68186_Lecture 11-A

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Dr Saad Al-ShahraniChE 334: Separat ion Process es

Limiting condition

D

LRD

McCabe Thiele Graphical

Equilibrium-Stage

If the reflux ratio ( ) is increased to very large value, the

operating lines become the 45o

line. The infinite reflux ratio occursin real life when the column is operated under what are called (total

reflux) condition

a) Minimum number of plates:

Under these conditions, no feed is added to the column (F=0) and

no products are withdrawn (D=0, B=0), but the vapor is boil up andcondensed to the column. So the column is just circulating vapor

and liquid up and down. Most columns are started up under total

reflux conditions.

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Distillation of Binary Mixture

Since the liquid flow rate in the column is same as the vapor flow

rate, and0.1V

L

V

L

The operating line

nD

nn xDL

Dxx

DL

Ly

1 m

B

mm xBL

Bxx

BL

Ly

1

,

The composition in the base of the column under total reflux = xB, and

the composition of the liquid in the reflux drum = xD

In this case the number of ideal plates is minimum.

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Binary Multistage Distillation

The minimum number of ideal plates can be done by:

a) Graphically as shown in the figure

y

XDXFXB

XB

x

y1

y2

y3

y4

X1

X2

X3

Operating lines

as total reflux

Composition of liquid

in reflux drum

Composition of liquid

in re-boiler

y1 = xD

y2 = x1

y3 = x2

y4 = x3

xB = xB

Minimum number ofplates = 3+reboiler

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b) Analytically (using Fenske Equation)

This equation gives the number of plates required under total reflux at

constant .

It is applicable to multi-component system as well as binary system(= constant, total reflux, ideal system).

It is very useful for getting quick estimates of the size of a column.

Derivation of Fenske Equation

Consider two component (A,B) forming ideal solution

productbottominrationmole

productin topratiomole

/

/

/

/

BA

BA

BB

AA

B

A

ABxx

yy

xy

xy

K

K (1)

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An ideal mixture follows Raoults law and = vapor product ratio


P

xP

P

PxPP A

sat

AAA

sat

AA Ay

P

xP

P

P

xPP

B

sat

BB

B

sat

BB

By

sat

B

sat

A

BB

sat

B

AA

sat

A

BB

AA

B

A

ABP

P

PxxP

PxxP

xy

xy

K

K

/

/

/

/

sat

B

sat

A PP / does not change much over the range of temperatureencountered,ABconstant

1,

1 A

A

B

A

A

A

B

A

x

x

x

x

y

y

y

y

(2)

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Substitute (2) in (1)

1

1

1

1

1

1

n

n

AB

n

n

x

x

y

y

AB

A

A

A

A

A

A

ABx

x

y

y

y

y

11,

1

DL

Dxx

DL

Ly Dnn

1

For plate n+1

Since D = 0 (total reflux), L / V= 1.0 ,

Then yn+1= xn and1

1

1

1

n

n

AB

n

n

x

x

x

x

zero

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7/19Dr Saad Al-ShahraniChE 334: Separat ion Process es


At the top of the column, if a totalcondenser is used y1= xD, n = 0

Substitute in (2)

1

1

1

1 x

x

x

x

ABD

D

For plate (1)

xD

y1

water

Re-boiler

Vbyb

Lb, xb

steam

x1

x2

x3

xn

xn-1

x0

y2

y3

y4

yn-1

yn

yrB

B

AB

n

n

x

x

x

x

1

1

For plate (n)

. . . .

. . . .

. . . .

n

n

AB

n

n

x

x

x

x

1

1 1

1

For re-boiler plate

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If all equations are multiplied together and all the intermediate terms

canceled,

B

Bn

AB

D

D

x

x

x

x

1)(

1

B

BN

AB

D

D

x

x

x

x

1)(

1

1min

AB

BBDD xxxxNln

)]1//()1/ln[(1min

AB

BBADBA

AB

BD xxxxN

ln

)//()/ln(

ln

])rationmole/()rationmoleln[(1or min

)1

/()1()(

B

B

D

Dn

ABx

x

x

x

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Equilibrium-Stage

Example:

Calculate the minimum number of trays required to achieved a

separate from 5 mole % bottoms to 90 moles % distillate in a binary

column with =2.5

solutionxB= 0.05 , xD= 0.9

61.419163.0

14.5,

5.2ln

)]05.01/05.0/()9.01/9.0ln[(1 minmin

NN

AB

BBDD xxxxNln

)]1//()1/ln[(1min

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Equilibrium-Stage

Example: in a mixture to be fed to a continuous distillation column, the molefraction of phenol is 0.35, of o-cresol 0.15, of m-cresol 0.3 and of xylenes

0.2. it is hoped to obtain a product with a mole fraction of phenol 0.952, of

o-cresol 0.0474, of m-cresol 0.0.0006. if p-o= 1.26, m-o=0.7, estimate how

many theoretical plates would be required at total reflux.

Solution:

A light component (o-cresol) B heavy component (m-cresol)

Total balance 100=D + B

For phenol 100*0.35=D*0.952+B*xB,p = zero

D= 36.8 Kmol, B = 63.2 Kmol

For ocresol

100*0.15=0.0474*36.8+xB,o

*63.2 xB,o=0.21

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Equilibrium-Stage

o-m= 1/0.7=1.4343.1ln

)]474.0/21.0/()0006.0/0474.0ln[(1min N

5.13min N

component Feed top Bottms

phenol 0.35 0.952 0 p-o= 1.26

o-cresol 0.15 0.0474 0.21 oo-o= 1.0

m-cresol 0.3 0.0006 0.474 m-o=0.7

xylenes 0.2 0 0.316

For mcresol

100*0.3=0.0006*36.8+xB,m*63.2 xB,m=0.474xB,X=0.316

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Equilibrium-Stage

b) Minimum Reflux Ratio

The next figure shows how changing the reflux ratio affects the

operating lines: the lower the reflux ratio, the closer the operating

line moves toward the equilibrium curve, and the larger the number

of plates.

If the reflux ratio finally reduced to the point where either operating

line intersects or becomes tangent to the VLE curve, an infinite

number of plates will be required and the reflux ratio is minimum.

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To obtain the RDmin


Equilibrium-Stage

111

D

D

n

D

D

nR

xx

R

Ry

xx

yx

R

R

D

D

D

D

1min

min

xy

yxR DD

min

or1

interceptabmin

D

D

R

x

y`

1m in D

D

R

x

x

a

bx`

(xD,xD)

y

xD

xD

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Equilibrium-Stage

If the equilibrium curve

has a cavity upward, e.g.,

the curve for water-

ethanol shown in the

figure in this case the

minimum reflux ratiomust be computed from

the slope of the operating

line (ac) that is tangent to

the equilibrium

xy

yxR DD

min

x`

1min D

D

Rx

a

b

y`

Feed lineNon-idealLine VLE

cy`

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0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x

y

.

Feed line

21

3

4

5

6

7

8

9

10

R

xD=0.974xB=0.0235

1D

D

R

x

xF=0.44

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68186_Lecture 11-A

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