Derive From Mass Balance the Rate of Reaction for an Open Continuous Reactor
4. Algorithm for Isothermal Reactor Design*
Topics
Part 1: Mole Balances in Terms of Conversion- Algorithm for Isothermal Reactor Design
- Applications/Examples of CRE Algorithm
- Reversible Reactions
- ODE (Polymath) Solutions to CRE Problems
- General Guidelines for California Problems
- PBR with Pressure Drop
- Engineering Analysis
- Measures Other Than Conversion
- Membrane Reactors
- Semibatch Reactors
Part 1: Mole Balances in Terms of Conversion
Algorithm for Isothermal Reactor Design | top |
French Menu Analogy
The reaction (2A+B-->C) carried out in a CSTR, PFR and a Batch Reactor.
Labratory Experiment
Semilog plot to find reaction rate constant
to a closed ended PFR/CSTR example.
CHEMKIN Reactor Models
Example: The elementary liquid phase reaction
is carried out isothermally in a CSTR. Pure A enters at a volumetric flow rate of 25 dm3/s and at a concentration of 0.2 mol/dm3.
What CSTR volume is necessary to achieve a 90% conversion when k = 10 dm3/(mol*s)?
Mole Balance | |
Rate Law | |
Stoichiometry | liquid phase (v = vo) |
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Combine | |
Evaluate | at X = 0.9, |
V = 1125 dm3 | |
Space Time |
Here are some links to example problems. You could also use these problems as self tests.
CSTR Type 1 Home Problem
CSTR Type 2 Home Problem
CSTR Type 3 Home Problem
Critical Thinking Questions for CSTR
The following movies were made by the students of Professor Alan Lane's chemical reaction engineering class at the University of Alabama Tuscaloosa
Gas Phase Elementary Reaction | Additional Information | ||
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| only A fed | P0 = 8.2 atm | |
T0 = 500 K | CA0 = 0.2 mol/dm3 | ||
k = 0.5 dm3/mol-s | vo = 2.5 dm3/s |
Solve for X = 0.9
Applying the algorithm to the above reaction occurring in a Batch, CSTR, and PFR.
Batch | CSTR | PFR | |
---|---|---|---|
Mole Balance: | | | |
Rate Law: | | | |
Stoichiometry: | Gas: V = V0 | Gas: T =T0, P =P0 | Gas: T = T0, P = P0 |
Per Mole of A: | Per Mole of A: | ||
| | | |
| | | |
| | | |
| | | |
Combine: | | | |
Integrate | | | |
Evaluate | | | |
For X = 0.9: | | V = 680.6 dm3 | V = 90.7 dm3 |
| | |
Visual Encyclopedia of Reaction Engineering Equipment
Gas Phase Reaction Example
CSTR and PFR Example
Calculate V for a Zero-Order Reaction
To determine the conversion or reactor volume for reversible reactions, one must first calculate the maximum conversion that can be achieved at the isothermal reaction temperature, which is the equilibrium conversion. (See Example 3-8 in the text for additional coverage of equilibrium conversion in isothermal reactor design.)
Equilibrium Conversion, Xe
From Appendix C:
Calculate Equilibrium Conversion (Xe) for a Constant Volume System
Example: Determine Xe for a PFR with no pressure drop, P = P0
Given that the system is gas phase and isothermal, determine the reactor volume when X = 0.8 Xe.
Reaction | Additional Information | |
---|---|---|
| CA0 = 0.2 mol/dm3 | k = 2 dm3/mol-min |
First calculate Xe:
Xe = 0.89
X = 0.8Xe = 0.711
One could then use Polymath to determine the volume of the PFR. The corresponding Polymath program is shown below.
ODE (Polymath) Solutions to CRE Problems | top |
Algorithm Steps | Polymath Equations | ||
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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 |
EquationsPlot of X vs. V
Results in Tabular Form
A volume of 94 dm3 (rounding up from slightly more than 93 dm3) appears to be our answer.
Batch Reactor With a Reversible Reaction
General Guidelines for California Problems | top |
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 equationexample:
[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
California Professional Engineers Registration Problem
Batch Reactor Optimization
PBR with Pressure Drop | top |
Note: Pressure drop does NOT affect liquid phase reactions
Sample Question:
Analyze the following second order gas phase reaction that occurs isothermally in a PBR:
Mole Balance
Must use the differential form of the mole balance to separate variables
Rate Law
Second order in A and irreversible:
Stoichiometry
Isothermal, T = T0
Combine
Need to find (P/P0) as a function of W (or V if you have a PFR).
Pressure Drop in Packed Bed Reactors
Ergun Equation | |
Variable Gas Density | |
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let | |
Catalyst Weight | |
where | |
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let | |
then | |
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English-
Espanol-
Svenska-
We will use this form for multiple reactions: | ||
We will use this form for single reactions: | ||
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Isothermal Operation | | |
recall that | | |
notice that | | |
The two expressions are coupled ordinary differential equations. We can solve them simultaneously using an ODE solver such as Polymath. For the special case of isothermal operation and epsilon = 0, we can obtain an analytical solution. Polymath will combine the mole balance, rate law and stoichiometry. |
Analytical Solution, [e], PFR with
Could now solve for X given W, or for W given X.
For gas phase reactions, as the pressure drop increases, the concentration decreases, resulting in a decreased rate of reaction, hence a lower conversion when compared to a reactor without a pressure drop.
Pressure Drop in a Packed Bed Reactor
Pressure and Reaction Orders
Formation of Ethyl Acetate
Here are some links to example problems dealing with packed bed reactors. You could also use these problems as self tests.
PBR Type 1 Home Problem
PBR Type 2 Home Problem
PBR Type 3 Home Problem
POLYMATH
Consider the following gas phase reaction carried out isothermally in a packed bed reactor. Pure A is fed at a rate of 2.5 moles/s and with
, and α = 0.0002 kg-1.
2AB
Mole Balance
Rate Law
Elementary
Stoichiometry
Gas with T = T0
AB/2
POLYMATH will combine everything - You do not need the combine step. Thank you POLYMATH
"What Four Things are Wrong with this Solution?"
Optimum Paritcle Diameter
Laminar Flow, Fix P0, ρ0,
ρ0 = P0(MW)/RT0
ρ0P0∼P0 2
Increasing the particle diameter descreases the pressure drop and increases the rate and conversion.
However, there is a competing effect. The specific reaction rate decreases as the particle size increases, therefore so deos the conversion.
k ∼ 1/Dp
| DP1 > DP2 Higher k, higher conversion |
The larger the particle, the more time it takes the reactant to get in and out of the catalyst particle. For a given catalyst weight, there is a greater external surgace area for smaller particles than larger particles. Therefore, there are more entry ways into the catalyst particle.
In CD-ROM chapter 12, we will learn that effectiveness factor decreases as the particle size increases
Engineering Analysis - Critical Thinking and Creative Thinking | top |
We want to learn how the various parameters (particle diameter, porosity, etc.) affect the pressure drop and hence conversion. We need to know how to respond to "What if" questions, such as:
"If we double the particle size, decrease the porosity by a factor of 3, and double the pipe size, what will happen to D P and X?"
(See Critical Thinking in Preface page xx. e.g., Questions the probe consenquences)
To answer these questions we need to see how a varies with these parameters.
Turbulent Flow
Compare Case 1 and Case 2:
For example, Case 1 might be our current situation and Case 2 might be the parameters we want to change to.
For constant mass flow through the system= constant
Laminar Flow
Effect of Reducing Particle Size on Conversion in a PBR
Here are more links to example problems dealing with packed bed reactors. Again, you could also use these problems as self tests.
PBR Type 5 Home Problem
PBR Type 7 Home Problem
PBR Type 8 Home Problem
Part 2: Measures Other Than Conversion
Measures Other Than Conversion | top |
Uses:
A. Membrane reactors
B. Multiple reaction
Liquids: Use concentrations, I.E. CA
1. For the elementary liquid phase reaction carried out in a CSTR, where V, vo, CAo, k, and Kc are given and the feed is pure A, the combined mole balance, rate laws, and stoichiometry are:
There are two equations, two unknowns, CA and CB
Gases: Use Molar Flow Rates, I.E. FI
2. If the above reaction, ,carried out in the gas phase in a PFR, where V, vo,CAo,k, and Kc are given and the feed is pure A, the combined mole balance, rate laws, and stoichiometry yield, for isothermal operation (T=To) and no pressure drop (DP=0) are:
Use Polymath to plot FA and FB down the length of the reactor.
Stoichiometry for Measures Other than Conversion
Gas Phase PFR
Liquid Phase CSTR
Use Creative and then Critical Thinking
What Four Things are Wrong With this solution?
Microreactors
For isothermal microreactors, we use the same equations as a PFR as long as the flow is not laminar. If the flow is laminar, we must use the techniques discussed in chapter 13. See example 4.8 of the text.
Membrane reactors can be used to achieve conversions greater than the original equilibrium value. These higher conversions are the result of Le Chatelier's Principle; you can remove one of the reaction products and drive the reaction to the right. To accomplish this, a membrane that is permeable to that reaction product, but is impermeable to all other species, is placed around the reacting mixture.
Example: The following reaction is to be carried out isothermally in a membrane reactor with no pressure drop. The membrane is permeable to Product C, but it is impermeable to all other species.
For membrane reactors, we cannot use conversion. We have to work in terms of the molar flow rates FA, FB, FC.
Polymath Program
Rate Laws | |
Stoichiometry | |
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Combine | Polymath will combine for you-- Thanks Polymath...you rock! |
Parameters | |
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Solve | Polymath |
"What four things are wrong with this membrane reactor solution?"
Here are links to example problems dealing with membrane reactors. You could also use these problems as self tests.
Membrane Type 4 Home Problem (Heterogeneous)
Membrane Type 4 Home Problem (Homogeneous)
Membrane Type 7 Home Problem
Membrane Type 8 Home Problem
Semibatch Reactors p. 190 | top |
Semibatch reactors can be very effective in maximizing selectivity in liquid phase reactions.
to Selectivity
The reactant that starts in the reactor is always the limiting reactant.
Three Forms of the Mole Balance Applied to Semibatch Reactors:
1. Molar Basis | | ||
2. Concentration Basis | | | |
3. Conversion | | |
For constant molar feed: | | |
For constant density: | | |
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|
Use the algorithm to solve the remainder of the problem.
Example: Elementary Irreversible Reaction
Consider the following irreversible elementary reaction:
-rA = kCACB
The combined mole balance, rate law, and stoichiometry may be written in terms of number of moles, conversion, and/or concentration:
Polymath Equations:
Conversion | Concentration | Moles |
d(X)/d(t) = -ra*V/Nao | d(Ca)/d(t) = ra - (Ca*vo)/V | d(Na)/d(t) = ra*V |
ra = -k*Ca*Cb | d(Cb)/d(t) = rb + ((Cbo-Cb)*vo)/V | d(Nb)/d(t) = rb*V + Fbo |
Ca = Nao*(1 - X)/V | ra = -k*Ca*Cb | ra = -k*Ca*Cb |
Cb = (Nbi + Fbo*t - Nao*X)/V | rb = ra | rb = ra |
V = Vo + vo*t | V = Vo + vo*t | V = Vo + vo*t |
Vo = 100 | Vo = 100 | Vo = 100 |
vo = 2 | vo = 2 | vo = 2 |
Nao = 100 | Fbo = 5 | Fbo = 5 |
Fbo = 5 | Nao = 100 | Ca = Na/V |
Nbi = 0 | Cbo = Fbo/vo | Cb = Nb/V |
k = 0.1 | k = 0.01 | k = 0.01 |
Na = Ca*V | ||
X = (Nao-Na)/Nao |
Conversion | Concentration |
Polymath Equations | Polymath Equations |
SummaryTable | Summary Table |
Conversion vs.Time | Conversion vs.Time |
Concentration vs.Time | Concentration vs.Time |
Volume vs.Time | Volume vs.Time |
Critical Thinking Questions
Equilibrium Conversion in Semibatch Reactors with Reversible Reactions
Consider the following reversible reaction:
Everything is the same as for the irreversible case, except for the rate law:
Where:
At equilibrium, -rA=0, then
Semibatch: A → B Acid Catalyzed
See Also:
Web Module on Reactive Distillation
Web Module on Wetlands
You Rate Some Wetlands Critical Thinking Questions
Object Assessment of Chapter 4
* All chapter references are for the 4th Edition of the text Elements of Chemical Reaction Engineering .
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Source: http://websites.umich.edu/~elements/course/lectures/four/index.htm
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