The first step to knowledge is to know that we are ignorant.
—Socrates (470–399 B.C.)
The Wide, Wild World of Chemical Reaction Engineering
Chemical kinetics is the study of chemical reaction rates and reaction mechanisms. The study of chemical reaction engineering (CRE) combines the study of chemical kinetics with the reactors in which the reactions occur. Chemical kinetics and reactor design are at the heart of producing almost all industrial chemicals, such as the manufacture of phthalic anhydride shown in Figure 11. It is primarily a knowledge of chemical kinetics and reactor design that distinguishes the chemical engineer from other engineers. The selection of a reaction system that operates in the safest and most efficient manner can be the key to the economic success or failure of a chemical plant. For example, if a reaction system produces a large amount of undesirable product, subsequent purification and separation of the desired product could make the entire process economically unfeasible.
How is a chemical engineer different from other engineers?
The chemical reaction engineering (CRE) principles learned here can also be applied in many areas, such as waste water treatment, microelectronics, nanoparticles fabrication, and pharmacokinetics of living systems, in addition to the more traditional areas of the manufacture of chemicals and pharmaceuticals. Some of the examples that illustrate the wide application of CRE principles in this book are shown in Figure 12. These examples, which can be found either in the text or as Web modules, include modeling smog in the Los Angeles (L.A.) basin (Chapter 1 Web module), the digestive system of a hippopotamus (Chapter 2 Web module) on the CRE Web site, (www.umich.edu/~elements/6e/index.html), and molecular CRE (Chapter 3 Web module). Also shown are the manufacture of ethylene glycol (antifreeze), where three of the most common types of industrial reactors are used (Chapters 5 and 6), and the use of wetlands to degrade toxic chemicals (Chapter 7 on the CRE Web site). Other examples shown are the solid–liquid kinetics of acid–rock interactions to improve oil recovery (Chapter 7); pharmacokinetics of cobra bites (Chapter 8 Web module); freeradical scavengers used in the design of motor oils (Chapter 9); enzyme kinetics (Chapter 9) and drug delivery pharmacokinetics (Chapter 9 on the CRE Web site); heat effects, runaway reactions, and plant safety (Chapters 11–13); and increasing the octane number of gasoline and the manufacture of computer chips (Chapter 10).
Identify
– Kind
– Number
– Configuration
The rate of reaction tells us how fast the number of moles of one chemical species are being consumed to form another chemical species. The term chemical species refers to any chemical component or element with a given identity. The identity of a chemical species is determined by the kind, number, and configuration of that species’ atoms. For example, the species paraxylene is made up of a
fixed number of specific atoms in a definite molecular arrangement or configuration. The structure shown illustrates the kind, number, and configuration of atoms on a molecular level. Even though two chemical compounds have exactly the same kind and number of atoms of each element, they could still be different species because of different configurations. For example, 2butene has four carbon atoms and eight hydrogen atoms; however, the atoms in this compound can form two different arrangements.
As a consequence of the different configurations, these two isomers display different chemical and physical properties. Therefore, we consider them as two different species, even though each has the same number of atoms of each element.
When has a chemical reaction taken place?
We say that a chemical reaction has taken place when a detectable number of molecules of one or more species have lost their identity and assumed a new form by a change in the kind or number of atoms in the compound and/or by a change in structure or configuration of these atoms. In this classical approach to chemical change, it is assumed that the total mass is neither created nor destroyed when a chemical reaction occurs. The mass referred to is the total collective mass of all the different species in the system. However, when considering the individual species involved in a particular reaction, we do speak of the rate of disappearance of mass of a particular species. The rate of disappearance of a species, say species A, is the number of A molecules that lose their chemical identity per unit time per unit volume through the breaking and subsequent reforming of chemical bonds during the course of the reaction. In order for a particular species to “appear” in the system, some prescribed fraction of another species must lose its chemical identity.
Definition of Rate of Reaction
There are three basic ways a species may lose its chemical identity: decomposition, combination, and isomerization. In decomposition, the molecule loses its identity by being broken down into smaller molecules, atoms, or atom fragments. For example, if benzene and propylene are formed from a cumene molecule,
A species can lose its identity by
Decomposition
Combination
Isomerization
the cumene molecule has lost its identity (i.e., disappeared) by breaking its bonds to form these molecules. A second way that a molecule may lose its chemical identity is through combination with another molecule or atom. In the above reaction, the propylene molecule would lose its chemical identity if the reaction were carried out in the reverse direction, so that it combined with benzene to form cumene. The third way a species may lose its chemical identity is through isomerization, such as the reaction
Here, although the molecule neither adds other molecules to itself nor breaks into smaller molecules, it still loses its identity through a change in configuration.
To summarize this point, we say that a given number of molecules (i.e., moles) of a particular chemical species have reacted or disappeared when the molecules have lost their chemical identity.
The rate at which a given chemical reaction proceeds can be expressed in different ways by referring it to different chemical species in the reaction. To illustrate, consider the reaction of chlorobenzene with chloral in the presence of fuming sulfuric acid to produce the banned insecticide DDT (dichlorodiphenyl trichloroethane).
CCl_{3}CHO + 2C_{6}H_{5}Cl → (C_{6}H_{4}Cl)_{2}CHCCl_{3} + H_{2}O
Letting the symbol A represent chloral, B be chlorobenzene, C be DDT, and D be H_{2}O, we obtain
A + 2B → C + D
What is –r_{A}?
The rate of reaction, –r_{A}, is the number of moles of A (e.g., chloral) reacting (disappearing) per unit time per unit volume (mol/dm^{3}·s).
The numerical value of the rate of disappearance of reactant A, –r_{A}, is a positive number.
Chloral is being consumed at a rate of 10 moles per second per m^{3} when reacting with chlorobenzene to form DDT and water in the reaction described above. In symbol form, the reaction is written as
A + 2B → C + D
Write the rates of disappearance and formation (i.e., generation; mol/m^{3}·s) for each species in this reaction when the rate of reaction of chloral [A] (–r_{A}) is as 10 mol/m^{3}·s.
NFPA Diamond
DDT See Section 2.7
Solution
(a) Chloral[A]: 
Rate of disappearance of A = –r_{A} = 10 mol/m^{3}·s Rate of formation of A = r_{A} = –10 mol/m^{3}·s 
(b) Chlorobenzene[B]: 
For every mole of chloral that disappears, two moles of chlorobenzene [B] also disappear. Rate of disappearance of B = –r_{B} = –2r_{A} = 20 mol/m^{3}·s Rate of formation of B = r_{B} = –20 mol/m^{3}·s 
(c) DDT[C]: 
For every mole of chloral that disappears, one mole of DDT [C] appears. r_{C} = –r_{A} Rate of disappearance of C = –r_{C} = –10 mol/m^{3}·s Rate of formation of C = r_{C} = –r_{A} = –(–10 mol/m^{3}·s) = 10 mol/m^{3}·s 
(d) Water[D]: 
Same relationship to chloral as the relationship to DDT Rate of formation of D = r_{D} = 10 mol/m^{3}·s Rate of disappearance of D = –r_{D} = –10 mol/m^{3}·s 
–r_{A} = 10 mol A/m^{3}s^{†}
r_{A} = –10 mol A/m^{3}·s
Then
rA–1=rB–2=rC1=rD1
r_{B} = 2(r_{A}) = –20 mol B/m^{3}·s
–r_{B} = 20 mol B/m^{3}·s
r_{C} = –r_{A} = 10 mol C/m^{3}·s
r_{D} = –r_{A} = 10 mol D/m^{3}·s
^{†}Tutorial Video: https://www.youtube.com/watch?v=6mAqX31RRJU
A + 2B → C + D
The convention
–r_{A} = 10 mol A/m^{3}·s
r_{A} = –10 mol A/m^{3}·s
–r_{B} = 20 mol B/m^{3}·s
r_{B} = –20 mol B/m^{3}·s
r_{C} = 10 mol C/m^{3}·s
Analysis: The purpose of this example is to better understand the convention for the rate of reaction. The symbol r_{j} is the rate of formation (generation) of species j. If species j is a reactant, the numerical value of r_{j} will be a negative number. If species j is a product, then r_{j} will be a positive number. The rate of reaction, –r_{A}, is the rate of disappearance of reactant A and must be a positive number. A mnemonic relationship to help remember how to obtain relative rates of reaction of A to B, and so on, is given by Equation (31) on page 77.
In Equation (31) in Chapter 3, we will delineate the prescribed relationship between the rate of formation of one species, r_{j} (e.g., DDT [C]), and the rate of disappearance of another species, – r_{i} (e.g., chlorobenzene [B]), in a chemical reaction.
Heterogeneous reactions involve more than one phase. In heterogeneous reaction systems, the rate of reaction is usually expressed in measures other than volume, such as reaction surface area or catalyst weight. For a gas–solid catalytic reaction, the gas molecules must interact with the solid catalyst surface for the reaction to take place, as described in Chapter 10.
What is –r′A?
The dimensions of this heterogeneous reaction rate, –r′A (prime), are the number of moles of A reacting per unit time per unit mass of catalyst (e.g., mol/s·g catalyst).
Definition of r_{j}
Most of the introductory discussions on chemical reaction engineering in this book focus on homogeneous systems, in which case we simply say that r_{j} is the rate of formation of species j per unit volume. It is the number of moles of species j generated per unit volume per unit time.
We can say four things about the reaction rate r_{j}: r_{j} is
The rate law does not depend on the type of reactor used!!
The rate of formation of species j (mole/time/volume)
An algebraic equation
Independent of the type of reactor (e.g., batch or continuous flow) in which the reaction is carried out
Solely a function of the properties of the reacting materials and reaction conditions (e.g., species concentration, temperature, pressure, or type of catalyst, if any) at a point in the system
What is –r_{A} a function of?
However, because the properties and reaction conditions of the reacting materials may vary with position in a chemical reactor, r_{j} can in turn be a function of position and can vary from point to point in the system. This concept is utilized in flow reactors.
The reactionrate law relates the rate of reaction to species concentration and temperature as will be shown in Chapter 3. The chemical reactionratelaw is essentially an algebraic equation involving concentration, not a differential equation.^{1} For example, the algebraic form of the rate law for –r_{A} for the reaction
A → Products
may be a linear function of concentration,
^{1} For further elaboration on this point, see Chem. Eng. Sci., 25, 337 (1970); B. L. Crynes and H. S. Fogler, eds., AIChE Modular Instruction Series E: Kinetics, 1, 1 New York: AIChE, 1981; and R. L. Kabel, “Rates,” Chem. Eng. Commun., 9, 15 (1981).
or it may be some other algebraic function of concentration, such as Equation 36 shown in Chapter 3,
or
−rA=k1CA1+k2CA
The rate law is an algebraic equation.
For a given reaction, the particular concentration dependence that the rate law follows (i.e., −r_{A} = kC_{A} or −rA=kC2A or ...) must be determined from experimental observation. Equation (12) states that the rate of disappearance of A is equal to a rate constant k (which is a function of temperature) times the square of the concentration of A. As noted earlier, by convention, r_{A} is the rate of formation of A; consequently, –r_{A} is the rate of disappearance of A. Throughout this book, the phrase rate of generation means exactly the same as the phrase rate of formation, and these phrases are used interchangeably.
The convention
To perform a mole balance on any system, the system boundaries must first be specified. The volume enclosed by these boundaries is referred to as the system volume. We shall perform a mole balance on species j in a system volume, where species j represents the particular chemical species of interest, such as water or NaOH (Figure 13).
A mole balance on species j at any instant in time, t, yields the following equation:
Mole balance
Accumulation: In this equation, N_{j} represents the number of moles of species j in the system at time t and (dNjdt) is the rate of accumulation of species j within the system volume.
Generation: If all the system variables (e.g., temperature, catalytic activity, and concentration of the chemical species) are spatially uniform throughout the system volume, the rate of generation of species j, G_{j} (moles/time) is just the product of the reaction volume, V, and the rate of formation of species j, r_{j}.
Gj=rj⋅V
molestime=molestime⋅volume⋅volume
Now suppose that the rate of formation of species j for the reaction varies with position in the system volume. That is, it has a value r_{j1} at location 1, which is surrounded by a small volume, ΔV_{1}, within which the rate is uniform; similarly, the reaction rate has a value at location 2 and an associated volume, r_{j2}, and so on (Figure 14).
The rate of generation, ΔG_{j1}, in terms of r_{j1} and subvolume ΔV_{1}, is
ΔG_{j1} = r_{j1} ΔV_{1}
Similar expressions can be written for ΔG_{j2} and the other system subvolumes, ΔV_{i}. The total rate of generation within the system volume is the sum of all the rates of generation in each of the subvolumes. If the total system volume is divided into M subvolumes, the total rate of generation is
${\mathit{G}}_{\mathit{j}}={\displaystyle \sum}_{\mathit{i}=\mathrm{1}}^{M}\Delta {\mathit{G}}_{\mathit{j}\mathit{i}}={\displaystyle \sum}_{\mathit{i}=\mathrm{1}}^{M}{\mathit{r}}_{\mathit{j}\mathit{i}}\Delta {\mathit{V}}_{\mathit{i}}$
By taking the appropriate limits (i.e., let M → ∞ and ΔV → 0) and making use of the definition of an integral, we can rewrite the foregoing equation in the form
${\mathit{G}}_{\mathit{j}}={{\displaystyle \int}}^{v}{\mathit{r}}_{\mathit{j}}\text{\hspace{0.17em}}\mathit{d}\mathit{V}$
From this equation, we see that r_{j} will be an indirect function of position, since the properties of the reacting materials and reaction conditions (e.g., concentration, temperature) can have different values at different locations in the reactor volume.
We now replace G_{j} in Equation (13), that is,
by its integral form to yield a form of the general mole balance equation for any chemical species j that is entering, leaving, reacting, and/or accumulating within any system volume V.
This is a basic equation for chemical reaction engineering.
From this general mole balance equation, we can develop the design equations for the various types of industrial reactors: batch, semibatch, and continuousflow. Upon evaluation of these equations, we can determine the time (batch), reactor volume or catalyst weight (continuousflow) necessary to convert a specified amount of the reactants into products.
When is a batch reactor used?
A batch reactor is used for smallscale operation, for testing new processes that have not been fully developed, for the manufacture of expensive products, and for processes that are difficult to convert to continuous operations. The reactor can be charged (i.e., filled) through the holes at the top (see Figure 15(a)). The batch reactor has the advantage of high conversions that can be obtained by leaving the reactant in the reactor for long periods of time, but it also has the disadvantages of high labor costs per batch, the variability of products from batch to batch, and the difficulty of largescale production (see Industrial Reactor Photos in Professional Reference Shelf [PRS] (http://www.umich.edu/~elements/6e/01chap/profreactors.html) on the CRE Web sites, www.umich.edu/~elements/6e/index.html). Also see http://encyclopedia.che.engin.umich.edu/Pages/Reactors/menu.html.
Also see http://encyclopedia.che.engin.umich.edu/Pages/Reactors/Batch/Batch.html.
A batch reactor has neither inflow nor outflow of reactants or products while the reaction is being carried out: F_{j}_{0} = F_{j} = 0. The resulting general mole balance on species j is
$\frac{\mathit{d}{N}_{j}}{\mathit{d}\mathit{t}}={{\displaystyle \int}}^{V}{\mathit{r}}_{\mathit{j}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathit{d}\mathit{V}$
If the reaction mixture is perfectly mixed (Figure 15(b)) so that there is no variation in the rate of reaction throughout the reactor volume, we can take r_{j} out of the integral, integrate, and write the differential form of the mole balance, that is,
Perfect mixing
Let’s consider the isomerization of species A in a batch reactor
$\text{A}\to \text{B}$
As the reaction proceeds, the number of moles of A decreases and the number of moles of B increases, as shown in Figure 16.
We might ask what time, t_{1}, is necessary to reduce the initial number of moles from N_{A0} to a final desired number N_{A1}. Applying Equation (15) to the isomerization
$\frac{\mathit{d}{N}_{\text{A}}}{\mathit{d}\mathit{t}}={\mathit{r}}_{\text{A}}\mathit{V}$
rearranging,
$\mathit{d}\mathit{t}=\frac{\mathit{d}{N}_{\text{A}}}{{r}_{\text{A}}\mathit{V}}$
and integrating with limits that at t = 0, then N_{A} = N_{A0}, and at t = t_{1}, then N_{A} = N_{A1}, we obtain
This equation is the integral form of the mole balance on a batch reactor. It gives the time, t_{1}, necessary to reduce the number of moles from N_{A0} to N_{A1} and also to form N_{B1} moles of B.
Continuousflow reactors are almost always operated at steady state. We will consider three types: the continuousstirred tank reactor (CSTR), the plugflow reactor (PFR), and the packedbed reactor (PBR). Detailed physical descriptions of these reactors can be found in both the Professional Reference Shelf (PRS), (http://www.umich.edu/~elements/6e/01chap/prof.html) for Chapter 1 and in the Visual Encyclopedia of Equipment, http://encyclopedia.che.engin.umich.edu/Pages/Reactors/CSTR/CSTR.html, and on the CRE Web site.
What is a CSTR used for?
A type of reactor commonly used in industrial processing is the stirred tank operated continuously (Figure 17). It is referred to as the continuousstirred tank reactor (CSTR) or vat, or backmix reactor, and is primarily used for
Also see http://encyclopedia.che.engin.umich.edu/Pages/Reactors/CSTR/CSTR.html.
liquidphase reactions. It is normally operated at steady state and is assumed to be perfectly mixed; consequently, there is no time dependence or position dependence of the temperature, concentration, or reaction rate inside the CSTR. That is, every variable is the same at every point inside the reactor. Because the temperature and concentration are identical everywhere within the reaction vessel, they are the same at the exit point as they are elsewhere in the tank. Thus, the temperature and concentration in the exit stream are modeled as being the same as those inside the reactor. In systems where mixing is highly nonideal, the wellmixed model is inadequate, and we must resort to other modeling techniques, such as residence time distributions, to obtain meaningful results. This topic of nonideal mixing is discussed in Chapters 16 and 17, while nonideal flow reactors are discussed in Chapter 18.
When the general mole balance equation
GMBE
is applied to a CSTR operated at steady state (i.e., conditions do not change with time),
$\frac{d{N}_{j}}{dt}=0$
in which there are no spatial variations in the rate of reaction (i.e., perfect mixing),
${{\displaystyle \int}}^{V}{\mathit{r}}_{j}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathit{d}\mathit{V}=\mathit{V}{\mathit{r}}_{j}$
The ideal CSTR is assumed to be perfectly mixed.
it takes the familiar form known sometimes called the design equation for a CSTR
The CSTR design equation gives the reactor volume V necessary to reduce the entering molar flow rate of species j from F_{j}_{0} to the exit molar flow rate F_{j}, when species j is disappearing at a rate of –r_{j}. We note that the CSTR is modeled such that the conditions in the exit stream (e.g., concentration and temperature) are identical to those in the tank. The molar flow rate F_{j} is just the product of the concentration of species j and the volumetric flow rate υ
Applying Equation (18) at the entrance of the reactor one obtains
F_{j}_{0} = C_{j}_{0} · υ_{0}
Consequently, we can substitute for F_{j}_{0} and F_{j} into Equation (17) to write a balance on species A in terms of concentration, as
The ideal CSTR mole balance equation is an algebraic equation, not a differential equation.
In addition to the CSTR and batch reactors, another type of reactor commonly used in industry is the tubular reactor. It consists of a cylindrical pipe and is normally operated at steady state, as is the CSTR. Tubular reactors are used most often for gasphase reactions. A schematic and a photograph of industrial tubular reactors are shown in Figure 18.
When is a tubular reactor most often used?
Also see http://encyclopedia.che.engin.umich.edu/Pages/Reactors/PFR/PFR.html.
In the tubular reactor, the reactants are continually consumed as they flow down the length of the reactor. In modeling the tubular reactor, we assume that the concentration varies continuously in the axial direction through the reactor. Consequently, the reaction rate, which is a function of concentration for all but zeroorder reactions (cf. Equation 32), will also vary axially. For the purposes of the material presented here, we consider systems in which the flow field may be modeled by that of a plugflow profile (e.g., uniform radial velocity as in turbulent flow), as shown in Figure 19. That is, there is no radial variation in reaction rate, and the reactor is referred to as a plugflow reactor (PFR). (The laminarflow reactor (LFR) is discussed in Chapters 16–18, along with a discussion of nonideal reactors.)
Also see PRS and Visual Encyclopedia of Equipment.
The general mole balance equation is given by Equation (14):
The equation we will use to design PFRs at steady state can be developed in two ways: (1) directly from Equation (14) by differentiating with respect to volume V, and then rearranging the result or (2) from a mole balance on species j in a differential segment of the reactor volume ΔV. Let’s choose the second way to arrive at the differential form of the PFR mole balance. The differential volume, ΔV, shown in Figure 110, will be chosen sufficiently small such that there are no spatial variations in reaction rate within this volume. Thus the generation term, ΔG_{j}, is
$\Delta {\mathit{G}}_{\mathit{j}}={{\displaystyle \int}}^{\Delta V}{\mathit{r}}_{\mathit{j}}\text{}\mathit{d}\mathit{V}={\mathit{r}}_{\mathit{j}}\text{\Delta}\mathit{V}$
Dividing Equation (110) by ΔV and rearranging
$\mathrm{\left[}\frac{{F}_{\mathit{j}}{\mathrm{}}_{V+\Delta {V}^{\mathrm{}}}{F}_{\mathit{j}}{\mathrm{}}_{V}}{\Delta \mathit{V}}\mathrm{\right]}={\mathit{r}}_{\mathit{j}}$
we note the term in brackets resembles the definition of a derivative
$\underset{\Delta x\to 0}{\mathrm{lim}}\left[\frac{f(x+\Delta x)f\left(x\right)}{\Delta x}\right]=\frac{df}{dx}$
Taking the limit as ΔV approaches zero, we obtain the differential form of steadystate mole balance on a PFR
We could have made the cylindrical reactor on which we carried out our mole balance an irregularly shaped reactor, such as the one shown in Figure 111 for reactant species A. However, we see that by applying Equation (110), the result would yield the same equation (i.e., Equation (111)). For species A, the mole balance is
Picasso’s reactor
Consequently, we see that Equation (111) applies equally well to our model of tubular reactors of variable and constant crosssectional area, although it is doubtful that one would find a reactor of the shape shown in Figure 111 unless it were designed by Pablo Picasso or perhaps one of his followers.
The conclusion drawn from the application of the design equation to Picasso’s reactor is an important one: the degree of completion of a reaction achieved in an ideal plugflow reactor (PFR) does not depend on its shape, only on its total volume.
Lets again consider the isomerization A → B, this time in a PFR. As the reactants proceed down the reactor, A is consumed by chemical reaction and B is produced. Consequently, the molar flow rate F_{A} decreases as shown in Figure 112(a) while F_{B} increases as the reactor volume V increases, as shown in Figure 112(b).
$V={\int}_{{\text{F}}_{\text{A}}}^{{\text{F}}_{\text{A0}}}\frac{{dF}_{\text{A}}}{\u2013{r}_{\text{A}}}$
We now ask, “What is the reactor volume V_{1} necessary to reduce the entering molar flow rate of A from F_{A0} to an exit flow rate F_{A1}?” Rearranging Equation (112) in the form
$\mathit{d}\mathit{V}=\frac{\mathit{d}{F}_{\text{A}}}{{\mathit{r}}_{\text{A}}}$
and integrating with limits at V = 0, then F_{A} = F_{A0}, and at V = V_{1}, then F_{A} = F_{A1}
V_{1} is the volume necessary to reduce the entering molar flow rate F_{A0} to some specified value F_{A1} and also the volume necessary to produce a molar flow rate of B of F_{B1}.
The principal difference between reactor design calculations involving homogeneous reactions and those involving fluid–solid heterogeneous reactions is that for the latter, the reaction takes place on the surface of the catalyst (see Figure 105). The greater the mass of a given catalyst, the greater the reactive surface area. Consequently, the reaction rate is based on mass of solid catalyst, W, rather than on reactor volume, V. For a fluid–solid heterogeneous system, the rate of reaction of a species A, ${r}_{\text{A}}^{\prime}$, is defined as
$\mathrm{}{r}_{\text{A}}^{\prime}=\text{mol Areacted/}(\text{time}\times \text{massofcatalyst})$
The mass of solid catalyst is used because the amount of catalyst is what is important to the rate of product formation. We note that by multiplying the heterogeneous reaction rate, ${r}_{\text{A}}^{\prime}$, by the bulk catalyst density, $\text{}{\rho}_{b}\left(\frac{\text{mass}}{\text{volume}}\right)$, we can obtain the reaction rate per unit volume, –r_{A}.
${r}_{\text{A}}={\rho}_{\text{b}}({r}_{\text{A}}^{\prime})$
$\left(\frac{\text{m}\text{o}\text{l}}{\text{d}{\text{m}}^{\mathrm{3}}\mathrm{\cdot}\text{s}}\right)\text{=}\left(\frac{\text{g}}{\text{d}{\text{m}}^{3}}\right)\left(\frac{\text{m}\text{o}\text{l}}{\text{g}\mathrm{\cdot}\text{s}}\right)$
The reactor volume that contains the catalyst is of secondary significance. Figure 113 shows a schematic of an industrial catalytic reactor with vertical tubes packed with solid catalyst.
In the three idealized types of reactors just discussed (the perfectly mixed batch reactor [BR], the plugflow tubular reactor [PFR]), and the perfectly mixed continuousstirred tank reactor [CSTR]), the design equations (i.e., mole balances) were developed based on reactor volume. The derivation of the design equation for a packedbed catalytic reactor (PBR) will be carried out in a manner analogous to the development of the tubular design equation. To accomplish this derivation, we simply replace the volume coordinate, V, in Equation (110) with the catalyst mass (i.e., weight) coordinate W (Figure 114).
PBR Mole Balance
As with the PFR, the PBR is assumed to have no radial gradients in concentration, temperature, or reaction rate. The generalized mole balance on species A over catalyst weight ΔW results in the equation
The dimensions of the generation term in Equation (114) are
$\left({r}_{\text{A}}^{\prime}\right)\Delta \mathit{W}\mathrm{\equiv}\frac{\text{moles A}}{\left(\text{time}\right)\left(\text{mass of catalyst}\right)}\cdot \left(\text{mass of catalyst}\right)\equiv \frac{\text{moles}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{A}}{\text{time}}$
which are, as expected, the same dimensions of the molar flow rate F_{A}. After dividing Equation (114) by ΔW and taking the limit as ΔW → 0, we arrive at the differential form of the mole balance for a packedbed reactor:
Use the differential form of design equation for catalyst decay and pressure drop.
When pressure drop through the reactor (see Section 5.5) and catalyst decay (see Section 10.7 in Chapter 10) are neglected, the integral form of the packedcatalystbed design equation can be used to calculate the catalyst weight
You can use the integral form only when there is no ΔP and no catalyst decay.
W is the catalyst weight necessary to reduce the entering molar flow rate of species A, F_{A0}, down to a molar flow rate F_{A}.
For particulate catalytic gasphase systems, the fluidized bed is also in common use. Depending of the flow regime, it can be modeled anywhere between a straight through transport reactor (Chapter 10) to a fluidized bed that is analogous to a CSTR (section 1.4.1), which is shown in Figure 115.
A mole balance on species A in a wellmixed “fluidized” bed is
Dividing by the catalyst weight W, we arrive at the Equation (118) that gives the catalyst weight necessary to reduce the molar rate entering from, F_{A0} (mol/s) to the molar rate leaving, F_{A}, (mol/s) when species A is disappearing at a rate, ${r}_{\text{A}}^{\prime}$ (mol/s·gcat) the design equation
For some insight into “things to come,” consider the following example of how one can use the tubular reactor design in Equation (111).
Let’s consider the liquid phase cis–trans isomerization of 2–butene
which we will write symbolically as
A → B
The reaction is first order in A (–r_{A} = kC_{A}) and is carried out in a tubular reactor in which the volumetric flow rate, υ, is constant, that is, υ = υ_{0}.
Without solving any equations, sketch what you think the concentration profile (C_{A} as a function of V) would look like.
Derive an equation relating the reactor volume to the entering and exiting concentrations of A, the rate constant k, and the volumetric flow rate υ_{0}.
Determine the reactor volume, V_{1}, necessary to reduce the exiting concentration to 10% of the entering concentration, that is, C_{A} = 0.1C_{A0}, when the volumetric flow rate υ_{0} is 10 dm^{3}/min (i.e., liters/min) and the specific reaction rate, k, is 0.23. min^{–1}.
Solution
Sketch C_{A} as a function of V.
Species A is consumed as we move down the reactor, and as a result, both the molar flow rate of A and the concentration of A will decrease. Because the volumetric flow rate is constant, υ = υ_{0}, one can use Equation (18) to obtain the concentration of A, C_{A} = F_{A}/υ_{0}, and then by comparison with the plot in Figure 112, obtain the concentration of A as a function of reactor volume, as shown in Figure E12.1.
Derive an equation relating V, υ_{0}, k, C_{A0}, and C_{A}.
For a tubular reactor, the mole balance on species A (j = A) was shown to be given by Equation (111). Then for species A (j = A)
Mole Balance:
$\begin{array}{cc}\frac{\mathit{d}{F}_{\text{A}}}{\mathit{d}V}={\mathit{r}}_{\text{A}}& \left(\mathrm{112}\right)\end{array}$
For a firstorder reaction, the rate law (as will be discussed in Chapter 3, Equation (35)) is
Rate Law:
$\begin{array}{cc}{r}_{\text{A}}=k{C}_{\text{A}}& \left(\text{E12.1}\right)\end{array}$
Because the volumetric flow rate, υ, is constant (υ = υ_{0}), as it is for virtually all liquidphase reactions,
$\begin{array}{cc}\frac{\mathit{d}{F}_{\text{A}}}{\mathit{d}V}=\frac{\mathit{d}({C}_{\text{A}}\mathit{v})}{\mathit{d}V}=\frac{\mathit{d}({C}_{\text{A}}{\mathit{v}}_{\mathrm{0}})}{\mathit{d}V}={\mathit{v}}_{\mathrm{0}}\frac{\mathit{d}{C}_{\text{A}}}{\mathit{d}V}={\mathit{r}}_{\text{A}}& \left(\text{E12.2}\right)\end{array}$
Reactor sizing
Multiplying both sides of Equation (E12.2) by minus one and then substituting Equation (E12.1) yields
Combine:
$\begin{array}{cc}\frac{{\upsilon}_{0}d{C}_{\text{A}}}{\mathit{d}V}=\mathrm{}{\mathit{r}}_{\text{A}}=k{C}_{\text{A}}& \left(\text{E12.3}\right)\end{array}$
Separating the variables and rearranging gives
$\mathrm{}\frac{{\mathit{v}}_{\mathrm{0}}}{k}\left(\frac{\mathit{d}{C}_{\text{A}}}{{C}_{\text{A}}}\right)=\mathit{d}V$
Using the conditions at the entrance of the reactor that when V = 0, then C_{A} = C_{A0}
$\begin{array}{cc}\mathrm{}\frac{{\mathit{v}}_{\mathrm{0}}}{k}{{\displaystyle \int}}_{{C}_{\text{A}\mathrm{0}}}^{{C}_{\text{A}}}\frac{\mathit{d}{C}_{\text{A}}}{{C}_{\text{A}}}={{\displaystyle \int}}_{\mathrm{0}}^{v}\mathit{d}V& \left(\text{E12.4}\right)\end{array}$
Carrying out the integration of Equation (E12.4) gives
Solve:
We can also rearrange Equation (E12.5) to solve for the concentration of A as a function of reactor volume to obtain
C_{A} = C_{A0}exp(–kV/υ_{0})
Concentration profile
Calculate the volume V. We want to find the volume, V_{1}, at which ${\mathit{C}}_{\text{A}}=\frac{1}{10}{\mathit{C}}_{\text{A}\mathrm{0}}$ when k = 0.23 min^{–1} and υ_{0} = 10 dm^{3}/min.
Evaluate:
Substituting C_{A0}, C_{A}, υ_{0}, and k in Equation (E12.5), we have $\text{V=}\frac{10\text{\hspace{0.17em}}{\text{dm}}^{3}/\mathrm{min}}{0.23\text{\hspace{0.17em}}{\text{min}}^{1}}\mathrm{ln}\text{\hspace{0.17em}}\frac{{C}_{\text{A0}}}{0.1{C}_{\text{A0}}}=\frac{10{\text{dm}}^{3}}{0.23}\mathrm{ln}\text{\hspace{0.17em}}10=100{\text{dm}}^{3}\left({\text{i.e.,100L;0.1m}}^{3}\right)$ Ans. We see that a reactor volume of 0.1 m^{3} is necessary to convert 90% of species A entering (i.e., C_{A} = 0.1 C_{A0}) into product B for the parameters given.
Let’s now calculate the reactor volume necessary to an even smaller concenttration say (1/100)^{th} of the entering concentration, that is,
C_{A} = 0.01C_{0}
$\text{V=}\frac{10\text{\hspace{0.17em}}{\text{dm}}^{3}/\mathrm{min}}{0.23\text{\hspace{0.17em}}{\text{min}}^{1}}\mathrm{ln}\text{\hspace{0.17em}}\frac{{C}_{\text{A0}}}{0.01{C}_{\text{A0}}}=\frac{10{\text{dm}}^{3}}{0.23}\mathrm{ln}\text{\hspace{0.17em}}100=200{\text{dm}}^{3}$ Ans.
Note: We see that a larger reactor (200 dm^{3}) is needed to reduce the exit concentration to a smaller fraction of the entering concentration (e.g., C_{A} = 0.01 C_{A0}).
Analysis: For this irreversible liquidphase firstorder reaction (i.e., –r_{A} = kC_{A}) being carried out in a PFR, the concentration of the reactant decreases exponentially down the length (i.e., volume V) of the reactor. The more that species A is consumed and converted to product B, the larger must be the reactor volume. The purpose of the example was to give a vision of the types of calculations we will be carrying out as we study chemical reaction engineering (CRE).
Now we will turn Example 12 into a Living Example Problem (LEP) where we can vary parameters to learn their effect on the volume and/or exit concentrations. We could use Polymath, Wolfram, or Python to solve the combined mole balance and rate law to determine the concentration profile. In this example, we will use Polymath.
We begin by rewriting the mole balance, Equation (E12.2), in Polymath notation form
Mole Balances
Rate Law
k = 0.23
v_{0} = 10
Polymath Formulation
A Polymath tutorial to solve the ordinary differential equations (ODEs) can be found on the Web site (http://www.umich.edu/~elements/6e/tutorials/ODE_Equation_Tutorial.pdf and http://www.umich.edu/~elements/6e/tutorials/Polymath_tutorials.html).
The parameter values are k = 0.23 min^{–1}, υ_{0} = 10 dm^{3}/min and C_{A} = 10 mole/dm^{3}. The initial and final values for the integration wrt the volume V are V = 0 and V = 100 dm^{3}.
The output from the Polymath solution is given in Table E13.1 and the axial concentration profiles from species A and B are shown in Figure E13.1.
LEP Sliders
Analysis: Because Polymath will be used extensively in later chapters to solve nonlinear ordinary differential equations (ODEs), we introduce it here so that the reader can start to become familiar with it. Figure E13.1 shows how the concentrations of species A and B vary down the length of the PFR. In order to become familiar with Polymath, the reader is encouraged to solve the foxes and rabbits problem using Polymath (P13_{B}(a)) and then study the dynamics of the reaction using Wolfram or Python (P13_{B}(b)).
Be sure to view the actual photographs of industrial reactors on the CRE Web site so you will know them when you run into them. There are also links to view reactors on different Web sites. The CRE Web site also includes a portion of the Visual Encyclopedia of Equipment, encyclopedia.che.engin.umich.edu, “Chemical Reactors” developed by Dr. Susan Montgomery and her students at the University of Michigan. Also see Professional Reference Shelf on the CRE Web site for “Reactors for LiquidPhase and GasPhase Reactions,” along with photos of industrial reactors, and Expanded Material on the CRE Web site.^{2}
^{2} Chem. Eng., 63(10), 211 (1956). See also AIChE Modular Instruction Series E, 5 (1984).
In this chapter, and on the CRE Web site, we’ve introduced each of the major types of industrial reactors: batch, stirred tank, tubular, and fixed bed (packed bed). Many variations and modifications of these commercial reactors (e.g., semibatch, fluidized bed) are in current use and these reactors will be discussed in Chapters 6 and 10, respectively. For further elaboration, refer to the detailed discussion of industrial reactors given by Walas.^{3}
^{3} S. M. Walas, Reaction Kinetics for Chemical Engineers, New York: McGrawHill, 1959, Chap. 11.
The CRE Web site describes industrial reactors, along with typical feed and operating conditions. In addition, two solved example problems for Chapter 1 can be found on the CRE Web site, http://www.umich.edu/~elements/6e.
A note to students: In this sixth edition of Elements of Chemical Reaction Engineering, I am including a section at the end of each chapter to bring a greater awareness to process safety. A critical aspect of process safety is “anticipating” what could go wrong in a chemical process and ensuring it won’t go wrong. Equipment and processes involving exothermic chemical reactions are some of the most at risk in a chemical plant. Consequently, each chapter will end with a segment “And Now... A Word From Our Sponsor–Safety” (AWFOS–S). In addition, to highlight process safety across the chemical engineering curriculum, a Web site (http://umich.edu/~safeche/) has been developed that features a safety module specific to every core chemical engineering lecture course plus lab safety. In this chapter, we define process safety along with a very brief discussion on why it is important to study process safety.
Chemical process safety is a blend of engineering and management practices focused on preventing accidents, namely explosions, fires, and toxic releases that result in loss of life and property.
Industrial disasters such as UCIL Bhopal, T2 Laboratories, BP Texas City, and Flixborough have collectively killed and injured thousands of people and caused billions of dollars in damage to chemical plants and nearby communities. Accidents such as these occur because chemical engineering processes are some of the most potentially dangerous due to extreme operating conditions and the use of explosive, reactive, and flammable materials. What surprises people is that most of these chemical engineering accidents, such as those listed in the Chemical Safety Board Videos on the companion Web site (http://umich.edu/~safeche/) were preventable. They were the result of poor engineering decisions, made by people who lacked fundamental understanding of basic chemical engineering concepts and chemical engineering safety. Thus, knowing the fundamentals of chemical engineering and process safety may save your life and the lives of innocent people, and prevent the loss of millions of dollars of material and equipment.
Engineers have an ethical and professional obligation to work only in areas for which they are competent and qualified. The best way to prevent future industrial disasters is to understand how to effectively and safely design, operate, and troubleshoot chemical processes. To prepare a prevention plan, we must take the time and effort to understand chemical processes and chemical process safety. To help achieve this understanding, the last section of every chapter has a tutorial, AWFOS–S, that can help you prevent accidents.
A comparison of process safety and personal safety is very succinctly given on the Web site (http://www.energysafetycanada.com/files/pdf/Personal_vs_Process_Safety_v3.pdf).
Each chapter summary gives the key points of the chapter that need to be remembered and carried into succeeding chapters.
A mole balance on species j, which enters, leaves, reacts, and accumulates in a system volume V, is
$\begin{array}{cc}{F}_{\mathit{j}\mathrm{0}}\mathrm{}{F}_{\mathit{j}}+{{\displaystyle \int}}^{\text{V}}{r}_{j}dV=\frac{\mathit{d}{N}_{j}}{\mathit{d}\mathit{t}}& \left(\text{S11}\right)\end{array}$
If, and only if, the contents of the reactor are well mixed will the mole balance (Equation (S11)) on species A give
$\begin{array}{cc}{F}_{\text{A}\mathrm{0}}\mathrm{}{F}_{\text{A}}+{\mathit{r}}_{\text{A}}\mathit{V}=\frac{\mathit{d}{N}_{\text{A}}}{\mathit{d}\mathit{t}}& \left(\text{S12}\right)\end{array}$
The kinetic rate law for r_{j} is
The rate of formation of species j per unit volume (e.g., mol/s·dm^{3})
Solely a function of the properties of reacting materials and reaction conditions (e.g., concentration [activities], temperature, pressure, catalyst, or solvent [if any]) and does not depend on reactor type
An intensive quantity (i.e., it does not depend on the total amount)
An algebraic equation, not a differential equation (e.g., −r_{A} = kC_{A} or ${r}_{\text{A}}=k{C}_{\text{A}}^{2}$)
For homogeneous catalytic systems, typical units of –r_{j} may be gram moles per second per liter; for heterogeneous systems, typical units of ${r}_{j}^{\prime}$ may be gram moles per second per gram of catalyst. By convention, –r_{A} is the rate of disappearance of species A and r_{A} is the rate of formation of species A.
Mole balances on species A in four common reactors are shown in Table S11.
TABLE S11 SUMMARY OF REACTOR MOLE BALANCES

Reactor 
Comment 
Mole Balance Differential Form 
Algebraic Form 
Integral Form 
BR 
No spatial variations 
$\frac{{dN}_{\text{A}}}{dt}={r}_{\text{A}}V$ 
${t}_{1}={\int}_{{N}_{\text{A1}}}^{{N}_{\text{A0}}}\frac{{dN}_{\text{A}}}{\u2013{r}_{\text{A}}V}$ 

CSTR 
No spatial variations, steady state 
— 
$V=\frac{{F}_{\text{A0}}\u2013{F}_{\text{A}}}{\u2013{r}_{\text{A}}}$ 
— 

PFR 
Steady state 
$\frac{{dF}_{\text{A}}}{dV}={r}_{\text{A}}$ 
${V}_{1}={\int}_{{F}_{\text{A1}}}^{{F}_{\text{A0}}}\frac{d{F}_{\text{A}}}{\u2013{r}_{\text{A}}}$ 

PBR 
Steady state 
$\frac{{dF}_{\text{A}}}{dV}={r}_{\text{A}}^{\prime}$ 
${W}_{1}={\int}_{{F}_{\text{A1}}}^{{F}_{\text{A0}}}\frac{d{F}_{\text{A}}}{\u2013{r}_{\text{A}}^{\prime}}$ 

Fluidized CSTR 
Steady state 
— 
$W=\frac{{F}_{\text{A0}}\u2013{F}_{\text{A}}}{\u2013{r}_{\text{A}}^{\prime}}$ 
— 
(http://www.umich.edu/~elements/6e/01chap/obj.html#/)
Getting Unstuck on a Problem
(http://www.umich.edu/~elements/6e/01chap/iclicker_ch1_q1.html)
Smog in L.A. Web Module
Photograph by Radoslaw Lecyk/Shutterstock
(http://www.umich.edu/~elements/6e/web_mod/la_basin/index.htm)
Living Example Problem:
http://www.umich.edu/~elements/6e/01chap/live.html
Interactive Computer Games (http://www.umich.edu/~elements/6e/icm/index.html)
A. Quiz Show I (http://www.umich.edu/~elements/6e/icm/kinchal1.html)
This game could help prepare you for the AIChE student chapter Jeopardy Competition held each year at the Annual AIChE meeting.
I wish I had an answer for that, because I’m getting tired of answering that question.
—Yogi Berra, New York Yankees
Sports Illustrated, June 11, 1984
The subscript to each of the problem numbers indicates the level of difficulty, that is, A, least difficult; B, moderate difficulty; C, fairly difficult; D, (double black diamond), most difficult. A = • B = ▪ C = ♦ D = ♦♦ For example, P15_{B} means “1” is the Chapter number, “5” is the problem number, “_{B}” is the problem difficulty, in this case B means moderate difficulty.
Before solving the problems, state or sketch qualitatively the expected results or trends.
Q11_{A} QBR Questions Before Reading. Research has shown (J. Exp. Psychol. Learn. Mem. Cogn., 40, 106–114 (2014)) that if you ask a question of the material before reading the material you will have greater retention. Consequently, the first question of every chapter will have such a question on that chapter’s material. For Chapter 1, the question is “Is the generation term, G, the only term in the mole balance that varies for each type of reactor?”
Q12_{A} Go to Chapter 1 Evaluation on the Web site. Click on i>Clicker Questions (http://www.umich.edu/~elements/6e/01chap/iclicker_ch1_q1.html) and view at least five i>clicker questions. Choose one that could be used as is, or a variation thereof, to be included on the next exam. You also could consider the opposite case: explaining why the question should not be on the next exam. In either case, explain your reasoning.
Q13_{A} What if... the PFR in Example 12 were replaced by a CSTR, what would be its volume?
Q14_{A} What if... you were asked to rework Example 12 to calculate the time to reduce the number of moles of A to 1% if its initial value for a constant volume BR, what would you say? Would you do it? If your answer is “yes,” go ahead and calculate it; if your answer is “NO, I won’t do it!” then suggest two ways to work this problem incorrectly.
Q15_{A} Read through the Introduction. Write a paragraph describing both the content goals and the intellectual goals of the course and text. Also describe what’s on the Web site and how the Web site can be used with the text and course.
Q16_{A} Go to Chapter 1 Useful Links (http://www.umich.edu/~elements/6e/01chap/obj.html#/) and click on Professional Reference Shelf to view the photos and schematics of real reactors. Write a paragraph describing two or more of the reactors. What similarities and differences do you observe between the reactors on the Web site (e.g., www.loebequipment.com), and in the text? How do the used reactor prices compare with those in Table 11?
Q17_{A} What assumptions were made in the derivation of the design equation for: (a) The batch reactor (BR)? (b) The CSTR? (c) The plugflow reactor (PFR)? (d) The packedbed reactor (PBR)? (e) State in words the meanings of –r_{A} and ${r}_{\text{A}}^{\prime}$.
Q18_{A} Fill out the following table for each of the reactors discussed in this chapter, BR, CSTR, PBR, and Fluidized Bed:
Type of Reactor 
Characteristics 
Phases Present 
Usage 
Advantages 
Disadvantages 












Q19_{A} Define Chemical Process Safety and list four reasons we need to study it and why it is particularly relevant to CRE (http://umich.edu/~safeche/index.html).
Q110_{A} Go to Chapter 1 Extra Help on the Web site and click on LearnChemE ScreenCasts (http://www.umich.edu/~elements/6e/01chap/learnchemevideos.html). Choose one of the LearnChemE videos and critique it for such things as (a) value, (b) clarity, (c) visuals, and (d) how well it held your interest. (Score 1–7; 7 = outstanding, 1 = poor)
Q111_{A} Go to Chapter 1 Extra Help on the Web site and click on LearnChemE ScreenCasts (http://www.umich.edu/~elements/6e/01chap/learnchemevideos.html) the How to Study screencast and list three ways that screencasts can help you learn the material.
Q112_{A} Go to Extra Help then click on Videos of Tips on Studying and Learning. Go to Chapter 1 (http://www.umich.edu/~elements/6e/01chap/obj.html#/videotips/).
View one of the 5 to 6minute video tutorials and list two of the most important points in the video. List what two things you think this screencast did well?
After viewing the three screencasts on How to Study (http://www.learncheme.com/studentresources/howtostudyresources), describe the most efficient way to study. In video 3 How to Study, the author of this book has a very different view of one of the points suggested. What do you think it is?
View the video 13 Study Tips**** (4 Stars) (https://www.youtube.com/watch?v=eVlvxHJdql8&feature=youtu.be). List four of the tips that you think might help your study habits.
Rate each of the sites on video tips, (1) Not Helpful, (5) Very Helpful.
Before running your experiments, stop a moment and try to predict how your curves will change shape as you change a variable (cf. Q11_{A}).
P11_{A}
Revisit Example 13.
Wolfram and Python
Describe how C_{A} and C_{B} change when you experiment with varying the volumetric flow rate, υ_{0}, and the specific reaction rate, k, and then write a conclusion about your experiments.
Click on the description of reversible reaction A ⇄ B to understand how the rate law becomes ${r}_{\text{A}}=k[{C}_{\text{A}}\frac{{C}_{B}}{{K}_{e}}]$. Set K_{e} at its minimum value and vary k and υ_{0}. Next, set K_{e} at its maximum value and vary k and υ_{0}. Write a couple sentences describing how varying k, υ_{0}, and K_{e} affect the concentration profiles. We will learn more about K_{e} in Section 3.2.
After reviewing Generating Ideas and Solutions on the Web site (http://www.umich.edu/~elements/6e/toc/SCPS,3rdEdBook(Ch07).pdf), choose one of the brainstorming techniques (e.g., lateral thinking) to suggest two questions that should be included in this problem.
Polymath
Modify the Polymath program to consider the case where the reaction is reversible as discussed in part (ii) above with K_{e} = 3. How do your results (i.e., C_{A}) compare with the irreversible reaction case?
P12_{B} Schematic diagrams of the Los Angeles basin are shown in Figure P12_{B}. The basin floor covers approximately 700 square miles (2 × 10^{10} ft^{2}) and is almost completely surrounded by mountain ranges. If one assumes an inversion height in the basin of 2,000 ft, the corresponding volume of air in the basin is 4 × 10^{13} ft^{3}. We shall use this system volume to model the accumulation and depletion of air pollutants. As a very rough first approximation, we shall treat the Los Angeles basin as a wellmixed container (analogous to a CSTR) in which there are no spatial variations in pollutant concentrations.
We shall perform an unsteadystate mole balance (Equation (14)) on CO as it is depleted from the basin area by a Santa Ana wind. Santa Ana winds are highvelocity winds that originate in the Mojave Desert just to the northeast of Los Angeles. Load the Smog in Los Angeles Basin Web Module. Use the data in the module to work parts 1–12 (a) through (h) given in the module. Load the Living Example Polymath code and explore the problem. For part (i), vary the parameters υ_{0}, a, and b, and write a paragraph describing what you find.
There is heavier traffic in the L.A. basin in the mornings and in the evenings as workers go to and from work in downtown L.A. Consequently, the flow of CO into the L.A. basin might be better represented by the sine function over a 24hour period.
P13_{B} This problem focuses on using Polymath, an ordinary differential equation (ODE) solver, and also a nonlinear equation (NLE) solver. These equation solvers will be used extensively in later chapters. Information on how to obtain and load the Polymath Software is given in Appendix D and on the CRE Web site.
Professor Sven Köttlov has a soninlaw, Štěpán Dolež, who has a farm near Riça, Jofostan where there are initially 500 rabbits (x) and 200 foxes (y). Use Polymath or MATLAB to plot the concentration of foxes and rabbits as a function of time for a period of up to 500 days. The predator–prey relationships are given by the following set of coupled ordinary differential equations:
$\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}={k}_{\mathrm{1}}\mathit{x}\mathrm{}{k}_{\mathrm{2}}\mathit{x}\mathrm{\cdot}\mathit{y}$
$\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}={k}_{3}\mathit{x}\mathrm{\cdot}\mathit{y}\mathrm{}{k}_{\mathrm{4}}y$
Constant for growth of rabbits k_{1} = 0.02 day^{–1}
Constant for death of rabbits k_{2} = 0.00004/(day × no. of foxes)
Constant for growth of foxes after eating rabbits k_{3} = 0.0004/(day × no. of rabbits)
Constant for death of foxes k_{4} = 0.04 day^{–1}
What do your results look like for the case of k_{3} = 0.00004/(day × no. of rabbits) and t_{final} = 800 days? Also, plot the number of foxes versus the number of rabbits. Explain why the curves look the way they do. Polymath Tutorial (https://www.youtube.com/watch?v=nyJmt6cTiL4)
Using Wolfram and/or Python in the Chapter 1 LEP on the Web site, what parameters would you change to convert the foxes versus rabbits plot from an oval to a circle? Suggest reasons that could cause this shape change to occur.
We will now consider the situation in which the rabbits contracted a deadly virus also called rabbit measles (measlii). The death rate is r_{Death} = k_{D}x with k_{D} = 0.005 day^{–1}. Now plot the fox and rabbit concentrations as a function of time and also plot the foxes versus rabbits. Describe, if possible, the minimum growth rate at which the death rate does not contribute to the net decrease in the total rabbit population.
Use Polymath or MATLAB to solve the following set of nonlinear algebraic equations
x^{3}y – 4y^{2} + 3x = 1
6y^{2} – 9xy = 5
with inital guesses of x = 2, y = 2. Try to become familiar with the edit keys in Polymath and MATLAB. See the CRE Web site for instructions. You will need to know how to use this solver in later chapters involving CSTRs.
Screen shots on how to run Polymath are shown at the end of Summary Notes for Chapter 1 or on the CRE Web site, www.umich.edu/~elements/6e/software/polymathtutorial.html.
P14_{A} Find the Interactive Computer Games (ICG) on the CRE Web site (http://www.umich.edu/~elements/6e/icg/index.html). Read the description of the Kinetic Challenge module (http://www.umich.edu/~elements/6e/icm/kinchal1.html) and then go to the installation instructions (http://www.umich.edu/~elements/6e/icm/install.html) to install the module on your computer. Play this game and then record your performance number, which indicates your mastery of the material.
Jeopardy Game Held at Annual AIChE Student Chapter Meeting
ICG Kinetics Challenge 1 Performance # ___________________________
P15_{A} OEQ (Old Exam Question) The reaction
A + B → 2C
takes place in an unsteady CSTR. The feed is only A and B in equimolar proportions. Which of the following sets of equations gives the correct set of mole balances on A, B, and C? Species A and B are disappearing and species C is being formed. Circle the correct answer where all the mole balances are correct. (a) (b)
$\begin{array}{c}{F}_{B\mathrm{0}}\mathrm{}{F}_{\text{A}}\mathrm{}{{\displaystyle \int}}^{V}{r}_{\text{A}}\mathit{d}V=\frac{\mathit{d}{N}_{\text{A}}}{\mathit{d}\mathit{t}}\phantom{\rule{0ex}{0ex}}\\ {F}_{B\mathrm{0}}\mathrm{}{F}_{\text{B}}\mathrm{}{{\displaystyle \int}}^{V}{r}_{\text{A}}\mathit{d}V=\frac{\mathit{d}{N}_{\text{B}}}{\mathit{d}\mathit{t}}\phantom{\rule{0ex}{0ex}}\\ {F}_{\text{C}}+2{{\displaystyle \int}}^{V}{r}_{\text{A}}\mathit{d}V=\frac{\mathit{d}{N}_{\text{C}}}{\mathit{d}\mathit{t}}\phantom{\rule{0ex}{0ex}}\end{array}$
$\begin{array}{c}{F}_{\text{A}\mathrm{0}}\mathrm{}{F}_{\text{A}}\mathrm{+}{{\displaystyle \int}}^{V}{r}_{\text{A}}\mathit{d}V=\frac{\mathit{d}{N}_{\text{A}}}{\mathit{d}\mathit{t}}\phantom{\rule{0ex}{0ex}}\\ {F}_{\text{A}\mathrm{0}}\mathrm{}{F}_{\text{B}}\mathrm{+}{{\displaystyle \int}}^{V}{r}_{\text{A}}\mathit{d}V=\frac{\mathit{d}{N}_{\text{B}}}{\mathit{d}\mathit{t}}\phantom{\rule{0ex}{0ex}}\\ {F}_{\text{C}}\u20132{{\displaystyle \int}}^{V}{r}_{\text{A}}\mathit{d}V=\frac{\mathit{d}{N}_{\text{C}}}{\mathit{d}\mathit{t}}\phantom{\rule{0ex}{0ex}}\end{array}$
$\begin{array}{c}{F}_{\text{A}\mathrm{0}}\mathrm{}{F}_{\text{A}}\mathrm{+}{{{\displaystyle \int}}^{V}}_{0}{r}_{\text{A}}\mathit{d}V=\frac{\mathit{d}{N}_{\text{A}}}{\mathit{d}\mathit{t}}\phantom{\rule{0ex}{0ex}}\\ {F}_{\text{A}\mathrm{0}}\mathrm{}{F}_{\text{B}}\mathrm{+}{{\displaystyle \int}}^{V}{r}_{\text{A}}\mathit{d}V=\frac{\mathit{d}{N}_{\text{B}}}{\mathit{d}\mathit{t}}\phantom{\rule{0ex}{0ex}}\\ {F}_{\text{C}}+{{\displaystyle \int}}^{V}{r}_{\text{C}}\mathit{d}V=\frac{\mathit{d}{N}_{\text{C}}}{\mathit{d}\mathit{t}}\phantom{\rule{0ex}{0ex}}\end{array}$
$\begin{array}{c}{F}_{\text{B}\mathrm{0}}\mathrm{}{F}_{\text{A}}\mathrm{}{{\displaystyle \int}}^{V}{r}_{\text{A}}\mathit{d}V=\frac{\mathit{d}{N}_{\text{A}}}{\mathit{d}\mathit{t}}\phantom{\rule{0ex}{0ex}}\\ {F}_{\text{B}\mathrm{0}}\mathrm{}{F}_{\text{A0}}\mathrm{}{{\displaystyle \int}}^{V}{r}_{\text{A}}\mathit{d}V=\frac{\mathit{d}{N}_{\text{B}}}{\mathit{d}\mathit{t}}\phantom{\rule{0ex}{0ex}}\\ {F}_{\text{C}}+{{\displaystyle \int}}^{V}{r}_{\text{C}}\mathit{d}V=\frac{\mathit{d}{N}_{\text{C}}}{\mathit{d}\mathit{t}}\phantom{\rule{0ex}{0ex}}\end{array}$
None of the above.
This problem was written in honor of Ann Arbor, Michigan’s own Grammy winning artist, Bob Seger (https://www.youtube.com/channel/UComKJVf5rNLl_RfC_rbt7qg/videos).
P16_{B} The reaction
A → B
is to be carried out isothermally in a continuousflow reactor. The entering volumetric flow rate υ_{0} is 10 dm^{3}/h. Note: F_{A} = C_{A}υ. For a constant volumetric flow rate υ = υ_{0}, then F_{A} = C_{A}υ_{0}. Also, C_{A0} = F_{A0}/υ_{0} = ([15 mol/h]/[10 dm^{3}/h])0.5 mol/dm^{3}.
Calculate both the CSTR and PFR volumes necessary to consume 99% of A (i.e., C_{A} = 0.01C_{A0}) when the entering molar flow rate is 5 mol/h, assuming the reaction rate –r_{A} is
−r_{A} = k with $k=0.05\frac{\text{mol}}{\text{h}\cdot {\text{dm}}^{3}}$ [Ans.: V_{CSTR} = 99 dm^{3}]
−r_{A} = k_{CA} with 0.0001 s^{−1}
−r_{A} = ${kC}_{\text{A}}^{2}$ with $k=300\frac{{\text{dm}}^{3}}{\text{mol}\cdot \text{h}}$ [Ans.: V_{CSTR} = 660 dm^{3}]
Repeat (a), (b), and/or (c) to calculate the time necessary to consume 99.9% of species A in a 1000 dm^{3} constantvolume batch reactor with C_{A0} = 0.5 mol/dm^{3}.
P17_{A} Enrico Fermi (1901–1954) Problems (EFP). Enrico Fermi was an Italian physicist who received the Nobel Prize for his work on nuclear processes. Fermi was famous for his “Back of the Envelope Order of Magnitude Calculation” to obtain an estimate of the answer through logic and then to make reasonable assumptions. He used a process to set bounds on the answer by saying it is probably larger than one number and smaller than another, and arrived at an answer that was within a factor of 10.
See http://mathforum.org/workshops/sum96/interdisc/sheila2.html.
Enrico Fermi Problem
EFP #1. How many piano tuners are there in the city of Chicago? Show the steps in your reasoning.
Population of Chicago __________
Number of people per household __________
And so on, __________
An answer is given on the CRE Web site under Summary Notes for Chapter 1.
EFP #2. How many square meters of pizza were eaten by an undergraduate student body population of 20,000 during the Fall term 2016?
EFP #3. How many bathtubs of water will the average person drink in a lifetime?
P18_{A} What is wrong with this solution? The irreversible liquidphase secondorder reaction $(\mathrm{}\mathit{r}=k{C}_{\text{A}}^{\mathrm{2}})$
$\mathrm{2}A\stackrel{{k}_{1}}{\to}\text{B}\mathit{k}=\mathrm{0}\mathrm{.}\mathrm{0}\mathrm{3}{\text{dm}}^{\mathrm{3}}\mathrm{/}\text{mol}\mathrm{\cdot}\text{s}$
is carried out in a CSTR. The entering concentration of A, C_{A0}, is 2 molar, and the exit concentration of A, C_{A} is 0.1 molar. The volumetric flow rate, υ_{0}, is constant at 3 dm^{3}/s. What is the corresponding reactor volume?
Solution
Mole Balance
$\mathit{V}=\frac{{F}_{\text{A}\mathrm{0}}\mathrm{}{F}_{\text{A}}}{{r}_{\text{A}}}$
Rate Law (second order)
${r}_{\text{A}}=k{C}_{\text{A}}^{2}$
Combine
$\text{V=}\frac{{F}_{\text{A0}}{F}_{\text{A}}}{k{C}_{\text{A}}^{2}}$
${F}_{\text{A}\mathrm{0}}={v}_{\mathit{0}}{C}_{\text{A}\mathrm{0}}=\frac{\mathrm{3}\text{d}{\text{m}}^{\mathrm{3}}}{\text{s}}\mathrm{\cdot}\frac{\mathrm{2}\text{m}\text{o}\mathrm{l}\text{A}}{\text{d}{\text{m}}^{\mathrm{3}}}\text{=}\frac{\mathrm{6}\text{m}\text{o}\text{l}\text{A}}{\text{s}}$
${F}_{\text{A}}={v}_{\mathit{0}}{C}_{\text{A}}=\frac{\mathrm{3}\text{d}{\text{m}}^{\mathrm{3}}}{\text{s}}\mathrm{\cdot}\frac{\mathrm{0}\mathrm{.}\mathrm{1}\text{m}\text{o}\mathrm{l}\text{A}}{\text{d}{\text{m}}^{\mathrm{3}}}\text{=}\frac{\mathrm{0}\mathrm{.}\mathrm{3}\text{m}\text{o}\text{l}\text{A}}{\text{s}}$
$\mathit{V}\text{=}\frac{\text{(}\mathrm{6}\mathrm{}\mathrm{0}\mathrm{.}\mathrm{3}\text{)}\frac{\text{m}\text{o}\mathrm{l}}{\text{s}}}{(\mathrm{0}\mathrm{.}\mathrm{0}\mathrm{3}\frac{\text{d}{\text{m}}^{\mathrm{3}}}{\text{m}\text{o}\mathrm{l}\cdot \text{s}})(\mathrm{2}\frac{\text{m}\text{o}\mathrm{l}}{\text{d}{\text{m}}^{\mathrm{3}}}{)}^{2}}\text{=}\mathrm{4}\mathrm{7}\mathrm{.}\mathrm{5}\text{d}{\text{m}}^{\mathrm{3}}$
If you like the Puzzle Problems in “What is wrong with the solutions”^{†} you can find more for later chapters on the Web site under Additional Material for that chapter.
^{†}Puzzle Problems for each chapter can be found on the CRE Web site under Expanded Material.
For more puzzles on what’s “wrong with this solution,” see additional material for each chapter on the CRE Web site home page, under “Expanded Material.”
NOTE TO INSTRUCTORS: Additional problems (cf. those from the preceding editions) can be found in the solutions manual and on the CRE Web site. These problems could be photocopied and used to help reinforce the fundamental principles discussed in this chapter.
For further elaboration of the development of the general balance equation, see not only the Web site www.umich.edu/~elements/6e/index.html but also
R. M. FELDER and R. W. ROUSSEAU, Elementary Principles of Chemical Processes, 3rd ed. New York: Wiley, 2000, Chap. 4.
R. J. SANDERS, The Anatomy of Skiing. Denver, CO: Golden Bell Press, 1976.
A detailed explanation of a number of topics in this chapter can be found in the tutorials.
B. L. CRYNES and H. S. FOGLER, eds., AIChE Modular Instruction Series E: Kinetics, Vols. 1 and 2. New York: AIChE, 1981.
A discussion of some of the most important industrial processes is presented by
G. T. AUSTIN, Shreve’s Chemical Process Industries, 5th ed. New York: McGrawHill, 1984.
Short instructional videos (6–9 minutes) that correspond to the topics in this book can be found at http://www.learncheme.com/.
See the Web site, “Process Safety Across the Chemical Engineering Curriculum,” (http://umich.edu/~safeche/index.html).