Chemical Reaction Engineering Asynchronous Video Series

1. Chemical Reaction EngineeringAsynchronous Video Series Chapter 4, Part 2: 1. Applying the Algorithm to a Batch Reactor, CSTR, and PFR 2. Calculating the Equilibrium Conversion H. Scott Fogler, Ph.D.

2. Using the Algorithm for Isothermal Reactor Design • Now we apply the algorithm to the reaction below occurring in a Batch Reactor, CSTR, and PFR. Gas Phase Elementary Reaction Additional Information only A fed P = 8.2 atm 0 3 T = 500 K C = 0.2 mol/dm 0 A0 3 3 k = 0.5 dm /mol-s v0 = 2.5 dm /s

3. Isothermal Reactor Design Batch CSTR PFR Mole Balance:

4. Isothermal Reactor Design Batch CSTR PFR Mole Balance: Rate Law:

5. Flow Isothermal Reactor Design Batch CSTR PFR Mole Balance: Rate Law: Stoichiometry: Gas: V = V0 Gas: T =T0, P =P0 Gas: T = T0, P = P0 (e.g., constant volume steel container) Per Mole of A: Per Mole of A: Batch V=V0

6. Flow Isothermal Reactor Design Batch CSTR PFR Mole Balance: Rate Law: Stoichiometry: Gas: V = V0 Gas: T =T0, P =P0 Gas: T = T0, P = P0 (e.g., constant volume steel container) Per Mole of A: Per Mole of A: Batch V=V0 V=V0

7. Isothermal Reactor Design Batch CSTR PFR Stoichiometry (continued):

8. Isothermal Reactor Design Batch CSTR PFR Stoichiometry (continued): Combine:

9. Isothermal Reactor Design Batch CSTR PFR Stoichiometry (continued): Combine: Integrate:

10. Isothermal Reactor Design Batch CSTR PFR Stoichiometry (continued): Combine: Integrate: Evaluate: Batch CSTR PFR For X=0.9:

11. Example 1 Reversible Reaction, Constant Volume Determine Xe for a batch system with constant volume, V=V0 Reaction: Additional Information: CA0 = 0.2 mol/dm3KC = 100 dm3/mol

12. Example 1 Reversible Reaction, Constant Volume Determine Xe for a batch system with constant volume, V=V0 Reaction: Additional Information: For constant volume: CA0 = 0.2 mol/dm3KC = 100 dm3/mol

13. Example 1 Reversible Reaction, Constant Volume Determine Xe for a batch system with constant volume, V=V0 Reaction: Additional Information: For constant volume: Solving for the equilibrium conversion:   Xe = 0.83 CA0 = 0.2 mol/dm3KC = 100 dm3/mol

14. Example 2 Reversible Reaction, Variable Volumetric Flow Rate Determine Xe for a PFR with no pressure drop, P=P0 Given: The system is gas phase and isothermal. Find: The reactor volume when X=0.8Xe Reaction: Additional Information: CA0 = 0.2 mol/dm3 k = 2 dm3/mol-min KC = 100 dm3/mol FA0 = 5 mol/min

15. Example 2 Reversible Reaction, Variable Volumetric Flow Rate Determine Xe for a PFR with no pressure drop, P=P0 Given: The system is gas phase and isothermal. Find: The reactor volume when X=0.8Xe Reaction: Additional Information: First Calculate Xe: CA0 = 0.2 mol/dm3 k = 2 dm3/mol-min KC = 100 dm3/mol FA0 = 5 mol/min

16. Example 2 Reversible Reaction, Variable Volumetric Flow Rate Determine Xe for a PFR with no pressure drop, P=P0 Given: The system is gas phase and isothermal. Find: The reactor volume when X=0.8Xe Reaction: Additional Information: First Calculate Xe: CA0 = 0.2 mol/dm3 k = 2 dm3/mol-min KC = 100 dm3/mol FA0 = 5 mol/min

17. Example 2 Reversible Reaction, Variable Volumetric Flow Rate Determine Xe for a PFR with no pressure drop, P=P0 Given: The system is gas phase and isothermal. Find: The reactor volume when X=0.8Xe Reaction: Additional Information: First Calculate Xe: Solving for Xe: CA0 = 0.2 mol/dm3 k = 2 dm3/mol-min KC = 100 dm3/mol FA0 = 5 mol/min Xe = 0.89 (vs. Xe= 0.83 in Example 1) X = 0.8Xe = 0.711    

18. Using Polymath Algorithm Steps Polymath Equations

19. Using Polymath Algorithm Steps Polymath Equations Mole Balance d(X)/d(V) = -rA/FA0

20. Using Polymath Algorithm Steps Polymath Equations Mole Balance d(X)/d(V) = -rA/FA0 Rate Law rA = -k*((CA**2)-(CB/KC))

21. Using Polymath Algorithm Steps Polymath Equations Mole Balance d(X)/d(V) = -rA/FA0 Rate Law rA = -k*((CA**2)-(CB/KC)) Stoichiometry CA = (CA0*(1-X))/(1+eps*X) CB = (CA0*X)/(2*(1+eps*X))

22. Using Polymath Algorithm Steps Polymath Equations Mole Balance d(X)/d(V) = -rA/FA0 Rate Law rA = -k*((CA**2)-(CB/KC)) Stoichiometry CA = (CA0*(1-X))/(1+eps*X) CB = (CA0*X)/(2*(1+eps*X)) Parameter Evaluation eps = -0.5 CA0 = 0.2 k = 2 FA0 = 5 KC = 100

23. Using Polymath Algorithm Steps Polymath Equations Mole Balance d(X)/d(V) = -rA/FA0 Rate Law rA = -k*((CA**2)-(CB/KC)) Stoichiometry CA = (CA0*(1-X))/(1+eps*X) CB = (CA0*X)/(2*(1+eps*X)) Parameter Evaluation eps = -0.5 CA0 = 0.2 k = 2 FA0 = 5 KC = 100 Initial and Final Values X0 = 0 V0 = 0 Vf = 500

24. General Guidelines for California Problems

25. General Guidelines for California Problems Every state has an examination engineers must pass to become a registered professional engineer.  In the past there have typically been six problems in a three hour segment of the California Professional Engineers Exam. Consequently one should be able to work each problem in 30 minutes or less. Many of these problems involve an intermediate calculation to determine the final answer.

26. General Guidelines for California Problems Every state has an examination engineers must pass to become a registered professional engineer.  In the past there have typically been six problems in a three hour segment of the California Professional Engineers Exam. Consequently one should be able to work each problem in 30 minutes or less. Many of these problems involve an intermediate calculation to determine the final answer. Some Hints: • Group unknown parameters/values on the same side of the equation example: [unknowns] = [knowns]

27. General Guidelines for California Problems Every state has an examination engineers must pass to become a registered professional engineer.  In the past there have typically been six problems in a three hour segment of the California Professional Engineers Exam. Consequently one should be able to work each problem in 30 minutes or less. Many of these problems involve an intermediate calculation to determine the final answer. Some Hints: • Group unknown parameters/values on the same side of the equation example: [unknowns] = [knowns] • Look for a Case 1 and a Case 2 (usually two data points) to make intermediate calculations

28. General Guidelines for California Problems Every state has an examination engineers must pass to become a registered professional engineer.  In the past there have typically been six problems in a three hour segment of the California Professional Engineers Exam. Consequently one should be able to work each problem in 30 minutes or less. Many of these problems involve an intermediate calculation to determine the final answer. Some Hints: • Group unknown parameters/values on the same side of the equation example: [unknowns] = [knowns] • Look for a Case 1 and a Case 2 (usually two data points) to make intermediate calculations • Take ratios of Case 1 and Case 2 to cancel as many unknowns as possible

29. General Guidelines for California Problems Every state has an examination engineers must pass to become a registered professional engineer.  In the past there have typically been six problems in a three hour segment of the California Professional Engineers Exam. Consequently one should be able to work each problem in 30 minutes or less. Many of these problems involve an intermediate calculation to determine the final answer. Some Hints: • Group unknown parameters/values on the same side of the equation example: [unknowns] = [knowns] • Look for a Case 1 and a Case 2 (usually two data points) to make intermediate calculations • Take ratios of Case 1 and Case 2 to cancel as many unknowns as possible • Carry all symbols to the end of the manipulation before evaluating, UNLESS THEY ARE ZERO