Chemical Reaction Engineering

Asynchronous Video Series

Chapter 4, Part 1:

Applying the Algorithm to a CSTR

H. Scott Fogler, Ph.D.

Summary

At the start of the chapter we saw we needed -rA=f(X). This result is achieved in two steps.

1. Rate Laws

– -rA=k f(Ci)

– 1st order A--> B or 1st order

– 2nd order A+B --> C

– Rate laws are found by experiment

2. Stoichiometry

– Liquid:

– Gas:

-rA=kCA

€

−rA = k C A −C B

KC

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

-rA=kACACB

€

C A = CA 0 1− X( )

€

C A =CA 0 1− X( )

1+ εX( )PP0

T0

T

€

C A = CT 0

FT

FT 0

PP0

T0

T

Algorithm for Isothermal Reactor Design

1. Mole Balance and Design Equation

2. Rate Law

3. Stoichiometry

4. Combine

5. Evaluate

Algorithm for Isothermal Reactor Design

1. Mole Balance and Design Equation

2. Rate Law

3. Stoichiometry

4. Combine

5. Evaluate

The Evaluate step can be carried out:

Algorithm for Isothermal Reactor Design

1. Mole Balance and Design Equation

2. Rate Law

3. Stoichiometry

4. Combine

5. Evaluate

The Evaluate step can be carried out:1. Graphically (Chapter 2 plots)

Algorithm for Isothermal Reactor Design

1. Mole Balance and Design Equation

2. Rate Law

3. Stoichiometry

4. Combine

5. Evaluate

The Evaluate step can be carried out:1. Graphically (Chapter 2 plots)

2. Numerically (Quadrature formulas: Chapter 2 and Appendices)

Algorithm for Isothermal Reactor Design

1. Mole Balance and Design Equation

2. Rate Law

3. Stoichiometry

4. Combine

5. Evaluate

The Evaluate step can be carried out:1. Graphically (Chapter 2 plots)

2. Numerically (Quadrature formulas: Chapter 2 and Appendices)

3. Analytically (Integral tables in Appendix)

Algorithm for Isothermal Reactor Design

1. Mole Balance and Design Equation

2. Rate Law

3. Stoichiometry

4. Combine

5. Evaluate

The Evaluate step can be carried out:1. Graphically (Chapter 2 plots)

2. Numerically (Quadrature formulas: Chapter 2 and Appendices)

3. Analytically (Integral tables in Appendix)

4. Software packages (Appendix - Polymath)

French Menu Analogy

French Menu Analogy

French Menu Analogy

French Menu Analogy

French Menu Analogy

French Menu Analogy

French Menu Analogy

French Menu Analogy

French Menu Analogy

Example: The elementary gas phase reaction takes place in a CSTR at constant temperature (500 K) and constant pressure (16.4 atm). The feed is equal molar in A and B.

French Menu Analogy

Example: The elementary gas phase reaction takes place in a CSTR at constant temperature (500 K) and constant pressure (16.4 atm). The feed is equal molar in A and B.

Mole Balance:

French Menu Analogy

Example: The elementary gas phase reaction takes place in a CSTR at constant temperature (500 K) and constant pressure (16.4 atm). The feed is equal molar in A and B.

Mole Balance:

Rate Law:

Example: The elementary gas phase reaction takes place in a CSTR at constant temperature (500 K) and constant pressure (16.4 atm). The feed is equal molar in A and B.

Mole Balance:

Rate Law:

Stoichiometry: gas phase, isothermal (T=T0), no pressure drop (P=P0)

French Menu Analogy

French Menu Analogy

Remember the French Menu reaction:

For a gas phase system:

Deriving CA and CB:

French Menu Analogy

Remember the French Menu reaction:

For a gas phase system:

If the conditions are isothermal (T = T0) and isobaric (P =P0):

Deriving CA and CB:

French Menu Analogy

Remember the French Menu reaction:

For a gas phase system:

If the conditions are isothermal (T = T0) and isobaric (P =P0):

We must divide by the stoichiometric coefficient of our basis of calculation yielding:

Deriving CA and CB:

French Menu Analogy

Remember the French Menu reaction:

For a gas phase system:

If the conditions are isothermal (T = T0) and isobaric (P =P0):

We must divide by the stoichiometric coefficient of our basis of calculation yielding:

And if the feed is equal molar, then:

Deriving CA and CB:

French Menu Analogy

This leaves us with CA as a function of conversion alone:

Deriving CA and CB:

French Menu Analogy

This leaves us with CA as a function of conversion alone:

Similarly for CB:

€

FB0 = FA0

Therefore, Θ B =1

Deriving CA and CB:

Example: The elementary gas phase reaction takes place in a CSTR at constant temperature (500 K) and constant pressure (16.4 atm). The feed is equal molar in A and B.

Mole Balance:

Rate Law:

Stoichiometry: gas phase, isothermal (T=T0), no pressure drop (P=P0)

[Why do you think CB is constant, when B is consumed?]

French Menu Analogy

Example: The elementary gas phase reaction takes place in a CSTR at constant temperature (500 K) and constant pressure (16.4 atm). The feed is equal molar in A and B.

Mole Balance:

Rate Law:

Stoichiometry: gas phase, isothermal (T=T0), no pressure drop (P=P0)

[Why do you think CB is constant, when B is consumed?]

French Menu Analogy

Example: The elementary gas phase reaction takes place in a CSTR at constant temperature (500 K) and constant pressure (16.4 atm). The feed is equal molar in A and B.

Mole Balance:

Rate Law:

Stoichiometry: gas phase, isothermal (T=T0), no pressure drop (P=P0)

[Why do you think CB is constant, when B is consumed?]

Combine:

French Menu Analogy

Example: The elementary gas phase reaction takes place in a CSTR at constant temperature (500 K) and constant pressure (16.4 atm). The feed is equal molar in A and B.

Mole Balance:

Rate Law:

Stoichiometry: gas phase, isothermal (T=T0), no pressure drop (P=P0)

[Why do you think CB is constant, when B is consumed?]

Combine:

Evaluate:

French Menu Analogy