CHAPTER 6 CHAPTER 6 RECYCLE STRUCTURE RECYCLE STRUCTURE

OF THE FLOWSHEETOF THE FLOWSHEET

6.1 DECISIONS THAT DETERMINE THE 6.1 DECISIONS THAT DETERMINE THE RECYCLE STRUCTURERECYCLE STRUCTURE

The decisions that fix the recycle structure of the flowsheet are listed in

Table 6.1-1.

TABLE 6.1-1 Decision for recycle structureTABLE 6.1-1 Decision for recycle structure

1. How many reactor system are required? Is there any separation between the reactor system?

2. How many recycle stream are required?3. Do we want to used an excess of one

reactant at the reactor inlet?4. Is a gas compressor required? What are

the cost?5. Should the reactor operated

adiabatically, with direct heating or cooling, or is a diluent or heat carries required?

TABLE 6.1-1 Decision for recycle structureTABLE 6.1-1 Decision for recycle structure (continuous)(continuous)

6. Do we want to shift the equilibrium conversion?

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

Number of reactor SystemsNumber of reactor Systems

Toluene+H2 Benzene+CH4

2Benzene Diphenyl+H2 }1150-1300F,

500 psia (6.1-1)

For example HDA process.

Both take place at the same T. and P. without catalyst. one reactor requied.

Number of reactor SystemsNumber of reactor Systems

Acetone Ketene+CH4

Ketene CO+1/2C2H4 }700C, 1 atm

(6.1-2)

two reactor requied.

Ketene + Acetic Acid Acetic Anhydride 80C, 1 atm

The anhydride process

Number of recycle streamNumber of recycle stream

From the anhydride process

Reactor

R1

Reactor

R2Acetone feed

Acid feed

Acid recycle

Acetone recycle

FIGURE 6.1-1 Acetic anhydride

-List of all components leaving the reactor -List of all components leaving the reactor that has been orderedthat has been ordered normal boiling normal boiling points. e.g., Table 5.1-4 points. e.g., Table 5.1-4 -List the reactor number as the destination -List the reactor number as the destination code for each recycle stream.code for each recycle stream.Next group recycle component having Next group recycle component having neighboring boiling points if they have the neighboring boiling points if they have the same reactor destination. same reactor destination.

Do not separate two components and then remix them at a reactor inlet.

(6.1-3)

Example 6.1-1 Number of recycle stream. Consider Example 6.1-1 Number of recycle stream. Consider the components and the destinations given below in the components and the destinations given below in order of their normal boiling point:order of their normal boiling point:

A. Waste by-product

B. Waste by-product

C. Reactant-recycle to R1

D. Fuel by-product

E. Fuel by-product

F. Primary product

G. Reactant-recycle to R2

H. Reactant-recycle to R2

I. Reactant-recycle to R1

J. Valuable by-product

4 product stream4 product stream A+B, D+E, F and JA+B, D+E, F and J3 recycle stream3 recycle stream C and I (go to R1), G+H (go to R2) C and I (go to R1), G+H (go to R2)

TABLE 6.1-2 HDA process. The component and their TABLE 6.1-2 HDA process. The component and their destination for the HDA process are as follows:destination for the HDA process are as follows:

Component NBP, C Destination

H2 -253 Recycle + purge-gas

CH4 -161 Recycle + purge-gas

Benzene 80 Primary product

Toluene 111 Recycle-liquid

Diphenyl 255 Fuel by-product

Product stream3

1) Purge

2) Benzene

3) Diphenyl

Recycle stream2

1) H2 and CH4

2) Toluene

A recycle flowsheet is given in Fig. 6.1-2A recycle flowsheet is given in Fig. 6.1-2

Reactor Separator

H2 feed Benzene

Toluene recycle

FIGURE 6.1-2 HDA recycle structure

Toluene feed

Compressor

Diphenyl

Purge

TABLE 6.1-3 Anhydride process. The component and TABLE 6.1-3 Anhydride process. The component and their destination for the anhydride process are given:their destination for the anhydride process are given:

Component NBP, F Destination

CO -312.6 Fuel by-product

CH4 -258.6 Fuel by-product

C2H4-154.8 Fuel by-product

Ketene -42.1Unstable 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

Product stream2

1) CH4 + CO + C2H4

2) anhydride

Liquid-recycle stream2

1) Acetone

2) Acetic Acid

Reactor

R1

Reactor

R2Acetone feed

Acid feed

Acetic Acid recycle

Acetone recycle

FIGURE 6.1-3 Acetic anhydride recycle

Reactor

R1

Anhyd.

CO, CH4, C2H4

Excess reactantsExcess reactants

In some cases the use of an excess reactant can shift the product distribution.

For example, the production of isooctane via butane alkylation as:

Butene + Isobutane Isooctane

Butene + Isooctane C12

6.1-4

If the kinetics match the stoichiometry, the use of excess of isobutane leads to an impoved selectivity to produce isooctane. The larger the excess, greater the improvement in the selectivity, but the larger cost to recover and recycle isobutane. Thus, an optimum amount of excess must be determined from economic analysis.

The use of an excess component can The use of an excess component can also be used to force another component also be used to force another component to be close to to be close to complete conversion.complete conversion. For example, the production of phosgene

CO +Cl2 COCl2 6.1-5

Which is an intermediate in the production of di-isocyanate, the product must be free of Cl2. Thus, an excess of CO is used to force Cl2 conversion to be very high.

Similarly, an excess can be used to Similarly, an excess can be used to shift shift equilibrium conversionequilibrium conversion

For example, the production of cyclohexane

Benzene + 3H2 cyclohexane 6.1-6

Molar ratio of reactant inlet is often a design variable

6.2 RECYCLE AND MATERIAL BALANCE 6.2 RECYCLE AND MATERIAL BALANCE

Our goal is to obtain a quick estimate of the recycle flow.

We have not specified any detail of separation system as yet, and therefore we assume that greater than 99% recoveries of reactants are equivalent 100% recoveries.

Limiting ReactantLimiting Reactant

For HDA process.(Fig. 6.2-1)

Reactor

R1

Toluene feed

H2, feed

FT(1-x)

Acetone recycle

FIGURE 6.2-1 HDA, liquid recycle

Benzene,PB

Purge

Separation

systemDiphenyl

FT(1-x)

FFT

FT(1-x)

FT

Limiting Reactant Limiting Reactant

FFT + FT(1-x) = FT (6.2-1)

x

FF FT

T (6.2-2)

Thus,the feed to the reactor is

In some case, some of limiting reactant might In some case, some of limiting reactant might leave leave the process in a gas recycle and purge streamthe process in a gas recycle and purge stream (ammonia synthesis), or it may (ammonia synthesis), or it may leave with the productleave with the product (ethanol synthesis).(ethanol synthesis).

OHOCHCHOHCHCH

OHCHCHOHCHCH

222323

23222

)(2

(6.2-3)

If we consider a simplified version of the ethanol process, the reaction are

If we want to produce 783 mol/hr of an EtOH-H2O If we want to produce 783 mol/hr of an EtOH-H2O azeotrope that contains 85.4 mol% EtOH, from an azeotrope that contains 85.4 mol% EtOH, from an ethylene feed stream containing 4%CHethylene feed stream containing 4%CH44 and pure and pure

waterwater

Reactor

Ether

FIGURE 6.2-2 Ethanol synthesis

EtOHH2O

Aceotrope

C2H4, CH4

Separation

systemH2O

C2H4, CH4

H2O

Overall material balances start with the Overall material balances start with the production rate of the azeotropeproduction rate of the azeotrope

hrmolPazeo / 783

EtOHazeoazeo PPy

EtOHhrmolPEtOH / 669)783(854.0

(6.2-4)

This contains This contains

(6.2-5)

(6.2-6)

The amount of water in the product stream is The amount of water in the product stream is

O / 114 669783 22HhrmolPPP EtOHazeoOH

Thus, from Eq. 6.2-3 and the result Thus, from Eq. 6.2-3 and the result above ,the required feed rate of water, which above ,the required feed rate of water, which is the limiting reactant, isis the limiting reactant, is

hrmol

)P-y ( PyF azeooazeazeoazeoOH

/ 783114669

1 2

(6.2-4)

We let the amount entering the reactor be FWe let the amount entering the reactor be Fww, the amount leaving the , the amount leaving the

reactor be Freactor be Fww(1-x), the amount leaving with the product be F(1-x), the amount leaving with the product be Fw,Pw,P, and , and

amount recycled beamount recycled be FFww(1-x)-F(1-x)-Fw,Pw,P Then a balance at the mixing point Then a balance at the mixing point

before the reactor givesbefore the reactor gives

xFF

FFxFFF

Rww

wPwwRwPw

/

])1([)(

,

,,,

(6.2-8)

(6.2-9)

Suppose that we let the water leaving with the Suppose that we let the water leaving with the product be Fproduct be Fw,pw,p = 114 and the fresh feed water = 114 and the fresh feed water

required for the reaction be Frequired for the reaction be Fw,Rw,R.. Now refering to Now refering to

the schematic in Fig. 6.2-3 for waterthe schematic in Fig. 6.2-3 for water

Reactor SeparatorFP+FR

F(1-x)-FP

FFP

F(1-x)

Other Reactants :For example , HDA process Other Reactants :For example , HDA process

x

FMRRyFy FT

GPHGFH (6.2-10)

(6.2-11)

PHHF

HP

PH

BG yy

y

x

MR

Sxy

PR

Reactor

R1

Toluene feed

H2, feed

95%H2, 5% CH4

FG

Benzene,PBSeparation

systemDiphenyl

RG ,yPHPurge, H2 CH4

FT

MR

FIGURE 6.2-4 gas recycle

Design HeuristicsDesign Heuristics

For the case single reactions, choose x= 0.96 or x=0.98xeq as a first guess.

This rule of thumb is discussed in Exercise 3.5-8

(6.2-12)

Reversible By-products Reversible By-products

• 2Benzene Diphenyl+H2

If we recycle a by-product formed by a reversible reactions and let the component build up to its equilibrium level. Such as the diphenyl in the HDA process.

Or the diethylether in ethanol synthesis (Eq.6.2-3)

We find the recycle flow by using the equilibrium relationship at the reactor exit.

22 ))((

Benzene

HDiphenylKeq (6.2-13)

Reactor Heat Load Reactor Heat Load

Rate FeedFresh Reaction ofHeat loadHeat Reactor

For the single reaction

FreshRR FHQ (6.3-1)

6.3 REACTOR HEAT EFFECTS 6.3 REACTOR HEAT EFFECTS

Example 6.3-1 HDA process. Example 6.3-1 HDA process.

If we want to estimate the reactor heat load for a case where x=0.75, PB=265, and FFT=273 mol/hr, we might neglect the second reaction and write

hrBtuFHQ FTRR /10878,5)273)(530,21( 6

Where HR is the heat of reaction at 1200 F and 500 psia and heat is liberated by the reaction.

(6.3-1)

Example 6.3-2Example 6.3-2 Acetone can be produced by the Acetone can be produced by the dehydrogenation of isopropanol dehydrogenation of isopropanol

hrBtuQR /10324.1)3.51)(800,25( 6

Heat is consumed by the endothermic reaction.

(CH3)2CHOH(CH3)2CO+H2 (6.3-2)

If we desire to produce 51.3 mol/hr of acetone and 51.3 mol/hr of IPA is required. The heat of reaction at 570 F and 1 atm is 25,800 Btu/mol,

So that the reactor heat load is

(6.3-3)

Adiabatic Temperature change Adiabatic Temperature change

Estimate the adiabatic Temp. change from the expression:

)( ,, outRinRPR TTFCQ (6.3-4)

Example 6.3-3Example 6.3-3 HDA process.The flow and heat HDA process.The flow and heat capacities of the reactor feed stream for case where capacities of the reactor feed stream for case where x=0.75 and yx=0.75 and yPHPH=0.4 are given below.=0.4 are given below.

From, from Ex.6.3-1 and Eq. 6.3-4 with TR,in=1,150 F

Stream Flow, mol/hr Cp,Btu/(molF)

Makeup gas 496 0.95(7)+0.05(10.1)=7.16

Recycle gas 3371 0.4(7)+0.6(10.1)=8.86

Toluene feed 273 48.7

Toluene recycle 91 48.7

FT

TT

Q

outR

outRinR

R

12651151150

)(86.8337116.74967.4891273

10878,5

,

,,

6

(6.3-5)

Example 6.3-4Example 6.3-4 IPA process. If the feed stream to acetone IPA process. If the feed stream to acetone process described Eq. 6.3-2 is an IPA-Hprocess described Eq. 6.3-2 is an IPA-H22O azeotrope 70% IPA) O azeotrope 70% IPA)

and if we recycle and azeotropic mixture, then it is to show that and if we recycle and azeotropic mixture, then it is to show that 22.0 mol/hr of water enters with the feed. Also, for a conversion 22.0 mol/hr of water enters with the feed. Also, for a conversion of x=0.96, the recycle flow will be 2.1 mol/hr of IPA and 0.9 of x=0.96, the recycle flow will be 2.1 mol/hr of IPA and 0.9 mol/hr of water. If the reactor inlet Temp. is 572 mol/hr of water. If the reactor inlet Temp. is 572 F, then from F, then from Eqs.6.3-1 and 6.3-4Eqs.6.3-1 and 6.3-4 the adiabatic Temp. change isthe adiabatic Temp. change is

FT

T

Q

outR

outR

R

216788572

)572(0.229.01.20.223.51

10324.1

,

,

6

(6.3-6)

This is unreasonable result. Thus, instead of using an adiabatic reactor, we attempt to achieve isothermal operation by supplying heat of the reaction to the process. In particular, we might attempt to pack the tubes of a heat exchanger with a catalyst.

Heuristic for Heat Loads Heuristic for Heat Loads

If adiabatic operation is not feasible, such as in the isopropanol example, then we attempt to use direct heating or cooling. However, in many cases there is limit to the amount of heat-transfer surface area that we can fit into a reactor.

Consider the of high T. gas-phase reaction

Let U=20 Btu/(hr ft2 F) and T= 5 F ,Then for the heat load 1x106 Btu/hr

2

6

10005020

101ft

TU

QA R

(6.3-7)

Example 6.4-1Example 6.4-1 Cyclohexane production. Cyclohexane production. Cyclohexane can be produced by the reactionCyclohexane can be produced by the reaction

126266 3 HCHHC (6.4-1)

We consider a case where we desire to produce 100 mol/hr of C6H12 with a 99.9%purity. A pure benzene feed stream is available, and the hydrogen makeup stream contains 97.5 %H2 , 2%CH4, 0.5 %N2. A flowsheet for recycle structure is shown in Fig. 6.4-1 for a case where we recycle some of the benzene(which is not necessarily the best flowsheet).

6.4 EQUILIBRIUM LIMITATIONS6.4 EQUILIBRIUM LIMITATIONS Equilibrium Conversion

SolutionSolutionOverall material balancees. Assume no losses. ThenOverall material balancees. Assume no losses. Then

(6.4-2)

Assume we use a gas recycle and a purge stream.Let

Production of C6H12: Pc=100

Benzene fresh feed: FFB=Pc=100

FE=Excess H2 Fed to process

Total H2 Feed =3Pc+FE=0.975FG

(6.4-3)

(6.4-4)

(6.4-5)GE

EPH2 0.025FF

Fy :H ofn compositio

Purge

(6.4-8)

Recycle balances

(6.4-9)

(6.4-6)

GE FFP 025.0P Rate urge G

xB c

B

PF:reactor tofeed enzene

(6.4-7)

G

c

PH

Fx

MRP

ycycle 975.0

1R:flow gas Re G

PH

PHG 0.975

y-13F Rate gas

yPMakeup G

Let molar ratio of H2 to Benzene be MR. THen

(6.4-11)

(6.4-13)

cP eCyclohexan

x

xB

1P enzene c

ccB Px

MRPMRH

332

(6.4-10)

cPH

PHGPHG P

x

MR

y

yRyFI

3

11025.0nerts

Reactor exit flows

(6.4-12)

PHc yx

MR

xPTotal

13

1flow (6.4-14)

(6.4-16)

(6.4-17)

3333e HBHBtot

CC

HB

C

yyp

y

ff

fK

13.1 1B

C

Hv

(6.4-15)

cPHe

e

e

eetot yxMR

xMR

x

xKP

3

31

113.1 3

Equilibrium relationship

Then

icomponentofFugacityfi :

icomponentoftcoefficienFugacityi :

Separator Reactors

If one of the product can be removed while the reaction is taking place, then an apparently equilibrium-limited reaction can be forced to go to complete conversion.

(6.4-19)

Example 6.4-2 Acetone production. Acetone can be produced by dehydrogenation of isopropanol

IsopropanolAcetone+H2

In the liquid phase as well as gas phase. At 300 F the equilibrium conversion for the liquid-phase process is

about xeq=0.32. However, by suspending the catalyst in a high-boiling solvent and operating the reactor at a

Temp. above the boiling point of Acetone, both H2 and Acetone can be removed as a vapor from reactor.

Thus equilibrium conversion is shift to right. A series of three continuous stirred tank reactor with a pump around loop containing a heating system that supplies the endothermic heat to reaction, can be used for process.

(6.4-19)

Example 6.4-3 Production of ethyl acrylate. Ethyl acrylate can be produced by the reaction

Acrylic Acid + Ethanol Ethyl Acrylate + H2O

Acrylic Acid , Ethanol are monomers, which tend to polymerize in the reboilers of distillation columns. We can eliminate a column required to purify and recycle acrylic Acid from the process if we can force the equilibrium-limited reaction to completion, by removing the water. Hence we use an excess of ethanol to shift the equilibrium to the right, and we carry out the reaction in the reboiler of retifying column. With this approach, the ethanol, water, and ethyl acrylate are taken overhead, and acrylic acid conversion approaches unity.

(6.4-21)

Reversible Exothermic Reactions

Sulfuric acid process : SO2+1/2H2O SO3

In ammonia synthesis

For Example

Water-gas shift : CO +H2O CO2+H2 (6.4-22)

ammonia synthesis: N2+3H2O 2NH3+H2 (6.4-22)

x ↓ as T

(6.4-24)

Diluents

Ethylbenzene styrene +H2

In some case an extraneous component (a diluent) is added which also causes a shift in the equilibrium conversion. For example, styrene can be produced by the reactions

(6.4-25)

(6.4-26)

Where the reactions take place at about 1100 F and 20 psia. The addition of steam so decrease the reverse reaction rate in Eq. 6.4-24. The stream serves in part as a heat carrier to supply endothermic heat of reaction.

Ethylbenzene Benzene+ethylene

Ethylbenzene Toluene+Methane

6.5 COMPRESSOR DESIGN AND COSTWhenever a gas-recycle stream is present, we will need a gas recycle compressor.

The design equation for the theoretical horsepower(hp) for a centrifugal gas compressor is

(6.5-1)

Where Pin=lbf/ft2 , Qin=ft3/min and =(Cp/Cv-1)/Cp/Cv

11003.3 5

in

outinin P

PQPhp

The exit Temp. from the compression stage is

in

out

in

out

P

P

T

T(6.5-2)

TABLE 6.51 Values of

Monotonic gases 0.40

Diatomic gases 0.29

More complex gases(CO2,CH4)

0.23

Other gases R/Cp

Efficiency

For the first designs, we assume a compressor efficiency of 90% to account for the fluid friction in suction and discharge values, ports, friction of moving metal surface fluid turbulence, etc. also we assume a driver efficiency of of 90% to account for the conversion of the input energy to shaft work.

Multistage Compressors

For a three-stage compressor with intercooling, the work required is

1

3

4

2

3

1

2

P

P

P

P

P

PMRTWork in

The intermediate pressures that minimize the work are determined from

032

P

Work

P

Work

(6.5-3)

(6.5-4)

Which lead to the results

3

4

2

3

1

2

P

P

P

P

P

P

Design heuristic: The compression ratios for each stage in a gas compressor should be equal.

(6.5-5)

Annualized Install Cost

The brake horsepower bph is obtained by introducing the compressor efficiency in to Eq. 6.5-1:

9.0

hpbhp (6.5-7)

Then, Guthrie’s correlation(page.573)

)11.2(517280

& 82.0

cFbhpSM

CostInstalled

(6.5-7)

(6.5-8)

Operating Cost

By dividing the brake horsepower by the driver efficiency. We can calculate the utility requirement.

Then from utility cast and using 8150 hr/yr, we can calculate the operating cost.