Lecture 6

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.

2

Lecture 6 – Tuesday 1/29/2013

    Block 1:

Mole Balances

Block 2:

Rate Laws

Block 3:

Stoichiometry

Block 4:

Combine Review of Blocks 1, 2 and 3 Examples : Undergraduate Reactor Experiments CSTR PFR BR Gas Phase Reaction with Change in the Total Number of Moles

3 Review Lecture 2

Building Block 1:

in terms of conversion, X

Mole Balances

Reactor

Batch

Differential

N A

0

dX dt

 

r A V

Algebraic

t



Integral

N A

0

X

0 

dX



r A V

X CSTR

V



F A

0

X



r A

PFR

F A

0

dX dV

 

r A



F A

0

dX dW V



F A

0 0 

X dX



r A W



F A

0 0 

X dX



r



A

X t W 

4 Review Lecture 3

Building Block 2: Rate Laws

Power Law Model: 

r A



kC



A C B



α

order in A 2

A



B

 3

C β

order in B Overall Rection Order 

α



β

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 

r A



k A C

2

A C B

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

5 Review Lecture 4

Building Block 3: Stoichiometry

6 Review Lecture 5

Building Block 4: Combine

Review Lecture 5 Building Block 4: Combine 7

8

Today’s lecture

 Example for Liquid Phase Undergraduate Laboratory Experiment (CH 2 CO) 2 O + H 2 O  2CH 3 COOH A + B  2C Entering Volumetric flow rate Acetic Anhydride Water Elementary with k’ v 0 = 0.0033 dm 7.8% (1M) 3 /s 92.2% (51.2M) 1.95x10

-4 dm 3 /(mol.s) Case I Case II CSTR PFR V = 1dm 3 V = 0.311 dm 3

9

Today’s lecture

 Example for Gas Phase : PFR and Batch 2NOCl  2A  2NO + Cl 2 2B + C Calculation Pure NOCl fed with C NOCl,0 = 0.2 mol/dm 3 elementary rate law with k = 0.29 dm 3 /(mol.s) follows an Case I Case II PFR with v 0 = 10 dm3/s Find space time, 𝜏 with X = 0.9 Find reactor volume, V for X = 0.9

Batch constant volume Find the time, t, necessary to achieve 90% conversion. Compare 𝜏 and t.

10 Part 1: Mole Balances in terms of Conversion

Algorithm for Isothermal Reactor Design 1.

Mole Balances

and Design Equation 2.

3.

Rate Laws Stoichiometry 4.

5.

Combine 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)

11

CSTR Laboratory Experiment Example: CH 3 CO 2 + H 2 0  2CH 3 OOH

C A

0  1

M C B

0  51 .

2

M

X  ?

V

 0  1

dm

3  3.3

 10  3

dm

3

s

A + B  2C CSTR :

V



F A

 0

r A X

12

CSTR Laboratory Experiment 2) Rate Law: 

r A



k A C A C B

3) Stoichiometry: A B F A0 F A0 Θ B C 0 -F A0 X -F A0 X 2F A0 X F A =F A0 (1-X) F B =F A0 (Θ B -X) F C =2F A0 X

13 CSTR Laboratory Experiment

C A



F A

 

F A

0  1   0

X

 

C A

0  1 

X



C B



F A

0   

B

0 

X

 

C A

0  

B



X

 

B

 51 .

2 1  51 .

2

C B



C A

0  51 .

2 

X

 

C A

0  51 .

2  

C B

0

14 CSTR Laboratory Experiment 

r A

 '  0

C A

0  1 

k X

 

kC A

0  1 

X



V



C

 0

A

0

kC A

0  1 

X X

 

V

 0   1

kX



X

   

V

 0   1

kX



X



X

 1  

k



k X

 3 .

03 4 .

03  0 .

75

15

PFR Laboratory Experiment A + B  2C 0 .

00324

dm

3

s

0.311

dm

3

1) Mole Balance:



dX dV

 

r A F A

0

X

 ?

2) Rate Law: 

r A



kC A C B

3) Stoichiometry:

C A



C A

0  1 

X



C B



C B

0

16

PFR Laboratory Experiment 4) Combine: 

r A



k

'

C B

0

C A

0  1 

X

 

kC A

0  1 

dX dV



kC A

0

C

 1

A

0   0

X



X

  1

dX



X

 

k

 0

dV

ln 1  1

X X

  1 

e



k



k

 

kd

  

V

 0  0 .

311

dm

0 .

00324

dm

3 3 sec  96 .

0 sec

k

 0 .

01

s

 1

X

 0 .

61

17

Gas Flow PFR Example  0 2 NOCl  2 NO + Cl 2 2A  2B + C  10

dm

3

s T



T

0

P k

 0 .

29

dm

3

mol



s



P

0

X

 0.9

C A

0  0 .

2

mol L V

 ?

1) Mole Balance:

  2) Rate Law:

dX dV

 

r A F A

0 

r A



kC A

2

18`1

Gas Flow PFR Example 3) Stoichiometry: (Gas Flow) 4) Combine: 

C A

  0  1  

X

 

C



A

0 1   1  

X X

  A  

r A

 B + ½C

kC A

2  1 0   1  

X dX dV



C

2

kC A

0

A

0  0

X

 2   1 1    2

X



X

 2  2 

X

0    1 1   

X X

  2 2

dX



V

0 

kC A

0  0

dV



kC A

 0 0

V

  

kC A

0 

Gas Flow PFR Example

19

kC A

0   2   1     1 

X

   2

X

  1  1    2

X X

 

y A

0   1 2  1 2

kC A

0   17 .

02   17 .

02

kC A

0  294 sec

V



V

0   2940

L

20

Constant Volume Batch Example Gas Phase 2A  2B + C t=?

1) Mole Balance:

2) Rate Law:

dX dt

 

r A V

0

N A

0  

N A

0

r A V

0  

r C A

0

A



r A



kC A

2 3) Stoichiometry: (Gas Flow)

V



V

0

C A



r A



N A

0  1  

V

0

kC

2

A

0  1 

X



X



C A

0  1   2

X



Constant Volume Batch Example

21

4) Combine:

dX dt

 2

kC A

0  1 

X

 2

C A

0 

kC A

0  1 

X

 2

dX



kC A

0  1 

X

 2

dt

 1 

dX X

 2 

kC A

0

dt

1  1

X



kC A

0

t t

 155 sec

22

Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance

23

End of Lecture 6