CHAPTER 1: DISTILLATION 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
1.1: Definition & process 1.1: Definition & process descriptiondescription
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
All components presents in both phases
Familiar examples include 1)distillation of crude fermentation broths
into alcoholic spirits such as gin and vodka2)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.
1.1: Definition & process 1.1: Definition & process descriptiondescription
Other definitionDistillation 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
Laboratory / TestingLaboratory / Testing
1.2: Physical Concept of 1.2: Physical Concept of distillationdistillation
Carried out by either 2 principal methodsFirst 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)
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)
1.3: Vapor – liquid 1.3: Vapor – liquid equilibrium 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
a) VAPOR – LIQUID EQUILIBRIUM OF a) VAPOR – LIQUID EQUILIBRIUM OF AN ORDINARY BINARY LIQUID AN ORDINARY BINARY LIQUID MIXTUREMIXTUREEquilibrium 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.
VAPOR – LIQUID EQUILIBRIUM VAPOR – LIQUID EQUILIBRIUM CURVECURVE
b) Prediction of vapor-liquid b) Prediction of vapor-liquid equilibrium compositions for equilibrium compositions for ordinary binary mixturesordinary 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
Prediction of vapor-liquid Prediction of vapor-liquid equilibrium compositions for equilibrium compositions for ordinary binary mixturesordinary binary mixturesFor 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 phaseyA = mole fraction of A in vapor phasePT = total pressure of the system
Phase Rule
Example: 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
SolutionSolution
1.4: Relative1.4: Relative volatility ( volatility (αα) ) of a of a mixturemixture
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.
Relative volatility (Relative volatility (αα) of a ) of a mixturemixtureThe 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:
Relative volatility (Relative volatility (αα) of a ) of a mixturemixture
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
xAAB xy
xy
y
y
B
A
Relative volatility (Relative volatility (αα) of a ) of a mixturemixtureSeparation is possible for > 1.0For 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.
Relative volatility (Relative volatility (αα) of a ) of a mixturemixture
Example:Example:
Using the data from table below, determine
the relative volatility for the benzene-
toluene system at 85°C and 105°C
Exercise 1Exercise 1A 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 temperatureii. Plot a boiling point diagram for n-hexaneiii. Plot an equilibrium diagram for the mixtureiv. Calculate the αAB at 100 °C
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.
SolutionSolutionVapor 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
PART 2PART 2
Flash & batch distillation Flash & batch distillation
Flash (equilibrium) distillationSimple batch distillation
Flash (Equilibrium) DistillationFlash (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
Flash (Equilibrium) DistillationFlash (Equilibrium) DistillationAs 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.
Flash (Equilibrium) DistillationFlash (Equilibrium) Distillation
Flash (Equilibrium) DistillationFlash (Equilibrium) Distillation
ExampleExampleA 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:
T (K) xA yA
371.6 1 1374 0.825 0.92377 0.647 0.784380 0.504 0.669383 0.387 0.558386 0.288 0.449389 0.204 0.342392 0.132 0.236395 0.068 0.132398.2 0 0
Solution Solution Basis = 100 moles of liquid feed (F)Given, xF = 0.7V = 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* = 0.62equilibrium mol fraction of n-heptane in vapor, y* = 0.76the temperature of the separator at equilibrium ≈ 378oC
Determination of vapor-liquid equilibrium Determination of vapor-liquid equilibrium composition for a flash distillation of n-composition for a flash distillation of n-heptane/n-octane mixtureheptane/n-octane mixture
Figure: Equilibrium curve and operating line
Determination of equilibrium temperature Determination of equilibrium temperature for a flash distillation of n-heptane-n-octane for a flash distillation of n-heptane-n-octane mixturemixture
x
y
Exercise 11.2-1 (page Exercise 11.2-1 (page 752)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.
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
Answer:xA = 0.430
yA = 0.854
Simple batch distillationSimple batch distillationSimple 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.
Simple batch distillationSimple batch distillation
Simple batch distillationSimple batch distillation
1. Raleigh equation for ideal and non-ideal mixtures
Consider a typical differential distillation at an instant time, t1 as shown below:
Simple batch distillationSimple batch distillationNow 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:
Simple batch distillationSimple batch distillation
Applicable for ideal and non-ideal solution
Simple batch distillationSimple batch distillation
The average composition of total material
distilled yav can be obtained by material
balance: (integrate from Rayleigh equation)
avyLLxLxL )( 212211
avVyxLxL 2211
Simple batch distillationSimple batch distillation
2. Simplified Raleigh equation for ideal mixture
Consider a simple batch distillation process at
an initial time, t1, as shown below
Simple batch distillationSimple 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 + dBy
Simple batch distillationSimple batch distillationWe know from definitions,
Since is constant for an ideal mixture,
After simplifying,
Rearranging,
BA
Bx
BA
Ax
dBdA
dBy
dBdA
dAy
BA
BA
AB
BAA
dBdAdB
BAB
dBdAdA
xy
xy
BB
AAAB
/
/
dBA
dABAB
A
dA
B
dBAB
Simple batch distillationSimple batch distillationIntegrating 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
AB A
dA
B
dB
2
1
2
1
2
1
2
1
lnln AA
BBAB
A
A
B
B
AB ABA
dA
B
dB
AB
)5.(lnln1
2
1
2 EqA
A
B
BAB
Simple batch distillationSimple batch distillationExample 1A 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
SolutionSolution
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 -
Solution Solution Given L1 = 100 mol V (mol distilled) = 40
molFrom 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
SolutionSolutionBy 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
835.040
60277.01005.022112211
av
avavav
y
yV
LxLxyVyLxLx
PART 3Continuous distillation
Continuous / retrification Continuous / retrification distillation distillation
Continuous / retrification Continuous / retrification distillation 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.
Continuous / retrification Continuous / retrification distillation 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.
Continuous / retrification Continuous / retrification distillation distillation
Continuous / retrification Continuous / retrification distillation 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
Continuous / retrification Continuous / retrification distillation 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.
Continuous / retrification Continuous / retrification distillation 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
Continuous / retrification Continuous / retrification distillation distillation
Mc-Cabe Thiele MethodMc-Cabe Thiele Method
The intersection of operating The intersection of operating lines, lines, qqFeed 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.
The intersection of operating The intersection of operating lines, lines, qq
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
The intersection of operating The intersection of operating lines, lines, qqSubstituting: , 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
The intersection of operating The intersection of operating lines, lines, qq
•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)
Animation of the construction Animation of the construction of enriching, stripping & q of enriching, stripping & q operating linesoperating lines
http://www.separationprocesses.com/Distillation/DT_Animation/McCabeThiele.html
ExerciseExercise
1) 11.4-12) 11.4-2
Steps of McCabe Thiele Steps of McCabe Thiele MethodMethod1. 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 line5. Plot stripping line6. Determine the stages7. Feed tray location
TutorialTutorial
11.4-5
Exercise 11.4-6Exercise 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
Example 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.
Example Example
Solution Solution Feed, xF = 0.4Product, xD = 0.9Bottom, xw = 0.10Taking 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
Solution 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
SolutionSolutionSince 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
Example 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
Solution Solution xF = 0.5xD = 0.94Xw = 0.02From 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
SolutionSolutionFind q-line. Feed enters at 75%
vapor.
lineforqandPlot
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
SolutionSolutionFrom 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
1 mmm
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
28.31
1
1
28.3)64.1(22
11
1
1
1
min
SolutionSolutionThe number of theoretical stages
required = 10.5 stages including boiler
Feed plate location: 5 from top.
Q1 Final Exam Jan 2012Q1 Final Exam Jan 2012A 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 compositionof 2.0 mole% ethanol are desired. The externalreflux ratio is 5/3 and it is returned as a saturated liquid. It is assumed the condition is at constant molal overflow.
Given;Enthalpy of feed at dew point, Hv =485 kJ/kg.molEnthalpy 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 trayii)The total number of equilibrium trayiii)The feed location
Q2, Final Exam Jan 2013Q2, Final Exam Jan 2013A 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.
Determinea)The flowrates of distillate and
bottom product in kg/hrb)The minimum reflux ratioc)The number of theoretical stages
if the reflux ratio used is 1.3 times the minimum reflux ratio.
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
AZEOTROPIC DISTILLATION AZEOTROPIC DISTILLATION
Azeotrope mixturesMinimum boiling pointMaximum boiling pointAzeotropic Distillation
Azeotrope mixturesAzeotrope mixturesLiquid and vapor are exactly the same
at a certain temperatureIt 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
Azeotropic DistillationAzeotropic DistillationAn 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
Azeotropic DistillationAzeotropic DistillationAzeotropic 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.
Azeotropic DistillationAzeotropic DistillationA 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
Azeotropic DistillationAzeotropic DistillationThe 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)
Azeotropic DistillationAzeotropic DistillationThe characteristic of such mixture is
boiling point curve goes through maximum phase diagram
Example: Acetone-chloroform
Azeotropic DistillationAzeotropic DistillationThe 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)
Azeotropic DistillationAzeotropic DistillationLet say binary mixture: A-B formed an
azeotrope mixtureEntrainer 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
Azeotropic DistillationAzeotropic DistillationExample: Acetic acid-water using entrainer n-butyl acetateBoiling point of acetic acid is 118.1 oC, water is 100 oC & n-butyl acetate is 125 oCThe 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 productThe distillate is condensed and collected in a decanter where it forms 2 insoluble layers
Azeotropic DistillationAzeotropic DistillationExample: Acetic acid-water using entrainer n-butyl acetateTop layer consist of nearly pure n-butyl acetate in water, whereas bottom layer of nearly pure water saturated with butyl acetateThe liquid from top layer is returned to column as reflux and entrainerThe liquid from bottom layer is sent to another column to recover the entrainer (by stream stripping)
Determination of Boiling Point Determination of Boiling Point Temperature in multi component Temperature in multi component distillationdistillationThe calculation is a trial and error
process where1.T is assumed2.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)
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)
Example 11.7-1Example 11.7-1
Bubble point@boiling Bubble point@boiling pointpointBubble point@boiling point=temperature at which liquid
begins to vaporizeDew point=temperature at which liquid
begins to condense out of the vapor