Isothermal Reactor Design  Reactor Engineering

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CH4: Isothermal Reactor DesignRE4Chemical Engineering Guy
www. Chemical Engineering Guy .comChemical Reaction Engineering Methodology
www. Chemical Engineering Guy .comCH3: Elements of Chemical Reaction EngineeringH. Scott Fogler (4th Edition)Chemical Reaction Engineering Methodology
www. Chemical Engineering Guy .comCH3: Elements of Chemical Reaction EngineeringH. Scott Fogler (4th Edition)ContentSection 1 Reactor Engineering MethodologyIn terms of ConversionFlow Rates and ConcentrationSection 2 Batch ReactorBatch Reactor & time of cycleSection 3 CSTR Design1 CSTR and the Dahmkhler numberSeriesParallelSection 4 PFR DesignLiquidphase PFRGaseousphase PFRSection 5 PBR DesignPressure Drop in a PBR (one reaction)Section 6 SemiContinuous ReactorsStartup of a CSTR (Unsteady state!)SemiBatch Reactors (Multiple Reactions ONLY)www. Chemical Engineering Guy .comSection 1
Reactor Engineering Methodologywww. Chemical Engineering Guy .comReactor Engineering MethodologyUsing Conversion in our Design EquationsCSTR (Single Reaction)Batch (Single Reaction)PFR (Single Reaction)PBR (Single Reaction)Using Flow/Concentration for DesignSemiContinuous (Single and Multiple Reactions)Due to the Differential Equations and many species involved
Its easier to calculate conversion at the endwww. Chemical Engineering Guy .com
Methodology for Batch, CSTR, PFRGeneral Mole Balance EquationDesign Equation for Reactor Typera = f(X) given?Determine Rate Law f(CA)Use Stoichiometry TablesGasPhase with Pressure DropStartCombine:Mole BalanceDesign EquationRate Law + TablesPressure DropSolve themIf no change in moles and no Pressure Drop:Combine rate law and Stoichiometry TablesGet ra = f(X)Evaluate Equations. Solve. Analyze Data.Get Final AnswerEndYesNoPwww. Chemical Engineering Guy .comMethodology for PBR and SemiCont.General Mole Balance EquationDesign Equation for Reactor TypeDetermine Rate Law f(CA)Use Stoichiometry TablesGasPhase with Pressure DropStartCombine:Mole BalanceDesign EquationRate Law + TablesPressure DropSolve (Software)Analyze Data.Get Final AnswerEndPwww. Chemical Engineering Guy .comRelate Rates of ReactionSection 2
Batch ReactorIsothermal Designwww. Chemical Engineering Guy .comMethodology for Batch, CSTR, PFRGeneral Mole Balance EquationDesign Equation for Reactor Typera = f(X) given?Determine Rate Law f(CA)Use Stoichiometry TablesGasPhase with Pressure DropStartCombine:Mole BalanceDesign EquationRate Law + TablesPressure DropSolve themIf no change in moles and no Pressure Drop:Combine rate law and Stoichiometry TablesGet ra = f(X)Evaluate Equations. Solve. Analyze Data.Get Final AnswerEndYesNoPwww. Chemical Engineering Guy .comRevisiting the BatchIf liquidphaseTypical change in density may be neglectedIf gasphaseThe volume of the vessel is fixed, no change in volumeAssumptionsWell mixedReactants enter at the same timeNo side reactionsFilling time may be neglected (tf = 0)Isothermal Operation
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Revisiting the BatchFor 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 chapterwww. Chemical Engineering Guy .com
Example of time required for ConversionFirst OrderGiven a First Order Elementary Reactionra = kCACalculate the time needed to achieve certain conversion XA=90%www. Chemical Engineering Guy .com
Example of time required for ConversionFirst Orderwww. Chemical Engineering Guy .com
Example of time required for ConversionFirst Orderwww. Chemical Engineering Guy .com
Example of time required for ConversionFirst Orderwww. Chemical Engineering Guy .com
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Example of time required for ConversionFirst Orderwww. Chemical Engineering Guy .com
Substitute all valuesYou DONT need initial concentration!Note K values!Example of time required for ConversionSecond OrderGiven a Second Order Elementary Reactionra = kCA2Calculate the time needed to achieve certain conversion XA=90%www. Chemical Engineering Guy .com
Example of time required for ConversionSecond Orderwww. Chemical Engineering Guy .com
Example of time required for ConversionSecond Orderwww. Chemical Engineering Guy .com
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Example of time required for ConversionSecond Orderwww. Chemical Engineering Guy .com
Example of time required for ConversionSecond Orderwww. Chemical Engineering Guy .com
Substitute all ValuesIt DOES depends on the Initial Concentration!Note the K valuesCompare First vs. Second Order reaction times (Batch Reactor)Note on Constant Values!Time of reaction decreasesK value is adapted to the rate of reactionMust match dimensions (time, concentration, volume, moles, etc)www. Chemical Engineering Guy .com
Compare First vs. Second Order reaction times (Batch Reactor)Time depends on initial concentration only for 2nd orderWhy does the 1st order does not depends on concentration!?www. Chemical Engineering Guy .com
Compare First vs. Second Order reaction times (Batch Reactor)ProcedureGet the Design Equation for the Batch in terms of Concentration (or Conversion for 1 rxn)If no rate of reaction vs. Conversion is givenYou need a rate lawSubstitute the rate law in the Design EquationDevelop MathematicallyAnalytical solution if possible!Get the answer!
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Example of time required for a BatchImagine a third order, or even a nonelementary orderTry those examples to practice and compare!The more you practice the math behind this, the more you learn about reactions and reactorswww. Chemical Engineering Guy .com
Exercise 41
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Questions and ProblemsThere are 33 problems in this Chapter 4.I also included some extra problems and exercisesAll problems are solved in the next webpagewww.ChemicalEngineeringGuy.comCoursesReactor EngineeringSolved Problems SectionCH4 Isothermal Reactor Designwww. Chemical Engineering Guy .comReaction TimeWe speak of the Batch and all the time required to the reactor to react the materialsThis time is actually the reaction timeIt is not the TOTAL time needed to perform a cycleCheck out the Course for more problems like this!www.ChemicalEngineeringGuy.com
www. Chemical Engineering Guy .comCycle of a Batchwww. Chemical Engineering Guy .com
TimetfTimethCycle of a Batchwww. Chemical Engineering Guy .com
TimetrTimeteCycle of a Batchwww. Chemical Engineering Guy .com
TimetcTimetkCycle TimeA normal cycle goes as:tf: time necessary to feedth: time necessary to heat/cool before RXNtr: time necessary to react that reactionte: time necessary to empty the reactortk: time necessary to heat/cool after RXNtc: time necessary to clean the reactor
The cycle starts again for a new batch:Cycle Time = tf + th + tr + te + tk+ tcwww. Chemical Engineering Guy .com
Cycle TimeThe fraction of time required to do the actual reaction vs. the total time
Must be near 1.0 as possibleIf 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 dataReaction time, feeding time, time needed to clean, time required to heat, time spent in maintenance, etc.Ratio = Time of Reaction/Time of cyclewww. Chemical Engineering Guy .comCycle Time ExerciseIftf: 25 minth: 2 hrtr: 6.7 hrte: 23 mintk: 1.2 hrtc: 30 min
What is the total Cycle Time?What is the fraction of time of that reaction vs. batch time?
Ratio = Time of Reaction/Time of cyclewww. Chemical Engineering Guy .com
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Cycle Time ExerciseIftf: 25 minth: 2 hrtr: 6.7 hrte: 23 mintk: 1.2 hrtc: 30 min
What is the total Cycle Time?What is the fraction of time of that reaction vs. batch time?
www. Chemical Engineering Guy .comRatio = Time of Reaction/Time of cycle
25+120+402+23+72+30 = 672 min
402/672 = 0.598This Material is only Available at
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Cycle Time ExerciseIftf: 25 minth: 2 hrtr: 6.7 hrte: 23 mintk: 1.2 hrtc: 30 min
What is the total Cycle Time?What is the fraction of time of that reaction vs. batch time?
www. Chemical Engineering Guy .comRatio = Time of Reaction/Time of cycle
25+120+402+23+72+30 = 672 min
402/672 = 0.598
60% of time the reactor is having a reaction40% is dead timeThis Material is only Available at
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Questions and ProblemsThere are 33 problems in this Chapter 4.I also included some extra problems and exercisesAll problems are solved in the next webpagewww.ChemicalEngineeringGuy.comCoursesReactor EngineeringSolved Problems SectionCH4 Isothermal Reactor Designwww. Chemical Engineering Guy .comSection 3
Continuous Stirred Tank Reactor Isothermal Designwww. Chemical Engineering Guy .comMethodology for Batch, CSTR, PFRGeneral Mole Balance EquationDesign Equation for Reactor Typera = f(X) given?Determine Rate Law f(CA)Use Stoichiometry TablesGasPhase with Pressure DropStartCombine:Mole BalanceDesign EquationRate Law + TablesPressure DropSolve themIf no change in moles and no Pressure Drop:Combine rate law and Stoichiometry TablesGet ra = f(X)Evaluate Equations. Solve. Analyze Data.Get Final AnswerEndYesNoPwww. Chemical Engineering Guy .comRevisiting the CSTRTypical liquidphase reactions!We will make the next assumptions:Well mixedNo change in volume/densityReactants enter at the same timeNo side reactionsFilling time may be neglected (tf = 0)Isothermal Operation
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Space time + CSTRLets force Space Time into our Design Equations in the CSTR
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FirstOrder Single CSTRwww. Chemical Engineering Guy .com
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The Dahmkhler NumberDimensionless numberquick estimate to know the degree of conversionRatio of Rate of reaction at entrance vs. Entering Flow Rate of AAlso ratio of rate of reaction vs. convection ratewww. Chemical Engineering Guy .com
Dahmkhler for CSTR 1st Order
www. Chemical Engineering Guy .comDahmkhler for CSTR 2nd Order
www. Chemical Engineering Guy .comDahmkhler for CSTR nth OrderVerify it by yourselfTry zeroth, third, and higher order
www. Chemical Engineering Guy .comThe Dahmkhler NumberRule of Thumb for DaIf Da > 10 Conversion may achieve 90%If Da < 0.1 Conversion will me max 10%Conversion in terms of Da Number www. Chemical Engineering Guy .com
CSTR and Da NumberWe will be using this for further analysis
The Dahmkhler Number help us analyze faster and easier 1st and 2nd Order reactions
Specially for Series or Parallel CSTReactors of the same characteristicsSizeTemperature
CSTR in SeriesSuppose we got 2 CSTRSame Size (Volume)Same Temperature of OperationSame k or constant rate Series Arrangement (dependent of previous)
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CSTR in SeriesNow lets suppose there are n reactors of same characteristics
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CSTR in Serieswww. Chemical Engineering Guy .com
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CSTR in Serieswww. Chemical Engineering Guy .com
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CSTR in Series
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CSTR in Series
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CSTR in SeriesWe get this equation
Obviously, as n increases, the conversion increases
If Da increases, conversion also increases!www. Chemical Engineering Guy .com
CSTR in Serieswww. Chemical Engineering Guy .com
Excel Spread Sheet Download in WebPage!Analysis of Number of ReactorsWe actually want Da increase, not nDa Number dependsVolume of tank (generally fixed)k Constant We can Increase Temperature!Volumetric Flow Rate We can Adjust it
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Analysis of Number of ReactorsTo increase conversion the most normal operation technique is:decrease volumetric flow rate (increase time in reactor)increase temperature
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CSTR in Parallel ArrangementsSuppose we got n CSTRSame Size (Volume)Same Temperature of OperationSame k or constant rate Parallel Arrangement (independent of each other)www. Chemical Engineering Guy .com
CSTR in Parallel ArrangementsConversion WILL be the same (same reactors)Rate of Reaction WILL be the sameTherefore, you need N tanks to get the total VolumeV = nVnFt = nFA0www. Chemical Engineering Guy .com
CSTR in Parallel ArrangementsIts like having 3 identical reactorsSame Volumes, Same Volumetric Flow, Same Flow Rates, Same k ConstantThe Total Volume 3 times that RKT volumeTotal Flow Rate 3 times that RKT Flow Rate
For n reactors n times that
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CSTR in Parallel ArrangementsWeve proved then that the parallel arrangement would be the same if we would actually have one LARGE reactor of that Volume
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CSTR in Parallel Arrangements
Exercise
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Questions and ProblemsThere are 33 problems in this Chapter 4.I also included some extra problems and exercisesAll problems are solved in the next webpagewww.ChemicalEngineeringGuy.comCoursesReactor EngineeringSolved Problems SectionCH4 Isothermal Reactor Designwww. Chemical Engineering Guy .comSection 4
Plug Flow ReactorIsothermal Designwww. Chemical Engineering Guy .comMethodology for Batch, CSTR, PFRGeneral Mole Balance EquationDesign Equation for Reactor Typera = f(X) given?Determine Rate Law f(CA)Use Stoichiometry TablesGasPhase with Pressure DropStartCombine:Mole BalanceDesign EquationRate Law + TablesPressure DropSolve themIf no change in moles and no Pressure Drop:Combine rate law and Stoichiometry TablesGet ra = f(X)Evaluate Equations. Solve. Analyze Data.Get Final AnswerEndYesNoPwww. Chemical Engineering Guy .comLiquid Phase PFRWe will analyze two casesFirst Order Rate LawSecond Order Rate LawGet equations in terms of Conversion & Da!www. Chemical Engineering Guy .com
Liquid Phase PFRAssumptionsPlug Flow ProfileNo dispersion or radial gradients in Temp, Vel, Conc.No Pressure Drop and Isothermal OperationSteady StateConstant Volume/Densitywww. Chemical Engineering Guy .com
Liquid Phase PFR: First Orderwww. Chemical Engineering Guy .com
Liquid Phase PFR: First Orderwww. Chemical Engineering Guy .com
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Liquid Phase PFR: Second Orderwww. Chemical Engineering Guy .com
Liquid Phase PFR: Second Orderwww. Chemical Engineering Guy .com
Liquid Phase PFR: ConclusionIts easy because Volume is ConstantMain problem the integral
Check out for zero and third order!www. Chemical Engineering Guy .com
Questions and ProblemsThere are 33 problems in this Chapter 4.I also included some extra problems and exercisesAll problems are solved in the next webpagewww.ChemicalEngineeringGuy.comCoursesReactor EngineeringSolved Problems SectionCH4 Isothermal Reactor Designwww. Chemical Engineering Guy .comGas Phase PFRTypical Gasphase operationAssumptionsTurbulent FlowPlug Flow ProfileNo dispersionNo radial gradients in Temp, Vel, Conc.No Pressure DropIsothermal OperationSteady State
www. Chemical Engineering Guy .comGas Phase PFRWe will analyze:First OrderSecond OrderNew Model for Concentration of Awww. Chemical Engineering Guy .com
Effect of Change in moles : First OrderExpress this Equations in terms of Concentration
We analyze the effect of which is f()www. Chemical Engineering Guy .com
Gas Phase PFR: First Orderwww. Chemical Engineering Guy .com
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Gas Phase PFR: First Orderwww. Chemical Engineering Guy .com
Integral from Appendix A5Gas Phase PFR: First Orderwww. Chemical Engineering Guy .com
Conversion of AEffect of Change in moles : Second OrderExpress this Equations in terms of Concentration
We analyze the effect of which is f()www. Chemical Engineering Guy .com
Gas Phase PFR: Second Order
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Gas Phase PFR: Second Order
Integral from Appendix A7www. Chemical Engineering Guy .comGas Phase PFR: Second Order
www. Chemical Engineering Guy .comConversion of AConclusion of PFR with change in Volumewww. Chemical Engineering Guy .comFor negative changes in volumeYou need less volume for the same conversionEconomics favor this type of reactionsIf you are producing moles you will need to invest in a larger reactorVolume of Reactor Changes dramatically when Second orderDue to the exponent (square) in Concentration!
Conclusion of PFR with change in Volumewww. Chemical Engineering Guy .comFor higher conversion, the Volume Required goes exponential
Change in Volume punished by factor of 1Change in Volume punished by factor of 2 twice!Exercise 43 for PFR43
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Questions and ProblemsThere are 33 problems in this Chapter 4.I also included some extra problems and exercisesAll problems are solved in the next webpagewww.ChemicalEngineeringGuy.comCoursesReactor EngineeringSolved Problems SectionCH4 Isothermal Reactor Designwww. Chemical Engineering Guy .comSection 5
Packed Bed ReactorIsothermal Designwww. Chemical Engineering Guy .comMethodology for PBR and SemiCont.General Mole Balance EquationDesign Equation for Reactor TypeDetermine Rate Law f(CA)Use Stoichiometry TablesGasPhase with Pressure DropStartCombine:Mole BalanceDesign EquationRate Law + TablesPressure DropSolve (Software)Analyze Data.Get Final AnswerEndPwww. Chemical Engineering Guy .comRelate Rates of ReactionPBR RevisitedTypical gassolid phase reactionsPacked Bed (catalyst on it)Liquidsolid may also be used there is no Pressure Dropwww. Chemical Engineering Guy .com
PBR RevisitedGasSolid interaction Drop of PressureDrop Pressure due to the friction of solidgasThe higher the velocity, the higher the Pwww. Chemical Engineering Guy .com
PBR Mole Balance + First OrderLets Suppose we have a 1st order rate law.Get the Design Equation of a PBR in terms of Conversion/Mass of Catalyst
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PBR Mole Balance + First Order
www. Chemical Engineering Guy .comPBR Mole Balance + First Order
www. Chemical Engineering Guy .comPBR Mole Balance + Second OrderLets Suppose we have a 2nd order rate law.Get the Design Equation of a PBR in terms of Conversion/Mass of Catalyst
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PBR Mole Balance + Second Order
www. Chemical Engineering Guy .comPBR Mole Balance + Second Order
PBR RevisitedOne small detailP changes as Conversion advancesThis conversion is dependent of the mass of catalystwww. Chemical Engineering Guy .com
PBR RevisitedPressure is a key factor now!We will need to model the Pressure Drop so we can accurately use this equationThis means Simultaneous Solving!www. Chemical Engineering Guy .com
Accounting for Pressure DropAs you may remember
We have a change in PressureIf pressure changes Concentration changesIf concentration changes rate of reaction changesIf rate of reaction changes concentration changeswww. Chemical Engineering Guy .com
First Order PBRSecond Order PBRErgun EquationThis is used in a packed bed or fluidized bed reactors/towersModels the Pressure change vs. Length of reactor/tower
Laminar (term 1)Turbulent (term 2)Only gas density changes with Pressure Drop
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Ergun EquationDefinition of variables
dP: Differential (change) of PressureDz: differential (change) of bed lengthG: mass flux (mass flow per unit area)gc: 1 for SI units (Forceweight ratio)DP: Particle/Pellet Diameter: viscosity of gas: gas density: Free space / Bed volume1: Volume of solids / Bed volume150 Laminar Correction Factor1.75 Turbulent Correction Factor
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Ergun Equation for PBRFrom Steady State Mass balancewww. Chemical Engineering Guy .com
Ergun Equation for PBR
Ergun Equation for PBRMake a single constant!
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Ergun Equation for PBRChange length of catalyst vs. mass of catalyst
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Bulk Density vs. Solid Density of CatalystSolid Density is the normal density you are used toBulk Density includes volume spacesBulk Density is ALWAYS less than Solid DensityPorosity is taken into consideration in Bulk Density
Ergun Equation for PBRwww. Chemical Engineering Guy .com
Ergun Equation for PBROnce again, use a constant
*We will revisit this model for Multiple Reactions!www. Chemical Engineering Guy .com
Ergun Equation for PBRLet y be the P/P0www. Chemical Engineering Guy .com
Ergun Equation for PBRChanging Flow Rates to Conversion
www. Chemical Engineering Guy .comErgun Equation for PBRWe get this equation for One Reaction!
For Isothermal Design
www. Chemical Engineering Guy .comErgun Equation for PBRWe 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 Conversionwww. Chemical Engineering Guy .com
Our PBR modelWe got a Two Coupled Differential Equation SystemdX/dW = F1 (X,P)dP/dW = F2 (X,P)
Two Equations, Two Variables Can be solved!They need initial conditions eachHow to solve:Analytical Methods By hand (not common)With Software (common and easier)
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Analytical Methods: PBRIf X may be approximated to 0We get thiswww. Chemical Engineering Guy .com
Analytical Methods: PBR
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Analytical Methods: PBRTake in mind the constant values, alpha and beta
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Numerical Methods: PBRIf X may NOT be approximated to 0Use Euler Methodwww. Chemical Engineering Guy .com
Numerical Methods: PBREULER Method small reviewYou should know by now that method!If you dont know it check your numerical method courseCheck out Topic:Typical Numerical Methods for Solving Differential Equations (1st Order)www. Chemical Engineering Guy .com
Numerical Methods: PBREuler methods RungeKutta
Software Solving: PBRThis is just an overviewI see this type of problems in other course
Computer Solving in Chemical Engineeringwww. Chemical Engineering Guy .com
Software Solving: PBREssentially:Set all constants with valuesR = 8.314Mass Flow = 4.5Set all variables to equationsVolumetric Flow = Ideal Gas law Mass Flux = Mass Flow / AreaSet a First Order Differential equationF1 (Rate Law + Design Equation + Stoichiometry)Set initial Point (e.g. X=0, W = W0)Set a Second Order Differential EquationF2 (Ergun Equation for PBR)Set Initial Point (e.g. P = P0, W = W0)Click Run to Solve in the Softwarewww. Chemical Engineering Guy .com
Exercise 44 of TextBook
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Exercise 45 of TextBook
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Exercise 46 of TextBook
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Exercise44, 45, and 46Are in the course are of the webpage
www.ChemicalEngineeringGuy.com/CoursesCheck the Reactor Engineering CourseSolved Problems Sectionwww. Chemical Engineering Guy .comQuestions and ProblemsThere are 33 problems in this Chapter 4.I also included some extra problems and exercisesAll problems are solved in the next webpagewww.ChemicalEngineeringGuy.comCoursesReactor EngineeringSolved Problems SectionCH4 Isothermal Reactor Designwww. Chemical Engineering Guy .comSection 6
SemiContinuous ReactorIsothermal Designwww. Chemical Engineering Guy .comMethodology for PBR and SemiCont.General Mole Balance EquationDesign Equation for Reactor TypeDetermine Rate Law f(CA)Use Stoichiometry TablesGasPhase with Pressure DropStartCombine:Mole BalanceDesign EquationRate Law + TablesPressure DropSolve (Software)Analyze Data.Get Final AnswerEndPwww. Chemical Engineering Guy .comRelate Rates of ReactionQuick Notes on SemiContinuous ReactorsStartup of a CSTRSeen in this ChapterHelps to see the basicsSemibatch Operation for Multiple ReactionsNot shown in this ChapterYou need to know the fundamentals of Multiple Reactions!Multiple Reactions CH6www. Chemical Engineering Guy .com
Startup of a CSTRCSTR are always continuous operationTo get to this steady state you need to starupThis process meansStart from some initial conditions to final conditionsThe final conditions are the steady state conditionswww. Chemical Engineering Guy .com
Startup of a CSTRThis is done becauseNew ProcessNew Equipment InstalledQuality/Maintenance shut downElectrical Failure Shut DownScaleUp or ScaleDown
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Startup of a CSTRTime necessary to achieve Steady stateConcentration and Conversion function of time!Analytical Solutions Zeroth and First Order RatesODE Superior Orders (2nd and up)www. Chemical Engineering Guy .com
Startup of a CSTRThe 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 SS
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Startup of a CSTR: 1st OrderFor a First Order Reactionwww. Chemical Engineering Guy .com
Startup of a CSTR: 1st OrderFor First Order Reactions
Startup of a CSTR: 1st Order
Time needed for Steady StateWe will suppose 99% of Steady State Concentration is when we achieve SS
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Time needed for Steady State
Time needed for Steady Statek>>1 then model as ts= 4.6/
k