chemical reactor design, optimization, and scaleup

Weoften write this equation using volumetric flow rates and volumes rather thanmass flow rates and mass inventories:bal-In reactor design, we are interested in chemical reactions that tran

CHEMICAL REACTOR DESIGN, OPTIMIZATION,

AND SCALEUP

CHEMICAL REACTOR DESIGN, OPTIMIZATION,

AND SCALEUP

E Bruce Nauman

Rensselaer Polytechnic Institute

Troy, New York

Copyright © 2002 by The McGraw-Hill Companies All rights reserved Manufactured in the United States of America Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form

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0-07-139558-X

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DOI: 10.1036/007139558X

abc

McGraw-Hill

Copyright 2002 The McGraw-Hill Companies, Inc Click Here for Terms of Use

Reference 76

5.3 Scaleup of Nonisothermal Reactors 173

8.4 Slit Flow and Rectangular Coordinates 285

10.5 Experimental Determination of Intrinsic Kinetics 371

13 Polymer Reaction Engineering 463

15.1.1 Inert Tracer Experiments 540

This book is an outgrowth of an earlier book, Chemical Reactor Design, JohnWiley & Sons, 1987 The title is different and reflects a new emphasis on optimi-zation and particularly on scaleup, a topic rarely covered in undergraduate orgraduate education but of paramount importance to many practicing engineers.The treatment of biochemical and polymer reaction engineering is also moreextensive than normal

Practitioners are the primary audience for the new book Here, in one spot,you will find a reasonably comprehensive treatment of reactor design, optimiza-tion and scaleup Spend a few minutes becoming comfortable with the notation(anyone bothering to read a preface obviously has the inclination), and you willfind practical answers to many design problems

The book is also useful for undergraduate and graduate courses in chemicalengineering Some faults of the old book have been eliminated One fault was itslevel of difficulty It was too hard for undergraduates at most U.S universities.The new book is better Known rough spots have been smoothed, and it is easier

to skip advanced material without loss of continuity However, the new bookremains terse and somewhat more advanced in its level of treatment than isthe current U.S standard Its goal as a text is not to train students in the appli-cation of existing solutions but to educate them for the solution of new pro-blems Thus, the reader should be prepared to work out the details of someexamples rather than expect a complete solution

There is a continuing emphasis on numerical solutions Numerical solutionsare needed for most practical problems in chemical reactor design, but sophisti-cated numerical techniques are rarely necessary given the speed of modern com-puters The goal is to make the techniques understandable and easily accessibleand to allow continued focus on the chemistry and physics of the problem.Computational elegance and efficiency are gladly sacrificed for simplicity.Too many engineers are completely in the dark when faced with variablephysical properties, and tend to assume them away without full knowledge

of whether the effects are important They are often unimportant, but a realdesign problem—as opposed to an undergraduate exercise or preliminary pro-cess synthesis—deserves careful assembly of data and a rigorous solution.Thus, the book gives simple but general techniques for dealing with varyingphysical properties in CSTRs and PFRs Random searches are used for optimi-zation and least-squares analysis These are appallingly inefficient but mar-velously robust and easy to implement The method of lines is used forsolving the partial differential equations that govern real tubular reactors andpacked beds This technique is adequate for most problems in reactor design

xiii

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No CD ROM is supplied with the book Many of the numerical problems can

be solved with canned ODE and PDE solvers, but most of the solutions are quitesimple to code Creative engineers must occasionally write their own code tosolve engineering problems Due to their varied nature, the solutions requireuse of a general-purpose language rather than a specific program Computa-tional examples in the book are illustrated using Basic This choice was madebecause Basic is indeed basic enough that it can be sight-read by anyone alreadyfamiliar with another general-purpose language and because the ubiquitousspreadsheet, Excel, uses Basic macros Excel provides input/output, plotting,and formatting routines as part of its structure so that coding efforts can beconcentrated on the actual calculations This makes it particularly well suitedfor students who have not yet become comfortable with another language.Those who prefer another language such as C or Fortran or a mathematicalprogramming system such as Mathematica, Maple, Mathcad, or Matlabshould be able to translate quite easily

I continue with a few eccentricities in notation, using a, b, c, to denotemolar concentrations of components A, B, C, I have tried to avoid acro-nyms and other abbreviations unless the usage is common and there is a trueeconomy of syllables Equations are numbered when the results are referenced

or the equations are important enough to deserve some emphasis The problems

at the back of each chapter are generally arranged to follow the flow of the textrather than level of difficulty Thus, some low-numbered problems can be fairlydifficult

Bruce NaumanTroy, New York

Roman Characters

Equationwhere used

tubular reactor

14.14

second-order ODEs to first second-order

Exam 9.6

emulsion phase

11.45

becomes full during a startup

Exam 14.3

gas phase

11.4

located at point (r, z)

Sec 10.4.1

al(l, r, z) Concentration at location l in a pore, the mouth of

which is located at point (r, z)

amix Mixing-cup average concentration 8.4

component A

Exam 14.1

as(r, z) Concentration on surface of catalyst at location (r, z)

in the reactor

Sec 10.1

polymerization

Sec 13.1

bubble phase

11.46

[B] Concentration of component B 1.8

jth tray

Exam 11.7

liquid on the jth tray

DA Diffusion coefficient for component A 8.3

reactor

Fig 9.9

model

Sec 9.1

model

Sec 9.1

fTðtTÞ Differential distribution function for thermal times 15.54

of tube radius for a monotonic velocity profile

15.29

Roman Characters—Continued

Equationwhere used

I Refers to inert component I 3.13

that end with an X group

Sec 13.4.4

that end with a Y group

Sec 13.4.4

step

Exam 10.2

Roman Characters—Continued

Equationwhere used

kg Mass transfer coefficient based on gas-phase

driving force

11.5

driving force

11.5

polymer chain ending with an X unit

Sec 13.4.4

polymer chain ending with an X unit

Sec 13.4.4

polymer chain ending with a Y unit

Sec 13.4.4

polymer chain ending with a Y unit

driving force

11.1

liquid-phase driving force

11.2

Roman Characters—Continued

Equationwhere used

Km Mass transfer coefficient between the emulsion and

bubble phases in a gas fluidized bed

11.45

polymerization

Sec 13.1

cell mass per time

12.15

N Vector of component moles (N  1) 2.39

polymerization

Sec 13.1

a pore

10.7

the catalyst

10.8

lengths

13.39

equilibrium with the oxygen dissolved in the

liquid phase

Exam 11.9

Roman Characters—Continued

Equationwhere used

q Transpiration volumetric flow per unit length 3.46

model

Exam 15.7

R

A 3.9

phase

11.11

variable cross section

3.8

Roman Characters—Continued

Equationwhere used

capacity model

Exam 15.7

vacant

Exam 10.1

at onset of polymerization

13.3

S2residual Sum of squares after data fit Sec 7.1.1

Stubes Scaleup factor for the number of tubes 3.31

ttn Residence time in the nth zone Exam 6.5

catalyst particle

10.4.3

Roman Characters—Continued

Equationwhere used

Vfull Full volume of reactor Exam 14.3

Roman Characters—Continued

Equationwhere used

YP/S Product mass produced per substrate mass 12.14

substrate

12.16

effectiveness model

Sec 10.4.3

Roman Characters—Continued

Equationwhere used

max Maximum growth rate for cell mass 12.8

when the 0th moment is cpolymer

13.9

2

residual Standard deviation after data fit Sec 7.1.1

consecutive chain lengths

Sec 13.4.2

Special Characters

Equationwhere used

. Denotes free radical 13.39

Subscripts

Equationwhere used

polymer have the same composition

particular time in a batch reactor or location in a

tubular reactor

Exam 13.9

Special Characters—Continued

Equationwhere used

large Refers to large vessel 1.60

molar units

1.3

components

7.20surface area Refers to reaction rate based on catalyst

surface area

Exam 10.9

diameter

9.6

Subscripts—Continued

Equationwhere used

Equationwhere used

Basic Language Code

Program segments and occasional variables within the text are set in a fixed-widthfont to indicate that they represent computer code

The Basic language does not allow continuation statements Instead, long ments give long lines of code Margin requirements for printing require continua-tions These are denoted by a plus sign, þ, in the first column of the code To run

state-a progrstate-am, delete the þ state-and move the code to the end of the previous line SeeAppendix 7 for examples

Rate at which mass enters the volume

¼ Rate at which mass leaves the volume

þ Rate at which mass accumulates within the volume

ð1:1Þ

where ‘‘entering’’ and ‘‘leaving’’ apply to the flow of material across the aries See Figure 1.1 Equation (1.1) is an overall mass balance that applies to thetotal mass within the control volume, as measured in kilograms or pounds Itcan be written as

bound-ðQmassÞin¼ ðQmassÞoutþdI

1

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where Qmass is the mass flow rate and I is the mass inventory in the system Weoften write this equation using volumetric flow rates and volumes rather thanmass flow rates and mass inventories:

bal-In reactor design, we are interested in chemical reactions that transform onekind of mass into another A material balance can be written for each compo-nent; however, since chemical reactions are possible, the rate of formation ofthe component within the control volume must now be considered The compo-nent balancefor some substance A is

Rate at which component A enters the volume

þ net rate at which component A is formed by reaction

¼ rate at which component A leaves the volume

þ rate at which component A accumulates within the volume ð1:4Þ

or, more briefly,

See Figure 1.2 A component balance can be expressed in mass units, and this isdone for materials such as polymers that have ill-defined molecular weights.Usually, however, component A will be a distinct molecular species, and it ismore convenient to use molar units:

Qinainþ ^RAV ¼ QoutaoutþdðV^aaÞ

where a is the concentration or molar density of component A in moles pervolume, and ^RA is the net rate of formation of component A in moles pervolume per time There may be several chemical reactions occurring simulta-neously, some of which generate A while others consume it ^RA is the netrate and will be positive if there is net production of component A and negative

if there is net consumption Unless the system is very well mixed, concentrationsand reaction rates will vary from point to point within the control volume Thecomponent balance applies to the entire control volume so that^aa and ^RAdenotespatial averages

A version of Equation (1.4) can be written for each component, A, B, C, :

If these equations are written in terms of mass and then summed over all ponents, the sum must equal Equation (1.1) since the net rate of mass formationmust be zero When written in molar units as in Equation (1.6), the sum need not

com-be zero since chemical reactions can cause a net increase or decrease in thenumber of moles

To design a chemical reactor, the average concentrations, ^aa, ^bb, ^cc, , or atleast the spatial distribution of concentrations, must be found Doing this issimple for a few special cases of elementary reactions and ideal reactors that

Average concentration = a Inventory = Va

Average reaction rate = 4A

Accumulation =

ˆ

ˆ ˆ

d (Va) dt

are considered here in Chapter 1 We begin by discussing elementary reactions ofwhich there are just a few basic types.

A and B molecules should be proportional to their concentrations, a and b Notall collisions cause a reaction, but at constant environmental conditions (e.g.,temperature) some definite fraction should react Thus, we expect

where k is a constant of proportionality known as the rate constant

Example 1.1: Use the kinetic theory of gases to rationalize the functionalform of Equation (1.8)

Solution: We suppose that a collision between an A and a B molecule isnecessary but not sufficient for reaction to occur Thus, we expect

R ¼CABfR

where CABis the collision rate (collisions per volume per time) and fRis thereaction efficiency Avogadro’s number, Av, has been included in Equation(1.9) so thatR will have normal units, mol/(m3Es), rather than units of mole-cules/(m3Es) By hypothesis, 0 < fR< 1

The molecules are treated as rigid spheres having radii rAand rB.Theycollide if they approach each other within a distance rAþ rB A result fromkinetic theory is

proportional to the product of the concentrations as postulated in Equation(1.8) The reaction rate constant is

Note that the rate constant k is positive so thatR is positive R is the rate ofthe reaction, not the rate at which a particular component reacts Components Aand B are consumed by the reaction of Equation (1.7) and thus are ‘‘formed’’ at

a negative rate:

RA¼ RB¼  kabwhile P is formed at a positive rate:

RP¼ þ kabThe sign convention we have adopted is that the rate of a reaction is always posi-tive The rate of formation of a component is positive when the component isformed by the reaction and is negative when the component is consumed

A general expression for any single reaction is

0M! AA þBB þ


þRR þSS þ


ð1:12Þ

As an example, the reaction 2H2þ O2 ! 2H2O can be written as

0M! 2H2 O2þ 2H2OThis form is obtained by setting all participating species, whether products orreactants, on the right-hand side of the stoichiometric equation The remainingterm on the left is the zero molecule, which is denoted by 0Mto avoid confusionwith atomic oxygen TheA,B, terms are the stoichiometric coefficients forthe reaction They are positive for products and negative for reactants Usingthem, the general relationship between the rate of the reaction and the rate offormation of component A is given by

The stoichiometric coefficients can be fractions However, for elementary tions, they must be small integers, of magnitude 2, 1, or 0 If the reaction of

reac-Equation (1.12) were reversible and elementary, its rate would be

R ¼ kf½A A½B B


 kr½R R½S S ð1:14Þand it would have an equilibrium constant

K ¼kf

kr

¼ ½A A½B B ½R R½S S ¼ ½RR½SS

½A A½B B ð1:15Þwhere A, B, are reactants; R, S, are products; kfis the rate constant for theforward reaction; and kris the rate constant for the reverse reaction

The functional form of the reaction rate in Equation (1.14) is dictated by thereaction stoichiometry, Equation (1.12) Only the constants kfand kr can beadjusted to fit the specific reaction This is the hallmark of an elementary reac-tion; its rate is consistent with the reaction stoichiometry However, reactionscan have the form of Equation (1.14) without being elementary

As a shorthand notation for indicating that a reaction is elementary, we shallinclude the rate constants in the stoichiometric equation Thus, the reaction

1.2.1 First-Order, Unimolecular Reactions

SinceR has units of moles per volume per time and a has units of moles pervolume, the rate constant for a first-order reaction has units of reciprocaltime: e.g., s1 The best example of a truly first-order reaction is radioactivedecay; for example,

U238! Th234þ He4

since it occurs spontaneously as a single-body event Among strictly chemicalreactions, thermal decompositions such as

follow first-order kinetics at normal gas densities The student of chemistry willrecognize that the complete decomposition of dimethyl ether into methane,carbon monoxide, and hydrogen is unlikely to occur in a single step Short-lived intermediates will exist; however, since the reaction is irreversible, theywill not affect the rate of the forward reaction, which is first order and hasthe form of Equation (1.16) The decomposition does require energy, and colli-sions between the reactant and other molecules are the usual mechanism foracquiring this energy Thus, a second-order dependence may be observed forthe pure gas at very low densities since reactant molecules must collide withthemselves to acquire energy.

1.2.2 Second-Order Reactions, One Reactant

where k has units of m3mol1s1 It is important to note that RA¼ 2ka2

according to the convention of Equation (1.13)

A gas-phase reaction believed to be elementary and second order is

2HI ! H2þ I2Here, collisions between two HI molecules supply energy and also supply thereactants needed to satisfy the observed stoichiometry

1.2.3 Second-Order Reactions, Two Reactants

A homogeneous gas-phase reaction that follows a third-order kinetic scheme is

2NO þ O2! 2NO2 R ¼ k½NO2½O2although the mechanism is believed to involve two steps1 and thus is notelementary

1.3 REACTION ORDER AND MECHANISM

As suggested by these examples, the order of a reaction is the sum of the nents m, n, in

expo-R ¼ kambn Reaction order ¼ m þ n þ


ð1:20ÞThis definition for reaction order is directly meaningful only for irreversible orforward reactions that have rate expressions in the form of Equation (1.20).Components A, B, are consumed by the reaction and have negative stoichio-metric coefficients so that m ¼ A, n ¼ B, are positive For elementaryreactions, m and n must be integers of 2 or less and must sum to 2 or less.Equation (1.20) is frequently used to correlate data from complex reactions.Complex reactions can give rise to rate expressions that have the form ofEquation (1.20), but with fractional or even negative exponents Complex reac-tions with observed orders of 1/2 or 3/2 can be explained theoretically based onmechanisms discussed in Chapter 2 Negative orders arise when a compoundretards a reaction—say, by competing for active sites in a heterogeneously cat-alyzed reaction—or when the reaction is reversible Observed reaction ordersabove 3 are occasionally reported An example is the reaction of styrene withnitric acid, where an overall order of 4 has been observed.2The likely explana-tion is that the acid serves both as a catalyst and as a reactant The reaction is farfrom elementary

Complex reactions can be broken into a number of series and parallel tary steps, possibly involving short-lived intermediates such as free radicals.These individual reactions collectively constitute the mechanism of the complexreaction The individual reactions are usually second order, and the number ofreactions needed to explain an observed, complex reaction can be surprisinglylarge For example, a good model for

elemen-CH4þ 2O2! CO2þ 2H2Owill involve 20 or more elementary reactions, even assuming that the indicatedproducts are the only ones formed in significant quantities A detailed modelfor the oxidation of toluene involves 141 chemical species in 743 elementaryreactions.3

As a simpler example of a complex reaction, consider (abstractly, not mentally) the nitration of toluene to give trinitrotoluene:

or, in shorthand,

A þ 3B ! C þ 3DThis reaction cannot be elementary We can hardly expect three nitric acid mole-cules to react at all three toluene sites (these are the ortho and para sites; metasubstitution is not favored) in a glorious, four-body collision Thus, thefourth-order rate expression R ¼ kab3 is implausible Instead, the mechanism

of the TNT reaction involves at least seven steps (two reactions leading toortho- or para-nitrotoluene, three reactions leading to 2,4- or 2,6-dinitrotoluene,and two reactions leading to 2,4,6-trinitrotoluene) Each step would require only

a two-body collision, could be elementary, and could be governed by a order rate equation Chapter 2 shows how the component balance equations can

second-be solved for multiple reactions so that an assumed mechanism can second-be testedexperimentally For the toluene nitration, even the set of seven series and parallelreactions may not constitute an adequate mechanism since an experimentalstudy4 found the reaction to be 1.3 order in toluene and 1.2 order in nitricacid for an overall order of 2.5 rather than the expected value of 2

An irreversible, elementary reaction must have Equation (1.20) as its rateexpression A complex reaction may have an empirical rate equation with theform of Equation (1.20) and with integral values for n and m, without being ele-mentary The classic example of this statement is a second-order reaction whereone of the reactants is present in great excess Consider the slow hydrolysis of

an organic compound in water A rate expression of the form

R ¼ k½water½organic

is plausible, at least for the first step of a possibly complex mechanism Suppose[organic]  [water] so that the concentration of water does not change appreci-ably during the course of the reaction Then the water concentration can be com-bined with k to give a composite rate constant that is approximately constant.The rate expression appears to be first order in [organic]:

R ¼ k½water½organic ¼ k0½organic ¼ k0awhere k0¼ k½water is a pseudo-first-order rate constant From an experimentalviewpoint, the reaction cannot be distinguished from first order even thoughthe actual mechanism is second order Gas-phase reactions also appear firstorder when one reactant is dilute Kinetic theory still predicts the collisionrates of Equation (1.10), but the concentration of one species, call it B, remainsapproximately constant The observed rate constant is

The only reactions that are strictly first order are radioactive decay reactions.Among chemical reactions, thermal decompositions may seem first order, but

an external energy source is generally required to excite the reaction As notedearlier, this energy is usually acquired by intermolecular collisions Thus, thereaction rate could be written as

R ¼ k½reactant molecules½all molecules

The concentration of all molecules is normally much higher than the tion of reactant molecules, so that it remains essentially constant during thecourse of the reaction Thus, what is truly a second-order reaction appears to

concentra-be first order

1.4 IDEAL, ISOTHERMAL REACTORS

There are four kinds of ideal reactors:

1 The batch reactor

2 The piston flow reactor (PFR)

3 The perfectly mixed, continuous-flow stirred tank reactor (CSTR)

4 The completely segregated, continuous-flow stirred tank reactor

This chapter discusses the first three types, which are overwhelmingly the mostimportant The fourth type is interesting theoretically, but has limited practicalimportance It is discussed in Chapter 15

1.4.1 The Ideal Batch Reactor

This is the classic reactor used by organic chemists The typical volume in ware is a few hundred milliliters Reactants are charged to the system, rapidlymixed, and rapidly brought up to temperature so that reaction conditions arewell defined Heating is carried out with an oil bath or an electric heatingmantle Mixing is carried out with a magnetic stirrer or a small mechanical agi-tator Temperature is controlled by regulating the bath temperature or by allow-ing a solvent to reflux

glass-Batch reactors are the most common type of industrial reactor and may havevolumes well in excess of 100,000 liters They tend to be used for small-volumespecialty products (e.g., an organic dye) rather than large-volume commoditychemicals (e.g., ethylene oxide) that are normally reacted in continuous-flowequipment Industrial-scale batch reactors can be heated or cooled by externalcoils or a jacket, by internal coils, or by an external heat exchanger in apump-around loop Reactants are often preheated by passing them throughheat exchangers as they are charged to the vessel Heat generation due to the

bal-In reactor design,. .. ISOTHERMAL REACTORS

There are four kinds of ideal reactors:

1 The batch reactor

2 The piston flow reactor (PFR)

3 The perfectly mixed, continuous-flow stirred tank reactor. .. the sum need not

com-be zero since chemical reactions can cause a net increase or decrease in thenumber of moles

To design a chemical reactor, the average concentrations, ^aa, ^bb,