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Chemical Reaction Engineering Asynchronous Video Series

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At the start of the chapter we saw we needed -rA=f(X). This result is achieved in two steps. ... Graphically (Chapter 2 plots) Algorithm for Isothermal Reactor Design ... – PowerPoint PPT presentation

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Title: Chemical Reaction Engineering Asynchronous Video Series


1
Chemical Reaction Engineering Asynchronous
Video Series
  • Chapter 4, Part 1
  • Applying the Algorithm to a CSTR
  • H. Scott Fogler, Ph.D.

2
Summary
  • At the start of the chapter we saw we needed
    -rAf(X). This result is achieved in two steps.
  • Rate Laws
  • -rAk f(Ci)
  • 1st order A--gt B or 1st order
  • 2nd order AB --gt C
  • Rate laws are found by experiment
  • Stoichiometry
  • Liquid
  • Gas

-rAkCA
-rAkACACB
3
Algorithm for Isothermal Reactor Design
  1. Mole Balance and Design Equation
  2. Rate Law
  3. Stoichiometry
  4. Combine
  5. Evaluate

4
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
5
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
  • Graphically (Chapter 2 plots)

6
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
  • Graphically (Chapter 2 plots)
  • Numerically (Quadrature formulas Chapter 2 and
    Appendices)

7
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
  • Graphically (Chapter 2 plots)
  • Numerically (Quadrature formulas Chapter 2 and
    Appendices)
  • Analytically (Integral tables in Appendix)

8
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
  • Graphically (Chapter 2 plots)
  • Numerically (Quadrature formulas Chapter 2 and
    Appendices)
  • Analytically (Integral tables in Appendix)
  • Software packages (Appendix - Polymath)

9
French Menu Analogy
10
French Menu Analogy
11
French Menu Analogy
12
French Menu Analogy
13
French Menu Analogy
14
French Menu Analogy
15
French Menu Analogy
16
French Menu Analogy
17
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.

18
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

19
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

20
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 (TT0), no
    pressure drop (PP0)

21
French Menu Analogy
Deriving CA and CB
  • Remember the French Menu reaction
  • For a gas phase system

22
French Menu Analogy
Deriving CA and CB
  • Remember the French Menu reaction
  • For a gas phase system
  • If the conditions are isothermal (T T0) and
    isobaric (P P0)

23
French Menu Analogy
Deriving CA and CB
  • 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

24
French Menu Analogy
Deriving CA and CB
  • 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

25
French Menu Analogy
Deriving CA and CB
  • This leaves us with CA as a function of
    conversion alone

26
French Menu Analogy
Deriving CA and CB
  • This leaves us with CA as a function of
    conversion alone
  • Similarly for CB

27
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 (TT0), no
    pressure drop (PP0)
  • Why do you think CB is constant, when B is
    consumed?

28
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 (TT0), no
    pressure drop (PP0)
  • Why do you think CB is constant, when B is
    consumed?

29
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 (TT0), no
    pressure drop (PP0)
  • Why do you think CB is constant, when B is
    consumed?
  • Combine

30
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 (TT0), no
    pressure drop (PP0)
  • Why do you think CB is constant, when B is
    consumed?
  • Combine
  • Evaluate
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