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

rXFV

CSTR

A0A rdVdXF

X

0 A0A r

dXFVPFR

VrdtdXN A0A

VrdXNt

X

0 A0A Batch

X

t

A0A rdWdXF

X

0 A0A r

dXFWPBRX

W3

Gas Phase Flow System:

Concentration Flow System:

0

00A

0

00

0AAA P

PTT

X1X1C

PP

TTX1

X1FFC

A

AFC

PP

TTX1 0

00

0

0B0A

0

00

B0AB

B PP

TT

X1

XabC

PP

TTX1

XabF

FC

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 rdWdXF

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

PP0

T0

TStoichiometry:

CA CA 01 X 1X

PP0

Isothermal, T=T0

2

02

2

0A

20A

PP

X1X1

FkC

dWdX

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

G75.1D11501

DgG

dzdPErgun Equation:

Pressure Drop in Packed Bed Reactors

7 0

00 T

TPP)X1(

0

0

0T

T0 T

TPP

FF

00

00

0mm Constant mass flow:

0T

T

0

0

p3

pc0 FF

TT

PPG75.1

D11501

DgG

dzdP

T

0T0

00 F

FTT

PP

Variable Density

G75.1

D11501

DgG

p3

pc00Let

Pressure Drop in Packed Bed Reactors

8

0T

T

0

0

cc

0

FF

TT

PP

1AdWdP

ccbc 1zAzAW Catalyst Weight

b bulk density

c solid catalyst density

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

Where

0cc

0

P1

1A2

Let

Pressure Drop in Packed Bed Reactors

9

We will use this form for single reactions:

X1TT

PP1

2dWPPd

00

0

0T

T

0 FF

TT

y2dWdy

0P

Py

X1TT

y2dWdy

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

X1FX1kC

dWdX

P,XfdWdX

P,XfdWdP

X,yfdWdy

and or

Pressure Drop in Packed Bed Reactors

11

PBR

1) Mole Balance:0A

A

Fr

dWdX

2) Rate Law:

22

20AA y

X1X1kCr

AB

y

X1X1C

PP

X1X1CC 0A

00AA

12

PBR

13

2/1

2

2

)W1(y

)W1(y

dWdy

1y0WWhen

y2dWdy

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

TPP)X1(

18

PPy,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 moldm5.1k

6

1kg 0099.0

kg 100W ?X ?P

1PP D2D 0102 P21P Case 2:

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

21

Case 1:

?X ?P

1) Mole Balance:0A

A

F'r

dWdX

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

y2dWdy

5) dWydy2

W1y2

21W1y

W1X1kCyX1kCr 220A

2220AA

0A

220A

FW1X1kC

dWdX

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

23

dWW1

FkC

X1dX

0A

20A

2

2WW

FkC

X1X 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-1

27

A + 2B C1) Mole Balance:

0A

A

F'r

dWdX

2) Rate Law: 2BAA CkC'r

3) Stoichiometry: Gas, Isothermal

PPX1VV 0

0

y

X1X1CC 0AA

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

28

4) y

X1X22CC 0AB

5) X1y2dW

dy

6) y

X1f 0

7) 02.0 6k 2F 2C 32

0A0A

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

1ATT

PP

FF

dWdP

cc0

00T

T0

1AyPTT

FF

dWdy

CC0

0

1AP2

00T

T

TT

FF

y2dWdy

Use for heat effects, multiple rxns

X1FF

0T

T Isothermal: T = T0 X1y2dW

dX

36

Mole BalanceRate Laws

StoichiometryIsothermal Design

Heat Effects

37

End of Lecture 8

38