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Design insights for reactive distillation

Arthur W. WesterbergDept. of Chemical Engineering and

Institute for Complex Engineered SystemsCarnegie Mellon University

Pittsburgh, PA 15213

Acknowledgements: NSF and Eastman Chemicals

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Collaborators• Steinar Hauan (was here, joined CMU Aug. 1999)• Jae Woo Lee (my Ph.D. student - finished May, 2000)

• Kristian Lien• Amy Ciric (Univ. of Cincinnati)

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Outline• Design in transport space• Four reactive distillation examples

+ diluted ternary: MTBE+ quaternary (using projection): methyl acetate+ binary: McCabe-Thiele+ binary: bypassing an azeotrope

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Transport space rather than unit operation space

ProposalUse transport processes as the building blocks for synthesis rather than existing unit operations

Transport spaceBalances for• material ==> composition• energy ==> molar enthalpy• momentum ==> molar momentum??

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Useful geometry in “transport space”

Straight linesGiven a set of balances of the form

If

Then we see

au bv cw+=

uii∑ vi

i∑ wi

i∑= =

a b c+=

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And

i.e., points satisfying these balances are on a straight line in this vector space

Augmenting this spaceAdding another dimension that satisfies

leads to straight lines in this augmented space

u bb c+------------ v⋅ c

b c+------------ w⋅+ α v⋅ 1 α–( ) w⋅+= =

aunew bvnew cwnew+=

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Consider the top of a columnComposition space

Augmented with molar enthalpy

Thus balances in a composition space and in an augmented molar enthalpy + composition space both give rise to points on straight lines. (Is there a way to include momentum?)

D

LnVn+1

QC

Vn 1+ yn 1+⋅ Ln xn⋅ D xD⋅+=

Vn 1+ Hn 1+⋅ Ln hn⋅ D hD

QC

D-------+

⋅+=

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Difference points - the geometry of balances• Vn+1 is the mixing of two positive flows, Ln and D• The lever rule indicates relative amounts

D (~pure A)

LnVn+1

B

C

LnVn+1

mat’l balance

Vn+1 = Ln + D

DA

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Reaction as a “flow” and “composition”

A + B <--> C <==> -1 A -1 B + 1 C = 0; ν’ =[-1, -1, 1]• Flow: Every time reaction “turns over” (ξ mol/s), we lose one

mol/s of materialflow = (-1 + (-1) + 1) ξ = −1 ξ (mol/s) = νT ξ

• Composition: We lose one mol/s of A, one mol/s of B and gain one mol/s of C

composition =

Note: composition sums to one but lies outside composition trian-gle

1–1–

1

νT( )⁄

1

11–

=

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The reaction difference pointD

LnVn+1

Vn+1 = Ln + D - νΤξ = Ln + D + ξ

∆ = D + ξ

Ln

Vn+1mat’l

D

νΤξ = −ξ

A + B <--> C

balance

111–varies with ξ

∆

δr

ξ

D

Ln

D + ξ

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Steering tray compositions in column

Ln

D

Vn

Vn+1

LnVn+1

Vn

n

n+1Vn+1

Note: altered material balance can maketemperature changes reverse in column

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Ex 1: MeOH + IBUT <==> MTBE

MeOHExcess IBUT

MTBE

IBUT

Why did industry settle on this design?

catalyst on ionexchange resin

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Reaction difference point (MeOH + IBUT <==> MTBE)

MeOH (128.8)

MTBE (136.8)

AZ (121.7)

IBUT (62.0)

AZ (60.1)

δr111–

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Two distillation regions

MeOH (128.8)

MTBE (136.8)

AZ (121.7)

IBUT (62.0)

AZ (60.1)

I

II

residue curve

B D

F

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Liquid and vapor (rxn) equilibrium curves

MeOH

MTBE IBUT

I

II

vapor equil

liquid equilresidue curve

curve

curve

x y*(x)

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Reactive diff pt and lever rule ==> little reaction in bottom

MeOH

MTBE IBUT

δr

infeasible

negative

positiveξ

B

Ln = Vn+1 + ∆∆ = B + ξ

B

Ln between Vn+1and ∆

Want ξ positive==> ∆ positive

Cannot be muchreaction in

bottom

In bottom

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Nonreactive bottom section required

MeOH (128.8)

MTBE (136.8)

AZ (121.7)

IBUT (62.0)

AZ (60.1)

Lowest trays+ nonreactive+ in region II

to recover pure MTBE

II

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Major reaction possible and needed in top section

MeOH

MTBE IBUT (62.0)

AZ (60.1)

δr

∆

D

ξ

(excess IBUT)Vn+1 = Ln + ∆

∆ = D + ξ

D

with no rxn

with rxn

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A “split” lever rule yn+1Vn+1 =xnLn + xDD + (νR-νP)2ξ

==> x∆∆ = xDD + (νR-νP)2ξ

BA

C

νR2ξ

νP2ξ

xDD

νP

νR

xD

D

2ξ

∆

xnLnyn+1Vn+1

δr

νT = -1-1+2 = 0

A+C <==> 2B

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Ex 2: Methyl acetate, MeOH + AC = MA + W

MA

W

AC

MeOH

2 moles produce 2 moles so reaction point at infinity

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Four components - use following projection• Set acetic acid concentration to a low value (say 5%) and fix it• Map to acetic acid free basis• All equations transform nicely• On acetic acid free basis

+ MeOH = MA + W+ reaction difference point moves to finite location

Question: Why does the amount of reaction that can occur in the column decrease with large reflux ratios? This is counterintuitive.

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Geometry

1.00.80.60.40.20.0

1.0

0.8

0.6

0.4

0.2

TMd~

Bx~

Dx~

nx~

1n+y~Ftx~

MAM

W

* *1

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2 *

n1n yy ~~ =+

ξn

Ln

D

Vn+1

~

~

Vn+1~

Ln~ + ξn = + D

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Maximum in conversion vs. reflux ratio

0

20

40

60

80

100

1 2 3 4 5 6 7 8

Molar reflux ratio

Ove

rall

con

vers

ion

(%

)

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Ex. 3: McCabe-Thiele: H-->L - Rxn in top

yP1

xP1

12

3

4

5

67

xD xB

Operating lines

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zF

D?

x 3D −

D?

x 2D −

Top

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Reaction in bottom - makes separation harder

yP1

xP1

123

xD xB

Operating lines

Pinch point

zF

B?

x 1sB

+−B?

x sB −

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Ex. 4: Breaking an azeotrope

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

zF

xB

xD

yA

xA

2 1

3

4

5

6

7

δ2

δ3

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Conclusions• Reactive distillation is rich with geometry• Simple projections allow one to visualize behavior for four com-

ponent system• We have even been able to see geometrically some general results

about where to place reaction• This stuff can be taught