introduction to multicomponent distillation

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

Multicomponent distillation in tray towersMulticomponent distillation 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 scale often leads to all the components lighter then the light key components accumulating in the distillate and all the components heavier then the heavy key component accumulating in the bottoms product: 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: 0.04n-hexane:0.40n-heptane: 0.50n-octane: 0.06

100kmol/h




F, zf

condenser

boiler

n-pentane: 0.04n-hexane:0.40n-heptane: 0.50n-octane: 0.06

100kmol/h

What is design ofa column?

- P (pressure)- N (stages)- R (reflux)- D (diameter)- auxilary equipment (condenser, boiler)




Rigorous methods (Aspen)

Stage j

Fj

Fj

Fjjij TPhzF ,,,, , jQ

11

11,1

,

,,,

−−

−−−

jj

Ljjij

TP

hxL

11

11,1

,

,,,

++

+++

jj

Vjjij

TP

hyV

jj

Ljjij

TP

hxL

,

,,, ,

jj

Vjjij

TP

hyV

,

,,, ,

jU

jW

(See my notes on the web)

Short-cut methods:

Fenske-Underwood-Gilliland(+Kirkbride)

Complete short cut designComplete short cut design


xF F xF Moles in D xD Moles in B xB Ki

Pentane 0.04 4 3.62

Hexane 0.4 40 1.39

Heptane 0.5 50 0.56

Octane 0.06 6 0.23

100

I. Material balance conisderations




Pentane 0.04 4 3.62

HexaneLK

0.4 40 39.2 1.39

HeptaneHK

0.5 50 0.5 0.56

Octane 0.06 6 0.23

100

I. Material balance considerations




Pentane 0.04 4 4 3.62

HexaneLK

0.4 40 39.2 1.39

HeptaneHK

0.5 50 0.5 0.56

Octane 0.06 6 0 0.23

100

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





Pentane 0.04 4 4 0.092 3.62

HexaneLK

0.4 40 39.2 0.897 1.39

HeptaneHK

0.5 50 0.5 0.011 0.56

Octane 0.06 6 0 0 0.23

100 D=43.7





Pentane 0.04 4 4 0.092 3.62

HexaneLK

0.4 40 39.2 0.897 0.8 1.39

HeptaneHK

0.5 50 0.5 0.011 49.5 0.56

Octane 0.06 6 0 0 0.23

100 D=43.7


iFii DyFxBx −=




Pentane 0.04 4 4 0.092 0 3.62

HexaneLK

0.4 40 39.2 0.897 0.8 1.39

HeptaneHK

0.5 50 0.5 0.011 49.5 0.56

Octane 0.06 6 0 0 6 0.23

100 D=43.7 B=56.3

- Sharp split: components heavier than the Heavy Key (HK) will end up completely in the overheads





Pentane 0.04 4 4 0.092 0 0 3.62

HexaneLK

0.4 40 39.2 0.897 0.8 0.014 1.39

HeptaneHK

0.5 50 0.5 0.011 49.5 0.879 0.56

Octane 0.06 6 0 0 6 0.107 0.23

100 D=43.7 B=56.3

- Sharp split: components heavier than the Heavy Key (HK) will end up completely in the overheads




II. 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 distillatehas to be at least 40C.

- If not, will need higher pressure

III. Gilliland correlation: Number of ideal III. Gilliland correlation: Number of ideal plates at the operating refluxplates at the operating reflux

+

−=+

−11

min

D

DmD

R

RRf

N

NN

III. Gilliland correlation: Number of ideal III. 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

Nmin

IV. Fenske equation for multicomponentIV. Fenske equation for multicomponentdistillationsdistillations

Assumption: relative volatilities of components remain constantthroughout the column

1ln

ln

,

,

,

,

,

min −

=HKLK

HKD

HKB

LKB

LKD

x

x

x

x

Nα

LK – light componentHK – heavy component

)(

)()(, TK

TKT

HK

LKHKLK =α

V. Minimum reflux ratio analysisV. Minimum 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

xzf

zf

xB xD

y1

yB

xN

V. Minimum reflux ratio analysis:V. Minimum reflux ratio analysis:Underwood equationsUnderwood equations

∑ −=−

i HKi

iFHKi

A

xq

,

,,)1(αα

∑ −==+

i HKi

iDHKim A

x

D

VR

,

,,1αα

For a given q, and the feed composition we are looking for A satisfies this equation(usually A is between αLK and αHK)

Once A is found, we can calculate theminimum reflux ratio

VI. Kirkbride equation: Feed stage locationVI. Kirkbride equation: 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: Fenske-Underwood-Gilliland methodFenske-Underwood-Gilliland methodGiven 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 compositionsof 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

VII. Stage efficiency analysisVII. Stage 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 happenson the tray

Point efficiency


Depending on the location on the tray the point efficiency will vary

high concentrationgradients

low concentrationgradients

stagnation points

The overall plate efficiency can be characterized by the Murphreeplate efficiency:

1*

1

+

+

−−=

nn

nnmV yy

yyE

When both the vapour and liquidphases are perfectly mixed the plateefficiency 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 liquidon the tray

VII. Stage efficiency analysis: O’Connell (1946)VII. Stage efficiency analysis: O’Connell (1946)

(Sinnott)

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

(Sinnott)


- AICHE method- AICHE method

- Fair-Chan- Fair-Chan

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)

VIII. Column diameter, etcVIII. Column diameter, etc

Sinnott,

Jim Douglas, Conceptual design of chemical process

introduction to multicomponent distillation

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