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Chemical Reaction Engineering Asynchronous Video Series

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 -r A =f(X). This result is achieved in two steps. Rate Laws -r A =k f(C i )

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Chemical Reaction Engineering Asynchronous Video Series

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  1. Chemical Reaction EngineeringAsynchronous 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 -rA=f(X). This result is achieved in two steps. • Rate Laws • -rA=k f(Ci) • 1st order A--> B or 1st order • 2nd order A+B --> C • Rate laws are found by experiment • Stoichiometry • Liquid: • Gas: -rA=kCA -rA=kACACB

  3. Algorithm for Isothermal Reactor Design • Mole Balance and Design Equation • Rate Law • Stoichiometry • Combine • Evaluate

  4. Algorithm for Isothermal Reactor Design • Mole Balance and Design Equation • Rate Law • Stoichiometry • Combine • Evaluate The Evaluate step can be carried out:

  5. Algorithm for Isothermal Reactor Design • Mole Balance and Design Equation • Rate Law • Stoichiometry • Combine • Evaluate • The Evaluate step can be carried out: • Graphically (Chapter 2 plots)

  6. Algorithm for Isothermal Reactor Design • Mole Balance and Design Equation • Rate Law • Stoichiometry • Combine • 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 • Mole Balance and Design Equation • Rate Law • Stoichiometry • Combine • 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 • Mole Balance and Design Equation • Rate Law • Stoichiometry • Combine • 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 (T=T0), no pressure drop (P=P0)

  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 (T=T0), no pressure drop (P=P0) [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 (T=T0), no pressure drop (P=P0) [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 (T=T0), no pressure drop (P=P0) [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 (T=T0), no pressure drop (P=P0) [Why do you think CB is constant, when B is consumed?] Combine: Evaluate:

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