assignment 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

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)

Fenske equation for multicomponentFenske equation for multicomponentdistillationsdistillations

Assumption: relative volatilities of components remain constantthroughout the column

1ln

ln

,

,

,

,

,

min

HKLK

HKD

HKB

LKB

LKD

x

x

x

x

N

LK – light componentHK – heavy component

)(

)()(, TK

TKT

HK

LKHKLK

Fenske equation for multicomponentFenske equation for multicomponentdistillationsdistillations

)(

)()(, TK

TKT

HK

LKHKLK

Choices for relative volatility:

D

B

T

1) Relative volatility at saturated feed condition

)(,, F

FHKLK T

HKLK

2) Geometric mean relative volatility

)()(,,, B

BD

DHKLK TT

HKLKHKLK

3, )()()(

,,, BB

DD

FF

HKLK TTTHKLKHKLKHKLK

why geometric mean?

Non key component distribution from Non key component distribution from the Fenske equationthe Fenske equation

HKB

HKDNHKi

iB

iD

x

x

x

x

,

,1,

,

, minHK

iHKi K

K,

HKB

HKDNHKi

HKB

HKDNHKiiF

iD

Bx

Dx

Bx

DxFx

Dx

,

,1,

,

,1,,

,

min

min

1

Convince yourself andderive 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

xzf

zf

xB xD

y1

yB

xN

* Relative volatility of each component has to be the same for each invariant zone

* Constant molar overflow

* αi=Ki/Kref (Usually Kref=KHK)

The operating line equations for each section of the column become:

Underwood method

Bimi

REFimi

DiniREFi

ni

BxyK

LyV

DxyK

LVy

,,1,

,,1,

rectifying section

stripping section


Bimi

REFimi

DiniREFi

ni

BxyK

LyV

DxyK

LVy

,,1,

,,1,

rectifying section

stripping section

In the invariant zones: ,,1, inini yyy

Bii

REFi

DiiREFi

BxyK

LV

DxyK

LV

,,

,,

A

x

VKL

xy

D

V

A

x

VKL

xy

D

V

i

Bii

REFi

Bii

i

i

Dii

REFi

Diii

,,,

,,,

Underwood method


A

x

B

Vy

D

V

A

x

D

Vy

D

V

i

Bii

i

i

Diii

,,

,, We are looking for a condition where

this is correct. In general there are multiple solutions

But consider the following

)1(,, qFA

xB

A

xDVV

i

Bii

i

Dii

Underwood method


)1(,, qFA

xB

A

xDVV

i

Bii

i

Dii

In other words:

A

xB

A

xD

A

xB

A

xDqF BDBD

2

,22

2

,22

1

,11

1

,11)1(

Under Underwood conditions: A=Ā, ii

A

x

A

x

A

xq

i

FiiFF

,

2

,22

1

,11)1(

Underwood method


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

i HKi

iFHKi

A

xq

,

,,)1(

i HKi

iDHKim A

x

D

VR

,

,,1

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

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

11min

D

DmD

R

RRf

N

NN

Kirkbride equation: Feed stage locationKirkbride 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) Calculate distribution of non key components

f) Use Underwood method to find RDm

g) Use Gilliland correlation to find actual number of ideal stages given operating reflux

h) Use Kirkbride equation to locate the feed stage

HKLK ,

Complete short cut design: exampleComplete short cut design: example

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) Find minimum number of stages and minimum reflux ratiob) Given operating reflux of 1.5 of the minimum find the operatingnumber of ideal stages

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

Stage efficiency analysisStage efficiency analysis

Step 1: Thermodynamics data and methods to predict equilibrium phase compositions

Step 2: Design of equilibrium stage separation

Step 3: Develop an actual design by applying the stage efficiency analysis to equilibrium stage design


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

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

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


- 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)


In addition we need to take in account effects of entrainment

Entrained liquid droplets

Dry Murphree efficiency can be corrected for theentrainment effects by Colburn equation:

1

1 mV

mVa

E

EE entrainment fraction =

entrained liquid/gross liquid flow


1

1 mV

mVa

E

EE


Finally the overall efficiency of the process defined as

ltheoretica

actualO N

NE

Types of traysTypes of trays

1. Sieve plates

2. Bubble-cap plates

3. Valve plates

Types of traysTypes of trays