multicomponent distillation _1

Introduction to multicomponent distillationIntroduction to multicomponent distillation

• Most of the distillation processes deal with multicomponent mixtures

• Multicomponent phase behaviour is much more complex than that

for the binary mixtures

• Rigorous design requires computers

• Short cut methods exist to outline the scope and limitations of a

particular process

Rigorous methods (Aspen)

Stage j

F

j

F

j

F

jjij TPhzF ,,,, , jQ

11

11,1

,

,,,

−−

−−−

jj

L

jjij

TP

hxL

11

11,1

,

,,,

++

+++

jj

V

jjij

TP

hyV

jj

L

jjij

TP

hxL

,

,,, ,

jj

V

jjij

TP

hyV

,

,,, ,

jU

jW

(See my notes on the web)

Short-cut methods:

Fenske-Underwood-Gilliland

(+Kirkbride)

Introduction to multicomponent distillationIntroduction to multicomponent distillation

MulticomponentMulticomponent distillation in tray towersdistillation in tray towers

• Objective of any distillation process is

to recover pure products

• In case of multicomponent mixtures we

may be interested in one, two

or more components

• Unlike in binary distillation, fixing mole

fraction of one of the components in a

product does not fix the mole fraction of other

components

• On the other hand fixing compositions of all

the components in the distillate and the bottoms

product, makes almost impossible to meet

specifications exactly

D

B

y1,y2,y3,y4�

Key componentsKey components

• In practice we usually choose two components

separation of which serves as an good indication

that a desired degree of separation is achieved

These two components are called key components

- light key

- heavy key

• There are different strategies to select these key

components

• Choosing two components that are next to each other

on the relative volatility: sharp separation

Distributed and undistributed componentsDistributed and undistributed components

• Components that are present in both the distillate and

the bottoms product are called distributed components

- The key components are always distributed components

• Components with negligible concentration (<10-6) in one

of the products are called undistributed

A B C D E G

key

components

heavy non-distributed components

(will end up in bottoms product)light non-distributed components

(will end up in the overhead product)

Complete designComplete design

A mixture of 4% n-pentane, 40% n-hexane, 50% n-heptane and 6%

n-octane is distilled at 1 atm. The goal is to recover 98%of hexane and

1% of heptane in the distillate. The feed is boiling liquid.

a) Design a distillation process

F, zf

condenser

boiler

n-pentane: 0.04

n-hexane:0.40

n-heptane: 0.50

n-octane: 0.06

100kmol/h





- Design a distillation process

F, zf

condenser

boiler

n-pentane: 0.04

n-hexane:0.40

n-heptane: 0.50

n-octane: 0.06

100kmol/h

What is design of

a column?

- P (pressure)

- N (stages)

- R (reflux)

- D (diameter)

- auxilary

equipment

(condenser, boiler)

Complete short cut designComplete short cut design




- Pressure consideration:

- what if you were not given P=1atm, how would you choose it?

- how do you validate that P=1atm is appropriate?

F, zf

condenser

boiler

- condenser uses cooling water

(20C). Let say the exit water

temperature is 30C.

- To maintain the temperature delta

at 10C, the dew point can not be

lower than 40C.

- Thus, the dew point of the distillate

has to be at least 40C.

- If not, will need higher pressure





- Design a distillation process

F, zf

condenser

boiler

n-pentane: 0.04

n-hexane:0.40

n-heptane: 0.50

n-octane: 0.06

100kmol/h

LK

HK

HNK

LNK





0.2360.06Octane

100

0.56500.5Heptane

1.39400.4Hexane

3.6240.04Pentane

KixBMoles in BxDMoles in DF xFxF

- Material balance conisderations

LK

HK

HNK

LNK





0.2360.06Octane

100

0.560.5500.5Heptane

1.3939.2400.4Hexane

3.6240.04Pentane


- Material balance considerations

LK

HK

HNK

LNK





0.23060.06Octane

100

0.560.5500.5Heptane

HK

1.3939.2400.4Hexane

LK

3.62440.04Pentane


- Sharp split: components lighter than the Light Key (LK) will end up completely

in the overheads


LK

HK

HNK

LNK





0.230060.06Octane

D=43.7100

0.560.0110.5500.5Heptane

HK

1.390.89739.2400.4Hexane

LK

3.620.092440.04Pentane



LK

HK

HNK

LNK





0.230060.06Octane

D=43.7100

0.5649.50.0110.5500.5Heptane

HK

1.390.80.89739.2400.4Hexane

LK

3.620.092440.04Pentane



i

F

ii DyFxBx −=

LK

HK

HNK

LNK





0.2360060.06Octane

B=56.3D=43.7100

0.5649.50.0110.5500.5Heptane

HK

1.390.80.89739.2400.4Hexane

LK

3.6200.092440.04Pentane


- Sharp split: components heavier than the Heavy Key (HK) will

end up completely in the bottoms


LK

HK

HNK

LNK





0.230.10760060.06Octane

B=56.3D=43.7100

0.560.87949.50.0110.5500.5Heptane

HK

1.390.0140.80.89739.2400.4Hexane

LK

3.62000.092440.04Pentane


- Sharp split: components heavier than the Heavy Key (HK) will

end up completely in the bottoms


LK

HK

HNK

LNK

Gilliland correlation: Number of ideal Gilliland correlation: Number of ideal

plates at the operating refluxplates at the operating reflux

+−

=+

−11

min

D

DmD

R

RRf

N

NN

Gilliland correlation: Number of ideal Gilliland correlation: Number of ideal

plates at the operating refluxplates at the operating reflux

+−

=+

−11

min

D

DmD

R

RRf

N

NN

Nmin

Rmin

R=1.5Rmin

N

FenskeFenske equation for multicomponentequation for multicomponent

distillationsdistillations

Assumption: relative volatilities of components remain constant

throughout the column

1ln

ln

,

,

,

,

,

min −

=HKLK

HKD

HKB

LKB

LKD

x

x

x

x

Nα

LK – light component

HK – heavy component

)(

)()(,

TK

TKT

HK

LKHKLK =α

FenskeFenske equation for multicomponentequation for multicomponent

distillationsdistillations

)(

)()(,

TK

TKT

HK

LKHKLK =α

Choices for relative volatility:

D

B

T

1) Relative volatility at saturated feed condition

)(,, F

F

HKLK THKLK

αα =

2) Geometric mean relative volatility

)()(,,, B

B

D

D

HKLK TTHKLKHKLK

ααα =

3, )()()(

,,, B

B

D

D

F

F

HKLK TTTHKLKHKLKHKLK

αααα =

why geometric mean?

Non key component distribution from Non key component distribution from

the the FenskeFenske equationequation

= +

HKB

HKDN

HKi

iB

iD

x

x

x

x

,

,1

,

,

, minαHK

iHKi

K

K=,α

⋅+

⋅⋅

=+

+

HKB

HKDN

HKi

HKB

HKDN

HKiiF

iD

Bx

Dx

Bx

DxFx

Dx

,

,1

,

,

,1

,,

,

min

min

1 α

α

Convince yourself and

derive for

iBBx ,

Minimum reflux ratio analysisMinimum reflux ratio analysis

• At the minimum reflux ratio condition

there are invariant zones that occur

above and below the feed plate, where

the number of plates is infinite and the

liquid and vapour compositions do not

change from plate to plate

• Unlike in binary distillations, in

multicomponent mixtures these zones

are not necessarily adjacent to the feed

plate location

y

x

zf

zf

xB xD

y1

yB

xN


F, zf

condenser

boiler

Invariant zones: presence of heavy and light non-distributed

components


F, zf

condenser

boiler

Invariant zones: only light non-distributed

components


F, zf

condenser

boiler

Invariant zones: only heavy non-distributed

components


F, zf

condenser

boiler

Invariant zones: no non-distributed

components

Minimum reflux ratio analysis:Minimum reflux ratio analysis:

Underwood equationsUnderwood equations

∑ −=−

i HKi

iFHKi xq

φαα

,

,,)1(

∑ −==+

i HKi

iDHKi

m

x

D

VR

φαα

,

,,1

For a given q, and the feed composition

we are looking for A satisfies this equation

(usually is between αLK and αHK)

Once is found, we can calculate the

minimum reflux ratio

φ

φ


Underwood equationsUnderwood equations∑ −

=−i HKi

iFHKi xq

φαα

,

,,)1(



=i HKi

iFHKi x

φαα

,

,,0

1.48



==+i HKi

iDHKi

m

x

D

VR

φαα

,

,,1



==+i HKi

iDHKi

m

x

D

VR

φαα

,

,,1

1.48

2.33

xi

1 2 3 4 5 6 7 8 9 10

hexane LKheptane HK

octanepentane

Feed stage

Distribution of components in Distribution of components in

multicomponentmulticomponent distillation processdistillation process

Non-distributed

heavy non-key

component

Non-distributed

Light non-key

component

KirkbrideKirkbride equation: Feed stage locationequation: Feed stage location

206.02

,

,

,

,

=

D

B

x

x

x

x

N

N

HKD

LKB

LKF

HKF

S

R

Complete short cut design: Complete short cut design:

FenskeFenske--UnderwoodUnderwood--Gilliland methodGilliland method

Given a multicomponent distillation problem:Given a multicomponent distillation problem:

a) Identify light and heavy key components

b) Guess splits of the non-key components and compositions

of the distillate and bottoms products

c) Calculate

d) Use Fenske equation to find Nmin

e) Use Underwood method to find RDm

f) Use Gilliland correlation to find actual number of ideal stages

given operating reflux

g) Use Kirkbride equation to locate the feed stage

HKLK ,α

Stage efficiency analysisStage efficiency analysis

In general the overall efficiency will depend:

1) Geometry and design of contact stages

2) Flow rates and patterns on the tray

3) Composition and properties of vapour and

liquid streams


Lin,xin

Lout,xout

Vout,yout

Vin,yin

Local efficiency

1

*

1

+

+

−−′

=nn

nnmv

yy

yyE

Actual separation

Separation that

would have been

achieved on an

ideal tray

What are the sources of inefficiencies?

For this we need to look at what actually happens

on the tray

Point efficiency


Depending on the location on the tray

the point efficiency will vary

high concentration

gradients

low concentration

gradients

stagnation points

The overall plate efficiency can

be characterized by the Murphree

plate efficiency:

1

*

1

+

+

−−

=nn

nnmV

yy

yyE

When both the vapour and liquid

phases are perfectly mixed the plate

efficiency is equal to the point

efficiency

mvmV EE =


In general a number of

empirical correlations exist

that relate point and plate

efficiencies

ce

LPe

tD

ZN

2

=

Peclet number

length of liquid

flow path

eddy diffusivity residence time of liquid

on the tray

Stage efficiency analysis: OStage efficiency analysis: O’’Connell (1946)Connell (1946)

(Sinnott)

Stage efficiency analysis: Van Winkle (1972)Stage efficiency analysis: Van Winkle (1972)

(Sinnott)


-- AICHE methodAICHE method

-- FairFair--ChanChan

Chan, H., J.R. Fair,” Prediction of Point Efficiencies for Sieve Trays, 1. Binary Systems”,

Ind Eng. Chem. .Process Des. Dev., 23, 814-819 (1984)

Chan, H., J.R. Fair, ,” Prediction of Point Efficiencies for Sieve Trays, 1. Multi-component Systems”,

Ind Eng. Chem. .Process Des. Dev., 23, 820-827 (1984)

(Sinnott)


Finally the overall efficiency of the process defined as

actual

ltheoreticaO

N

NE =

If no access to the data: E0=0.5 (i.e. double the number of plates)

Column diameter, etcColumn diameter, etc

Sinnott,

Jim Douglas, Conceptual design of chemical process

Column diameter, etcColumn diameter, etc

multicomponent distillation _1

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