multi component distillation 20130408

transcript

Multicomponent Distillation

Introduction 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

Short‐cut methods:

Fenske‐Underwood‐Gilliland(+Kirkbride)

Introduction to multicomponent distillation

Multicomponent 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 allthe components in the distillate and the bottomsproduct, makes almost impossible to meet specifications exactly

y1,y2,y3,y4…

Key components• In practice we usually choose two componentsseparation of which serves as a good indicationthat 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 otheron the relative volatility: sharp separation

Distributed 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 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

condenser

boiler

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

100kmol/h

Complete design

‐ Design a distillation process

condenser

boiler

100kmol/h

What is design ofa column?

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

Complete 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?

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

Complete design

‐ Design a distillation process

condenser

boiler

100kmol/h

LKHKHNK

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

‐Material balance conisderations

Pentane 0.04 4 3.62

Hexane 0.4 40 39.2 1.39

Heptane 0.5 50 0.5 0.56

Octane 0.06 6 0.23

‐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

‐ 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

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 bottoms

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 bottoms

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

• 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

condenser

boiler

Invariant zones: presence of heavy and light non‐distributed components

condenser

boiler

Invariant zones: only light non‐distributed components

condenser

boiler

Invariant zones: only heavy non‐distributed components

condenser

boiler

Invariant zones: no non‐distributed components

1 2 3 4 5 6 7 8 9 10

hexane LKheptane HK

octanepentane

Feed stage

Distribution of components in multicomponent distillation process

Non‐distributedheavy non‐keycomponent

Non‐distributedLight non‐keycomponent

Gilliland correlation: Number of ideal plates at the operating reflux GP‐A

Fenske equation for multicomponentDistillations GP‐B

Non key component distribution from the Fenske equation GP‐C

Minimum reflux ratio analysis:Underwood equations GP‐D

Kirkbride equation: Feed stage location GP‐E

Stage efficiency analysis GP‐FStage efficiency analysis: O’Connell (1946)

Stage efficiency analysis: Van Winkle (1972)

Column diameter GP‐G

Gilliland correlation: Number of ideal plates at the operating reflux

R=1.5Rmin

Fenske equation for multicomponentdistillations

Assumption: relative volatilities of components remain constantthroughout the column

LK – light componentHK – heavy component

Fenske equation for multicomponentdistillations

Choices for relative volatility: D

1) Relative volatility at saturated feed condition

2) Geometric mean relative volatility

why geometric mean?

Non key component distribution from the Fenske equation

Convince yourself andderive for

Minimum reflux ratio analysis:Underwood equations

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 theminimum reflux ratio

Minimum reflux ratio analysis:Underwood equations

1 2 3 4 5 6 7 8 9 10

hexane LKheptane HK

octanepentane

Feed stage

Distribution of components in multicomponent distillation process

Non‐distributedheavy non‐keycomponent

Non‐distributedLight non‐keycomponent

Kirkbride equation: Feed stage location

Complete short cut design: Fenske‐Underwood‐Gilliland methodGiven 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 stagesgiven operating reflux

g) Use Kirkbride equation to locate the feed stage

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

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:

When both the vapour and liquidphases are perfectly mixed the plateefficiency is equal to the point efficiency

In general a number of empirical correlations exist that relate point and plate efficiencies

Peclet number

length of liquid flow path

eddy diffusivity residence time of liquidon the tray

Stage efficiency analysis: O’Connell (1946)

(Sinnott)

Stage efficiency analysis: Van Winkle (1972)

(Sinnott)

‐ AICHE method

‐ Fair‐ChanChan, 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

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

Column diameter, etc

Sinnott,

Jim Douglas, Conceptual design of chemical process

Column diameter, etc

Multi‐Component Distillation Phase equilibrium For multi‐components

Bubble point calculations Given the total pressure and liquid mole fraction ‐‐> T and y

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

component xi Ki αι α ιxi

Butane Pentane Hexane Heptane

0.4 0.25 0.2 0.15

1.68 0.63 0.245 0.093

6.857 2.571 1.0 0.38

2.743 0.643 0.2 0.057

Total 3.643

A liquid stream at 405.3 kPa with 40% butane, 25% pentane, 30% hexane and 15% heptane 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

A liquid stream at 405.3 kPa and 100 mole/h with 40% butane (A), 25% pentane (B), 30% hexane (C) and 15% hepatne (D) is fed to a distillation column. 90% of B is recovered in distillate and 90% of C is 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 balance mole balance on butane 25 = nBD + nBBQ nBD = 0.9(25) = 22.5 mole ‐‐> nDD = 2.5 mole mole balance on hexane 30 = nCD + nCB

• Q ncB = 0.9(20) = 18 mole ‐‐> nCD = 2 mole • Assume all A in distillate ‐‐> nAD = 40 mole ‐‐> nAB = 0 • Assume all D in bottom ‐‐> nAD = 40 mole ‐‐> nAB = 0

• Part b)

• part ( c) • let B light component; let C heavy component

• αLD = 2.5; αLB = 2.04

part (d)

Solve by trial‐and‐error θ = 1.2096

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

part (e)

Ne/Ns = 1.184 Ne+Ns = 11

Ne = 6 ; Ns = 5

multi component distillation 20130408

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