Isothermal Reactor Design - Reactor Engineering

CH4: Isothermal Reactor Design
RE4
Chemical Engineering Guy
www. Chemical Engineering Guy .com

Chemical Reaction Engineering Methodology
CH3: Elements of Chemical Reaction Engineering
H. Scott Fogler (4th Edition)

Content
• Section 1 Reactor Engineering Methodology
– In terms of Conversion
– Flow Rates and Concentration
• Section 2 Batch Reactor
– Batch Reactor & time of cycle
• Section 3 CSTR Design
– 1 CSTR and the Dahmköhler number
– Series
– Parallel
• Section 4 PFR Design
– Liquid-phase PFR
– Gaseous-phase PFR
• Section 5 PBR Design
– Pressure Drop in a PBR (one reaction)
• Section 6 Semi-Continuous Reactors
– Start-up of a CSTR (Unsteady state!)
– Semi-Batch Reactors (Multiple Reactions ONLY)

Section 1
Reactor Engineering Methodology

Reactor Engineering Methodology
• Using Conversion in our Design Equations
– CSTR (Single Reaction)
– Batch (Single Reaction)
– PFR (Single Reaction)
– PBR (Single Reaction)
• Using Flow/Concentration for Design…
– Semi-Continuous (Single and Multiple Reactions)
Due to the “Differential Equations” and many species involved…
Its easier to calculate conversion at the end

Methodology for Batch, CSTR, PFR
General Mole Balance Equation
Design Equation for Reactor Type
-ra = f(X)
given?
Determine Rate Law f(CA)
Use Stoichiometry Tables
Gas-Phase with Pressure Drop
Start
Combine:
• Mole Balance
• Design Equation
• Rate Law + Tables
• Pressure Drop
Solve them
If no change in moles and no Pressure Drop:
• Combine rate law and Stoichiometry Tables
• Get –ra = f(X)
Evaluate Equations. Solve. Analyze Data.
Get Final Answer
End
Yes
No
-ΔP

Methodology for PBR and SemiCont.
General Mole Balance Equation
Design Equation for Reactor Type
Determine Rate Law f(CA)
Use Stoichiometry Tables
Gas-Phase with Pressure Drop
Start
Combine:
• Mole Balance
• Design Equation
• Rate Law + Tables
• Pressure Drop
Solve (Software)
Analyze Data.
Get Final Answer
End
-ΔP
Relate Rates of Reaction

Section 2
Batch Reactor
Isothermal Design

Revisiting the Batch
• If liquid-phase
– Typical change in density may be neglected
• If gas-phase
– The volume of the vessel is fixed, no change in volume
• Assumptions
– Well mixed
– Reactants enter at the same time
– No side reactions
– Filling time may be neglected (tf = 0)
– Isothermal Operation

Revisiting the Batch
• For both the cases we use
constant volume:
– We will use Concentrations!
• This is the form we will use
for analyzing rate of
reaction data in the next
chapter

Example of time required for Conversion
First Order
• Given a First Order Elementary Reaction
-ra = k·CA
• Calculate the time needed to achieve certain
conversion  XA=90%

First Order

First Order
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First Order
• Substitute all values
• You DON’T need initial concentration!
• Note K values!

Second Order
• Given a Second Order Elementary Reaction
-ra = k·CA
2
• Calculate the time needed to achieve certain
conversion  XA=90%

Second Order

Second Order
• Substitute all Values
• It DOES depends on the Initial Concentration!
• Note the K values

Compare First vs. Second Order
reaction times (Batch Reactor)
• Note on Constant Values!
• Time of reaction decreases
• K value is adapted to the rate of reaction
– Must match dimensions (time, concentration,
volume, moles, etc)

• Time depends on initial concentration only for
2nd order
• Why does the 1st order does not depends on
concentration!?

• Procedure
– Get the “Design Equation” for the
Batch in terms of Concentration (or
Conversion for 1 rxn)
– If no rate of reaction vs. Conversion is
given
• You need a rate law
– Substitute the rate law in the Design
Equation
– Develop Mathematically
• Analytical solution if possible!
– Get the answer!

Example of time required for a Batch
• Imagine a third order, or even
a non-elementary order…
• Try those examples to practice
and compare!
• The more you practice the
math behind this, the more
you learn about reactions and
reactors

Exercise 4-1

Questions and Problems
• There are 33 problems in this Chapter 4.
• I also included some extra problems and
exercises
• All problems are solved in the next webpage
– www.ChemicalEngineeringGuy.com
• Courses
–Reactor Engineering
»Solved Problems Section
• CH4 – Isothermal Reactor Design

Reaction Time
• We speak of the Batch and all the time
required to the reactor to “react” the
materials
• This “time” is actually the “reaction time”
• It is not the TOTAL time needed to perform a
cycle
• Check out the Course for more problems like
this!
www.ChemicalEngineeringGuy.com

Cycle of a Batch
Time
tf
Time
th

Cycle of a Batch
Time
tr
Time
te

Cycle of a Batch
Time
tc
Time
tk

Cycle Time
• A normal cycle goes as:
– tf: time necessary to feed
– th: time necessary to heat/cool before RXN
– tr: time necessary to react that reaction
– te: time necessary to empty the reactor
– tk: time necessary to heat/cool after RXN
– tc: time necessary to clean the reactor
• The cycle starts again for a new batch:
Cycle Time = tf + th + tr + te + tk+ tc

Cycle Time
• The fraction of time required to do the actual
reaction vs. the total time
– Must be near 1.0 as possible
– If near 0, then the time we spend is mainly to
“prepare” the reactor for that specific reaction
• Be sure not to mix the times when given in data
– Reaction time, feeding time, time needed to clean,
time required to heat, time spent in maintenance, etc.
Ratio = Time of Reaction/Time of cycle

Cycle Time Exercise
• If
– tf: 25 min
– th: 2 hr
– tr: 6.7 hr
– te: 23 min
– tk: 1.2 hr
– tc: 30 min
• What is the total Cycle Time?
• What is the fraction of time of that reaction vs. batch time?

Cycle Time Exercise
• If
– tf: 25 min
– th: 2 hr
– tr: 6.7 hr
– te: 23 min
– tk: 1.2 hr
– tc: 30 min
a) 25+120+402+23+72+30 = 672 min
b) 402/672 = 0.598

Cycle Time Exercise
• If
– tf: 25 min
– th: 2 hr
– tr: 6.7 hr
– te: 23 min
– tk: 1.2 hr
– tc: 30 min
a) 25+120+402+23+72+30 = 672 min
b) 402/672 = 0.598
60% of time the reactor is having a reaction
40% is dead time

Section 3
Continuous Stirred Tank Reactor
Isothermal Design

Revisiting the CSTR
• Typical liquid-phase reactions!
• We will make the next
assumptions:
– Well mixed
– No change in volume/density
– Reactants enter at the same time
– No side reactions
– Filling time may be neglected (tf
= 0)

Space time + CSTR
• Lets force “Space Time” into our Design
Equations in the CSTR

First-Order Single CSTR

The Dahmköhler Number
• Dimensionless number
• “quick” estimate to know the degree of
conversion
• Ratio of “Rate of reaction at entrance” vs.
“Entering Flow Rate of A”
• Also ratio of “rate of reaction vs. convection
rate”

Dahmköhler for CSTR 1st Order

Dahmköhler for CSTR 2nd Order

Dahmköhler for CSTR nth Order
• Verify it by yourself…
• Try zeroth, third, and higher order…

The Dahmköhler Number
• Rule of Thumb for Da
– If Da > 10  Conversion may
achieve 90%
– If Da < 0.1  Conversion will
me max 10%
• Conversion in terms of Da Number 

CSTR and Da Number
• We will be using this for further analysis
• The Dahmköhler Number help us analyze “faster” and
“easier” 1st and 2nd Order reactions
• Specially for Series or Parallel CST-Reactors of the same
characteristics
– Size
– Temperature

CSTR in Series
• Suppose we got 2 CSTR
• Same Size (Volume)
• Same Temperature of Operation
• Same “k” or constant rate
• Series Arrangement (dependent of previous)

CSTR in Series
• Now lets suppose there are “n” reactors of
same characteristics

CSTR in Series

CSTR in Series
• We get this equation
• Obviously, as n increases, the conversion
increases
• If Da increases, conversion also increases!

CSTR in Series
Excel Spread Sheet Download in Web-Page!

Analysis of Number of Reactors
• We actually want Da increase, not n
• Da Number depends
– Volume of tank (generally fixed)
– “k” Constant… We can Increase Temperature!
– Volumetric Flow Rate  We can Adjust it

Analysis of Number of Reactors
• To increase conversion… the most normal
operation technique is:
– decrease volumetric flow rate (increase time in
reactor)
– increase temperature

CSTR in Parallel Arrangements
• Suppose we got n CSTR
• Same Size (Volume)
• Same Temperature of Operation
• Same “k” or constant rate
• Parallel Arrangement (independent of each other)

• Conversion WILL be the same (same reactors)
• Rate of Reaction WILL be the same
• Therefore, you need N tanks to get the total
Volume
– V = n·Vn
– Ft = n·FA0

• Its like having 3 identical reactors
• Same Volumes, Same Volumetric Flow, Same Flow Rates,
Same “k” Constant
• The Total Volume  3 times that RKT volume
• Total Flow Rate  3 times that RKT Flow Rate
• For n reactors  n times that…

• We’ve proved then that the parallel
arrangement would be the same if we would
actually have one LARGE reactor of that
Volume

Exercise

Section 4
Plug Flow Reactor
Isothermal Design

Liquid Phase PFR
• We will analyze two cases
– First Order Rate Law
– Second Order Rate Law
• Get equations in terms of Conversion & Da!

Liquid Phase PFR
• Assumptions
– Plug Flow Profile
– No dispersion or radial
gradients in Temp, Vel, Conc.
– No Pressure Drop and
Isothermal Operation
– Steady State
– Constant Volume/Density

Liquid Phase PFR: First Order

Liquid Phase PFR: Second Order

Liquid Phase PFR: Conclusion
• Its easy because Volume is Constant
• Main “problem”  the integral
• Check out for zero and third order!

Gas Phase PFR
• Typical Gas-phase operation
• Assumptions
– Turbulent Flow
– Plug Flow Profile
– No dispersion
– No radial gradients in Temp, Vel, Conc.
– No Pressure Drop
– Steady State

Gas Phase PFR
• We will analyze:
– First Order
– Second Order
• New Model for Concentration of A

Effect of Change in moles δ:
First Order
• Express this Equations in terms of
Concentration
• We analyze the effect of ε which is f(δ)

Gas Phase PFR: First Order

Integral from Appendix A-5

Conversion of A

Effect of Change in moles δ:
Second Order
• Express this Equations in terms of
Concentration
• We analyze the effect of ε which is f(δ)

Gas Phase PFR: Second Order

Integral from Appendix A-7

Conversion of A

Conclusion of PFR with change in Volume
• For negative changes in volume…
– You need less volume for the same conversion
• Economics favor this type of reactions
• If you are producing moles… you will need to
invest in a larger reactor
• Volume of Reactor Changes dramatically when
Second order
– Due to the exponent (square) in Concentration!

Conclusion of PFR with change in Volume
• For higher conversion, the Volume Required goes
exponential
Change in Volume
“punished” by factor of 1
Change in Volume
“punished” by factor of 2
twice!

Exercise 4-3 for PFR
• 4-3

Section 5
Packed Bed Reactor
Isothermal Design

PBR Revisited
• Typical gas-solid phase
reactions
• Packed Bed (catalyst on it)
• Liquid-solid may also be
used… there is no Pressure
Drop

PBR Revisited
• Gas-Solid interaction Drop of Pressure
• Drop Pressure due to the friction of solid-gas
• The higher the velocity, the higher the -ΔP

PBR Mole Balance + First Order
• Let’s Suppose we have a 1st order rate law.
• Get the Design Equation of a PBR in terms of
Conversion/Mass of Catalyst

PBR Mole Balance + First Order

PBR Mole Balance + Second Order
• Let’s Suppose we have a 2nd order rate law.
• Get the Design Equation of a PBR in terms of
Conversion/Mass of Catalyst

PBR Revisited
• One small detail…
• P changes as Conversion advances
• This conversion is dependent of the mass of
catalyst

PBR Revisited
• Pressure is a key factor now!
• We will need to model the
Pressure Drop so we can
accurately use this equation
• This means  Simultaneous
Solving!

Accounting for Pressure Drop
• As you may remember…
We have a change in Pressure
• If pressure changes  Concentration changes
• If concentration changes  rate of reaction changes
• If rate of reaction changes  concentration changes
First Order PBR Second Order PBR

Ergun Equation
• This is used in a packed bed or fluidized bed
reactors/towers
• Models the Pressure change vs. Length of
reactor/tower
• Laminar (term 1)
• Turbulent (term 2)
• Only gas density changes with Pressure Drop

Ergun Equation
• Definition of variables
• dP: Differential (change) of Pressure
• Dz: differential (change) of bed length
• G: mass flux (mass flow per unit area)
• gc: 1 for SI units (Force-weight ratio)
• DP: Particle/Pellet Diameter
• µ: viscosity of gas
• Ρ: gas density
• ϕ: Free space / Bed volume
• 1-ϕ: Volume of solids / Bed volume
• 150  Laminar Correction Factor
• 1.75  Turbulent Correction Factor

Ergun Equation for PBR
• From Steady State Mass balance

• Make a single constant!

• Change “length” of catalyst vs. “mass” of
catalyst

Bulk Density vs. Solid Density of Catalyst
• Solid Density is the “normal”
density you are used to
• Bulk Density includes “volume
spaces”
• Bulk Density is ALWAYS less than
Solid Density
• Porosity is taken into consideration
in Bulk Density

• Once again, use a constant
*We will revisit this model for Multiple Reactions!

• Let “y” be the P/P0

• Changing Flow Rates to Conversion

• We get this equation for One Reaction!
• For Isothermal Design

• We got this equation for Pressure drop vs.
mass of catalyst
• It is a Differential Equation!
• And this equation is dP/dW=F2(X,P)
– Depends on Conversion

Our PBR model
• We got a Two Coupled Differential Equation System
dX/dW = F1 (X,P)
dP/dW = F2 (X,P)
• Two Equations, Two Variables  Can be solved!
• They need “initial conditions” each
• How to solve:
– Analytical Methods “By hand” (not common)
– With Software (common and easier)

Analytical Methods: PBR
• If εX may be approximated to 0…
• We get this…

• Take in mind the constant values, alpha and beta

Numerical Methods: PBR
• If εX may NOT be approximated to 0…
• Use Euler Method

• EULER Method small review
• You should know by now that method!
• If you don’t know it… check your numerical
method course
• Check out Topic:
– Typical Numerical Methods for Solving Differential
Equations (1st Order)

• Euler methods…
• Runge-Kutta

Software Solving: PBR
• This is just an overview
• I see this type of problems in other course
• Computer Solving in Chemical Engineering

Software Solving: PBR
• Essentially:
– Set all constants with values
• R = 8.314
• Mass Flow = 4.5
– Set all variables to equations
• Volumetric Flow = Ideal Gas law
• Mass Flux = Mass Flow / Area
– Set a First Order Differential equation
• F1 (Rate Law + Design Equation + Stoichiometry)
• Set initial Point (e.g. X=0, W = W0)
– Set a Second Order Differential Equation
• F2 (Ergun Equation for PBR)
• Set Initial Point (e.g. P = P0, W = W0)
– Click “Run” to Solve in the Software

Exercise 4-4 of TextBook

Exercise
• 4-4, 4-5, and 4-6
• Are in the course are of the web-page
– www.ChemicalEngineeringGuy.com/Courses
• Check the Reactor Engineering Course
– Solved Problems Section

Section 6
Semi-Continuous Reactor
Isothermal Design

Quick Notes on Semi-Continuous Reactors
• Start-up of a CSTR
– Seen in this Chapter
– Helps to see the “basics”
• Semi-batch Operation for Multiple Reactions
– Not shown in this Chapter
– You need to know the fundamentals of Multiple
Reactions!
– Multiple Reactions  CH6

Start-up of a CSTR
• CSTR are always continuous operation
• To get to this “steady state” you need to “star-up”
• This process means
– Start from some initial conditions to final conditions
– The final conditions are the “steady state” conditions

Start-up of a CSTR
• This is done because
– New Process
– New Equipment Installed
– Quality/Maintenance shut down
– Electrical Failure  Shut Down
– Scale-Up or Scale-Down

Start-up of a CSTR
• Time necessary to achieve Steady state
• Concentration and Conversion  function of time!
• Analytical Solutions  Zeroth and First Order Rates
• ODE  Superior Orders (2nd and up)

Start-up of a CSTR
• The Mole Balance Equation “Modified”
• Conversion  we cannot account it because
of the accumulation!
• Use concentration (Methodology 2)
• We will suppose 99% of Steady State
Concentration is when we achieve S-S

Start-up of a CSTR: 1st Order
• For a First Order Reaction

Start-up of a CSTR: 1st Order
• For First Order Reactions 

Time needed for Steady State
• We will suppose 99% of Steady State
Concentration is when we achieve S-S

Time needed for Steady State
• k·τ>>1 then model as ts= 4.6/τ
• k·τ<<1 then model as ts = 4.6τ

Exercise: First Order
• Reaction: AB
– Elementary Rate of Reaction
– k = 2.2 dm3/ s·mol
– V rate = 0.05 dm3/s
– V0 = 2.5 dm3
– CA0 = 0.05 mol/dm3
• Calculate the time needed to achieve Steady State

Semi-batch Reactor
• Will be analyzed after CH6: Multiple Reactions

End of Block RE4
• You are now prepared to model Isothermal
Reactors with one reaction!
• You understand now when to apply the Design
Equations
• You know why is it important to study rates of
rection laws
• You know the methodology and why we need
a Design Equation + Reaction Rate Data

End of Block RE4
• You know when it is convinient to work with
conversion, mole flow, concentration
• You can model Batch in 1st, 2nd and you could
model higher rates!
• Now you know Dahmlköhler Number and its
importance in Reactor Engineering
• You can model CSTR with Da Number
• You can Model Multiple CSTR in Series and
Parallel Arrangements

End of Block RE4
• You can model PFR in liquid and gas phase!
• You know that in PBR there is a drop in
pressure
• You can model that drop of pressure with
Ergun Equation!
• You know how to solve a ODE for a PBR in
terms of conversion and pressure drop

End of Block RE4
• You know now how to calculate stability times
for CSTR Starting up!
• For next chapter, we will analyze the Data for
Rate of Reactions!

More Information…
• Get extra information here!
– Directly on the WebPage:
• www.ChemicalEngineeringGuy.com/courses
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• www.facebook.com/Chemical.Engineering.Guy
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Text Book & Reference
Essentials of Chemical
Reaction Engineering
H. Scott Fogler (1st Edition)
Chemical Reactor
Analysis and Design
Fundamentals
J.B. Rawlings and J.G.
Ekerdt (1st Edition)
Elements of Chemical
Reaction Engineering

Bibliography
Elements of Chemical Reaction Engineering
We’ve seen  CH4

Isothermal Reactor Design - Reactor Engineering

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Isothermal Reactor Design - Reactor Engineering