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Chemical Reaction Engineering Chapter 4, Part 2: 1. Applying the Algorithm to a Batch Reactor, CSTR, and PFR 2. Calculating the Equilibrium Conversion 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 P0 = 8.2 atm T0 = 500 K CA0 = 0.2 mol/dm 3 k = 0.5 dm/mol-s 3 3 v0 = 2.5 dm /s Isothermal Reactor Design Batch Mole Balance: CSTR PFR Isothermal Reactor Design Batch Mole Balance: Rate Law: CSTR PFR 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) Batch V=V0 P T v v 0 1 X 0 v0 1X P T0 Per Mole of A: Flow Per Mole of A: 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) Batch V=V0 P T v v 0 1 X 0 v0 1X P T0 Per Mole of A: Flow Per Mole of A: Isothermal Reactor Design Batch CSTR PFR Stoichiometry (continued): 1 1 FA 0 X FA 0 X F 2 F 2 CB B CB B v v v 0 1 X v 0 1 X Isothermal Reactor Design Batch CSTR PFR Stoichiometry (continued): 1 1 FA 0 X FA 0 X F 2 F 2 CB B CB B v v v 0 1 X v 0 1 X Combine: Isothermal Reactor Design Batch CSTR PFR Stoichiometry (continued): 1 1 FA 0 X FA 0 X F 2 F 2 CB B CB B v v v 0 1 X v 0 1 X Combine: Integrate: Isothermal Reactor Design Batch CSTR PFR Stoichiometry (continued): 1 1 FA 0 X FA 0 X F 2 F 2 CB B CB B v v v 0 1 X v 0 1 X Combine: Integrate: Batch Evaluate: For X=0.9: CSTR PFR Example 1 Reversible Reaction, Constant Volume Determine Xe for a batch system with constant volume, V=V0 Reaction: Additional Information: CA0 = 0.2 mol/dm3 KC = 100 dm3/mol KC C Be C2Ae Example 1 Reversible Reaction, Constant Volume Determine Xe for a batch system with constant volume, V=V0 Reaction: Additional Information: CA0 = 0.2 mol/dm3 KC = 100 dm3/mol KC For constant volume: C Be C2Ae C Ae C A0 1 Xe C Xe C Be A0 2 Example 1 Reversible Reaction, Constant Volume Determine Xe for a batch system with constant volume, V=V0 Reaction: Additional Information: CA0 = 0.2 mol/dm3 KC = 100 dm3/mol KC For constant volume: C Be C2Ae C Ae C A0 1 Xe C Xe C Be A0 2 Solving for the equilibrium conversion: Xe = 0.83 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: CA0 = 0.2 mol/dm3 k = 2 dm3/mol-min Additional Information: KC = 100 dm3/mol FA0 = 5 mol/min 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: CA0 = 0.2 mol/dm3 k = 2 dm3/mol-min Additional Information: KC = 100 dm3/mol FA0 = 5 mol/min First Calculate Xe: KC C Be C 2Ae C Ae C A 0 C Be 1 Xe 1 X e CA0 X e 2 1 X e 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: CA0 = 0.2 mol/dm3 k = 2 dm3/mol-min Additional Information: KC = 100 dm3/mol FA0 = 5 mol/min First Calculate Xe: KC C Be C 2Ae C Ae C A 0 C Be 1 Xe 1 X e CA0 X e 2 1 X e A B 2 1 1 1 y A0 1 2 2 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: CA0 = 0.2 mol/dm3 k = 2 dm3/mol-min Additional Information: KC = 100 dm3/mol FA0 = 5 mol/min First Calculate Xe: KC C Be C 2Ae C Ae C A 0 C Be 1 Xe 1 X e CA0 X e 2 1 X e A B 2 1 1 1 y A0 1 2 2 Solving for Xe: Xe = 0.89 (vs. Xe= 0.83 in Example 1) X = 0.8Xe = 0.711 Using Polymath Algorithm Steps Polymath Equations Using Polymath Algorithm Steps Mole Balance Polymath Equations d(X)/d(V) = -rA/FA0 Using Polymath Algorithm Steps Mole Balance Rate Law Polymath Equations d(X)/d(V) = -rA/FA0 rA = -k*((CA**2)-(CB/KC)) Using Polymath Algorithm Steps Mole Balance Rate Law Stoichiometry Polymath Equations d(X)/d(V) = -rA/FA0 rA = -k*((CA**2)-(CB/KC)) CA = (CA0*(1-X))/(1+eps*X) CB = (CA0*X)/(2*(1+eps*X)) Using Polymath Algorithm Steps Mole Balance Rate Law Stoichiometry Polymath Equations d(X)/d(V) = -rA/FA0 rA = -k*((CA**2)-(CB/KC)) 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 Using Polymath Algorithm Steps Polymath Equations Mole Balance Rate Law Stoichiometry d(X)/d(V) = -rA/FA0 rA = -k*((CA**2)-(CB/KC)) 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 General Guidelines for California Problems 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. 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: 1. Group unknown parameters/values on the same side of the equation example: [unknowns] = [knowns] 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: 1. Group unknown parameters/values on the same side of the equation example: [unknowns] = [knowns] 2. Look for a Case 1 and a Case 2 (usually two data points) to make intermediate calculations 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: 1. Group unknown parameters/values on the same side of the equation example: [unknowns] = [knowns] 2. Look for a Case 1 and a Case 2 (usually two data points) to make intermediate calculations 3. Take ratios of Case 1 and Case 2 to cancel as many unknowns as possible 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: 1. Group unknown parameters/values on the same side of the equation example: [unknowns] = [knowns] 2. Look for a Case 1 and a Case 2 (usually two data points) to make intermediate calculations 3. Take ratios of Case 1 and Case 2 to cancel as many unknowns as possible 4. Carry all symbols to the end of the manipulation before evaluating, UNLESS THEY ARE ZERO