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After this topic, you should be able to:
Analysis of Continuous Reactors Topic 6 Ch.E. 422 Part 1- Basic Concepts and Graphical Method
Review of Continuous Reactors
Mixed Flow Reactor (MFR)/Constant Stirred Tank Tank Reactor (CSTR) •
•
Contents are well mixed so uniform concentration concentration inside the tank, and equal to product concentrations May be connected in series to i ncrease the the conversion
Plug Flow (PFR)/Tubular Reactor (TR) •
•
Concentration varies with length of the reactor, but not radially Has higher conversion per volume than the MFR or CSTR
Apply the mole balance and its applications to steady state: •
•
•
Constant Stirred Tank Reactor (CSTR) or Mixed Flow Reactor (MFR) Plug Flow Reactor (PFR) or Tubular Reactor (TR) Combination of CSTR and PFR
Use both graphical and numerical or algebraic methods to solve the applications
MFR or CSTR Reactants are continuously fed to a well mixed tank and the products are also withdrawn continuously. At steady state, the concentration of any material in the tank is also the same as that in the product stream.
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PFR or TR
Common Nomenclature
Reactants are continuously fed to a tube so that what goes in first also comes out first. Reaction proceeds across the reactor length and concentrations concentrations vary from entrance to exit as there is no lateral mixing. Radial concentrations are however assumed constant.
MFR or CSTR
FJo = input molar rate of J = v oCJo FJf = output molar rate of J = v f CJf vo, vf = volumetric feed & exit rates CJo;CJf = inlet & outlet outl et concentrations -rJ = rate of reaction of J XJ = (FJo – FJf )/FJo = fract. conv. of J τ = Space Time = time to process one reactor volume of feed = V/v o V = reactor volume θ = residence time = V/v f
Plug Flow Reactor (PFR)
Steady State Flow: Steady State Flow:
No spatial variation:
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Volume Formulas for Steady State
Plug Flow Reactor (PFR)
Single MFR or CSTR V1
FJo FJ1
rJ1
Vm
FJm1 FJm
Determination of Reactor Volume •
•
•
•
•
Determination of Reaction Kinetics •
Unknown Order and Rate Constant
v0 (CJo CJ1) Jo
r J1
1
CJo CJ1
r J1
FJo ( XJm1 X Jm )
r Jm
Vm
v 0 (CJm1 CJm )
r Jm
CJm1 CJm
m
r Jm
FJf
0
dFJ r J
Vr v o
CJf
C Ao
dCJ r J
r
CJf
C Ao
dCJ r J
Solution Methods
Single MFR or CSTR Equivalent PFR Combination of Reactors Single MFR or CSTR Equivalent PFR Combination of Reactors
r J1
V1
PFR or TR
Graphical •
•
Determination of Over-all Conversion •
rJm
Vr
FJo X Ji
Multiple MFR or CSTR in Series
Covered Applicati Applications ons
Applicable if data of –rJ and CJ or XJ are available Uses Levenspiel plots (Plot of 1/-rJ vs CJ or XJ
Numerical or Algebraic Methods •
Applicable if the kinetics of the reaction is known or unknown, meaning the order, rate constants are available or unknown
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Graphical Methods
By Levenspiel •
Graphical Method
Volume of a CSTR is represented partly by the area of rectangles bounded by the plot of 1/-rJ vs XJ, with height equal to 1/-rJ m and
width as ∆XJm •
Volume of a PFR is represented partly by the area under the curve bounded by a plot of 1/-rJ vs XJ between 0 and XJ.
Graphical Method
Given the following
By Walas (For CSTR only) A plot of –rJ vs CJ is used. Each CSTR is represented by a straight line with an x-intercept of CJm-1 and slope -1/τm For multiple CSTRs of same size, the slopes of each line must be the same.
Levenspiel Plot (1/-rA vs XA)
–rA-CA Data
Fao = vo Cao = 12.5
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Examples:
Walas Plot
1. Find the the volume volume of a single single CSTR CSTR needed needed to achieve 80% conversion of A. 2. Determine Determine the the volume of a PFR to to achieve achieve the same conversion in (1). 3. Suppose Suppose 2 CSTRs CSTRs are connec connected ted in series. series. What total volume is needed for the same conversion? 4. Suppose Suppose equal equal sized CSTR CSTR and PFR PFR are connected in series. What total volume i s needed for the same conversion if CSTR is ahead and if PFR is ahead? 5. Repeat Repeat 1-4 1-4 if the total total volume volume is to to be 60 ft3 and the fi nal conversion conversion is unknown.
Solution: Problem 1 F FJ1 FJo XJ1 V1 Jo rJ1 r J1
Solution: Problem 1 V1
FJo FJ1 FJo XJ1 rJ1 r J1
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Calculations for Problem 1
Solution to Problem 2: Vr
Solution to Problem 2: Vr
F Af
0
Xaf dX dF A A FAo 0 r A r A
F Af
0
Xaf dX dF A A FAo 0 r A r A
Solution to Problem 2: Vr
F Af
0
Xaf dX dF A A FAo 0 r A r A
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Calculations for Problem 2
Solution: Problem 3 (Lev Plot) V1
Solution: Problem 3 (Lev Plot) V1
FJo XJ1 F F F (1 XJ1) FJo(1 X J2 ) FJo ( XJ 2 X j1) V2 J1 J2 Jo rJ1 rJ2 rJ2 rJ2
FJo XJ1
rJ1
V2
FJ1 FJ2
rJ2
FJo(1 X J1) FJo(1 X J2 )
rJ2
FJo ( XJ 2 X j1)
rJ2
Solution: Problem 3 (Lev Plot) V1
FJo XJ1 F F F (1 X J1) FJo(1 X J2 ) FJo ( XJ 2 X j1) V2 J1 J2 Jo rJ1 rJ2 rJ2 rJ2
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Calculations (Levenspiel Method):
Another Method (by Walas)
Another Method (by Walas)
Another Method (by Walas)
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Another Method (by Walas)
Calculations Calculatio ns (Walas Method)
Solution: Problem 4a(CSTR to PFR)
Solution: Problem 4a(CSTR to PFR)
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Solution: Problem 4a(CSTR to PFR)
Calculations (Problem 4a)
Solution: Problem 4b(PFR to CSTR)
Solution: Problem 4b(PFR to CSTR)
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Solution: Problem 4b(PFR to CSTR)
Solution: Problem 4b(PFR to CSTR)
Calculations (Problem 4b)
Solution Algorithm for Prob 5 (For application and submission)
SAME as CSTR to PFR!
Prob 1: Assume Xa1, get -ra1 from curve; check: Xa1=V1(-ra1)/Xa1 Prob 2: Assume Xa1, get area under curve of Levenspiel plot from 0 to Xa1. Check if V1=Area(Fao), V1=Area(Fao), adjust Xa1 accordingly if trial is not correct. Prob 3:V1=V2=V/2=30 . Solve slope of line = -vo/V1. Using Walas Walas Plot, from Cao at X axis, draw 2 lines with same slope to get Ca2. Solve Xa2. Prob 4: Knowing V1=V2=30, find appropriate Xa1 and Xa2 so that Area1=Area2 depending on which reactor comes fi rst.