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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

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**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 T = 500 K 3 C = 0.2 mol/dm A0 3 3 k = 0.5 dm /mol-s v0 = 2.5 dm /s

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**Isothermal Reactor Design**

Batch CSTR PFR Mole Balance:

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**Isothermal Reactor Design**

Batch CSTR PFR Mole Balance: Rate Law:

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**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: Flow Batch V=V0

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**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: Flow Batch V=V0 V=V0

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**Isothermal Reactor Design**

Batch CSTR PFR Stoichiometry (continued):

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**Isothermal Reactor Design**

Batch CSTR PFR Stoichiometry (continued): Combine:

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**Isothermal Reactor Design**

Batch CSTR PFR Stoichiometry (continued): Combine: Integrate:

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**Isothermal Reactor Design**

Batch CSTR PFR Stoichiometry (continued): Combine: Integrate: Evaluate: Batch CSTR PFR For X=0.9:

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**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

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**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/dm3 KC = 100 dm3/mol

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**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/dm3 KC = 100 dm3/mol

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**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

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**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

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**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

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**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 =

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Using Polymath Algorithm Steps Polymath Equations

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**Using Polymath Algorithm Steps Polymath Equations**

Mole Balance d(X)/d(V) = -rA/FA0

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**Using Polymath Algorithm Steps Polymath Equations**

Mole Balance d(X)/d(V) = -rA/FA0 Rate Law rA = -k*((CA**2)-(CB/KC))

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**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))

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**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 = CA0 = k = 2 FA0 = KC = 100

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**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 = CA0 = k = 2 FA0 = KC = 100 Initial and Final Values X0 = V0 = Vf = 500

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**General Guidelines for California Problems**

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**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.

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**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]

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**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

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**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

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**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

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Lecture 24 Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors.

Lecture 24 Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors.

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