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Departm

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ngineering

Lecture 6

Kjemisk reaksjonsteknikk

Chemical Reaction Engineering

Review of previous lectures Pressure drop in fixed bed reactor PFR reactor design with pressure drop (ε=0)

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Departm

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These topics build upon one another

Mole

Bala

nce

Rate

Law

s

Sto

ich

iom

etr

y

Reaction Engineering

2

Isothermal reactor design

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Reactor Mole Balances in terms of conversion

Reactor Differential Algebraic Integral

A

0A

r

XFV

CSTR

A0A rdV

dXF

X

0 A0A r

dXFVPFR

Vrdt

dXN A0A

Vr

dXNt

X

0 A0A Batch

X

t

A0A rdW

dXF

X

0 A0A r

dXFWPBR

X

W3

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Gas Phase Flow System:

Concentration Flow System:

0

00A

0

00

0AAA P

P

T

T

X1

X1C

PP

TT

X1

X1FFC

A

A

FC

P

P

T

TX1 0

00

0

0B0A

0

00

B0AB

B P

P

T

T

X1

Xab

C

P

P

TT

X1

Xab

FF

C

4

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Note: Pressure drop does NOT affect liquid phase reactionsSample Question:

Analyze the following second order gas phase reaction that occurs isothermally in a PBR:

2AB

A0A rdW

dXF

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

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CA FA

CA 0

1 X 1X

P

P0

T0

TStoichiometry:

CA CA 0

1 X 1X

P

P0

Isothermal, T=T0

2

02

2

0A

20A

P

P

X1

X1

F

kC

dW

dX

Combine:

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

Pressure Drop in Packed Bed Reactors

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TURBULENT

LAMINAR

p3

pc

G75.1D

11501

Dg

G

dz

dPErgun Equation:

Pressure Drop in Packed Bed Reactors

7

P pressure, kPa

Φ porosity (volume of void/total bed volume)

1- Φ (volume of solid/total bed volume)

gc conversion factor. 1.0 for metric system

Dp diameter of particle in bed m

μ viscosity of gas passing through the bed kg/m.s

Z length down the packed bed m

u, superficial velocity m/s

ρ gas density kg/m3

G= ρu superficial mass velocity kg/m2,s

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TURBULENT

LAMINAR

p3

pc

G75.1D

11501

Dg

G

dz

dPErgun Equation:

Pressure Drop in Packed Bed Reactors

8 0

00 T

T

P

P)X1(

0

0

0T

T0 T

T

P

P

F

F

00

00

0mm Constant mass flow:

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0T

T

0

0

p3

pc0 F

F

T

T

P

PG75.1

D

11501

Dg

G

dz

dP

T

0T0

00 F

F

T

T

P

PVariable

Density

G75.1

D

11501

Dg

G

p3

pc00Let

Pressure Drop in Packed Bed Reactors

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0T

T

0

0

cc

0

F

F

T

T

P

P

1AdW

dP

ccbc 1zAzAW Catalyst Weight

0cc

0

P

1

1A

2

Let

Pressure Drop in Packed Bed Reactors

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b bulk density

c solid catalyst density

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

Where

Ac, cross section area

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We will use this form for single reactions:

X1

T

T

PP

1

2dW

PPd

00

0

0T

T

0 F

F

T

T

y2dW

dy

0P

Py

X1T

T

y2dW

dy

0

X1y2dW

dy

Isothermal

case

Pressure Drop in Packed Bed Reactors

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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.

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0A

220A y

X1F

X1kC

dW

dX

P,XfdW

dX P,Xf

dW

dP X,yf

dW

dyan

do

r

Pressure Drop in Packed Bed Reactors

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PBR

1) Mole Balance:0A

A

F

r

dW

dX

2) Rate Law:

2

2

20AA y

X1

X1kCr

AB

y

X1

X1C

P

P

X1

X1CC 0A

00AA

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PBR

14

2/1

2

2

)W1(y

)W1(y

dWdy

1y0WWhen

y2dW

dy

0For

X1T

T

y2dW

dy

0

Initial condition

0cc

0

P

1

1A

2

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W

21W1y

P1

15

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CA CA 0 1 X PP0

CA

2

W

P

No P

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No P

rA kCA2

-rA

3

P

W17

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No P

X4

W

P

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0

00 T

T

P

P)X1(

19

P

Py,TT 0

0

y)X1(

1f 0

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No P

5

W

P

1.0

20

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

Gas Phase Reaction in PBR with δ = 0 (Polymath Solution) A + B 2C

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

Find X at 100 kg

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A + B 2C

min kg mol

dm5.1k

6

1kg 0099.0

kg 100W ?X ?P

1PP D2D 0102 P2

1P Case 2:

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

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Case 1:

?X ?P

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1) Mole Balance:0A

A

F

'r

dW

dX

2) Rate Law: BAA CkC'r

3) yX1CC 0AA

4) yX1CC 0AB

0W 1y

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

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y2dW

dy 5) dWydy2

W1y2

21W1y

W1X1kCyX1kCr 220A

2220AA

0A

220A

F

W1X1kC

dW

dX

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

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dWW1

F

kC

X1

dX

0A

20A

2

2

WW

F

kC

X1

X 2

0A

20A

XX ,WW ,0X ,0W

0.e.i,droppressurewithout 75.0X

droppressurewith 6.0X

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

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26

Example A + B → 2C

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Example A + B → 2C