Chapter 4

Isothermal Reactor Design

Overview

• Chapter 1 and 2 focus on mole balances on reactors to predict the volume

• Chapter 3 focuses on reactions • Cahpter 4 combine previous chapters to

obtain optimum reactor design

Design Algorithm

1. Mole balance (reactor type)2. Reaction rate law (reaction type, orders)

3. Stoichiometry (reaction coefficients)

4. Combine steps 1, 2 and 35. Evaluate (integrate) either

Analytically GraphicallyNumerically Polymath

,...),( BAA CCfr

)(XfCi

Liquid Phase Batch

For the irrev, 2nd order reaction • Mole balance step• Rate law step • Stoichiometry step• Combine step• Evaluate step

00 Vrdt

dXN AA

BA

2AA kCr

)1(0 XCC AA

20 )1( XkC

dt

dXA

X

X

kCt

X

dX

kCdt

A

X

A

t

1

1

)1(

1

0

02

00

4.3 CSTRFor 1st order and irrev reaction• Mole balance step• Rate law step • Stoichiometry step• Combine step• Evaluate step• Damkohler number Da• Da gives the degree of conversion in flow

reactor

exitA

A

exitA

A

exitA

A

r

XCV

r

XC

r

XFV

)(

)()(

0

0

000

AA kCr

)1(0 XCC AA

k

CC

k

kX

X

X

k

AA

1

1

1

1

0

101.00

0

A

A

F

VrDa

4.3.2 CSTRs in Series• For equal size CSTRs τ1=τ2=τ operate at the same T

k1=k2=k and constant ν0

• For n equal size CSTRs τ1=τ2=…=τn=τ operate at the same T k1=k2=…=kn=k

20

1122

0

22

12

22

210

2

212

11

01

)1()1)(1(1

)(

1

k

C

kk

C

k

CC

Ck

CC

r

FFV

k

CC

AAAA

A

AA

A

AA

AA

n

AnA

An

kX

XCk

CC

)1(

11

)1()1( 0

0

4.3.3 CSTRs in Parallel• For identical individual reactor volume, Vi,

conversion, Xi, and reaction rate -rAi

• The conversion by each reactor is the same as if the total feed is charged to one large reactor of volume V

A

A

A

A

Ai

iiAi

r

XFV

r

XnF

r

XF

n

VV

0

00 )/(

4.3.4 2nd order reaction in a CSTR

• For 2nd order, liquid phase reaction in a CSTR

chosenissignXkC

kCkCX

XkC

XV

XCXFXCCA

kC

XF

r

XFV

A

AA

A

AAA

A

A

A

A

12

41)21(

)1(

)1(

0

00

200

00000

200

4.4 Tubular Reactors

• Consider 2nd order reaction in PFRFor liquid phase

For constant T and P gas phase0

0

0

0 0

022

0

0

0

0

1

1)1(

A

A

X

AA

AX

AA

kC

kCX

V

X

X

kCX

dX

kC

F

r

dXFV

X

XXX

kCV

dXX

X

kC

FV

X

XCC

A

X

A

AAA

1

)1()1ln()1(2

)1(

)1(

)1(

)1(

22

0

0

02

2

20

00

• Three reaction types A→nB– n<1, ε<0 (δ<0) → ν↓, the molecules will spend

longer time and ↑X than if v=v0– n>1, ε>0 (δ>0) ν ↑, the molecules will spend less

time and ↓ X than if v=v0– n=1, ε=0 (δ=0) v=v0

4.5 Pressure Drop in Reactors

• For liquid phase reactions the pressure drop can be ignored because the effect of pressure on the concs is small.

• For gas phase reactions the conc. of the reacting species is directly proportional to the total pressure

• Accounting for the pressure drop is a key factor in the proper reactor operation

4.5.1 Pressure drop and the rate law

• To account for pressure drop differential form design equation must be used

• For gas phase 2nd order reaction in PBR

0

2

0

2

0

0

0

00

2

0

1

1

1

1

TTP

P

X

XkC

dW

dX

T

T

P

P

X

XCC

kCr

F

r

dW

dX

A

AA

AA

A

A

4.5.2 Flow through a packed beds

• If y is defined as y=P/P0

• For a gas phase reactions in PBR of catalyst particles

• α is the bed characteristics

0

)1(2 T

TX

ydW

dy

),,,,,,( 0PdAGf pcb

4.5.4 Analytical solution

• For 2nd order isothermal reaction with ε=0 in PBR

dWWX

dX

kC

F

WXF

kC

dW

dX

Wy

yXCC

kCr

rdW

dXF

A

A

A

A

AA

AA

AA

)1()1(

])1[()1(

)1(

)1(

220

0

22/12

0

20

2/1

0

2

0

Integrating with X=0 @ W=0 andSolving for X and W gives

000 AA CF

2/100

0

0

0

0

0

0

)]}1/(]{/)2[(1{1

211

21

21

1

XXkCW

WWkC

WWkC

X

WW

X

X

kC

A

A

A

A