Azegjuggotropic and MC Distillation

8/19/2019 Azegjuggotropic and MC Distillation

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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 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


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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


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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.


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› 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


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› 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)


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› The characteristic of such mixture is boiling pointcurve goes through maximum phase diagram

› Example: Acetone-chloroform


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› 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)


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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


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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


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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)


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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)


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Discuss

example 11.7-1


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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


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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


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›

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


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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


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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,


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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)


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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


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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)


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The equilibrium K values graph


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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


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› 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


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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


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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.


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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


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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


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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.

Azegjuggotropic and MC Distillation

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