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 lecture �  Block 1: Mole Balances on PFRs and PBR

� Must Use the Differential Form �  Block 2: Rate Laws �  Block 3: Stoichiometry

� Pressure 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 Isothermal P=f(W) Combine then Separate Variables (X,W) and Integrate 2

Reactor Mole Balances in terms of conversion

Reactor

Differential

Algebraic

Integral

A

0A

rXFV

−=CSTR

A0A rdVdXF −= ∫ −=

X

0 A0A rdXFVPFR

VrdtdXN A0A −=

VrdXNt

X

0 A0A ∫ −=Batch

X

t

A0A rdWdXF ′−= ∫ ′−

=X

0 A0A rdXFWPBR

X

W3

Gas Phase Flow System:

Concentration Flow System:

( )( )

( )( ) 0

00A

0

00

0AAA P

PTT

X1X1C

PP

TTX1

X1FFCε+−=

ε+υ

−=υ

=

υ= A

AFC

( )PP

TTX1 0

00 ε+υ=υ

( ) ( ) 0

0B0A

0

00

B0AB

B PP

TT

X1

XabC

PP

TTX1

XabF

FCε+

⎟⎠⎞⎜

⎝⎛ −Θ

=ε+υ

⎟⎠⎞⎜

⎝⎛ −Θ

=υ

=

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:

AàB

′−= 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 =FAυ

=CA01− X( )1+εX( )

PP0T0T

Stoichiometry:

CA = CA01− X( )1+εX( )

PP0

Isothermal, T=T0

( )( )

2

02

2

0A

20A

PP

X1X1

FkC

dWdX

⎟⎟⎠

⎞⎜⎜⎝

⎛ε+−=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

( )

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+µφ−⎟⎟⎠

⎞⎜⎜⎝

⎛φφ−

ρ−=

TURBULENT

LAMINAR

p3

pc

G75.1D11501

DgG

dzdPErgun Equation:

Pressure Drop in Packed Bed Reactors

7 0

00 T

TPP)X1( ε+υ=υ0

0

0T

T0 T

TPP

FFυ=υ

υυρ=ρ

υρ=ρυ=

00

00

0mm Constant mass flow:

( )0T

T

0

0

p3

pc0 FF

TT

PPG75.1

D11501

DgG

dzdP

⎥⎥⎦

⎤

⎢⎢⎣

⎡+µφ−

⎟⎟⎠

⎞⎜⎜⎝

⎛φφ−

ρ−=

T

0T0

00 F

FTT

PPρ=ρVariable Density

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡+µφ−

⎟⎟⎠

⎞⎜⎜⎝

⎛φφ−

ρ=β G75.1

D11501

DgG

p3

pc00Let

Pressure Drop in Packed Bed Reactors

8

( ) 0T

T

0

0

cc

0

FF

TT

PP

1AdWdP

ρφ−β−=

( ) ccbc 1zAzAW ρφ−=ρ=Catalyst Weight

ρb = bulk density

ρc = solid catalyst density

φ = porosity (a.k.a., void fraction)

Where

( ) 0cc

0

P1

1A2

ρφ−β=αLet

Pressure Drop in Packed Bed Reactors

9

We will use this form for single reactions: ( )

( ) ( )X1TT

PP1

2dWPPd

00

0 ε+α−=

0T

T

0 FF

TT

y2dWdy α−=

0PPy =

( )X1TT

y2dWdy

0

ε+α−=

( )X1y2dW

dy ε+α−= 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

ε+−=

( )P,XfdWdX = ( )P,Xf

dWdP = ( )X,yf

dWdy =and or

Pressure Drop in Packed Bed Reactors

11

PBR

1) Mole Balance: 0A

A

Fr

dWdX −=

2) Rate Law: ( )( )

22

20AA y

X1X1kCr ⎥

⎦

⎤⎢⎣

⎡ε+−=−

AàB

( )( )

( )( )yX1

X1CPP

X1X1CC 0A

00AA ε+

−=⎟⎟⎠

⎞⎜⎜⎝

⎛ε+−=

12

PBR

13

2/1

2

2

)W1(y

)W1(y

dWdy

1y0WWhen

y2dWdy

0For

α−=

α−=

α−=

==

α−=

=ε

W

( ) 21W1y α−=

P 1

14

CA = CA0 1− X( ) PP0

CA 2

W

ΔP

No ΔP

15

No ΔP

−rA = kCA2

-rA 3

ΔP

W 16

No ΔP

X 4

W

ΔP

17

0

00 T

TPP)X1( ε+υ=υ

18

PPy,TT 0

0 ==

y)X1(1f 0

ε+=

υυ=

No ΔP

5

W

ΔP

1.0

υ

19

Example 1: Gas Phase Reaction in PBR for δ = 0 Gas Phase Reaction in PBR with δ = 0 (Polymath Solution) A + B à 2C

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

Find X at 100 kg

20

A + B à 2C

min kg moldm5.1k

6

= 1kg 0099.0 −=α

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 α−=5) dWydy2 α−=

W1y2 α−=

( ) 21W1y α−=

( ) ( ) ( )W1X1kCyX1kCr 220A

2220AA α−−=−=−

( ) ( )0A

220A

FW1X1kC

dWdX α−−=

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

23

( ) ( )dWW1FkC

X1dX

0A

20A

2 α−=−

⎟⎟⎠

⎞⎜⎜⎝

⎛ α−=− 2

WWFkC

X1X 2

0A

20A

XX ,WW ,0X ,0W ====

( )( )0.e.i,droppressurewithout 75.0X

droppressurewith 6.0X

=α=

=

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 ≠ 0 Polymath Solution A + 2B à 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 Information kA = 6dm9/mol2/kg/min α = 0.02 kg-1

27

A + 2B à C

1) Mole Balance: 0A

A

F'r

dWdX −=

2) Rate Law: 2BAA CkC'r =−

3) Stoichiometry: Gas, Isothermal

( )PPX1VV 0

0 ε+=

( )( )yX1

X1CC 0AA ε+−=

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

28

4) CB = CA0!B ! 2X( )1+ !X( ) y

5) ( )X1y2dW

dy ε+α−=

6) f = !!0

=1+ "X( )y

7) 02.0 6k 2F 2C 32

0A0A =α===−=ε

Initial values: W=0, X=0, y=1 à 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

1ATT

PP

FF

dWdP

ρϕ−ρ

β−=

( ) cc0

00T

T0

1AyPTT

FF

dWdy

ρϕ−

β−= ( ) CC0

0

1AP2

ρϕ−β=α

00T

T

TT

FF

y2dWdy α−= Use for heat effects, multiple rxns

( )X1FF

0T

T ε+= Isothermal: T = T0 ( )X1y2dW

dX ε+α−=36

Mole Balance

Rate Laws

Stoichiometry

Isothermal Design

Heat Effects

37

Gas Phase Flow System:

Concentration Flow System:

( )( )

( )( ) 0

00A

0

00

0AAA P

PTT

X1X1C

PP

TTX1

X1FFCε+−=

ε+υ

−=υ

=

υ= A

AFC

( )PP

TTX1 0

00 ε+υ=υ

( ) ( ) 0

0B0A

0

00

B0AB

B PP

TT

X1

XabC

PP

TTX1

XabF

FCε+

⎟⎠⎞⎜

⎝⎛ −Θ

=ε+υ

⎟⎠⎞⎜

⎝⎛ −Θ

=υ

=

38

Gas Phase Reaction with Pressure Drop

A + 2B à C

1) Mole Balance:

dXdW

= ! " r AFA0

2) Rate Law:

! " r A = kCACB2

3) Stoichiometry: Gas, Isothermal

! = !0 1+ "X( ) P0P

CA = CA01!X( )1+ "X( )

y

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

39

End of Lecture 8

40