chemical reaction engineering asynchronous video series
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Chemical Reaction Engineering Asynchronous 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. Using the Algorithm for Isothermal Reactor Design. - PowerPoint PPT PresentationTRANSCRIPT
Chemical Reaction Engineering
Asynchronous 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.
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 = v 0 1+ εX( )P0
PTT0
= v0 1+εX( )
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 = v 0 1+ εX( )P0
PTT0
= v0 1+εX( )
Flow
V=V0
Batch
Isothermal Reactor Design
Stoichiometry (continued):
Batch CSTR PFR
€
C B =FB
v=
FA 0 +12
X ⎛ ⎝ ⎜
⎞ ⎠ ⎟
v 0 1+ εX( )
€
C B =FB
v=
FA 0 +12
X ⎛ ⎝ ⎜
⎞ ⎠ ⎟
v 0 1+ εX( )
Isothermal Reactor Design
Stoichiometry (continued):
Combine:
Batch CSTR PFR
€
C B =FB
v=
FA 0 +12
X ⎛ ⎝ ⎜
⎞ ⎠ ⎟
v 0 1+ εX( )
€
C B =FB
v=
FA 0 +12
X ⎛ ⎝ ⎜
⎞ ⎠ ⎟
v 0 1+ εX( )
Isothermal Reactor Design
Stoichiometry (continued):
Combine:
Integrate:
Batch CSTR PFR
€
C B =FB
v=
FA 0 +12
X ⎛ ⎝ ⎜
⎞ ⎠ ⎟
v 0 1+ εX( )
€
C B =FB
v=
FA 0 +12
X ⎛ ⎝ ⎜
⎞ ⎠ ⎟
v 0 1+ εX( )
Isothermal Reactor Design
Stoichiometry (continued):
Combine:
Integrate:
Evaluate:
Batch CSTR PFR
€
C B =FB
v=
FA 0 +12
X ⎛ ⎝ ⎜
⎞ ⎠ ⎟
v 0 1+ εX( )
€
C B =FB
v=
FA 0 +12
X ⎛ ⎝ ⎜
⎞ ⎠ ⎟
v 0 1+ εX( )
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( )
1+εXe( )
CBe =CA0Xe
2 1+ εXe( )
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
ε = yA0δ = 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( )
1+εXe( )
CBe =CA0Xe
2 1+ εXe( )
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
ε = yA0δ = 1( ) 12 −1 ⎛
⎝ ⎞ ⎠ = −1
2
€
KC =CBe
CAe2
CAe =CA0
1− Xe( )
1+εXe( )
CBe =CA0Xe
2 1+ εXe( )
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