lecture 8

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 2/5/2013

2

Block 1: Mole BalancesBlock 2: Rate LawsBlock 3: StoichiometryBlock 4: Combine

Pressure DropLiquid Phase ReactionsGas Phase Reactions

Engineering Analysis of Pressure Drop

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

Pressure Drop in PBRs

3

Pressure Drop in PBRs

4

Note: Pressure Drop does NOT affect liquid phase reactions

Sample Question:Analyze the following second order gas phase reaction

that occurs isothermally in a PBR:AB

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

Rate LawsSecond order in A and irreversible:

AA rdWdXF 0

2AA kCr

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 PBRs

TURBULENT

LAMINAR

ppc

GDDg

GdzdP 75.111501

3

Ergun Equation:

60

00 )1(

TT

PPX

0

0

00 T

TPP

FF

T

T

00

00

0

mm Constant mass flow:

Pressure Drop in PBRs

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 PBRs

00

00

1 T

T

cc FF

TT

PP

AdWdP

ccbc zAzAW 1Catalyst Weight

0

0 112

PA cc

Let

8

Pressure Drop in PBRs

b bulk density

c solid catalyst density

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

Where

(1 ) solid fraction

We will use this form for single reactions:

X

TT

PPdWPPd

112 00

0

002 T

T

FF

TT

ydWdy

0PPy

XTT

ydWdy

12 0

XydW

dy 1

2 Isothermal case9

Pressure Drop in PBRs

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 Balances, Rate Laws and Stoichiometry.

22

0

220

11 y

XFXkC

dWdX

A

A

PXfdWdX , PXf

dWdP , Xyf

dWdy ,and or

10

Pressure Drop in PBRs

11

2/1

2

2

)1(

)1(

10

)1(2

0

Wy

Wy

dWdy

yWWhen

XydW

dyFor

Packed Bed Reactors

W

P

1

12

Pressure Drop in a PBR

13

Concentration Profile in a PBR2

No P

CA CA 0 1 X PP0

CA

W

P

14

Reaction Rate in a PBR3

No P

2

2 2

0

(1 )A APr kC k XP

-rA

P

W

15

Conversion in a PBR4

No P

X

W

P

16

Flow Rate in a PBR5

No P

W

P

1

PP

For

00

:0

0

f

17

0TT

0

00 1

TT

PPX

0 1(1 )

fX y

PPy 0

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

18

Gas 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

00 BA CC

0AC

0BC

?X

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

19

1) Mole Balance0

'

A

A

Fr

dWdX

2) Rate Law BAA CkCr '

3) Stoichiometry yXCC AA 10

yXCC AB 10

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

20

ydWdy

2

, dWydy 2

Wy 12

211 Wy

WXkCyXkCr AAA 111 220

2220

0

220 11

A

A

FWXkC

dWdX

0W 1y,

4) Combine

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

21

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

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

22

The reaction A + 2B C

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

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

Additional InformationkA = 6 dm9/mol2/kg/minα = 0.02 kg-1

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

23

A + 2B C

1) Mole Balance0A

A

Fr

dWdX

2) Rate Law 2BAA CkCr

3) Stoichiometry: Gas, Isothermal

PPX 0

0 1

yX

XCC AA

11

0

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

24

4) y

XXCC B

AB

1

20

5) XydW

dy 1

2

6) yXf

1

0

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

7)

02.0,6,2,232]2[

31]211[

00

0

kFC

y

AA

A

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

25

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

26

27

T = T0

Robert the Worrier wonders: What if we increase the catalyst size by a factor of 2?

Pressure DropEngineering Analysis

29

𝜌0=𝑀𝑊∗𝐶𝑇 0=𝑀𝑊∗𝑃0

𝑅𝑇0

𝛼=2𝑅𝑇0

𝐴𝐶𝜌𝐶𝑔𝐶 𝑃02 𝐷𝑃 𝜙3𝑀𝑊

𝐺[150 (1−𝜙 )𝜇𝐷𝑃

+1.75𝐺]𝛼≈ ( 1

𝑃0 )2

Pressure DropEngineering Analysis

30

Pressure DropEngineering Analysis

31

Mole Balance

Rate Laws

Stoichiometry

Isothermal Design

Heat Effects

32

End of Lecture 8

33

Pressure Drop - Summary

34

Pressure DropLiquid Phase Reactions

Pressure Drop does not affect concentrations in lquid phase reactions.

Gas Phase ReactionsEpsilon does not equal to zero

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

Epsilon = 0 and isothermalP=f(W)Combine then separate variables (X,W) and integrate

Engineering Analysis of Pressure Drop

Pressure Change – Molar Flow Rate

35

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

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

36

A + B 2C

Case 1:

Case 2:

0AC

0BC??

XP

001

6

,0099.0,min

5.1 AB CCkgkgmoldmk

?,?,100 PXkgW

?,?,21,2 01021 PXPPDD PP

PBR

37

0

1/20

'

0 and (1 )

A A

A A B

AA

T

A B

dXF rdW

r kC CFC yF

C C

T T y W