I

Abstract: The design of process to produce Acetic Anhydride from Acetone has

been discussed in this project. The purpose of this project is simply to

collect and present a set of simple calculation for this case study. For

calculation and design we use App A&C&E of Conceptual design of

chemical process by Douglas.

II

Table of Contents

Input Information ....................................................................................................................................................... 1

Decision Batch Versus continuous ............................................................................................................................. 5

Input-Out put structure ............................................................................................................................................... 7

Recycle structure ..................................................................................................................................................... 13

Separation system .................................................................................................................................................... 22

1

Level 0 - Input Information

The definition of reaction and importance information are as follow:

2

Input Information –level 0:

atmHCCoKetene

CHKeteneAcetoneC 1,700

2

10

42

4

+→

+→

atmdrideAceticAnhyAceticAcidKetene C 1,800→+

Product: Acetic Anhydride

Production rate: P = 16.5 mol/hr with 99% Purity

Cost Data:

Acetone:15.66 $/mol

Acetic Acide: 15 $/mol

Acetic Anhydride: 44.41 $/mol

Fuel: 4 $/106Btu

Reaction Data: ∆HR,acetone=34700 Btu/mol

∆HR,ketene=-27000 Btu/mol

∆HR,anhydride=20700 Btu/mol

Heat Values: Co: 0.122*106 Btu/mol CH4: 0.383*106 Btu/mol C2H4: 0.608*106 Btu/mol

S = moles of ketene leaving the pyrolysis reactor

moles of Acetone converted

3

selectivity data for this process is given in the 1958 AICHE Student contest problem, is as follows:

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 X

0.13 0.19 0.28 0.38 0.49 0.62 0.75 0.88 S

S=1-4/3 X

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

S

X

S=1- 4/3 X

4

Heat capacity data:

Component heat capacity

T=700oc Btu/mol T=80oc Btu/mol Co 7.89 6.97

16.97 9.15 CH4

11.55 C2H4 22.56 Acetic Anhydride 53.48 49.02 Acetone 38.25 19.98 Acetic Acid 31.66 23.43 Ketene 21.03 12.04

Vapor pressure Equation constants:

),()(

ln 0 KTKPaPCT

BAP sat ==++=

COMPONENT A B C

CO 41.66 -1109.88 5.46 CH4 31.35 -1307.5 -3.26 C2H4 48.11 -2473.7 -5.74 Acetic Anhydride 98.15 -8897.7 -12.16 Acetone 71.30 -5952 -8.53 Acetic Acid 61.34 -6768.8 -6.73 Ketene 14.00 -1849.2 -35.15

5

Level 1

Decision Batch Versus continuous

6

Choose a continuous process.

7

level 2

Input-Out put structure

We must answer these questions:

1. Should we purify the feed streams before they inter the

process?

2. Should we remove or recycle a reversible byproduct?

3. Should we use a gas recycle and purge stream?

4. Should we not bother to recover and recycle some reactants?

5. How many product streams will there be?

6. What are the design variables for the input-output structure,

and what economic trade-offs are associated whit these

variables?

8

Input-Out put structure-level 2:

1. Purify feed streams: the acetone and acetic acid feed streams are pure.

2. Reversible byproduct: we do not have such by product in our process.

3. Recycle and purge: hence we do not have any light component in the feed streams,

so no recycle and purge are needed.

4. Excess reactants: in the first reactor, as we have only one feed stream entering the

reactor, so there would be no case of using excess, but in the second reactor, we

consider using of excess acetic acid and try to monitor its effect on economic potential

of the process (trade-off).

5. Number of product streams:

The boiling point of the components and their destination are as follows:

9

COMPONENT NBP,(oF) DESTINATION

CO -312.6 Fuel byproduct CH4 -258.6 Fuel byproduct C2H4 -158.4 Fuel byproduct

Ketene -42.1 unstable reactant-completely converted

Acetone 133.2 Reactant-recycle to R1 – liquid

Acetic Acid 244.3 Reactant-recycle to R2 – liquid

Acetic Anhydride 281.9 Primary product So we have:

Two product streams: (CH4+C2H4+CO) and acetic anhydride.

6. Material balance and stream costs: by considering the production rate of

anhydride (16.58 mol/hr), we have:

S

PF =1 Acetone:

PF =2 Acetic acid:

We assume that the ketene is totally converted in the reactor.

10

=

=

=

−=−=−+−+=

−+−+=⇒

−=

−=

−=−=

=

hr

molF

hr

mol

SF

hr

molP

hr

mols

SS

S

PSS

S

PP

SS

PS

S

P

S

PP

SS

PP

S

PYPHC

SS

PP

S

PYPCO

S

PYPCH

HC

CO

CH

58.16

58.16

58.16

)2

3

2

5(

58.16)

2

3

2

5())1(

2

1)1(1(

)1(2

)1(

)1(22

1*:

)1(*:

*:

2

1

1

1

142

1

14

42

4

As we know the relation between s and x, and we also have the cost data of each

component, we can write the economic potential of this level and plot it versus the

conversion (or selectivity).

EP2 = Product value + byproduct value- raw material costs ($/yr) which for anhydride

process would be:

(We assume working hours per year= 8150 hr/yr)

−−−+−++= )*()*()**)1(2

()**)1(*()**()*(815021424244 ,,,2 FFHCHCVCOCOVCHCHVP CPC

S

PCHS

S

PCHS

S

PCH

S

PCPEP

−++= −− )10*4*10*122.0*)1(58.16

()10*4*10*383.0*58.16

(14.44*58.16*8150 6666

2 SSS

EP

)15*58.16()66.15*58.16

()10*4*10*608.0*)1(2

58.16( 66 −−−+ −

SS

S

)()()( 22 xhEPxgSsfEP =→=→=

11

X s EP2($/yr)*10^-6

0.1 0.88 1.836081107

0.2 0.75 1.505404865

0.3 0.62 1.036057939

0.4 0.49 0.317669789

0.5 0.38 -0.674113043

0.6 0.28 -2.251949367

0.7 0.19 -5.092054749

0.8 0.13 -9.170154784

We see that EP decrease with X increase Because more Acetone convert to by product.

Economic Potential vs. Conversion

-10

-8

-6

-4

-2

0

2

4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

conversion (X)

EP2($/yr)*10^-6

12

LEVEL 2 – Alternatives:

There are no recycle and purge streams and we do not have any reversible

by product; so, we have no alternatives at this level of design.

13

LEVEL 3

Recycle structure

Questions that should be answered at this level

1. How many reactor systems are required? Is there any

Separation between the reactor systems?

2. How many recycle streams are required?

3. Do we want to use an excess of one reactant at reactor inlet?

4. Is a gas compressor required? What are the costs?

5. Should the reactor be operated adiabatically, with direct

heating or cooling or a diluents or heat carrier is required?

6. Do we want to shift the equilibrium conversion? How?

7. How do the reactor costs affect the economic potential?

14

LEVEL 3 – Recycle structure:

1. Because we have two set of reactions, with two different temperature

conditions, we must use two reactors.

2. There are two recycle streams:

a) Acetone recycle stream

b) Acetic acid recycle stream

3. We use excess amount of acetic acid at the inlet of the second reactor.

4. Because the recycle stream is totally liquid, so we do not need to use

any compressor. Thus, we need pumps in the recycle loops which their

cost is neglectable in comparison with other equipment costs.

15

Recycle material balances:

Acetone:

Acetone leave the first reactor=R1

Total acetone enters the first reactor=F1+R1

X

XFRRXRF

)1()1)(( 1

1111

−=→=−+⇒

Total flow entering the first reactor=FR1

SX

P

X

F

X

XFFRFFR ==

−+=+= 11

1111

)1(

(1-X)Acetone enters the second reactor= P SX

16

Other materials which enter the second reactor are:

−=

−=−=

=

=

)1(2

)1(

42

4

SS

PHC

SS

PP

S

PCO

S

PCH

PKeten

)(*

, 2222

MRPacidacetic

P

FR

ketene

acidaceticMRFRacidacetic

=→

+==+=

Total flow entering the second reactor=FR2

+−+−

=

+−+−+++−=

2

5)12(

2

)1(

)()1(2

)1()1(

2

2

MRS

X

X

S

PF

PMRSS

PS

S

P

S

PPX

SX

PF

R

R

Reactor heat effects:

a. First reactor: two reactors are taken place in the first reactor (we have

the cp of materials from the information data part).

−=∆=

∆−+∆=

)1292(**

*)1(*

111

1

OUTacetonePRRR

keteneAcetoneR

TCPSX

PTCFQ

HSS

PH

S

PQ

−=

−−+=

)1292(*25.38*58.16

)27000(*)1(58.16

34700*58.16

1

1

OUTR

R

TSX

Q

SSS

Q

17

−=

−−

+=

)1292(*185.634

)1(447660575326

1

1

OUTR

R

TSX

Q

SSS

Q

[ ])882.705307.201(1292

3.11

SXT

XS

OUT +−=

⇒−=

b. The second reactor:

∆=

∆=

TCFQ

HpQ

PRR

AnhydridR

22

2

)176(*)*)(*()*2

)1(*()*)1(*()*()*()*

)1((

4242 OUTacidHCCOCHketeneacetonR TCPMRPCPS

SPCPS

S

PCP

s

pcppCP

SX

XPQ −

+−

+−+++−

=

34320620700*58.162 ==RQ

)176(*)43.23*)(*58.16()55.11*2

)1(58.16()97.6*)1(*

58.16()15.9*

58.16()4.12*58.16()98.19*

)1(58.16(2 OUTR TMR

S

SS

SsSX

XQ −

+−

+−+++−

=

NOTE: the heat capacity values for each component in each reactor, is used at

the reactor condition.

Now we can plot the reactors effluent stream temperatures versus conversion

for each reactor:

18

X s Tout ( F)

0.1 0.88 1209.75

0.2 0.75 1145.86

0.3 0.62 1100.31

0.4 0.49 1073.12

0.5 0.38 1057.23

0.6 0.28 1052.63

0.7 0.19 1057.2

0.8 0.13 1057.54

Because reactions are endothermic so T decrease with reaction progress.

Reactor 1 exit temperature vs. conversion

1040

1060

1080

1100

1120

1140

1160

1180

1200

1220

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

conversion (x)

T out (F)

19

MR=2 MR=3 MR=5

X s T out (F) T out (F) T out (F)

0.1 0.88 100.9284733 106.8078924 116.178103

0.2 0.75 62.43109104 75.36714614 94.03872714

0.3 0.62 44.17186029 61.28847212 84.93602812

0.4 0.49 40.13863227 58.24668457 83.02951251

0.5 0.38 43.90993531 61.09019948 84.81112102

0.6 0.28 55.84760157 70.23193138 90.66427054

0.7 0.19 75.91147661 86.09653086 101.2996208

0.8 0.13 94.29716962 101.2133133 112.041488

We see that with high MR T has changed little because asetic asid treat like heat carrier.

0

20

40

60

80

100

120

140

0 0.2 0.4 0.6 0.8 1

Tout

X

MR=2

MR=3

MR=4

20

REACTOR COST:

We assume that the pyrolysis reactor (first reactor of the process) cost is

calculated as a furnace cost, and if anhydride reactor cost is neglectable

in comparison with the first reactor cost, then we would have:

)27.1(*)10*52.5)(280

&($,cos 85.03

CFQSM

tInstalled +=

=

=

=

++=

pressurecatmospheriinrunsprocesstheasF

materialtuberadiantassteelcarbonuseweifF

typepyrolysisaforF

FFFF

P

m

d

pmdC

)7.14(0

0

1.1

Also we need M&S factor in our calculation, we assume M&S factor=1350

[ ]

+=

−−+=

→∆−+∆==

SS

Q

SSS

PQ

HSS

PH

S

PQQ keteneAcetoneR

27000770058.16

)27000(*)1(*58.16

)34700(*58.16

*)1(*1

)37.2(*10

)270007700(*58.16

)10*52.5)(280

1350($,cos

85.0

6

3

+=

SStInstalled

In order to write installed cost in annualized basis we should calculate capital charge factor (ccf): We assume: interest rate: i=0.2 Number of operating years: 10 yr

[ ] [ ]1

10

10

24.01)2.01(

)2.01(2.0

1)1(

)1( −=−+

+=

−+

+= yr

i

iiCCF

n

n

21

)*cos$,cos. CCFtinstalledtInstalledAnn =

tactorEPEP cosRe23 −=

X s EP2($/yr)*10^-6 EP3($/yr)*10^-6

0.1 0.88 1.836081107 1.826378412

0.2 0.75 1.505404865 1.49535338

0.3 0.62 1.036057939 1.025515021

0.4 0.49 0.317669789 0.306382362

0.5 0.38 -0.674113043 -0.686414373

0.6 0.28 -2.251949367 -2.265834331

0.7 0.19 -5.092054749 -5.108714167

0.8 0.13 -9.170154784 -9.190662659

By subtracting EP2 from reactor cost we can calculate EP3 as function

of design variable. Of course the optimum from this figure is not true

because we don’t consider distillation column cost.

Economic Potential vs. conversion

-10

-8

-6

-4

-2

0

2

4

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

conversion (x)

EP3 ($/yr)*10^-6

85.02700077000

*308.1/$,cos.

+=

S

SyrtInstalledAnn

22

level 4

Separation system

We don’t have gas recovery system so we must design liquid sepration

system.

To synthesize the liquid separation system the following decisions should

be made:

1. How should the light ends be removed if they might

contaminate the product?

2. What should be the destination of the light ends?

3. Do we recycle components that form zoetrope’s with the

reactants, or do we split the azeotropes?

4. What sequence of columns do we use?

5. How should we accomplish separations if distillation is not

feasible?

23

Separation system-level 4:

Light ends: We assume most of light ends product (CH4, C2H4 , CO)

departed in flash, and with neglecting amount of light ends product at

liquid part.

Azoetrope’s: we don’t have any zoetrope’s.

Sequence of columns: we consider two atmospheric distillation column

in separation system. The heuristic of lightest first, most plentiful first,

and favor equimolar splits all suggest direct sequence.

Note: we are neglecting Acetone, Acetic Acid, Anhydride losses in flash.

With assuming X=.3 , s=.6 , MR=2 we design two distillation columns.

(P=16.58)

P/S(1-X) = 19.3:Fresh acetone Feed

P(MR)-P=16.58:Fresh Acetic Acid Feed

P=16.58:Produced Anhydride

24

653.0)58.16*2(4.62

4.62=

+=FX

Temperature at top is N.P of Acetone 329.2K , and Temperature at

bottom is mean N.P of Anhydride and Acid Asetic:

97.4

34.36.61

206:401@

4.758.5

56.43:2.329@

=→

=→

=

=

=→

=

=

avg

bottomsat

acidacitic

sat

acetone

topsat

Acetone

sat

Acetone

P

PK

P

PK

α

α

α

With assuming 99.5% recovery Acetone and 99.5% Anhydride and Acid

Asetic we have:

62.4*0.995= 62.088 :Acetone at top

16.58*(1-0.995) = 0.0829 Acetic acid at top:

9986.00829.0088.62

088.62=

+=DX

00934.058.16497.16312.0

312.0=

++=WX

25

We consider feed is saturated liquid:

386.0653.0*)197.4(

1

*1

1=

−=

−≈

F

mX

Rα

463.0386.0*2.12.1 === mRR

006.7

)97.4ln(

)00934.0

00934.01*

10*4.1

9986.0ln(

)ln(

)1

*1

ln(3

=→

−

=

−

−=

−

m

avg

W

W

D

D

m

N

X

X

X

X

EqNFenskeα

528.19

)463.1

386.0463.0(175.0

1

006.7

)1

(175.01

:

5688.0

5688.0

=→

−−=

+

−

+

−−=

+

−

N

N

N

R

RR

N

NNEqGilland mm

Tower height:

With considering: 2 ft towers space and 15ft at top and bottom end we

have:

.H=2*40+15=95 ft

401)5.0

528.19(

5.0

528.19

0

=

+=→== actact NE

NN

26

Tower Cross – sectional Area and Diameter:

VTbMA *)460(*10*124.2 4 += −

For bottom of distillation column:

075.812

1.10205.60

2=

+=

+= anhydrideacidacetic

avg

MMM

FTb o

avg 1.2632

9.2813.244=

+=

ftA

D

ftA

DRV

44.24

677.4

95.90)0829.0088.62(*463.1)1(

2

==

=

→=+=+=

π

yrtAnn

$3741824.0*)118.2(*)95(*)44.2(*)9.101(*)

280

1350($,cos. 802.0066.1 =+=

Acetone column condenser:

)90120(***** −=∆=∆= PCmCCVc CWTAUVHQ

We consider water at 900F enter and exit at 1200F:

.

95.90

12530)..(

10002

=

=∆

=

V

H

ffthr

BTUU

V

C

[ ]FTm

082.8)1202.133()902.133(ln

90120=

−−−

−=∆

21292

82.8*100

95.90*12530ftAC ==

27

Install cost for column:

)29.2(*)3.101)(280

&($, 65.0

CFASM

+=

yrtAnn

$4059924.0*)29.3(*)1292)(3.101)(

280

1350($,cos 65.0 ==

Cooling Water cost & cooling tower :

yryr

hr

hr

lb

lb

gal

galtAnn

$2227)8150(*

30

95.90*12530)

341.8

1)(

1000

$06.0($,cos =

=

Acetone column reboiler:

mRRSSVR TAUHWVHQ ∆=∆=∆= **

The boiling point of mixture at bottom is 4100k (assuming ∆� = 30�� ) so

we can use stream with 67 PSI pressure and 4120k temperature.

223.11411250

95.90*14130

910,11250,14130

ftA

lb

BtuHUH

R

m

SR

==

=∆==∆

yrtAnn

$839024.0*)29.3(*)23.114)(3.101)(

280

1350($,cos 65.0 ==

The stream cost:

yryr

hr

hr

lb

lbtAnn

$25897)8150(*

910

95.90*14130*)

1000

$25.2($,cos =

=

28

Anhydride column (second column):

Acetone =0.005*62.4=0.312

Acetic Acid=0.995*16.58=16.49

Anhydride =16.58

494.0312.058.16497.16

497.16=

++=FX

With assuming 99.5% recovery Anhydride and 99% Anhydride we have:

Anhydride at bottom: 16.58*0.995=16.497

Acetic Acid at bottom:(16.497)/(0.99)*(1-0.99)=0.167

Acid Acetic at top : 16.497-0.167=16.33 Anhydride at top of column: 16.58-16.497=0.083 Acetone at top : 0.312

976.0312.0083.033.16

33.16=

++=FX

310*84.9497.16167.0

167.0 −=+

=WX

Temperature at Top is N.P of Acetic Acid (319K) and anhydride at

bottom(412K):

29

85.2

86.26.80

16.28:412@

84.268.16

42.47:391@

=→

=→

=

=

=→

=

=

avg

bottomsat

acidacitic

sat

ahydrid

topsat

Anhydrid

sat

acidAcetic

P

PK

P

PK

α

α

α

We consider feed is saturated liquid:

094.1494.0*)185.2(

1=

−=mR

313.1)094.1(*2.1 ==R

94.7)85.2ln(

)10*84.9

)10*84.9(1*

976.01

976.0ln(

:3

3

=

−

−=−

−

mNEqFenske

035.19)313.2

094.1313.1(175.0

1

94.7: 5688.0 =

−−=

+

−

N

NEqGilland

3915.0

035.19=

+=actN

Tower height:

H=2*39+15=93 ft

30

Tower Cross-sectional Area and Diameter:

24

4

26.2)685.38(*)4609.281(*1.102*10*124.2

685.38)083.033.16312.0(*)1313.1()1(

*)460(*10*124.2

ftA

DRV

VTbMA

=+=

=+++=+=

+=

−

−

ftA

D 7.14

==π

yrtAnn

$2502524.0*)18.3(*)93(*)7.1(*)9.101(*)

280

1350($,cos. 802.0066.1 ==

Anhydride column condenser:

)90120(***** −=∆=∆= PCmCCVc CWTAUVHQ

[ ]FTm

082.8)1203.244()903.244(ln

90120=

−−−

−=∆

244082.8*100

685.38*10030ftAC ==

yrtAnn

$2015824.0*)29.3(*)440)(3.101)(

280

1350($,cos 65.0 ==

Cooling Water cost & cooling tower :

yryr

hr

hr

lb

lb

gal

galtAnn

$758)8150(*

30

685.38*10030)

341.8

1)(

1000

$06.0($,cos =

=

Anhydride column reboiler:

The boiling point of mixture at bottom is 4120k (assuming ∆� = 30�� ) so

we can use stream with 90 PSI pressure and 4330k temperature.

31

235.6011250

685.38*17550

895,11250,17550

ftA

lb

BtuHUH

R

m

SR

==

=∆==∆

yrtAnn

$552524.0*)29.3(*)35.60)(3.101)(

280

1350($,cos 65.0 ==

stream cost:

yryr

hr

hr

lb

galtAnn

$13910)8150(*

895

685.38*17550)

1000

$25.2($,cos =

=

Acetone column($/yr) Anhydride Column($/yr) cost

37418 25.25 Ann. Install column cost

40599 20158 Ann. Condenser cost

2227 758 Ann. Cooling water cost

8390 5525 Ann. Reboiler cost

25897 13910 Ann. Stream cost

114531 65376 Total cost

Total cost of two column : 179907 $/yr

For x=0.3 we have:

EP4= EP3-Costs of two towers=1025515-179907

EP4=845608 $/yr for x=0.3

It seems that for getting good profit we have to consider x below 0.3

NOTE: We have numerous alternative for this part like condenser instead

flash .

32

THE END