Success is a journey, not a destination.
—Ben Sweetland
The overall goal is to use the following equation:
$\begin{array}{c}\hline \text{RTDData}+\text{Model}+\text{Kinetics}=\text{Prediction}\\ \hline\end{array}$
By Kinetics we usually mean the rate law and ratelaw parameters. The choice of the particular model to be used depends largely on the chemical reaction engineer’s judgment and analysis of the reaction system. It is your job to choose the model that best combines the conflicting goals of mathematical simplicity and physical realism. There is a certain amount of art in the development of a model for a particular reactor, and the examples presented here can only point toward a direction that a chemical reaction engineer’s thinking might follow.
Conflicting goals
A Model must
Fit the data
Be able to extrapolate theory and experiment
Have realistic parameters
For a given real reactor, it is not uncommon to use all the models discussed previously to predict conversion and then make comparisons. Usually, the real conversion will be bounded by the model calculations.
The following guidelines are suggested when developing models for nonideal reactors:
The model must be mathematically tractable. The equations used to describe a chemical reactor should be solved without an inordinate expenditure of human or computer time.
The model must realistically describe the characteristics of the nonideal reactor. The phenomena occurring in the nonideal reactor must be reasonably described physically, chemically, and mathematically.
The model should not have more than two adjustable parameters. This constraint is often used because an expression with more than two adjustable parameters can be fitted to a great variety of experimental data, and the modeling process in this circumstance is nothing more than an exercise in curve fitting. The statement “Give me four adjustable parameters and I can fit an elephant; give me five and I can include his tail!” is one that I have heard from many colleagues. Unless one is into modern art, a substantially larger number of adjustable parameters is necessary to draw a reasonablelooking elephant.^{1} A oneparameter model is, of course, superior to a twoparameter model if the oneparameter model is sufficiently realistic. To be fair, however, in complex systems (e.g., internal diffusion and conduction, mass transfer limitations) where other parameters may be measured independently, then more than two parameters are quite acceptable.
^{1} J. Wei, CHEMTECH, 5, 128 (1975).
Table 181 gives some guidelines that will help your analysis and model building of nonideal reaction systems.
The Guidelines
TABLE 181 A PROCEDURE FOR CHOOSING A MODEL TO PREDICT THE OUTLET CONCENTRATIONS AND THE CONVERSION

When using the algorithm in Table 181, we classify a model as being either a oneparameter model (e.g., tanksinseries model or dispersion model) or a twoparameter model (e.g., reactor with bypassing and dead volume or combinations of ideal reactors). In Sections 18.1.1 and 18.1.2, we give an overview of these models, which will be discussed in greater detail later in the chapter.
Here, we use a single parameter to account for the nonideality of our reactor. This parameter is most always evaluated by analyzing the RTD determined from a tracer test.
CSTRs. Examples of oneparameter models for nonideal CSTRs include either a reactor dead volume, V_{D}, where no reaction takes place, or volumetric flow rate with part of the fluid bypassing the reactor, υ_{b}, thereby exiting unreacted. Tubular Reactors. Examples of oneparameter models for tubular reactors include the tanksinseries model and the dispersion model. For the tanksinseries model, the one parameter is the number of tanks, n, and for the dispersion model, the one parameter is the dispersion coefficient, D_{a}.^{†} Knowing these parameter values, we then proceed to determine the conversion and/or effluent concentrations for the reactor.
^{†} Nomenclature note: Da_{1} (or Da_{2}) is the Damköhler number and D_{a} is the dispersion coefficient.
Nonideal tubular reactors
First let’s consider nonideal tubular reactors. Tubular reactors may be empty (i.e., no packing), or they may be packed with some material that acts as a catalyst, heattransfer medium, or means of promoting interphase contact. Until Chapters 16–18, we have modeled the fluid to move through the reactor in a pistonlike flow where every atom spends an identical length of time in the reaction environment. Here, the velocity profile is flat, and there is no axial mixing. Both of these assumptions are false, to some extent in every tubular reactor; frequently, they are sufficiently false to warrant some modification of our CRE algorithm. Tubular reactor models need to have the flexibility to allow for failure of the plugflow assumption and the insignificant axial mixing assumption. Cases where these failures occur include the unpacked laminarflow tubular reactor, the unpacked turbulent flow reactor, and packedbed reactors. One of two approaches is usually taken to compensate for failure of either or both of the ideal assumptions. One approach involves modeling the nonideal tubular reactor as a series of identically sized CSTRs, which was discussed in Section 17.4. The other approach (the dispersion model) involves a modification of the ideal reactor by including axial and radial dispersion on the flow and is discussed in this chapter.
The premise for the twoparameter model is that we can use a combination of ideal reactors to model the real reactor (cf. Section 18.8). For example, consider a packedbed reactor with channeling and dead zone. Here, the response to a pulse tracer input would show two dispersed pulses in the output as shown in Figure 161 and Figure 181.
Here, we could model the real reactor as two ideal PBRs in parallel, with the two parameters being the volumetric flow rate that channels or by passes, υ_{b}, and the reactor dead volume, V_{D}. The real reactor volume is V = V_{D} + V_{S} with entering volumetric flow rate υ_{0} = υ_{b} + υ_{S}.
The dispersion model is also often used to describe nonideal tubular reactors. In this model, there is an axial dispersion of the material, which is governed by an analogy to Fick’s law of diffusion, superimposed on the flow as shown in Figure 182. So in addition to transport by bulk flow, UA_{c}C, every component in the mixture is transported through any cross section of the reactor at a rate equal to [–D_{a}A_{c}(dC/dz)] resulting from molecular and convective diffusion. By convective diffusion (i.e., dispersion), we mean either ArisTaylor dispersion in laminarflow reactors or turbulent diffusion resulting from turbulent eddies. Radial concentration profiles for plug flow (a) and a representative axial and radial profile for dispersive flow (b) are shown in Figure 182. Some molecules will diffuse forward ahead of the molar average velocity, while others will lag behind.
Tracer pulse with dispersion
To illustrate how dispersion affects the concentration profile in a tubular reactor, we consider the injection of a perfect tracer pulse. Figure 183 shows how dispersion causes the pulse to broaden as it moves down the reactor and becomes less concentrated.
Why does the tracer pulse broaden?
Recall Equation (1414). The molar flow rate of tracer (F_{T}) by both convection and dispersion is
$\begin{array}{cc}{F}_{T}=[{D}_{a}\frac{\partial {C}_{\text{T}}}{\partial z}+U{C}_{\text{T}}]{A}_{c}& \left(14\text{14}\right)\end{array}$
In this expression, D_{a} is the effective dispersion coefficient (m^{2}/s) and U (m/s) is the superficial velocity. To better understand how the pulse broadens, we refer to the concentration peaks t_{2} and t_{3} in Figure 184. We see that there is a concentration gradient on both sides of the peak causing molecules to diffuse away from the peak and thus broaden the pulse. The pulse then continues to broaden as it moves through the reactor.
Correlations for the dispersion coefficients in both liquid and gas systems may be found in Levenspiel.^{2} Some of these correlations are given in Section 18.4.
^{2} O. Levenspiel, Chemical Reaction Engineering. New York: Wiley, 1962, pp. 290–293.
An unsteadystate mole balance on the inert tracer T gives
$\begin{array}{cc}\frac{\partial {F}_{\text{T}}}{\partial z}={A}_{c}\frac{\partial {C}_{\text{T}}}{\partial t}& \left(18\text{1}\right)\end{array}$
Substituting for F_{T} and dividing by the crosssectional area A_{c}, we have
Pulse tracer balance with dispersion
$\begin{array}{c}\hline {D}_{a}\frac{{\partial}^{2}{C}_{\text{T}}}{\partial {z}^{2}}=\frac{\partial \left(U{C}_{\text{T}}\right)}{\partial z}=\frac{\partial {C}_{\text{T}}}{\partial t}\\ \hline\end{array}\begin{array}{c}\left(18\text{2}\right)\end{array}$
Letting ψ = C_{T}/C_{T0} and λ = z/L and Θ = Ut/L can put Equation (182) in dimensionless form as
$\begin{array}{cc}\hline \frac{1}{{P}_{{e}_{r}}}\frac{{\partial}^{2}\psi}{\partial {\lambda}^{2}}\frac{\partial \psi}{\partial \lambda}=\frac{\partial \psi}{\partial \mathrm{\Theta}}& \left(18\text{3}\right)\\ \hline\end{array}$
where
$\begin{array}{c}\hline {P}_{{e}_{r}}=\frac{Rate\text{}of\text{}Transport\text{}by\text{}Convection}{Rate\text{}of\text{}Transport\text{}by\text{}Diffusion\text{}or\text{}Dispersion}=\frac{Ul}{{D}_{a}}\\ \hline\end{array}$
Here ℓ is the characteristic length, for example, d_{p} for spherical particle or for the case at hand, L, the length of the reactor. The Péclet number, Pe_{r} gives a measure of the degree of dispersion. Small Péclet numbers indicate significant dispersion and high Pe_{r} indicates little dispersion. In the following sections, we will discuss how the dispersion coefficient, D_{a}, and Péclet number can be obtained either from correlations in the literature (cf. Section 18.4) or from the analysis obtained from a tracer experiment on the reactor that yields the RTD curve.
Small Pe_{r}
High Dispersion
Large Pe_{r}
Small Dispersion
To solve this equation, we need the initial and boundary conditions. The boundary conditions will now be discussed in detail.
For open tubes
Pe_{r} ∼ 10^{6},
Pe_{f} ∼ 10^{4}
For packed beds
Pe_{r} ∼ 10^{3},
Pe_{f} ∼ 10^{1}
There are two different types of Péclet numbers in common use. One is the reactor Péclet number, Pe_{r}: It uses the reactor length, L, for the characteristic length, so Pe_{r} ≡ UL/D_{a}. It is Pe_{r} that appears in Equation (1817). The reactor Péclet number, Pe_{r}, for mass dispersion is often referred to as the Bodenstein number, Bo, in reacting systems rather than the Péclet number. The other type of Péclet number can be called the fluid Péclet number, Pe_{f}; it uses the characteristic length that determines the fluid’s mechanical behavior. In a packed bed, this length is the particle diameter d_{p}, and Pe_{f} ≡ Ud_{p}/ϕD_{a}. (The term U is the empty tube or superficial velocity. For packed beds, we often wish to use the average interstitial velocity, and thus U/ϕ is commonly used for the packedbed velocity term.) In an empty tube, the fluid behavior is determined by the tube diameter d_{t}, and Pe_{f} = Ud_{t} /D_{a}. The fluid Péclet number, Pe_{f}, is given in virtually all literature correlations relating the Péclet number to the Reynolds number because both are directly related to the fluid mechanical behavior. It is, of course, very simple to convert Pe_{f} to Pe_{r}: Multiply by the ratio L/d_{p} or L/d_{t}. The reciprocal of Pe_{r}, D_{a}/UL, is sometimes called the vessel dispersion number.
There are two cases that we need to consider: boundary conditions for closedclosed vessels and for openopen vessels. In the case of closedclosed vessels, we assume that there is no dispersion or radial variation in concentration either upstream (closed) or downstream (closed) of the reaction section; hence, this is a closedclosed vessel, as shown in Figure 185(a). In an openopen vessel, dispersion occurs both upstream (open) and downstream (open) of the reaction section; hence, this is an openopen vessel as shown in Figure 185(b). These two cases are shown in Figure 185, where fluctuations in concentration due to dispersion are superimposed on the plugflow velocity profile. A closedopen vessel boundary condition is one in which there is no dispersion in the entrance section but there is dispersion in the reaction and exit sections.
Two types of boundary conditions
For a closedclosed vessel, we have plug flow (no dispersion) to the immediate left of the entrance line (z = 0^{–}) (closed) and to the immediate right of the exit z = L (z = L^{+}) (closed). However, between z = 0^{+} and z = L^{–}, we have dispersion and reaction. The corresponding entrance boundary condition for either reactant A or tracer T is
$\begin{array}{cccc}\text{At z}=0:& & & {F}_{\text{A}}\left({0}^{}\right)={F}_{\text{A}}\left({0}^{+}\right)\end{array}$
Substituting for F_{A} yields
$\begin{array}{cc}U{A}_{c}{C}_{\text{A}}\left({0}^{}\right)={A}_{c}{D}_{a}(\frac{d{C}_{\text{A}}}{dz}{)}_{z={0}^{+}}+U{A}_{c}{C}_{\text{A}}\left({0}^{+}\right)& \left(18\text{}4\right)\end{array}$
Solving for the entering concentration C_{A}(0^{–}) = C_{A0}
Concentration boundary conditions at the entrance
$\begin{array}{c}\hline {C}_{\text{A0}}=\frac{{D}_{a}}{U}(\frac{d{C}_{\text{A}}}{dz}{)}_{z={0}^{+}}+{C}_{\text{A}}\left({0}^{+}\right)\\ \hline\end{array}\begin{array}{c}\left(18\text{5}\right)\end{array}$
At the exit to the reaction section, the concentration is continuous, and there is no gradient in tracer concentration.
Concentration boundary conditions at the exit
At z = L:
$\begin{array}{c}\begin{array}{cc}\hline {C}_{\text{A}}\left({L}^{}\right)={C}_{\text{A}}\left({L}^{+}\right)& \\ \frac{d{c}_{\text{A}}}{dz}=0& \\ \hline\end{array}\end{array}\begin{array}{ccc}\left(\begin{array}{c}18\text{6}\end{array}\right)& & \end{array}$
Danckwerts boundary conditions
These two boundary conditions, Equations (185) and (186), first stated by Danckwerts, have become known as the famous Danckwerts boundary conditions.^{3} Bischoff has given a rigorous derivation by solving the differential equations governing the dispersion of component A in the entrance and exit sections, and taking the limit as the dispersion coefficient, D_{a} in the entrance and exit sections approaches zero.^{4} From the solutions, he obtained boundary conditions on the reaction section identical with those Danckwerts proposed.
^{3} P. V. Danckwerts, Chem. Eng. Sci., 2, 1 (1953).
^{4} K. B. Bischoff, Chem. Eng. Sci., 16, 131 (1961).
The closedclosed concentration boundary condition at the entrance is shown schematically in Figure 186. One should not be uncomfortable with the discontinuity in concentration at z = 0 because if you recall for an ideal CSTR, the concentration drops immediately on entering from C_{A0} to C_{Aexit}. For the other boundary condition at the exit z = L, we see the concentration gradient, (dC_{A}/dz), has gone to zero. At steady state, it can be shown that this Danckwerts boundary condition at z = L also applies to the openopen system at steady state.
ClosedClosed Boundary Condition
Prof. P. V. Danckwerts, Cambridge University, United Kingdom
For an openopen system, there is continuity of flux at the boundaries at z = 0
F_{A}(0^{–}) = F_{A}(0^{+})
Openopen boundary condition
$\begin{array}{c}\hline {D}_{a}\frac{\partial {C}_{\text{A}}}{\partial z}{)}_{z={0}^{}}+U{C}_{\text{A}}\left({0}^{}\right)={D}_{a}\frac{\partial {C}_{\text{A}}}{\partial z}{)}_{z={0}^{+}}+U{C}_{\text{A}}\left({0}^{+}\right)\\ \hline\end{array}\begin{array}{c}\left(18\text{7}\right)\end{array}$
At z = L, we have continuity of concentration and
$\begin{array}{c}\hline \frac{d{C}_{\text{A}}}{dz}=0\\ \hline\end{array}\begin{array}{c}\left(18\text{8}\right)\end{array}$
Now that we have an intuitive feel for how dispersion affects the transport of molecules in a tubular reactor, we shall consider two types of dispersion in a tubular reactor, laminar and turbulent.
In Chapter 14, we showed that the mole balance on reacting species A flowing in a tubular reactor was
$\begin{array}{c}\hline {D}_{a}\begin{array}{cc}\frac{{d}^{2}{C}_{\text{A}}}{d{z}^{2}}U\frac{d{C}_{\text{A}}}{dz}+{r}_{\text{A}}=0& \end{array}\\ \hline\end{array}\begin{array}{c}\left(14\text{17}\right)\end{array}$
Rearranging Equation (1417) we obtain
$\frac{{D}_{a}}{U}\begin{array}{cc}\frac{{d}^{2}{C}_{\text{A}}}{d{z}^{2}}\frac{d{C}_{\text{A}}}{dz}+\frac{{r}_{\text{A}}}{U}=0& \left(18\text{}9\right)\end{array}$
This equation is a secondorder ordinary differential equation. It is nonlinear when r_{A} is other than zero or first order.
When the reaction rate r_{A} is first order, r_{A} = –kC_{A}, then Equation (1810)
Flow, reaction, and dispersion
$\begin{array}{c}\hline \frac{{D}_{a}}{U}\begin{array}{cc}\frac{{d}^{2}{C}_{\text{A}}}{d{z}^{2}}\frac{d{C}_{\text{A}}}{dz}\frac{k{C}_{\text{A}}}{U}=0& \end{array}\\ \hline\end{array}\begin{array}{c}\left(\text{1810}\right)\end{array}$
is amenable to an analytical solution. However, before obtaining a solution, we put our Equation (1810) describing dispersion and reaction in dimensionless form by letting ψ = C_{A}/C_{A0} and λ = z/L
D_{a} = Dispersion coefficient
$\begin{array}{c}\hline \frac{1}{P{e}_{r}}\frac{{d}^{2}\psi}{d{\lambda}^{2}}\frac{d\psi}{d\lambda}{\mathit{\text{D}}}_{{\mathit{\text{a}}}_{\mathbf{1}}}\cdot \psi =0\\ \hline\end{array}\begin{array}{c}\left(18\text{11}\right)\end{array}$
Da_{1} = Damköhler number
We recall from Chapter 5 that the quantity Da_{1} appearing in Equation (1811) is called the Damköhler number for a firstorder conversion and physically represents the ratio
Damköhler number for a firstorder reaction
$\begin{array}{c}\hline {\mathit{D}}_{{\mathit{a}}_{\mathbf{1}}}=\frac{\text{Rate of consumption of A by reaction}}{\text{Rate of transport of A by convection}}=k\tau \\ \hline\end{array}\begin{array}{c}\left(18\text{}12\right)\end{array}$
We now shall solve the dispersion reaction balance for a firstorder reaction
$\begin{array}{cc}\frac{1}{P{e}_{r}}\frac{{d}^{2}\psi}{d{\lambda}^{2}}\frac{d\psi}{d\lambda}{\mathit{\text{D}}}_{{\mathit{\text{a}}}_{\mathbf{1}}}\cdot \psi =0& \left(18\text{11}\right)\end{array}$
For the closedclosed system, the Danckwerts boundary conditions in dimensionless form are
$\begin{array}{cc}\text{At}\lambda =0\text{then}1=\frac{1}{P{e}_{r}}\frac{d\psi}{d\lambda}{)}_{\lambda ={0}^{+}}+\psi \left({0}^{+}\right)& \left(18\text{}13\right)\end{array}$
$\begin{array}{cc}\text{At}\lambda =1\text{then}\frac{d\psi}{d\lambda}=0& \left(18\text{}14\right)\end{array}$
Da_{1} = τk
Pe_{r} = UL/D_{a}
Danckwerts^{†} solved the closedclosed system to give the dimensionless concentration profile as
^{†} Lit cit
Dimensionless concentration profile for dispersion and reaction in a tubular reactor
$\begin{array}{cc}\hline \begin{array}{l}\begin{array}{c}\psi =2*\text{exp}\left(\frac{{P}_{{e}_{r}}\lambda}{2}\right)*\frac{(1+q)\text{exp}\left(\frac{{P}_{{e}_{r}}}{2}q(1\lambda )\right)(1q)\text{exp}(\frac{{P}_{{e}_{r}}}{2}q(1\lambda ))}{(1+q{)}^{2}\text{exp}\left(\frac{{P}_{{e}_{r}}q}{2}\right){(1q)}^{2}\text{}\text{exp}(\frac{{P}_{{e}_{r}}q}{2})}\end{array}\\ \text{where}\\ \text{}q=\sqrt{1+\frac{4\mathit{D}{\mathit{a}}_{\mathbf{1}}}{{P}_{{e}_{r}}}}\end{array}& \\ \\ \hline\end{array}\begin{array}{c}\left(\text{1815}\right)\end{array}$
The conversion profile is
$\begin{array}{c}\hline X=1\psi \\ \hline\end{array}\begin{array}{c}\left(18\text{16}\right)\end{array}$
To find the conversion leaving the reactor, X, we simple set λ = 1 to obtain
Nomenclature note:
Da_{1} is the Damköhler number for a firstorder reaction, τk. D_{a} is the dispersion coefficient in cm^{2}/s and Pe_{r} = UL/D_{a}.
$\begin{array}{c}\hline \begin{array}{l}{\psi}_{L}=\frac{{C}_{\text{A}L}}{{C}_{\text{A}0}}=1X\\ \text{=}\frac{4q\text{exp}(P{e}_{r}/2)}{{(1+q)}^{2}\text{exp}(P{e}_{r}q/2){(1q)}^{2}\text{exp}(P{e}_{r}q/2)}\\ \text{where}q=\sqrt{1+4\mathit{D}{\mathit{a}}_{\mathit{1}}/P{e}_{r}}\end{array}\\ \hline\end{array}\begin{array}{c}\left(\text{1817}\right)\end{array}$
This solution was first obtained by Danckwerts and has been published in many places (e.g., Levenspiel).^{5,6} With a slight rearrangement of Equation (1817), we obtain the conversion as a function of Da_{1} and Pe_{r}.
^{5} P. V. Danckwerts, Chem. Eng. Sci., 2, 1 (1953).
^{6} Levenspiel, Chemical Reaction Engineering, 3rd ed. New York: Wiley, 1999.
$\begin{array}{c}\hline X=1\frac{4q\text{exp}(P{e}_{r}/2)}{(1+q{)}^{2}\text{exp}(P{e}_{r}q/2)(1q{)}^{2}\text{exp}(P{e}_{r}q/2)}\\ \hline\end{array}\begin{array}{c}\left(18\text{18}\right)\end{array}$
Outside the limited case of a firstorder reaction, a numerical solution of the equation is required, and because this is a splitboundaryvalue problem, for example, conditions given at z = 0 and z = L, an iterative technique is needed. We have to be a little cautious here with conversion defined as [(C_{A0} – C_{A})/C_{A0}]. We see that at entrance to the reactor λ = 0, C_{A} is less than C_{A0} meaning X > 0. This value of X greater than zero is a result of A being dispersed at the entrance, not because of reaction.
To evaluate the exit concentration given by Equation (1817) or the conversion given by Equation (1818), we need to know the Damköhler and Péclet numbers. The firstorder reactionrate constant, k, and hence Da_{1} = τk, can be found using the techniques in Chapter 7. In Section 18.4, we discuss three methods to determine the Péclet number and dispersion coefficient.
A firstorder reaction with k = 0.25 min^{–1} is occurring in a tubular reactor with dispersion where the space time is τ = 5.15 min and the Péclet number is Pe_{r} = 7.5. In Example 182, we will show how to use this tracer data to calculate the Damköler number and Péclet number.
Plot the concentration and conversion profiles for a closedclosed system.
Solution
The dimensionless concentration profile is given by Equation (1815)
$\begin{array}{cc}\psi =2*\text{exp}\left(\frac{P{e}_{r}\lambda}{2}\right)*\frac{(1+q)\text{exp}\left(\frac{P{e}_{r}}{2}q(1\lambda )\right)(1q)\text{exp}\left(\frac{P{e}_{r}}{2}q(1\lambda )\right)}{(1+q{)}^{2}\text{exp}\left(\frac{P{e}_{r}q}{2}\right)(1q{)}^{2}\text{exp}(\frac{P{e}_{r}q}{2})}& \left(18\text{15}\right)\end{array}$
where
$q=\sqrt{1+\frac{4\mathit{D}{\mathit{a}}_{1}}{P{e}_{r}}}$
The conversion profile is
$\begin{array}{cc}X=1\psi & \left(18\text{16}\right)\end{array}$
The Damköhler number for this firstorder reaction is
$\begin{array}{c}\hline {\begin{array}{c}\mathit{D}\mathit{a}\end{array}}_{1}=\tau k=\left(5.15\text{min}\right)\left(0.25{\text{min}}^{1}\right)=1.29\\ \hline\end{array}\begin{array}{c}\left(\text{E181.1}\right)\end{array}$
$\begin{array}{cc}\text{Given:}\text{}P{e}_{r}=7.5& \left(\text{E181.2}\right)\end{array}$
$\begin{array}{c}\hline q=\sqrt{1+4\frac{D{a}_{1}}{P{e}_{r}}}=\sqrt{1+4\frac{\left(1.29\right)}{7.5}}=1.30\\ \hline\end{array}\begin{array}{c}\left(\text{E181.3}\right)\end{array}$
$\begin{array}{c}\hline \frac{P{e}_{r}q}{2}=\frac{\left(7.5\right)\left(1.3\right)}{2}=4.87\\ \hline\end{array}\begin{array}{c}\left(\text{E181.4}\right)\end{array}$
Substituting for q, Da_{1}, and Pe_{r}, the following concentration profiles can be obtained using Polymath, MatLab, Python, or Wolfram simulations in the Living Example Problem (LEP) on the Web site.
Be sure to use the LEP Sliders to get a full understanding of how the parameters affect the profile.
Figure E181.1 shows the concentration profiles for three Péclet numbers. The dimensionless residence time function E(θ) is shown in the upper righthand corner for the three corresponding Péclet numbers.
Analysis: Figure E181.1 shows the comparative concentration profiles for a PFR and a reactor with dispersion. For small values of the Péclet number, Figure (a), the dispersion is large and the concentration is distributed throughout the tube and the concentration profiles approach that of a CSTR, one again notes that for a CSTR the entering feed concentration immediately drops form C_{A0} to C_{A}. For large values of Pe_{r}, Figure (c), there is virtually no dispersion and the profile approaches that of a PFR. Figure (b) shows the profiles for Péclet numbers in between the extremes of (a) and (c).
As we will see in Example 182, the solution to find the exit reactant concentration or the conversion requires both D_{a} and Pe_{r} which will be discussed in Section 18.5.
Three ways to find D_{a}
Nemo in the Disney movie is a fish, but here our fish Meno is the dispersion coefficient, D_{a}. There are three ways we can use to find D_{a} and hence Pe_{r}:
Laminar flow with radial and axial molecular diffusion theory
Correlations from the literature for pipes and packed beds
Experimental tracer data
Finding Meno from 1, 2, or 3
At first sight, simple models described by Equation (1811) appear to have the capability of accounting only for axial mixing effects. It will be shown, however, that this approach can compensate not only for problems caused by axial mixing, but also for those caused by radial mixing and other nonflat velocity profiles.^{7} These fluctuations in concentration can result from different flow velocities and pathways and from molecular and turbulent diffusion.
^{7} R. Aris, Proc. R. Soc. (London), A235, 67 (1956).
Let’s first discuss a qualitative description of how dispersion can occur in a laminarflow tubular reactor. We know that the axial velocity varies in the radial direction according to the wellknown parabolic velocity profile:
$u\left(r\right)=2U[1{\left(\frac{r}{R}\right)}^{2}]$
where U is the average velocity. For laminar flow, we saw that the RTD function E(t) was given by
$\begin{array}{cc}E\left(t\right)=\{\begin{array}{ccc}0& \text{for}t\frac{\tau}{2}& (\tau =\frac{L}{U})\\ \frac{{\tau}^{2}}{2{t}^{3}}& \text{for}t\ge \frac{\tau}{2}& \end{array}& (1647)\end{array}$
In arriving at this distribution E(t), it was assumed that there was no transfer of molecules in the radial direction between streamlines. Consequently, with the aid of Equation (1647), we know that the molecules on the center streamline (r = 0) exited the reactor at a time t = τ/2, and molecules traveling on the streamline at r = 3R/4 exited the reactor at time
$\begin{array}{lll}t& =& \frac{L}{u}=\frac{L}{2U[1(r/R{)}^{2}]}=\frac{\tau}{2[1(3/4{)}^{2}]}\\ & =& \frac{8}{7}\cdot \tau \end{array}$
Qualitative description of dispersion
The question now arises: What would happen if some of the molecules traveling on the streamline at r = 3R/4 jumped (i.e., diffused) onto the streamline at r = 0? The answer is that they would exit sooner than if they had stayed on the streamline at r = 3R/4. Analogously, if some of the molecules from the faster streamline at r = 0 jumped (i.e., diffused) onto the streamline at r = 3R/4, they would take a longer time to exit (Figure 187). In addition to the molecules diffusing between streamlines, they can also move forward or backward relative to the average fluid velocity by molecular diffusion (Fick’s law). With both axial and radial diffusion occurring, the question arises as to what will be the distribution of residence times when molecules are transported between and along streamlines by diffusion. To answer this question, we will derive an equation for the axial dispersion coefficient, D_{a}, that accounts for the axial and radial diffusion mechanisms. In deriving D_{a}, which is often referred to as the ArisTaylor dispersion coefficient, we closely follow the development given by Brenner and Edwards.^{8}
^{8} H. Brenner and D. A. Edwards, Macrotransport Processes. Boston: ButterworthHeinemann, 1993.
Molecules diffusing back and forth between streamlines
The convective–diffusion equation for solute (e.g., tracer) transport in both the axial and radial direction can be obtained by combining Equation (142) with the diffusion equation (cf. Equation (1411)) applied to the tracer concentration, c, and transformed to radial coordinates
$\begin{array}{c}\hline \frac{\partial c}{\partial t}+u\left(r\right)\frac{\partial c}{\partial z}={D}_{\text{AB}}\{\frac{1}{r}\frac{\partial \left[r\right(\partial c/\partial r\left)\right]}{\partial r}+\frac{{\partial}^{2}c}{\partial {z}^{2}}\}\\ \hline\end{array}\begin{array}{c}\left(18\text{19}\right)\end{array}$
where c is the solute concentration at a particular r, z, and t, and D_{AB} is the molecular diffusion coefficient of species A in B.
We are going to change the variable in the axial direction z to z*, which corresponds to an observer moving with the fluid
$\begin{array}{cc}{z}^{*}=zUt& \left(18\text{20}\right)\end{array}$
A value of z* = 0 corresponds to an observer moving with the average velocity of the fluid, U. Using the chain rule, we obtain
$\begin{array}{cc}(\frac{\partial c}{\partial t}{)}_{z*}+[u\left(r\right)U]\frac{\partial c}{\partial z*}={D}_{\text{AB}}\left[\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial c}{\partial r}\right)+\frac{{\partial}^{2}c}{\partial {z}^{*2}}\right]& \left(18\text{21}\right)\end{array}$
Because we want to know the concentrations and conversions at the exit to the reactor, we are really only interested in the average axial concentration, $\overline{C}$, which is given by
$\begin{array}{cc}\overline{C}(z,t)=\frac{1}{\pi {R}^{2}}{\int}_{0}^{R}\text{}c(r,z,t)2\pi r\text{}dr& \left(\begin{array}{c}18\text{22}\end{array}\right)\end{array}$
Consequently, we are going to solve Equation (1821) for the solution concentration as a function of r and then substitute the solution c (r, z, t) into Equation (1822) to find $\overline{C}$ (z, t). All the intermediate steps are given on the CRE Web site in the Professional Reference Shelf, and the partial differential equation describing the variation of the average axial concentration with time and distance is
$\begin{array}{cc}\frac{\partial \overline{C}}{\partial t}+U\frac{\partial \overline{C}}{\partial {z}^{*}}=D*\frac{{\partial}^{2}\overline{C}}{\partial {z}^{*2}}& \left(\begin{array}{c}18\text{}23\end{array}\right)\end{array}$
where D* is the ArisTaylor dispersion coefficient
ArisTaylor dispersion coefficient
$\begin{array}{c}\hline {D}^{*}={D}_{\text{AB}}+\frac{{U}^{2}{R}^{2}}{{48D}_{\text{AB}}}\\ \hline\end{array}\begin{array}{c}\left(18\text{}24\right)\end{array}$
That is, for laminar flow in a pipe
$\begin{array}{c}\hline {D}_{a}\equiv D*\\ \hline\end{array}$
Figure 188 shows the dispersion coefficient D* in terms of the ratio D* /U(2R) = D* /Ud_{t} as a function of the product of the Reynolds (Re) and Schmidt (Sc) numbers.
It was surprising to me how D* could account for both radial and axial dispersion in laminar flow.
Meno! 1.
We will use correlations from the literature to determine the dispersion coefficient D_{a} for flow in cylindrical tubes (pipes) and for flow in packed beds.
An estimate of the dispersion coefficient, D_{a}, for laminar and turbulent flow in pipes, can be determined from Figure 189. Here, d_{t} is the tube diameter and Sc is the Schmidt number discussed in Chapter 14. The flow is laminar (streamline) below 2,100, and we see the ratio (D_{a}/Ud_{t}) increases with increasing Schmidt and Reynolds numbers. Between Reynolds numbers of 2,100 and 30,000, one can put bounds on D_{a} by calculating the maximum and minimum values at the top and bottom of the shaded regions.
1. Meno, i.e., Pe and D_{a}, has been found
For the case of gas–solid and liquid–solid catalytic reactions that take place in packedbed reactors, the dispersion coefficient, D_{a}, can be estimated by using Figure 1810. Here, d_{p} is the particle diameter and ε is the porosity.
The dispersion coefficient can be determined from a pulse tracer experiment. Here, we will use t_{m} and σ^{2} to solve for the dispersion coefficient D_{a} and then
2. Finding Meno in packed beds
Once the Reynolds number is calculated, D_{a} can be found.
the Péclet number, Pe_{r}. Here the effluent concentration of the reactor is measured as a function of time. From the effluent concentration data, the mean residence time, t_{m}, and variance, σ^{2}, are calculated, and these values are then used to determine D_{a}. To show how this is accomplished, we will write the unsteadystate mass balance on the tracer flowing in a tubular reactor
$\begin{array}{cc}{D}_{a}\frac{{\partial}^{2}{C}_{\text{T}}}{\partial {z}^{2}}\frac{\partial \left(U{C}_{\text{T}}\right)}{\partial z}=\frac{\partial {C}_{\text{T}}}{\partial t}& \left(\begin{array}{c}\text{182}\end{array}\right)\end{array}$
in dimensionless form, discuss the different types of boundary conditions at the reactor entrance and exit, and then solve for the exit concentration as a function of dimensionless time (Θ = t / τ), and then relate D_{a}, σ^{2}, and τ.
If I cannot use any of the earlier correlations or graphs, I will just need to find Meno on my own. To do this, I will need to do some theoretical analysis and some experiments to determine the RTD, which we can then analyze to find t_{m} and σ^{2} and then calculate D_{a}(Pe_{r}). The first step to find Meno is to put Equation (182) in dimensionless form to arrive at the dimensionless group(s) that characterize the process. Again let
$\begin{array}{cccc}\psi =\frac{{C}_{\text{T}}}{{C}_{\text{T0}}},& \lambda =\frac{z}{L},& \text{and}& \mathrm{\Theta}=\frac{tU}{L}\end{array}$
For a pulse input, C_{T0} is defined as the mass of tracer injected, M, divided by the vessel volume, V. Then
$\begin{array}{c}\hline \frac{1}{P{e}_{r}}\frac{{\partial}^{2}\psi}{\partial {\lambda}^{2}}\frac{\partial \psi}{\partial \lambda}=\frac{\partial \psi}{\partial \mathrm{\Theta}}\\ \hline\end{array}\begin{array}{c}\left(18\text{}3\right)\end{array}$
The initial condition is
Initial condition
$\begin{array}{ccccc}\text{At}\text{}t=0,& z0,& {C}_{\text{T}}({0}^{+},0)=0,& \psi \left({0}^{+}\right)=0& \left(\text{1825}\right)\end{array}$
The mass of tracer injected, M, is
$M=U{A}_{c}\text{}{\int}_{0}^{\infty}\text{}{C}_{\text{T}}({0}^{},t)dt$
In dimensionless form, the Danckwerts boundary conditions are
$\begin{array}{cc}\hline \text{At}\text{}\lambda =0:& (\begin{array}{c}\frac{1}{P{e}_{r}}\frac{\partial \psi}{\partial \lambda}\end{array}{)}_{\lambda ={0}^{+}}+\psi \left({0}^{+}\right)=\frac{{C}_{\text{T}}({0}^{},t)}{{C}_{\text{T0}}}=1\\ \hline\end{array}\begin{array}{c}\left(18\text{26}\right)\end{array}$
$\begin{array}{ccc}\hline \text{At}\lambda =1:& \frac{\partial \psi}{\partial \lambda}=0& \\ \hline\end{array}\begin{array}{c}\left(\text{1827}\right)\end{array}$
Equation (183) has been solved numerically for a pulse injection, and the resulting dimensionless effluent tracer concentration at the exit of the reactor, ψ_{exit}, is shown as a function of the dimensionless time Θ in Figure 1811 for various Péclet numbers. Although analytical solutions for ψ can be found, the result is an infinite series. The corresponding equations for the mean residence time, t_{m}, and the variance, σ^{2}, are^{9}
^{9} See K. Bischoff and O. Levenspiel, Adv. Chem. Eng., 4, 95 (1963).
$\begin{array}{c}\hline {t}_{m}=\tau \\ \hline\end{array}\begin{array}{c}\left(18\text{28}\right)\end{array}$
Effects of dispersion on the effluent tracer concentration
^{10} O. Levenspiel, Chemical Reaction Engineering, 2nd ed. New York: Wiley, 1972, p. 277.
and
Meno 3!
$\frac{{\sigma}^{2}}{{t}_{m}^{2}}=\frac{1}{{\tau}^{2}}{\int}_{0}^{\infty}\text{}(t\tau {)}^{2}E\left(t\right)dt$
which can be used with the solution to Equation (183) to obtain
Calculating Pe_{r} using t_{m} and σ^{2} determined from RTD data for a closedclosed system
$\begin{array}{c}\hline \frac{{\sigma}^{2}}{{t}_{m}^{2}}=\frac{2}{P{e}_{r}}\frac{2}{P{e}_{r}^{2}}(1{e}^{P{e}_{r}})\\ \hline\end{array}\begin{array}{c}\left(\text{1829}\right)\end{array}$
Consequently, we see that the Péclet number, Pe_{r} (and hence D_{a}), can be found experimentally by determining t_{m} and σ^{2} from the RTD data and then solving Equation (1829) for Pe_{r}.
When a tracer is injected into a packed bed at a location more than two or three particle diameters downstream from the entrance and measured some distance upstream from the exit, the openopen vessel boundary conditions apply. For an openopen system, an analytical solution to Equation (1811) can be obtained for a pulse tracer input.
For an openopen system, the boundary conditions at the entrance are
${F}_{\text{T}}({0}^{},t)={F}_{\text{T}}({0}^{+},t)$
Then, for the case when the dispersion coefficient is the same in the entrance and reaction sections
Open at the entrance
$\begin{array}{cc}{D}_{a}{(\frac{\partial {C}_{\text{T}}}{\partial z}{)}_{z={0}^{}}+U{C}_{\text{T}}({0}^{},t)={D}_{a}\left(\frac{\partial {C}_{\text{T}}}{\partial z}\right)}_{z={0}^{+}}+U{C}_{\text{T}}({0}^{+},t)& \left(18\text{}30\right)\end{array}$
Because there are no discontinuities across the boundary at z = 0
$\begin{array}{cc}{C}_{\text{T}}({0}^{},t)={C}_{\text{T}}({0}^{+},t)& \left(18\text{}\text{31}\right)\end{array}$
At the exit
Open at the exit
$\begin{array}{cc}{D}_{a}(\frac{\partial {C}_{\text{T}}}{\partial z}{)}_{z={L}^{}}+U{C}_{\text{T}}({L}^{},t)={D}_{a}{\left(\frac{\partial {C}_{\text{T}}}{\partial z}\right)}_{z={L}^{+}}+U{C}_{\text{T}}({L}^{+},t)& \left(18\text{32}\right)\end{array}$
$\begin{array}{cc}{C}_{\text{T}}({L}^{},t)={C}_{\text{T}}({L}^{+},t)& \left(\text{1833}\right)\end{array}$
There are a number of perturbations of these boundary conditions that can be applied. The dispersion coefficient can take on different values in each of the three regions (z < 0, 0 ≤ z ≤ and z > L), and the tracer can also be injected at some point z_{1} rather than at the boundary, z = 0. These cases and others can be found in the supplementary readings cited at the end of this chapter. We shall consider the case when there is no variation in the dispersion coefficient for all z and an impulse of tracer is injected at z = 0 at t = 0.
Again, we let ψ = C_{A}/C_{A0} and θ = t/τ.
For long tubes (Pe_{r} > 100) in which the concentration gradient at ±∞ will be zero, the solution to Equation (183) giving the dimensionless tracer concentration at the exit is^{11}
^{11} W. Jost, Diffusion in Solids, Liquids and Gases. New York: Academic Press, 1960, pp. 17, 47.
Valid for Pe_{r} > 100
$\begin{array}{c}\hline \psi (1,\mathrm{\Theta})=\frac{{C}_{\text{T}}(L,t)}{{C}_{\text{T0}}}=\frac{1}{2\sqrt{\pi \mathrm{\Theta}/P{e}_{r}}}\text{exp}\left[\frac{(1\mathrm{\Theta}{)}^{2}}{4\mathrm{\Theta}/P{e}_{r}}\right]\\ \hline\end{array}\begin{array}{c}\left(\text{1834}\right)\end{array}$
The mean residence time for an openopen system is
Calculate τ for an openopen system.
$\begin{array}{c}\hline {t}_{m}=(1+\frac{2}{P{e}_{r}})\tau \\ \hline\end{array}\begin{array}{c}\left(\text{1835}\right)\end{array}$
Diffuse in and out.
where τ is based on the volume between z = 0 and z = L (i.e., reactor volume measured with a yardstick). We note that the mean residence time for an open system is greater than that for a closed system. The reason is that the molecules can diffuse out of the reactor at the entrance and then diffuse, or be carried, back into the reactor. The variance for an openopen system is
Calculate Pe_{r} for an open–open system.
$\begin{array}{c}\hline \frac{{\sigma}^{2}}{{\tau}^{2}}=\frac{2}{P{e}_{r}}+\frac{8}{P{e}_{r}^{2}}\\ \hline\end{array}\begin{array}{c}\left(\text{1836}\right)\end{array}$
We found Meno 3.
We now consider two cases for which we can use Equations (1829) and (1836) to determine the system parameters:
Case 1. The space time τ is known. That is, V and υ_{0} are measured independently. Here, we can determine the Péclet number by determining t_{m} and σ^{2} from the concentration–time data and then use Equation (1836) to calculate Pe_{r}. We can also calculate t_{m} and then use Equation (1835) as a check, but this is usually less accurate.
Case 2. The space time τ is unknown. This situation arises when there are dead or stagnant pockets that exist in the reactor along with the dispersion effects. To analyze this situation, we first calculate mean residence time, t_{m}, and the variance, σ^{2}, from the data as in case 1. Then, we use Equation (1835) to eliminate τ^{2} from Equation (1836) to arrive at
$\begin{array}{c}\hline \frac{{\sigma}^{2}}{{t}_{m}^{2}}=\frac{2P{e}_{r}+8}{P{e}_{r}^{2}+4P{e}_{r}+4}\\ \hline\end{array}\begin{array}{c}\left(18\text{37}\right)\end{array}$
Finding the effective reactor voume
We now can solve for the Péclet number in terms of our experimentally determined variables σ^{2} and ${t}_{m}^{2}$. Knowing Pe_{r}, we can solve Equation (1835) for τ, and hence V. The dead volume is the difference between the measured volume (i.e., with a yardstick) and the effective volume calculated from the RTD.
Meno 3!
Once we have determined D_{a} and Pe_{r} from the RTD for our real reactor (i.e., found Meno), we can solve for the conversion from an analytical solution, or for a firstorder reaction as in Example 181, or solve numerically for other reaction orders.
The firstorder reaction
A → B
is carried out in a 10cmdiameter tubular reactor 6.36 m in length. The specific reaction rate is 0.25 min^{–1}. The results of a tracer test carried out on this reactor are shown in Table E182.1.
TABLE E182.1 EFFLUENT TRACER CONCENTRATION AS A FUNCTION OF TIME
t (min) 
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
12 
14 
C (mg/L) 
0 
1.4 
5 
8 
10 
8 
6 
4 
3 
2.2 
1.6 
0.6 
0 
Calculate the conversion using (a) the closed vessel dispersion model, (b) PFR, (c) the tanksinseries model, and (d) a single CSTR.
Solution
We will use Equation (1818) to calculate the conversion at the exit
$\begin{array}{cc}X=1\frac{4q\text{exp}(P{e}_{r}/2)}{(1+q{)}^{2}\text{exp}(P{e}_{r}/2)(1q{)}^{2}\text{exp}(P{e}_{r}q/2)}& \left(18\text{18}\right)\end{array}$
where $q=\sqrt{1+4\mathit{D}{\mathit{a}}_{1}/P{e}_{r}}\text{}\mathit{D}{\mathit{a}}_{1}=\tau k,\text{and}P{e}_{r}=UL/{D}_{a}\mathrm{.}$
(1) Parameter evaluation using the RTD data to evaluate Pe_{r}:
We can calculate Pe_{r} from Equation (1829)
$\begin{array}{cc}\frac{{\sigma}^{2}}{{\tau}^{2}}=\frac{2}{P{e}_{r}}\frac{2}{P{e}_{r}^{2}}(1{e}^{P{e}_{r}})& \left(\text{1829}\right)\end{array}$
First calculate t_{m} and σ^{2} from RTD data.
However, we must find τ^{2} and σ^{2} from the tracer concentration data first.
$\begin{array}{cc}\tau =\frac{V}{\upsilon}={\int}_{0}^{\infty}\text{}tE\left(t\right)dt& \left(\text{E182.1}\right)\end{array}$
$\begin{array}{cc}{\sigma}^{2}={\int}_{0}^{\infty}\text{}(t\tau {)}^{2}E\left(t\right)dt& \left(\text{E182.2}\right)\end{array}$
Here again, spreadsheets can be used to calculate τ^{2} and σ^{2}.
Don’t fall asleep. These are calculations we need to know how to carry out.
We note that the data in Table E182.1 is the same data set used in Examples 161 and 162 where we found
and
$\begin{array}{cc}\begin{array}{c}\hline {t}_{m}=5.15\text{minutes}\\ {\sigma}^{2}=6.2{\text{minutes}}^{2}\\ \hline\end{array}& \end{array}$
Are we lucky or what? We will use these values in Equation (1829) to calculate Pe_{r}. Dispersion in a closed vessel is represented by
$\begin{array}{llll}\frac{{\sigma}^{2}}{{\tau}^{2}}& =& \frac{\begin{array}{c}2\end{array}}{P{e}_{r}^{2}}(P{e}_{r}1+{e}^{P{e}_{r}})& \left(18\text{29}\right)\\ & =& \frac{6.2}{(5.15{)}^{2}}=0.23=\frac{2}{P{e}_{r}^{2}}(P{e}_{r}1+{e}^{P{e}_{r}})& \end{array}$
Calculate Pe_{r} from t_{m} and σ^{2}.
Solving for Pe_{r} either by trial and error or using Polymath, we obtain
$\begin{array}{c}\hline P{e}_{r}=7.5\\ \hline\end{array}\begin{array}{c}\left(\text{E182.3}\right)\end{array}$
Next, calculate Da_{1}, q, and X.
(2) Next, we calculate Da_{1} and q:
$\begin{array}{c}\hline \mathit{D}{\mathit{a}}_{\mathbf{1}}=\tau k=\left(5.15\text{min}\right)\left(0.25{\text{min}}^{1}\right)=1.29\\ \hline\end{array}\begin{array}{c}\left(\text{E182.4}\right)\end{array}$
Using the equations for q and X gives
$\begin{array}{c}\hline q=\sqrt{1+\frac{4\mathit{D}{\mathit{a}}_{1}}{P{e}_{r}}}=\sqrt{1+\frac{4\left(1.29\right)}{7.5}}=1.30\\ \hline\end{array}\begin{array}{c}\left(\text{E182.5}\right)\end{array}$
Then
$\begin{array}{cc}\frac{P{e}_{r}q}{2}=\frac{\left(7.5\right)\left(1.3\right)}{2}=4.87& \left(\begin{array}{c}\text{E182.6}\end{array}\right)\text{}\end{array}$
(3) Finally, we calculate the conversion:
Substitution into Equation (1818) yields
Dispersion model
$\begin{array}{lll}X& =& 1\frac{4\left(1.30\right){e}^{(7.5/2)}}{(2.3{)}^{2}\text{exp}\left(4.87\right)(0.3{)}^{2}\text{exp}(4.87)}\\ X& =& 0.68\text{68\% conversion for the dispersion model}\end{array}$
When dispersion effects are present in this tubular reactor, 68% conversion is achieved.
(b) If the reactor were operating ideally as a plugflow reactor, the conversion would be
PFR
$\begin{array}{c}\hline X=1{e}^{\tau k}=1{e}^{D{a}_{1}}=1{e}^{1.29}=0.725\\ \hline\end{array}\begin{array}{c}\left(\text{E182.7}\right)\end{array}$
That is, 72.5% conversion would be achieved in an ideal plugflow reactor.
Tanksinseries model
(c) Conversion using the tanksinseries model: We recall Equation (1725) to calculate the number of tanks in series:
$\begin{array}{cc}n=\frac{{\tau}^{2}}{{\sigma}^{2}}=\frac{{\left(5.15\right)}^{2}}{6.1}=4.35& \left(\text{E182.8}\right)\end{array}$
To calculate the conversion for the TIS model, we recall Equation (515). For a firstorder reaction for n tanks in series, the conversion is
$\begin{array}{c}\hline \begin{array}{l}X=1\frac{1}{(1+{\tau}_{i}k{)}^{n}}=1\frac{1}{[1+(\tau /n)k{]}^{n}}=1\frac{1}{(1+1.29/4.35{)}^{4.35}}\\ X=\mathbf{67.7}\mathbf{\%}\mathbf{\text{forthetanksinseriesmodel}}\end{array}\\ \hline\end{array}\begin{array}{c}\left(\text{E182.9}\right)\end{array}$
(d) For a single CSTR
CSTR
$\begin{array}{cc}X=\frac{\tau k}{1+\tau k}=\frac{1.29}{2.29}=0.563& \left(\text{E182.10}\right)\end{array}$
So, 56.3% conversion would be achieved in a single ideal tank. Summary:
Summary
$\begin{array}{rrr}\hline \text{PFR:}X& =& 72.5\%\\ \text{Dispersion:}X& =& 68.0\%\\ \text{Tanks in series:}X& =& 67.7\%\\ \text{Single CSTR:}X& =& 56.3\%\\ \hline\end{array}$
In this example, correction for finite dispersion, whether by a dispersion model or a tanksinseries model, is significant when compared with a PFR.
Analysis: This example is a very important and comprehensive one. We showed how to calculate the conversion by (1) choosing a model, (2) using the RTD to evaluate the model parameters, and (3) substituting the reactionrate parameters in the chosen model. As expected, the dispersion and TIS model gave essentially the same result and this result fell between the limits predicted by an ideal PFR and an ideal CSTR.
We have seen that we can apply both of these oneparameter models to tubular reactors using the variance of the RTD. For firstorder reactions, the two models can be applied with equal ease. However, the tanksinseries model is mathematically easier to use to obtain the effluent concentration and conversion for reaction orders other than one, and for multiple reactions. However, we need to ask what would be the accuracy of using the tanksinseries model over the dispersion model. These two models are equivalent when the Péclet–Bodenstein number is related to the number of tanks in series, n, by the equation^{12}
^{12} K. Elgeti, Chem. Eng. Sci., 51, 5077 (1996).
$\begin{array}{cc}Bo=2(n1)& \left(18\text{38}\right)\end{array}$
Equivalency between models of tanksinseries and dispersion
or
$\begin{array}{c}\hline n=\frac{Bo}{2}+1\\ \hline\end{array}\begin{array}{c}\left(\begin{array}{c}18\text{39}\end{array}\right)\end{array}$
where
$\begin{array}{c}\hline Bo=UL/Da\\ \hline\end{array}\begin{array}{c}\left(18\text{40}\right)\end{array}$
where U is the superficial velocity, L the reactor length, and D_{a} the dispersion coefficient.
For the conditions in Example 182, we see that the number of tanks calculated from the Bodenstein number, Bo (i.e., Pe_{r}), Equation (1839), is 4.75, which is very close to the value of 4.35 calculated from Equation (1725). Consequently, for reactions other than first order, one would solve successively for the exit concentration and conversion from each tank in series for both a battery of four tanks in series and for five tanks in series in order to bound the expected values.
In addition to the oneparameter models of tanksinseries and dispersion, many other oneparameter models exist when a combination of ideal reactors is used to model the real reactor shown in Section 18.8 for reactors with bypassing and dead volume. Another example of a oneparameter model would be to model the real reactor as a PFR and a CSTR in series with the one parameter being the fraction of the total volume that behaves as a CSTR. We can dream up many other situations that would alter the behavior of ideal reactors in a way that adequately describes a real reactor. However, it may be that one parameter is not sufficient to yield an adequate comparison between theory and practice. We explore these situations with combinations of ideal reactors in the section on twoparameter models.
The reactionrate parameters are usually known (e.g., Da), but the Péclet number is usually not known because it depends on the flow and the vessel. Consequently, we need to find Pe_{r} using one of the three techniques discussed earlier in the chapter.
We now consider dispersion and reaction in a tubular reactor. We first write our mole balance on species A in cylindrical coordinates by recalling Equation (1819) and including the rate of formation of A, r_{A}. At steady state we obtain
$\begin{array}{cc}{D}_{\text{AB}}[\frac{1}{r}\frac{\partial \left(r\frac{\partial {C}_{\text{A}}}{\partial r}\right)}{\partial r}+\frac{{\partial}^{2}{C}_{\text{A}}}{\partial {z}^{2}}]u\left(r\right)\frac{\partial {C}_{\text{A}}}{\partial z}+{r}_{\text{A}}=0& \left(18\text{41}\right)\end{array}$
Analytical solutions to dispersion with reaction can only be obtained for isothermal zero and firstorder reactions. We are now going to use COMSOL to solve the flow with reaction and dispersion with reaction.
We are going to compare two solutions: one which uses the ArisTaylor approach and one in which we numerically solve for both the axial and radial concentration using COMSOL. These solutions are on the CRE Web site.
Case A. ArisTaylor Analysis for Laminar Flow
For the case of an nthorder reaction, Equation (189) becomes
$\begin{array}{cc}\frac{{D}_{a}}{U}\frac{{d}^{2}{\overline{C}}_{\text{A}}}{d{z}^{2}}\frac{d{\overline{C}}_{\text{A}}}{dz}\frac{k{\overline{C}}_{\text{A}}^{n}}{U}=0& \left(18\text{}42\right)\end{array}$
where ${\overline{C}}_{\text{A}}$ is the radially averaged axial concentration from r = 0 to r = R, that is,
${\overline{C}}_{\text{A}}\left(z\right)=\frac{{\int}_{0}^{R}2\pi r{C}_{\text{A}}(z,r)dr}{\pi {R}^{2}}$
If we use the ArisTaylor analysis, we can use Equation (189) with a caveat that $\overline{\psi}={\overline{C}}_{\text{A}}/{C}_{\text{A0}}$ and λ = z/L we obtain
$\begin{array}{cc}\frac{1}{P{e}_{r}}\frac{{\text{d}}^{2}\overline{\psi}}{d{\lambda}^{2}}\frac{d\overline{\psi}}{d\lambda}\mathit{D}{\mathit{a}}_{\mathit{n}}{\overline{\psi}}^{n}=0& (1843)\end{array}$
where
$P{e}_{r}=\frac{UL}{{D}_{a}}\text{and}\mathit{D}{\mathit{a}}_{\mathit{n}}=\tau k{C}_{\text{A0}}^{n1}$
For the closedclosed boundary conditions we have
$\begin{array}{llll}\text{At}& \lambda =0:& \frac{1}{P{e}_{r}}\frac{d\overline{\psi}}{d\lambda}{}_{\lambda ={0}^{+}}+\overline{\psi}\left({0}^{+}\right)=1& \left(\begin{array}{c}\text{1844}\end{array}\right)\text{}\\ \text{At}& \lambda =1:& \frac{d\overline{\psi}}{d\lambda}=0& \end{array}$
Danckwerts boundary conditions
For the openopen boundary conditions we have
$\begin{array}{cccc}\text{At}& \lambda =0:& \overline{\psi}\left({0}^{}\right)\frac{1}{P{e}_{r}}\frac{d\overline{\psi}}{d\lambda}{}_{\lambda ={0}^{}}=\overline{\psi}\left({0}^{+}\right)\frac{1}{P{e}_{r}}\frac{d\overline{\psi}}{d\lambda}{}_{\lambda ={0}^{+}}& \left(\begin{array}{c}\text{1845}\end{array}\right)\text{}\end{array}$
$\begin{array}{ccc}\text{At}& \lambda =1:& \frac{d\overline{\psi}}{d\lambda}\end{array}=0$
Equation (1843) gives the dimensionless concentration profiles for dispersion and reaction in a laminarflow reactor. For a firstorder reaction, we can use Equation (1826) and (1827) to obtain the concentration and conversion profiles. For a secondorder reaction, the equation becomes nonlinear that needs to be solved numerically. We have used COMSOL in the LEP on the Web site (see http://www.umich.edu/~elements/6e/18chap/expanded_ch18_example2comsol.pdf) to solve for the concentration profiles.
Case B. Full Numerical Solution
To obtain profiles, C_{A}(r, z), we now solve Equation (1841)
$\begin{array}{cc}{D}_{\text{AB}}[\frac{1}{r}\frac{\partial \left(r\frac{\partial {C}_{\text{A}}}{\partial r}\right)}{\partial r}+\frac{{\partial}^{2}{C}_{\text{A}}}{{{\partial}_{z}}^{2}}]u\left(r\right)\frac{\partial {C}_{\text{A}}}{\partial z}+{r}_{\text{A}}=0& \left(18\text{46}\right)\end{array}$
First, we will put the equations in dimensionless form by letting ψ = C_{A}/C_{A0}, λ = z/L, and ϕ = r/R. Following our earlier transformation of variables, Equation (1846) becomes
$\begin{array}{c}\hline \left(\frac{L}{R}\right)\frac{1}{P{e}_{r}}\left[\frac{1}{\varphi}\frac{\partial \left(\varphi \frac{\partial \psi}{\partial \varphi}\right)}{\partial \varphi}\right]+\frac{1}{P{e}_{r}}\frac{{d}^{2}\psi}{d{\lambda}^{2}}2(1{\varphi}^{2})\frac{d\psi}{d\lambda}\mathit{D}{\mathit{a}}_{\mathit{n}}{\psi}^{n}=0\\ \hline\end{array}\begin{array}{c}\left(18\text{47}\right)\end{array}$
COMSOL
We now consider flow, reaction and dispersion in cylindrical pipes. The parameter values (e.g., k, U_{0}) will be varied using a COMSOL program. Initially we will use the following values, k = 8.33 × 10^{–6} m^{3}/mol · s, C_{A0} = 500 mol/m^{3}, U_{0} = 0.021 m/s, and D_{A0} = 1.25 × 10^{–6} m^{2}/s.
Plot the concentration surface for to and ϕ = 0 to ϕ = 1 and λ = 0 to λ = 1. To make this plot, go to the CRE Web site and load the COMSOL LEP instruction on How to Access COMSOL (http://www.umich.edu/~elements/6e/comsol/comsol_access_instructions.html). Vary the parameters and run the simulation.
Plot the radial concentration profile.
Plot the axial profiles at ϕ = 0 and ϕ = 1/2.
Plot the radially averaged axial concentration profiles C_{A}(z) (i.e., ϕ(λ)).
Solution
The dimensionless form of the equation describing flow, reaction, and dispersion was given in Equation (1847)
$\begin{array}{cc}\left(\frac{L}{R}\right)\frac{1}{P{e}_{r}}\left[\frac{1}{\varphi}\frac{\partial \left(\varphi \frac{\partial \psi}{\partial \varphi}\right)}{\partial \varphi}\right]+\frac{1}{P{e}_{r}}\frac{{d}^{2}\psi}{d{\lambda}^{2}}2(1{\varphi}^{2})\frac{d\psi}{d\lambda}\mathit{D}{\mathit{a}}_{\mathit{n}}{\psi}^{n}=0& \left(18\text{47}\right)\end{array}$
Boundary and Initial Conditions
A. Boundary conditions
Radial
At r = 0, we have symmetry ∂C_{i}/∂r = 0; in dimensionless form,
$\begin{array}{cc}\partial \psi /\partial \lambda =0\text{at}\varphi =0& \left(\text{E183.1}\right)\end{array}$
There is no mass flow through the tube walls therefore ∂C_{i}/∂_{r} = 0, at r = R, in dimensionless form
$\begin{array}{cc}\partial \psi /\partial \lambda =0\text{at}\varphi =1& \left(\text{E183.2}\right)\end{array}$
Radially averaged axial concentration is given by
${\overline{C}}_{\text{A}}\left(z\right)=\frac{{\int}_{0}^{\text{R}}2\pi r{C}_{\text{A}}(z,r)dr}{\pi {\text{R}}^{2}}$
Axial
At the entrance to the reactor z = 0, λ = 0 for all r (i.e., ϕ)
$\begin{array}{cc}{C}_{i}={C}_{i0},(\text{i.e.,}\psi =1\text{for all}\varphi )& \left(\text{E183.3}\right)\end{array}$
COMSOL
At the exit of the reactor z = L, i.e., λ = 1
$\begin{array}{cc}\frac{\partial {C}_{i}}{\partial z}=0\text{and therefore}\frac{\partial \psi}{\partial \lambda}=0& \left(\text{E183.4}\right)\end{array}$
#COMSOL
Equation (1847) and Equations (E183.1)–(E183.4) were solved using COMSOL. Go to the COMSOL module (http://www.umich.edu/~elements/6e/18chap/obj.html#/comsol/) and run the module and find the following surface plots and profiles. It will really be a great experience for you.
A COMSOL tutorial for this example is given on the Web site in the LEP section for Chapter 18.
The results for a reactor of length 4 m from running the COMSOL simulation are shown in Figure E183.1(a), (b), (c), and (d).
In Figure E183.1(b) we observe a concentration profiles at the entrance, halfway down the reactor and at the end of the reactor.
Figure E183.1(c) shows the axial concentration at different radius while Figure E183.1(d) shows the radially averaged axial concentration profile ${\overline{C}}_{\text{A}}\left(z\right)$.
Analysis: We observe in Figure E183.1(a) and (b) that the concentration profile is flat (plug flow like) at the reactor entrance and then develops into a parabolic shape. This profile shape remains approximately the same from the halfway point until the fluid exits the reactor. In Figure E183.1(c) the steepest axial profile was near the wall. One notes the axial profile in Figure E183.1(c) is similar to the radial averaged concentration profile shown in Figure E183.1(d). Why do you think they are similar?
In the previous sections, we have assumed that there were no radial variations in temperature in the tubular and packedbed reactors. In this section, we will consider the nonisothermal case where we have both axial and radial variations in the system variables. Here, we will again use COMSOL for solving partial differential equation such as shown in the Radial Effects COMSOL Web Module on the CRE Web site (http://umich.edu/~elements/6e/web_mod/radialeffects/index.htm).^{13}
^{13} An introductory webinar on COMSOL can be found on the AIChE webinar Web site: http://www.aiche.org/resources/chemeondemand/webinars/modelingnonidealreactorsandmixers.
COMSOL application
We are going to carry out differential mole and energy balances on the differential cylindrical annulus shown in Figure 1812.
In order to derive the governing equations, we need to define a couple of terms. The first is the molar flux of species i, W_{i} (mol/m^{2} · s). The molar flux has two components, the radial component, W_{ir}, and the axial component, W_{iz}.
The molar flow rates are just the product of the molar fluxes and the crosssectional areas normal to their direction of flow A_{cz}. For example, for species i flowing in the axial (i.e., z) direction
F_{iz} = W_{iz} A_{cz}
where W_{iz} is the molar flux in the axial z direction (mol/m^{2}/s), and A_{cz} (m^{2}) is the crosssectional area of the tubular reactor.
In Chapter 14 we discussed the molar fluxes in some detail, but for now let’s just say they consist of a diffusional component, –D_{e}(∂C_{i}/∂z), and a convectiveflow component, U_{z}C_{i}, so that the flux W_{iz} in the axial direction is
$\begin{array}{cc}{W}_{iz}={D}_{e}\frac{\partial {C}_{i}}{\partial z}+{U}_{z}{C}_{i}& \left(14\text{8a}\right)\end{array}$
where D_{e} is the effective diffusivity (or dispersion coefficient) (m^{2}/s), and U_{z} is the axial molar average velocity (m/s). Similarly, the flux W_{ir} in the radial direction, r, is
$\begin{array}{cc}{W}_{iz}={D}_{e}\frac{\partial {C}_{i}}{\partial r}+{U}_{r}{C}_{i}& \left(18\text{48}\right)\end{array}$
Radial direction
where U_{r} (m/s) is the average velocity in the radial direction. For now, we will neglect the velocity in the radial direction, that is, U_{r} = 0.
We first recall the mole balance for cylindrical coordinates in Chapter 14.
$\begin{array}{c}\hline \frac{1}{r}\frac{\partial \left(r{W}_{ir}\right)}{\partial r}\frac{\partial {W}_{iz}}{\partial z}+{r}_{i}=0\\ \hline\end{array}\begin{array}{c}\left(14\text{2}\right)\end{array}$
Using Equations (148b) and (1848) to substitute for W_{iz} and W_{ir} in Equation (142) and then setting the radial velocity to zero, U_{r} = 0, we obtain
$\begin{array}{cc}\frac{1\partial}{r\partial r}\left[({D}_{e}\frac{\partial {C}_{i}}{\partial r}r)\right]+(\frac{\partial}{\partial z}[{D}_{e}\frac{\partial {C}_{i}}{\partial z}+{U}_{z}{C}_{i}])+{r}_{i}=0& (1849)\end{array}$
For steadystate conditions and assuming U_{z} does not vary in the axial direction
$\begin{array}{c}\hline {D}_{e}\frac{{\partial}^{2}{C}_{i}}{\partial {r}^{2}}+\frac{{D}_{e}}{r}\frac{\partial {C}_{i}}{\partial r}+{D}_{e}\frac{{\partial}^{2}{C}_{i}}{\partial {z}^{2}}{U}_{z}\frac{\partial {C}_{i}}{\partial z}+{r}_{i}=0\\ \hline\end{array}\begin{array}{c}\left(18\text{50}\right)\end{array}$
When we applied the first law of thermodynamics to a reactor to relate either temperature and conversion or molar flow rates and concentration, we arrived at Equation (1110). Neglecting the work term we have for steadystate conditions
$\begin{array}{cc}\begin{array}{cc}\begin{array}{c}\text{Conduction}\end{array}\text{}\text{}& \text{Convection}\\ \overbrace{\dot{Q}}& +\underset{i=1}{\overset{n}{\mathrm{\Sigma}}}\overbrace{{F}_{i0}{H}_{i0}\underset{i=1}{\mathrm{\Sigma}}{F}_{i}{H}_{i}}=0\end{array}& \left(18\text{51}\right)\end{array}$
In terms of the molar fluxes, F_{i} = W_{i}A_{C}, the crosssectional area, A_{C}, and $(\mathit{q}=\dot{Q}/{A}_{c})$
$\begin{array}{cc}{\text{A}}_{c}[\mathbf{\text{q}}+(\mathrm{\Sigma}{\mathbf{\text{W}}}_{i0}{H}_{io}\mathrm{\Sigma}{\mathbf{\text{W}}}_{i}{H}_{i})]=0& \left(18\text{52}\right)\end{array}$
The q term is the heat added to the system and almost always includes a conduction component of some form. We now define an energy flux vector, e, (J/m^{2} · s), to include both the conduction and convection of energy.
e = energy flux J/s·m^{2}
e = Conduction + Convection
$\begin{array}{c}\hline \mathbf{\text{e =}}\mathbf{\text{q}}+\mathrm{\Sigma}{\mathbf{\text{W}}}_{i}{H}_{i}\\ \hline\end{array}\begin{array}{c}\left(18\text{53}\right)\end{array}$
where the conduction term q (kJ/m^{2} · s) is given by Fourier’s law. For axial and radial conduction, Fourier’s laws are
$\begin{array}{ccc}{q}_{z}={k}_{e}\frac{\partial T}{\partial z}& \text{and}& {q}_{r}={k}_{e}\frac{\partial T}{\partial r}\end{array}$
where k_{c} is the thermal conductivity (J/m·s·K). The energy transfer (flow) is the flux vector times the crosssectional area, A_{c}, normal to the energy flux
Energy flow = e · A_{c}
Using the energy flux, e, to carry out an energy balance on our annulus (Figure 1215) with system volume 2πrΔrΔz, we have
(Energy flow in at r) = e_{r} A_{cr} = e_{r} · 2πrΔz
(Energy flow in at r) = e_{r} A_{cz} = e_{z} · 2πrΔr
$\left(\begin{array}{c}\text{Energyflow}\\ \text{inat}r\end{array}\right)\left(\begin{array}{c}\text{Energyflow}\\ \text{outat}r+\mathrm{\Delta}r\end{array}\right)+\left(\begin{array}{c}\text{Energyflow}\\ \text{inat}z\end{array}\right)+\left(\begin{array}{c}\text{Energyflow}\\ \text{out}z+\mathrm{\Delta}z\end{array}\right)=\left(\begin{array}{c}\text{Accumulation}\\ \text{ofenergyin}\\ \text{volume(2}\pi r\mathrm{\Delta}r\mathrm{\Delta}z)\end{array}\right)$
$\left({e}_{r}2\pi r\mathrm{\Delta}z\right){}_{r}\left({e}_{r}2\pi r\mathrm{\Delta}z\right){}_{r+\mathrm{\Delta}}+{e}_{z}2\pi r\mathrm{\Delta}r{}_{z}{e}_{z}\pi r\mathrm{\Delta}r{}_{z+\mathrm{\Delta}z}=0$
Dividing by 2πrΔrΔz and taking the limit as Δr and Δz → 0
$\begin{array}{c}\hline \frac{1}{r}\frac{\partial \left(r{e}_{r}\right)}{\partial r}\frac{\partial {e}_{z}}{\partial z}=0\\ \hline\end{array}\begin{array}{c}\left(18\text{54}\right)\end{array}$
The radial and axial energy fluxes are
$\begin{array}{ccc}{e}_{r}& =& {q}_{r}+\mathrm{\Sigma}{W}_{ir}{H}_{i}\\ {e}_{z}& =& {q}_{z}+\mathrm{\Sigma}{W}_{iz}{H}_{i}\end{array}$
Substituting for the energy fluxes, e_{r} and e_{z}
$\begin{array}{cc}\frac{1}{r}\frac{\partial \left[r\right[{q}_{r}+\mathrm{\Sigma}{W}_{ir}{H}_{i}\left]\right]}{\partial r}\frac{\partial [{q}_{z}+\mathrm{\Sigma}{W}_{iz}{H}_{i}]}{\partial z}=0& \left(18\text{55}\right)\end{array}$
and expanding the convective energy fluxes, $\mathrm{\Sigma}{W}_{i}{H}_{i}$,
$\begin{array}{cccc}\text{Axial:}& \frac{\begin{array}{c}\partial \left(\mathrm{\Sigma}{W}_{iz}{H}_{i}\right)\end{array}}{\partial z}=\mathrm{\Sigma}{H}_{i}\frac{\partial {W}_{iz}}{\partial z}+\mathrm{\Sigma}{W}_{iz}\frac{\partial {H}_{i}}{\partial z}& & \left(\begin{array}{c}18\text{57}\end{array}\right)\end{array}$
Because U_{r} and the gradient in the flux term W_{ir} are small, we can neglect the last term in Equation (1856) with regard to the other terms in the equation. Substituting Equations (1856) and (1857) into Equation (1855), we obtain upon rearrangement
Recognizing that the term in brackets is related to Equation (142) and is just the rate of formation of species i, r_{i}, for steadystate conditions we have
$\begin{array}{c}\hline \frac{1}{r}\frac{\partial}{\partial r}\left(r{q}_{r}\right)\frac{\partial {q}_{z}}{\partial z}\mathrm{\Sigma}{H}_{i}{r}_{i}\mathrm{\Sigma}{W}_{iz}\frac{\partial {H}_{i}}{\partial z}=0\\ \hline\end{array}\begin{array}{c}\left(18\text{58}\right)\end{array}$
Recalling
${q}_{r}={k}_{e}\frac{\partial T}{\partial r},{q}_{z}={k}_{e}\frac{\partial T}{\partial z},\frac{\partial {H}_{i}}{\partial z}={C}_{{P}_{i}}\frac{\partial T}{\partial z},$
and
$\begin{array}{c}{r}_{i}={v}_{i}({r}_{\text{A}})\\ \mathrm{\Sigma}{r}_{i}{H}_{i}=\mathrm{\Sigma}{v}_{i}{H}_{i}({r}_{\text{A}})=\mathrm{\Delta}{H}_{\text{Rx}}{r}_{\text{A}}\end{array}$
we have the energy in the form
$\begin{array}{c}\hline \frac{{k}_{e}}{r}\left[\frac{\partial}{\partial r}\left(\frac{r\partial T}{\partial r}\right)\right]+{k}_{e}\frac{{\partial}^{2}T}{\partial {z}^{2}}+\mathrm{\Delta}{H}_{\text{Rx}}{r}_{A}\left(\mathrm{\Sigma}{W}_{iz}{C}_{{P}_{i}}\right)\frac{\partial T}{\partial z}=0\\ \hline\end{array}\begin{array}{c}\left(18\text{}\text{59}\right)\end{array}$
where W_{iz} is given by Equation (148a). Equation (1859) would be coupled with the mole balance (Equation (1850)), rate law, and stoichiometric equations to solve for the radial and axial concentration gradients. However, a great amount of computing time would be required. Let’s see if we can make some approximations to simplify the solution.
Assumption 1. Neglect the diffusive term in the axial direction, wrt, the convective term in Equation (148a) in the expression involving heat capacities
$\mathrm{\Sigma}{C}_{{P}_{i}}{W}_{iz}=\mathrm{\Sigma}{C}_{{P}_{i}}(0+{U}_{Z}{C}_{i})=\mathrm{\Sigma}{C}_{{P}_{i}}{C}_{i}{U}_{z}$
With this assumption, Equation (1859) becomes
$\begin{array}{c}\hline \frac{{k}_{e}}{r}\frac{\partial}{\partial r}\left(\frac{r\partial T}{\partial r}\right)+{k}_{e}\frac{{\partial}^{2}T}{\partial {z}^{2}}+\mathrm{\Delta}{H}_{\text{Rx}}{r}_{\text{A}}\left({U}_{z}\mathrm{\Sigma}{C}_{{P}_{i}}{C}_{i}\right)\frac{\partial T}{\partial z}=0\\ \hline\end{array}\begin{array}{c}\left(\text{1860}\right)\end{array}$
For laminar flow, the velocity profile is
$\begin{array}{cc}{U}_{z}=2{U}_{0}[1(\frac{r}{R}{)}^{2}]& \left(18\text{61}\right)\end{array}$
where U_{0} is the average velocity inside the reactor.
Energy balance with radial and axial gradients
Assumption 2. Assume that the sum ${C}_{{P}_{m}}=\mathrm{\Sigma}{C}_{{P}_{i}}{C}_{i}={C}_{A0}\mathrm{\Sigma}{\mathrm{\Theta}}_{i}{C}_{{P}_{i}}$ is constant. The energy balance now becomes
$\begin{array}{c}\hline {k}_{e}\frac{{\partial}^{2}T}{\partial {z}^{2}}+\frac{{k}_{e}}{r}\frac{\partial}{\partial r}\left(r\frac{\partial T}{\partial r}\right)+\mathrm{\Delta}{H}_{\text{Rx}}{r}_{\text{A}}{U}_{z}{C}_{{P}_{m}}\frac{\partial T}{\partial z}=0\\ \hline\end{array}\begin{array}{c}\left(\text{1862}\right)\end{array}$
Equation (1861) is the form we will use in our COMSOL problem. In many instances, the term C_{Pm} is just the product of the solution density (kg/m^{3}) and the heat capacity of the solution (kJ/kg · K).
We also recall that a balance on the coolant gives the variation of coolant temperature with axial distance where U_{ht} is the overall heat transfer coefficient and R is the reactor wall radius
$\begin{array}{c}\hline {\dot{m}}_{c}{C}_{{P}_{c}}\frac{\partial {T}_{a}}{\partial z}={U}_{ht}2\pi R\left[T(R,z){T}_{a}\right]\\ \hline\end{array}\begin{array}{c}\left(\begin{array}{c}18\text{63}\end{array}\right)\end{array}$
A. Initial conditions if other than steady state (not considered here) t = 0, C_{i} = 0, T = T_{0}, for z > 0 all r
B. Boundary conditions
Radial
At r = 0, we have symmetry ∂T / ∂r = 0 and ∂C_{i} / ∂r = 0.
At the tube wall, r = R, the temperature flux to the wall on the reaction side equals the convective flux out of the reactor into the shell side of the heat exchanger.
${k}_{e}\frac{\partial T}{\partial r}{}_{\text{R}}=U\left(T(R,z){T}_{a}\right)$
There is no mass flow through the tube walls ∂C_{i} / ∂r = 0 at r = R
Axial
At the entrance to the reactor z = 0
T = T_{0} and C_{i} = C_{i0}
At the exit of the reactor z = L
$\frac{\partial T}{\partial z}=0\text{and}\frac{\partial {C}_{i}}{\partial z}=0$
The preceding equations were used to describe and analyze flow and reaction in a tubular reactor with heat exchange as described in the following example, which can be found in the Expanded Material on the CRE Web site (http://www.umich.edu/~elements/6e/12chap/expanded.html). What follows is only a brief outline of that example with a few results from the output of the COMSOL program.
The liquid phase reaction was analyzed using COMSOL to study both axial and radial variations, and the details can be found on the home page of the CRE Web site, www.umich.edu/~elements/6e/index.html, by clicking on the Additional Material for Chapter 18. The algorithm for this example can be found on the Web site: (http://umich.edu/~elements/6e/18chap/expanded_ch18_radial.pdf). Web Figure E128.1 is a screen shot from COMSOL of the base case reactor and reaction parameters. In the COMSOL LEP you are asked “What if…” questions about varying the base case parameters. Typical radial (a) and axial (b) temperature profiles for this example are shown in Figure E184.1.
Results. The graphical solutions to the COMSOL code are shown in Figure E184.2.
Results of the COMSOL simulation
COMSOL
The volumetric flow rate of water is 3.5 times the volumetric flow rate of the mixture of propylene oxide in methanol. Further details of this example are provided in the Chapter 12 Additional Material on the CRE Web site: (http://umich.edu/~elements/6e/software/software_comsol.html).
Analysis: One can observe from the temperature surface plot in Figure E184.2(a) how the temperature changes both axially and radially in the reactor from its entering temperature of 312 K. These same profiles can be found in color on the CRE Web site in the Web Modules. Be sure to note the predicted maximum and minimum in the temperature and conversion profiles in Figure E184.2(a) to (d). Near the wall, the temperature of the mixture is lower because of the cold wall temperature, which is cooled by the jacket. Consequently, the reaction rate will be lower, and thus the conversion will be lower. However, right next to the wall the flow velocity through the reactor is almost zero, due to the friction with the wall, so the reactants spend a longer time in the reactor; therefore, a greater conversion is achieved, as noted by the upturn right next to the wall.
In this interactive Web site and text, you will be able to use the LEP COMSOL program instead of having to write your own code. The CRE LEP COMSOL modules are shown in Figure 1813. To access these LEPs, you don’t need to have COMSOL installed on your computer as you can access it through the CRE Web site. We use the COMSOL codes in a similar manner to the Polymath LEPs where you are able to vary the parameters to explore the radial variations in temperature, conversion, and concentration, in addition to axial profiles.
COMSOL
Creativity and engineering judgment are necessary for model formulation.
A tracer experiment is used to evaluate the model parameters.
We now will see how a real reactor might be modeled by different combinations of ideal reactors. There are an almost unlimited number of combinations that could be made. However, if we limit the number of adjustable parameters to two (e.g., bypass flow rate, υ_{b}, and dead volume, V_{D}), the situation becomes much more tractable. After reviewing the steps in Table 181, choose a model and determine whether it is reasonable by qualitatively comparing it with the RTD and, if it is, determine the model parameters. Usually, the simplest means of obtaining the necessary data is some form of a tracer test. These tests have been described in Chapters 16 and 17, together with their uses in determining the RTD of a reactor system. Tracer tests can be used to determine the RTD, which can then be used in a similar manner to determine the suitability of the model and the value of its parameters.
In determining the suitability of a particular reactor model and the parameter values from tracer tests, it may not be necessary to calculate the RTD function E(t). The model parameters (e.g., V_{D}) may be acquired directly from measurements of effluent concentration in a tracer test. The theoretical prediction of the particular tracer test in the chosen model system is compared with the tracer measurements from the real reactor. The parameters in the model are chosen so as to obtain the closest possible agreement between the model and experiment. If the agreement is then sufficiently close, the model is deemed reasonable. If not, another model must be chosen.
What is sufficiently close?
The quality of the agreement necessary to fulfill the criterion “sufficiently close” again depends on creativity in developing the model and on engineering judgment. The most extreme demands are that the maximum error in the prediction not exceed the estimated error in the tracer test, and that there be no observable trends with time in the difference between prediction (the model) and observation (the real reactor). In the Expanded Material on the CRE Web site, we illustrate how the modeling is carried out. We will now consider two different models for a CSTR: first, a CSTR with a dead zone and bypassing; and second, a CSTR model as two CST in interchange. In each of these cases, we will first show how to model the system with two adjustable parameters α and β and then show how to calculate the exit conversion on concentrations.
A real CSTR is believed to be modeled as a combination of an ideal CSTR with a wellmixed volume V_{s}, a dead zone of volume V_{d}, and a bypass with a volumetric flow rate υ_{b} (Figure 1814). We have used a tracer experiment to evaluate the parameters of the model V_{s} and υ_{s}. Because the total volume and volumetric flow rate are known, once V_{s} and υ_{s} are found, υ_{b} and V_{d} can readily be calculated.
The model system
The bypass stream and effluent stream from the reaction volume are mixed at the junction point 2. From a balance on species A around this point
[In] = [Out]
$\begin{array}{cc}{\upsilon}_{0}{C}_{\text{A}}=({\upsilon}_{b}+{\upsilon}_{s}){C}_{\text{A}}={C}_{\text{A0}}{\upsilon}_{b}+{C}_{\text{As}}{\upsilon}_{s}={C}_{\text{A0}}{\upsilon}_{0}\beta +{C}_{\text{As}}{\upsilon}_{0}(1\beta )& (1864)\end{array}$
The Duct Tape Council of Jofostan would like to point out the new wrinkle: The Junction Balance.
In the absence of reaction (e.g., tracer) the concentration exiting the system is
$\begin{array}{cc}{C}_{\text{A}}={C}_{\text{A0}}(1X)& \left(18\text{65}\right)\end{array}$
where we will let α = V_{s} / V and β = υ_{b} / υ_{0}.
We will use this model to calculate the conversion for the firstorder reaction
$\text{A}\stackrel{{k}_{1}}{\to}\text{B}$
A mole balance on the wellmixed volume V_{s} gives
${\upsilon}_{s}{C}_{\text{A0}}{\upsilon}_{s}{C}_{\text{As}}k{C}_{\text{As}}{V}_{s}=0$
Mole balance on CSTR
As shown in the extended material on the CRE Web site the exit concentration of species A in terms of α and β
$\begin{array}{c}\hline \frac{{C}_{\text{A}}}{{C}_{\text{A0}}}=1X=\beta +\frac{(1\beta {)}^{2}}{(1+\beta )+\alpha \tau k}\\ \hline\end{array}\begin{array}{c}\left(18\text{66}\right)\end{array}$
Conversion as a function of model parameters
We have used the ideal reactor system shown in Figure 1814 to predict the conversion in the real reactor (i.e., Equation (1866)). The model has two parameters, α and β. The parameter α is the dead zone volume fraction and parameter β is the fraction of the volumetric flow rate that bypasses the reaction zone. If these parameters are known, we can readily predict the conversion. In the following section, we shall see how we can use tracer experiments and RTD data to evaluate the model parameters.
Model system
If we were to inject a positive step tracer into the system shown in Figure 1814, the unsteadystate tracer balance in a wellmixed reactor value V_{s} is
In – out = accumulation
$\begin{array}{c}\hline {\upsilon}_{s}{C}_{\text{T0}}{\upsilon}_{s}{C}_{\text{Ts}}=\frac{d{N}_{\text{Ts}}}{dt}={V}_{s}\frac{d{C}_{\text{Ts}}}{dt}\\ \hline\end{array}\begin{array}{c}\left(18\text{67}\right)\end{array}$
Tracer balance for step input
The conditions for the positivestep input are
$\begin{array}{cc}\text{At}t0& {C}_{\text{T}}=0\\ \text{At}t\ge 0& {C}_{\text{T}}={C}_{\text{T}0}\end{array}$
Solving the tracer balance equations and using a balance around junction point 2, we arrive at the following equation relating C_{T} and t:
$\begin{array}{cc}\frac{{C}_{\text{T}}}{{C}_{\text{T0}}}=1(1\beta )\text{exp}[\frac{1\beta}{\alpha}\left(\frac{t}{\tau}\right)]& \left(\text{1868}\right)\end{array}$
The following tracer concentration time data was obtained from a step input to the system in Figure 1814.
TABLE E185.1 TRACER DATA FOR STEP INPUT
C_{T} (mg/dm^{3}) 
1000 
1333 
1500 
1666 
1750 
1800 
t (min) 
4 
8 
10 
14 
16 
18 
The entering tracer concentration is C_{T0} = 2000 mg/dm^{3}.
Determine the model parameters α and β where α = V_{s}/V and β = υ_{b}/υ_{0}.
Determine the conversion for a secondorder reaction with C_{A0} = 2 kmol/m^{3}, τ = 10 min, and k = 0.28 m^{3}/kmol · min.
Solution
Calculating α and β
The model parameters, α and β, are obtained either by regression (Polymath/ MATLAB/Excel) or from the proper plot of the effluent tracer concentration as a function of time. Rearranging Equation (1868) yields
$\begin{array}{c}\hline \text{In}\frac{{C}_{\text{T0}}}{{C}_{\text{T0}}{C}_{\text{T}}}=\text{In}\frac{1}{1\beta}+\left(\frac{1\beta}{\alpha}\right)\frac{t}{\tau}\\ \hline\end{array}\begin{array}{c}\phantom{\rule[0.0ex]{8em}{0.0ex}}\left(\text{E185.1}\right)\end{array}$
Consequently, using Table E185.1 we plot ln[C_{T0}/(C_{T0} = C_{T})] as a function of t. If our model is correct, a straight line should result with a slope of (1 = β)/τα and an intercept of ln[1/(1 = β)].
TABLE E185.2 TRACER DATA FOR STEP INPUT
C_{T} (mg/dm^{3}) 
1000 
1333 
1500 
1666 
1750 
1800 
$\frac{C\text{}{\text{}}_{\text{T0}}}{{C}_{\text{}\text{T0}}{C}_{r}}$ 
2 
3 
4 
6 
8 
10 
t (min) 
4 
8 
10 
14 
16 
18 
When this data is regressed using Equation (E185.1), we obtain
$\begin{array}{c}\hline \alpha =0.7\text{and}\\ \beta =0.2\\ \hline\end{array}$
${\tau}_{s}=\frac{{V}_{s}}{{\upsilon}_{s}}=\frac{\alpha V}{(1\beta ){\upsilon}_{0}}=\frac{\alpha \tau}{(1\beta )}=\frac{\left(0.7\right)\left(10\text{min}\right)}{(10.2)}=8.7\text{min}$
Calculating the Conversion
We now can proceed applying our algorithm to this system. A mole balance on reactor V_{s} gives
$\begin{array}{cc}{\upsilon}_{s}{C}_{\text{A0}}{\upsilon}_{s}{C}_{\text{A}}k{C}_{\text{A}s}^{2}{V}_{s}=0& \left(\text{E185.2}\right)\end{array}$
For a secondorder reaction
${r}_{\text{A}}=k{C}_{\text{A}}^{2}\text{with}k=\frac{0.28{\text{m}}^{3}}{\text{kmol}\cdot \text{min}}$
The exit concentration is
$\begin{array}{cc}{C}_{\text{A}}=\beta {C}_{\text{A0}}+(1\beta ){C}_{\text{A}s}& \left(\text{E185.3}\right)\end{array}$
$\begin{array}{cc}\hline {C}_{\text{A}s}=\frac{1+\sqrt{1+4{\tau}_{s}k{C}_{\text{A0}}}}{2{\tau}_{s}k}& \\ \hline\end{array}\begin{array}{c}\left(\text{E185.4}\right)\end{array}$
Substituting parameter values we found that the exit concentration and the conversion predicted in the real (nonideal) reactor are
C_{A} = 0.979 kmol/m^{3} and
$\begin{array}{c}\hline {X}_{\text{model}}=0.51\\ \hline\end{array}$
X_{model} = 0.51
X_{Ideal} = 0.66
Using the same ratelaw parameter values for or an ideal CSTR we find
C_{A} = 0.685 kmol/m^{3} and
X_{Ideal} = 0.66
Analysis: In this example we used a combination of an ideal CSTR with a dead volume and bypassing to model a nonideal reactor. If the nonideal reactor behaved as an ideal CSTR, a conversion of 66% was expected. Because of the dead volume, not all the space would be available for reaction; also, some of the fluid did not enter the space where the reaction was taking place and, as a result, the conversion in this nonideal reactor was only 51%.
Other Models. In Section 18.8.1 it was shown how we formulated a model consisting of ideal reactors to represent a real reactor. First, we solved for the exit concentration and conversion for our model system in terms of two parameters, α and β. We next evaluated these parameters from data on tracer concentration as a function of time. Finally, we substituted these parameter values into the mole balance, rate law, and stoichiometric equations to predict the converstion in our real reactor.
To reinforce this concept, we will use one more example (yes, just one more, as given in Example 186).
In this particular model there is a highly agitated region in the vicinity of the impeller; outside this region, there is a region with less agitation (Figure 1815). There is considerable material transfer between the two regions. Both inlet and outlet flow channels connect to the highly agitated region. We shall model the highly agitated region as one CSTR, the quieter region as another CSTR, with material transfer between the two.
The model system
Let β represent that fraction of the total flow that is exchanged between reactors 1 and 2; υ_{1} = υ_{2} = β_{υ0} and let α represent that fraction of the total volume, V, occupied by the highly agitated region V_{1} = αV:
The space time is measured as the real reactor volume V divided by the total volumetric flow rate υ_{0}.
$\tau =\frac{\text{V}}{{\upsilon}_{0}}=\frac{{\text{V}}_{1}+{\text{V}}_{2}}{{\upsilon}_{0}}$
As shown on the CRE Web site Professional Reference Shelf R18.2, for a firstorder reaction, the exit concentration and conversion are
$\begin{array}{cc}{C}_{\text{A1}}=\frac{{C}_{\text{A0}}}{1+\beta +\alpha \tau k\{{\beta}^{2}/[\beta +(1\alpha )\tau k\left]\right\}}& \left(18\text{69}\right)\end{array}$
and
$\begin{array}{cc}\hline X=1\frac{{C}_{\text{A1}}}{{C}_{\text{A0}}}=\frac{(\beta +\alpha \tau k)[\beta +(1+\alpha )\tau k]{\beta}^{2}}{(1+\beta +\alpha \tau k)[\beta +(1\alpha )\tau k]{\beta}^{2}}& \\ \left(18\text{70}\right)\\ \hline\end{array}$
where C_{A1} is the reactor concentration exiting the first reactor in Figure 1815(b).
Conversion for twoCSTR model
The problem now is to evaluate the parameters α and β using the RTD data. C_{T1} is the measured tracer concentration exiting the real reactor. The tracer is initially dumped only into reactor 1, so that the initial conditions C_{T10} = N_{T0}/V_{1} and C_{T20} = 0.
Using a tracer balance in reactors 1 and 2 in terms of α, β, and τ, we arrive at two coupled differential equations describing the unsteady behavior of the tracer that must be solved simultaneously.
$\begin{array}{c}\hline \tau \alpha \frac{d{C}_{\text{T1}}}{dt}=\beta {C}_{\text{T2}}(1+\beta ){C}_{\text{T1}}\\ \hline\end{array}\begin{array}{c}\left(\begin{array}{c}18\text{71}\end{array}\right)\end{array}$
$\begin{array}{c}\hline \tau (1\alpha )\frac{d{C}_{\text{T2}}}{dt}=\beta {C}_{\text{T1}}\beta {C}_{\text{T2}}\\ \hline\end{array}\begin{array}{c}\left(\begin{array}{c}18\text{}\text{72}\end{array}\right)\end{array}$
See Appendix A.3 for method of solution.
Analytical solutions to Equations (1871) and (1872) are given on the CRE Web site, in Appendix A.3 and in Equation (1873), below. However, for more complicated systems, analytical solutions to evaluate the system parameters may not be possible.
$\begin{array}{cc}(\frac{{C}_{\text{T1}}}{{C}_{\text{T10}}}{)}_{\text{pulse}}=\frac{(\alpha {m}_{1}+\beta +1){e}^{{m}_{2}t/\tau}(\alpha {m}_{2}+\beta +1){e}^{{m}_{1}t/\tau}}{\alpha ({m}_{1}{m}_{2})}& \left(18\text{73}\right)\end{array}$
where
${m}_{1},{m}_{2}=\left[\frac{1\alpha +\beta}{2\alpha (1\alpha )}\right][1\pm \sqrt{\frac{4\alpha \beta (1\alpha )}{{(1\alpha +\beta )}^{2}}}]$
A pulse trace test was carried out on the model system shown in Figure 1815 (http://www.umich.edu/~elements/6e/18chap/Ch18_WebAdditional%20Material.pdf) with ${C}_{\text{T0}}=2000{\text{mg/dm}}^{3}\text{and}\tau =\frac{V}{{\upsilon}_{0}}=\frac{1000{\text{dm}}^{3}}{25{\text{dm}}^{3}/\text{min}}=40\text{min}$, and the results are shown in Table E186.1.
TABLE E186.1 TRACER DATA FOR PULSE INPUT
t (min) 
0.0 
20 
40 
60 
80 
120 
160 
200 
240 
280 
320 
θ = t/τ 
0.0 
0.5 
1.0 
1.5 
2.0 
3.0 
4.0 
5.0 
6.0 
7.0 
8.0 
C 
2000 
1050 
520 
280 
160 
61 
29 
16.4 
10.0 
6.4 
4.0 
C/C_{10} 
1.0 
0.525 
0.26 
0.14 
0.08 
0.03 
0.0145 
0.0082 
0.005 
0.0032 
0.002 
Find the interchange parameters α and β.
Find the conversion for a firstorder reaction with k = 0.03 min^{–1}.
Find the corresponding conversion in an ideal CSTR and in an ideal PFR.
Solution
Finding α and β
A pulse tracer experiment is used so that all the tracer, N_{T0}, goes into reactor 1 with volume, V_{1}, that is, (N_{T0} = C_{10} V_{1}).
α = V_{1}/V
β = υ_{1}/υ_{0}
A tracer balance yields
(mass added at t = 0) = (mass out over all time)
$\begin{array}{cc}{C}_{\text{10}}\alpha V={\upsilon}_{0}{\int}_{0}^{\infty}\text{}C\left(t\right)dt& \left(\text{E186.1}\right)\end{array}$
Solving for α
$\begin{array}{cc}\alpha =\frac{1}{{C}_{10}}{\int}_{0}^{\infty}\text{}C\left(\theta \right)d\theta ={\int}_{0}^{\infty}\frac{C\left(\theta \right)}{{C}_{10}}d\theta & \left(\text{E186.2}\right)\end{array}$
We see that α is just the area under the curve C(θ) in Figure E186.1 divided by C_{10}.
Evaluating the area we obtain
α = 0.80
As shown on the Web site (http://www.umich.edu/~elements/6e/18chap/Ch18_WebAdditional%20Material.pdf) that by plotting the ratio C(t)/C_{10} as a function of θ on semilog coordinates, we get the graph shown on the Web site in Figure Web E181.3. At long times, in Equation (1873) the first term containing m_{2} in the exponent is negligible with respect to the second term. Consequently, if we extrapolate the portion of the curve for long times back to θ = 0, we have
$\begin{array}{cc}\text{Intercept}=I=\frac{\alpha {m}_{2}+\beta +1}{\alpha ({m}_{1}{m}_{2})}=0.66& \text{(Web E182.4)}\end{array}$
The values of m1 and m2 are obtained from Figure Web E181.3.
$0.066=\frac{\left(0.8\right)(1.44)+\beta +1}{\left(0.8\right)[0.434(1.44)]}$
Two CSTRs with interchange
Solving for β, we obtain β = 0.1. The two parameters for this model are then
$\begin{array}{c}\hline \alpha =0.8\text{and}\beta =0.1\\ \hline\end{array}$
$\begin{array}{c}\hline \tau k=\left(40\text{min}\right)\left(0.03{\text{min}}^{1}\right)=1.2\\ \hline\end{array}$
Find the conversion for a firstorder reaction
For two CSTRs with interchange, recall Equation (1870)
$\begin{array}{c}\hline X=1\frac{{C}_{\text{A1}}}{{C}_{\text{A0}}}=\frac{(\beta +\alpha \tau k)[\beta +(1\alpha )\tau k]{\beta}^{2}}{(1+\beta +\alpha \tau k)[\beta +(1\alpha )\tau k]{\beta}^{2}}\\ \hline\end{array}\begin{array}{c}\left(\begin{array}{c}18\text{70}\end{array}\right)\end{array}$
Using regression we find
α = 0.8
β = 0.1
We now substitute α, β, τk into Equation (1870), and as also shown on the CRE Web site, Substituting for τk, α and β in Equation (1869) yields
$\begin{array}{cc}\frac{{C}_{\text{A}}}{{C}_{\text{A0}}}=1X\frac{1}{1+\beta +\alpha \tau k\frac{{\beta}^{2}}{\beta +(1\alpha )\tau k}}& \left(\text{E186.3}\right)\end{array}$
$\begin{array}{cc}1X=\frac{1}{1+0.1+\left(0.8\right)\left(1.2\right)\frac{(0.1{)}^{2}}{0.1+(10.8)\left(1.2\right)}}& \left(\text{E186.4}\right)\end{array}$
$\begin{array}{cc}X=0.51& \text{Ans.}\end{array}$
Finding conversion in ideal CSTR and ideal PFR
For a single ideal CSTR,
$\begin{array}{cc}X=\frac{\tau k}{1+\tau k}=\frac{1.2}{2.2}=0.55& \text{Ans.}\end{array}$
For a single ideal PFR,
$\begin{array}{cc}X=1{e}^{\tau k}=1{e}^{1.2}=0.7& \text{Ans.}\end{array}$
Two CSTRs with interchange
Comparing models, we find
(X_{model} = 0.51) < (X_{CSTR} = 0.55) < (X_{PFR} = 0.7)
Analysis: For the twoparameter model chosen, we used the RTD to determine the two parameters’ to find the conversion X, i.e., those parameters were the fraction of the larger fluid volume V_{1} = αV and the fraction β of fluid exchanged between the reactors, υ_{1} = βυ_{0}. We next calculated the exit conversion using an ideal CSTR. The CRE algorithm was then applied to model the reactor system where the conversion was found to be 51%, which is smaller than that for an ideal CSTR (x = 0.56).
Several reactor models have been discussed in the preceding pages. All are based on the physical observation that in almost all agitated tank reactors, there is a wellmixed zone in the vicinity of the agitator. This zone is usually represented by a CSTR. The region outside this wellmixed zone may then be modeled in various fashions. We have already considered the simplest models, which have the main CSTR combined with a deadspace volume; if some shortcircuiting of the feed to the outlet is suspected, a bypass stream can be added. The next step is to look at all possible combinations that we can use to model a nonideal reactor using only CSTRs, PFRs, dead volume, and bypassing. The rate of transfer between the two reactors is one of the model parameters. The positions of the inlet and outlet to the model reactor system depend on the physical layout of the real reactor.
Figure 1816(a) describes a real PFR or PBR with channeling that is modeled as two PFRs/PBRs in parallel. The two parameters are the fraction of flow to the reactors (i.e., β and (1 =β)) and the fractional volume (i.e., α and (1 – α)) of each reactor. Figure 1816(b) describes a real PFR/PBR that has a backmix region and is modeled as a PFR/PBR in parallel with a CSTR. Figures 1816(a) and (b) on page 973 show a real CSTR modeled as two CSTRs with interchange. In one case, the fluid exits from the top CSTR (a), and in the other case the fluid exits from the bottom CSTR (b). The parameter β represents the interchange volumetric flow rate, βυ_{0}, and = the fractional volume of the top reactor, =V, where the fluid exits the reaction system. We note that the reactor in Figure 1816(b) was found to describe extremely well a real reactor used in the production of terephthalic acid.^{14} A number of other combinations of ideal reactions can be found in Levenspiel.^{15}
^{14} Proc. Indian Inst. Chem. Eng. Golden Jubilee, a Congress, Delhi, 1997, p. 323.
^{15} Levenspiel, O. Chemical Reaction Engineering, 3rd ed. New York: Wiley, 1999, pp. 284–292.
A case history for terephthalic acid
Models for nonideal reactors
The use of combinations of ideal reactors to model metabolism and drug distribution in the human body is becoming commonplace. For example, one of the simplest models for drug adsorption and elimination is similar to that shown in Figure 1817(a). The drug is injected intravenously into a central compartment containing the blood (the top reactor). The blood distributes the drug back and forth to the tissue compartment (the bottom reactor) before being eliminated (top reactor). This model will give the familiar linear semilog plot found in pharmacokinetics textbooks. As can be seen in Chapter 9, in the figure for Professional Reference Shelf R9.8 on the CRE Web site on pharmacokinetics, and on pages 408–409, there are two different slopes, one for the drug distribution phase and one for the elimination phase.
As chemical engineers, it is important for us to understand the aspects of safety involved when we subject a process, plant equipment, or an operating procedure to a change. We must have the ability to look at not only what parameters improve as a result of this change but also at those which may be affected inadvertently and to take proper measures to minimize or at best, nullify the effect of these changes.
We must pay attention to how we define “change.” It is, “anything which is not a likeforlike replacement.”
True Case History
Actual Case History 1: John, Kötloff (Sven’s brother) head of Purchasing at Jofostan Chemical Co. (JCC) was contacted by a new company hoping to sell one of their raw materials to JCC at a cheaper price than their current supplier. They also assured him that they will deliver this highly hazardous liquid to their production facility in the same type of ISO tank (i.e., as per international standards) and with matching connection valves to unload the tank. John went ahead with the purchase without consulting the plant.
Epilogue: The first supply damaged the delivery gantry, because the new supplier used a higher trailer to carry the ISO tank.
Before reading further, let us define risk assessment. Risk assessment is the process of identifying all the possible consequences of an operation and evaluation of their hazards.
Proper and precise description of the change supported with crucial information along with all the paperwork is necessary in order to carry out risk assessment of the change. Before the change is made, the MoC system must identify possible effects and sideeffects in order to assess the hazardous consequences and provide a risk assessment of the change.
Actual Case History 2 (Only the names have been changed): In the early 70s, the Jofostan Mixers produced magnesium oxide (MgO) at one of their plants. The MgO was marketed under the name JofoNutri and was added as a nutritional supplement to the feed for dairy cows. The same plant also manufactured polybrominated biphenyls (PBBs) which were used as flame retardant under the name JofoFire.^{†} Sometime in 1973, the paper bags containing JofoFire reached the complex producing dairy feed and was added by the plant operator to the dairy feed assuming that it was JofoNutri.
^{†} M. Venier, A. Salamova, and R. A. Hites, “Halogenated flame retardants in the Great Lakes Environment,” Acc. Chem. Res., 48, 1853–1861 (2015).
Any activity that classifies as a “change” must go through the following, before implemented, as described in the third edition of “Strategies for Creative Problem Solving”:
Make the case for change: identify the need for change and the stakeholders involved.
Vision for change: what will it be like, after the change?
What skills are needed to implement the change as far as design, communication, etc. are concerned?
What is the incentive for the organization to undertake the change?
Do we have enough resources (personnel, knowledge, etc.) to implement the change?
Make an action plan (Gantt chart, timetables, etc.) for the change.
In light of the earlier discussion, answer:
What should a “change” include?
Any definition of change must clearly include changes in equipment, process, or software as well as addition of a new equipment or process. Any modification to organizational structure or any alteration to an operating procedure must also be defined as a change.
What safeguards could have prevented the Jofostan dairy feed incident?
For proper isolation of the harmful chemicals at the facility from the foodgrade products, different transport vehicles as well as different personnel should have been employed for JofoNutri and JofoFire.
A distinctive feature on the packaging of all the harmful chemicals at the facility would have allowed operators to distinguish foodgrade products from harmful chemicals.
The plant operators should have been sensitive to the presence of harmful chemicals that may be accidentally transported to the dairy feed producing mill.
Use of similar names like JofoNutri and JofoFire should have been avoided as it could have further led to the mixup.
As the chief of investigation after the Jofostan dairy feed incident, what will be the recommendations of your report to prevent such incidents in the future?
While some of the safeguards in (b) may be recommended, it is important to focus on the “management of change” aspect here.
When the company decided to shift to plain brown bags instead of printed, colorcoded bags, it should have flagged this as a “change.”
All stakeholders—Purchasing, Operations, and Logistics—must be notified of this proposal and consulted for their views.
A dedicated safety manager must be entrusted to coordinate with all stakeholders and submit his recommendations before such a proposal goes through. He must evaluate the proposal for change in the manner described earlier.
The models for predicting conversion from RTD data are:
Zero adjustable parameters
Segregation model
Maximum mixedness model
One adjustable parameter
Tanksinseries model
Dispersion model
Two adjustable parameters: real reactor modeled as combinations of ideal reactors
Dispersion model: For a firstorder reaction, use the Danckwerts boundary conditions
$\begin{array}{c}\hline X=1\frac{4q\text{exp}(P{e}_{r}/2)}{(1+q{)}^{2}\text{exp}(P{e}_{r}q/2)(1q{)}^{2}\text{exp}(P{e}_{r}q/2)}\\ \hline\end{array}\begin{array}{c}\left(\text{S181}\right)\end{array}$
where
$\begin{array}{cc}q=\sqrt{1+\frac{4\mathit{D}{\mathit{a}}_{1}}{P{e}_{r}}}& \left(\text{S182}\right)\end{array}$
$\begin{array}{cc}\mathit{D}{\mathit{a}}_{1}=\tau k& \left(\begin{array}{c}\text{S183}\end{array}\right)\end{array}$
For a firstorder reaction
$\begin{array}{cc}\begin{array}{cccc}P{e}_{r}=\frac{UL}{{D}_{a}}& & P{e}_{f}=\frac{U{d}_{p}}{{D}_{a}\varphi}& \end{array}& \left(\text{S184}\right)\end{array}$
Determine D_{a}
For laminar flow, the dispersion coefficient is
$\begin{array}{c}\hline D*={D}_{\text{AB}}+\frac{{U}^{2}{R}^{2}}{48{D}_{\text{AB}}}\\ \hline\end{array}\begin{array}{c}\left(\text{S185}\right)\end{array}$
Correlations. Use Figures 188 through 1810.
Experiment in RTD analysis to find t_{m} and σ^{2}.
For a closedclosed system, use Equation (S185) to calculate Pe_{r} from the RTD data
$\begin{array}{c}\hline \frac{{\sigma}^{2}}{{\tau}^{2}}=\frac{2}{P{e}_{r}}\frac{2}{P{e}_{r}^{2}}(1{e}^{P{e}_{r}})\\ \hline\end{array}\begin{array}{c}\left(\text{S186}\right)\end{array}$
For an openopen system, use
$\begin{array}{c}\hline \frac{{\sigma}^{2}}{{\text{t}}_{m}^{2}}=\frac{2P{e}_{r}+8}{P{e}_{r}^{2}+4P{e}_{r}+4}\\ \hline\end{array}\begin{array}{c}\left(\text{S187}\right)\end{array}$
If a real reactor is modeled as a combination of ideal reactors, the model should have at most two parameters.
The RTD is used to extract model parameters.
Comparison of conversions for a PFR and CSTR with the zeroparameter and twoparameter models. X_{seg} symbolizes the conversion obtained from the segregation model and X_{mm} is that from the maximum mixedness model for reaction orders greater than one.
$\begin{array}{c}\hline {X}_{\text{PFR}}>{X}_{\text{seg}}>{X}_{\text{mm}}>{X}_{\text{CSTR}}\\ \begin{array}{cccc}{X}_{\text{PFR}}>{X}_{\text{model}}& \text{with}& {X}_{\text{model}}<{X}_{\text{CSTR}}\text{or}& {X}_{\text{model}}{X}_{\text{CSTR}}\end{array}\\ \hline\end{array}$
Cautions: For rate laws with unusual concentration functionalities or for nonisothermal operation, these bounds may not be accurate for certain types of rate laws.
Axial or radial temperature and concentration gradients. The following coupled, partial differential equations were solved using COMSOL:
$\begin{array}{c}\hline {D}_{e}\frac{{\partial}^{2}{C}_{i}}{\partial {r}^{2}}+\frac{{D}_{e}}{r}\frac{\partial {C}_{i}}{\partial r}+{D}_{e}\frac{{\partial}^{2}{C}_{i}}{\partial {z}^{2}}{U}_{z}\frac{\partial {C}_{i}}{\partial z}+{r}_{i}=0\\ \hline\end{array}\begin{array}{c}\left(\begin{array}{c}\text{S188}\end{array}\right)\end{array}$
and
$\begin{array}{c}\hline {k}_{e}\frac{\partial 2T}{\partial {z}^{2}}+\frac{{k}_{e}\partial}{r\partial r}\left(r\frac{\partial T}{\partial r}\right)+\mathrm{\Delta}{H}_{\text{Rx}}{r}_{\text{A}}{U}_{Z}{C}_{{\text{P}}_{m}}\frac{\partial T}{\partial z}=0\\ \hline\end{array}\begin{array}{c}\left(\text{S189}\right)\end{array}$
(http://www.umich.edu/~elements/6e/18chap/obj.html#/)
Useful Links
Evaluation
The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.
A = • B = ■ C = ♦ D = ♦♦
Q181 QBR (Question Before Reading). What if you were asked to design a tubular vessel that would minimize dispersion? What would be your guidelines? How would you maximize the dispersion? How would your design change for a packed bed?
Q182_{B} Make up and solve an original problem. The guidelines are given in Problem
Q183_{A} Can you use Figure 182 to find the dispersion coefficient for a liquid with a Schmidt number, Sc = 0.35 and a gas porosity ϕ = 0.5, a particle diameter d_{p} = l cm, a gas velocity U = 10 cm/s, and a viscosity of a kinematic υ = 30 cm^{2}/s? If so, what is its value?
Q184 What if someone suggested you could use the solution to the flowdispersionreactor equation, Equation (1818), for a secondorder equation by linearizing the rate law by lettering ${r}_{\text{A}}={kC}_{\text{A}}^{2}\cong (k{C}_{\text{A0}}/2){C}_{\text{A}}\mathrm{.}\text{}\left(1\right)$ Under what circumstances might this be a good approximation? Would you divide C_{A0} by something other than 2? (2) What do you think of linearizing other nonfirstorder reactions and using Equation (1818)? (3) How could you test your results to learn whether the approximation is justified?
Q185 Go to the LearnChemE screencast link for Chapter 18 (http://umich.edu/~elements/6e/18chap/learnchemevideos.html). View one or more of the screencast 5 to 6minute videos and write a twosentence evaluation.
Q186 AWFOS–S18. An Algorithm for Management of Change (MoC). List three things that would not have occurred if the MoC algorithm had been followed.
P181_{b}
Example 181. Concentration and Conversion Profiles for Dispersion and Reaction in a Tubular Reactor
Wolfram and Python
Vary Péclet number, Pe_{r} from its largest to smallest value. Describe how the outlet concentration for “reaction with dispersion” deviates from that of “Ideal PFR” as Pe_{r} is varied.
What should be the minimum value of space time, τ, at which conversion reaches at least 95% for reaction with dispersion.
If the reaction rate constant is increased to 0.5 min^{–1}, what should be the value of Péclet number, Pe_{r}, so as to achieve conversion of 90%.
Vary the sliders and write a set of conclusions.
Example 182. Comparing Conversion Using Dispersion, PFR, CSTR, and TanksinSeries Models for Isothermal Reactors
Wolfram and Python
Which parameter will you vary so that conversion obtained by dispersion model approaches to conversion obtained using ideal PFR model.
What is the number of tanks in series required so that conversion obtained by the tankinseries model is more than conversion obtained by the dispersion model?
Write a set of conclusions after you have varied all the parameter values.
Example 183. Isothermal Reaction with Radial and Axial Dispersion in an LFR
Go to the COMSOL link LEP 183 in Chapter 18 on the CRE Web site.
Vary the velocity and describe how the radial concentration profiles as well as the surface plot changes.
Vary the reactor radius and describe what you find.
Vary D_{AB} and describe how the radial concentration profiles and the exit concentration change.
Compare the radial concentration profiles at the inlet, exit, and halfway down the reactor with the radially averaged axial concentration when you vary D_{AB} and U and describe what you find.
What combination of parameters, for example, (D_{a}/UL) do not significantly change the base case profile when varied over a wide range? For example, diameter is between 0.01 dm and 1 m and the velocity varies from U = 0.01 cm/s to 1 m/s with D_{AB} = 10^{–5} cm^{2}/s with υ = μ/ρ = 0.01 m^{2}/s. Is there a diameter that minimizes or maximizes your conversion?
To what parameters or groups of parameters (e.g., kL^{2}/D_{a}) would the conversion be most sensitive? What if the firstorder reaction were carried out in tubular reactors of different diameters, but with the space time, τ, remaining constant? The diameters would range from a diameter of 0.1 dm to a diameter of 1 m for a kinematic viscosity υ = μ/ρ = 0.01 cm^{2}/s, U = 0.1 cm/s, and D_{AB} = 10^{–5} cm^{2}/s. How would your conversion change? Is there a diameter that would maximize or minimize conversion in this range?
Which type of velocity profile gives the higher outlet concentration?
What two parameters would reduce the radial variation in concentration?
Which variable has very little effect on radial variation in concentration?
What is the effect of diffusivity on the outlet concentration?
Keep Da and L/R constant and vary the reaction order n, (0.5 ≤ n ≤ 5) for different Péclet numbers. Are there any combinations of n and Pe where dispersion is more important or less important on the exit concentration? What generalizations can you make? Hint: for n < 1 use ${r}_{\text{A}}=k\mathrm{.}\left(\text{Abs}\left({C}_{\text{A}}^{n}\right)\right)$.
Write a set of conclusions based on your experiments (i) through (xi).
Example 184. Tubular Reactor with Axial and Radial Temperature and Concentration Gradients
Use the COMSOL LEP on the Web site to explore radial effects in a tubular reactor. The Additional Material on the Web site describes this LEP in detail (http://umich.edu/~elements/6e/18chap/expanded_ch18_radial.pdf). Tutorials on how to access and use COMSOL can be found at (http://umich.edu/~elements/6e/software/software_comsol.html).
Vary the heat transfer coefficient, U, and observe the effect on both the conversion and temperature profiles. Why is there an effect of U on the profiles at low values of U, but not at higher values of U?
What is the effect of the different diffusivities, D_{a}, on the conversion? Should D_{a} be minimized or maximized?
What parameter would you vary so that the maximum of the radial conversion profile becomes 1? What is the value of that parameter?
How would your profiles change if the velocity profile was changed to plug flow (U = U_{0})? Based on your observations, which of the two profiles, namely, laminar flow and plug flow, would you recommend for a larger conversion? How does the change in velocity profile make an effect in the temperature profile?
Intuitively, how do you expect the radial conversion to vary with diffusion coefficient? Explain the reason behind your intuition. Verify this by varying the diffusion coefficient. Investigate axial and radial conversion and write two conclusions.
Example 185. Using a Tracer to Determine the Model Parameters in a CSTR with Dead Volume and Bypass Model
Wolfram and Python
For the case of no dead volume (α = 1), vary bypass volume fraction, β, and describe its effect on outlet concentration.
Repeat part (i) for the case of no bypass volume (β = 0) and vary = to observe its effect on outlet concentration.
From your observation in part (i) and part (ii), which of the two model parameters (α or β) have more impact on outlet concentration and conversion? Explain.
Write a set of conclusions based on your experiments in (i) through (iii).
Example 186. CSTR with Interchange
Wolfram and Python
For a fixed value of α (say α = 0.5), how does the conversion vary with increase in β? Explain.
Vary α and describe its impact on the exit concentration.
Vary rate constant, k, and describe what you find about its effect on exit concentration.
Write a set of conclusions based on your experiments in (i) through (iii).
P182_{B} The gasphase isomerization
A → B
is to be carried out in a flow reactor. Experiments were carried out at a volumetric flow rate of υ_{0} = 2 dm^{3}/min in a reactor that had the following RTD
E(t) = 10 e^{10t} min^{–1}
where t is in minutes.
When the volumetric flow rate was 2 dm^{3}/min, the conversion was 9.1%. What is the reactor volume?
When the volumetric flow rate was 0.2 dm^{3}/min, the conversion was 50%. When the volumetric flow rate was 0.02 dm^{3}/min, the conversion was 91%. Assuming the mixing patterns don’t change as the flow rate changes, what will the conversion be when the volumetric flow rate is 10 dm^{3}/min?
This reaction is now to be carried out in a 1dm^{3} plugflow reactor where volumetric flow rate has been changed to 1 dm^{3}/min. What will be the conversion?
It is proposed to carry out the reaction in a 10mdiameter pipe where the flow is highly turbulent (Re = 10^{6}). There are significant dispersion effects. The superficial gas velocity is 1 m/s. If the pipe is 6 m long, what conversion can be expected? If you were unable to determine the reaction order and the specific reactionrate constant in part (b), assume k = 1 min^{–1} and carry out the calculation!
P183_{b} OEQ (Old Exam Question). The secondorder liquidphase reaction
A → B + C
is to be carried out isothermally. The entering concentration of A is 1.0 mol/dm^{3}. The specific reaction rate is 1.0 dm^{3}/mol=min. A number of used reactors (shown below) are available, each of which has been characterized by an RTD. There are two crimson and white reactors, and three maize and blue reactors available.
Reactor 
σ(min) 
τ(min) 
Cost 
Maize and blue 
2 
2 
$25,000 
Green and white 
4 
4 
50,000 
Scarlet and gray 
3.05 
4 
50,000 
Orange and blue 
2.31 
4 
50,000 
Purple and white 
5.17 
4 
50,000 
Silver and black 
2.5 
4 
50,000 
Crimson and white 
2.5 
2 
25,000 
You have $50,000 available to spend. What is the greatest conversion you can achieve with the available money and reactors?
How would your answer to (a) change if you had an extra $75,000 available to spend?
From which cities do you think the various used reactors came from?
P184_{B} OEQ (Old Exam Question). The elementary liquidphase reaction
$\begin{array}{cc}\text{A}\stackrel{{k}_{1}}{\to \text{B,}}& {k}_{1}=1.0{\text{min}}^{1}\end{array}$
is carried out in a packedbed reactor in which dispersion is present.
What is the conversion?
Additional information:
Porosity = 50% 
Reactor length = 0.1 m 
Particle size = 0.1 cm 
Mean velocity = 1 cm/s 
Kinematic viscosity = 0.01 cm^{2}/S 
Bed fluidicity = 7.3 
(Ans: X = 0.15)
P185_{A} A gasphase reaction is being carried out in a 5cmdiameter tubular reactor that is 2 m in length. The velocity inside the pipe is 2 cm/s. As a very first approximation, the gas properties can be taken as those of air (kinematic viscosity = 0.01 cm^{2}/s), and the diffusivities of the reacting species are approximately 0.005 cm^{2}/s.
How many tanks in series would you suggest to model this reactor?
If the secondorder reaction A + B → C + D is carried out for the case of equimolar feed, and with C_{A0} = 0.01 mol/dm^{3}, what conversion can be expected at a temperature for which k = 25 dm^{3}/mol·s?
How would your answers to parts (a) and (b) change if the fluid velocity was reduced to 0.1 cm/s? Increased to 1 m/s?
How would your answers to parts (a) and (b) change if the superficial velocity was 4 cm/s through a packed bed of 0.2cmdiameter spheres?
How would your answers to parts (a) through (d) change if the fluid was a liquid with properties similar to water instead of a gas, and the diffusivity was 5 = 10^{–6} cm^{2}/s?
P186_{A} Use the data in Example 162 to make the following determinations. (The volumetric feed rate to this reactor was 60 dm^{3}/min.)
Calculate the Péclet numbers for both open and closed systems.
For an open system, determine the space time τ and then calculate the % dead volume in a reactor for which the manufacturer’s specifications give a volume of 420 dm^{3}.
Using the dispersion calculate the conversion for a closedclosed vessel and tanksinseries models, for the firstorder isomerization
A → B
with k = 0.18 min^{–1}.
Compare your results in part (c) with the conversion calculated from the tanksinseries model, a PFR, and a CSTR.
P187_{A} OEQ (Old Exam Question). A tubular reactor has been sized to obtain 98% conversion and to process 0.03 m^{3}/s. The reaction is a firstorder irreversible isomerization. The reactor is 3 m long, with a crosssectional area of 25 dm^{2}. After being built, a pulse tracer test on the reactor gave the following data: t_{m} = 10 s and =^{2} = 65 s^{2}. What conversion can be expected in the real reactor?
P188_{B} The following E(t) curve was obtained from a tracer test on a reactor.
$\begin{array}{llllll}E\left(t\right)& =& 0.25t& & & 0<t<2\\ & =& 10.25t& & & 2<t<4\\ & =& 0& & & t>4\end{array}$
t in minutes, and E(t) in min^{–1}.
The conversion predicted by the tanksinseries model for the isothermal elementary reaction
A → B
was 50% at 300 K.
If the temperature is to be raised 10°C (E = 25000 cal/mol) and the reaction carried out isothermally, what will be the conversion predicted by the maximum mixedness model? The TIS model?
The elementary reactions
$\begin{array}{l}\text{A}\stackrel{{k}_{1}}{\to}B\stackrel{{k}_{2}}{\to}\text{C}\\ \text{A}\stackrel{{k}_{3}}{\to}\text{D}\\ {k}_{1}={k}_{2}={k}_{3}=0.1{\text{min}}^{1}{\text{at300K,C}}_{\text{A0}}=1{\text{mol/d,}}^{3}\end{array}$
were carried out isothermally at 300 K in the same reactor. What is the concentration of B in the exit stream predicted by the maximum mixedness model?
For the multiple reactions given in part (b), what is the conversion of A predicted by the dispersion model in an isothermal closedclosed system?
P189_{B} Revisit Problem P163_{C} where the RTD function is a hemicircle. The liquidphase reaction is first order with k_{1} = 0.8 min^{–1}.
What is the conversion predicted by
The tanksinseries model? (Ans: X^{TIS} = 0.447)
The dispersion model? (Ans: X^{Dispersion} = 0.41)
P1810_{B} Revisit Problem P165_{B}. The liquidphase reaction A → B is third order with k_{3} = 0.3 dm^{6}/mol^{2} min and C_{A0} = 2M.
What combination of ideal reactors would you use to model the RTD?
What are the model parameters?
What is the conversion predicted for your model?
What is the conversion predicted by X_{mm}, X_{seg}, X_{TIS}, and X_{Dispersion?}
Repeat (a) through (d) for a liquid phase reaction A → B is second order with k_{2} C_{A0} = 0.1 min^{–1}.
P1811_{B} Below in Figure P1811_{b} are two COMSOL simulations for a laminarflow reactor with heat effects: Run 1 and Run 2. The following figures show the crosssection plot of concentration for species A at the middle of the reactor. Run 2 shows a minimum on the crosssection plot. This minimum could be the result of (circle all that apply and explain your reasoning for each suggestion (a) through (e)).
The thermal conductivity of reaction mixture decreases
Overall heat transfer coefficient increases
Overall heat transfer coefficient decreases
The coolant flow rate increases
The coolant flow rate decreases
Hint: Explore “Nonisothermal Reactor II” on the COMSOL LEP.
P1812_{D} Let’s continue Problem P1611_{D}. Where τ = 10 min and = 14 min^{2}
What would be the conversion for a secondorder reaction with kC_{A0} = 0.1 min^{–1} and C_{A0} = 1 mol/dm^{3} using the segregation model?
What would be the conversion for a secondorder reaction with kC_{A0} = 0.1 min^{–1} and C_{A0} = 1 mol/dm^{3} using the maximum mixedness model?
If the reactor is modeled as tanks in series, how many tanks are needed to represent this reactor? What is the conversion for a firstorder reaction with k = 0.1 min^{–1} ?
If the reactor is modeled by a dispersion model, what are the Péclet numbers for an open system and for a closed system? What is the conversion for a firstorder reaction with k = 0.1 min^{–1} for each case?
Use the dispersion model to estimate the conversion for a secondorder reaction with k = 0.1 dm^{3}/mol·s and C_{A0} = 1 mol/dm^{3}.
It is suspected that the reactor might be behaving as shown in Figure P1812_{D}, with perhaps (?) V_{1} = V_{2}. What is the “backflow” from the second to the first vessel, as a multiple of υ_{0}?
If the model above is correct, what would be the conversion for a secondorder reaction with k = 0.1 dm^{3}/mol=min if C_{A0} = 1.0 mol/dm^{3}?
Prepare a table comparing the conversion predicted by each of the models described above.
P1813_{B} A secondorder reaction is to be carried out in a real reactor that gives the following outlet concentration for a step input:
For 0 ≤ t < 10 min, then C_{T} = 10 (1 –e^{.1t})
For 10 min ≤ t, then C_{T} = 5 + 10 (1 – e^{.1t}
What model do you propose and what are your model parameters, α and β?
What conversion can be expected in the real reactor?
How would your model change and conversion change if your outlet tracer concentration was as follows?
For t ≤ 10 min, then C_{T} = 0
For t ≥ 10 min, then C_{T} = 5 + 10 (1 –e^{–0.2(t–10)}
υ_{0} = 1 dm^{3}/min, k = 0.1 dm^{3}/mol · min, C_{A0} = 1.25 mol/dm^{3}
P1814_{b} Suggest combinations of ideal reactors to model the real reactors given in Problem P162_{b}(b) for either E(θ), E(t), F(θ), F(t), or (1 – F(θ)).
P1815_{B} OEQ (Old Exam Question). The Fcurves for two tubular reactors are shown in Figure P1815_{B} for a closedclosed system.
Which curve has the higher Péclet number? Explain.
Which curve has the higher dispersion coefficient? Explain.
If this Fcurve is for the tanksinseries model applied to two different reactors, which curve has the largest number of TIS, (1) or (2)?
U of M, ChE528 MidTerm Exam
P1816_{C} Consider the following system in Figure P1816_{C} used to model a real reactor:
Describe how you would evaluate the parameters α and β.
Draw the F and Ecurves for this system of ideal reactors used to model a real reactor using β = 0.2 and α = 0.4. Identify the numerical values of the points on the Fcurve (e.g., t_{1}) as they relate to τ.
If the reaction A → B is second order with kC_{A0} = 0.5 min^{–1}, what is the conversion assuming the space time for the real reactor is 2 min?
U of M, ChE528 Final Exam
P1817_{B} OEQ (Old Exam Question). There is a 2m^{3} reactor in storage that is to be used to carry out the liquidphase secondorder reaction
A + B → C
A and B are to be fed in equimolar amounts at a volumetric rate of 1 m^{3}/min. The entering concentration of A is 2 molar, and the specific reaction rate is 1.5 m^{3}/kmol · min. A tracer experiment was carried out and reported in terms of F as a function of time in minutes as shown in Figure P1817_{B}.
Suggest a twoparameter model consistent with the data; evaluate the model parameters and the expected conversion.
U of M, ChE528 Final Exam
P1818_{b} OEQ (Old Exam Question). The following Ecurve shown in Figure P1818_{b} was obtained from a tracer test:
What is the mean residence time?
What is the Péclet number for a closedclosed system?
How many tanks in series are necessary to model this nonideal reactor?
U of M, Doctoral Qualifying Exam (DQE)
P1819_{B} OEQ (Old Exam Question). A firstorder reaction with k = 0.1 min^{–1} is to be carried out in the reactor whose RTD is shown in Figure 1819_{B}.
Fill in Table P1819_{B} with the conversion predicted by each type of model/reactor.
TABLE P1819_{B}. COMPARISONS OF CONVERSION PREDICTED BY VARIOUS MODELS
Ideal PFR 
Ideal CSTR 
Ideal LaminarFlow Reactor 
Segregation 
Maximum Mixedness 
Dispersion 
Tanks in Series 







P1820_{C} Consider a real tubular reactor in which dispersion is occurring.
For small deviations from plug flow, show that the conversion for a firstorder reaction is given approximately as
$\begin{array}{cc}X=1\text{exp}[tk+\frac{{\left(\tau k\right)}^{2}}{P{e}_{r}}]& \left(\text{P1820.1}\right)\end{array}$
Show that to achieve the same conversion, the relationship between the volume of a plugflow reactor, V_{P}_{}, and volume of a real reactor, V, in which dispersion occurs is
$\begin{array}{cc}\frac{{\text{V}}_{P}}{\text{}\text{V}}=1\frac{\left(\tau k\right)}{P{e}_{r}}=1\frac{k{D}_{e}}{{U}^{2}}& \left(\text{P1820.2}\right)\end{array}$
For a Péclet number of 0.1 based on the PFR length, how much bigger than a PFR must the real reactor be to achieve the 99% conversion predicted by the PFR?
For an nthorder reaction, the ratio of exit concentration for reactors of the same length has been suggested as
$\begin{array}{cc}\frac{{C}_{\text{A}}}{{C}_{{\text{A}}_{\text{plug}}}}=1+\frac{n}{P{e}_{r}}\left(\tau k{C}_{\text{A0}}^{n1}\right)\text{ln}\frac{{C}_{\text{A0}}}{{C}_{{\text{A}}_{\text{plug}}}}& \text{}\left(\text{P1820.3}\right)\end{array}$
What do you think of this suggestion?
What is the effect of dispersion on zeroorder reactions?
Excellent discussions of maximum mixedness can be found in
J. M. DOUGLAS, “The effect of mixing on reactor design,” AIChE Symp. Ser. 48, vol. 60, p. 1 (1964).
TH. N. ZWIETERING, Chem. Eng. Sci., 11, 1 (1959).
Modeling real reactors with a combination of ideal reactors is discussed together with axial dispersion in
O. LEVENSPIEL, Chemical Reaction Engineering, 3rd ed. New York: Wiley, 1999.
C. Y. WEN and L. T. FAN, Models for Flow Systems and Chemical Reactors. New York: Marcel Dekker, 1975.
Mixing and its effects on chemical reactor design have been receiving increasingly sophisticated treatment. See, for example:
K. B. BISCHOFF, “Mixing and contacting in chemical reactors,” Ind. Eng. Chem., 58 (11), 18 (1966).
E. B. NAUMAN, “Residence time distributions and micromixing,” Chem. Eng. Commun., 8, 53 (1981).
E. B. NAUMAN and B. A. BUFFHAM, Mixing in Continuous Flow Systems. New York: Wiley, 1983.
See also
M. DUDUKOVIC and R. FELDER, in CHEMI Modules on Chemical Reaction Engineering, vol. 4, ed. B. Crynes and H. S. Fogler. New York: AIChE, 1985.
Dispersion. A discussion of the boundary conditions for closedclosed, openopen, closedopen, and openclosed vessels can be found in
R. ARIS, Chem. Eng. Sci., 9, 266 (1959).
O. LEVENSPIEL and K. B. BISCHOFF, Adv. in Chem. Eng., 4, 95 (1963).
E. B. NAUMAN, Chem. Eng. Commun., 8, 53 (1981).
Now that you have finished this book, suggestions on what to do with the book can be posted on the kiosk in the town square in downtown Riça, Jofostan.