pressure drop

Chemical Reaction Engineering (CRE) is thefield that studies the rates and mechanisms of

chemical reactions and the design of the reactors inwhich they take place.

Lecture 5b

Lecture 5b

2

Block 1: Mole Balances Block 2: Rate Laws Block 3: Stoichiometry Block 4: Combine

Pressure Drop

Concentration Flow System:

( )( )

( )( ) 0

00

0

00

0

1

1

1

1

P

P

T

T

X

XC

P

P

T

TX

XFFC AAA

A +−=

+

−==

A

A

FC =

( )P

P

T

TX 0

00 1 +=

Pressure Drop in PBRs

3


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:

Mole BalancesMust use the differential form of the mole balance toseparate variables:

Rate LawsSecond order in A and irreversible:

€

CA =FA

=CA 01− X( )1+X( )

PP0T0T

Stoichiometry

€

CA = CA 01− X( )1+X( )

PP0

Isothermal, T=T0

5


( )

+−

−−= TURBULENT

LAMINAR

ppc

GDDg

G

dz

dP75.1

115013

Ergun Equation:

60

00 )1(

T

T

P

PX +=

Constant mass flow:


T

T

F

F

T

T

P

P 00

00 =Variable Density

( )

+−

−= GDDg

G

ppc

75.111501

30

0

Let

7


( ) 00

00

1 T

T

cc F

F

T

T

P

P

AdW

dP

−−=

( ) ccbc zAzAW −== 1Catalyst Weight

( ) 0

0 1

1

2

PA cc −

=Let8


We will use this form for single reactions:

( )( ) ( )X

T

T

PPdW

PPd +−= 11

2 00

0

002 T

T

F

F

T

T

ydW

dy −=0P

Py =

Isothermal case

9


The two expressions are coupled ordinary differentialequations. We can only solve them simultaneously usingan ODE solver such as Polymath. For the special case ofisothermal operation and epsilon = 0, we can obtain ananalytical solution.

Polymath will combine the Mole Balances, Rate Lawsand Stoichiometry.10


11

Packed Bed Reactors

Example 4-4 Calculating Pressure Dropin a Packed Bed

12

Calculate the pressure drop in a 60 ft length of 1 1/2-in.schedule 40 pipe packed with catalyst pellets 114-h. indiameter when 104.4 lb/h of gas is passing through the bed.The temperature is constant along the length of pipe at260°C. The void fraction is 45% and the properties of thegas are similar to those of air at this temperature. Theentering pressure is 10 atm.

W

P

1

14

Pressure Drop in a PBR

15

Concentration Profile in a PBR2

€

No ∆P

16

Reaction Rate in a PBR3

€

No ∆P

2

2 2

0

(1 )A A

Pr kC k X

P

− = = −

-rA

€

∆P

W

17

Conversion in a PBR4

€

No ∆P

X

W

€

∆P

18

Flow Rate in a PBR5

€

No∆P

W

€

∆P

10=f

19

0TT =

0 1

(1 )f

X y

= =+

P

Py 0=

Example 4.5: Effect of Pressure Dropon the Conversion

20

22

1. Mass transfer (diffusion) of the reactant(s) (e.g., species A) from the bulk fluid to theexternal surface of the catalyst pellet2. Diffusion of the reactant from the pore mouth through the catalyst pores to theimmediate vicinity of the internal catalytic surface3. Adsorption of reactant A onto the catalyst surface4. Reaction on the surface of the catalyst (e.g., A B)5.Desorption of the products (e.g., B) from the surface6. Diffusion of the products from the interior of the pellet to the pore mouth at theexternal surface.7. Mass transfer of the products from the external pellet surface to the bulk fluid

Finding the optimum particle diameter

23

Example 4-6: Calculating X in a Reactorwith Pressure Drop

24

Calculate the catalyst weight necessary to achieve 60% conversion when ethyleneoxide is to be made by the vapor-phase catalytic oxidation of ethylene with air.

Ethylene and oxygen are fed in stoichiometric proportions to a packed-bed reactoroperated isothermally at 260°C. Ethylene is fed at a rate of 0.30 Ib mol/s at apressure of 10 atm. It is proposed to use 10 banks of 1 f -in.-diameter schedule 40tubes packed with catalyst with 100 tubes per bank. Consequently, the molar flowrate to each tube is to be 3 X Ib mol/s. The properties of the reacting fluid are tobe considered identical to those of air at this temperature and pressure. Thedensity of the a -in.-catalyst particles is 120 lb/ft3 and the bed void fraction is0.45. The rate law is:

Example 1:Gas Phase Reaction in PBR for =0

27

Gas Phase reaction in PBR with δ = 0 (AnalyticalSolution)

Repeat the previous one with equimolar feed of Aand B and:

kA = 1.5dm6/mol/kg/minα = 0.0099 kg-1

Find X at 100 kg

00 BA CC =

0AC

0BC

?=X


28

1) Mole Balance0

'

A

A

F

r

dW

dX −=

2) Rate Law BAA CkCr =− '

( )yXCC AB −= 10


29

ydW

dy

2

−= , dWydy −=2

Wy −= 12

( ) ( )0

220 11

A

A

F

WXkC

dW

dX −−=

0=W 1=y,

4) Combine


30

( ) ( )dWWF

kC

X

dX

A

A −=−

11 0

20

2

( )( )0..,75.0

6.0

===

eidroppressurewithoutX

droppressurewithX

Example 2:Gas Phase Reaction in PBR for ≠0

31

The reaction

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

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

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


32

A + 2B C

1) Mole Balance0A

A

F

r

dW

dX ′−=

3) Stoichiometry: Gas, Isothermal

( )P

PX 0

0 1 +=


33

4)

5) ( )XydW

dy +−= 12

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

7)

02.0,6,2,23

2]2[

3

1]211[

00

0

====

−=−=−−=

kFC

y

AA

A


34


35

36

T = T0

Pressure DropEngineering Analysis

38

= ∗ = ∗= 2 150 1 − + 1.75≈ 1


39


40

Mole Balance

Rate Laws

Stoichiometry

Isothermal Design

Heat Effects

41

End of Lecture 5b

42

pressure drop

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