Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of

chemical reactions and the design of the reactors in which they take place.

Lecture 8

Today’s lectureBlock 1: Mole Balances on PFRs and PBRMust Use the Differential Form

Block 2: Rate LawsBlock 3: StoichiometryPressure Drop:Liquid Phase Reactions: Pressure Drop does not affect the concentrations in liquid phase rx.Gas Phase Reactions:Epsilon not Equal to Zero

d(P)/d(W)=..Polymath will combine with d(X)/f(W)=..for you

Epsilon = 0 and IsothermalP=f(W)Combine then Separate Variables (X,W) and Integrate

2

Reactor Mole Balances in terms of conversionReactor Differential Algebraic Integral

A

0A

r

XFV

CSTR

A0A rdV

dXF

X

0 A0A r

dXFVPFR

Vrdt

dXN A0A

Vr

dXNt

X

0 A0A Batch

X

t

A0A rdW

dXF

X

0 A0A r

dXFWPBR

X

W3

Gas Phase Flow System:

Concentration Flow System:

0

00A

0

00

0AAA P

P

T

T

X1

X1C

PP

TT

X1

X1FFC

A

A

FC

P

P

T

TX1 0

00

0

0B0A

0

00

B0AB

B P

P

T

T

X1

Xab

C

PP

TT

X1

Xab

FF

C

4

Note: Pressure drop does NOT affect liquid phase reactionsSample Question:

Analyze the following second order gas phase reaction that occurs isothermally in a PBR:

AB

A0A rdW

dXF

Mole Balance:Must use the differential form of the mole balance to separate variables:

2AA kCr Second order in A and

irreversible:

Rate Law:

Pressure Drop in Packed Bed Reactors

5

CA FACA 0

1 X 1X

P

P0

T0

TStoichiometry:

CA CA 0

1 X 1X

P

P0

Isothermal, T=T0

2

02

2

0A

20A

P

P

X1

X1

F

kC

dW

dX

Combine:

Need to find (P/P0) as a function of W (or V if you have a PFR)

Pressure Drop in Packed Bed Reactors

6

TURBULENT

LAMINAR

p3

pc

G75.1D

11501

Dg

G

dz

dPErgun Equation:

Pressure Drop in Packed Bed Reactors

7 0

00 T

T

P

P)X1(

0

0

0T

T0 T

T

P

P

F

F

00

00

0mm Constant mass flow:

0T

T

0

0

p3

pc0 F

F

T

T

P

PG75.1

D

11501

Dg

G

dz

dP

T

0T0

00 F

F

T

T

P

PVariable

Density

G75.1

D

11501

Dg

G

p3

pc00Let

Pressure Drop in Packed Bed Reactors

8

0T

T

0

0

cc

0

F

F

T

T

P

P

1AdW

dP

ccbc 1zAzAW Catalyst Weight

b bulk density

c solid catalyst density

porosity (a.k.a., void fraction)

Where

0cc

0

P

1

1A

2

Let

Pressure Drop in Packed Bed Reactors

9

We will use this form for single reactions:

X1

T

T

PP

1

2dW

PPd

00

0

0T

T

0 F

F

T

T

y2dW

dy

0P

Py

X1T

T

y2dW

dy

0

X1y2dW

dy

Isothermal case

Pressure Drop in Packed Bed Reactors

10

The two expressions are coupled ordinary differential equations. We can only solve them simultaneously using an ODE solver such as Polymath. For the special case of isothermal operation and epsilon = 0, we can obtain an analytical solution.

Polymath will combine the mole balance, rate law and stoichiometry.

22

0A

220A y

X1F

X1kC

dW

dX

P,XfdW

dX P,Xf

dW

dP X,yf

dW

dyand or

Pressure Drop in Packed Bed Reactors

11

PBR

1) Mole Balance:0A

A

F

r

dW

dX

2) Rate Law:

2

2

20AA y

X1

X1kCr

AB

y

X1

X1C

P

P

X1

X1CC 0A

00AA

12

PBR

13

2/1

2

2

)W1(y

)W1(y

dWdy

1y0WWhen

y2dW

dy

0For

W

21W1y

P1

14

CA CA 0 1 X PP0

CA

2

W

P

No P

15

No P

rA kCA2

-rA

3

P

W16

No P

X4

W

P

17

0

00 T

T

P

P)X1(

18

P

Py,TT 0

0

y)X1(

1f 0

No P

5

W

P

1.0

19

Example 1: Gas Phase Reaction in PBR for δ = 0Gas Phase Reaction in PBR with δ = 0

(Polymath Solution) A + B 2C

Repeat the previous one with equil molar feed of A and B and kA = 1.5dm9/mol2/kg/minα = 0.0099 kg-1

Find X at 100 kg

20

A + B 2C

min kg mol

dm5.1k

6

1kg 0099.0

kg 100W ?X ?P

1PP D2D 0102 P2

1P Case 2:

Example 1: Gas Phase Reaction in PBR for δ = 0

21

Case 1:

?X ?P

1) Mole Balance:0A

A

F

'r

dW

dX

2) Rate Law: BAA CkC'r

3) yX1CC 0AA

4) yX1CC 0AB

0W 1y

Example 1: Gas Phase Reaction in PBR for δ = 0

22

y2dW

dy 5) dWydy2

W1y2

21W1y

W1X1kCyX1kCr 220A

2220AA

0A

220A

F

W1X1kC

dW

dX

Example 1: Gas Phase Reaction in PBR for δ = 0

23

dWW1

F

kC

X1

dX

0A

20A

2

2

WW

F

kC

X1

X 2

0A

20A

XX ,WW ,0X ,0W

0.e.i,droppressurewithout 75.0X

droppressurewith 6.0X

Example 1: Gas Phase Reaction in PBR for δ = 0

24

25

Example A + B → 2C

26

Example A + B → 2C

Example 2: Gas Phase Reaction in PBR for δ ≠ 0Polymath Solution A + 2B C

is carried out in a packed bed reactor in which there is pressure drop.The fed is stoichiometric in A and B.

Plot the conversion and pressure ratio y = P/P0 as a function of catalyst weight upto 100 kg.

Additional InformationkA = 6dm9/mol2/kg/minα = 0.02 kg-127

A + 2B C

1) Mole Balance:0A

A

F

'r

dW

dX

2) Rate Law: 2BAA CkC'r

3) Stoichiometry: Gas, Isothermal

P

PX1VV 0

0

y

X1

X1CC 0AA

Example 2: Gas Phase Reaction in PBR for δ ≠ 0

28

4)

5) X1y2dW

dy

6)

7) 02.0 6k 2F 2C 3

20A0A

Initial values: W=0, X=0, y=1 W=100

Combine with Polymath.

If δ≠0, polymath must be used to solve.

Example 2: Gas Phase Reaction in PBR for δ ≠ 0

29

30

Example 2: Gas Phase Reaction in PBR for δ ≠ 0

31

Example 2: Gas Phase Reaction in PBR for δ ≠ 0

32

T = T0

Engineering Analysis

33

Engineering Analysis

34

Engineering Analysis

35

Pressure Change – Molar Flow Rate

cc

0

0

0T

T0

1A

TT

P

P

FF

dW

dP

cc0

00T

T0

1AyP

TT

FF

dW

dy

CC0

0

1AP

2

00T

T

T

T

F

F

y2dW

dy Use for heat effects,

multiple rxns

X1F

F

0T

T Isothermal: T = T0 X1y2dW

dX

36

Mole Balance

Rate Laws

Stoichiometry

Isothermal Design

Heat Effects

37

Gas Phase Flow System:

Concentration Flow System:

0

00A

0

00

0AAA P

P

T

T

X1

X1C

PP

TT

X1

X1FFC

A

A

FC

P

P

T

TX1 0

00

0

0B0A

0

00

B0AB

B P

P

T

T

X1

Xab

C

PP

TT

X1

Xab

FF

C

38

Gas Phase Reaction with Pressure Drop

A + 2B C

1) Mole Balance:

2) Rate Law:

3) Stoichiometry: Gas, Isothermal

Example 2: Gas Phase Reaction in PBR for δ ≠ 0

39

End of Lecture 8

40