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

Lecture 8 – Tuesday 2/1/2011Block 1: Mole Balances on PFRs and PBR

Must Use the Differential Form of Mole Balance

Block 2: Rate LawsBlock 3: StoichiometryBlock 4: CombinePressure Drop: Liquid Phase Reactions:

Pressure Drop does not affect the concentrations in liquid phase rxn. 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 Isothermal P=f(W) Combine then Separate Variables (X,W) and IntegrateEngineering Analysis of Pressure Drop

2

Gas Phase Flow System:

Concentration Flow System:

0

00

0

00

0

11

1

1PP

TT

XXC

PP

TTX

XFFC AAAA

A

AFC

PP

TTX 0

00 1

0

00

0

00

0

11 PP

TT

X

XabC

PP

TTX

XabF

FCBABA

BB

3

Pressure Drop in Packed Bed Reactors

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

AA rdWdXF 0

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

4

CA FA

CA 0

1 X 1X

PP0

T0

TStoichiometry:

CA CA 01 X 1X

PP0

Isothermal, T=T0

2

02

2

0

20

11

PP

XX

FkC

dWdX

A

A

Combine:

Need to find (P/P0) as a function of W (or V if you have a PFR)5

Pressure Drop in Packed Bed Reactors

TURBULENT

LAMINAR

ppc

GDDg

GdzdP 75.111501

3

Ergun Equation:

6 0

00 )1(

TT

PPX

0

0

00 T

TPP

FF

T

T

00

00

0

mm Constant mass flow:

Pressure Drop in Packed Bed Reactors

00

03

0

75.111501

T

T

ppc FF

TT

PPG

DDgG

dzdP

T

T

FF

TT

PP 00

00 Variable

Density

G

DDgG

ppc

75.1115013

00

Let

7

Pressure Drop in Packed Bed Reactors

00

00

1 T

T

cc FF

TT

PP

AdWdP

ccbc zAzAW 1Catalyst Weight

b bulk density

c solid catalyst density

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

Where

0

0 112

PA cc

Let8

Pressure Drop in Packed Bed Reactors

We will use this form for single reactions:

XTT

PPdWPPd

112 00

0

002 T

T

FF

TT

ydWdy

0PPy

XTT

ydWdy

12 0

XydW

dy 1

2Isothermal case

9

Pressure Drop in Packed Bed Reactors

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

0

220

11 y

XFXkC

dWdX

A

A

PXfdWdX , PXf

dWdP , Xyf

dWdy ,and or

10

Pressure Drop in Packed Bed Reactors

PBR

11

2/1

2

2

)1(

)1(

10

)1(2

0

Wy

Wy

dWdy

yWWhen

XydW

dyFor

W

P

1

12

Pressure Drop in a PBR

No P

CA CA 0 1 X PP0

CA

W

P

13

Concentration Profile in a PBR2

No P

rA kCA2

-rA

P

W14

Reaction Rate in a PBR3

No P

X

W

P

15

Conversion in a PBR4

No P

W

P

1.0

16

PP

For

00

:0

Flow rate in a PBR5

17

0TT

0

00 1

TT

PPX

yXf

)1(10

PPy 0

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

(Analytical Solution) A + B 2C Repeat the previous one with equimolar feed of A and B and

kA = 1.5dm6/mol/kg/minα = 0.0099 kg-1

Find X at 100 kg

18

00 BA CC

0AC

0BC?X

1) Mole Balance:0

'

A

A

Fr

dWdX

2) Rate Law: BAA CkCr '

3) Stoichiometry: yXCC AA 10

yXCC AB 10

19

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

ydWdy

2

, dWydy 2

Wy 12

211 Wy

WXkCyXkCr AAA 111 220

2220

0

220 11

A

A

FWXkC

dWdX

20

0W 1y,

4) Combine:

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

dWW

FkC

XdX

A

A

11 0

20

2

21

2

0

20 WW

FkC

XX

A

A

XXWWXW , ,0 ,0

0.., 75.0

6.0

eidroppressurewithoutX

droppressurewithX

21

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

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

22

A + 2B C1) Mole Balance:

0A

A

Fr

dWdX

2) Rate Law: 2BAA CkCr

3) Stoichiometry: Gas, Isothermal

PPX 0

0 1

yX

XCC AA

11

0

23

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

4) y

XXCC B

AB

1

20

5) XydW

dy 1

2

6) yXf

1

0

7) 02.0 , 6 , 2 , 2 , 32

00 kFC AA

Initial values: W=0, X=0, y=1 W=100Combine with Polymath.If δ≠0, polymath must be used to solve.24

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

25

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

26

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

27

T = T0

Pressure Drop

Pressure Drop Engineering Analysis

29

30

Pressure Drop Engineering Analysis

31

Pressure Drop Engineering Analysis

Mole BalanceRate Laws

StoichiometryIsothermal Design

Heat Effects

32

End of Lecture 8

33

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

34

Example 1: Gas Phase Reaction in PBR for δ = 0A + B 2C

Case 1:

Case 2:

35

0AC

0BC?X

001

6

,0099.0,min

5.1 AB CCkgkgmoldmk

?,?,100 PXkgW

?,?,21,2 01021 PXPPDD PP

PBR

36

ParametersWy

CC

yFFC

CkCr

rdWdXF

BA

T

AA

BAA

AA

2/1

0

)1(

'