The breakfast of champions is not a cereal, it’s your opposition.
—Nick Steitz
The concentration in the internal surface of the pellet is less than that of the external surface.
In Chapter 14, we discussed reactions limited by mass transfer from the bulk fluid through a boundary layer to a surface where the reaction rapidly takes place. In Chapter 14, the reaction did not take place as the reactants diffused, however, in this chapter we consider diffusion with reaction. Here we will model Steps 2 and 6 in Figure 151 (which is the same as Figure 141), where the molecules are reacting as they diffuse.
For homogeneous systems, the mole balance on species A, Equation (141), for onedimensional diffusion at steady state is
$\begin{array}{cc}{\displaystyle \frac{\mathit{d}{W}_{\mathrm{A}z}}{\mathit{d}z}}+{r}_{\mathrm{A}}=0& \left(15\text{1}\right)\end{array}$
For diffusion through a stagnant film at dilute concentrations (i.e., y_{A} <<1), Equation (149) becomes
${W}_{\text{A}z}={D}_{\text{AB}}{\displaystyle \frac{d{C}_{\text{A}}}{dz}}$
Substituting in Equation (141) one obtains
$\begin{array}{c}\hline {D}_{\text{AB}}{\displaystyle \frac{{d}^{2}{C}_{\text{A}}}{{dz}^{2}}}+{r}_{\text{A}}=0\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}2\right)\end{array}$
Homogeneous diffusion and reaction
Understanding and modeling diffusion with chemical reaction is not only important for industrial catalysts but also has many other applications. These applications include medicine, cancer treatment using druglaced particulates, and, as shown in the Additional Material on the CRE Web site (http://www.umich.edu/~elements/6e/15chap/expanded.html), tissue engineering. In Problem P1518_{B}, we discuss the diffusion and reaction of oxygen in cartilage.
We will now discuss solid–gas catalytic reactions and diffusion limitation in catalyst pellets.
The following sections of this chapter will focus solely on the transport and reaction in heterogeneous systems with catalyst pellets. In a heterogeneous reaction sequence, mass transfer of reactants must first take place from the bulk fluid to the external surface of the pellet. The reactants then diffuse from the external surface into and through the pores within the pellet (C_{A}_{s} > C_{A}(r)), with reaction taking place only on the catalytic surface of the pores. A schematic representation of this twostep diffusion process is shown in Figures 106, 141, and 152.
In Chapter 14 we discussed external diffusion. In this section we will discuss internal diffusion and develop the internal effectiveness factor for spherical catalyst pellets. The development of models that treat individual pores and pellets of different shapes is undertaken in the problems at the end of this chapter. We will first look at the internal mass transfer resistance to either the reaction products or reactants that occurs between the external pellet surface and the interior of the pellet. To illustrate the salient principles of this model, we consider the irreversible isomerization
$\mathrm{A}\to \mathrm{B}$
that occurs on the surface of the pore walls within the spherical pellet of radius R.
In order to reach and be catalyzed on the inner surface of the pellet, the reactant A must diffuse from a higher reactant concentration at the pellet external surface into and through the pores of pellets which are at a lower concentration as shown in Figure 152.
The pores in the pellet are not straight and cylindrical; rather, they are a series of tortuous, interconnecting paths of pore bodies and pore throats with varying crosssectional areas. It would not be fruitful to describe diffusion within each and every one of the tortuous pathways individually; consequently, we shall define an effective diffusion coefficient so as to describe the average diffusion taking place at any interior position r in the pellet. We shall consider only radial variations in the concentration in a spherical catalyst pellet; the radial flux W_{A}_{r} will be based on the total area (voids and solid) normal to diffusion transport (i.e., 4πr^{2}) rather than void area alone. This basis for W_{A}_{r} is made possible by proper definition of the effective diffusivity D_{e}.
The effective diffusivity accounts for the fact that:
Not all of the area normal to the direction of the flux is available (i.e., the area occupied by solids) for the molecules to diffuse.
The paths are tortuous.
The pores are of varying crosssectional areas.
The effective diffusivity
An equation that relates the effective diffusivity D_{e} to either the bulk diffusivity D_{AB} or the Knudsen diffusivity D_{K} is
$\begin{array}{cc}{D}_{e}={\displaystyle \frac{{D}_{\text{AB}}{\mathit{\varphi}}_{p}{\sigma}_{c}}{\tilde{\tau}}}& \hspace{1em}\left(15\text{}3\right)\end{array}$
where
$\begin{array}{ccc}\tilde{\tau}& =& {\text{tortuosity}}^{1}={\displaystyle \frac{\text{Actualdistanceamoleculetravelsbetweentwopoints}}{\text{Shortestdistancebetweenthosetwopoints}}}\hfill \\ {\mathit{\varphi}}_{p}& =& \text{pelletporosity}={\displaystyle \frac{\text{Volumeofvoidspace}}{\text{Totalvolume}\left(\text{voidsandsolids}\right)}}\hfill \\ {\sigma}_{c}& =& \text{Constrictionfactor,seeFigure}15\text{}3\left(\text{a}\right)\hfill \end{array}$
^{1} Some investigators lump the constriction factor and tortuosity into one factor, called the tortuosity factor, and set it equal to $\tilde{\tau}/{\sigma}_{c}$. See C. N. Satterfield, Mass Transfer in Heterogeneous Catalysis, Cambridge, MA: MIT Press, 1970, pp. 33–47.
The constriction factor, σ_{c}, accounts for the variation in the crosssectional area that is normal to diffusion.^{2} It is a function of the ratio of maximum to minimum pore areas (Figure 153(a)). When the two areas, A_{1} and A_{2}, are equal, the constriction factor is unity, and when β = 10, the constriction factor is approximately 0.5.
^{2} See E. E. Petersen, Chemical Reaction Analysis, Upper Saddle River, NJ: Prentice Hall, 1965, Chap. 3; C. N. Satterfield and T. K. Sherwood, The Role of Diffusion in Catalysis, Reading, MA: AddisonWesley, 1963, Chap. 1.
Using typical values of D_{AB}, ϕ_{p}, σ_{c}, and τ, estimate the effective diffusivity, D_{e}.
Solution
First, calculate the tortuosity for the hypothetical pore of length, L (Figure 153(b)), from the definition of $\tilde{\tau}$.
The shortest distance between points A and B for the idealized pore shown in Figure 153(b) is $\sqrt{2}L$. The actual distance the molecule travels from A to B is 2L.
Although this value is reasonable for $\tilde{\tau}$, values for $\tilde{\tau}=6\text{\u2212}10$ are not uncommon. Typical values of the constriction factor, the tortuosity, and the pellet porosity are, respectively, σ_{c} = 0.8, $\tilde{\tau}=3.0$, and ϕ_{p} = 0.40. A typical value of the gasphase diffusivity is D_{AB} = 10^{−6} m^{2}/s.
Using these values in Equation (153)
$\begin{array}{cc}{D}_{e}={\displaystyle \frac{{\varphi}_{p}{\sigma}_{c}}{\tilde{\tau}}}{D}_{\text{AB}}& \hspace{1em}\left(15\text{}3\right)\end{array}$
${D}_{e}={\displaystyle \frac{\left(0.4\right)0.8}{\left(3\right)}}{D}_{\text{AB}}=0.106{D}_{\text{AB}}$
therefore
D_{e} = 0.1 • 10^{−6} m^{2}/s = 10^{−7} m^{2}/s
Analysis: The purpose of this example was to describe tortuosity and constriction factor in order to help the reader understand how these parameters decrease the effective diffusivity, D_{e}. We also see that a representative value of the effective diffusivity in the porous pellet is 10% of the gasphase diffusivity.
We now perform a steadystate mole balance on species A as it enters, leaves, and reacts in a spherical shell of inner radius r and outer radius r + Δr of the pellet (Figure 154). Note that even though A is diffusing inward toward the center of the pellet, the convention of our shell balance dictates that the flux be in the direction of increasing r. We choose the flux of A to be positive in the direction of increasing r (i.e., the outward direction). Because A is actually diffusing inward, the flux of A will have some negative numerical value, such as (–10 mol/m^{2}·s), indicating that the flux is actually in the direction of decreasing r.
First, we will derive the concentration profile of reactant A in the pellet.
We now proceed to perform our shell balance on A. The area that appears in the balance equation is the total area (voids and solids) normal to the direction of the molar flux shown by the arrows in Figure 153.
$\begin{array}{ccccc}\hfill \text{RateofAinat}\text{}\text{}r& =\hfill & {\text{W}}_{\text{A}r}\cdot \text{Area}\hfill & =\hfill & {\text{W}}_{\text{A}r}\times 4\pi {r}^{2}{}_{r}\hfill \\ \hfill \text{Rate of}\text{A out at}(r+\mathrm{\Delta}r)& =\hfill & {\text{W}}_{\text{A}r}\cdot \text{Area}\hfill & =\hfill & {\text{W}}_{\text{A}r}\times 4\pi {r}^{2}{}_{r+\mathrm{\Delta}r}\hfill \\ \text{}\hfill & \text{}\hfill & \text{}\hfill & \text{}& \text{}\end{array}$
$\begin{array}{ccccccc}\left[\begin{array}{c}\text{Rateof}\\ \text{generation}\\ \text{ofAwithina}\\ \text{shellofthickness}\\ \mathrm{\Delta}r\end{array}\right]& =& \left[{\displaystyle \frac{\text{Rateofreaction}}{\text{Massofcatalyst}}}\right]& \times & \left[{\displaystyle \frac{\text{Masscatalyst}}{\text{Volume}}}\right]& \times & \left[\text{Volumeofshell}\right]\\ & =& {r}_{\text{A}}^{\prime}& \times & {\rho}_{c}& \times & 4{\pi r}_{\text{m}}^{2}\mathrm{\Delta}r\\ & & & & & & \hspace{1em}\left(15\text{}4\right)\end{array}$
where r_{m} is some mean radius between r and r + Δr that is used to approximate the volume ΔV of the shell, and ρ_{c} is the density of the pellet.
Mole balance for diffusion and reaction inside the catalyst pellet
Mole balance
The mole balance over the shell thickness Δr is
$\begin{array}{cc}\begin{array}{ccccccc}\left(\mathbf{\text{lnat}}\mathit{r}\right)& & (\mathbf{\text{Outat}}\mathit{r}\mathbf{+}\mathbf{\Delta}\mathit{r})& +& \left(\mathbf{\text{Generationwithin}}\mathbf{\Delta}\mathit{r}\right)& =& 0\\ ({W}_{\text{Ar}}\times 4\pi {r}^{2}{}_{r})& & ({W}_{\text{Ar}}\times 4\pi {r}^{2}{}_{r+\mathrm{\Delta}r})& +& ({r}_{\text{A}}^{\prime}{\rho}_{c}\times \pi {r}_{\text{m}}^{2}\mathrm{\Delta}r)& =& 0\end{array}& \left(15\text{}5\right)\end{array}$
After dividing by (–4πΔr) and taking the limit as Δr → 0, we obtain the following differential equation
$\begin{array}{c}\hline {\displaystyle \frac{d\left({W}_{\text{A}r}{r}^{2}\right)}{dr}}{r}_{\text{A}}^{\prime}{\rho}_{c}{r}^{2}=0\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}6\right)\end{array}$
Because 1 mol of A reacts under conditions of constant temperature and pressure to form 1 mol of B, we have Equimolar Counter Diffusion (EMCD) at constant total molar concentration (Section 14.2.1) and, therefore
The flux equation
where C_{A} is the number of moles of A per dm^{3} of openpore volume (i.e., volume of gas) as opposed to (mol/vol of gas and solids). In systems where we do not have EMCD in catalyst pores, it may still be possible to use Equation (157) if the reactant gases are present in dilute concentrations.
$\begin{array}{cc}{W}_{\text{Ar}}={D}_{e}{\displaystyle \frac{d{C}_{\text{A}}}{dr}}& \hspace{1em}\left(15\text{}7\right)\end{array}$
After substituting Equation (157) into Equation (156), we arrive at the following differential equation describing diffusion with reaction in a catalyst pellet
$\begin{array}{c}\hline {\displaystyle \frac{d[{D}_{e}(d{C}_{\text{A}}/dr){r}^{2}]}{dr}}{r}^{2}{p}_{c}{r}_{\text{A}}^{\prime}=0\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}8\right)\end{array}$
We now need to incorporate the rate law. In the past, we have based the rate of reaction in terms of either per unit volume
Inside the Pellet
$\begin{array}{ccc}{r}_{\text{A}}^{\prime}& =& {S}_{a}\left({r}_{\text{A}}^{\u2033}\right)\hfill \\ {r}_{\text{A}}& =& {\rho}_{c}\left({r}_{\text{A}}^{\prime}\right)\hfill \\ {r}_{\text{A}}& =& {\rho}_{c}{S}_{a}\left({r}_{\text{A}}^{\u2033}\right)\hfill \end{array}$
or per unit mass of catalyst
When we study reactions on the internal surface area of catalysts, the rate of reaction and rate law are often based on per unit surface area
As a result, the surface area of the catalyst per unit mass of catalyst
S_{a} [=] (m^{2}/gcat)
is an important property of the catalyst. The rate of reaction per unit mass of catalyst, ${r}_{\text{A}}^{\prime},$ and the rate of reaction per unit surface area of catalyst are related through the equation
S_{a}: 10 grams of catalyst may cover as much surface area as a football field.
A typical value of S_{a} might be 150 m^{2}/g of catalyst.
The rate law
As mentioned previously, at high temperatures, the denominator of the catalytic rate law often approaches 1 as discussed in Section 10.3.7. Consequently, for the moment, it is reasonable to assume that the surface reaction is of nth order in the gasphase concentration of A within the pellet
$\begin{array}{cc}{r}_{\text{A}}^{\u2033}={k}_{n}^{\u2033}{C}_{\text{A}}^{n}& \hspace{1em}\left(15\text{}9\right)\end{array}$
where the units of the rate constants for r_{A}, ${r}_{\text{A}}^{\prime}$, and ${r}_{\text{A}}^{\u2033}$ are
Similarly,
For a firstorder catalytic reactor
per unit pellet surface area:  ${k}_{1}^{\u2033}=\left[\text{m/s}\right]$ 
per unit mass of single pellet catalyst:  ${k}_{1}^{\prime}={k}_{1}^{\u2033}{S}_{a}=[{\text{m}}^{3}/\text{kg}\cdot \text{s}]$ 
per unit single pellet volume:  ${k}_{1}={k}_{1}^{\u2033}{S}_{a}{\rho}_{c}=\left[{\text{s}}^{1}\right]$ 
Substituting the ratelaw equation (159) into Equation (158) gives
$\begin{array}{cc}{\displaystyle \frac{d\left[{r}^{2}({D}_{e}d{C}_{\text{A}}/dr)\right]}{dr}}+{r}^{2}\stackrel{{k}_{n}}{\overbrace{{k}_{n}^{\u2033}{S}_{a}{\rho}_{c}}}{C}_{\text{A}}^{n}=0& \hspace{1em}\left(15\text{}10\right)\end{array}$
Letting k_{n} represent the terms under the bracket, differentiating the first term and dividing through by −r^{2}De, Equation (1510) becomes
$\begin{array}{c}\hline {\displaystyle \frac{{d}^{2}{C}_{\text{A}}}{d{r}^{2}}}+{\displaystyle \frac{2}{r}}\left({\displaystyle \frac{d{C}_{\text{A}}}{dr}}\right){\displaystyle \frac{{k}_{n}}{{D}_{e}}}{C}_{\text{A}}^{n}=0\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}11\right)\end{array}$
Differential equation and boundary conditions describing diffusion and reaction in a catalyst pellet
The boundary conditions are:
The concentration remains finite at the center of the pellet
At the external surface of the catalyst pellet, the concentration is C_{A}_{s}
As engineers, we often put our governing equations in dimensionless form as it makes calculations much easier when we change numbers and units. Consequently, we now introduce dimensionless variables ψ and λ so that we may arrive at a parameter that is frequently discussed in catalytic reactions, the Thiele modulus. Let
$\begin{array}{cc}\mathit{\psi}={\displaystyle \frac{{C}_{\text{A}}}{{C}_{\text{A}s}}}& \hspace{1em}\left(15\text{}12\right)\end{array}$
$\begin{array}{cc}\lambda ={\displaystyle \frac{r}{R}}& \hspace{1em}\left(15\text{}13\right)\end{array}$
With the transformation of variables, the boundary condition
becomes
and the boundary condition
becomes
We now rewrite the differential equation for the molar flux in terms of our dimensionless variables. Starting with
$\begin{array}{cc}{W}_{\text{A}r}={D}_{e}{\displaystyle \frac{d{C}_{\text{A}}}{dr}}& \hspace{1em}\left(15\text{}7\right)\end{array}$
we use the chain rule to write
$\begin{array}{cc}{\displaystyle \frac{d{C}_{\text{A}}}{dr}}=\left({\displaystyle \frac{d{C}_{\text{A}}}{d\lambda}}\right){\displaystyle \frac{d\lambda}{dr}}={\displaystyle \frac{d\psi}{d\lambda}}\left({\displaystyle \frac{d{C}_{\text{A}}}{d\psi}}\right){\displaystyle \frac{d\lambda}{dr}}& \hspace{1em}\left(15\text{}14\right)\end{array}$
Then differentiate Equation (1512) with respect to y and Equation (1513) with respect to r, and substitute the resulting expressions
into Equation (1514) to obtain
$\begin{array}{cc}\begin{array}{c}{\displaystyle \frac{d{C}_{\text{A}}}{dr}}={\displaystyle \frac{d\psi}{d\lambda}}\end{array}{\displaystyle \frac{{C}_{\text{A}s}}{R}}& \hspace{1em}\left(15\text{}15\right)\end{array}$
The flux of A in terms of the dimensionless variables, ψ and λ, is
The total rate of consumption of A inside the pellet, M_{A} (mol/s)
${W}_{\text{A}r}={D}_{e}\begin{array}{cc}\begin{array}{c}{\displaystyle \frac{d{C}_{\text{A}}}{dr}}={\displaystyle \frac{{D}_{e}{C}_{\text{A}s}}{R}}\left({\displaystyle \frac{d\psi}{d\lambda}}\right)\end{array}& \hspace{1em}\left(15\text{}16\right)\end{array}$
At steady state, the net flow of species A that enters into the pellet at the external pellet surface reacts completely within the pellet. The overall rate of reaction is therefore equal to the total molar flow of A into the catalyst pellet. The overall rate of reaction, M_{A} (mol/s), can be obtained by multiplying the molar flux into the pellet at the outer surface by the external surface area of the pellet, 4πR^{2}
All the reactant that diffuses into the pellet is consumed (a black hole?).
$\begin{array}{cc}{M}_{\text{A}}=4\pi {R}^{2}{W}_{\text{A}r}{}_{r=R}=+4\pi {R}^{2}{D}_{e}{\displaystyle \frac{d{C}_{\text{A}}}{dr}}{}_{r=R}=4\pi R{D}_{e}{C}_{\text{A}s}{\displaystyle \frac{d\psi}{d\lambda}}{}_{\lambda =1}& \hspace{1em}\left(15\text{}17\right)\end{array}$
Consequently, to determine the overall rate of reaction, which is given by Equation (1517), we first solve Equation (1511) for C_{A}, differentiate C_{A} with respect to r, and then substitute the resulting expression into Equation (1517).
Differentiating the concentration gradient, Equation (1515), yields
$\begin{array}{cc}{\displaystyle \frac{{d}^{2}{C}_{\text{A}}}{{dr}^{2}}}={\displaystyle \frac{d}{dr}}\left({\displaystyle \frac{d{C}_{\text{A}}}{dr}}\right)={\displaystyle \frac{d}{d\lambda}}\left({\displaystyle \frac{d\psi}{d\lambda}}{\displaystyle \frac{{C}_{\text{A}s}}{R}}\right){\displaystyle \frac{d\lambda}{dr}}={\displaystyle \frac{{d}^{2}\psi}{d{\lambda}^{2}}}\left({\displaystyle \frac{{C}_{\text{A}s}}{R}}\right)& \hspace{1em}\left(15\text{}18\right)\end{array}$
After dividing by C_{A}_{s} /R^{2}, the dimensionless form of Equation (1511) is written as
Then
Dimensionless form of equations describing diffusion and reaction
$\begin{array}{c}\hline \text{}{\displaystyle \frac{{d}^{2}\psi}{d{\lambda}^{2}}}+{\displaystyle \frac{2}{\lambda}}\left({\displaystyle \frac{d\psi}{d\lambda}}\right){\varphi}_{n}^{2}{\psi}^{n}=0\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}19\right)\end{array}$
where
$\begin{array}{c}\hline {\varphi}_{n}^{2}={\displaystyle \frac{{k}_{n}{R}^{2}{C}_{\text{A}s}^{n1}}{{D}_{e}}}\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}20\right)\end{array}$
The Thiele Modulus
The square root of the coefficient of ψ^{n} in Equation 1519 (i.e., ϕ_{n}) is called the Thiele modulus (pronounced th ē lē). The Thiele modulus, ϕ_{n}, will always contain a subscript (e.g., n), which refers to the reaction order and distinguishes this symbol from the symbol for porosity, ϕ, used in the Ergun pressure drop equation and defined in Chapter 5, which has no subscript. The quantity ${\varphi}_{n}^{2}$ is a measure of the ratio of “a” surface reaction rate to “a” rate of diffusion through the catalyst pellet
$\begin{array}{c}\hline {\varphi}_{n}^{2}={\displaystyle \frac{{k}_{n}{R}^{2}{C}_{\text{A}s}^{n1}}{{D}_{e}}}={\displaystyle \frac{{k}_{n}R{C}_{\text{A}s}^{n}}{{D}_{e}[({C}_{\text{A}s}0)/R]}}={\displaystyle \frac{\u201ca\u201d\text{}\text{surfacereactionrate}}{\u201ca\u201d\text{}\text{diffusionrate}}}\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}20\right)\end{array}$
Limiting conditions:
When the Thiele modulus is large, internal diffusion usually limits the overall rate of reaction; when ϕ_{n} is small, the surface reaction is usually ratelimiting. If, for the firstorder reaction
A → B
the reaction were surface reaction rate limited with respect to the adsorption of A and desorption of B, and if A and B were weakly adsorbed (i.e., low surface coverage), we can write the apparent firstorder reaction rate law per unit volume of pellet as
–r_{A} = k_{1}C_{A}
where k_{1} is the rate constant for a single catalyst pellet.
Recalling k_{1} = S_{a}ρ_{c}k″ we could also write the rate in terms of pellet catalytic surface area (mol/m^{2} • s)
$\begin{array}{cc}{r}_{\text{A}}^{\u2033}={k}_{1}^{\u2033}{C}_{\text{A}}\left({\displaystyle \frac{\text{mol}}{{\text{m}}^{2}\cdot \text{s}}}\right)& \hspace{1em}\left(15\text{}21\right)\end{array}$
The units of ${k}_{1}^{\u2033}$ are m^{3}/m^{2}s (= m/s).
For a firstorder reaction, Equation (1519) becomes
$\begin{array}{c}\hline \text{}{\displaystyle \frac{{d}^{2}\psi}{d{\lambda}^{2}}}+{\displaystyle \frac{2}{\lambda}}{\displaystyle \frac{d\psi}{d\lambda}}{\varphi}_{1}^{2}\psi =0\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}22\right)\end{array}$
the Thiele modulus for this firstorder reaction is
$\begin{array}{c}\hline {\varphi}_{1}=R\sqrt{{\displaystyle \frac{{k}_{1}^{\u2033}{\rho}_{c}{S}_{a}}{{D}_{e}}}}=R\sqrt{{\displaystyle \frac{{k}_{1}}{{D}_{e}}}}\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}23\right)\end{array}$
where
Always check to make sure the units are consistent.
The boundary conditions are
$\begin{array}{cc}\begin{array}{c}\hline \begin{array}{cc}\text{B}\mathrm{.}\text{C}\mathrm{.}\text{}1:\psi =1\hfill & \text{at}\lambda =1\\ \text{B.}\text{C}\mathrm{.}\text{}2:\psi \text{isfinite}\hfill & \text{at}\lambda =0\end{array}\\ \hline\end{array}& \begin{array}{c}\hspace{1em}\left(15\text{}24\right)\\ \hspace{1em}\left(15\text{}25\right)\end{array}\end{array}$
Differential equation (1522) is readily solved with the aid of the transformation y = ψλ:
With these transformations, Equation (1522) reduces to
$\begin{array}{cc}{\displaystyle \frac{{d}^{2}y}{d{\lambda}^{2}}}{\varphi}_{1}^{2}y=0& \hspace{1em}\left(15\text{}26\right)\end{array}$
This differential equation has the following solution (see Appendix A.3):
$y={\text{A}}_{1}\text{}\text{cosh}\text{}{\varphi}_{1}\lambda +{\text{B}}_{1}\text{sinh}\text{}{\varphi}_{1}\lambda $
In terms of ψ,
$\psi ={\displaystyle \frac{{A}_{1}}{\lambda}}\text{}\text{cosh}\text{}{\varphi}_{1}\lambda +{\displaystyle \frac{{B}_{1}}{\lambda}}\text{}\text{sinh}\text{}{\varphi}_{1}\lambda $
The arbitrary constants A_{1} and B_{1} can easily be evaluated with the aid of the boundary conditions. At λ = 0, cosh ϕ_{1}λ → 1, (1/λ) → ∞, and sinh ϕ_{1}λ → 0. The second boundary condition requires ψ to be finite at the center (i.e., λ = 0), therefore A_{1} must be zero.
The constant B_{1} is evaluated from B.C. 1 (i.e., ψ = 1, λ = 1) and the dimensionless concentration profile is
Dimensionless concentration profile
$\begin{array}{c}\hline \psi ={\displaystyle \frac{{C}_{\text{A}}}{{C}_{\text{A}s}}}={\displaystyle \frac{1}{\lambda}}\left({\displaystyle \frac{\text{sinh}{\varphi}_{1}\lambda}{\text{sinh}{\varphi}_{1}}}\right)\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}27\right)\end{array}$
Figure 155 shows the concentration profile for three different values of the Thiele modulus, ϕ_{1}. Small values of the Thiele modulus indicate surface reaction controls and a significant amount of the reactant diffuses well into the pellet interior without reacting. As a result, the concentration profile is very shallow, with the concentration at the center of the pellet being close to that at the external surface. That is, virtually the entire internal surface is accessible to the reactant concentration C_{A}_{s}. Large values of the Thiele modulus indicate that the surface reaction is rapid and that the reactant is consumed very close to the external pellet surface and very little penetrates into the interior of the pellet. Consequently, if the porous pellet is to be plated with a precious metal catalyst (e.g., Pt), it should only be plated in the immediate vicinity of the external surface when large values of ϕ_{n} characterize the diffusion and reaction. That is, it would be a waste of the precious metal to plate the entire insides of the pellet when internal diffusion is limiting because the reacting gases are consumed near the outer pellet surface and never reach the metal catalyst near the pellet’s center. Consequently, the reacting gases would never contact the center portion of the pellet.
Dimensionless concentration profile
#WasteOfMoney
For large values of the Thiele modulus, internal diffusion limits the rate of reaction.
In Figure 155 we saw that the concentration varied with the pellet radius. Consequently, for all but zeroorder reactions, the rate will also vary throughout the pellet. In order to account for this variation in rates, we introduce the internal effectiveness factor.
The magnitude of the effectiveness factor (ranging from 0 to 1) indicates the relative importance of diffusion and reaction limitations. The internal effectiveness factor is defined as
η = is a measure of how far the reactant diffuses into the pellet before reacting.
$\begin{array}{c}\hline \eta ={\displaystyle \frac{\text{Actualoverallrateofreaction}}{\begin{array}{c}\text{Rateofreactionthatwouldresultifentireinteriorsurfacewere}\\ \text{exposedtotheexternalpelletsurfaceconditions}{C}_{\text{A}s},{T}_{s}\end{array}}}\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}28\right)\end{array}$
The overall rate, ${r}_{\text{A}}^{\prime}$, is also referred to as the observed rate of reaction [–r_{A}(obs)]. In terms of symbols, the effectiveness factor is
To derive the effectiveness factor for a firstorder reaction, it is easiest to work in reaction rates of moles per unit time, M_{A}, rather than in moles per unit time per volume of catalyst (i.e., −r_{A}). Consequently, we multiply the numerator and denominator of Equation (1528) by the volume, V, of the catalyst pellet.
First, we shall consider the denominator, M_{A}_{s}, the molar rate to the surface. If the entire surface were exposed to the concentration at the external surface of the pellet, C_{A}_{s}, the rate for a firstorder reaction would be
M_{As} = (Rate at external surface per unit volume) × (Volume of catalyst pellet)
$\begin{array}{cc}{M}_{\text{A}s}={r}_{\text{A}s}\times \left({\displaystyle \frac{4}{3}}\pi {R}^{3}\right)={k}_{1}{C}_{\text{A}s}\left({\displaystyle \frac{4}{3}}\pi {R}^{3}\right)& \hspace{1em}\left(15\text{}29\right)\end{array}$
The subscript s indicates that the rate –r_{A}_{s} is evaluated at the conditions (e.g., concentration, temperature) present at the external surface of the pellet (i.e., λ = 1).
The actual rate of reaction is the rate at which the reactant diffuses into the pellet at the outer surface; that is, all of A that diffuses into the pellet at the outer surface reacts and no A diffuses back out. (It behaves as a “black hole.”) We recall Equation (1517) on page 799 for the actual rate of reaction
The actual rate of reaction
$\begin{array}{cc}{M}_{\text{A}}=4\pi R{D}_{e}{C}_{\text{A}s}{\displaystyle \frac{d\psi}{d\lambda}}{}_{\lambda =1}& \hspace{1em}\left(15\text{}17\right)\end{array}$
Differentiating Equation (1527) and then evaluating the result at λ = 1 yields
$\begin{array}{cc}{\displaystyle \frac{d\psi}{d\lambda}}{}_{\lambda =1}=({\displaystyle \frac{{\varphi}_{1}\text{cosh}\lambda {\varphi}_{1}}{\lambda \text{sinh}{\varphi}_{1}}}{\displaystyle \frac{1}{{\lambda}^{2}}}{\displaystyle \frac{\text{sinh}\lambda {\varphi}_{1}}{\text{sinh}{\varphi}_{1}}}{)}_{\lambda =1}=({\varphi}_{1}\text{coth}{\varphi}_{1}1)& \hspace{1em}\left(15\text{}30\right)\end{array}$
Substituting Equation (1530) into (1517) gives us
$\begin{array}{cc}{M}_{\text{A}}=4\pi R{D}_{e}{C}_{\text{A}s}({\varphi}_{1}\text{coth}{\varphi}_{1}1)& \hspace{1em}\left(15\text{}31\right)\end{array}$
We now substitute Equations (1529) and (1531) into Equation (1528) to obtain an expression for the effectiveness factor
Internal effectiveness factor for a firstorder reaction in a spherical catalyst pellet
$\begin{array}{c}\hline \eta ={\displaystyle \frac{3}{{\varphi}_{1}^{2}}}({\varphi}_{1}\text{coth}{\varphi}_{1}1)\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}32\right)\end{array}$
A plot of the internal effectiveness factor as a function of the Thiele modulus is shown in Figure 156. Figure 156(a) shows η as a function of the Thiele modulus, ϕ_{s}, for a spherical catalyst pellet for reactions of zero, first, and second order. Figure 156(b) corresponds to a firstorder reaction occurring in three differently shaped pellets of volume V_{p} and external surface area A_{p}, and where the Thiele modulus for a firstorder reaction, ϕ_{1}, is defined differently for each shape. When volume change accompanies a reaction (i.e., ∊ ≠ 0), the corrections shown in Figure 157 apply to the effectiveness factor for a firstorder reaction.
The Thiele Modulus is a very important parameter that helps us identify when either diffusion is rate limiting (Large ϕ_{n}) or reaction is rate limiting (Small ϕ_{n}). We observe that as the particle diameter becomes very small, ϕ_{n} decreases, so that the internal effectiveness factor approaches 1 and the reaction is surface reactionlimited. On the other hand, when the Thiele modulus ϕ_{n} is large (e.g., ~30), the internal effectiveness factor η is small (e.g., ~0.1), and the reaction is diffusionlimited within the pellet. Consequently, factors influencing the rate of external mass transport such as fluid velocity will have a negligible effect on the overall reaction rate when the reaction is either internal surface reaction rate limited or internal diffusion limited. For large values of the Thiele modulus, one can expand coth =_{1} in Equation (1532) in a Taylor series and show that the effectiveness factor can be written as
If ϕ > 2
then $\eta \approx {\displaystyle \frac{3}{{\varphi}_{1}^{2}}}[{\varphi}_{1}1]$
If ϕ > 20
then $\eta \approx {\displaystyle \frac{3}{{\varphi}_{1}}}$
$\begin{array}{cc}\eta \backsimeq {\displaystyle \frac{3}{{\varphi}_{1}}}={\displaystyle \frac{3}{R}}\sqrt{{\displaystyle \frac{{D}_{e}}{{k}_{1}}}}& \hspace{1em}\left(15\text{}33\right)\end{array}$
To express the overall rate of reaction in terms of the Thiele modulus, we rearrange Equation (1528) and use the rate law for a firstorder reaction in Equation (1529)
$\begin{array}{ccc}{r}_{\text{A}}& =& \left({\displaystyle \frac{\text{Actualreactionrate}}{\text{Reactionrateat}{C}_{\text{A}s}}}\right)\times \left(\text{Reactionrateat}{C}_{\text{A}s}\right)\hfill \\ & =& \eta \left({r}_{\text{A}s}\right)\hfill \end{array}$
For a firstorder reaction
$\begin{array}{cc}{r}_{\text{A}}=\eta \left({k}_{1}{C}_{\text{A}s}\right)& \hspace{1em}\left(15\text{}34\right)\end{array}$
Limited regimes
Combining Equations (1533) and (1534), the overall rate of reaction for a firstorder, internaldiffusionlimited reaction is
Note that for catalytic packed beds the rate varies inversely with the particle diameter of the catalyst.
When there is volume change, ε ≠ 0, we use a correction factor to account for this change. The correction is obtained from a plot of the ratio of effective factors
as a function of ε for various values of the Thiele modulus. This plot is given in Figure 157.
Correction for volume change with reaction (i.e., ε ≠ 0)
For example, if the Thiele modulus were 10 for the gasphase reaction A → 2B with (ε = 1), then the effectiveness factor with volume change would be η′ = 0.8 η.
How can the rate of reaction be increased?
To increase the reaction rate, ${r}_{\text{A}}^{\prime}$, for internaldiffusionlimited reactions, we can (1) decrease the radius R (make pellets smaller); (2) increase the temperature; (3) increase the concentration; and (4) increase the internal surface area. For reactions of order n, we have from Equation (1520)
$\begin{array}{cc}{\text{\hspace{0.17em}}\varphi}_{n}^{2}={\displaystyle \frac{{k}_{n}^{\u2033}{S}_{a}{\rho}_{c}{R}^{2}{C}_{\text{A}s}^{n1}}{{D}_{e}}}={\displaystyle \frac{{k}_{n}{R}^{2}{C}_{\text{A}s}^{n1}}{{D}_{e}}}& \hspace{1em}\left(15\text{}20\right)\end{array}$
For large values of the Thiele modulus, the effectiveness factor is
$\begin{array}{c}\hline \eta ={\left({\displaystyle \frac{2}{n+1}}\right)}^{1/2}{\displaystyle \frac{3}{{\varphi}_{n}}}={\left({\displaystyle \frac{2}{n+1}}\right)}^{1/2}{\displaystyle \frac{3}{R}}\sqrt{{\displaystyle \frac{{D}_{e}}{{k}_{n}}}}{C}_{\text{A}s}^{(1n)/2}\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}35\right)\end{array}$
Consequently, for reaction orders greater than 1, the effectiveness factor decreases with increasing concentration at the external pellet surface. We will use this equation when we discuss Falsified Kinetics.
The Weisz–Prater criterion uses measured values of the rate of reaction, ${r}_{\text{A}}^{\prime}$ (obs), to determine whether internal diffusion is limiting the reaction. This criterion can be developed intuitively for a firstorder reaction by first rearranging Equation (1532) in the form
$\begin{array}{c}\hline \eta {\varphi}_{1}^{2}=3({\varphi}_{1}\text{coth}{\varphi}_{1}1)\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}36\right)\end{array}$
The lefthand side of Equation (1536) is the Weisz–Prater parameter, WP:
Showing where the Weisz–Prater comes from
$\begin{array}{cccc}\text{WP}& =& \eta \times {\varphi}_{1}^{2}\hfill & \hspace{1em}\left(15\text{}37\right)\\ \text{WP}& =& {\displaystyle \frac{\text{Observed}\left(\text{actual}\right)\text{reactionrate}}{\text{Reactionrateevaluatedat}{C}_{\text{A}s}}}\times {\displaystyle \frac{\text{Reactionrateevaluatedat}{C}_{\text{A}s}}{\u201ca\u201d\text{\hspace{0.17em}}\text{diffusionrate}}}\hfill & \\ \text{WP}& =& {\displaystyle \frac{\text{Actualreactionrate}}{\u201ca\u201d\text{\hspace{0.17em}}\text{diffusionrate}}}\hfill & \end{array}$
Explore WP using Wolfram LEP P153 on the Web site
Substituting for
in Equation (1559) we obtain
$\begin{array}{cc}\text{WP}={\displaystyle \frac{{r}_{\text{A}}^{\prime}\left(\text{obs}\right)}{{r}_{\text{A}s}^{\prime}}}\text{}\left({\displaystyle \frac{{r}_{\text{A}}^{\prime}{\rho}_{c}{R}^{2}}{{D}_{e}{C}_{\text{A}s}}}\right)& \hspace{1em}\left(15\text{}38\right)\end{array}$
$\begin{array}{c}\hline \text{WP}={\eta \varphi}_{1}^{2}={\displaystyle \frac{{r}_{\text{A}}^{\prime}\left(\text{obs}\right)\text{}{\rho}_{c}{R}^{2}}{{D}_{e}{C}_{\text{A}s}}}\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(15\text{}39\right)\end{array}$
Are there any internal diffusion limitations indicated from the Weisz–Prater criterion?
All the terms in the rhs of Equation (1539) are either measured or known. Consequently, we can calculate WP to learn whether there are any diffusion limitations.
If
No internal diffusion limitations
there are no diffusion limitations and consequently no concentration gradient exists within the pellet.
However, if
Severe internal diffusion limitations
internal diffusion limits the reaction severely. Ouch!
The firstorder reaction
was carried out over two differentsized pellets. The pellets were contained in a spinning basket reactor that was operated at sufficiently high rotation speeds that external mass transfer resistance was negligible. The results of two experimental runs made under identical conditions are given in Table E152.1.
These two experiments yield an enormous amount of information.
Estimate the Thiele modulus and effectiveness factor for each pellet.
How small should the pellets be made to virtually eliminate all internal diffusion resistance, for example, η = 0.95?
TABLE E152.1 DATA FROM A SPINNING BASKET REACTOR
Measured Rate (obs) (mol/gcat · s) × 10^{5} 
Pellet Radius (m) 

Run 1 
3.0 
0.01 
Run 2 
15.0 
0.001 
Solution
(a) Combining Equations (1536) and (1539), we obtain
$\begin{array}{cc}\text{WP}={\displaystyle \frac{{r}_{\text{A}}^{\prime}\left(\text{Obs}\right){R}^{2}{\rho}_{c}}{{D}_{e}{C}_{\text{A}s}}}={\eta \varphi}_{1}^{2}=3({\varphi}_{1}\text{coth}{\varphi}_{1}1)& \left(\text{E}15\text{}2.1\right)\end{array}$
Letting the subscripts 1 and 2 refer to runs 1 and 2, we apply Equation (E152.1) to runs 1 and 2 and then take the ratio to obtain
$\begin{array}{c}\hline {\displaystyle \frac{{r}_{{\text{A}}_{2}}^{\prime}{R}_{2}^{2}}{{r}_{{\text{A}}_{1}}^{\prime}{R}_{1}^{2}}}={\displaystyle \frac{{\varphi}_{12}\text{coth}{\varphi}_{12}1}{{\varphi}_{11}\text{coth}{\varphi}_{11}1}}\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(\mathrm{E15}\text{}2.2\right)\end{array}$
The terms ρ_{c}, D_{e}, and C_{A}_{s} cancel because the runs were carried out under identical conditions. The Thiele modulus is
$\begin{array}{cc}{\varphi}_{1}=R\sqrt{{\displaystyle \frac{{r}_{\text{A}s}^{\prime}{\rho}_{c}}{{D}_{e}{C}_{\text{A}s}}}}& \left(\text{E152.3}\right)\end{array}$
Taking the ratio of the Thiele moduli for runs 1 and 2, we obtain
$\begin{array}{c}\hline {\displaystyle \frac{{\varphi}_{11}}{{\Theta}_{12}}}={\displaystyle \frac{{R}_{1}}{{R}_{2}}}\\ \hline\end{array}\begin{array}{c}\hspace{1em}\left(\text{E152.4}\right)\end{array}$
or
$\begin{array}{cc}{\varphi}_{11}={\displaystyle \frac{{R}_{1}}{{R}_{2}}}\text{}{\varphi}_{12}={\displaystyle \frac{0.01\text{m}}{0.001\text{m}}}\text{}{\varphi}_{12}=10{\varphi}_{12}& \text{(E152.5)}\end{array}$
Substituting for ϕ_{11} in Equation (E152.2) above and evaluating and R for runs 1 and 2 gives us
$\begin{array}{cc}\left({\displaystyle \frac{15\times {10}^{5}}{3\times {10}^{5}}}\right){\displaystyle \frac{{\left(0.001\right)}^{2}}{{\left(0.01\right)}^{2}}}={\displaystyle \frac{{\varphi}_{12}\mathrm{coth}{\varphi}_{12}1}{10{\varphi}_{12}\text{}\text{coth}\left(10\text{}{\varphi}_{12}\right)1}}& \text{(E152.6)}\end{array}$
$\begin{array}{cc}\text{0.05}={\displaystyle \frac{{\varphi}_{12}\text{}\text{coth}\text{}{\varphi}_{12}1}{10{\varphi}_{12}\text{}\text{coth}\text{}\left(10{\varphi}_{12}\right)1}}& \text{(152.7)}\end{array}$
We now have one equation and one unknown. Solving Equation (E152.7) we find that
$\begin{array}{cc}\hline {\varphi}_{12}=1.65\hfill & {\text{for}\text{}R}_{2}=0.001\text{}\text{m}\\ \text{Then}\hfill & \text{}\\ {\varphi}_{11}=10{\varphi}_{12}=16.5\hfill & \text{for}\text{}{R}_{1}=0.01\text{}\text{m}\\ \hline\end{array}$
The corresponding effectiveness factors
Given two experimental points, one can predict the particle size where internal mass transfer does not limit the rate of reaction.
$\begin{array}{cc}\hline {\text{For}\text{}R}_{2}:\hfill & {\eta}_{2}={\displaystyle \frac{3({\varphi}_{12}\text{coth}{\varphi}_{12}1)}{{\varphi}_{12}^{2}}}={\displaystyle \frac{3(1.65\text{coth 1.65}1)}{{\left(1.65\right)}^{2}}}=0.856\hfill \\ {\text{For}\text{}R}_{1}:\hfill & {\eta}_{1}={\displaystyle \frac{3(16.5\text{coth 16.5}1)}{{\left(16.5\right)}^{2}}}\approx {\displaystyle \frac{3}{16.5}}=0.182\hfill \\ \hline\end{array}$
(b) Next we calculate the particle radius needed to virtually eliminate internal diffusion control (say, η = 0.95)
$\begin{array}{cc}0.95={\displaystyle \frac{3({\varphi}_{13}\text{coth}{\varphi}_{13}1)}{{\varphi}_{13}^{2}}}& \text{(E152.8)}\end{array}$
The solution to Equation (E152.8) yields ϕ_{13} = 0.9
${R}_{3}={R}_{1}{\displaystyle \frac{{\varphi}_{13}}{{\varphi}_{11}}}=\left(0.01\right)\text{}\left({\displaystyle \frac{0.9}{16.5}}\right)=5.5\times {10}^{4}\text{}\text{m}$
A particle size of 0.55 mm is necessary to virtually eliminate diffusion control (i.e., η = 0.95).
Only two data points were necessary.
Analysis: This example is important because it shows us how, with only two measurements and some assumptions, we can determine internal diffusion limitations for the two pellet sizes, and predict the pellet size necessary to completely eliminate internal diffusion.
You may not be measuring what you think you are.
There are circumstances under which the measured reaction order and activation energy are not the true values. Consider the case in which we obtain reactionrate data in a differential reactor at two different temperatures, T_{1} and T_{2}, where precautions are taken to virtually eliminate external mass transfer resistance (i.e., C_{A}_{s} = C_{A}_{b}). From these data, we construct a log–log plot of the measured rate of reaction ${r}_{\text{A}}^{\prime}$ as a function of the gasphase concentration, C_{As} (Figure 158). The slope of this plot is the apparent reaction order n′ and the measured rate law takes the form
Measured rate:
$\begin{array}{cc}{r}_{\text{A}\mathit{\text{m}}}^{\prime}={k}_{n}^{\prime}{C}_{\text{A}s}^{n\prime}& \text{(1540)}\end{array}$
We will now proceed to relate this measured reaction order n′ to the true reaction order n. Using the definition of the effectiveness factor, noting that the
Measured rate with apparent reaction order n^{′}
actual rate, ${r}_{\text{A}}^{\prime}$, is the product of η and the rate of reaction evaluated at the external surface, ${k}_{n}{\text{C}}_{\text{As}}^{n}$, for example,
Actual rate:
$\begin{array}{c}\hline {r}_{\text{A}}^{\prime}=\eta ({r}_{\text{A}s}^{\prime})=\eta \left({k}_{n}{C}_{\text{A}s}^{n}\right)\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}\hspace{1em}(1541)}\hspace{1em}\end{array}$
For large values of the Thiele modulus, ϕ_{n}, where internal mass transfer is limiting, we can use Equation (1535) to substitute into Equation (1541) to obtain
${r}_{\text{A}}^{\prime}={\displaystyle \frac{3}{{\varphi}_{n}}}\sqrt{{\displaystyle \frac{2}{n+1}}}{k}_{n}{\mathit{\text{C}}}_{\text{A}s}^{n}={\displaystyle \frac{3}{R}}\sqrt{{\displaystyle \frac{{\mathit{\text{D}}}_{e}}{{k}_{n}}}{\mathit{\text{C}}}_{\text{A}s}^{1n}}\sqrt{{\displaystyle \frac{2}{n+1}}}\text{}{k}_{n}{\mathit{\text{C}}}_{\text{A}s}^{n}$
Simplifying
$\begin{array}{c}\hline {r}_{\text{A}}^{\prime}={\displaystyle \frac{3}{R}}\sqrt{{\displaystyle \frac{2{D}_{e}}{(n+1)}}}\text{}{k}_{n}^{1/2}{C}_{\text{A}s}^{(n+1)/2}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}\hspace{1em}(1542)}\end{array}$
We equate the true reaction rate, Equation (1542), to the measured reaction rate, Equation (1540), to get
$\begin{array}{c}\hline {r}_{\text{A}}^{\prime}=\stackrel{\text{True}}{\overbrace{\sqrt{{\displaystyle \frac{2}{n+1}}}\left({\displaystyle \frac{3}{R}}\sqrt{{D}_{e}}\text{}{k}_{n}^{1/2}{C}_{\text{A}s}^{(n+1)/2}\right)}}=\stackrel{\underset{\text{Apparent}}{\text{Measured}}}{\overbrace{{k}_{n}^{\prime}{C}_{\text{A}s}^{n\prime}}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}\hspace{1em}(1543)}\end{array}$
The functional dependence of the reaction rate on concentration must be the same for both the measured rate and the theoretically predicted rate
${C}_{\text{A}s}^{(n+1)/2}={C}_{\text{A}s}^{n\prime}$
therefore the measured apparent reaction order n′ (n_{Apparent}) is related to the true reaction order n (n_{True}) by
The true and the apparent reaction order
$\begin{array}{c}\hline n\prime ={\displaystyle \frac{1+n}{2}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}\hspace{1em}(1544)}\end{array}$
In addition to an apparent reaction order, there is also an apparent activation energy, E_{App}. This value is the activation energy we would calculate using the experimental data from the slope of a plot of ln $({r}_{\text{A}}^{\prime})$ as a function of (1/T) at a fixed concentration of A. Substituting for the measured and true specific reaction rates in terms of the activation energy gives
into Equation (1543), we find that
$\begin{array}{c}\hline {n}_{\text{true}}={2n}_{\text{apparent}}1\\ \hline\end{array}$
${r}_{\text{A}}^{\prime}=\left({\displaystyle \frac{3}{R}}\sqrt{{\displaystyle \frac{2}{n+1}}{D}_{e}}\right){A}_{T}^{1/2}{\left[\text{exp}\left({\displaystyle \frac{{E}_{T}}{\mathit{\text{RT}}}}\right)\right]}^{1/2}{C}_{As}^{(n+1)/2}={A}_{\text{App}}\left[\text{exp}\left({\displaystyle \frac{{E}_{\text{App}}}{\mathit{\text{RT}}}}\right)\right]{C}_{As}^{n\prime}$
Taking the natural log of both sides gives us
$\begin{array}{cc}\text{ln}\text{}\left[{\displaystyle \frac{3}{R}}\sqrt{{\displaystyle \frac{2}{n+1}}{D}_{e}}\text{}{A}_{T}^{1/2}{C}_{As}^{(n+1)/2}\right]{\displaystyle \frac{{E}_{T}}{2\mathit{\text{RT}}}}=\text{ln}\text{}\left[{A}_{\text{App}}{C}_{As}^{n\prime}\right]{\displaystyle \frac{{E}_{\text{App}}}{\mathit{\text{RT}}}}& \text{(1545)}\end{array}$
where E_{T} is the true activation energy.
As with the dependence of reaction rate on concentration, the temperature dependence must be the same for the analytical rate. Comparing the temperaturedependent terms on the right and lefthand sides of Equation (1545), we see that the true activation energy is equal to twice the apparent activation energy.
The true activation energy
$\begin{array}{c}\hline {E}_{T}=2{E}_{\text{App}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}\hspace{1em}(1546)}\end{array}$
Important industrial consequence of falsified kinetic runaway reactions. Safety considerations!
This measurement of the apparent reaction order and activation energy results primarily when internal diffusion limitations are present and is referred to as disguised or falsified kinetics. Serious consequences could occur if the laboratory data were taken in the disguised regime and the reactor were operated in a different regime. For example, what if the particle size were reduced so that internal diffusion limitations became negligible? The higher activation energy, E_{T}, could cause the reaction to be much more temperature sensitive, and there is the possibility for runaway reaction conditions causing an explosion to occur.
For firstorder reactions, we can use an overall effectiveness factor to help us analyze diffusion, flow, and reaction in packed beds. We now consider a situation where external and internal resistance to mass transfer to and within the pellet are of the same order of magnitude (Figure 159). At steady state, the transport of the reactant(s) from the bulk fluid to the external surface of the catalyst is equal to the net rate of reaction of the reactant within and on the pellet.
Here, both internal and external diffusion are important.
The molar rate of mass transfer from the bulk fluid to the external surface is
$\begin{array}{ccc}\hfill \text{Molar rate}& =\hfill & \left(\text{Molar flux}{\text{C}}_{\text{A}b}\text{}\text{to}{C}_{\text{A}s}\right)\cdot \left(\text{External surface area}\right)\hfill \\ \hfill {M}_{\text{A}}& =\hfill & {W}_{\text{A}r}\cdot \left(\text{External surface area/Volume}\right)\left(\text{Reactor volume}\right)\hfill \\ \text{}\hfill & =\hfill & \begin{array}{cc}\begin{array}{c}{W}_{\text{Ar}}\cdot {a}_{c}\mathrm{\Delta}V\hfill \end{array}\hfill & \hspace{1em}\hspace{1em}\left(15\text{}47\right)\hfill \end{array}\hfill \end{array}$
where a_{c} is the external surface area per unit reactor volume (cf. Chapter 14) and ΔV is the reactor volume.
This molar rate of mass transfer to the surface, M_{A}, is also equal to the net (total) rate of reaction on and within the pellet
$\begin{array}{c}\hline {M}_{A}=({r}_{A}^{\u2033})\times (\text{External area}+\text{Internal area})\\ \hline\end{array}$
$\begin{array}{ccc}\hfill \text{External area}& =\hfill & {\displaystyle \frac{\text{Externalarea}}{\text{Reactorvolume}}}\times \text{Reactorvolume}\hfill \\ \text{}\hfill & =\hfill & {a}_{c}\text{}\text{}\mathrm{\Delta}V\hfill \\ \hfill \text{Internalarea}& =\hfill & {\displaystyle \frac{\text{Internalarea}}{\text{Massofcatalyst}}}\times {\displaystyle \frac{\text{Massofcatalyst}}{\text{Volumeofcatalyst}}}\times {\displaystyle \frac{\text{}\text{Volumeofcatalyst}}{\text{Reactorvolume}}}\times \text{Reactorvolume}\hfill \\ \text{}\hfill & =\hfill & {S}_{a}\times {\rho}_{c}\times (1\varphi )\times \mathrm{\Delta}V\hfill \\ \text{}\hfill & =\hfill & [{{\displaystyle \frac{\text{m}}{\text{g}}}}^{2}\times {\displaystyle \frac{\text{g}}{{\text{m}}^{3}}}\times {\text{m}}^{3}]\hfill \\ \text{}\hfill & =\hfill & {S}_{a}\text{}\text{}\stackrel{{\rho}_{b}}{\overbrace{{\rho}_{\text{}c}(1\varphi )}}\text{}\text{}\mathrm{\Delta}V\hfill \\ \text{}\hfill & =\hfill & {S}_{a}\text{}{\rho}_{b}\text{}\mathrm{\Delta}V\hfill \end{array}$
$\begin{array}{ccc}\hfill {\rho}_{b}& =& \text{Bulk}\text{density}\hfill \\ \hfill & =& {\rho}_{c}\text{}(1\varphi )\hfill \\ \hfill \varphi & =& \text{porosity}\hfill \end{array}$
See nomenclature note in Example 154.
Recall that ρ_{c} is the density of catalyst pellet, kg per volume of pellet and ρ_{b} is the bulk density of catalyst in the reactor, kgcat per reactor volume.
We now combine the above equations for external surface area and internal surface area with Equations (1547) to obtain total molar flow into all the catalyst in volume ΔV
$\begin{array}{c}\hline {M}_{\text{A}}={r}_{\text{A}}^{\u2033}[{a}_{\text{c}}\text{}\text{}\mathrm{\Delta}V+{S}_{a}{\rho}_{b}\text{}\mathrm{\Delta}V]\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1548)}\end{array}$
Combining Equations (1547) and (1548), and canceling the volume ΔV, we see the flux to the pellet surface, W_{A}_{z}a_{c}, is equal to the rate of consumption of A in and on the catalyst.
${W}_{\text{A}r}{a}_{c}={r}_{\text{A}}^{\u2033}\cdot ({a}_{c}+{S}_{a}{\rho}_{b})$
For most catalysts, the internal surface area is much greater than the external surface area (i.e., S_{a}ρ_{b} ≫ a_{c}), in which case we have
${W}_{\text{A}r}{a}_{c}={r}_{\text{A}}^{n}\text{}{S}_{a}\text{}{\rho}_{b}$
Comparing units on the r.h.s. and l.h.s. of Equation (1549), we find no inconsistencies, that is,
$[{\displaystyle \frac{\text{mol}}{{\text{m}}^{2}\cdot \text{s}}}\cdot {\displaystyle \frac{{\text{m}}^{2}}{3}}]=\left[{\displaystyle \frac{\text{mol}}{{\text{m}}^{2}\cdot \text{s}}}\left({\displaystyle \frac{{\text{m}}^{2}}{\text{g}}}\right){\displaystyle \frac{\text{g}}{{\text{m}}^{3}}}\right]$
where ${r}_{\text{A}}^{"}$ is the overall rate of reaction within and on the pellet per unit surface area, ${r}_{\text{A}}^{\prime}$ is the rate of reaction per mass of catalyst
${r}_{\text{A}}^{\prime}={r}_{\text{A}}^{\u2033}{S}_{a}$
and –r_{A} is the overall rate per volume of reactor, that is,
${r}_{\text{A}}={r}_{\text{A}}^{\u2033}\text{}{\rho}_{b}$
then
$\begin{array}{c}\hline {W}_{\text{A}z}{a}_{c}={r}_{\text{A}}={r}_{\text{A}}^{\u2033}\text{}{S}_{a}\text{}{\rho}_{b}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1549)}\end{array}$
with the corresponding units for each term in Equation (1549) shown below.
$\text{}{\displaystyle \frac{\text{mol}}{{\text{m}}^{2}\cdot \text{s}}}\times {\displaystyle \frac{{\text{m}}^{2}}{{\text{m}}^{3}}}={\displaystyle \frac{\text{mol}}{{\text{m}}^{3}\cdot \text{s}}}={\displaystyle \frac{\text{mol}}{{\text{m}}^{2}\cdot \text{s}}}\times {\displaystyle \frac{{\text{m}}^{2}}{\text{gcat}}}\times {\displaystyle \frac{\text{gcat}}{{\text{m}}^{3}}}$
Again, we find no inconsistencies.
The relationship for the rate of mass transport to the external catalyst surface is
$\begin{array}{c}\hline \begin{array}{c}\begin{array}{cc}\begin{array}{cc}\begin{array}{cc}{M}_{\text{A}}=& {W}_{\text{A}r}\end{array}{a}_{c}\text{}\mathrm{\Delta}V& ={k}_{c}({C}_{\text{A}b}{C}_{\text{A}s}){a}_{c}\text{}\mathrm{\Delta}V\end{array}& \text{}\end{array}\\ \text{}\end{array}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}\hspace{1em}(1550)}\hfill \end{array}$
$\begin{array}{c}\left[{\displaystyle \frac{\text{mol}}{\text{s}}}\right]=\left[\left({\displaystyle \frac{\text{mol}}{{\text{m}}^{2}\text{s}}}\right)\left({\displaystyle \frac{{\text{m}}^{2}}{{\text{m}}^{3}}}\right){\text{m}}^{3}\right]=\left[\left({\displaystyle \frac{\text{m}}{\text{s}}}\right)\left({\displaystyle \frac{\text{mol}}{{\text{m}}^{3}}}\right)\left({\displaystyle \frac{{\text{m}}^{2}}{{\text{m}}^{3}}}\right){\text{m}}^{3}\right]\end{array}$
Again, comparing units on the l.h.s. and r.h.s., no inconsistencies
where k_{c} is the external mass transfer coefficient (m/s) discussed in Section 14.4. Because internal diffusion resistance is also significant, not all of the interior surface of the pellet is accessible to the concentration at the external surface of the pellet, C_{A}_{s}. We have already learned that the effectiveness factor is a measure of the internal surface accessibility (see Equation (1541)):
$\begin{array}{c}{r}_{\text{A}}={r}_{\text{A}s}\eta \\ \left[\begin{array}{c}\text{Actual}\\ \text{rate}\end{array}\right]=\left[\begin{array}{c}\begin{array}{c}\text{Rate at}\\ \text{surface}\end{array}\\ \text{conditions}\end{array}\right]\left[{\displaystyle \frac{\text{Actual rate}}{\text{Rate at surface conditions}}}\right]\end{array}$
Assuming that the surface reaction is first order with respect to A, we can utilize the internal effectiveness factor to write
$\begin{array}{cc}{r}_{\text{A}}=\eta {k}_{1}{C}_{\text{A}s}& \text{(1551)}\end{array}$
Recall that
$({k}_{1}={k}_{1}^{\u2033}\text{}{S}_{a}{\rho}_{b})$
We need to eliminate the surface concentration from any equation involving the rate of reaction or rate of mass transfer, because C_{A}_{s} cannot be measured by standard techniques. To accomplish this elimination, we use Equations (1549), (1550), and (1551) in order to equate the mass transfer rate of A to the pellet surface, –W_{A}_{r}a_{c}, to the rate of consumption of A within the pellet, ηk_{1}C_{A}_{s}
${W}_{\text{A}r}{a}_{c}=\eta {k}_{1}{C}_{\text{A}s}$
Then substitute for W_{A}_{r}a_{c} using Equation (1550)
$\begin{array}{cc}{k}_{c}{a}_{c}({C}_{\text{A}b}{C}_{\text{A}s})=\eta {k}_{1}{C}_{\text{A}s}& \text{(1552)}\end{array}$
Solving for C_{A}_{s}, we obtain
$\begin{array}{cc}{C}_{\text{A}s}={\displaystyle \frac{{k}_{c}{a}_{c}}{{k}_{c}{a}_{c}+\eta {k}_{1}}}{C}_{\text{A}b}& \text{(1553)}\end{array}$
Concentration at the pellet surface as a function of bulk gas concentration
Substituting for C_{A}_{s} in Equation (1551) gives
$\begin{array}{c}\hline {r}_{\text{A}}={\displaystyle \frac{\eta {k}_{1}{k}_{c}{a}_{c}{C}_{\text{A}b}}{{k}_{c}{a}_{c}+\eta {k}_{1}}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1554)}\end{array}$
In discussing the surface accessibility, we defined the internal effectiveness factor η with respect to the concentration at the external surface of the pellet, C_{A}_{s}, as
$\begin{array}{c}\hline \eta ={\displaystyle \frac{\text{Actual overall rate of reaction}}{\begin{array}{c}\text{Rate of reaction that would result if entire interior surface were}\\ \text{exposed to the exteral pellet surface conditions,}{C}_{\text{A}s},{T}_{s}\end{array}}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1528)}\end{array}$
Two different effectiveness factors
We now define an overall effectiveness factor that is based on the bulk concentration
$\begin{array}{c}\hline \Omega ={\displaystyle \frac{\text{Actualoverallrateofreaction}}{\begin{array}{c}\text{Rate that would result if the entire surface were}\\ \text{exposed to the bulk conditions,}{C}_{\text{A}b},{T}_{b}\end{array}}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1555)}\end{array}$
Dividing the numerator and denominator of Equation (1554) by k_{c}a_{c}, we obtain the net rate of reaction, –r_{A}, in terms of the bulk fluid concentration, which is a measurable quantity:
$\begin{array}{cc}{r}_{\text{A}}={\displaystyle \frac{\eta}{1+{\displaystyle \frac{\eta {k}_{1}}{{k}_{c}{a}_{c}}}}}{k}_{1}{C}_{\text{A}b}& \text{(1556)}\end{array}$
The actual rate of reaction is related to the reaction rate evaluated at the bulk concentration of A. Consequently, the overall rate of reaction in terms of the bulk concentration C_{A}_{b} is
$\begin{array}{c}\hline {r}_{\text{A}}=\Omega ({r}_{\text{A}b})={\Omega k}_{1}{C}_{\text{A}b}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1557)}\end{array}$
where the overall effectiveness factor is
Overall effectiveness factor for a firstorder reaction
$\begin{array}{c}\hline \Omega ={\displaystyle \frac{\eta}{1+{\displaystyle \frac{\eta {k}_{1}}{{k}_{c}{a}_{c}}}}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1558)}\end{array}$
Let’s look at some limiting situations for the overall effectiveness factor. First note that the rates of reaction based on surface and bulk concentrations are related by
$\begin{array}{cc}{r}_{\text{A}}=\Omega ({r}_{\text{A}b})=\eta ({r}_{\text{A}s})& \text{(1559)}\end{array}$
where
$\begin{array}{c}{r}_{\text{A}s}={k}_{1}{C}_{\text{A}s}\\ {r}_{\text{A}b}={k}_{1}{C}_{\text{A}b}\end{array}$
The actual rate can be expressed in terms of the rate per unit volume, −r_{A}, the rate per unit mass, ${r}_{\text{A}}^{\prime},$ and the rate per unit surface area, ${r}_{\text{A}}^{\u2033}$, which are related by the equation
${r}_{\text{A}}={r}_{\text{A}}^{\prime}{\rho}_{b}={r}_{\text{A}}^{\u2033}{S}_{a}{\rho}_{b}$
Recall that ${k}_{1}^{\u2033}$ is given in terms of the catalyst surface area (m^{3}/m^{2}·s), ${k}_{1}^{\prime}$ is given in terms of catalyst mass (m^{3}/gcat · s), and k_{1} (1/s) is given in terms of reactor volume
${k}_{1}={\rho}_{b}\text{}{k}_{1}^{\prime}={\rho}_{b}\cdot {S}_{a}\cdot {k}_{1}^{\u2033}$
We saw in Chapter 14 that as the velocity of the fluid increases, the external mass transfer coefficient k_{c} increases (cf. Equation 1446). Consequently, for large flow rates resulting in large values of the external mass transfer coefficient k_{c}, we can neglect the ratio in the denominator
High flow rates of fluid
and the overall effectiveness factor approaches the internal effectiveness factor
$\begin{array}{c}\hline \begin{array}{c}\Omega \equiv \eta \end{array}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1560)}\end{array}$
Now let’s consider the case where the ratio of $\left({\displaystyle \frac{\eta {k}_{1}}{{k}_{c}{a}_{c}}}\right)$ is very small, then
External diffusion limits
The overall effectiveness factor for external diffusion control is
$\begin{array}{c}\hline \Omega ={\displaystyle \frac{{k}_{c}{a}_{c}}{{k}_{1}}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1561)}\end{array}$
Dial soap
In many instances it is of interest to obtain “quick and dirty” estimates to learn which is the ratelimiting step in a heterogeneous reaction.
The Mears criterion, like the Weisz–Prater criterion, uses the measured rate of reaction, ${r}_{\text{A}}^{\prime}$, (kmol/kgcat·s) to learn whether external mass transfer from the bulk gas phase to the catalyst surface can be neglected.^{3} The Mears number, MR, is
^{3} D. E. Mears, Ind. Eng. Chem. Process Des. Dev., 10, 541 (1971). Other interphase transportlimiting criteria can be found in AIChE Symp. Ser. 143 (S. W. Weller, ed.), 70 (1974).
Is external diffusion limiting?
$\begin{array}{c}\hline \text{MR}={\displaystyle \frac{{r}_{\text{A}}^{\prime}\text{}\left(\text{obs}\right)\text{}{\rho}_{b}\mathit{\text{Rn}}}{{k}_{c}{C}_{\text{A}b}}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1562)}\end{array}$
The Mears number can be used to establish external mass transfer limitations
Here, we measure ${r}_{\text{A}}^{\prime}$ (obs), C_{A}_{b}, ρ_{b}, R and n, and then calculate k_{c} to determine MR, where
$\begin{array}{c}\begin{array}{ccc}\hfill n& =& \text{reactionorder}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\hfill \\ \hfill R& =& \text{catalystparticleradius,m}\hfill \\ \hfill {\rho}_{b}& =& {\text{bulkdensityofcatalystbed,kg/m}}^{3}\hfill \\ \hfill \text{}& =& (1\varphi ){\rho}_{c}\text{}\text{}\text{}(\varphi =\text{porosity})\hfill \\ \hfill {\rho}_{c}& =& {\text{soliddensityofcatalystpellet,kg/m}}^{3}\hfill \\ \hfill {C}_{\text{A}b}& =& {\text{bulkreactantconcentration,mol/dm}}^{3}\hfill \\ \hfill {k}_{c}& =& \text{masstransfercoefficient,m/s}\hfill \end{array}\\ \text{}\end{array}$
Mears proposed that when
MR < 0.15
external mass transfer effects can be neglected and no concentration gradient exists between the bulk gas and external surface of the catalyst pellet. This proposal by Mears was endorsed unanimously by the Jofostan legislature. The mass transfer coefficient can be calculated from the appropriate correlation, such as that of Thoenes–Kramers, for the flow conditions through the bed.
Mears also proposed that the bulk fluid temperature, T, will be virtually the same as the temperature at the external surface of the pellet when
T_{b} ≌ T_{s}
$\begin{array}{c}\hline \left{\displaystyle \frac{\mathrm{\Delta}{H}_{\text{Rx}}({r}_{\text{A}}^{\prime}){\rho}_{b}\text{}\text{}\mathit{\text{RE}}}{h{T}^{2}{R}_{g}}}\right0.15\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1563)}\end{array}$
where
$\begin{array}{ccc}\hfill h& =& {\text{heat transfer coefficient between gas and pellet, kJ/m}}^{2}\cdot \text{s}\cdot \text{K}\hfill \\ \hfill {R}_{g}& =& \text{gas constant, 8.314 J/mol}\cdot \text{K}\hfill \\ \hfill {\begin{array}{c}\mathrm{\Delta}H\end{array}}_{\text{Rx}}& =& \text{heat of reaction, kJ/mol}\hfill \\ \hfill E& =& \text{activation energy, kJ kmol}\hfill \end{array}$
and the other symbols are as in Equation (1562).
We now consider the same isomerization taking place in a packed bed of catalyst pellets rather than on one single pellet (see Figure 1510). The concentration C_{A}_{b} is the bulk gasphase concentration of A at any point along the length of the bed.
We shall perform a balance on species A over the volume element, ΔV, neglecting any radial variations in concentration and assuming that the bed is operated at steady state. The following symbols will be used in developing our model:
$\begin{array}{ccc}\hfill {\text{A}}_{c}& =& {\text{crosssectional area of the tube containing the catalyst, dm}}^{2}\hfill \\ \hfill {C}_{\text{A}b}& =& {\text{bulk gas concentration of A, mol/dm}}^{3}\hfill \\ \hfill {\rho}_{b}& =& {\text{bulk density of the catalyst bed, g/dm}}^{3}\hfill \\ \hfill {\upsilon}_{0}& =& {\text{volumetric flow rate, dm}}^{3}/\text{s}\hfill \\ \hfill U& =& \text{superficial velocity}={\upsilon}_{0}/{\text{A}}_{c},\text{dm/s}\hfill \end{array}$
Mole Balance
A mole balance on ΔV, the volume element (A_{c}Δz), yields
$\begin{array}{cccccc}\left[\text{Rate in}\right]& & \left[\text{Rate out}\right]\hfill & +& \left[\text{Rate of formation of A}\right]& =0\\ {\text{A}}_{c}{W}_{\text{Az}}{}_{z}& & {\text{A}}_{c}{W}_{\text{Az}}{}_{z+\mathrm{\Delta}z}\hfill & +& {r}_{\text{A}}^{\prime}{\rho}_{b}{\text{A}}_{c}\text{}\mathrm{\Delta}z& =0\end{array}$
Dividing by A_{c} Δz and taking the limit as Δz → 0 yields
$\begin{array}{cc}{\displaystyle \frac{{dW}_{\text{Az}}}{dz}}+\stackrel{{r}_{\text{A}}}{\overbrace{{r}_{\text{A}}^{\prime}{\rho}_{b}}}=0& \text{(1564)}\end{array}$
Combining Equation (144) and (146), we get
${W}_{\text{Az}}={D}_{\text{AB}}{\displaystyle \frac{d{C}_{\text{A}b}}{dz}}+{y}_{\text{A}b}({W}_{\text{Az}}+{W}_{\text{Bz}})$
Also, writing the bulk flow term in the form
${B}_{\text{Az}}={y}_{\text{A}b}({W}_{\text{Az}}+{W}_{\text{Bz}})={y}_{\text{A}b}cU=U{C}_{\text{A}b}$
Equation (1564) can be written in the form
$\begin{array}{cc}{D}_{\text{AB}}{\displaystyle \frac{{d}^{2}{C}_{\text{A}b}}{{dz}^{2}}}U{\displaystyle \frac{d\text{}{C}_{\text{A}b}}{dz}}+{r}_{\text{A}}=0& \text{(1565)}\end{array}$
Now we will see how to use η and Ω to calculate conversion in a packed bed.
The term D_{AB}(d^{2}C_{A}_{b}/dz^{2}) is used to represent either diffusion and/or dispersion in the axial direction. Consequently, we shall use the symbol D_{a} for the dispersion coefficient to represent either or both of these cases. We will come back to this form of the diffusion equation when we discuss dispersion in Chapter 18 (cf. Equation (1816)). The overall reaction rate, –r_{A}, is a function of the reactant concentration within the catalyst. This overall rate can be related to the rate of reaction of A that would exist if the entire surface were exposed to the bulk concentration C_{A}_{b} through the overall effectiveness factor Ω
$\begin{array}{cc}{r}_{\text{A}}={r}_{\text{A}b}\Omega & \text{(1557)}\end{array}$
For the firstorder reaction considered here
$\begin{array}{cc}{r}_{\text{A}}=\Omega {k}_{1}{C}_{\text{A}b}& \text{(1566)}\end{array}$
Substituting this equation for –r_{A} into Equation (1565), we form the differential equation describing diffusion with a firstorder reaction in a catalyst bed
$\begin{array}{c}\hline {D}_{a}{\displaystyle \frac{{d}^{2}{C}_{\text{A}b}}{{dz}^{2}}}U{\displaystyle \frac{d{C}_{\text{A}b}}{dz}}\Omega {k}_{1}{C}_{\text{A}b}=0\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1567)}\end{array}$
Flow and firstorder reaction in a packed bed
As will be shown in Chapter 18, the solution to Equations (1567) and (1816) can be quite complicated depending on the boundary conditions and assumptions. Consequently for now, we will only discuss one approximate solution. We shall solve this equation for the case in which the flow rate through the bed is very large and the axial diffusion can be neglected. Young and Finlayson have shown that axial dispersion can be neglected when^{4}
^{4} L. C. Young and B. A. Finlayson, Ind. Eng. Chem. Fund., 12, 412.
Criterion for neglecting axial dispersion/diffusion
$\begin{array}{c}\hline {\displaystyle \frac{{r}_{\text{A}\left(\text{obs}\right){d}_{p}}}{{U}_{0}{C}_{\text{A}b}}}<{\displaystyle \frac{{U}_{0}{d}_{p}}{{D}_{a}}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1568)}\end{array}$
where U_{0} is the superficial velocity, d_{p} is the particle diameter, and D_{a} is the effective axial dispersion coefficient. In Chapter 18 we will consider solutions to the complete form of Equation (1567).
Neglecting axial dispersion with respect to forced axial convection
$\leftU{\displaystyle \frac{d{C}_{\text{A}b}}{dz}}\right\text{\u226b}\left{D}_{a}{\displaystyle \frac{{d}^{2}{C}_{\text{A}b}}{{dz}^{2}}}\right$
Equation (1567) can be arranged in the form
$\begin{array}{cc}{\displaystyle \frac{d{C}_{\text{A}b}}{dz}}={\displaystyle \frac{\Omega {k}_{1}{C}_{\text{A}b}}{U}}& \text{(1569)}\end{array}$
With the aid of the boundary condition at the entrance of the reactor
$\begin{array}{cc}{C}_{\text{A}b}={C}_{\text{A}b0}& \text{at}\text{}\text{}z=0\end{array}$
Bulk concentration in a packedbed reactor
$\begin{array}{c}\hline \begin{array}{cc}{C}_{\text{A}b}={C}_{\text{A}b0}\text{}\text{exp}[{\displaystyle \frac{\Omega {k}_{1}z}{U}}]& \text{}\end{array}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1570)}\end{array}$
Recall we can easily relate the distance z to the catalyst weight down the reactor, W, using the bulk density, ρ_{b}, that is,
W = ρ_{b}A_{c}z
The conversion at the reactor’s exit, z = L, is
$\begin{array}{c}\hline X=1{\displaystyle \frac{{C}_{\text{A}b}}{{C}_{\text{A}b0}}}=1\text{exp}[{\displaystyle \frac{{k}_{1}\Omega L}{U}}]\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1571)}\end{array}$
The sketch of the corresponding conversion profile is shown in the margin.
^{†}This is a Stop and Smell the Roses Example. Exploring this LEP Example with Wolfram and/or Python that will give you a fantastic intuitive feel of the parameters and conditions that result in internal and external diffusion limitations.
We have seen that Nitric Oxide (NO) plays an important role in smog formation and there are great incentives for reducing its concentration in the atmosphere. It is proposed to reduce the concentration of NO in an effluent stream from a plant by passing it through a packed bed of spherical, porous carbonaceous solid pellets. A 2% NO and 98% air mixture flows at a rate of 1 × 10^{–6} m^{3}/s (0.001 dm^{3}/s) through a 2in.ID tube packed with porous solid pellets at a temperature of 1173 K and a pressure of 101.3 kPa. The reaction
$\begin{array}{cc}\begin{array}{cc}\text{NO + C}& \to \end{array}& \text{CO}+\frac{1}{2}\end{array}{\text{N}}_{2}$
Green chemical reaction engineering
is pseudo first order in NO because of the excess of carbon surface area; that is,
${r}_{\text{NO}}^{\prime}={k}_{1}^{\u2033\text{}}{S}_{a}{C}_{\text{NO}}$
and occurs primarily in the pores inside the pellet, where
S_{a} = Internal surface area = 530 m^{2}/g
$\begin{array}{ccc}{k}_{1}^{\u2033}& =\hfill & 4.42\times {10}^{10}\text{}{\text{m}}^{3}/{\text{m}}^{2}\cdot \text{s}\hfill \end{array}$
$\begin{array}{ccc}{k}_{1}^{\prime}& =& {k}_{1}^{\u2033}\text{}{S}_{a}=[4.42\times {10}^{10}{\text{m}}^{3}/{\text{m}}^{2}\cdot \text{s}]\left[530{\displaystyle \frac{{\text{m}}^{2}}{\text{g}}}\right]=2.34\times {10}^{7}{\text{m}}^{3}/\text{g}/\text{s}\end{array}$
From those values of ${k}_{1}^{"}$ and ${k}_{1}^{\prime}$ we find
${k}_{1}={\rho}_{b}{k}_{1}^{\prime}=(1.4\times {10}^{6}{\text{g/m}}^{3})(2.34\times {10}^{7}{\text{m}}^{3}\text{/g/s})=0.328{\text{s}}^{1}$
Derive an equation to find the conversion profiles, X = f(W).
Calculate the Thiele modulus ϕ_{1} and the η internal effectiveness factor.
Calculate the Weisz–Prater parameter, WP. Is internal mass transfer limiting?
Calculate the overall effectiveness factor, Ω
Calculate the Mears parameter, MR. Is external mass transfer limiting?
Calculate the weight of solid porous catalyst necessary to reduce the NO concentration of 0.004%, which is below the Environmental Protection Agency’s limit.
Additional information:
At 1173 K, the fluid properties are
$\begin{array}{ccc}\hfill \nu & =& \text{Kinematicviscosity}=1.53\times {10}^{8}{\text{m}}^{2}\text{/s}\hfill \\ \hfill {D}_{e}& =& \text{Effectivediffusivity}=1.82\times {10}^{8}{\text{m}}^{2}\text{/s}\hfill \\ \hfill {D}_{\text{AB}}& =& \text{Gasdiffusivity}=\text{2.0}\times {\text{10}}^{8}\text{}{\text{m}}^{2}\text{/s}\hfill \end{array}$
Also see the Web site www.rowan.edu/greenengineering
The properties of the catalyst and bed are
$\begin{array}{ccc}\hfill {\rho}_{c}& =& \text{Densityofcatalystparticle}=2.8{\text{g/cm}}^{3}=2.8\times {10}^{6}{\text{g/m}}^{3}\hfill \\ \hfill \varphi & =& \text{Bedporsity}={0.5}_{\text{}}\hfill \\ \hfill {\rho}_{b}& =& \text{Bulkdensityofbed}=\text{}{\rho}_{c}(1\varphi )=1.4\times {10}^{6}{\text{g/m}}^{3}\hfill \\ \hfill R& =& \text{Pelletradius}=3\times {10}^{3}\text{}\text{m}\hfill \\ \hfill \gamma & =& \text{Sphericity}=1.0\hfill \end{array}\begin{array}{ccc}\text{}& \text{}& \text{}\\ \text{}& \text{}& \text{}\\ \text{}& \text{}& \text{}\end{array}\text{}$
Solution
1. Find X and the W = f(X)
It is desired to reduce the NO concentration from 2.0% to 0.004%. Neglecting any volume change at these low concentrations gives us the exit conversion
$\begin{array}{c}\hline X={\displaystyle \frac{{C}_{\text{A}b0}{C}_{\text{A}b}}{{C}_{\text{A}b0}}}={\displaystyle \frac{20.004}{2}}=0.998\\ \hline\end{array}$
where A represents NO.
The variation of NO down the length of the reactor is given by Equation (1569). Replacing k_{1} by ${k}_{1}^{\prime}{\rho}_{b}$
$\begin{array}{cc}{\displaystyle \frac{d{C}_{\text{A}b}}{dz}}={\displaystyle \frac{\Omega {k}_{1}{C}_{\text{A}b}}{U}}=\Omega {\displaystyle \frac{{k}_{1}^{\prime}{\rho}_{b}{C}_{\text{A}b}}{U}}& \text{(1569)}\end{array}$
Multiplying the denominator on the right and left hand sides of Equation (1569) by the crosssectional area, A_{c}, and realizing that the weight of the catalyst up to a point z in the bed is
(Mole balance)
+
(Rate law)
+
(Overall effectiveness factor)
W = ρ_{b}A_{c}z
the variation of NO concentration with solids is
$\begin{array}{cc}{\displaystyle \frac{d{C}_{\text{A}b}}{dW}}={\displaystyle \frac{\Omega {k}_{1}^{\prime}{C}_{\text{A}b}}{{\text{A}}_{c}U}}& \text{(E153.1)}\end{array}$
Because NO is present in dilute concentrations (i.e., y_{A0} ≪ 1), we shall take ε ≪ 1 and set A_{c}U = υ_{0}. We integrate Equation (E153.1) using the boundary condition that when W = 0, then C_{Ab} = C_{Ab0}
$\begin{array}{c}\hline X(1{\displaystyle \frac{{C}_{\text{A}b}}{{C}_{\text{A}b0}}})=1\text{exp}({\displaystyle \frac{\Omega {k}_{1}^{\prime}W}{{\upsilon}_{0}}})\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}\hspace{1em}(E153.2)}\end{array}$
where
$\begin{array}{cc}\Omega ={\displaystyle \frac{\eta}{1+\eta {\displaystyle \frac{{k}_{1}^{\prime}{\rho}_{b}}{{k}_{c}{a}_{c}}}}}& \text{(1558)}\end{array}$
Rearranging, we have
$\begin{array}{c}\hline W={\displaystyle \frac{{\upsilon}_{0}}{\Omega {k}_{1}^{\prime}}}\text{ln}{\displaystyle \frac{1}{1X}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}\hspace{1em}(E153.3)}\end{array}$
2. Calculating the internal effectiveness factor for spherical pellets in which a firstorder reaction is occurring, we obtained
$\begin{array}{cc}\eta ={\displaystyle \frac{3}{{\varphi}_{1}^{2}}}({\varphi}_{1}\text{coth}{\varphi}_{1}1)& \text{(1532)}\end{array}$
As a first approximation, we shall neglect any changes in the pellet size resulting from the reactions of NO with the porous carbon. The Thiele modulus for this system is^{5}
^{5} L. K. Chan, A. F. Sarofim, and J. M. Beer, Combust. Flame, 52, 37.
$\begin{array}{cc}{\varphi}_{1}=R\sqrt{{\displaystyle \frac{{k}_{1}^{\prime}\text{}{\rho}_{c}}{{D}_{e}}}}=R\sqrt{{\displaystyle \frac{{k}_{1}^{"}{S}_{a}{\rho}_{c}}{{D}_{e}}}}& \text{(E153.4)}\end{array}$
where
$\begin{array}{ccc}R& =& \text{pellet radius}=3\times {10}^{3}\text{\hspace{0.17em}}\text{m}\hfill \\ {D}_{e}& =& \text{effective diffusivity}\text{}=1.82\times {10}^{8}\text{}{\text{m}}^{2}\text{/s}\hfill \\ {\rho}_{c}& =& \text{catalyst pellet density}=2.8{\text{g/cm}}^{3}=2.8\times {10}^{6}\text{\hspace{0.17em}}{\text{g/m}}^{3}\hfill \\ {k}_{1}^{\u2033}& =& \text{specific reaction rate}\left(\text{m/s}\right)=4.42\times {10}^{10}\text{\hspace{0.17em}}{\text{m}}^{3}{\text{/m}}^{2}\text{/s}\hfill \end{array}$
Substituting in Equation (E153.4)
$\text{}\begin{array}{ccc}{\varphi}_{1}& =& 0.003\text{m}\sqrt{{\displaystyle \frac{(4.42\times {10}^{10}\text{}\text{m/s})(530\text{}{\text{m}}^{2}/\text{g})(2.8\times {10}^{6}\text{}{\text{g/m}}^{3})}{1.82\times {10}^{8}\text{}{\text{m}}^{2}/\text{s}}}}\hfill \end{array}$
$\begin{array}{ccc}\hline \varphi & =& 18\hfill \\ \hline\end{array}$
Because ϕ_{1} is large
$\begin{array}{c}\hline \eta \cong {\displaystyle \frac{3}{{\varphi}_{1}}}={\displaystyle \frac{3}{18}}=0.167\\ \hline\end{array}$
3. Calculate WP
$\begin{array}{ccc}\hfill \text{WP}& =\hfill & \eta {\varphi}_{1}^{2}\hfill \\ \text{}\hfill & =\hfill & \left(0.167\right){\left(18\right)}^{2}=54.\hfill \end{array}$
WP >> 1 and internal diffusion limits the reaction
4. Calculate the overall effectiveness factor (a) To calculate we first need to calculate the external mass transfer coefficient, we will use the Thoenes–Kramers correlation. From Chapter 14 we recall
(a) To calculate Ω we first need to calculate the external mass transfer coefficient, we will use the Thoenes–Kramers correlation. From Chapter 14 we recall
$\begin{array}{cc}\text{Sh}\prime ={(\text{Re}\prime )}^{1/2}{\text{Sc}}^{1/3}& \text{(1476)}\end{array}$
For a 2in.ID pipe, A_{c} = 2.03 × 10^{–3} m^{2}. The superficial velocity is
$U={\displaystyle \frac{{\upsilon}_{0}}{{\text{A}}_{c}}}={\displaystyle \frac{{10}^{6}\text{}\text{}{\text{m}}^{3}\text{/}s}{2.03\times {10}^{3}{\text{m}}^{2}}}=4.93\times {10}^{4}\text{m/s}$
Procedure
Calculate
Re′
Sc
Then
Sh′
Then
k_{c}
Assuming a shape factor of 1.0
$\begin{array}{c}\hline \text{Re}\prime ={\displaystyle \frac{U{d}_{p}}{(1\varphi )\text{\hspace{0.17em}}\nu}}={\displaystyle \frac{(4.93\times {10}^{4}\text{m/s})(6\times {10}^{3}\text{m})}{(10.5)(1.53\times {10}^{8}{\text{m}}^{2}\text{/s})}}=386.7\\ \hline\end{array}$
$\begin{array}{ccc}\hline \text{Nomenclaturenote:}& \varphi \text{withsubscript1,}{\varphi}_{1}& =\text{Thielemodulus}\\ \text{}& \varphi \text{withoutsubscript,}\varphi & =\text{porosity}\hfill \\ \hline\end{array}$
$\begin{array}{ccc}SC& =& {\displaystyle \frac{\nu}{{D}_{\text{AB}}}}={\displaystyle \frac{1.53\times {10}^{8}\text{}\text{}{\text{m}}^{2}\text{/S}}{2.0\times {10}^{8}\text{}{\text{m}}^{2}\text{/S}}}=0.765\hfill \\ \text{Sh}\prime & =& {\left(0.386\right)}^{1/2}{\left(0.765\right)}^{1/3}=\left(19.7\right)\left(0.915\right)=18.0\hfill \\ \begin{array}{c}\begin{array}{c}{k}_{c}\end{array}\\ \text{}\end{array}& =& {\displaystyle \frac{1\varphi}{\varphi}}\left({\displaystyle \frac{{D}_{\text{AB}}}{{d}_{p}}}\right)\text{Sh}\prime ={\displaystyle \frac{0.5}{0.5}}\left({\displaystyle \frac{2.0\times {10}^{8}\text{}\left({\text{m}}^{2}\text{/}\text{s}\right)}{6.0\times {10}^{3}\text{}\text{m}}}\right)\left(18.0\right)\hfill \\ {k}_{c}& =& 6\times {10}^{5}\text{}\text{m/s}\hfill \end{array}$
(b) Calculating the external area per unit reactor volume, we obtain
$\begin{array}{cc}\begin{array}{ccc}{a}_{c}& =& {\displaystyle \frac{6(1\varphi )}{{d}_{p}}}={\displaystyle \frac{6(10.5)}{6\times {10}^{3}\text{}\text{m}}}\\ \text{}& =& 500\text{}{\text{m}}^{2}{\text{/m}}^{3}\hfill \end{array}& \text{(E153.5)}\end{array}$
This example is long and detailed. Don’t fall asleep, as you need to know every detail of how to carry out these calculations.
5. Evaluating the overall effectiveness factor. Substituting into Equation (1558), we have
$\begin{array}{ccc}\Omega & =& {\displaystyle \frac{\eta}{1+\eta {k}_{1}^{\u2033}\text{}\text{}{S}_{a}{\rho}_{b}\text{/}{k}_{c}{a}_{c}}}\hfill \\ \Omega & =& {\displaystyle \frac{0.167}{1+{\displaystyle \frac{\left(0.167\right)(4.4\times {10}^{10}\text{}{\text{m}}^{3}{\text{/m}}^{2}\cdot \text{s})\left(530{\text{m}}^{2}\text{/g}\right)(1.4\times {10}^{6}{\text{g/m}}^{3})}{\left(\right(6\times {10}^{5}\left)\text{}\text{}\text{m/s}\right)(500\text{}{\text{m}}^{2}/{\text{m}}^{3})\text{}}}}}\end{array}$
$\begin{array}{ccc}\hline \Omega & =& {\displaystyle \frac{0.167}{1+1.83}}=0.059\hfill \\ \hline\end{array}$
In this example we see that both the external and internal resistances to mass transfer are significant.
6. Calculate the Mears criterion, MR, to see if mass transfer limits the reaction.
$\begin{array}{ccc}\text{MR}& =& {\displaystyle \frac{{r}_{\text{A}}^{\prime}\text{}{\rho}_{b}Rn}{{k}_{c}{C}_{\text{A}b}}}={\displaystyle \frac{\Omega ({r}_{\text{A}b})\mathit{Rn}}{{k}_{c}{C}_{\text{A}b}}}\hfill \end{array}$
$\begin{array}{cc}=& {\displaystyle \frac{\begin{array}{c}\Omega {k}_{1}^{\prime}{C}_{\text{A}b}{\rho}_{b}R\text{}\end{array}}{{k}_{c}{C}_{\text{A}b}}}={\displaystyle \frac{\Omega {k}_{1}^{\prime}{\rho}_{b}R}{{k}_{c}}}\hfill \end{array}$
$\begin{array}{cc}=& {\displaystyle \frac{\left(0.059\right)(2.34\times {10}^{7}{\text{m}}^{3}\text{/g/}\text{s})(1.4\times {10}^{6}{\displaystyle \frac{\text{g}}{{\text{m}}^{3}}})(3\times {10}^{3}\text{m})}{6\times {10}^{5}\text{m/s}}}\end{array}$
$\begin{array}{ccc}\hline \text{MR}& =& 0.97\hfill \\ \hline\end{array}$
MR (i.e., 0.97 > 0.15) and diffusion limits the reaction.
7. (a) Calculating the weight of solid catalyst, W, necessary to achieve 99.8% conversion. Substituting into Equation (E153.3), we obtain
$\begin{array}{ccc}W& =& {\displaystyle \frac{1\times {10}^{6}{\text{m}}^{3}\text{/s}}{\left(0.059\right)(4.42\times {10}^{10}{\text{m}}^{3}{\text{/m}}^{2}\cdot \text{s})(530{\text{m}}^{2}/\text{g})}}\text{ln}{\displaystyle \frac{1}{10.998}}\end{array}$
$\begin{array}{ccc}\hline W& =& 450\text{}\text{g}\hfill \\ \hline\end{array}$
(b) The reactor length, L, is
This catalyst weight and corresponding reactor length are rather small and as such we could easily increase the feed rate to the reactor.
$\begin{array}{ccc}L& =& {\displaystyle \frac{W}{{\text{A}}_{c}{\rho}_{b}}}={\displaystyle \frac{450\text{g}}{(2.03\times {10}^{3}{\text{m}}^{2})(1.4\times {10}^{6}{\text{g/m}}^{3})}}\end{array}$
$\begin{array}{ccc}\hline L& =& 0.16\text{m}\hfill \\ \hline\end{array}$
Analysis: One of the purposes of this example was to show how to carry out detailed calculations to size a reactor (i.e., calculate z or V or W) to achieve a specified conversion, when both external and internal diffusion resistances affect the rate of reaction. These calculations are tedious and detailed, and it was my feeling that we should show and know all the intermediate calculations, for example, a_{c}, η and Ω, so that the reader will have a better understanding of how to make such calculations in the future.
For external mass transfer–limited reactions in packed beds, the rate of reaction per unit mass of catalyst at a point in the bed is
$\begin{array}{cc}{r}_{\text{A}}^{\prime}={k}_{c}{a}_{c}{C}_{\text{A}}& \text{(1572)}\end{array}$
Variation of reaction rate with system variables
The correlation for the mass transfer coefficient, Equation (1477), shows that k_{c} is directly proportional to the square root of the velocity and inversely proportional to the square root of the particle diameter
$\begin{array}{c}\hline {k}_{c}\text{}\text{\u221d}{\displaystyle \frac{{U}^{1/2}}{{d}_{p}^{1/2}}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1573)}\end{array}$
We recall from Equation (E153.5), a_{c} = 6(1 – ϕ)/d_{p}, that the variation of external surface area with catalyst particle size is
${a}_{c}\text{\u221d}{\displaystyle \frac{1}{{d}_{p}}}$
We now combine Equations (1572) and (1573) to obtain
$\begin{array}{c}\hline {r}_{\text{A}\text{}}^{\prime}\text{\u221d}{\displaystyle \frac{{U}^{1/2}}{{d}_{p}^{3/2}}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(1574)}\end{array}$
Consequently, for external mass transfer–limited reactions, the rate is proportional to the velocity to the onehalf power and is inversely proportional to the particle diameter to the threehalves power.
From Equation (1483), we see that for gasphase external mass transfer– limited reactions, the rate increases approximately linearly with temperature.
Many heterogeneous reactions are diffusion limited.
When internal diffusion limits the rate of reaction, we observe from Equation (1542) that the rate of reaction varies inversely with particle diameter, is independent of velocity, and exhibits an exponential temperature dependence that is not as strong as that for surfacereactioncontrolling reactions. For surfacereactionlimited reactions, the rate is independent of particle size and is a strong function of temperature (exponential). Table 151 summarizes the dependence of the rate of reaction on the velocity through the bed, particle diameter, and temperature for the three types of limitations that we have been discussing.
Very important table
TABLE 151 LIMITING CONDITIONS
Type of Limitation 
Variation of Reaction Rate with: 

Velocity 
Particle Size 
Temperature 

External diffusion 
U^{1/2} 
(d_{p})^{−3/2} 
≈ Linear 
Internal diffusion 
Independent 
(d_{p})^{−1} 
Exponential 
Surface reaction 
Independent 
Independent 
Exponential 
The exponential temperature dependence for internal diffusion limitations is usually not as strong a function of temperature as is the dependence for surface reaction limitations (cf. Section 15.4). If we would calculate an activation energy between 8 and 24 kJ/mol, chances are that the reaction would be strongly diffusionlimited. An activation energy of 200 kJ/mol, however, suggests that the reaction is surface reaction rate–limited.
Multiphase reactors are reactors in which two or more phases are necessary to carry out the reaction. The majority of multiphase reactors involve gas and liquid phases that contact a solid. In the case of the slurry and trickle bed reactors, the reaction between the gas and the liquid takes place on a solid catalyst surface (see Table 152). However, in some reactors the liquid phase is an inert medium for the gas to contact the solid catalyst. The latter situation arises when a large heat sink is required for highly exothermic reactions. In many cases, the catalyst life is extended by these milder operating conditions.
The multiphase reactors discussed in the previous editions of this book are the slurry reactor, fluidized bed, and the trickle bed reactor. The material for the slurry reactors, trickle bed reactors, and fluidized bed reactors described below has been typeset for easy reading on the Web and can be easily printed out from the Professional Reference Shelf for this Chapter 15 (http://www.umich.edu/~elements/6e/15chap/prof.html). The trickle bed reactor, which has reaction and transport steps similar to the slurry reactor, is discussed in the first edition of this book and on the CRE Web site along with the bubbling fluidized bed. In slurry reactors, the catalyst is suspended in the liquid, and gas is bubbled through the liquid. A slurry reactor may be operated in either a semibatch or continuous mode.
TABLE 152 APPLICATIONS OF THREEPHASE REACTORS

Source: Satterfield, C. N. AIChE Journal., 21, 209 (1975); P. A. Ramachandran and R. V. Chaudhari, Chem. Eng., 87(24), 74 (1980); R. V. Chaudhari and P. A. Ramachandran, AIChE Journal., 26, 177 (1980). With permission of the American Institute of Chemical Engineers. Copyright © 1980 AIChE. All rights reserved.
A complete description of the slurry reactor and the transport and reaction steps is given on the CRE Web site, along with the design equations and a number of examples. Methods to determine which of the transport and reaction steps are rate limiting are included. See Professional Reference Shelf R12.1 (http://www.umich.edu/~elements/6e/15chap/pdf/slurry.pdf) for the complete text.
The CRE Web site (in Additional Material → Expanded Material) includes all the material on trickle bed reactors from the first edition of this book. A comprehensive example problem for trickle bed reactor design is included. See Professional Reference Shelf R15.2 (http://www.umich.edu/~elements/6e/15chap/pdf/trickle.pdf).
The Kunii–Levenspiel model for fluidization is given on the CRE Web site (in Additional Material → Expanded Material) along with a comprehensive example problem. The ratelimiting transport steps are also discussed. See Professional Reference Shelf R15.3 (http://www.umich.edu/~elements/6e/15chap/pdf/FluidizedBed.pdf).
Chemical vapor deposition in boat reactors is discussed and modeled (see in Additional Material → Expanded Material on the CRE Web site). The equations and parameters that affect wafer thickness and shape are derived and analyzed. This material is taken directly from the second edition of this book. See Professional Reference Shelf R15.4 (http://www.umich.edu/~elements/6e/15chap/pdf/CVD.pdf ).
The 2017 April issue of Chemical Engineering^{†} has a great article on safety and on a number of critical thinking and creative thinking questions that should be asked. Socratic questioning is at the heart of critical thinking and we shall use R.W. Paul’s Six Types of Socratic (i.e., Critical Thinking) Questions shown in Table 153 as a basis for our discussion.^{‡} A critical aspect of process safety is “anticipating” what could go wrong in a chemical process and ensuring it won’t go wrong. We are going to use an actual life example to discuss critical thinking.
^{‡} Paul, R. W., Critical Thinking, Foundation for Critical Thinking, Santa Rosa, CA, 1992.
Actual Case History:^{*} A large tank containing ethylene oxide has been insulated and is out in the plant. There is uncertainty as to whether or not corrosion has taken place under the insulation. To strip the insulation and check for corrosion would require shutting the plant down for 3 weeks. Because such a shutdown would affect the supply chain and many customers, the shutdown would be very costly, ca. 5 million dollars.
^{*} Written in conjunction with Joel Young, BASF, Wyandotte, MI.
Let’s apply R. W. Paul’s Six Types of Critical Thinking Questions to this situation to help us decide whether or not to strip the insulation.
TABLE 153 R.W. PAUL’S SIX TYPES OF CRITICAL THINKING QUESTIONS (CTQS) AND EXAMPLES** STORAGE TANK INSULATION
Type of CTQ 
Example Phrases of CTQ 
CTQ Safety Examples 
1. Questions about the question or problem statement: The purpose of this question is to determine why the question was asked, who asked it, and why the question or problem needs to be solved. 

Why do you think I questioned you about corrosion under the insulation, considering the storage tank is only 10 years old? 
2. Questions for clarification: The purpose of this question is to identify missing or unclear information in the problem statement or question. 

Are there industryidentified case histories about corrosion occurring under insulation? 
3. Questions that probe assumptions: The purpose of this question is to identify any misleading or false assumptions. 

How did you assume stripping the insulation is the only method to check for corrosion? 
4. Questions that probe reasons and evidence: The purpose of this question is to explore whether facts and observations support an assertion. · What would be an example? 

What evidence do you have that corrosion may have occurred in this tank in the last 10 years? 
5. Questions that probe viewpoints and perspectives: The purpose of this question is to learn how things are viewed or judged and consider things not only in a relative perspective but also as a whole. · What is a counterargument for ____? 

What are counterarguments for taking all the insulation off and inspecting the tank? 
6. Questions that probe implications and consequences: The purpose of this question is to help understand the inferences or deductions and the end result if the inferred action is carried out. · What are the consequences if that assumption turns out to be false? 

What are consequences of ethylene oxide leaking into the atmosphere on people, equipment, and the environment? 
^{**} See pages 58–59, H. S. Fogler, S. E. LeBlanc, and B. R. Rizzo, Strategies for Creative Problem Solving, 3rd ed. Boston, MA: Prentice Hall, 2014.
HSF favorite types of critical thinking questions are (1) Why do you think I asked this question? (2) What information do we need to answer this question? (3) Could you explain your reasoning for making that choice? (4) Can you give me an example? (5) What is a counterargument for your suggestion? (6) What are the consequences if your assumption is false?
1. The concentration profile for a firstorder reaction occurring in a spherical catalyst pellet is
$\begin{array}{cc}{\displaystyle \frac{{C}_{\text{A}}}{{C}_{\text{A}s}}}={\displaystyle \frac{R}{r}}\left[{\displaystyle \frac{\text{Sinh}\left({\varphi}_{1}r\text{/}R\right)}{\text{sinh}{\varphi}_{1}}}\right]& \text{(S151)}\end{array}$
where ϕ_{1} is the Thiele modulus. For a firstorder reaction
$\begin{array}{cc}{\varphi}_{1}^{2}={\displaystyle \frac{{k}_{1}}{{D}_{e}}}{R}^{2}& \text{(}\text{S152)}\end{array}$
2. The effectiveness factors are
$\begin{array}{ccc}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{Internal}\end{array}\\ \text{effectiveness}\end{array}\end{array}\\ \text{factor}\end{array}& =& \eta ={\displaystyle \frac{\text{Actual rate of reaction}}{\begin{array}{c}\text{Reaction rate if entire interior}\\ \begin{array}{c}\begin{array}{c}\text{surface is exposed to concentration}\end{array}\\ \text{at the external pellet surface}\end{array}\end{array}}}\\ {r}_{\text{A}}& =& \eta ({r}_{\text{A}s})\hfill \\ \begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\text{Overall}\end{array}\\ \text{effectiveness}\end{array}\end{array}\\ \text{factor}\end{array}& =& \Omega ={\displaystyle \frac{\text{Actual rate of reaction}}{\begin{array}{c}\text{Reaction rate if entire surface area}\\ \text{is exposed to bulk concentration}\end{array}}}\\ {r}_{\text{A}}& =& \Omega ({r}_{\text{A}b})\hfill \end{array}$
3. For large values of the Thiele modulus for an n^{th} order reaction
$\begin{array}{cc}\eta ={\left({\displaystyle \frac{2}{n+1}}\right)}^{1/2}{\displaystyle \frac{3}{{\varphi}_{n}}}& \text{(S153)}\end{array}$
4. For internal diffusion control, the true reaction order is related to the measured reaction order by
$\begin{array}{cc}{n}_{\text{true}}={2n}_{\text{apparent}}1& \text{(S154)}\end{array}$
The true and apparent activation energies are related by
$\begin{array}{cc}{E}_{\text{true}}=2{E}_{\text{app}}& \text{(S155)}\end{array}$
5. A. The Weisz–Prater Parameter
$\begin{array}{cc}\text{WP}={\varphi}_{1}^{2}\text{}\eta ={\displaystyle \frac{{r}_{\text{A}}^{\prime}\text{}\left(\text{observed}\right)\text{}{\rho}_{c}{R}^{2}}{{D}_{e}{C}_{\text{A}s}}}& \text{(S156)}\end{array}$
The Weisz–Prater criterion dictates that
$\begin{array}{cc}\text{If}\text{WP \u226a}& \text{no internal diffusion limitations present}\\ \text{IfWP \u226b}& \text{internal diffusion limitations present}\hfill \end{array}$
B. The Mears Criterion for Neglecting External Diffusion and Heat Transfer
The Mears number is
$\begin{array}{c}\hline \text{MR}={\displaystyle \frac{{r}_{\text{A}}^{\prime}\text{}\left(\text{obs}\right){\rho}_{b}R\text{n}}{{k}_{c}{C}_{\text{A}b}}}\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(S157)}\end{array}$
There will be no external diffusion limitations if
$\begin{array}{c}\hline \text{MR}<0.15\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(S158)}\end{array}$
And there will be no temperature gradients if
$\begin{array}{c}\hline \left{\displaystyle \frac{\mathrm{\Delta}{H}_{\text{Rx}}({r}_{\text{A}}^{\prime})\left({\rho}_{b}\mathit{\text{RE}}\right)}{h{T}^{2}{R}_{g}}}\right<0.15\\ \hline\end{array}\begin{array}{c}\text{\hspace{1em}\hspace{1em}(S159)}\end{array}$
• Professional Reference Shelf
R15.1. Slurry Reactors (http://www.umich.edu/~elements/6e/15chap/pdf/slurry.pdf)
Description of the Use of Slurry Reactors
Example R151 Industrial Slurry Reactor
Reaction and Transport Steps in a Slurry Reactor
$\begin{array}{c}\hline {\displaystyle \frac{{C}_{i}}{{R}_{\text{A}}}}={\displaystyle \frac{1}{{k}_{b}{a}_{b}}}+{\displaystyle \frac{1}{\text{m}}}({\displaystyle \frac{\stackrel{{r}_{c}}{\overbrace{1}}}{{k}_{b}{a}_{b}}}+{\displaystyle \frac{\stackrel{{r}_{r}}{\overbrace{1}}}{k\eta}})\\ \hline\end{array}$
Determining the RateLimiting Step
Effect of Loading, Particle Size, and Gas Adsorption
Effect of Shear Example R152 Determining the Controlling Resistance
Slurry Reactor Design
R15.2. Trickle Bed Reactors (http://www.umich.edu/~elements/6e/15chap/proftrickle.html#seca)
Fundamentals
Limiting Situations
Evaluating the Transport Coefficients
R15.3. Fluidized Bed Reactors (http://www.umich.edu/~elements/6e/15chap/proffluidized.html)
Descriptive Behavior of the Kunii–Levenspiel Bubbling Bed Model
Mechanics of Fluidized Beds
Example R154 Maximum Solids HoldUp
Mass Transfer in Fluidized Beds
Reaction in a Fluidized Bed
Solution to the Balance Equations for a FirstOrder Reaction
$\begin{array}{c}\hline W={\displaystyle \frac{{\rho}_{c}{\text{A}}_{c}{u}_{b}(1{\in}_{mf})(1\delta )}{{k}_{\text{cat}}{K}_{R}}}\text{ln}{\displaystyle \frac{1}{1X}}\\ \hline\end{array}$
$\begin{array}{c}\hline {K}_{R}={\gamma}_{b}+{\displaystyle \frac{1}{{\displaystyle \frac{{k}_{\text{cat}}}{{K}_{bc}}}+{\displaystyle \frac{1}{{\gamma}_{c}+{\displaystyle \frac{1}{{\displaystyle \frac{1}{{\gamma}_{e}}}+{\displaystyle \frac{{k}_{\text{cat}}}{{K}_{ce}}}}}}}}}\\ \hline\end{array}$
Example R155 Catalytic Oxidation of Ammonia
Limiting Situations
Example R156 Calculation of the Resistances
Example R157 Effect of Particle Size on Catalyst Weight for a Slow Reaction
Example R158 Effect of Catalyst Weight for a Rapid Reaction
R15.4. Chemical Vapor Deposition Reactors (http://www.umich.edu/~elements/6e/15chap/profcvd.html)
Chemical Reaction Engineering in Microelectronic Processing
Fundamentals of CVD
Effectiveness Factors for Boat Reactors
$\begin{array}{c}\hline \eta ={\displaystyle \frac{{2I}_{1}\left({\varphi}_{1}\right)}{{\varphi}_{1}{I}_{0}\left({\varphi}_{1}\right)}}\\ \hline\end{array}$
Example R159 Diffusion Between Wafers
Example R1510 CVD Boat Reactor
The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.
A = • B = ▪ C = ♦ D = ♦♦
Q151_{A} QBR (Question Before Reading). What are two factors that affect the rate of diffusion inside a porous catalyst pellet the most?
Q152 Make up an original problem using the concepts presented in Section ______ (your instructor will specify the section). Extra credit will be given if you obtain and use real data from the literature. (See Problem P51_{A} for the guidelines.)
Q153 What if you consider an exothermic reaction in which internal diffusion limits the rate of reaction? Can you explain how the internal effectiveness factor could be greater than one, that is, η > 1? Hint: See nonisothermal reactors on Web under Additional Material.
Q154 What if the temperature in the CRE Web site Professional Reference Shelf Example R15.2 were increased? How would the relative resistances in the slurry reactor change?
Q155 What if you were asked for all the things that could go wrong in the operation of a slurry reactor as described in the Professional Reference Shelf for Chapter 15? What would you say?
Q156 What if someone had used the falsified kinetics parameters (i.e., wrong E, wrong n)? Can you explain why one might have a runaway reaction? Under what circumstances would the catalyst weight be overdesigned or underdesigned? What are other positive or negative effects that could occur?
Q157_{A} Go to the LearnChemE screencast link for Chapter 15 (http://www.learncheme.com/screencasts/kineticsreactordesign). View the two screencasts: (1) “Diffusion and Reaction in a Cylindrical Porous Catalyst,” (a) What is the equation for the amount reacting in 1 cm; and (2) “Effectiveness Factor for spherical Catalyst,” (b) Write the effectiveness factor in terms of the Thiele Modulus. For these two screencasts list two of the most important points.
Q158 AWFOSS15 Six Types of Critical Thinking Questions
Write a Critical Thinking Question for each type of CTQ for the Monsanto Incident, Example 132.
Write another question for each CTQ for the case history concerning ethylene oxide in the insulated storage tank.
P151_{B}
Example 151: Finding the Effective Diffusivity of D_{e}
Wolfram
Can you think up a question about the sliders that would be somewhat interesting? I am not sure I can, that’s why there is no specific question here.
Example 153: Reducing Nitrous Oxides in a Plant’s Effluent. This is a Stop and Smell the Roses Simulation.
Wolfram
Which parameter has the larger impact in reducing internal diffusion limitations, pellet radius or internal surface area?
Vary and list at least three parameters which can be changed to achieve approximately 90% conversion for a new flow rate of 10^{–5} m^{3}/s.
Which combination of parameter values causes the surface reaction rate to become the closest to the internal diffusion rate, that is, WP ~ 1?
For the base case, calculate the percent of the total resistance for each of the individual resistances of external diffusion, internal diffusion, and surface reaction.
Qualitatively, how would each of your percentages change if the temperature were increased significantly?
Vary various parameters to see the effect of each parameter on Mears number, MR and write a set of conclusions.
What variable affects the internal effectiveness factor the most?
What variable affects the overall effectiveness factor the most?
What variable affects the Mears criteria (MR) the most?
What variable affects the Weisz–Prater (WP) the most?
Apply the Weisz–Prater criteria to a particle 0.005 m in diameter.
Write a set of conclusions for your experiments (i) through (xi).
Polymath
How would your answers in (i) through (v) change if the reaction were zero order with k_{0} = 9 × 10^{–4} mol/g/s. Hint: What are the new equations for η and Ω?
Overall Effectiveness Factor. Calculate the percent of the total resistance for the resistance of external diffusion, internal diffusion, and surface reaction. Qualitatively, how would each of your percentages change?
What if you applied the Mears and Weisz–Prater criteria to Examples 154 and 153? What would you find? What would you learn if $\mathrm{\Delta}{H}_{\text{Rx}}^{\circ}$ = –25 kcal/mol, h = 100 Btu/h • ft^{2} • °F, and E = 20 kcal/mol?
What if your internal surface area decreased with time because of sintering (see Section 10.7). Describe how your effectiveness factor would change and the rate of reaction change with time if k_{d} = 0.01 h^{–1} and η = 0.01 at t = 0? Explain, being as quantitative as possible when you can.
What if you were to assume that the resistance to gas absorption in the CRE Web site Professional Reference Shelf R15.1 was the same as in Professional Reference Shelf R15.3 and that the liquidphase reactor volume in Professional Reference Shelf R15.3 was 50% of the total? Could you determine the limiting resistance? If so, what is it? What other things could you calculate in Professional Reference Shelf R15.1 (e.g., selectivity, conversion, molar flow rates in and out)? Hint: Some of the other reactions that occur include
$\begin{array}{ccc}\text{CO}+{\text{3H}}_{2}& \to & {\text{CH}}_{4}+{\text{H}}_{2}\text{O}\\ {\text{H}}_{2}\text{O}+\text{CO}& \to & {\text{CO}}_{2}+{\text{H}}_{2}\end{array}$
P152_{B} OEQ (Old Exam Question). Concept problem:
The catalytic reaction
A → B
takes place within a fixed bed containing spherical porous catalyst X22. Figure P152_{B} shows the overall rates of reaction at a point in the reactor as a function of temperature for various entering total molar flow rates, F_{T}_{0}.
Is the reaction limited by external diffusion?
If your answer to part (a) was “yes,” under what conditions of those shown (i.e., T, F_{T}_{0}) is the reaction limited by external diffusion?
Is the reaction “reactionratelimited”?
If your answer to part (c) was “yes,” under what conditions of those shown (i.e., T, F_{T}_{0}) is the reaction limited by the rate of the surface reactions?
Is the reaction limited by internal diffusion?
If your answer to part (e) was “yes,” under what conditions of those shown (i.e., T, F_{T}_{0}) is the reaction limited by the rate of internal diffusion?
For a flow rate of 10 g mol/h, determine (if possible) the overall effectiveness factor, Ω = η, at 360 K.
Estimate (if possible) the internal effectiveness factor, η, at 367 K. (Ans: η = 0.86)
If the concentration at the external catalyst surface is 0.01 mol/dm^{3}, calculate (if possible) the concentration at r = R/2 inside the porous catalyst at 367 K. (Assume a firstorder reaction.)
Additional information:
Gas properties: 
Bed properties: 
Diffusivity: 0.1 cm^{2}/s 
Tortuosity of pellet: 1.414 
Density: 0.001 g/cm^{3} 
Bed permeability: 1 millidarcy 
Viscosity: 0.0001 g/cm·s 
Porosity = 0.3 
P153_{B} OEQ (Old Exam Question). Concept problem: The reaction
A → B
is carried out in a differential packedbed reactor at different temperatures, flow rates, and particle sizes. The results shown in Figure P153_{B} were obtained.
What regions (i.e., conditions d_{p}, T, F_{T}_{0}) are external mass transfer–limited?
What regions are reaction rate–limited?
What region is internaldiffusioncontrolled?
What is the internal effectiveness factor at T = 400 K and d_{p} = 0.8 cm? (Ans: η = 0.625)
P154_{A} OEQ (Old Exam Question). Concept problem: Curves A, B, and C in Figure P154_{A} show the variations in reaction rate for three different reactions catalyzed by solid catalyst pellets. What can you say about each reaction?
P155_{B} OEQ (Old Exam Question). A firstorder heterogeneous irreversible reaction is taking place within a spherical catalyst pellet that is plated with platinum throughout the pellet (see Figure 153). The reactant concentration halfway between the external surface and the center of the pellet (i.e., r = R/2) is equal to onetenth the concentration of the pellet’s external surface. The concentration at the external surface is 0.001 g mol/dm^{3}, the diameter (2R) is 2 × 10^{−3} cm, and the diffusion coefficient is 0.1 cm^{2}/s.
A → B
What is the concentration of reactant at a distance of 3 × 10^{–4} cm in from the external pellet surface? (Ans: C_{A} = 2.36 × 10^{–4} mol/dm^{3})
To what diameter should the pellet be reduced if the effectiveness factor is to be 0.8? (Ans: d_{p} = 6.8 × 10^{–4} cm. Critique this answer!)
If the catalyst support were not yet plated with platinum, how would you suggest that the catalyst support be plated after it had been reduced by grinding?
P156_{B} The swimming rate of a small organism (J. Theoret. Biol., 26, 11 (1970)) is related to the energy released by the hydrolysis of adenosine triphosphate (ATP) to adenosine diphosphate (ADP). The rate of hydrolysis is equal to the rate of diffusion of ATP from the midpiece to the tail (see Figure P156_{B}). The diffusion coefficient of ATP in the midpiece and tail is 3.6 × 10^{–6} cm^{2}/s. ADP is converted to ATP in the midsection, where its concentration is 4.36 × 10^{–5} mol/cm^{3}. The crosssectional area of the tail is 3 × 10^{–10} cm^{2}.
Derive an equation for diffusion and reaction in the tail.
Derive an equation for the effectiveness factor in the tail.
Taking the reaction in the tail to be of zero order, calculate the length of the tail. The rate of reaction in the tail is 23 × 10^{–18} mol/s.
Compare your answer with the average tail length of 41 μm. What are possible sources of error?
P157_{B} A firstorder, heterogeneous, irreversible reaction is taking place within a catalyst pore that is plated with platinum entirely along the length of the pore (Figure P157_{B}). The reactant concentration at the plane of symmetry (i.e., equal distance from the pore mouth) of the pore is equal to onetenth the concentration at the pore mouth. The concentration at the pore mouth is 0.001 mol/dm^{3}, the pore length (2L) is 2 × 10^{–3} cm, and the diffusion coefficient is 0.1 cm^{2}/s.
Derive an equation for the effectiveness factor.
What is the concentration of reactant at L/2?
To what length should the pore length be reduced if the effectiveness factor is to be 0.8?
If the catalyst support were not yet plated with platinum, how would you suggest the catalyst support be plated after the pore length, L, had been reduced by grinding?
P158_{B} Extension of Problem P157_{B}. The elementary isomerization reaction
A → B
is taking place on the walls of a cylindrical catalyst pore (see Figure P157_{B}.) In one run, a catalyst poison P entered the reactor together with the reactant A. To estimate the effect of poisoning, we assume that the poison renders the catalyst pore walls near the pore mouth ineffective up to a distance z_{1}, so that no reaction takes place on the walls in this entry region.
Show that before poisoning of the pore occurred, the effectiveness factor was given by
$\eta ={\displaystyle \frac{1}{\varphi}}\text{tanh}\text{}\varphi $
where
$\varphi =L\sqrt{{\displaystyle \frac{2k}{r{D}_{e}}}}$
with
$\begin{array}{ccc}\hfill k& =\hfill & \text{reactionrate constant}\left(\text{lenght/time}\right)\hfill \\ \hfill r& =\hfill & \text{pore radius}\left(\text{length}\right)\hfill \\ \hfill {D}_{e}& =\hfill & \text{effective molecular diffusivity}\left(\text{area/time}\right)\hfill \end{array}$
Derive an expression for the concentration profile and also for the molar flux of A in the ineffective region, 0 < z < z_{1}, in terms of z_{1}, D_{AB}, C_{A1}, and C_{A}_{s}. Without solving any further differential equations, obtain the new effectiveness factor η′ for the poisoned pore.
P159_{A} A firstorder reaction is taking place inside a porous catalyst. Assume dilute concentrations and neglect any variations in the axial (x) direction.
Derive an equation for both the internal and overall effectiveness factors for the rectangular porous slab shown in Figure P159_{A}.
Repeat part (a) for a cylindrical catalyst pellet where the reactants diffuse inward in the radial direction. (Clevel problem, i.e., P159_{C}(b).)
P1510_{B} The irreversible reaction
A → B
is taking place in the same porous catalyst slab shown in Figure P159_{A}.
The reaction is zero order in A.
Show that the concentration profile using the symmetry B.C. is
$\begin{array}{cc}{\displaystyle \frac{{C}_{\text{A}}}{{C}_{\text{A}s}}}=1+{\varphi}_{0}^{2}[{\left({\displaystyle \frac{z}{L}}\right)}^{2}1]& \text{(P1510.1)}\end{array}$
where
$\begin{array}{cc}{\varphi}_{0}^{2}={\displaystyle \frac{{kL}^{2}}{{2D}_{e}{C}_{\text{A}s}}}& \text{(P1510.2)}\end{array}$
For a Thiele modulus of 1.0, at what point in the slab is the concentration zero? For ϕ_{0} = 4?
What is the concentration you calculate at z = 0.1 L and ϕ_{0} = 10 using Equation (P1510.1)? What do you conclude about using this equation?
Plot the dimensionless concentration profile ψ = C_{A}/C_{A}_{s} as a function of λ = z/L for ϕ_{0} = 0.5, 1, 5 and 10. Hint: there are regions where the concentration is zero. Show that λ_{C} = (1 – 1/ϕ_{0}) is the start of this region where the gradient and concentration are both zero. (L. K. Jang, R. L. York, J. Chin, and L. R. Hile, Inst. Chem. Engr., 34, 319 (2003).)
Show that $\psi ={\varphi}_{0}^{2}{\lambda}^{2}2{\varphi}_{0}({\varphi}_{0}1)\lambda +({\varphi}_{0}1{)}^{2}$ for ${\lambda}_{\text{C}}\le \lambda <1$.
The effectiveness factor can be written as
$\begin{array}{cc}\eta ={\displaystyle \frac{{\int}_{0}^{L}\text{}{r}_{\text{A}}{\text{A}}_{\text{}c}\text{}dz}{{r}_{\text{A}s}{\text{A}}_{c}L}}={\displaystyle \frac{{\int}_{0}^{{z}_{c}}{r}_{\text{A}}{\text{A}}_{c}\text{}dz+{\int}_{{z}_{c}}^{L}{r}_{\text{A}}{\text{A}}_{c}\text{}dz}{{r}_{\text{A}s}{\text{A}}_{\text{}c}L}}& \text{(P150.3)}\end{array}$
where z_{C} (λ_{C}) is the point at which both the concentration gradients and flux go to zero, and A_{c} is the crosssectional area of the slab. Show for a zeroorder reaction that
Make a sketch for η versus ϕ_{0} similar to the one shown in Figure 155.
Repeat parts (a) through (f) for a spherical catalyst pellet.
What do you believe to be the point of this problem?
P1511_{C} The secondorder decomposition reaction
A → B + 2C
is carried out in a tubular reactor packed with catalyst pellets 0.4 cm in diameter. The reaction is internaldiffusionlimited. Pure A enters the reactor at a superficial velocity of 3 m/s, a temperature of 250°C, and a pressure of 500 kPa. Experiments carried out on smaller pellets where surface reaction is limiting yielded a specific reaction rate of 0.05 m^{6}/mol·gcat·s. Calculate the length of bed necessary to achieve 80% conversion. Critique the numerical answer. (Ans: L = 2.8 × 10^{–5} m)
Additional information:
Effective diffusivity: 2.66 × 10^{–8} m^{2}/s 
Pellet density: 2 × 10^{6} g/m^{3} 
Ineffective diffusivity: 0.00 m^{2}/s 
Internal surface area: 400 m^{2}/g 
Bed porosity: 0.4 
P1512_{C} Derive the concentration profile and effectiveness factor for cylindrical pellets 0.2 cm in diameter and 1.5 cm in length. Neglect diffusion through the ends of the pellet.
Assume that the reaction is a firstorder isomerization. Hint: Look for a Bessel function.
Rework Problem P1511_{C} for these pellets.
P1513_{C} OEQ (Old Exam Question). Falsified Kinetics. The irreversible gasphase dimerization
2A → A_{2}
is carried out at 8.2 atm in a stirred containedsolids reactor to which only pure A is fed. There are 40 g of catalyst in each of the four spinning baskets. The following runs were carried out at 227°C:
Total Molar Feed Rate, F_{T0} (g mol/min) 
1 
2 
4 
6 
11 
20 
Mole Fraction A in Exit, y_{A} 
0.21 
0.33 
0.40 
0.57 
0.70 
0.81 
The following experiment was carried out at 237°C:
$\begin{array}{cc}{F}_{\text{T0}}=9\text{g}\text{mol/min}& {y}_{\text{A}}=0.097\end{array}$
What are the apparent reaction order and the apparent activation energy?
Determine the true reaction order, specific reaction rate, and activation energy.
Calculate the Thiele modulus and effectiveness factor.
What pellet diameter should be used to make the catalyst more effective?
Calculate the rate of reaction on a rotating disk made of the catalytic material when the gasphase reactant concentration is 0.01 g mol/L and the temperature is 227°C. The disk is flat, nonporous, and 5 cm in diameter.
Additional information:
Effective diffusivity: 0.23 cm^{2}/s 
Radius of catalyst pellets: 1 cm 
Surface area of porous catalyst: 49 m^{2}/gcat 
Color of pellets: blushing peach 
Density of catalyst pellets: 2.3 g/cm^{3} 

P1514_{B} Derive Equation (1539). Hint: Multiply both sides of Equation (1525) for nth order reaction; that is,
$\frac{{\mathit{d}}^{\mathbf{2}}\mathit{y}}{{\mathit{d}\mathit{\lambda}}^{\mathbf{2}}}}\mathbf{}{\mathit{\varphi}}_{n}^{\mathbf{2}}{\mathit{y}}^{\text{n}}\mathbf{=}0$
by 2dy/dλ, rearrange to get
$\frac{d}{d\lambda}}{\left({\displaystyle \frac{dy}{d\lambda}}\right)}^{2}={\varphi}_{n}^{2}{y}^{n}2{\displaystyle \frac{dy}{d\lambda}$
and solve using the boundary conditions dy/dλ = 0 at λ = 0.
P1515_{C} You will need to read about slurry reactions on the Web site’s Additional Material. The following table was obtained from the data taken in a slurry reactor for the hydrogenation of methyl linoleate to form methyl oleate.
$\stackrel{\begin{array}{cc}\mathrm{L}+{\mathrm{H}}_{2}& \begin{array}{cc}\begin{array}{c}\to \end{array}& \text{O}\end{array}\end{array}}{\begin{array}{ccc}S& =& {\text{solubilityofH}}_{2}\text{}{\text{intheliquidmixture,mol/dm}}^{3}\hfill \\ \text{m}& =& {\text{catalystcharge,g/dm}}^{3}\hfill \\ {r}_{L}^{\prime}& =& {\text{rate ofreactionofmethyllinoleate,mol/dm}}^{3}\u2022\text{min}\hfill \end{array}}$
Catalyst Size 
$S\text{/}{r}_{L}^{\prime}$ (min) 
1/m (dm^{3}/g) 
A 
4.2 
0.01 
A 
7.5 
0.02 
B 
1.5 
0.01 
B 
2.5 
0.03 
B 
3.0 
0.04 
Which catalyst size has the smaller effectiveness factor?
If catalyst size A is to be used in the reactor at a concentration of 50 g/dm^{3}, would a significant increase in the reaction be obtained if a more efficient gas sparger were used?
If catalyst size B is to be used, what is the minimum catalyst charge that should be used to ensure that the combined diffusional resistances for the pellet are less than 50% of the total resistance?
P1516_{C} You will need to read about slurry reactions on the Web site’s Additional Material. The catalytic hydrogenation of methyl linoleate to methyl oleate was carried out in a laboratoryscale slurry reactor in which hydrogen gas was bubbled up through the liquid containing spherical catalyst pellets. The pellet density is 2 g/cm^{3}. The following experiments were carried out at 25°C:
Run 
Partial Pressure of H_{2} (atm) 
Solubility of H_{2} (g mol/dm^{3}) 
H_{2} Rate of Reaction (g mol/dm^{3} • min) 
Catalyst Charge (g/dm^{3}) 
Catalyst Particle Size (μm) 
1 
3 
0.007 
0.014 
3.0 
12 
2 
18 
0.042 
0.014 
0.5 
50 
3 
3 
0.007 
0.007 
1.5 
50 
It has been suggested that the overall reaction rate can be enhanced by increasing the agitation, decreasing the particle size, and installing a more efficient sparger. With which, if any, of these recommendations do you agree? Are there other ways that the overall rate of reaction might be increased? Support your decisions with calculations.
Is it possible to determine the effectiveness factor from the data above? If so, what is it?
For economical reasons concerning the entrainment of the small solid catalyst particles in the liquid, it is proposed to use particles an order of magnitude larger. The following data were obtained from these particles at 25°C:
Run 
Partial Pressure of H_{2} (atm) 
Solubility of H_{2} (g mol/dm^{3}) 
H_{2} Rate of Reaction (g mol/dm^{3} • min) 
Catalyst Charge (g/dm^{3}) 
Catalyst Particle Size (μm) 
4 
3 
0.007 
0.00233 
2.0 
750 
The Thiele modulus is 9.0 for the 750μm particle size in run 4. Determine (if possible) the external mass transfer coefficient, k_{c}, and the percent (of the overall) of the external mass transfer resistance to the catalyst pellet.
P1517_{B} Applications of Diffusion and Reaction to Tissue Engineering. The equations describing diffusion and reaction in porous catalysts also can be used to derive rates of tissue growth and have been studied by Professor Kristi Anseth and her students at the University of Colorado. One important area of tissue growth is in cartilage tissue in joints such as the knee. Over 200,000 patients per year receive knee joint replacements. Alternative strategies include the growth of cartilage to repair the damaged knee.
One approach is to deliver cartilageforming cells in a hydrogel to the damaged area such as the one shown in Figure WP151.1 in additional material on the CRE Web site.
Here, the patient’s own cells are obtained from a biopsy and embedded in a hydrogel, which is a crosslinked polymer network that is swollen in water. In order for the cells to survive and grow new tissue, many properties of the gel must be tuned to allow diffusion of important species in and out (e.g., nutrients in and cellsecreted extracellular molecules such as collagen out). Because there is no blood flow through the cartilage, oxygen transport to the cartilage cells is primarily by diffusion. Consequently, the design must be such that the gel can maintain the necessary rates of diffusion of nutrients (e.g., O_{2}) into the hydrogel. These rates of exchange in the gel depend on the geometry and the thickness of the gel. To illustrate the application of chemical reaction engineering principles to tissue engineering, we will examine the diffusion and consumption of one of the nutrients, oxygen.
Our examination of diffusion and reaction in catalyst pellets showed that in many cases the reactant concentration near the center of the particle was virtually zero. If this condition were to occur in a hydrogel, the cells at the center would die. Consequently, the gel thickness needs to be designed to allow rapid transport of oxygen.
Let’s consider the simple gel geometry shown in Figure P1517_{B}. We want to find the gel
thickness at which the minimum oxygen consumption rate is 10^{–13} mol/cell/h $(k={\displaystyle \frac{{10}^{3}{\text{mol O}}_{2}}{{\text{dm}}^{3}\text{h}}})$.
The cell density in the gel is 10^{10} cells/dm^{3}, the bulk concentration of oxygen C_{A0} (z = 0) is 2 × 10^{–4} mol/dm^{3}, and the diffusivity, D_{AB}, is 10^{–5} cm^{2}/s.
Show that the dimensionless form of concentration and length, ψ = C_{A}/C_{A0}, and λ = z/L, differential mole balance on O_{2} gives
$\frac{{d}^{2}\psi}{d{\lambda}^{2}}}{\displaystyle \frac{k{L}^{2}}{{D}_{\text{AB}}{C}_{\text{A0}}}}=0$
Show the dimensionless O_{2} concentration profile in the gel is
$\psi ={\displaystyle \frac{{C}_{\text{A}}}{{C}_{\text{A0}}}}={\varphi}_{0}\lambda (\lambda 2)+1$
where
$\begin{array}{ccc}\hfill \lambda & =\hfill & z\text{/L}\hfill \\ \hfill {\varphi}_{0}& =\hfill & \left({\displaystyle \frac{k}{2{D}_{\text{AB}}{C}_{\text{A0}}}}\right){L}^{2}\hfill \end{array}$
Solve the gel thickness when the concentration at z = 0 and C_{A} = 0.1 mmole/dm^{3}.
How would your answers change if the reaction kinetics were (1) first order in the O_{2} concentration with k_{1} = 10^{–2} h^{–1}? (D level of difficulty)
Carry out a quasisteadystate analysis using Equation (WE151.19) from the Web site (http://www.umich.edu/~elements/6e/15chap/expanded_ch15_tissue.pdf), along with the overall balance
$\frac{{{dN}_{\text{}}}_{w}}{dt}}={\upsilon}_{c}{W}_{{O}_{2}}{}_{z=0}{\text{A}}_{c$
to predict the O_{2} flux and collagen buildup as a function of time.
Sketch ψ versus λ at different times.
Sketch λ_{c} as a function of time. Hint: V = A_{c}L. Assume α = 10 and the stoichiometric coefficient for oxygen to collagen, v_{c}, is 0.05 mass fraction of cell/mol O_{2}. A_{c} = 2 cm^{2}.
Section 15.9 through 15.11 have been typeset and can be read and/or printed out from the Web site (http://www.umich.edu/~elements/6e/15chap/prof.html). Homework problems for slurry reactors, trickle bed reactors, fluidized bed reactors, and CVD boat reactors can be found at http://www.umich.edu/~elements/6e/15chap/add.html.
1. There are a number of books that discuss internal diffusion in catalyst pellets; however, one of the first books that should be consulted on this and other topics on heterogeneous catalysis is
L. LAPIDUS AND N. R. AMUNDSON, Chemical Reactor Theory: A Review, Upper Saddle River, NJ: Prentice Hall, 1977.
In addition, see
R. ARIS, Elementary Chemical Reactor Analysis. Upper Saddle River, NJ: Prentice Hall, 1989, Chap. 6. Old, but one should find the references listed at the end of this reading particularly useful.
JOSEPH J. FOGLER, AKA, JOFO, A Chemical Reaction Engineers Guide to the Country of Jofostan. To be self published, hopefully by 2025.
D. LUSS, “Diffusion—Reaction Interactions in Catalyst Pellets,” p. 239 in Chemical Reaction and Reactor Engineering. New York: Marcel Dekker, 1987.
The effects of mass transfer on reactor performance are also discussed in
FRANK C. COLLINS AND GEORGE E. KIMBALL, “Diffusion Controlled Reaction Rates,” Journal of Colloid Science, Vol. 4, Issue 4, August 1949, Pages 425437.
2. Diffusion with homogeneous reaction is discussed in
G. ASTARITA and R. OCONE, Special Topics in Transport Phenomena. New York: Elsevier, 2002.
Gasliquid reactor design is also discussed in
Y. T. SHAH, Gas–Liquid–Solid Reactor Design. New York: McGrawHill, 1979.
3. Modeling of CVD reactors is discussed in
DANIEL DOBKIN AND M. K. ZUKRAW, Principles of Chemical Vapor Deposition. The Netherlands: Kluwer Academic Publishers, 2003.
D. W. HESS, K. F. JENSEN, and T. J. ANDERSON, “Chemical Vapor Deposition: A Chemical Engineering Perspective,” Rev. Chem. Eng., 3, 97, 1985.
4. Multiphase reactors are discussed in
P. A. RAMACHANDRAN and R. V. CHAUDHARI, ThreePhase Catalytic Reactors. New York: Gordon and Breach, 1983.
A. E. RODRIGUES, J. M. COLO, and N. H. SWEED, eds., Multiphase Reactors, Vol. 1: Fundamentals. Alphen aan den Rijn, The Netherlands: Sitjhoff and Noordhoff, 1981.
A. E. RODRIGUES, J. M. COLO, and N. H. SWEED, eds., Multiphase Reactors, Vol. 2: Design Methods. Alphen aan den Rijn, The Netherlands: Sitjhoff and Noordhoff, 1981.
Y. T. SHAH, B. G. KELKAR, S. P. GODBOLE, and W. D. DECKWER, “Design Parameters Estimations for Bubble Column Reactors” (journal review), AIChE J., 28, 353 (1982).
The following Advances in Chemistry Series volume discusses a number of multiphase reactors:
H. S. FOGLER, ed., Chemical Reactors, ACS Symp. Ser. 168. Washington, DC: American Chemical Society, 1981, pp. 3–255.
5. Fluidization
DAIZO KUNII AND OCTAVE LEVENSPIEL, Fluidization Engineering, 2nd ed. (Butterworths Series in Chemical Engineering Deposition). Stoneham, MA: ButterworthHeinemann, 1991.
In addition to Kunii and Levenspiel’s book, many correlations can be found in
J. F. DAVIDSON, R. CLIFF, and D. HARRISON, Fluidization, 2nd ed. Orlando: Academic Press, 1985.
J. G. YATES, Fundamentals of FluidizedBed Chemical Processes, 3rd ed. London: Butterworth, 1983.