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 6

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

W

Previous Lectures

2

BAA CkCr

βα

β

α

OrderRection Overall

Bin order

Ain order

Rate Laws - Power Law Model

3

C3BA2 A reactor follows an elementary rate law if the reaction orders just happens to agree with the stoichiometric coefficients for the reaction as written.e.g. If the above reaction follows an elementary rate law

2nd order in A, 1st order in B, overall third order

B2AAA CCkr

Previous Lectures

4

Previous Lectures

5

Previous Lectures

6

Previous Lectures

Today’s lectureBlock 1: Mole Balances Block 2: Rate LawsBlock 3: Stoichiometry Block 4: Combine

7

Today’s lectureExamples:Liquid Phase ChE 460 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 dm38

Today’s lectureExamples: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 =

0.9Case II Batch constant volume

Find the time, t, necessary to achieve 90% conversion. Compare and t.

9

Part 1: Mole Balances in Terms of ConversionAlgorithm for Isothermal Reactor Design1. Mole Balance and Design Equation2. Rate Law3. Stoichiometry4. Combine5. 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)10

Lecture 6

CSTR Laboratory Experiment

Example: CH3CO2 + H20 2CH3OOH

1) Mole Balance:           CSTR:A

0A

r

XFV

M2.51C 0B

M1C 0A

?X

€

V =1 dm3

€

υ0 = 3.3⋅10−3 dm3

s

11

A + B 2C

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

12

X1CX1FF

C 0A0

0AAA

υ

υ

XCXF

C B0A0

B0AB

υ

2.511

2.51B

0B0A0AB C2.51CX2.51CC

CSTR Laboratory Experiment

13

X1kCX1C

k

C'kr 0A0A0BA

V υ0kCA 0X

CA 0 1 X V

υ0

kX

1 X V

υ0

kX

1 X

k1

kX

75.003.4

03.3X

CSTR Laboratory Experiment

14

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 15

A + B 2C

PFR Laboratory Experiment4) Combine: X1kCX1CC'kr 0A0A0BA

υ

υ

kddVk

X1

dX

C

X1kC

dV

dX

0

00A

0A

kX1

1ln

ke1X

sec 94 1s 01.0k

61.0X 16

Example (Gas Flow, PFR)2 NOCl 2 NO + Cl2

s

dm 10

3

0 υsmol

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

17

2A 2B + C

Example (Gas Flow, PFR)

3) Stoich: Gas X10 υυ

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

υ

υ

υ

18

Example (Gas Flow, PFR)

X1

X1XX1ln12kC

22

0A

2

1

2

11y 0A

02.17kC 0A

sec 294kC

02.17

0A

L 2940VV 0 19

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 0υυυ

V

20

Gas Phase 2A 2B + C

Example Constant Volume (Batch)

20A0A

220A X1kCC

X1kC

dt

dX

4) Combine:

dtkC

X1

dX0A2

tkCX1

10A

sec 155t

20A X1kCd

dX

21

Mole Balance

Rate Laws

Stoichiometry

Isothermal Design

Heat Effects

22

End of Lecture 6

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