- 1. CHAPTER 1: DISTILLATION Definition& General description
of the process Physicalconcept of distillation
Vapor-liquidequilibriumrelationship Flash& Batch distillation
Continuous Azeotropicdistillationdistillation
Multicomponentdistillation
2. Definition & general description of the process
Separating the various components of a liquid solution Depends upon
the distribution of these components between a vapor phase & a
liquid phase 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 3.
Laboratory / Testing 4. Physical Concept of distillationCarried 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) 5. 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) 6. Vapor liquid equilibrium DEFINITION: EVAPORATION:
The phase transformation processes from liquid to gas or vapor
phase VOLATILITY: The tendency of liquid to change form to gas
VAPOR LIQUID EQUILIBRIUM OF AN ORDINARY BINARY LIQUID MIXTURE
PREDICTION OF VAPOR LIQUID EQUILIBRIUM COMPOSITIONS FOR ORDINARY
BINARY MIXTURES RELATIVE VOLATILITY OF A MIXTURE 7. VAPOR LIQUID
EQUILIBRIUM OF AN ORDINARY BINARY LIQUID MIXTURE A binary liquid
mixture consist of 2 different liquids. Can be classified into
homogeneous mixtures or non-homogeneous (heterogeneous)
mixtures.Low boiler liquid (A) liquid that vaporized easily) low
boiling point or high vapor pressure)High boiler liquid (B) liquid
which have higher boiling point or low vapor pressureHomogeneous
mixtures mix at all proportions resulting in one continuous
phaseHeterogeneous mixtures do not mix uniformly resulting in more
than one distinct phases 8. 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. Equilibrium mole fraction of A in
vapor is larger than mole fraction of A in liquid phase. This is
expected since that A has lower boiling point than B, A would
vaporize more than B. 9. Prediction of vapor-liquid equilibrium
compositions for ordinary binary mixtures Raoults Law for ideal
solution & Daltons Law of partial pressure can be manipulated
in order to calculate compostions of liquid and vapor, which are in
equilibrium. Raoults 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 10. Prediction of vapor-liquid equilibrium compositions
for ordinary binary mixtures For a mixture of the different gases
inside a close container, Daltons 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 = y A PTpA = partial pressure of A in vapor phase yA = mole
fraction of A in vapor phase PT = total pressure of the system 11.
Example: Calculate the vapor and liquid compositions in equilibrium
at 95oC (368.2K) for benzenetoluene using the vapor pressure. The
vapor pressure of benzene = 155.7 kPa and toluene = 63.3 kPa. The
boiling point of benzene = 80oC whereas toluene = 110oC. The
pressure of the system is 101.32 kPa. 12. Relative volatility () 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. Indicates the ease or difficulty of using
distillation to separate the more volatile components from the less
volatile components in a mixture. 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. 13. Relative volatility () of a
mixture Useful in designing all types of distillation processes as
well as other separation or absorption processes which involve the
contacting of vapor and liquid phases in a series of equilibrium
stages. However, cannot be used in separation or absorption
processes that involve components reacting with each other (E.g.
Absorption of CO2 in aqueous solutions of sodium hydroxide). 14.
Relative volatility () of a mixture 15. Relative volatility () of a
mixtureThe 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, AB 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: 16. Relative volatility () of a mixture y A / xA ABy B /
xBy A / xA (1 y A ) /(1 x A )AB relative volatility of A with
respect to B in the binary system If the system obeys Raoults law
for an ideal system: yAPA x A PTyBPB xB PTABPA PByASeparation is
possible for > 1.0xA 1 ( 1)x A 17. Relative volatility () 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 doesnt change with temperature. For solution that
approaches ideal solution, its would fairly constant. 18. Exercise
1 A liquid mixture is formed by mixing n-heptane (A) & n-octane
(B) in a closed container at constant pressure of 1 atm (101.3kPa).
i. Calculate the equilibrium compositions of vapor & liquid ii.
Plot a boiling point diagram for the system iii. Plot an
equilibrium curve for the system iv. Calculate the ABv. What is the
condition of the mixture? Use the following list if vapor pressure
for pure n-heptane & noctane at various temperature. 19. Flash
& batch distillation Flash (equilibrium) distillation Simple
batch distillation 20. Concept of equilibrium Liquid molecules are
continually vaporizing, while vapor molecules are continually
condensing. If 2 chemical species are present, they will condense
& vaporize at the same rates. If not in equilibrium, the liquid
& vapor can be at different temperature and pressure &
present in different mole fraction. At equilibrium, temperature
& pressure & fractions stop to change. Although molecules
continue to evaporate & condense, the rate at which each
species condenses is equal to the rate yA, yB yA + yB = 1.0 at
which it evaporates. AFigure : Vapor-liquid contacting systemPvap,
TvapA Pliq, Tliq xA, xBBBxA + xB = 1.0 21. 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 22. 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. 23. Flash
(Equilibrium) Distillation 24. Flash (Equilibrium) Distillation 25.
Exercise 2 A liquid mixture containing 70 mol% nheptane (A) and 30
mol % n-octane at 30oC is to be continuously flash at the standard
atmospheric pressure vaporized 60 mol% of the feed. What will be
the compositions of vapor and liquid and the temperature of the T
(K) x separator for an equilibrium stage? y 371.6 374 377 380 383
386 389 392 395 398.21 0.825 0.647 0.504 0.387 0.288 0.204 0.132
0.068 01 0.92 0.784 0.669 0.558 0.449 0.342 0.236 0.132 0 26.
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 27. Determination of vapor-liquid equilibrium composition for
a flash distillation of n-heptane/noctane mixture y, mol fraction
of n-heptane in vapor1.210.80.60.40.20 00.20.40.60.81x, mol
fraction of n-heptane in liquidFigure: Equilibrium curve and
operating line1.2 28. Determination of equilibrium temperature for
a flash distillation of n-heptane-n-octane mixture
400395395390390385385380380375375370Temperature
(K)4003700x0.20.40.60.8Mol fraction of n-heptane in vapor (y) and
liquid (x)1y 29. Single-stage equilibrium contact for VL system A
single-stage process is defined as two different phases are brought
into intimate contact with each other and then are separated.
During the time contact, intimate mixing occurs and the various
components diffuse and redistribute themselves between the 2
phases. If the time contact is long enough, they will reach and
equilibrium condition. 30. Single-stage equilibrium contact for VL
system The two entering phases, Lo and V2, of known amounts and
compositions, enter the stage; mixing and equilibration occur; and
the two exit streams, L1 and V1, leave in equilibrium with each
other. If sensible heat effects are small and the latent heats of
both compounds are the same, then when 1 mol of A condenses, 1 mol
of B vaporize. Therefore, the total moles of vapor V2 = V1, also 2
= Lo. This case is 1called as L1 V V constant molal overflow. L1Lo
31. Single-stage equilibrium contact for VL system Total mass
balance: Lo + V2 = L1 + V1 Component balance; LoXAo + V2yA2 = L1xA1
+ V1yA1 Usually, need to find the xA1 and yA1 To solve, need to
plot an equilibrium curve and make trial an error in the
calculations. 32. Single-stage equilibrium contact for VL system
Exersice A vapor at the dew point and 101.32 kPa containing a mole
fraction of 0.40 benzene and 0.60 toluene and 100 kg mol total is
brought into contact with 110 kg mol of a liquid at the boiling
point containing a mole fraction of 0.30 benzene and 0.70 toluene.
The two streams are contacted in a single stage, and the outlet
streams leave in equilibrium with each other. Assume constant molal
overflow. Calculate the amounts and compositions of the exit
streams. 33. 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 so that the vapor and the liquid do not have sufficient
time to reach its equilibriumThe first portion of vapor condensed
will be richest in the more volatile component A. As the
vaporization proceeds, the vaporized product 34. Simple batch
distillation 35. Simple batch distillation Raleigh equation for
ideal and non-ideal mixtures Consider a typical differential
distillation at an instant time, t1 as shown below: 36. Simple
batch distillation 37. Simple batch distillation We know from
definitions, yA xASinceABdA dA dB AxBA By A / xA y B / xBB A BdA dA
dB dB dA dBAfter simplifying, ABdB dA dBis constant for an ideal
mixture, AByBRearranging, dB BB dA A dBABdA AB A B A A B 38. Simple
batch distillation Integrating within the limits of t1 and t2,
B2B1dB BAB A1dA ASince AB is constant, B2 AB B1dB B AB lnA2A2A1dA
AB2 B1ln B B2 AB B1lnA2 A1ln A A2 A1Eq .(5)Equation 5 known as
simplified Raleigh equation for simple batch distillation which
applicable for ideal solution. 39. 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
liquid left. The equilibrium data as follows, where x and y are
mole fractions of n-pentane: xA yA
1.0001.0000.8670.9840.5940.9250.3980.8360.2540.7010.1450.5210.0590.27100
40. Solution Given L1 = 100 mol V (mol distilled) = 40 mol From
material balance: L1L2 VL2L1 VL260 molSubstiting into Eq. (4) lnL1
L2x1x21 yxdx0.50.510 x2100 40ln 1yx100 600.5x21 yxdxdxThe unknown,
x2 is the composition n of the liquid L2 at the end of batch
distillation. 41. Solution0.511By plotting graphy x versus x, they
x dx x2 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: x1L1x 2L2y avy avx1L1 x 2L2 Vy av0.5 100 0.277
60 400.835y av VAverage composition of total vapor distilled, yav =
0.835 42. 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
43. 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, 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 (refulx)
will supply the bulk of liquid need to mix with vapor. 44.
Continuous / retrification distillation 45. 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 processdo not change as
time passes during operation. With constant molal overflow
assumption: Ln 1LnLn1....etc. Vn1VnVn1Conditions for constant molal
overflow:....etc. Heat loses negligible (achieved more easily in
industrial column) Negligible heat of mixing Equal or close heats
of vaporization 46. Continuous / retrification distillationNumber
of plates required in a distillation columnFour 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+1Action 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. 47. 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 48. Continuous /
retrification distillation 49. Mc-Cabe Thiele Method 50. 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. 51. The intersection of operating lines,
needed to vaporize 1 mol of feed at entering conditions q heat
qmolar latent heat of vaporizati on of feed qHv HvHF HLq 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 qF Lm Ln
flows above & below (1) entrance of feed: Vn Vm (1 q )F
(2)Rewrite the equations ofn enriching3& stripping Vn y L x Dx
D ( ) without the tray subscripts: Wxw (4) Vm y Lm xSubtractingVn
)yfrom m Ln )x (DxD Wxw ) (5) (Vm (3) (L (4) 52. The intersection
of operating lines, q Substituting:FxF will produce: y q q 1xDxD
WxwxF q 1, Eq. (1) & (2) into (5)(q line equation )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. 53. The intersection of operating lines, q Dependingon 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) 54. Animation of the construction of
enriching, stripping & q operating lines
http://www.separationprocesses.com/Distillation/DT_
Animation/McCabeThiele.html 55. 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.
56. Example 57. Solution Feed, xF = 0.4 Product, xD = 0.9 Bottom,
xw = 0.10 Taking basis; 100 kmol of feed. A total mass balance:
hence; W = 100 D (Eq. 1)F=D+WA balance on MVC (benzene); F xF 40D
xDW xw0.9D 0.1 W100(0.4)D(0.9) W (0.1)(Eq.2)From the calculations;
D = 37.5 kmol, W = 62.5 kmol 58. Solution Using notation from
reflux: R LnLn Ln D 112 .5Ln3DLn3(37.5)From material balance at the
top stage; VnRD1LnDVn150kmol1Thus, the operating line equation: yn
yn1Ln xn Vn 1D xD Vn 110.75 x n0.225yn1112 .5 xn 150D xD Vn 1 59.
Solution Since the feed is all liquid at its boiling point, it will
all run down as increased reflux to the plate below: LmFLm112.5
100Lm212 .5kmolThe material balance at the bottom:LmLnVm1WVm212 .5
62.5Vm150 kmolBottom operating line equation: ym ym1Lm xm Vm 1W xw
Vm 111.417 x m0.0417ym1212 .5 xm 15062.5 (0.1) 150Vn 60. Example
11,200 kg/h of equal parts (in wt) of BenzeneToluene 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 61. Solution xF = 0.5
xD = 0.94Xw = 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 62. Solution Find q-line. Feed enters at 75%
vapor. q0.75(qvapor ) 0.25(q liquid )q0.75(0) 0.25(1.0)q0.25q line
equation yq yq Let xq q 1xf q 1xq0.25 xq 0.25 1 0.3, y q0.54 0.25
1yq0.333 (0.3) 0.72Plot (0.54, 0.54 ) and (0.3, 0.62 ) forq0.333 x
qyq0.62line0.72 63. Solution From the graph, y intercept for q-line
= x 0.95 0.36 0.36 0.36 R 1.64 DRm1Rm1mThe number of theoretical
stages required if the reflux ratio used is 2 times R 2Rmin the
minimum2(1.64 ) Rratio reflux 3.28 yn1yn1yn1At xR 1 xn xD R 1 R 1
3.28 1 xn (0.95 ) 3.28 1 3.28 1 0.766 x n 0.222 0.5,yn10.766 (0.5)
0.222yn10.605Plot (0.95, 0.95 ) and (0.5, 0.605 ) for enriching OL
64. Solution The number of theoretical stages required = 10.5
stages including boiler Feed plate location: 5 from top. 65.
AZEOTROPIC DISTILLATION Azeotrope mixtures Minimum boiling point
Maximum boiling point Azeotropic Distillation 66. Azeotrope
mixtures Liquid and vapor are of 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 67. 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 68. 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 69. 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 70. 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 doesnt change in
composition This is called as minimum boiling point (positive
deviation) 71. Azeotropic Distillation The characteristic of such
mixture is boiling point curve goes through maximum phase diagram
Example: Acetone-chloroform 72. 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) 73. 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 seapration of
A and B can be achieved 74. 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 75. 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)