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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 only A fedP 0 = 8.2 atm T 0 = 500 K C A0 = 0.2 mol/dm 3 k = 0.5 dm 3 /mol-sv0v0 = 2.5 dm 3 /s Additional Information

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Isothermal Reactor Design Mole Balance: Batch CSTRPFR

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Isothermal Reactor Design Mole Balance: Rate Law: Batch CSTRPFR

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Isothermal Reactor Design Mole Balance: Rate Law: Stoichiometry: Gas: V = V 0 Gas: T =T 0, P =P 0 Gas: T = T 0, P = P 0 (e.g., constant volume steel container) Per Mole of A: Batch CSTRPFR V=V 0 Batch Flow

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Isothermal Reactor Design Mole Balance: Rate Law: Stoichiometry: Gas: V = V 0 Gas: T =T 0, P =P 0 Gas: T = T 0, P = P 0 (e.g., constant volume steel container) Per Mole of A: Batch CSTRPFR V=V 0 Flow V=V 0 Batch

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Isothermal Reactor Design Stoichiometry (continued): Batch CSTRPFR

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Isothermal Reactor Design Stoichiometry (continued): Combine: Batch CSTRPFR

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Isothermal Reactor Design Stoichiometry (continued): Combine: Integrate: Batch CSTRPFR

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Isothermal Reactor Design Stoichiometry (continued): Combine: Integrate: Evaluate: Batch CSTRPFR BatchCSTRPFR For X=0.9:

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Example 1 Reaction: Additional Information: C A0 = 0.2 mol/dm 3 K C = 100 dm 3 /mol Determine X e for a batch system with constant volume, V=V 0 Reversible Reaction, Constant Volume

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Example 1 Reaction: Additional Information: For constant volume: C A0 = 0.2 mol/dm 3 K C = 100 dm 3 /mol Determine X e for a batch system with constant volume, V=V 0 Reversible Reaction, Constant Volume

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Example 1 Reaction: Additional Information: For constant volume: Solving for the equilibrium conversion: X e = 0.83 C A0 = 0.2 mol/dm 3 K C = 100 dm 3 /mol Reversible Reaction, Constant Volume Determine X e for a batch system with constant volume, V=V 0

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Example 2 Given: The system is gas phase and isothermal. Find: The reactor volume when X=0.8X e Reaction: Additional Information: Determine X e for a PFR with no pressure drop, P=P 0 C A0 = 0.2 mol/dm 3 k = 2 dm 3 /mol-min K C = 100 dm 3 /mol F A0 = 5 mol/min Reversible Reaction, Variable Volumetric Flow Rate

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Example 2 Given: The system is gas phase and isothermal. Find: The reactor volume when X=0.8X e Reaction: Additional Information: First Calculate X e : C A0 = 0.2 mol/dm 3 k = 2 dm 3 /mol-min K C = 100 dm 3 /mol F A0 = 5 mol/min Determine X e for a PFR with no pressure drop, P=P 0 Reversible Reaction, Variable Volumetric Flow Rate

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Example 2 Given: The system is gas phase and isothermal. Find: The reactor volume when X=0.8X e Reaction: Additional Information: First Calculate X e : C A0 = 0.2 mol/dm 3 k = 2 dm 3 /mol-min K C = 100 dm 3 /mol F A0 = 5 mol/min Determine X e for a PFR with no pressure drop, P=P 0 Reversible Reaction, Variable Volumetric Flow Rate

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Example 2 Given: The system is gas phase and isothermal. Find: The reactor volume when X=0.8X e Reaction: Additional Information: First Calculate X e : Solving for X e : C A0 = 0.2 mol/dm 3 k = 2 dm 3 /mol-min K C = 100 dm 3 /mol F A0 = 5 mol/min X e = 0.89 (vs. X e = 0.83 in Example 1) X = 0.8X e = Determine X e for a PFR with no pressure drop, P=P 0 Reversible Reaction, Variable Volumetric Flow Rate

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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 = -0.5 CA0 = 0.2 k = 2 FA0 = 5 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 = -0.5 CA0 = 0.2 k = 2 FA0 = 5 KC = 100 Initial and Final Values X 0 = 0 V 0 = 0 V f = 500

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General Guidelines for California Problems

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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: 1.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: 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

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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: 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

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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: 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

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