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

only A fed P0 = 8.2 atm

T0 = 500 K CA0 = 0.2 mol/dm3

k = 0.5 dm3/mol-s v0 = 2.5 dm

3/s

Additional Information

Isothermal Reactor Design

Mole Balance:

Batch CSTR PFR

Isothermal Reactor Design

Mole Balance:

Rate Law:

Batch CSTR PFR

Isothermal Reactor Design

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 CSTR PFR

V=V0

v v0 1 X P0

PTT0

v0 1X

BatchFlow

Isothermal Reactor Design

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 CSTR PFR

V=V0

v v0 1 X P0

PTT0

v0 1X Flow

V=V0

Batch

Isothermal Reactor Design

Stoichiometry (continued):

Batch CSTR PFR

C B FB

v

FA 0 12

X

v 0 1 X

C B FB

v

FA 0 12

X

v 0 1 X

Isothermal Reactor Design

Stoichiometry (continued):

Combine:

Batch CSTR PFR

C B FB

v

FA 0 12

X

v 0 1 X

C B FB

v

FA 0 12

X

v 0 1 X

Isothermal Reactor Design

Stoichiometry (continued):

Combine:

Integrate:

Batch CSTR PFR

C B FB

v

FA 0 12

X

v 0 1 X

C B FB

v

FA 0 12

X

v 0 1 X

Isothermal Reactor Design

Stoichiometry (continued):

Combine:

Integrate:

Evaluate:

Batch CSTR PFR

C B FB

v

FA 0 12

X

v 0 1 X

C B FB

v

FA 0 12

X

v 0 1 X

Batch CSTR PFR

For X=0.9:

Example 1

Reaction:

Additional Information:CA0 = 0.2 mol/dm3

KC = 100 dm3/mol

KC CBe

CAe2

Determine Xe for a batch system with constant volume, V=V0

Reversible Reaction, Constant Volume

Example 1

Reaction:

Additional Information:

For constant volume:

CA0 = 0.2 mol/dm3

KC = 100 dm3/mol

CAe CA0 1 Xe CBe

CA0Xe

2

KC CBe

CAe2

Determine Xe for a batch system with constant volume, V=V0

Reversible Reaction, Constant Volume

Example 1

Reaction:

Additional Information:

For constant volume:

Solving for the equilibrium conversion:

  Xe = 0.83

CA0 = 0.2 mol/dm3

KC = 100 dm3/mol

CAe CA0 1 Xe CBe

CA0Xe

2

KC CBe

CAe2

Reversible Reaction, Constant Volume

Determine Xe for a batch system with constant volume, V=V0

Example 2

Given: The system is gas phase and isothermal.

Find: The reactor volume when X=0.8Xe

Reaction:

Additional Information:

Determine Xe for a PFR with no pressure drop, P=P0

CA0 = 0.2 mol/dm3 k = 2 dm3/mol-min

KC = 100 dm3/mol FA0 = 5 mol/min

Reversible Reaction, Variable Volumetric Flow Rate

Example 2

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

KC CBe

CAe2

CAe CA0

1 Xe 1Xe

CBe CA0Xe

2 1 Xe

Determine Xe for a PFR with no pressure drop, P=P0

Reversible Reaction, Variable Volumetric Flow Rate

Example 2

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

A B2

yA0 1 12 1

1

2

Determine Xe for a PFR with no pressure drop, P=P0

Reversible Reaction, Variable Volumetric Flow Rate

KC CBe

CAe2

CAe CA0

1 Xe 1Xe

CBe CA0Xe

2 1 Xe

Example 2

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    

Determine Xe for a PFR with no pressure drop, P=P0

Reversible Reaction, Variable Volumetric Flow Rate

A B2

yA0 1 12 1

1

2

KC CBe

CAe2

CAe CA0

1 Xe 1Xe

CBe CA0Xe

2 1 Xe

Using Polymath

Algorithm Steps Polymath Equations

Using Polymath

Algorithm Steps Polymath Equations

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

Using Polymath

Algorithm Steps Polymath Equations

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

Rate Law rA = -k*((CA**2)-(CB/KC))

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))

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

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

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