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 Balances �  Block 2: Rate Laws �  Block 3: Stoichiometry �  Block 4: Combine

�  Pressure Drop � Liquid Phase Reactions � Gas 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 εευυ +

−=

+

−==

υA

AFC =

( )PP

TTX 0

00 1 ευυ +=

( ) ( ) 0

00

0

00

0

11 PP

TT

X

XabC

PP

TTX

XabF

FCBABA

BB εευυ +

⎟⎠

⎞⎜⎝

⎛ −Θ=

+

⎟⎠

⎞⎜⎝

⎛ −Θ==

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: AàB

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

Rate Laws Second order in A and irreversible:

ʹ′−= AA rdWdXF 0

2AA kCr =ʹ′−

€

CA =FAυ

=CA01− X( )1+εX( )

PP0T0T

Stoichiometry

€

CA =CA01− X( )1+εX( )

PP0

Isothermal, T=T0

( )( )

2

02

2

0

20

11

⎟⎟⎠

⎞⎜⎜⎝

⎛

+

−=

PP

XX

FkC

dWdX

A

A

εCombine:

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

µφφφ

ρErgun Equation:

6 0

00 )1(

TT

PPXευυ +=

0

0

00 T

TPP

FF

T

Tυυ =

υυ

ρρ

υρρυ

00

00

0

=

=

= mm !!Constant mass flow:

Pressure Drop in PBRs

( )00

03

0

75.111501

T

T

ppc FF

TT

PPG

DDgG

dzdP

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

−⎟⎟⎠

⎞⎜⎜⎝

⎛ −−=

µφφφ

ρ

T

T

FF

TT

PP 00

00ρρ =Variable Density

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡+

−⎟⎟⎠

⎞⎜⎜⎝

⎛ −= G

DDgG

ppc

75.1115013

00

µφφφ

ρβLet

7

Pressure Drop in PBRs

( ) 00

00

1 T

T

cc FF

TT

PP

AdWdP

ρφβ−

−=

( ) ccbc zAzAW ρφρ −== 1Catalyst Weight

( ) 0

0 112

PA cc ρφβ

α−

=Let 8

Pressure Drop in PBRs

€

ρb = bulk density

€

ρc = solid catalyst density

€

φ = porosity (a.k.a., void fraction)

Where

(1 ) solid fractionφ− =

We will use this form for single reactions:

( )( )

( )XTT

PPdWPPd

εα

+−= 112 00

0

dpdW

= −α2p

TT0

FTFT 0

p = PP0

dpdW

= −α2p

TT01+εX( )

dpdW

= −α2p

1+εX( ) Isothermal case

9

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.

dXdW

=kCA0

2 1− X( )2

FA0 1+εX( )2p2

( )PXfdWdX ,= ( )PXf

dWdP ,=

dpdW

= f p,X( )and or

10

Pressure Drop in PBRs

11

For ε = 0dpdW

=−α2p(1+εX)

When W = 0 p =1dp2 = −α dW

p2 = (1−αW )p = (1−αW )1/2

Packed Bed Reactors

W

P

1

12

Pressure Drop in a PBR

13

Concentration Profile in a PBR2

€

No ΔP

€

CA =CA0 1− X( ) PP0

CA

W

€

ΔP

14

Reaction Rate in a PBR3

€

No ΔP

22 2

0

(1 )A APr kC k XP

⎛ ⎞− = = − ⎜ ⎟

⎝ ⎠

-rA

€

ΔP

W

15

Conversion in a PBR4

€

No ΔP

X

W

€

ΔP

16

Flow Rate in a PBR5

€

No ΔP

W

€

ΔP

1

⎟⎠

⎞⎜⎝

⎛=

=

PP

For

00

:0

υυ

ε

0υυ

=f

17

0TT =

( )0

00 1 T

TPPXευυ +=

f = υ0υ=

1(1+εX )p

p = P0P

Example 1: Gas Phase Reaction in PBR for =0

18

Gas Phase reaction in PBR with δ = 0 (Analytical Solution)

A + B à 2C Repeat the previous one with equimolar feed of A and B and: kA = 1.5dm6/mol/kg/min α = 0.0099 kg-1

Find X at 100 kg

00 BA CC =

0AC

0BC?=X

Example 1: Gas Phase Reaction in PBR for =0

19

1) Mole Balance 0

'

A

A

Fr

dWdX −

=

2) Rate Law BAA CkCr =− '

3) Stoichiometry CA =CA0 1− X( ) p

CB =CA0 1− X( ) p

Example 1: Gas Phase Reaction in PBR for =0

20

dpdW

= −α2p

2pdp = −αdW

p2 =1−αW

p = 1−αW( )1 2

−rA = kCA02 1− X( )2 p2 = kCA0

2 1− X( )2 1−αW( )

( ) ( )0

220 11

A

A

FWXkC

dWdX α−−

=

0=W p =1,

4) Combine

Example 1: Gas Phase Reaction in PBR for =0

21

( )( )dWW

FkC

XdX

A

A α−=−

11 0

20

2

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

− 21

2

0

20 WW

FkC

XX

A

A α

XXWWXW ==== , ,0 ,0

( )( )0.., 75.0

6.0==

=

αeidroppressurewithoutXdroppressurewithX

Example 2: Gas Phase Reaction in PBR for ≠0

22

The reaction A + 2B à 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 Information kA = 6 dm9/mol2/kg/min α = 0.02 kg-1

Example 2: Gas Phase Reaction in PBR for ≠0

23

A + 2B à C

1) Mole Balance 0A

A

Fr

dWdX ʹ′−

=

2) Rate Law 2BAA CkCr =ʹ′−

3) Stoichiometry: Gas, Isothermal

( )PPX 0

0 1 ευυ +=

CA =CA01− X( )1+εX( )

p

Example 2: Gas Phase Reaction in PBR for ≠0

24

4) CB =CA0ΘB − 2X( )1+εX( )

p

5) dpdW

= −α2p

1+εX( )

6) f = υυ0

=1+εX( )p

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

7)

02.0,6,2,232]2[

31]211[

00

0

====

−=−=−−=

α

ε

kFC

y

AA

A

Example 2: Gas Phase Reaction in PBR for ≠0

25

Example 2: Gas Phase Reaction in PBR for ≠0

26

27

T = T0

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

Pressure Drop Engineering Analysis

29

𝜌↓0 =𝑀𝑊∗𝐶↓𝑇0 = 𝑀𝑊∗  𝑃↓0 /𝑅𝑇↓0  

𝛼= 2𝑅𝑇↓0 /𝐴↓𝐶 𝜌↓𝐶 𝑔↓𝐶 𝑃↓0↑2 𝐷↓𝑃 𝜙↑3 𝑀𝑊 𝐺[150(1−𝜙)𝜇/𝐷↓𝑃  +1.75𝐺]

𝛼≈(1/𝑃↓0  )↑2 

Pressure Drop Engineering Analysis

30

Pressure Drop Engineering Analysis

31

Mole Balance

Rate Laws

Stoichiometry

Isothermal Design

Heat Effects

32

End of Lecture 8

33

Pressure Drop - Summary

34

�  Pressure Drop � Liquid Phase Reactions

�  Pressure Drop does not affect concentrations in liquid phase reactions.

� Gas Phase Reactions �  Epsilon does not equal to zero

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

�  Epsilon = 0 and isothermal P=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

ρϕ−ρ

β−=

dpdW

= −β0

FTFT 0

TT0

pP0Ac 1−φ( )ρc ( ) CC0

0

1AP2

ρϕ−

β=α

dydW

= −α2p

FTFT 0

TT0

Use for heat effects, multiple rxns

( )X1FF

0T

T ε+= Isothermal: T = T0 dXdW

= −α2p

1+εX( )

Example 1: Gas Phase Reaction in PBR for =0

36

A + B à 2C

Case 1:

Case 2: 0AC

0BC??

XP=

=

001

6

,0099.0,min

5.1 AB CCkgkgmoldmk ==

⋅⋅= −α

?,?,100 === PXkgW

?,?,21,2 01021 ==== PXPPDD PP

PBR

37

FA0dXdW

= −r 'A

rA = −kCACB

CA =FAFTp

CA =CBδ = 0 and T =T0 ∴ p = (1−αW )1/2