Chemical Reaction Engineeringcbe.snu.ac.kr/sites/cbe.snu.ac.kr/files/board/Lecture... · 2016-12-23 · Levenspiel Plot Table 2-3 Fig. 2-2 Seoul National University. For vs. X, the

Chemical Reaction Engineering

Youn-Woo LeeSchool of Chemical and Biological Engineering

Seoul National University155-741, 599 Gwanangro, Gwanak-gu, Seoul, Korea [email protected] http://sfpl.snu.ac.kr

Lecture #3

Reaction Engineering 1

第2章Conversion

and Reactor Sizing

反應工學 I

Seoul National University


개요. 제1장에서는 반응이 일어나는 영역(반응기)에서 일반 몰수지식

(GMBE)을 유도하였고 이를 4가지 이상적인 반응기에 적용하여 각 반응기에 대

하여 설계방정식을 유도하였다. 제2장에서는 어떻게 이런 반응기들의 크기를 구

하고 개념적으로 어떻게 배열하는지 보여줄 것이다. 이번 단원에서는

전화율(X)을 정의하고,

4종류의 이상적인 반응기의 설계방정식들을 전화율 X의 항으로 다시 쓰며,

일단 반응속도와 전화율 사이의 관계가 주어진 경우(즉, -rA=f(x))에 Levenspiel

Plot을 그려보고,

Levenspiel plot으로 부터 반응기의 크기를 구해보고,

어떻게 CSTR과 PFR의 크기를 비교하는지 보여주고,

어떻게 반응기들을 직렬로 최적 배열하는지 보여 줄 것이다.

더욱이, 반응속도와 전화율 사이의 관계가 주어진 경우에, CSTR과 PFR의 크기를

구할 수 있고 직렬로 배열된 반응기들의 총괄전화율과 각각의 반응기 부피들을

계산할 수 있을 것이다.

Follow the Reaction Design Algorithm

Follow the Yellow Brick Road

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Objectives

After completing Chapter 2, reader will be able to:

Define conversion.

Write the mole balances in terms of conversionfor a batch reactor, CSTR, PFR, and PBR.

Size reactors either alone or in series once given the molar flow rate of A, and the rate of reaction, - rA, as a function of conversion, X.


왜 전화율을사용하려고 하는가?

편리한 전화율 사용!

X=0.5일 때 반응기크기?X=0.9일 때 반응시간은?

2.1 Definition of ConversionConsider the general equation

Choose A as our basis of calculation(The basis of calculation is most always the limiting reactant )

Questions- How can we quantify how far a reaction has progressed ?- How many moles of C are formed for every mole A consumed ?

The convenient way to answer these question is to define conversion.

DCBA dcba

DCBAad

ac

ab

fedAofmolereactedAofmoleX

(2-2)

(2-1)


2.2 Batch Design Equations

the longer a reactant is in the reactor, the more reactant is converted toproduct the reactant is exhausted. Consequently, in batch system, theconversion X is a function of reaction time the reactants spend in thereactor.

If NA0 is the number of moles of A initially in the reactor, then the totalnumber of moles of A that have reached after a time t is [NA0 X]

XNconsumed

Aofmole

fedAofmolereactedAofmoles

fedAofmole

consumedAofmole

A

0

In most batch reactors,

(2-3)

The number of moles of A that remain in the reactor after a time t, NA, can beexpress in terms of NA0 and X:

XNNNreactionchemicalby

consumedbeenhavethatAofmoles

tatreactortofedinitially

Aofmoles

ttimeatreactorin

Aofmoles

AAA 00

0

(2-4)


The mole balance on species A for a batch system

The number of moles of A in the reactor after a conversion X

In term of conversion by differentiating equation

The design equation for a batch reactor in differential form is

)1(000 XNXNNN AAAA

Vrdt

dNA

A

dtdXN

dtdN

AA

00

VrdtdXN AA 0

The differential formfor a batch reactor

(2-4)

(1-5)

(2-5)



The design equation for a batch reactor in differential form

VrdtdXN AA 0

Write the mole balances in terms of conversion


Vrdt

dNA

A

(2-6)

(2-5)


회분식반응기의설계방정식


A

AAAA rdt

dCdt

VNddt

dNVdt

dNV

0

0

/11


Vrdt

dNA

A

Constant volume, V=V0

dtdCr A

A (2-7)



VrdtdXN AA 0


Vrdt

dNA

A

The differential forms of the batch reactor mole balances, Eqs(2-5) and (2-6), are often usedin the interpretation of reaction rate data (Chapter 7)and for reactors with heat effects (Chapter 11-13), respectively.

(2-6)(2-5)



Batch reactors are frequently used in industry forboth gas-phase and liquid-phase reactions.

The lab bomb calorimeter reactor is widely used forobtaining reaction rate data.

Liquid-phase reactions are frequently carried out inbatch reactors when small-scale production is desiredor operating difficulties rule out the use of continuousflow systems.


For constant-volume batch reactor, V=V0

For the most common batch reactors where volume is notpredetermined function of time, the time necessary to achieve aconversion X is

A

AAA rdt

dCdt

VNddt

dNV

0

0

/1

tX

AA Vr

dXNt00

The integral form for a batch reactor

0VNC A

A


(2-7)

If FA0 is the molar flow rate of species A fed to a system at steady state, the molar rate at which species A is reacting within the entire system will be FA0X.

time

reactedAofmolesXF

fedAofmolesreactedAofmoles

timefedAofmolesXF

A

A

0

0

2.3 Design Equations for Flow Reactors

FA0 FA


The molar flow rate

Rearranging gives

XFF AA 10

AAA FXFFsystemtheleaves

Awhichatrateflowmolar

systemthewithinconsumed

whichatratemolar

systemthetofedisAwhichat

rateflowmolar

00

2.3 Design Equations for Flow Reactors

FA0 FA

(2-8)


- The design equation for a CSTR

- conversion of flow system

- Combining (2-12) with (2-11)

2.3.1 CSTR or Backmix Reactor

XFFF AAA 00

A

AA

rFFV

0

exitA

A

rXFV

0

(2-11)

(2-12)

Equation to determine the CSTR volume necessary to achieve a specifiedconversion X. Since the exit composition from the reactor is identical to thecomposition inside the reactor, the rate of reaction is evaluated at the exit condition.

FA0

FA

(2-13) design equation for a CSTR

DCBAad

ac

ab


- General mole balance equation


- The differential form of the design equation

- Volume to achieve a specified conversion X

2.3.2 Tubular Flow Reactor (PFR)

AA r

dVdF

XFFF AAA 00

AA rdVdXF 0

X

AA r

dXFV00

FA0 FA

(1-12)

(2-15)

(2-16)


(2-12)

- General mole balance equation


- The differential form of the design equation with P 0

2.3.3 Packed-Bed Reactor (PBR)

'0 AA r

dWdXF

X

AA r

dXFW0 '0

XFFF AAA 00

'A

A rdWdF

FA0 FA

(2-17)

(2-18)

-The catalyst weight W to achieve a specified conversion X with P=0

(1-15)


tX

AA Vr

dXNt00

Design equationfor a batch reactor

Summary of Design Equation

exitA

A

rXFV

0

FA0

FA

Design equation for a CSTR

X

AA r

dXFV00FA0 FA

Design equation for a PFR

X

AA r

dXFW0 '0

FA0 FA

Design equation for a PBR

공통점?

NA0


tX

AA Vr

dXNt00

Summary of Design Equation Reaction time~ NA0

~ X~ 1/rAV

exitA

A

rXFV

0

FA0

FA

X

AA r

dXFV00FA0 FA

X

AA r

dXFW0 '0

FA0 FA

Reactor volume (Catalyst weight)

~ FA0

~ X~ 1/rA’




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2.4 Applications of the design equation for continuous-flow reactor

XkCkCr AAA 10For a first-order reaction :

The rate of disappear of A, -rA, is almost always a function of theconcentrations of the various species present. When a single reactionis occurring, each of the concentrations can be expressed as afunction of the conversion x; consequently, -rA, can be expressed asa function of X.

X

AA r

dXFV00FA0 FA


Conversion, X0.0 0.2 0.4 0.6 0.8 1.0

-rA

(mol

/m3 s)

0.0

0.1

0.2

0.3

0.4

0.5

Consider the isothermal gas-phase isomerization

A B

How to use the raw data of chemical reaction rate?

The laboratory measurements give the chemical reaction rate as a function of conversion.

(at T=500K, 8.2atm)

Greatest rate

Smallest rate

raw data


Conversion, X0.0 0.2 0.4 0.6 0.8 1.0

1/-r

A (m

3 s/m

ol)

0

5

10

15

20

25

30

- rate data convert reciprocal rates, 1/- rA

- plot of 1/- rA as a function of X

Levenspiel Plot

Greatest rate

Small rate


Conversion0.0 0.2 0.4 0.6 0.8 1.0

F A/-r

A (m

3 )

0

2

4

6

8

10

12

- plot of [FA/- rA] as a function of [X]

Levenspiel Plot

Table 2-3

Fig. 2-2 Seoul National University

For vs. X, the volume of a CSTR and the volume of a PFR

can be represented as the shaded areas in the Levenspiel plots.A

A0

rF

• Given –rA as a function of conversion.

• Constructing a Levenspiel plot.

• Here we plot either or as a function of X.Ar

1A

A0

rF

Reactor Size


The reaction described by the data in Table 2-3 (below)

A B

is to be carried out in a CSTR. Species A enters the reactor at a molar flow rate of 0.4 mol/s.

(a) Using the data in Table 2-3, or Fig. 2-1, calculate the volume necessary to achieve 80% conversion in a CSTR.

(b) Shade the area in Fig. 2-2 that would give the CSTR volume necessary to achieve 80% conversion.

Example 2-1 Sizing a CSTR

Table 2-3


Calculate the volume necessary to achieve 80% conversion in a CSTR


lmmol

sms

molr

XFV

exitA

A 64004.6)20)(8.0)(4.0( 33

0

Conversion0.0 0.2 0.4 0.6 0.8 1.0

F A/-r

A (m

3 )

0

2

4

6

8

10

12

VCSTR

= 8 x 0.8= 6.4 m3

In CSTR, C, T, P, and X of the effluentstream are identical to that of the fluidwithin the reactor, because perfect mixing isassumed.

EXIT

FA0=0.4 mol/s

FA

1.5m

3.6m(a)

(b)


The volume necessary to achieve 80% conversion in a CSTR is 6.4m3.


FA0=0.4 mol/s

FA

1.5m

3.6m

FA0=0.4 mol/s

FA

2.01m

2.01m

It’s a large CSTR, but this is a gas-phase reaction, and CSTRs arenormally not used for gas-phase reaction, and CSTRs are usedprimarily for liquid-phase reactions.


Calculate the volume necessary to achieve 80% conversion in a PFR.We shall use the five point quadrature formula (A-23) in Appendix A.4.

Example 2-2 Sizing a PFR

smol /4.0FA0

333

00000

8.0

00

165.2)47.32(32.0)00.8()54.3(4)05.2(2)33.1(489.0

32.0

)8.0()6.0(4

)4.0(2

)2.0(4

)0(3

mmm

XrF

XrF

XrF

XrF

XrFXrdXF

V

A

A

A

A

A

A

A

A

A

A

X

A

A

V = 2.165 m3

= 2165 dm3


Conversion0.0 0.2 0.4 0.6 0.8 1.0

F A0/-

r A (m

3 )

0

2

4

6

8

10

12

Calculate the volume necessary to achieve 80% conversion in a PFR


FA0 FA

dXr

FV

X

A

A

8.0

00

= area under the curvebetween X=0 and X=0.8

= 2165 dm3 (2.165 m3)

(see appropriate shadedarea in Fig. E2-3.1)

Graphic Method

VPFR=2.165 m3


Sketch the profile of –rA and X down the length of the reactor.


FA0 FASolutionAs we proceed down the reactor and more and more of reactant isconsumed, the concentration of reactant decreases, as does therate of disappearance of A. However, the conversion increases asmore and more reactant is converted to product.

Simpson’s rule (Appendix A.4 Eq. A-21)

X=0.2, X=0.1

3333

0002.0

00

218218.0)54.6(31.033.1)08.1(489.0

31.0

)2.0()1.0(4

)0(3

dmmmm

XrF

XrF

XrFX

rdXFV

A

A

A

A

A

AX

AA




FA0 FASolution

3333

0004.0

00

551551.0)26.8(32.005.2)33.1(489.0

32.0

)4.0()2.0(4

)0(3

dmmmm

XrF

XrF

XrFX

rdXFV

A

A

A

A

A

AX

AA


X=0.4, X=0.2




FA0 FASolution

3333

0006.0

00

1093093.1)93.10(33.054.3)625.1(489.0

33.0

)6.0()3.0(4

)0(3

dmmmm

XrF

XrF

XrFX

rdXFV

A

A

A

A

A

AX

AA


X=0.6, X=0.3




FA0 FASolution

3333

0008.0

00

2279279.2)09.17(34.00.8)05.2(489.0

34.0

)8.0()4.0(4

)0(3

dmmmm

XrF

XrF

XrFX

rdXFV

A

A

A

A

A

AX

AA


X=0.8, X=0.4




X

-rA (mol/m3·s)

V (dm3)

0

0.45

0

0.2

0.30

218

0.4

0.195

551

0.6

0.113

1093

0.8

0.05

2279


V (dm3)

0 500 1000 1500 2000 2500

0.0

0.2

0.4

0.6

0.8

1.0

X



X=0.8X=0.6

X=0.2

X=0.4

V=2165 LV=1093 L

V=551 L

V=218 L

Seoul National University반응기를 따라 내려감에 따라서 전화율은 증가한다.


X0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.1

0.2

0.3

0.4

0.5

-rA

0.5 1.0 1.5 2.0 2.5

V (m3)

(mol/m3s)



반응기를 따라 내려감에 따라서 전화율은 증가하는 한편 반응속도 rA는 감소한다.

Conversion0.0 0.2 0.4 0.6 0.8 1.0

F A/-r

A (m

3 )

0

2

4

6

8

10

12

Calculate the volume necessary to achieve 80% conversion in a CSTR and a PFR

Example 2-3 Comparing CSTR and PFR Sizes

smol /4.0FA0

FA0

FA

V=6.4 m3

FA0 FA

V=2.2 m3

For isothermal reaction ofgreater than zero order, thePFR will always require asmaller volume than the CSTRto achieve.


0차보다 더 큰 차수의 등온반응의 경우에, 동일한 전화율과 동일한 반응조건들(온도, 유량 등)에 대해서CSTR 부피가 PFR 부피보다 일반적으로 더 크다.

X0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.1

0.2

0.3

0.4

0.5

-rA

Example 2-3 Comparing CSTR and PFR Sizes

FA0 FA

FA0

FA

V=6.4 dm3

V=2.2 dm3The isothermal CSTR volume isusually greater than the PFRvolume is that the CSTR isalways operating at the lowestreaction rate (-rA=0.05).

The PFR start at the higher rateat the entrance and graduallydecreases to the exit rate,thereby requiring less volumebecause the volume is inverselyproportional to the rate.


Chemical Reaction Engineering

Youn-Woo LeeSchool of Chemical and Biological Engineering

Seoul National University155-741, 599 Gwanangro, Gwanak-gu, Seoul, Korea [email protected] http://sfpl.snu.ac.kr

Lecture #4



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2.5 Reactors in series

Define conversion

The conversion X defined as the “total number of moles” of A thathave reacted up to that point per mole of A fed to the “first” reactor.(assumption : no side stream withdrawn and the feed stream entersonly the first reactor in the series)

reactor first to fed A of molesi point to up reacted A of moles totalX i

PFR-CSTR-PFR in series

The relationships between conversion and molar flow rate

V1

X=0FA0

X1

FA1

V3

X2

FA2V2

X3

FA3

FA1 = FA0 - FA0 X1

FA2 = FA0 - FA0 X2

FA3 = FA0 - FA0 X3

reactor first to fed A of moles2 point to up reacted A of moles totalX 2 where

similar definitions exist for X1 and X3

1

001

X

AA r

dXFV

00.

2221

VrFFgenoutin

AAA

2

1202

)(

A

A

rXXFV

3

203

X

XA

A rdXFV

V1

X=0FA0

X1FA1

V3

X2FA2V2

X3FA3

Reactor 1:

Reactor 2 :

Reactor 3 :

FA1 = FA0 - FA0 X1

FA2 = FA0 - FA0 X2

FA3 = FA0 - FA0 X3

-rA2 is evaluatedat X2 for the CSTRIn this seriesarrangement

-rA2 -rA

-rA

FAe

X2=0.8

Four different schemes of reactors in series

Two CSTRs in series

Two PFRs in series

a PFR and CSTR in series

FA0

X1=0.4

FAe

X2=0.8

FAe

X2=0.8

FA0X1=0.4

FA0

FAe

X2=0.8

X1=0.5

X1=0.5

FA0

a CSTR and PFR in series

11

011 Xr

FVA

A

2

1202

)(

A

A

rXXFV

2.5.1 Two CSTRs in seriesFA0

X1=0.4

FAe

X2=0.8

Reactor 1

Reactor 2

-rA1

-rA2

(2-21)

(2-24)

Example 2-5: Two CSTRs in SeriesFA0

X1=0.4

FAe

X2=0.8

What is the volume of each of two CSTR reactors?

Reactor 1

[FAo/-rA]x=0.4=2.05 m3

V1=([FAo/-rA]x=0.4)(X1-X0)=(2.05)(0.4-0)=0.82 m3

Reactor 2

[FAo/-rA]x=0.8=8.0 m3

V1=([FAo/-rA]x=0.8)(X2-X1)=(8.0)(0.8-0.4)=3.2 m3

XA

[FAo/-rA] (m3)

0.0 0.1 0.2 0.4 0.6 0.7 0.8

0.89 1.09 1.33 2.05 3.54 5.06 8.0

Example 2-4: Two CSTRs in Series

Therefore, V1 + V2 = 0.82 + 3.2 = 4.02 m3

What is the reactor volume to achieve 80% conversion in a single CSTR?

[FAo/-rA]x=0.8 = 8.0 m3

V1 = ([FAo/-rA]x=0.8) (X1-X0)= (8.0)(0.8-0) = 6.4 m3

The sum of the two CSTR reactor volumes (4.02 m3) inseries is less than the volume of one CSTR (6.4 m3) toachieve the same conversion (X=0.8)

Conversion, X0.0 0.2 0.4 0.6 0.8 1.0

0

2

4

6

8

10

12

FAO/-rA

[m3]

Conversion, X0.0 0.2 0.4 0.6 0.8 1.0

0

2

4

6

8

10

12

FAO/-rA

[m3]

One CSTR vs Two CSTRsExample 2-4

FA0

X1=0.4

FAe

X2=0.8

FA0

FA

X=0.8

Vtotal = 4.02 m3

Vtotal = 6.4 m3

The sum of the two CSTR reactor volumes (4.02 m3) in series is less than the volume of one CSTR (6.4 m3) to achieve the same conversion (X=0.8)

0.82 m3

3.20 m3

V1 = 0.82 m3

V1 = 3.2 m3

6.4 m3

Approximating a PFR

Approximating a PFR with a number of small, equal-volume CSTRs of Vi in series

Then, compare the volume of all the CSTRs with the volume of one plug-flow reactor for the same conversion, say 80%

54321

1 2 3 4 5

As we make the volume of each CSTR smaller and increase the number ofCSTRs, the total volume of the CSTRs and the PFR will become identical!

80

60

40

20

A

A

rF

0

.35 .53 .65 .74 .8X

We can model a PFR as a number of CSTRs in series

1 2 3 4 554321

V1

V2

V3

V4

V5

Modeling of a PFR with a large number of CSTRs in series.

1

001

X

AA r

dXFV

2

102

X

XA

A rdXFV

2.5.2 Two PFRs in series

Reactor 1

Reactor 2

FAe

X2=0.8

FA0 FA1

X1=0.4

2

1

12

00

00

0X

X AA

X

AA

X

AAtotal r

dXFr

dXFr

dXFV

Two PFRs in Series

0X1

dX +-rA

FA0VTotal= V1 + V2= X2

-rA

FA0 dX = X1

X2

-rA

FA0

0

Conversion, X0.0 0.2 0.4 0.6 0.8 1.0

0

2

4

6

8

10

12

FAO/-rA

[m3]

V1V2

Sizing PFR in Series

FAe

X2=0.8

FA0X1=0.4

What is the volume of each of two reactors?Molar flow rate of A is 0.4 mol/s

XA[FAo/-rA] (m3)

0.0 0.1 0.2 0.4 0.6 0.7 0.80.89 1.09 1.33 2.05 3.54 5.06 8.0

Reactor 1By applying Simpson’s rule in Appendix A.4 (Text page 60),

30.2( )V1= [0.89+4(1.33)+2.05] =0.551 m3=551 dm3

Reactor 2

By applying Simpson’s rule in Appendix A.4 (Text page 60),

0.2( )V2= [2.05+4(3.54)+8.0] =1.614 m3=1614 dm33

Therefore, V1 + V2=0.551 m3 + 1.614 m3=2.165 m3 < 4.02 m3 (Two CSTR in Series)

2.5.3 Combination of CSTR and PFR in Series

FA0X=0 FA1

X1

FA2X2

FA3X3

V2

V1

V3

An industrial example of reactors in seriesfor using dimerization of propylene into isohexane

CH3

2 CH3-CH=CH2 CH3C=CH-CH2 -CH3

Conversion, X0.0 0.2 0.4 0.6 0.8 1.0

0

2

4

6

8

10

12

FAO/-rA

[m3]

2.5.3 Combination of CSTR and PFR in Series

0

CSTR 1

CSTR 2

PFR

1

0101

)(

A

A

rXXF

V

dXr

FV

X

X A

A

3

2

03

2

1202

)(

A

A

rXXF

V

FA0X=0 FA1

X1

FA2X2

FA3X3

V1

V2

V3

V2

V1

V3

Dimerization propylene into isohexanes

Dimersol G unit (Two –CSTR and one PFR in series)Institute Français du Petrόle Process

The finishing reactor (“the snake”) to comply with LPG specification in the USA (less than 5% olefins)

Plug-flow reactor for Dimersol™ process

The Dimersol process is used to dimerize light olefins such asethylene, propylene and butylene.

The process typically begins with the pretreatment of the propane/propylene or butane/butene feed prior to entering the reactorsection of the process. Pretreatment can include the use ofmolecular sieve dryers, sand filters, etc. to remove water and/orH2S. Water in the feed stream can deactivate the catalysts used inthe Dimersol process.

After drying the feed is combined with a liquid nickelcarboxylate/ethyl aluminum dichloride (EADC) catalyst prior toentering the first of a series of three reactors.

Description of Dimersol Process

The first two are continuous stirred tank reactors and the third is aplug-flow tubular reactor.

The reactor feed is converted to the process product, dimate,primarily in the first reactor, and additional conversion is achievedin the last two reactors. The final reactor effluent consists ofdimate product, unreacted C3/C4s, and liquid catalyst.

Immediately following the last reactor, the liquid catalyst isremoved from the reactor effluent by treating the reactor effluentwith caustic, subsequent water washing, and filtering to removesolids.


Spent caustic residuals are typically reused or reclaimed on- oroff-site, and as a result, do not constitute solid wastes.

After filtering, the product stream enters a "Dimersol stabilizer," adistillation unit that removes unreacted LPG from the dimateproduct. In some cases, the product stream is also further treatedby drying.

LPG from the stabilizer overhead is typically sent to another unitof the refinery for further processing. The dimate product from thebottom of the stabilizer is sent to storage or product blending.


Application : C3 or C4 Olefins Dimerization (Dimersol®)Type : nickel carboxylate/ethyl aluminum dichloride (EADC)Shape : Liquid Catalyst

LC 1252 catalyst is used in the Dimersol process licensed by Axens.

High octane value motor gasoline is obtained from olefinic C3 cuts from FCCs or steam crackers.

Oligomerization of C3 or C4 olefins produces, with high selectivity, hexenes, heptenes and higher olefins up to dodecenes

LC 1252 catalyst

Dimersol G Process25 plants, 3,000,000MT/year

C3H6 (l) [C3H6]2 Ho298=-89.1 kJ/mol (-21.30 kcal/mol)

NaOHNH3

LPGUnreacted C3

=

Isohexanebp=60oC

X1=0.7 X2=0.9X3=0.97

T=57oCP=17bar

C3 67%C3

= 33%

5% max. propylenein propane(US LPG

specificationas a fuel)

Dimersol stabilizer

Weak acid process for producing dinitrotolueneEP 0 903 336 A2, AIR PRODUCTS 1998

Dinitrotoluene is an important intermediate in producing toluenediisocyanate based polyurethanes.

TNT Production PlantEP 0 903 336 A2, AIR PRODUCTS 1998

TNT plant in Hiroshima, Japan

Conversion, X0.0 0.2 0.4 0.6 0.8 1.0

0

2

4

6

8

10

12

FAO/-rA

[m3]

Isothermal vs. Adiabatic

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.70.0

0.5

1.0

1.5

2.0

2.5

Conversion, X

FA0/-rA

(m3)

Isothermal Adiabatic

Example 2-5: An Adiabatic Liquid-Phase Isomerization

n-C4H10 i-C4H10

Calculate the volume ofeach of the three reactorsfor an entering molar flowrate n-butane of 50 kmol/hr.

Isomerization of butane

X 0.0 0.2 0.4 0.6 0.65-rA (kmol/m3-h) 39 53 59 38 25

FA0X=0

FA1X1=0.2

V3

V1V2

FA2X2=0.6

FA3X3=0.65rA1

rA3

FAo = 50 kmol/h

(a) CSTR 1 (X1=0.2)

0

(b) PFR (X2=0.6)

X 0.0 0.2 0.4 0.6 0.65

-rA (kmol/m3-h) 39 53 59 38 25

[FAo/-rA](m3) 1.28 0.94 0.85 1.32 2.0

331

1

0

1

0101 188.0)2.0)(94.0(

)(mmX

rF

rXXF

VA

A

A

A

36.0

0

4.0

0

2.0

06.0

2.00

2

38.032.1)85.0(494.032.0

43

m

rF

rF

rFXdX

rF

VXA

A

XA

A

XA

A

A

A

(c) CSTR 2 (X3=0.65)33

233

03 1.0)6.065.0)(2()( mmXX

rF

VA

A

Example 2-5

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.70.0

0.5

1.0

1.5

2.0

2.5

Conversion, X

FA0/-rA

(m3)

V1 =0.188 m3 V2 =0.38 m3

V3=0.1 m3

Example 2-5

Conversion0.0 0.2 0.4 0.6 0.8 1.0

F A/-r

A (m

3 )

0

2

4

6

8

10

12

Conversion0.0 0.2 0.4 0.6 0.8 1.0

F A/-r

A (m

3 )

0

2

4

6

8

10

12

Comparing CSTR and PFR Sizes

smol /4.0FA0

FA0

FA

V=6.4 dm3FA0 FA

V=2.2 dm3

For isothermal reaction of greater than zero order, the PFR will always require a smaller volume than the CSTR to achieve.

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.70.0

0.5

1.0

1.5

2.0

2.5

Conversion, X

FA0/-rA

(m3)

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.70.0

0.5

1.0

1.5

2.0

2.5

Conversion, X

FA0/-rA

(m3)FA0

FA

V=0.188 m3

FA0 FA

V=0.207 m3

>

<

For adabatic reaction, the CSTR may require a smaller volume than the PFR to achieve.

Which reactor should go first to give the highest overall conversion?

Which arrangement is best? “It depends.”

그때 그때달라요

FAe

X2

FA0

X1

FAe

X2

FAe

X2

FA0

X1

FA0

FAe

X2

X1

X1

FA0

An Adiabatic Liquid-Phase Isomerization

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.70.0

0.5

1.0

1.5

2.0

2.5

Conversion, X

FA0/-rA

(m3)

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.70.0

0.5

1.0

1.5

2.0

2.5

Conversion, X

FA0/-rA

(m3)

FAe

X2 =0.65

X1=0.4

FA0FA0

FAe

X2 =0.65

X1 =0.4

Best arrangement Worst arrangement

Laboratory and Full-scale operating conditions must be identical.

-If we know the molar flow rate to the reactor and the reaction rate as a function of conversion, then we can calculate the reactor volume necessary to achieve a specific conversion.

-However, the rate does not depend on conversion alone. It is also affected by the initial concentrations of the reactants, the temperature, and the pressure.

-Consequently, the experimental data obtained in the laboratory are useful only in the design of full-scale reactors that are to be operated at the same conditions as the laboratory experiments (T, P, CA0).

-This conditional relationship is generally true; i.e., to use laboratory data directly for sizing reactors, the laboratory and full-scale operating conditions must be identical.

-Usually, such circumstances are seldom encountered and we must revert to the methods described in Chapter 3 to obtain –rA as a function of X.


To size flow reactor, only need -rA=ƒ(X),

It is important to understand that

if the rate of reaction is available or can be obtained solely as afunction of conversion, -rA=ƒ(X), or

if it can be generated by some intermediate calculations,

one can design a variety of reactor or a combination of reactors.

In Chapter 3, we show how we obtain the relationship betweenreaction rate and conversion from rate law and reactionstoichiometry.


timeconditionspecifiedat

measuredfeedofvolumereactoroneprocesstorequiredtime

vV

0

X

AA

X

A

A

rdXC

rdX

vF

vV

0000

0

0

2.6. Space timeSpace-time :

The time necessary to process one reactor volume of fluid based onentrance conditions. Also called the holding time or mean residence time.

A space-time of 2 min means that every 2 min one reactor volume offeed at specified condition is being treated by the reactor.

Space timeThe time necessary to process one reactor volume of fluid based onentrance conditions. Also called the holding time or mean residence time.

Consider the tubular reactor, which is 20m long and 0.2 m3 in volume. Thedashed line represents 0.2 m3 of fluid directly upstream of the reactor. Thetime it takes for this fluid to enter the reactor completely is the space time.For example, if the volumetric flow rate were 0.01 m3/s, it would take theupstream volume shown by the dash lines a time

To enter the reactor. It take 20s for the fluid at point “a” to move point “b”

20m 20m

ssm

mV 20/01.0

2.03

3

0

a b

Space time

In the absence of dispersion, which is discussed inChapter 14, the space time is equal to the meanresidence time in the reactor, tm.

This time is the average time the molecules spend inthe reactor.

Table 2-4 Typical Space time for industrial reactor

Reactor type Production capacity

Batch 15 min ~ 20 h Few kg/day ~ 100,000 tons/year

CSTR 10 min ~ 4 h 10 ~ 3,000,000 tons/year

Tubular 0.5 s ~ 1 h 50 ~ 5,000,000 tons/year

Table 2-5 shows space times for six industrial reactions and reactors. (page 67)

• A space-velocity of 5 hr-1 means that five reactor volumes of feed at specified condition are being fed into the reactor per hour.

• Difference in the definitions of SV and

- space time : the entering volumetric flow rate is measured at the entrance condition- space velocity : other conditions are often used

10 1

time

volumeunitintreatedbecanwhichconditionspecifiedatfeed

ofvolumesreactorofnumber

Vv

SV

Space velocityDefinition of Space-velocity

• LHSV ( liquid hourly space velocity)- v0 is frequently measured as that of a liquid at 60 or 75 0F, even though the feed to the reactor may be a vapor at some higher temperature.

• GHSV ( gas hourly space velocity)- v0 is normally measured at standard temperature and

pressure (STP).

LHSV and GHSV

Vv

GHSVV

vLHSV STPliquid 00

Calculate the space time and space velocity for each of the reactors in Examples 2-2 and 2-3

Example 2-6 Reactor Space Times and Space Velocity

From Examples 2-2,v0=0.002 m3/s, Volume of CSTR=6.4m3

13

3

0125.1

89.011;89.03200

/002.04.6

h

hSVhs

smmV

From Examples 2-3,v0=0.002 m3/s, Volume of PFR=2.165m3

13

3

03.3

30.011;30.01083

/002.0165.2

h

hSVhs

smmV

In the design of reactors that are to be operated at conditions (e.g.,temperature and initial concentration) identical to those at whichthe reaction rate data were obtained, we can size (determine thereactor volume) both CSTRs and PFRs alone or in variouscombinations.

In principle, it may be possible to scale up a laboratory-bench orpilot-plant reaction system solely from knowledge of –rA as afunction of X or CA.

However, for most reactor systems in industry, a scale-up processcannot be achieved in this manner because knowledge of –rAsolely as a function of X is seldom, if ever, available underidentical conditions.

To summarized these last examples….


In Chapter 3, we shall see how we can obtain -rA=ƒ(X) frominformation obtained either in the laboratory or from the literature.This relationship will be developed in a two-step process.

In Step 1, we will find the rate law that gives the rate as a functionof concentration and in Step 2, we will find the concentrations as afunction of conversion. Combining Step 1 and 2 in Chapter 3, weobtain -rA=ƒ(X).

To summarized these last examples….


1. P2-5C2. P2-7B3. P2-10B4. P2-13A

Due date: one week

Homework #3

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Chemical Reaction Engineeringcbe.snu.ac.kr/sites/cbe.snu.ac.kr/files/board/Lecture... · 2016-12-23 · Levenspiel Plot Table 2-3 Fig. 2-2 Seoul National University. For vs. X, the