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Departm

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Lecture 5

Kjemisk reaksjonsteknikk

Chemical Reaction Engineering

Isothermal reaction design algorithm Algorithm to estimate the time or conversion for

different reactor Examples

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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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For Gas Phase Flow Systems

0

00

0

00

0

11 P

P

T

T

X

XC

PP

TT

X

XFFC jjAjjAjj

Cj=f(Fj, T, P) =f(x, T,P)

Da

dC

a

cB

a

bA

νj stochiomentric number, b/a, c/a, d/a

Isothermal T=T0

Liquid phase: ε=0

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Batch reactor

1) Mole balance

AA kCr

Vrdt

dXN A0A

1

1ln

1

xkt

2) Rate law

X

tFirst-order Second-order

22 AA Ckr

3) Stoichiometry: X1CC 0AA

4) Combine: Xkdt

dx 1 2

02 1 XCkdt

dxA

5. Evaluation)1(02 xCk

xt

A

)exp(1 1tkx 02

02

1 A

A

Ctk

Ctkx

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CSTR reactor

1) Mole balance

AA kCr 2) Rate law First-order Second-order

22 AA Ckr

3) Stoichiometry: XFF AA 10

4) Combine:

)1(0

0

xkC

xCV

A

A

5. Evaluation

k

kx

1

A

0A

r

XFV

X1CX1FF

C 0A0

0AAA

20 )1( xkC

xV

A

0

00

2

4121

A

AA

kC

kCkCx

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PFR reactor

1) Mole balance

AA kCr 2) Rate law First-order Second-order

22 AA Ckr

3) Stoichiometry: XFF AA 10

4) Combine:

5. Evaluation

xx

k

1

1ln

A0A rdV

dXF

X1

X1CC 0A

A

2

202

1

1

)/( X

XCk

Vd

dx A

X1

X1XX1ln12kC

22

0A

X

Xk

Vd

dx

1

1

)/(

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Examples:

Liquid Phase Laboratory Experiment(CH2CO)2O + H2O 2CH3COOH

A + B 2CEnteringVolumetric flow rate v0 = 0.0033 dm3/sAcetic Anhydride 7.8% (1M)Water 92.2% (51.2M)Elementary with k’ 1.95x10-4 dm3/(mol.s)

Case I CSTR V = 1dm3

Case II PFR V = 0.311 dm3

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Part 1: Mole Balances in Terms of Conversion

Algorithm for Isothermal Reactor Design

1. Mole Balance and Design Equation

2. Rate Law

3. Stoichiometry

4. Combine

5. Evaluate

A. Graphically (Chapter 2 plots)

B. Numerical (Quadrature Formulas Chapter 2 and appendices)

C. Analytical (Integral Tables in Appendix)

D. Software Packages (Appendix- Polymath)

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CSTR Laboratory Experiment

Example: CH3CO2 + H20 2CH3OOH

1) Mole Balance:           CSTR:A

0A

r

XFV

M2.51C 0B

M1C 0A

?X

V 1 dm3

0 3.310 3 dm3

s

12

A + B 2C

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1) Rate Law:BAAA CCkr

1) Stoichiometry:

B FA0ΘB -FA0X FB=FA0(ΘB-X)

A FA0 -FA0X FA=FA0(1-X)

C 0 2FA0X FC=2FA0X

CSTR Laboratory Experiment

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X1CX1FF

C 0A0

0AAA

XCXF

C B0A0

B0AB

2.511

2.51B

0B0A0AB C2.51CX2.51CC

CSTR Laboratory Experiment

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X1kCX1C

k

C'kr 0A0A0BA

k1

kX

75.003.4

03.3X

CSTR Laboratory Experiment

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V

0

kCA 0XC

A0 1 X V0

X1 X V

0

kX1 X k

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PFR Laboratory Experiment

X

0.311 dm3

1) Mole Balance:0A

A

F

r

dV

dX

2) Rate Law:BAA CkCr

3) Stoichiometry: X1CC 0AA

0BB CC 16

A + B 2C

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PFR Laboratory Experiment

4) Combine: X1kCX1CC'kr 0A0A0BA

kddVk

X1

dX

C

X1kC

dV

dX

0

00A

0A

kX1

1ln

ke1X

sec 94 1s 01.0k

61.0X 17

4) Combine: X1kCX1CC'kr 0A0A0BA

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Examples:Gas Phase : PFR and Batch Calculation 2NOCl 2NO + Cl2

2A 2B + C

Pure NOCl fed with CNOCl,0 = 0.2 mol/dm3 follows an

elementary rate law with k = 0.29 dm3/(mol.s)

Case I PFR with v0 = 10 dm3/sFind space time, with X = 0.9 Find reactor volume, V for X = 0.9

Case II Batch constant volumeFind the time, t, necessary to achieve 90% conversion.

Compare and t.

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Example (Gas Flow, PFR)

2 NOCl 2 NO + Cl2

s

dm 10

3

0 smol

dm29.0k

3

L

mol2.0C 0A

0TT

P P0

X 0.9

1) Mole Balance:0A

A

F

r

dV

dX

2) Rate Law: 2AA kCr

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2A 2B + C

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Example (Gas Flow, PFR)

3) Stoich: Gas X10

4) Combine: 2

220A

AX1

X1kCr

X1

X1CC 0A

A

A B + C/2

DakC

VkCdV

kCdX

X1

X1

X1C

X1kC

dV

dX

0A0

0AV

0 0

0AX

02

2

200A

220A

20

Damköhler number

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Example (Gas Flow, PFR)

X1

X1XX1ln12kC

22

0A

2

1

2

11y 0A

02.17kC 0A

sec 294kC

02.17

0A

L 2940VV 0 21

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Example Constant Volume (Batch)

1) Mole Balance:0A

A

00A

A

0A

0A

C

r

VN

r

N

Vr

dt

dX

2) Rate Law: 2AA kCr

X1CV

X1NC 0A

0

0AA

220AA X1kCr

3) Stoich: Gas 0VV 0

V

22

Gas Phase 2A 2B + C

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Example Constant Volume (Batch)

20A

0A

220A X1kCC

X1kC

dt

dX

4) Combine:

dtkC

X1

dX0A2

tkCX1

10A

sec 155t

20A X1kC

d

dX

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