fogler chapter 4 notes

8182019 Fogler chapter 4 notes

httpslidepdfcomreaderfullfogler-chapter-4-notes 119

292016 Chapter 4 Summary Notes

httpwwwumichedu~elementsfoglerampgurmenhtmlcourselectur esfourindexhtmtest1b

4 ALGORITHM FOR ISOTHERMAL R EACTOR DESIGN

Topics

Part 1 Mole Balances in Terms of Conversion

1 Algorithm for Isothermal Reactor Design2 ApplicationsExamples of CRE Algorithm3 Reversible Reactions

4 ODE (Polymath) Solutions to CRE Problems5 General Guidelines for California Problems6 PBR with Pressure Drop 7 Engineering Analysis

Part 2 Measures Other Than Conversion

1 Measures Other Than Conversion2 Membrane Reactors 3 Semibatch Reactors

Part 1 Mole Balances in Terms of Conversion

1 Algorithm for Isothermal Reactor Design top




httpwwwumichedu~elementsfoglerampgurmenhtmlcourselectur esfourindexhtmtest1b 2

French Menu Analogy

41 The reaction (2A+B--gtC) carried out in a CSTR PFR and a Batch Reactor

42 Labratory Experiment

43 Semilog plot to find reaction rate constant

to a closed ended PFRCSTR 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 02





moldm3

What CSTR volume is necessary to achieve a 90 conversion when k = 10 dm 3 (mols)

Mole Balance

Rate Law

Stoichiometry liquid phase (v = vo)

Combine

Evaluate at X = 09

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



Critical Thinking Questions for CSTR

The following movies were made by the students of Professor Alan Lanes chemical reaction engineering class at theUniversity of Alabama Tuscaloosa





2 ApplicationsExamples of the CRE Algorithm top

Gas Phase Elementary Reaction Additional Information

only A fed P0 = 82 atm

T0 = 500 K CA0 = 02 moldm3

k = 05 dm3 mol-s vo = 25 dm3 s

Solve for X = 09

Applying the algorithm to the above reaction occurring in a Batch CSTR and PFR

Batch CSTR PFR

Mole Balance

Rate Law

StoichiometryGas V = V0

(eg constant volumesteel container)

Gas T =T0 P =P0 Gas T = T0 P = P0





Per Mole of A Per Mole of A

Combine

Integrate

Evaluate

For X = 09 V = 6806 dm3 V = 907 dm3

Visual Encyclopedia of Reaction Engineering Equipment

44 Gas Phase Reaction Example

CSTR and PFR Example

41 Calculate V for a Zero-Order Reaction

3 Reversible Reactions top

To determine the conversion or reactor volume for reversible reactions one must first calculate the maximum conversionthat can be achieved at the isothermal reaction temperature which is the equilibrium conversion (See Example 3-8 in thetext for additional coverage of equilibrium conversion in isothermal reactor design)

Equilibrium Conversion Xe

From Appendix C





45 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 = 08 Xe

Reaction Additional Information

CA0 = 02 moldm3

KC = 100 dm3 mol

k = 2 dm3 mol-min

FA0 = 5 molmin

First calculate Xe

Xe = 089

X = 08Xe = 0711

One could then use Polymath to determine the volume of the PFR The corresponding Polymath program is

shown below

4 ODE (Polymath) Solutions to CRE Problems top

Algorithm Steps Polymath Equations

Mole Balance d(X)d(V) = -rAF A0

Rate Law rA = -k((CA2)-(CBKC))

Stoichiometry CA = (CA0(1-X))(1+epsX)

CB = (CA0X)(2(1+epsX))

Parameter Evaluation eps = -05 CA0 = 02 k = 2

FA0 = 5 KC = 100

Initial and Final Values X0 = 0 V0 = 0 Vf = 500

Polymath Screen Shots

Equations

Plot of X vs V

Results in Tabular Form

A volume of 94 dm

3

(rounding up from slightly more than 93 dm

3

) appears to be our answer

46 Batch Reactor With a Reversible Reaction

5 General Guidelines for California Problems top

Every state has an examination engineers must pass to become a registered professional engineer In the past there havetypically been six problems in a three hour segment of the California Professional Engineers Exam Consequently one shouldbe able to work each problem in 30 minutes or less Many of these problems involve an intermediate calculation todetermine the final answer

Some Hints





1 group unknown parametersvalues 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 calculations3 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

California Professional Engineers Registration Problem

47 Batch Reactor Optimization

6 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 (PP0) as a function of W (or V if you have a PFR)

Pressure Drop in Packed Bed Reactors

Ergun Equation

Variable Gas Density





let

Catalyst Weight

where

let

then

English-

Espanol-

Svenska-

We will use this form for multiple reactions

We will use this form for single reactions





IsothermalOperation

recall that

notice that

The two expressions are coupled ordinary differential equations We can solve them simultaneously usingan ODE solver such as Polymath For the special case of isothermal operation and epsilon = 0 we canobtain an analytical solution

Polymath will combine the mole balance rate law and stoichiometry

Analytical Solution [e] PFR with

CAUTION Never use this form if

Combine

Solve

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





42 Pressure Drop in a Packed Bed Reactor

43 Pressure and Reaction Orders

48 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



POLYMATH

Consider the following gas phase reaction carried out isothermally in a packed bed reactor Pure A is fed at a rate of 25

moless and with and α = 00002 kg-1

2A B

Mole Balance

Rate Law

Elementary

Stoichiometry

Gas with T = T0

A B2

POLYMATH will combine everything - You do not need the combine step Thank you POLYMATH





Profiles

44 What Four Things are Wrong with this Solution

Optimum Paritcle Diameter

Laminar Flow Fix P0 ρ0

ρ0 = P0(MW)RT0

ρ0P0simP02





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 deosthe conversion

k sim 1Dp

DP1 gt DP2

k1 gt k2

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 entryways into the catalyst particle

In CD-ROM chapter 12 we will learn that effectiveness factor decreases as the particle size increases

7 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 henceconversion 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 eg 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

45 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 asself tests









1 Measures Other Than Conversion top

Uses

A Membrane reactorsB Multiple reaction

Liquids Use concentrations IE 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 IE FI

2 If the above reaction carried out in the gas phase in a PFR where V voCAok and Kc are given and the feed ispure A the combined mole balance rate laws and stoichiometry yield for isothermal operation (T=To) and no pressuredrop (P=0) are

Use Polymath to plot FA and FB down the length of the reactor

49 Stoichiometry for Measures Other than Conversion

46 Gas Phase PFR

47 Liquid Phase CSTR





Use Creative and then Critical Thinking

48 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 laminarwe must use the techniques discussed in chapter 13 See example 48 of the text

University of Washington Transport Effects in Microreactors site

Institut fr Mikrotechnik Mainz GmbH

2 Membrane Reactors top

Membrane reactors can be used to achieve conversions greater than the original equilibrium value These higher conversionsare the result of Le Chateliers Principle you can remove one of the reaction products and drive the reaction to the right Toaccomplish this a membrane that is permeable to that reaction product but is impermeable to all other species is placedaround the reacting mixture

Example The following reaction is to be carried out isothermally in a membrane reactor with no pressure drop Themembrane 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 F A FB FC

Polymath Program

Mole Balances

Rate Laws

StoichiometryIsothermal no pressure drop

Combine Polymath will combine for you-- Thanks Polymathyou rock

Parameters

Solve Polymath

49 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


3 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

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 andorconcentration

Conversion Concentration Number of Moles





Polymath Equations

Conversion Concentration Moles

d(X)d(t) = -raVNao d(Ca)d(t) = ra - (Cavo)V d(Na)d(t) = raV

ra = -kCaCb d(Cb)d(t) = rb + ((Cbo-Cb)vo)V d(Nb)d(t) = rbV + Fbo

Ca = Nao(1 - X)V ra = -kCaCb ra = -kCaCb

Cb = (Nbi + Fbot - NaoX)V rb = ra rb = ra

V = Vo + vot V = Vo + vot V = Vo + vot

Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5

Fbo = 5 Nao = 100 Ca = NaV

Nbi = 0 Cbo = Fbovo Cb = NbV

k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao

Polymath Screenshots

C onv er sion C onc entr ation

Polymath Equations Polymath Equations

SummaryTable Summary Table

Conversion vsTime Conversion vsTime

Concentration vsTime Concentration vsTime

Volume vsTime Volume vsTime




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

410 Semibatch A rarr 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

top

Fogler amp Gurmencopy 2007 University of Michigan





French Menu Analogy

41 The reaction (2A+B--gtC) carried out in a CSTR PFR and a Batch Reactor

42 Labratory Experiment

43 Semilog plot to find reaction rate constant

to a closed ended PFRCSTR 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 02





moldm3


Mole Balance

Rate Law


Combine

Evaluate at X = 09

V = 1125 dm3

Space Time














T0 = 500 K CA0 = 02 moldm3


Solve for X = 09


Batch CSTR PFR

Mole Balance

Rate Law









Combine

Integrate

Evaluate

For X = 09 V = 6806 dm3 V = 907 dm3








From Appendix C









CA0 = 02 moldm3

KC = 100 dm3 mol

k = 2 dm3 mol-min

FA0 = 5 molmin

First calculate Xe

Xe = 089

X = 08Xe = 0711


shown below








FA0 = 5 KC = 100



Equations

Plot of X vs V


A volume of 94 dm

3


3





Some Hints













Sample Question


Mole Balance


Rate Law


Stoichiometry

Isothermal T = T0

Combine



Ergun Equation






let

Catalyst Weight

where

let

then

English-

Espanol-

Svenska-







IsothermalOperation

recall that

notice that





Combine

Solve














POLYMATH



2A B

Mole Balance

Rate Law

Elementary

Stoichiometry

Gas with T = T0

A B2






Profiles




ρ0 = P0(MW)RT0

ρ0P0simP02







k sim 1Dp

DP1 gt DP2

k1 gt k2












to D P and X



Turbulent Flow




Laminar Flow












Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top






moldm3


Mole Balance

Rate Law


Combine

Evaluate at X = 09

V = 1125 dm3

Space Time














T0 = 500 K CA0 = 02 moldm3


Solve for X = 09


Batch CSTR PFR

Mole Balance

Rate Law









Combine

Integrate

Evaluate

For X = 09 V = 6806 dm3 V = 907 dm3








From Appendix C









CA0 = 02 moldm3

KC = 100 dm3 mol

k = 2 dm3 mol-min

FA0 = 5 molmin

First calculate Xe

Xe = 089

X = 08Xe = 0711


shown below








FA0 = 5 KC = 100



Equations

Plot of X vs V


A volume of 94 dm

3


3





Some Hints













Sample Question


Mole Balance


Rate Law


Stoichiometry

Isothermal T = T0

Combine



Ergun Equation






let

Catalyst Weight

where

let

then

English-

Espanol-

Svenska-







IsothermalOperation

recall that

notice that





Combine

Solve














POLYMATH



2A B

Mole Balance

Rate Law

Elementary

Stoichiometry

Gas with T = T0

A B2






Profiles




ρ0 = P0(MW)RT0

ρ0P0simP02







k sim 1Dp

DP1 gt DP2

k1 gt k2












to D P and X



Turbulent Flow




Laminar Flow












Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top









T0 = 500 K CA0 = 02 moldm3


Solve for X = 09


Batch CSTR PFR

Mole Balance

Rate Law









Combine

Integrate

Evaluate

For X = 09 V = 6806 dm3 V = 907 dm3








From Appendix C









CA0 = 02 moldm3

KC = 100 dm3 mol

k = 2 dm3 mol-min

FA0 = 5 molmin

First calculate Xe

Xe = 089

X = 08Xe = 0711


shown below








FA0 = 5 KC = 100



Equations

Plot of X vs V


A volume of 94 dm

3


3





Some Hints













Sample Question


Mole Balance


Rate Law


Stoichiometry

Isothermal T = T0

Combine



Ergun Equation






let

Catalyst Weight

where

let

then

English-

Espanol-

Svenska-







IsothermalOperation

recall that

notice that





Combine

Solve














POLYMATH



2A B

Mole Balance

Rate Law

Elementary

Stoichiometry

Gas with T = T0

A B2






Profiles




ρ0 = P0(MW)RT0

ρ0P0simP02







k sim 1Dp

DP1 gt DP2

k1 gt k2












to D P and X



Turbulent Flow




Laminar Flow












Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top







Combine

Integrate

Evaluate

For X = 09 V = 6806 dm3 V = 907 dm3








From Appendix C









CA0 = 02 moldm3

KC = 100 dm3 mol

k = 2 dm3 mol-min

FA0 = 5 molmin

First calculate Xe

Xe = 089

X = 08Xe = 0711


shown below








FA0 = 5 KC = 100



Equations

Plot of X vs V


A volume of 94 dm

3


3





Some Hints













Sample Question


Mole Balance


Rate Law


Stoichiometry

Isothermal T = T0

Combine



Ergun Equation






let

Catalyst Weight

where

let

then

English-

Espanol-

Svenska-







IsothermalOperation

recall that

notice that





Combine

Solve














POLYMATH



2A B

Mole Balance

Rate Law

Elementary

Stoichiometry

Gas with T = T0

A B2






Profiles




ρ0 = P0(MW)RT0

ρ0P0simP02







k sim 1Dp

DP1 gt DP2

k1 gt k2












to D P and X



Turbulent Flow




Laminar Flow












Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top










CA0 = 02 moldm3

KC = 100 dm3 mol

k = 2 dm3 mol-min

FA0 = 5 molmin

First calculate Xe

Xe = 089

X = 08Xe = 0711


shown below








FA0 = 5 KC = 100



Equations

Plot of X vs V


A volume of 94 dm

3


3





Some Hints













Sample Question


Mole Balance


Rate Law


Stoichiometry

Isothermal T = T0

Combine



Ergun Equation






let

Catalyst Weight

where

let

then

English-

Espanol-

Svenska-







IsothermalOperation

recall that

notice that





Combine

Solve














POLYMATH



2A B

Mole Balance

Rate Law

Elementary

Stoichiometry

Gas with T = T0

A B2






Profiles




ρ0 = P0(MW)RT0

ρ0P0simP02







k sim 1Dp

DP1 gt DP2

k1 gt k2












to D P and X



Turbulent Flow




Laminar Flow












Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top














Sample Question


Mole Balance


Rate Law


Stoichiometry

Isothermal T = T0

Combine



Ergun Equation






let

Catalyst Weight

where

let

then

English-

Espanol-

Svenska-







IsothermalOperation

recall that

notice that





Combine

Solve














POLYMATH



2A B

Mole Balance

Rate Law

Elementary

Stoichiometry

Gas with T = T0

A B2






Profiles




ρ0 = P0(MW)RT0

ρ0P0simP02







k sim 1Dp

DP1 gt DP2

k1 gt k2












to D P and X



Turbulent Flow




Laminar Flow












Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top






let

Catalyst Weight

where

let

then

English-

Espanol-

Svenska-







IsothermalOperation

recall that

notice that





Combine

Solve














POLYMATH



2A B

Mole Balance

Rate Law

Elementary

Stoichiometry

Gas with T = T0

A B2






Profiles




ρ0 = P0(MW)RT0

ρ0P0simP02







k sim 1Dp

DP1 gt DP2

k1 gt k2












to D P and X



Turbulent Flow




Laminar Flow












Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top






IsothermalOperation

recall that

notice that





Combine

Solve














POLYMATH



2A B

Mole Balance

Rate Law

Elementary

Stoichiometry

Gas with T = T0

A B2






Profiles




ρ0 = P0(MW)RT0

ρ0P0simP02







k sim 1Dp

DP1 gt DP2

k1 gt k2












to D P and X



Turbulent Flow




Laminar Flow












Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top













POLYMATH



2A B

Mole Balance

Rate Law

Elementary

Stoichiometry

Gas with T = T0

A B2






Profiles




ρ0 = P0(MW)RT0

ρ0P0simP02







k sim 1Dp

DP1 gt DP2

k1 gt k2












to D P and X



Turbulent Flow




Laminar Flow












Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top






Profiles




ρ0 = P0(MW)RT0

ρ0P0simP02







k sim 1Dp

DP1 gt DP2

k1 gt k2












to D P and X



Turbulent Flow




Laminar Flow












Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top








k sim 1Dp

DP1 gt DP2

k1 gt k2












to D P and X



Turbulent Flow




Laminar Flow












Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top











to D P and X



Turbulent Flow




Laminar Flow












Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top







Uses










46 Gas Phase PFR








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top








Microreactors












Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






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Polymath Program

Mole Balances

Rate Laws



Parameters

Solve Polymath













to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top










to Selectivity



1 Molar Basis


3 Conversion






-rA = kCACB







Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top






Polymath Equations







Vo = 100 Vo = 100 Vo = 100

vo = 2 vo = 2 vo = 2

Nao = 100 Fbo = 5 Fbo = 5



k = 01 k = 001 k = 001

Na = CaV

X = (Nao-Na)Nao















Where



See Also






top









Where



See Also






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fogler chapter 4 notes

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