use a feed that
is saturated vapor at dew point. Determine
(a) Minimum reflux ratio, Rm
(b) Minimum number of theoretical plates at total
reflux
(c) Theoretical number of trays at an operating
reflux ratio of 1.5Rm
![Page 80: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/80.jpg)
Example
A mixture of benzene and toluene containing 40
mole% benzene is to be separated to give a product
of 90 mole% benzene at the top, and a bottom
product with not more than 10 mole% of benzene.
The feed is heated so that it enters the column at its
boiling point, and the vapor leaving the column is
condensed but not cooled, and provides reflux and
product.
It is proposed to operate the unit with a reflux ratio
of 3 kmol/kmol product. It is required to find the
number of theoretical stages needed and the
position of entry for the feed.
![Page 81: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/81.jpg)
Example
![Page 82: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/82.jpg)
Solution
Feed, xF = 0.4 Product, xD = 0.9 Bottom, xw = 0.10 Taking basis; 100 kmol of feed. A total mass balance: F = D + W hence; W = 100 – D (Eq. 1)
A balance on MVC (benzene);
From the calculations; D = 37.5 kmol, W = 62.5 kmol
)2.(1.09.040
)1.0()9.0()4.0(100
EqWD
WDxWxDxF wDF
![Page 83: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/83.jpg)
Solution
Using notation from reflux:
From material balance at the top stage;
Thus, the operating line equation:
5.112
)5.37(33
n
nnnn
L
LDLRDLD
LR
kmolVDLV nnn 15011
225.075.0
150
5.112
1
11
111
nn
Dn
nnDn
nn
nn
xy
xV
Dxyx
V
Dx
V
Ly
![Page 84: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/84.jpg)
Solution
Since the feed is all liquid at its boiling point, it will
all run down as increased reflux to the plate
below:
The material balance at the bottom:
Bottom operating line equation:
nmmmm VkmolVVWVL 1505.625.2121
kmolLLFLL mmnm 5.2121005.112
0417.0417.1
)1.0(150
5.62
150
5.212
1
111
1
mm
mmwm
mm
mm
xy
xyxV
Wx
V
Ly
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Example 11,200 kg/h of equal parts (in wt) of Benzene-Toluene
solution is to be distilled in a fractionating tower at atmospheric pressure.
The liquid is fed as a liquid-vapor mixture in which the feed consist of 75% vapor. The distillate contains 94 wt% Benzene whereas the bottom products contains 98 wt% toluene. Determine;
◦ The flowrate of distillate and bottom product (kg/h)
◦ The minimum reflux ratio, Rm.
◦ The number of theoretical stages required if the reflux ratio used is 1.5 times the minimum reflux ratio
◦ The position of the feed tray The MW of Benzene = 78
The MW of Toluene = 92
![Page 86: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/86.jpg)
Solution
xF = 0.5
xD = 0.94
Xw = 0.02
From the total & component material balance:
D = 5739.1 kg/h, W = 5260.9 kg/h
Convert mass fraction to mol fraction. (Basis of
calculation = 100kg)
Mol fraction: xF = 0.54, xD = 0.95, xw = 0.03
![Page 87: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/87.jpg)
Solution Find q-line. Feed enters at 75% vapor.
l ineforqandPlot
yyxLet
xyxy
q
xx
q
qy
equationlineq
qqq
qqqq
fqq
liquidvapor
)62.0,3.0()54.0,54.0(
62.072.0)3.0(333.0,3.0
72.0333.0125.0
54.0
125.0
25.0
11
25.0)0.1(25.0)0(75.0
)(25.0)(75.0
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Solution
From the graph, y intercept for q-line = 0.36
The number of theoretical stages required if the
reflux ratio used is 2 times the minimum reflux ratio
64.136.01
95.036.0
1m
mm
D RRR
x
OLenrichingforandPlot
yyxAt
xy
xy
xR
xR
Ry
RRR
nn
nn
nn
Dnn
)605.0,5.0()95.0,95.0(
605.0222.0)5.0(766.0,5.0
222.0766.0
)95.0(128.3
1
128.3
28.3
1
1
1
28.3)64.1(22
11
1
1
1
min
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Solution
The number of theoretical stages required = 10.5
stages including boiler
Feed plate location: 5 from top.
![Page 90: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/90.jpg)
Q1 Final Exam Jan 2012 A distillation column with a total condenser
and partial reboiler is used to separate an
ethanol-water mixture. The feed containing
20 mole% ethanol enters the column at feed
rate 1000 kg.moles/hr. A distillate composition
of 80 mole% ethanol and bottom composition
of 2.0 mole% ethanol are desired. The external
reflux ratio is 5/3 and it is returned as a
saturated liquid. It is assumed the condition is
at constant molal overflow.
![Page 91: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/91.jpg)
Given;
Enthalpy of feed at dew point, Hv =485 kJ/kg.mol
Enthalpy of feed at boiling point, HL =70 kJ/kg.mol
Enthalpy of feed at entrance condition, HF =15 kJ/kg.mol
The equilibrium curve of ethanol-water is provided in
Appendix 1.
Determine;
i) The minimum number of tray
ii) The total number of equilibrium tray
iii) The feed location
![Page 92: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/92.jpg)
Q2, Final Exam Jan 2013
A total feed of 500 kmol/hr having an
overall composition of 55 mol% heptane
and 45mol% ethyl benzene is to be
fractionated at 1.0 bar to give a distillate
containing 95 mol% heptane and bottom
containing 2 mol% heptane. The feed enters
the tower at equimolar vapor and liquid.
The molecular weight of heptane = 100.2
kg/kmol and ethyl benzene =106.6 kg/kmol.
Equilibrium data for heptane-ethylbenzene is
given in Table 1.
![Page 93: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/93.jpg)
Determine
a) The flowrates of distillate and bottom product in
kg/hr
b) The minimum reflux ratio
c) The number of theoretical stages if the reflux
ratio used is 1.3 times the minimum reflux ratio.
![Page 94: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/94.jpg)
Table 1: Equilibrium data for heptane
ethylbenzene
Temperature (°C) Mole fraction
XH YH
98.3 1.00 1.00
102.8 0.79 0.90
110.6 0.49 0.73
119.4 0.25 0.51
129.4 0.08 0.23
136.1 0.00 0.00
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AZEOTROPIC DISTILLATION
Azeotrope mixtures
Minimum boiling point
Maximum boiling point
Azeotropic Distillation
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Azeotrope mixtures
Liquid and vapor are exactly the same at a
certain temperature
It is a special class of liquid mixture that boils at
a constant temperature at a certain composition
Cannot be separated by a simple/conventional
distillation
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Azeotropic Distillation
An introduction of a new component called entrainer is added to the original mixture to form an azeotrope with one or more of feed component
The azeotrope is then removed as either the distillate or bottoms
The purpose of the introduction of entrainer is to break an azeotrope from being formed by the original feed mixture
Function of entrainer:
◦ To separate one component of a closely boiling point
◦ To separate one component of an azeotrope
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Azeotropic Distillation
Azeotropic distillation is a widely practiced process for the dehydration of a wide range of materials including acetic acid, chloroform, ethanol, and many higher alcohols.
The technique involves separating close boiling components by adding a third component, called an entrainer, to form a minimum boiling.
Normally ternary azeotrope which carries the water overhead and leaves dry product in the bottom.
The overhead is condensed to two liquid phases; the organic, "entrainer rich" phase being refluxed while the aqueous phase is decanted.
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Azeotropic Distillation
A common example of distillation with an azeotrope
is the distillation of ethanol and water.
Using normal distillation techniques, ethanol can only
be purified to approximately 89.4%
Further conventional distillation is ineffective.
Other separation methods may be used are
azeotropic distillation or solvent extraction
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Azeotropic Distillation
The concentration in the vapor phase is the same as
the concentration in the liquid phase (y=x)
At this point, the mixture boils at constant
temperature and doesn’t change in composition
This is called as minimum boiling point (positive
deviation)
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Azeotropic Distillation
The characteristic of such mixture is boiling point
curve goes through maximum phase diagram
Example: Acetone-chloroform
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Azeotropic Distillation
The most common examples:
Ethanol-water (89.4 mole%, 78.25 oC, 1 atm)
Carbon Disulfide-acetone (61 mol% CS2, 39.25oC, 1
atm)
Benzene-water (29.6 mol% water, 69.25 oC, 1 atm)
![Page 103: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/103.jpg)
Azeotropic Distillation
Let say binary mixture: A-B formed an azeotrope
mixture
Entrainer C is added to form a new azeotrope with
the original components, often in the LVC, say A
The new azeotrope (A-C) is separated from the
other original component B
This new azeotrope is then separated into entrainer
C and original component A.
Hence the separation of A and B can be achieved
![Page 104: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/104.jpg)
Azeotropic Distillation
Example: Acetic acid-water using entrainer n-butyl
acetate
Boiling point of acetic acid is 118.1 oC, water is 100 oC
& n-butyl acetate is 125 oC
The addition of the entrainer results in the formation of a
minimum boiling point azeotrope with water with a
boiling point = 90.2 oC.
The azeotropic mixture therefore be distilled over as a
vapor product & acetic acid as a bottom product
The distillate is condensed and collected in a decanter
where it forms 2 insoluble layers
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Azeotropic Distillation
Example: Acetic acid-water using entrainer n-butyl
acetate
Top layer consist of nearly pure n-butyl acetate in water,
whereas bottom layer of nearly pure water saturated
with butyl acetate
The liquid from top layer is returned to column as reflux
and entrainer
The liquid from bottom layer is sent to another column to
recover the entrainer (by stream stripping)
![Page 106: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/106.jpg)
Determination of Boiling Point Temperature in multi
component distillation
The calculation is a trial and error process
where
1. T is assumed
2. Value of relative volatility of each component
are then calculated using K values at the
assume T.
3. Then calculate value of Kc where
Kc= 1/(∑relative volatility x liq mole
fraction)
![Page 107: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/107.jpg)
4. Find the T that corresponds to the calculated
value of Kc
5. Compare with T value read from table that
corresponds to the Kc.
6. If value is differ, the calculated T is used for the
next iteration.
7. After the final T is known, the vapor composition
is calculated from
Yi= (relative volatility x liq mole fraction)/∑(relative
volatility x liq mole fraction)
![Page 108: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/108.jpg)
Example 11.7-1
![Page 109: Chapter 1 Dc](https://reader034.fdocuments.net/reader034/viewer/2022042501/55cf8fac550346703b9ea929/html5/thumbnails/109.jpg)
Bubble point@boiling point
Bubble point@boiling point
=temperature at which liquid begins to vaporize
Dew point
=temperature at which liquid begins to condense out
of the vapor