Reactor

Differential

Algebraic

Integral

A

A

r

XFV

0

CSTR

AA rdV

dXF 0

X

A

Ar

dXFV

0

0PFR

Vrdt

dXN AA 0

0

0

X

A

AVr

dXNtBatch

X

t

AA rdW

dXF 0

X

A

Ar

dXFW

0

0PBR

X

W 1

How to find

rA f X

rA g Ci Step 1: Rate Law

Ci h X Step 2: Stoichiometry

rA f X Step 3: Combine to get

2

These topics build upon one another

Mole Balance Rate Laws Stoichiometry

3

Species Symbol Reactor Feed Change Reactor Effluent

A A FA0 -FA0X FA=FA0(1-X)

B B FB0=FA0ΘB -b/aFA0X FB=FA0(ΘB-b/aX)

C C FC0=FA0ΘC +c/aFA0X FC=FA0(ΘC+c/aX)

D D FD0=FA0ΘD +d/aFA0X FD=FA0(ΘD+d/aX)

Inert I FI0=FA0ΘI ---------- FI=FA0ΘI

FT0 FT=FT0+δFA0X

0

0

0

0

00

00

0

0

A

i

A

i

A

i

A

ii

y

y

C

C

C

C

F

F

1

a

b

a

c

a

dWhere: and

A

A

FC Concentration – Flow System

4

A

A

FC Concentration Flow System:

0 Liquid Phase Flow System:

XC

XFFC A

AAA

1

10

0

0

X

a

bCX

a

b

V

N

V

NC BAB

ABB 0

0

0

Flow Liquid Phase

etc.

5

BAA CkCr If the rate of reaction were

X

a

bXCr BAA 1

2

0then we would have

XfrA This gives us

A

A

r

F

0

X6

We obtain:

0

0

0

0T

T

P

P

F

F

T

T

Combining the compressibility factor equation of

state with Z = Z0

000

000

00

TT

TT

T

T

CF

CF

TRZ

PC

ZRT

PC

Stoichiometry:

7

T

T

P

PF

T

T

P

P

F

F

FFC T

T

AAA

0

00

00

0

0

0

T

T

P

P

F

FCC

T

BTB

0

0

0

000 TT FC

8

XFFF ATT 00 The total molar flow rate is:

P

P

T

T

F

XFF

T

AT 0

00

000

P

P

T

TX

F

F

T

A 0

00

00 1

P

P

T

TXyA

0

0

00 1

P

P

T

TX 0

0

0 1

Substituting FT gives:

9

Gas Phase Flow System:

Concentration Flow System:

0

00

0

0

0

0

1

1

1

1

P

P

T

T

X

XC

P

P

T

TX

XFFC AAA

A

A

A

FC

P

P

T

TX 0

0

0 1

0

0

0

0

0

0

0

11

P

P

T

T

X

Xa

bC

P

P

T

TX

Xa

bF

FC

BABA

BB

10

2

0

0

2

011

1

T

T

P

P

X

Xa

b

X

XCkr

B

AAA

If –rA=kCACB

This gives us

FA0/-rA

X 11

12

Consider the following elementary reaction with KC=20 dm3/mol

and CA0=0.2 mol/dm3.

Calculate Equilibrium Conversion or both a batch reactor (Xeb)

and a flow reactor (Xef).

Xef

C

BAAA

K

CCkr

2

BA2

13

2AB

B2

1A

Xef ?

Rate law:

C

BAAA

K

CCkr 2

Xeb 0.703

14

Solution:

Species Fed Change Remaining

A FA0 -FA0X FA=FA0(1-X)

B 0 +FA0X/2 FB=FA0X/2

FT0=FA0 FT=FA0-FA0X/2

15

A FA0 -FA0X FA=FA0(1-X)

B 0 FA0X/2 FB=FA0X/2

Stoichiometry: Gas isothermal T=T0, isobaric P=P0

V V0 1X

CA FA0 1 X V0 1X

CA0 1 X 1X

CB FA 0 X 2

V0 1X CA 0 1 X 2 1X

16

Pure A yA0=1, CA0=yA0P0/RT0, CA0=P0/RT0

C

AAAA

KX

XC

X

XCkr

121

1 0

2

0

201

12

e

eeAC

X

XXCK

2

11

2

110

Ay

@ eq: -rA=0

17

82.020223

3

0

dm

mol

mol

dmCK AC

yA0 11

21

1

2

8 Xe 0.5Xe

2

1 2Xe Xe2

8.5Xe217Xe 8 0

Xef 0.757Flow: Recall

Xeb 0.70Batch: 18

19

20

21

22