Upload
fikrie-muhd
View
215
Download
0
Embed Size (px)
Citation preview
8/19/2019 Azegjuggotropic and MC Distillation
1/31
AZEOTROPIC DISTILLATION› Azeotrope mixtures
› Minimum boiling point
› Maximum boiling point
› Azeotropic Distillation
8/19/2019 Azegjuggotropic and MC Distillation
2/31
Azeotrope mixtures
› Liquid and vapor are exactly the same at acertain temperature
› It is a special class of liquid mixture that boils ata constant temperature at a certain composition
› Cannot be separated by a simple/conventionaldistillation
8/19/2019 Azegjuggotropic and MC Distillation
3/31
Azeotropic Distillation
An introduction of a new component calledentrainer is added to the original mixture to forman azeotrope with one or more of feed component
The azeotrope is then removed as either thedistillate or bottoms
The purpose of the introduction of entrainer is tobreak an azeotrope from being formed by theoriginal feed mixture
Function of entrainer:
◦ To separate one component of a closely boiling point◦ To separate one component of an azeotrope
8/19/2019 Azegjuggotropic and MC Distillation
4/31
Azeotropic Distillation Azeotropic distillation is a widely practiced process
for the dehydration of a wide range of materialsincluding acetic acid, chloroform, ethanol, and manyhigher alcohols.
The technique involves separating close boilingcomponents by adding a third component, called anentrainer, to form a minimum boiling.
Normally ternary azeotrope which carries the wateroverhead and leaves dry product in the bottom.
The overhead is condensed to two liquid phases; theorganic, "entrainer rich" phase being refluxed whilethe aqueous phase is decanted.
8/19/2019 Azegjuggotropic and MC Distillation
5/31
Azeotropic Distillation
› A common example of distillation with an azeotropeis the distillation of ethanol and water.
› Using normal distillation techniques, ethanol canonly be purified to approximately 89.4%
›
Further conventional distillation is ineffective.› Other separation methods may be used are
azeotropic distillation or solvent extraction
8/19/2019 Azegjuggotropic and MC Distillation
6/31
Azeotropic Distillation
› The concentration in the vapor phase is the same asthe concentration in the liquid phase (y=x)
› At this point, the mixture boils at constanttemperature and doesn’t change in composition
› This is called as minimum boiling point (positivedeviation)
8/19/2019 Azegjuggotropic and MC Distillation
7/31
Azeotropic Distillation
› The characteristic of such mixture is boiling pointcurve goes through maximum phase diagram
› Example: Acetone-chloroform
8/19/2019 Azegjuggotropic and MC Distillation
8/31
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)
8/19/2019 Azegjuggotropic and MC Distillation
9/31
Azeotropic Distillation› Let say binary mixture: A-B formed an azeotrope
mixture
› Entrainer C is added to form a new azeotrope withthe original components, often in the LVC, say A
›
The new azeotrope (A-C) is separated from theother original component B
› This new azeotrope is then separated into entrainerC and original component A.
› Hence the separation of A and B can be achieved
8/19/2019 Azegjuggotropic and MC Distillation
10/31
Azeotropic Distillation
Example: Acetic acid-water using entrainer n-butylacetate
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 ofa minimum boiling point azeotrope with water with aboiling point = 90.2 oC.
The azeotropic mixture therefore be distilled over as avapor product & acetic acid as a bottom product
The distillate is condensed and collected in a decanterwhere it forms 2 insoluble layers
8/19/2019 Azegjuggotropic and MC Distillation
11/31
Azeotropic Distillation
Example: Acetic acid-water using entrainer n-butylacetate
Top layer consist of nearly pure n-butyl acetate inwater, whereas bottom layer of nearly pure watersaturated with butyl acetate
The liquid from top layer is returned to column as refluxand entrainer
The liquid from bottom layer is sent to another columnto recover the entrainer (by stream stripping)
8/19/2019 Azegjuggotropic and MC Distillation
12/31
8/19/2019 Azegjuggotropic and MC Distillation
13/31
4. Find the T that corresponds to the calculatedvalue of Kc
5. Compare with T value read from table thatcorresponds to the Kc.
6. If value is differ, the calculated T is used for thenext iteration.
7. After the final T is known, the vapor compositionis calculated from
Yi= (relative volatility x liq mole fraction)/∑(relativevolatility x liq mole fraction)
8/19/2019 Azegjuggotropic and MC Distillation
14/31
Discuss
example 11.7-1
8/19/2019 Azegjuggotropic and MC Distillation
15/31
Bubble point@boiling pointBubble point@boiling point
=temperature at which liquid begins to vaporize
Dew point
=temperature at which liquid begins to condense out of thevapor
8/19/2019 Azegjuggotropic and MC Distillation
16/31
Fractionation of multicomponent mixtures› Three or more components in the products
› The designer generally choose two components whoseconcentrations or fractional recoveries in the distillate andbottom products are a good index of the separation achieved
› These two components are called key components
› They must differ in volatility
› Light key (L) is more volatile (vapor)
› The heavy key (H) is less volatile (liquid
8/19/2019 Azegjuggotropic and MC Distillation
17/31
›
For the compositions, mole fractions ; xH xDH andxL xBL
› xBL means most of the light key ends up at the bottom
› xDH means most of the heavy key ends up at thedistillate
› Normally the key components are adjacent in therank order of volatility. Such choice is called a sharpseparation
› In sharp separations the keys are only componentsthat appear in both products in appreciable
concentrations
8/19/2019 Azegjuggotropic and MC Distillation
18/31
Multi Component Distillation
ie
ie
x
y
p
p
pp
pp
x
yK i
i
ii
i
ii
.
The vapor-liquid equilibria for a mixture aredescribed by distribution coefficients or K factors
where; K
K is the ratio of mole fractions in the vapor and liquid
phases at equilibrium If Raoult’s Law and Dalton’s Law hold, Ki can be
calculated from the vapor pressure and total pressure of
the system.
Pi = xiPi˚ ; yi = Pi/P
8/19/2019 Azegjuggotropic and MC Distillation
19/31
j
i
j j
iiij
K
K
xy
xy
/
/
j
iij
P
P
K factors are strongly temperature dependent
---- ration of K factors is the same as
the relative volatility of the
components
When Raoult’s Law applies,
8/19/2019 Azegjuggotropic and MC Distillation
20/31
Bubble Point and Dew Point Calculation
0.111
i
N
ii
N
ii xKy
0.111
N
i i
iN
ii
K
yx
1. For bubble point (initially boiling point of a liquidmixture)
3. For dew point(intial condensation temperature)
4. A temperature is assumed
5. Obtain Ki values from chart
-----(1)2.
----(2)
8/19/2019 Azegjuggotropic and MC Distillation
21/31
7. Calculate Σ Kixi (for bubble point) or Σ(yi/ Ki) (for dewpoint)
8. If Σ Kixi > 1.00, a lower temperature is chosen untilequation (1) is satisfied
9. If Σ(yi/ Ki) > 1.00, a higher temperature is chosen untilequaition (2) is satisfied
10.The bubble point and dew point are obtained whenequation (1) and (2) is satisfied
11.The compositions can be determined with the formula:
i
N
ii
iii
i
N
ii
iii
Ky
Kyx
xK
xKy
11
/
/
or
8/19/2019 Azegjuggotropic and MC Distillation
22/31
ExampleFind the bubble-point and the dew point temperatures andthe corresponding vapor and liquid compositions for a mixtureof 33 mole % n-hexane, 37 mole % n-heptane and 30 mole %n-octane at 1.2 atm total pressure. (The equilibrium K values
for hydrocarbon system is given)
8/19/2019 Azegjuggotropic and MC Distillation
23/31
The equilibrium K values graph
8/19/2019 Azegjuggotropic and MC Distillation
24/31
For bubble point:Choose T = 105 C, where the vapor pressure of the
middle component at 1.2 atm. Construct and fill in thetable:
Component Pi˚ at
105˚C,1.2 atm
K i = Pi˚/P xi yi =K ixi
1 Hexane 2.68 2.23 0.33 0.7359
2 Heptane 1.21 1.01 0.37 0.3737
3 Octane 0.554 0.462 0.30 0.1386
8/19/2019 Azegjuggotropic and MC Distillation
25/31
› Since Σ Kixi > 1.00, try a lower temperature. The major
contribution comes from hexane, pick a temperaturewhere Ki is lower by a factor (1/1.248)
› Example;
1/1.248 x 2.23 = Pi˚/1.2
Pi˚ = 2.14 atm where T = 96 ˚C
8/19/2019 Azegjuggotropic and MC Distillation
26/31
Component Pi˚ at
96˚C, 1.2
atm
K i =
Pi˚/P
xi yi =K ixi yi
calculated
1 Hexane 2.14 1.8 0.33 0.5940 0.604
2 Heptane 0.93 0.755 0.37 0.2868 0.292
3 Octane 0.41 0.342 0.30 0.1025 0.104
Σ =
0.9833
Σ =
1.00
Repeat the iteration until Σ Kixi = 0.9833
8/19/2019 Azegjuggotropic and MC Distillation
27/31
The yi calculated is using the formula;
1i
N
ii
iii
xK
xKy
By interpolation, the bubble point = 97 ˚C
Therefore, the vapor in equilibrium with the liquid is 60.4
mol % n-hexane, 29.2 mole % n-heptane and 10.4 mole %
n-octane.
8/19/2019 Azegjuggotropic and MC Distillation
28/31
For dew point;
Choose a higher T = 105 ˚C as first guess
Component Pi˚ at
105˚C, 1.2
atm
K i = Pi˚/P xi xi =yi/K i
1 Hexane 2.68 2.23 0.33 0.1482 Heptane 1.21 1.01 0.37 0.366
3 Octane 0.554 0.462 0.30 0.655
Σ =
1.169
8/19/2019 Azegjuggotropic and MC Distillation
29/31
8/19/2019 Azegjuggotropic and MC Distillation
30/31
Componen
t
Pi˚ K i yi yi/K i xi calculated
Hexane 3.0 2.5 0.33 0.132 0.130
Heptane 1.38 1.15 0.37 0.3217 1.317
Octane 0.64 0.535 0.30 0.5625
Σ = 1.0162 Σ = 1.00
8/19/2019 Azegjuggotropic and MC Distillation
31/31
By extrapolation, the dew point = 110.5 and the
composition of the liquid in the equilibrium with thevapor is calculated by formula;
i
N
ii
iii
Ky
Kyx
1
/
/
which 13.0 mole % n-hexane, 31.7 mole % n-
heptane and 55.3 mole % n-octane.