# McCabe Thiele Graphical Equilibrium-Stage - 12.pdf · McCabe Thiele Graphical Equilibrium-Stage ...

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Dr Saad Al-Shahrani ChE 334: Separation Processes

Limiting condition

McCabe Thiele Graphical

Equilibrium-Stage

When analyzing or designing a process, it is useful to look at limiting

cases to assess the possible values of process parameters. In

distillation analysis, separation of a pair of components can be improved

by increasing the number of stages while holding reflux constant, or by

increasing the reflux flow for a given number of stages. This tradeoff

sets up two limiting cases:

1. Total Reflux (minimum ideal stages)

2. Minimum Reflux (infinite ideal stages)

The design tradeoff between reflux and stages is the standard economic

optimization problem chemical engineers always face -- balancing capital costs

(the number of trays to be built) vs. the operating cost (the amount of reflux to be

recirculated). A good design will operate near a cost optimum reflux ratio.

Dr Saad Al-Shahrani ChE 334: Separation Processes

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 occurs

in 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no products are withdrawn (D=0, B=0), but the vapor is raised up

and condensed to the column. So the column is just circulating

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

total reflux conditions.

Dr Saad Al-Shahrani ChE 334: Separation Processes

Distillation of Binary Mixture

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

rate, and

0.1V

L

V

L

The operating line

nD

nn xDL

Dxx

DL

Ly

1 m

Bmm x

BL

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.

Dr Saad Al-Shahrani ChE 334: Separation Processes

Binary Multistage Distillation

The minimum number of ideal plates can be done by:

a) Graphically as shown in the figure

y

XD XF

XB

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 of

plates = 3+reboiler

Dr Saad Al-Shahrani ChE 334: Separation Processes

Binary Multistage Distillation

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

product bottomin ration mole

product in top ratio mole

/

/

/

/

BA

BA

BB

AA

B

AAB

xx

yy

xy

xy

K

K

(1)

Dr Saad Al-Shahrani ChE 334: Separation Processes

An ideal mixture follows Raoults law and = vapor product ratio

Binary Multistage Distillation

P

xP

P

PxPP A

sat

AAA

sat

AA Ay

P

xP

P

PxPP B

sat

BBB

sat

BB By

sat

B

sat

A

BB

sat

B

AA

sat

A

BB

AA

B

AAB

P

P

PxxP

PxxP

xy

xy

K

K

/

/

/

/

sat

B

sat

A PP /

does not change much over the range of temperature

encountered,AB constant

1

, 1 A

A

B

A

A

A

B

A

x

x

x

x

y

y

y

y

(2)

Dr Saad Al-Shahrani ChE 334: Separation Processes

Binary Multistage Distillation

Substitute (2) in (1)

1

1

1

1

1

1

n

n

AB

n

n

x

x

y

y

AB

A

A

A

A

x

x

y

y

11

DL

Dxx

DL

Ly Dnn

1

For plate n+1

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

Then yn+1 = xn and 1

1

1

1

n

n

AB

n

n

x

x

x

x

zero

Dr Saad Al-Shahrani ChE 334: Separation Processes

Binary Multistage Distillation

At the top of the column, if a total

condenser is used y1 = xD , n = 0

Substitute in (2)

1

1

1

1 x

x

x

xAB

D

D

For plate (1)

xD

y1

water

Re-boiler

Vb yb

Lb, xb

steam

x1

x2

x3

xn

xn-1

x0

y2

y3

y4

yn-1

yn

yr

B

BAB

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

For plate (2) 2

2

1

1

1

1 x

x

x

xAB

Dr Saad Al-Shahrani ChE 334: Separation Processes

Binary Multistage Distillation

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

])ration mole/()ration moleln[(1or min

Where n= Nmin + reboiler

Dr Saad Al-Shahrani ChE 334: Separation Processes

McCabe Thiele Graphical

Equilibrium-Stage

Example:

Calculate the minimum number of trays required to achieve a

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

column with =2.5

solution xB = 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

Dr Saad Al-Shahrani ChE 334: Separation Processes

McCabe Thiele Graphical

Equilibrium-Stage

Example: in a mixture to be fed to a continuous distillation column, the mole

fraction 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. Assume no phenol in the bottoms.

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 o cresol

100*0.15=0.0474*36.8+xB,o*63.2 xB,o=0.21

Dr Saad Al-Shahrani ChE 334: Separation Processes

McCabe Thiele Graphical

Equilibrium-Stage

o-m= 1/0.7=1.43 43.1ln

)]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 o-o= 1.0

m-cresol 0.3 0.0006 0.474 m-o=0.7

xylenes 0.2 0 0.316

For m cresol

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

xB,X=0.316

Dr Saad Al-Shahrani ChE 334: Separation Processes

McCabe Thiele Graphical

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.

Dr Saad Al-Shahrani ChE 334: Separation Processes

To obtain the RDmin

McCabe Thiele Graphical

Equilibrium-Stage

111

D

Dn

D

Dn

R

xx

R

Ry

1intercept ab

min

D

D

R

x

1min D

D

R

x

x

a

b x`

(xD,xD)

y

xD

xD

Dr Saad Al-Shahrani ChE 334: Separation Processes

McCabe Thiele Graphical

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 ratio

must be computed from

the slope of the operating

line (ac

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