Mutli Component Distillation

7/29/2019 Mutli Component Distillation

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Multi-Component Distillation

Phase equilibrium For multi-components

iii

s

i pPyxP == 12

iii

s

i

i xKxP

P

y == 13

Taking 1 as the reference component

i

i

i

iiiii

x

x

y

y

x

x

y

y

xy

xy

K

K

====

1

1

/

/ 1

1

1

1111

14

Bubble point calculations

Given the total pressure and liquid mole fraction T and y

i

n

i

iii

n

i

i

s

in

i

i

n

i

xKxKxP

Py

====

====1

1

111

1 15

If vapor pressure data is available solve:

011

== =

i

s

in

i

xP

Pf

If k-values data is available solve:

011

== =

ii

n

i

xKf

If relative volatility data is available solve:

011

1 == =

ii

n

i

xKf

Iterate over temperature till solution is found calculate the vapor mole fraction from

i

s

ii x

P

Py =

iii xKy =

32


2/13

=

==

n

i

ii

iiiii

x

xxKy

1

1 ; note that iixK = /11

See Example in Geankoplis BookA liquid stream at 405.3 kPa with 40% butane, 25% bentane, 30% hexane and 15%hepatne is fed to a distillation column. Find the bubble point and the corresponding vapor

composition in equilibrium with the liquid.

Using the K-value diagram we build this Table:

Let K-hexane be the reference component

Let T = 65 oC

component xi Ki xi

Butane

Pentane

0.4

0.25

1.68

0.63

6.857

2.571

2.743

0.643

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Hexane

Heptane

0.2

0.15

0.245

0.093

1.0

0.38

0.2

0.057

Total 3.643

According to Equation 18: 0107.0)643.3(245.01111

===

=ii

n

i

xKf

Let T = 70 oC

component xi Ki xi yi

Butane

BentaneHexane

Heptane

0.40.25

0.2

0.15

1.860.71

0.2815

0.11

6.6072.522

1.0

0.391

2.6430.631

0.2

0.059

0.748

0.178

0.0570.017

Total 3.533 1.000

According to Equation 18: 00547.0)533.3(2815.0111

1 === = iin

ixKf

Dew point calculations

Given P and vapor mole fraction Tandx

i

in

is

i

in

i

i

n

i K

y

PP

yx

===

===111 /

1 16

Thus iterate over temperature till the following is solved:

011

== = i

in

i K

yf

17

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4/13

Flash calculations

iiiLxVyFz += 18

iii xVFVyFz )( += 19

iii xF

Vy

F

Vz )1( +=

20

iiixffyz )1( += 21

iii xf

fz

fy

)1(1 =

22

F, z

V, y

L, x

P, T

Using the equilibrium relation:

iiiiix

f

fz

fxKy

)1(1 ==

23

Solving forx:

1)1( +=

i

ii

Kf

zx

24

11)1(11 =+= == ii

n

ii

n

i Kf

z

x

25

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5/13

iii LxVyFz +=

iiii KLyVyFz /+=

( ) iii yKLVFz /+=

( ) iii VyVKLFz /1+=

i

iii

VKL

FzVyV

/1+==

36


6/13

37


7/13

Another approach

VLF += 1 L 1

iii VyLxFz += n-1 V 1

= 1ix 1 X n =1iy 1 Y n

iiixKy = n T 1

vLF VHLhFh += 1

2n+3 2n+3

Non-ideal systems (activity coefficient)

For non-ideal systems the phase equilibrium:

i

s

iii

xP

Py

=

where is the liquid phase activity coefficient and depends on temperature, pressure andcomposition.

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8/13

Distillation Column design

Stage to stage calculation using computer program Short cut methods

Key component

A single distillation column is designed to separate binary components. Therefore an

important species to be concentrated in the top product Is known as the light key while

the one to be concentrated in the bottom product is the heavy key.

Total reflux minimum number of stages

[ ]log ( / )( / )

log( )av

LD HD HB LB

m

L

x D x D x B x BN

=

LBLDLav=

Bx

Dx

Bx

Dx

HB

HDN

avei

iB

iD m,=

Minimum Reflux

=i

ifixq1

=+i

iDim

xR 1

is the relative volatility at average column temperature

39


9/13

Erbar-Maddox Chart

Feed position

= 2)()(log206.0log

HD

LB

Lf

Hf

x

x

D

B

x

x

Ns

Ne

Example:

A liquid stream at 405.3 kPa and 100 mole/h with 40% butane (A), 25% bentane (B),30% hexane (C) and 15% hepatne (D) is fed to a distillation column.

90% ofB is recovered in distillate and 90% ofCis the bottom, calculate

(a)mole and composition of distillate and bottom(b)Dew point of the distillate and bubble point of the bottom(c)minimum stages for total reflux(d)minimum reflux ratio(e)number of theoretical stage working at 1.5Rm(f) location of feed tray

(a)solve the material balancemole balance on butane

25 = nBD + nBB

Q nBD = 0.9(25) = 22.5 mole nDD = 2.5 mole

mole balance on hexane

30 = nCD + nCB

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10/13

Q ncB = 0.9(20) = 18 mole nCD = 2 mole

Assume allA in distillate nAD= 40 mole nAB = 0Assume allD in bottom nAD= 40 mole nAB = 0

comp nf nD yD nb xB

A 25 40 0.62 0 0

B 25 22.5 0.349 2.5 0.07

C 20 2 0.031 18 0.507

D 15 0 0 15 0.423

total 100 64.5 1.0 35.5 1.0

part (b)

Dew point calculations:i

in

is

i

in

i

i

n

i K

y

PP

yx

===

===111 /

1

start at T = 67oC

comp yD Ki yD/Ki A 0.62 1.75 0.354 6.73

B 0.349 0.65 0.537 2.5

C 0.031 0.26 0.119 1.0D 0 0.1 0 0.385

total 1.0 1.01

Bubble point calculation

i

n

i

iii

n

i

i

s

in

i

i

n

i

xKxKxP

Py

====

====1

1

111

1

start at T = 132 oC

comp xB Ki xBKi A 0 5.0 0 4.384

B 0.07 2.35 0.165 2.043

C 0.507 1.15 0.583 1.0

D 0.423 0.61 0.258 0.53

total 1.0 1.006

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11/13

part ( c)

letB light componentlet C heavy component

LD = 2.5LB = 2.04

258.2)043.2(5.2 === LBLDave

[ ] [ ]4.5

)258.2log(

)07.0/507.0)(031.0/349.0(log

)log(

)/)(/(log==

=

avL

HBLBHDLDm

BxBxDxDxN

Distribution of component A:

43.5)384.4(73.6 === ABADave

103218

243.5 4.5 ==

AB

AD

n

n

40 = nAD + nAB = 1033nAB nAB = 0.039 mole/h nAD = 39.961 mole/h

Distribution of component D:

45.0)385.0(53.0 === DBDDave

0015.018

245.0 4.5 ==

DB

DD

n

n

15 = nDD + nDB = 1.0015nDB nDB = 14.97 mole/h nDD = 0.022 mole/h

comp nD yD nb xB

A 40 0.62 0.024 0.0007

B 22.5 0.349 2.5 0.07C 2 0.031 18 0.507

D 0.022 0.00034 15 0.423

total 64.522 1.0 35.5 1.0

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part (d)

=i

ifixq1

=+i

iDim

xR 1

The average tower temperature = (67 + 123)/2 = 99.5oC

comp xB Ki (99.5) (99.5)A 0.4 3.12 5.2

B 0.25 1.38 2.3

C 0.2 0.6 1.0

D 0.15 0.28 0.467total 1.0

For feed at boiling point q =1

+

+

+

===

467.0

)15.0(467.0

0.1

)2.0(0.1

3.2

)25.0(3.2

2.5

)4.0(2.50111 q

Solve by trial-and-error (see book)

= 1.2096

12096.1467.0

)0(467.0

2096.11

)031.0(0.1

2096.13.2

)35.0(3.2

2096.12.5

)62.0(2.51

+

+

+

=

=i

iDim

xR

Rm = 0.395

part (d)

R = 1.5Rm = 1.5(0.395) = 0.593

R/(R+1) = 0.3723

Rm/(Rm+1) = 0.283

Using the chart Nm/N = 0.49 N = Nm/0.49 N = 5.4/0.49 = 11

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part (e)

= 2)()(log206.0log

HD

LB

Lf

Hf

xx

DB

xx

NsNe

07344.0)031.0

0704.0(

5.64

5.35)

25.0

2.0(log206.0log 2 =

=Ns

Ne

Ne/Ns = 1.184 Ne+Ns = 11

Ne = 6Ns = 5

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Mutli Component Distillation