08/12/2015 1 Azeotropic Distillation AZEOTROPE / AZEOTROPIC MIXTURE MINIMUM BOILING POINT (POSITIVE DEVIATION) MAXIMUM BOILING POINT (NEGATIVE DEVIATION) AZEOTROPIC DISTILLATION Very large deviations from ideally lead to a special class of mixtures known as azeotropes. Azeotrope is a special class of liquid mixture that boils at a constant temperature at a certain composition. It acts if it was one component with one constant boiling point. In other words, vapor and liquid compositions are equal. Cannot be separated by conventional distillation. For example ethanol –water system, one cannot recover more than 89.4 mol% ethanol by ordinary distillation because the mixture forms azeotropes at this point. Other separation methods (for eg solvent extraction) or azeotropic distillation may be used. Two types of azeotropes known are: (i) Minimum boiling (ii) Maximum boiling (less common) AZEOTROPE / AZEOTROPIC MIXTURE
Transcript
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Azeotropic Distillation
AZEOTROPE / AZEOTROPIC MIXTURE
MINIMUM BOILING POINT (POSITIVE DEVIATION)
MAXIMUM BOILING POINT (NEGATIVE DEVIATION)
AZEOTROPIC DISTILLATION
Very large deviations from ideally lead to a special class of mixtures known as azeotropes.
Azeotrope is a special class of liquid mixture that boils at a constant
temperature at a certain composition. It acts if it was one component with one constant boiling point. In other words, vapor and liquid compositions are equal. Cannot be separated by conventional distillation. For example ethanol –water system, one cannot recover more than 89.4 mol%
ethanol by ordinary distillation because the mixture forms azeotropes at this point. Other separation methods (for eg solvent extraction) or azeotropic distillation
may be used. Two types of azeotropes known are:
(i) Minimum boiling (ii) Maximum boiling (less common)
AZEOTROPE / AZEOTROPIC MIXTURE
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At point Z2, the concentration in the vapor phase is the same as the concentration in the liquid phase (y=x) and =1.
At this point, the mixture boils at constant temperature and doesn’t change
in composition. On the equilibrium diagram, the equilibrium crossed at 45O diagonal line.
MINIMUM BOILING POINT (POSITIVE DEVIATION)
Z2
The most common examples of this type of azeotrope are:
Benzene – water (29.6 mole% water, 69.25OC, 1 atm)
MINIMUM BOILING POINT (POSITIVE DEVIATION)
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At point Z2, the characteristic of such mixture is boiling point curve goes through maximum phase diagram.
Example of this type of azeotrope is:
Acetone – chloroform
Z2
MAXIMUM BOILING POINT (NEGATIVE DEVIATION)
The distillation of a mixture that is azeotropic begins similarly to conventional distillation.
The difference is that, as the process continues. A temperature is reached at which the compositions of the vapor phase and the liquid phase become the SAME.
Once this azeotrope has been formed, the individual components can no longer be separated by conventional distillation. *Note: the description of azeotrope distillation only refers to minimum boiling distillation of azeo. only.
Recall the mixture of ethanol – water, that demonstrates azeo behaviour at 89.4 mol% at 78.15OC and 1 atm.
AZEOTROPIC DISTILLATION
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Therefore, by conventional distillation, the highest concentration of ethanol achieved is the azeotropic concentration, i.e 89.4 mol%.
Azeotropic distillation refers to 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.
This type of distillation also refers to a new components (entrainer) is added to
break an azeotrope from being formed by the original feed mixture.
Entrainer – as described above has the following functions that are:
To separate one component of a closely boiling point, or To separate one component of an azeotrope.
AZEOTROPIC DISTILLATION
Recall the binary mixture, A – B that formed an ezeotropic mixture, the 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.
Alternative method to azeo – distillation: is when no entainer is used, instead the column is operated at different pressure, since changing the operating P can eliminate azeotropic behaviour in some systems.
AZEOTROPIC DISTILLATION
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Example of Azeotropic Distillation:
AZEOTROPIC DISTILLATION
Distillation acetic – acid water using entrainer: n-butyl acetate.
Boiling point (B.P) of acetic acid is 118.1OC while water is 100OC, whereas n-butyl acetate is 125OC.
The addition of n-butyl acetate results in the formation of a minimum boiling azeotrope with water with a B.P of 90.2OC.
The azeotropic mixture will therefore be
distilled over as a vapor product and acetic acid as bottom product.
The distillate is condensed and collected in
a decanter where it forms two insoluble layers.
Example of Azeotropic Distillation:
AZEOTROPIC DISTILLATION
Distillation acetic – acid water using entrainer: n-butyl acetate.
Top layer consists of nearly pure 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).
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AZEOTROPIC DISTILLATION
Distillation ethanol-water system.
Entainer: Benzene with Boiling point (B.P) at 80.2OC that forms a new-minimum boiling point azeotrope with 22.8 mol% ethanol (E), 23.3 mol% water (W) and 53.9 mol% benzene (B) (three comp. azeotrope).
The B.P of the azeotrope mixture is 64.86OC which leaves top product and
bottom consist of nearly pure ethanol.
The condensed overhead product forms two layer in the decanter – top and bottom layer).
The bottom layer is sent to a 2nd column where benzene is stripped and returned to the 1st column.
The bottom product (E-W) are fed to a 3rd Distill.
for further distillation.
Top layer *Rich in benzene. *Returned to column as reflux and entrainer
Bottom layer *Rich in water with a small amount of ethanol.
EQUILIBRIUM DATA IN MULTI. DISTILLATION
BOILING POINT, DEW POINT AND FLASH DISTILLATION
LIGHT AND HEAVY KEY COMPONENT
Multi – component Distillation
INTRODUCTION OF MULTI – COMP. DISTILLATION
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For binary mixture for constant P – vapor liquid equilibrium curve rather unique.
With a ternary mixture, the conditions of equilibrium curve are more complex
for constant P where mol fraction of the two components in liq. phase must be given before composition of vapor can be fixed/determine.
Therefore, the mol fraction yA not only depends on xA but also on the
proportions of the other two components. Graphical method are not generally suitable. The possible method used – plate to plate analysis. For many systems consisting particularly similar chemically substances, the
relative volatilities of the components remain constant over a wide range of T and composition.
INTRODUCTION OF MULTI – COMP. DISTILLATION
This is illustrated by the table below:
INTRODUCTION OF MULTI – COMP. DISTILLATION
Temperature (K)
Component 353 393 453
Phenol 1.25 1.25 1.25
o – Cresol 1 1 1
M – Cresol 0.57 0.62 0.70
Xylenols 0.30 0.38 0.42
An alternative method is to use the simple relation.
Particularly useful for hydrocarbon processing in petroleum industry The K values can be measured .
ratio) onVaporizati mEquilibriu (or
constant mequilibriu K where K AA xy
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The feed and Product Compositions
Gilliand and Reed have determined the no. of freedom, F for multi – component Distillation (MCD).
For the common case:
where Feed composition, nature of feed and operating pressure are given, there remain only four variables to be selected.
If the reflux ratio, R is fixed, the no. of plate above and below the feed plate
are chosen to give the best used of plates, then only two variables remains. This means that the top and bottom product composition has been
determined by trial and error method. This problem of Trial and Error method can rather be simplified by
assuming MVC appears at top and LVC appears on bottom product.
INTRODUCTION OF MULTI – COMP. DISTILLATION
Consider as ideal mixture, we can apply Raoult’s Law to determine the composition of vapor in equilibrium with liquid.
Consider a four – component (A,B,C,D) mixture:
In hydrocarbon system, because nonidealities, the equilibrium data are often represented by:
KA is the vapor – liquid equilibrium constant or distribution coefficient for
component A. for component i individual component.
EQUILIBRIUM DATA IN MULTI. DISTILLATION
D
oD
DC
oC
CB
oB
BA
oAA
A
oDDD
oCCC
oBBB
oAAA
xP
Pyx
P
Pyx
P
Pyx
P
P
P
py
PxpPxpPxpPxp
DDDCCCBBBAAA xKyxKyxKyxKy
P
PK
oi
i
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K value is easily determined (refer K-chart) for hydrocarbons system but need to be calculated for other system.
The relative volatility, i for each individual component in a multi component mixture can be defined in a similar manner to that of a binary mixture.
Need to select a base component, say component C:
EQUILIBRIUM DATA IN MULTI. DISTILLATION
C
DD
C
CC
C
BB
C
AA
C
ii
K
Kα
K
Kα
K
Kα
K
Kα,
K
Kα 1.0 hence
The values of Ki depend strongly on T.
Boiling Point.
xαKxKy iiCiii 1.0
T is assumed. Values of i are then calculated using K values at the assume T. Then calculate value of KC where
Find the T that corresponds to the calculated value of KC Compare with T value read from table that corresponds to the KC
(calculated)
BOILING POINT, DEW POINT AND FLASH DISTILLATION
At a specified pressure, the boiling point or bubble point of a given multi – component mixture satisfy the relation yi = 1.0. For a mixture A, B, C, and D, with C as the base component,
The calculation is a trial – and – error process where:
xα
Kii
C
1.0
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Boiling Point.
If values is differ, the calculated T is used for next iteration. After the final T is known, the vapor composition is calculated from:
BOILING POINT, DEW POINT AND FLASH DISTILLATION
xα
xαy
ii
iii
Dew Point.
Also based on the trial and error method:
α
y
KK
yx
i
i
Ci
ii 1.0
1
The values of KC is calculated from:
ii
Cxα
K1.0
After the final T is known, the liquid composition is calculated from:
ii
iii
αy
αyx
BOILING POINT, DEW POINT AND FLASH DISTILLATION
Example 3
A liquid feed to a distillation tower at 405.3 kPa abs is fed to a distillation tower. The composition in mol fractions is as follows:
n – butane (xA = 0.40) n –pentane (xB = 0.25) n – hexane (xC = 0.20) n – heptane (xD = 0.15)
Calculate the boiling point and the vapor in equilibrium with liquid.
Solution 3
Assume first trial and error temperature, let say T=65OC. Find the values of K from Figure B1
Using component C (n – hexane) as base component, the following values
are calculated for the first trial:
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BOILING POINT, DEW POINT AND FLASH DISTILLATION
Figure B1: Equilibrium K values for light hydrocarbon system at 405.3 kPa absolute.
BOILING POINT, DEW POINT AND FLASH DISTILLATION
Component xi Ki (from chart) i=Ki/KC ixi
n-butane (xA) 0.40 1.68 6.857 2.743
n-pentane (xB) 0.25 0.63 2.571 0.643
n-hexane (xC) 0.20 0.245 1.000 0.200
n-heptane (xD) 0.15 0.093 0.380 0.057
1.00 ixi = 3.643
Assume first trial and error temperature, let say T=65OC.
Set n-hexane as base component. Initial Kc = Ki (hexane) = 0.245
6.8570.245
1.68α butane,-n For
C
i
C
ii
K
K
K
Kα
Do same calculation for n-pentane, n-hexane and n-heptane For the first trial (T=65OC),
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BOILING POINT, DEW POINT AND FLASH DISTILLATION
Component xi Ki (from chart) i=Ki/KC ixi yi
n-butane (xA) 0.40 1.86 6.607 2.643 0.748
n-pentane (xB) 0.25 0.710 2.522 0.631 0.178
n-hexane (xC) 0.20 0.2815 1.000 0.200 0.057
n-heptane (xD) 0.15 0.110 0.391 0.059 0.017
1.00 ixi = 3.533 1.000
0.27453.643
11K
iC
ix From table above,
From graph, for n-hexane at K = 0.275, T = 69OC Using T = 69OC for trial 2, a temperature of 70OC is obtained. Using T = 70OC for trial 3, the calculation shown final value of 70OC,
which is bubble point. Values y is calculated from: From trial 3,
yi=ixi/ixi
The essential requirement in Multi – comp. Distillation is to separate the two components.
These components are called key component. Suppose a four component mixture A-B-C-D where:
A is MVC B is LVC
LIGHT AND HEAVY KEY COMPONENT
where the mixture is to be separated as below:
Feed Top Product Bottom Product
A A
B B B
C C C
D D
Then; B – is the lightest comp. in bottom and called Light Key Component (LKC)
C – is the heaviest comp. in top / distillate and called Heavy Key Component (HKC) The main purpose of fractionating is to separate B from C
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The calculation of No. of plates required.
LIGHT AND HEAVY KEY COMPONENT
Developed by Lewis and Matheson (which is based on Lewis – Sorel method).
This L – M method illustrates that:
if the comp. of liquid on any plate is known, then the vapor component can be calculated using the vapor pressure @ of individual components.
Suppose: a component mixture A-B-C-D with liquid composition xA, xB, xC, xD,& etc. and vapor comp. yA, yB, yC, yD ,& etc. Then…..
The calculation of No. of plates required.
From the relation of vapor pressure, mole fraction, relative volatility (α) and the K value:
ii
DBDD
ii
CBCC
ii
ABAA
ii
BB
x
xy
x
xy
x
xy
x
xy
entce) compone (referenis the basLet say B
Example 4
A mixture of ortho (o), meta (m), and para (p) mononitrotoluenes with mole percentage as below:
ortho (o) (xo = 0.60) meta (m) (xm = 0.04)
The mixture is to be distilled at temperature of 410 K requiring a pressure in the boiler of about 0.06 bar.
LIGHT AND HEAVY KEY COMPONENT
para (p) (xp = 0.36)
is to be continuously distilled to give a top product of 98 mol% ortho, and the bottom is to contain 12.5 mol% ortho.
If a reflux ratio of 5 is used, calculate: a) The number of plates will be required b) The approximate compositions of product streams
The volatility of ortho relative to para isomer is taken as 1.70 and the volatility of meta relative to para is 1.16 over the temperature range of 380 to 415 K
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Solution 4
As a first estimate, suppose the distillate to contain 0.6 mol% meta and 1.4 mol% para.
Material balance will give the composition of the bottom.
LIGHT AND HEAVY KEY COMPONENT
Do overall material balance to find the amount of Distillate and Bottom Product
The equation of operating lines below the feed plate can be written as:
Ortho
Meta
Para
w
m
m
m
mm x
V
Wx
V
Ly 1
083.036.333
44.44
36.333
8.3771 mmm xy
125.036.333
44.44
36.333
8.3771 mmo xy 0167.0133.1 1 mmo xy
011.0133.1 1 mmm xy
792.036.333
44.44
36.333
8.3771 mmp xy 105.0133.1 1 mmp xy
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Solution 4
LIGHT AND HEAVY KEY COMPONENT
The equation of operating lines above the feed plate can be written as:
Ortho
Meta
Para
d
n
n
n
nn x
V
Dx
V
Ly 1
163.0833.0 1 nno xy 98.036.333
56.55
36.333
8.2771 nno xy
006.036.333
56.55
36.333
8.2771 nno xy 001.0833.0 1 nnm xy
014.036.333
56.55
36.333
8.2771 nnp xy 003.0833.0 1 nnp xy
Solution 4
LIGHT AND HEAVY KEY COMPONENT
Start the calculation from bottom plate (below the feed plate)
Component xs xs ys x1
Ortho 1.7 0.125 0.213 0.193 0.185
Meta 1.16 0.083 0.096 0.087 0.087
Para 1 0.792 0.792 0.719 0.728
1.101 1.000 1.000
Component x1 x1 y1 x2
Ortho 0.185 0.315 0.275 0.258
Meta 0.087 0.101 0.088 0.088
Para 0.728 0.728 0.637 0.655
1.000 1.143 1.000 1.000
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Solution 4
LIGHT AND HEAVY KEY COMPONENT
Continue the calculation until x > xf
Component
Ortho
Meta
Para
x6 x6 y6 x7
0.564 0.96 0.68 0.617
0.064 0.07 0.05 0.056
0.372 0.37 0.26 0.326
1.000 1.40 1.00 1.000
The feed will be introduce at plate 7 since the liquid on this plate has a composition with the ratio of concentration ortho and para about that in the feed. Continue the Calculation above the feed plate.
Solution 4
LIGHT AND HEAVY KEY COMPONENT
Continue the Calculation above the feed plate
Component x7 x7 y7 x8
Ortho 1.7 0.617 1.049 0.728 0.678
Meta 1.16 0.056 0.065 0.045 0.053
Para 1 0.326 0.326 0.227 0.270
1.000 1.441 1.000 1.001
Continue the Calculation until x > xd
Component y15 x16 x16 y16
Ortho 0.970 0.969 1.647 0.980
Meta 0.011 0.012 0.014 0.008
Para 0.019 0.020 0.020 0.012
1.000 1.001 1.681 1.000
The vapor from the sixteenth plate has the required concentration of the ortho isomer. 16 ideal stages is required for the separation.
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