Giving up is the ultimate tragedy.
—Robert J. Donovan
or
It ain’t over ’til it’s over.
—Yogi Berra NY Yankees
The Algorithm
1. Mole balance
2. Rate law
3. Stoichiometry
4. Combine
5. Evaluate
The first step in our CRE algorithm is the mole balance, which we now need to extend to include the molar flux, W_{A}_{z}, and diffusional effects. The molar flow rate of A in a given direction, such as the z direction down the length of a tubular reactor, is just the product of the flux, W_{A}_{z} (mol/m^{2} • s), and the crosssectional area, A_{c} (m^{2}); that is,
F_{A}_{z} = A_{c} W_{A}_{z}
In the previous chapters, we have only considered plug flow with no diffusion superimposed, in which case
We now drop the plugflow assumption and extend our discussion of mass transfer in catalytic and other mass transferlimited reactions. In Chapter 10, we focused on the middle three steps (3, 4, and 5) in a catalytic reaction shown in Figure 141 and neglected steps (1), (2), (6), and (7) by assuming the reaction was surfacereaction limited. In this chapter, we describe the first and last steps (1) and (7), as well as showing other applications in which mass transfer plays a role.
Where are we going??^{†}
^{†} “If you don’t know where you are going, you’ll probably wind up some place else.” Yogi Berra, NY Yankees
We want to arrive at the mole balance that incorporates both diffusion and reaction effects, such as Equation (1417) on page 746, that is,
We begin with Section 14.1.1 where we write the mole balance on Species A in three dimensions in terms of the molar flux, W_{A}. In Section 14.1.2 we write W_{A} in terms of the bulk flow of A in the fluid, B_{A} and the diffusion flux J_{A} of A that is superimposed on bulk flow. In Section 14.1.3 we use the previous two subsections as a basis to finally write the molar flux, W_{A}, in terms of concentration using Fick’s first law, J_{A}, and the bulk flow, B_{A}. Next, in Section 14.2 we combine diffusion convective transport and reaction in our mole balance.
Diffusion is the spontaneous intermingling or mixing of atoms or molecules by random thermal motion. It gives rise to motion of the species relative to motion of the mixture. In the absence of other gradients (such as temperature, electric potential, or gravitational potential), molecules of a given species within a single phase will always diffuse from regions of higher concentrations to regions of lower concentrations. This gradient results in a molar flux of the species (e.g., A), W_{A} (moles/area=time), in the direction of the concentration gradient. The flux of A, W_{A}, is relative to a fixed coordinate (e.g., the lab bench) and is a vector quantity with typical units of mol/m^{2} · s. In rectangular coordinates
W_{A} = iW_{A}_{x} + jW_{A}_{y} + kW_{A}_{z}
We now apply the mole balance to species A, which flows and reacts in an element of volume ΔV = ΔxΔyΔz to obtain the variation of the molar fluxes in three dimensions.
F_{Az} = W_{Az}ΔxΔy
F_{Ay} = W_{Ay}ΔxΔz
F_{Ax} = W_{Ax}ΔzΔy
Mole balance
where r_{A} is the rate of generation of A by reaction per unit volume (e.g., mol/m^{3}/h).
Dividing by ΔxΔyΔz and taking the limit as they go to zero, we obtain the molar flux balance in rectangular coordinates
$\begin{array}{c}\hline \begin{array}{c}\begin{array}{c}\frac{\partial {W}_{\text{A}x}}{\partial x}\frac{\partial {W}_{\text{Ay}}}{\partial y}\frac{\partial {W}_{\text{A}z}}{\partial z}+{r}_{\text{A}}=\frac{\partial {C}_{\text{A}}}{\partial t}\end{array}\end{array}\\ \hline\end{array}\begin{array}{c}\left(\text{141}\right)\end{array}$
The corresponding balance in cylindrical coordinates with no variation in the rotation about the zaxis is
COMSOL
$\begin{array}{c}\hline \frac{1}{r}\text{}\frac{\partial}{\partial r}\left(r{W}_{\text{A}r}\right)\frac{\partial {W}_{\text{A}z}}{\partial z}+{r}_{\text{A}}=\frac{\partial {C}_{\text{A}}}{\partial t}\\ \hline\end{array}\begin{array}{c}\left(\text{142}\right)\end{array}$
We will now evaluate the flux terms W_{A}. We have taken the time to derive the molar flux equations in this form because they are now in a form that is consistent with the partial differential equation (PDE) solver COMSOL, which is accessible from the CRE Web site.
The molar flux of A, W_{A}, is the result of two contributions: J_{A}, the molecular diffusion flux relative to the bulk motion of the fluid produced by a concentration gradient, and B_{A}, the flux resulting from the bulk motion of the fluid:
Total flux = diffusion + bulk motion
$\begin{array}{c}\hline {\mathbf{\text{W}}}_{\text{A}}={\mathbf{\text{J}}}_{\text{A}}+{\mathbf{\text{B}}}_{\text{A}}\\ \hline\end{array}\begin{array}{c}\left(\text{143}\right)\end{array}$
The bulkflow term for species A is the total flux of all molecules relative to a fixed coordinate times the mole fraction of A, y_{A}; that is, B_{A} = y_{A} Σ W_{i}.
For a twocomponent system of A diffusing in B, the flux of A is
The diffusional flux, J_{A}, is the flux of A molecules that is superimposed on the bulk flow. It tells how fast A is moving ahead of the bulk flow velocity, that is, the molar average velocity.
The flux of species A, W_{A}, is wrt a fixed coordinate system (e.g., the lab bench) and is just the concentration of A, C_{A}, times the particle velocity of species A, U_{A}, at that point
$\begin{array}{c}{\mathbf{\text{W}}}_{\text{A}}={\mathbf{\text{U}}}_{\text{A}}{C}_{\text{A}}\\ \frac{\text{mol}}{{\text{m}}^{2}\text{S}}=\left(\frac{\text{m}}{\text{s}}\right)\left(\frac{\text{mol}}{{\text{m}}^{3}}\right)\end{array}$
By particle velocities, we mean the vector average of millions of molecules of A at a given point. Similarly for species B: W_{B} = U_{B}C_{B}; substituting into the bulkflow term
B_{A} = y_{A} Σ W_{i} = y_{A} (W_{A} + W_{B}) = y_{A} (C_{A} U_{A} + C_{B}U_{B}
Writing the concentration of A and B in the generic form in terms of the mole fraction, y_{i}, and the total concentration, c, that is, C_{i} = y_{i}c, and then factoring out the total concentration, c, the bulk flow, B_{A}, is
B_{A} = (c y_{A})(y_{A} U_{A} + y_{B} U_{B}) = C_{A} U
Molar average velocity U
where U is the molar average velocity: U = Σy_{i} U_{i}. The molar flux of A can now be written as
$\begin{array}{c}\hline {\mathbf{\text{W}}}_{\text{A}}={\text{J}}_{\text{A}}+\text{}{C}_{\text{A}}\mathbf{\text{U}}\\ \hline\end{array}\begin{array}{c}\left(\text{145}\right)\end{array}$
We now need to determine the equation for the molar flux of A, J_{A}, that is superimposed on the molar average velocity.
Experimentation with frog legs led to Fick’s first law.
Our discussion on diffusion will be restricted primarily to binary systems containing only species A and B. We now wish to determine how the molar diffusive flux of a species (i.e., J_{A}) is related to its concentration gradient. As an aid in the discussion of the transport law that is ordinarily used to describe diffusion, recall similar laws from other transport processes. For example, in conductive heat transfer the constitutive equation relating the heat flux q and the temperature gradient is Fourier’s law, q = −k_{t} ∇ T, where k_{t} is the thermal conductivity.
Constitutive equations in heat, momentum, and mass transfer
In rectangular coordinates, the gradient is in the form
$\mathrm{\nabla}=i\frac{\partial}{\partial x}+j\frac{\partial}{\partial y}+k\frac{\partial}{\partial z}$
The mass transfer law for the diffusional flux of A resulting from a concentration gradient is analogous to Fourier’s law for heat transfer and is given by Fick’s first law^{†}
^{†} Adolf Fick was an interesting character as evidenced by his use of frogs to study diffusion.
$\begin{array}{cc}{{\mathbf{\text{J}}}_{\text{A}}={D}_{\text{AB}}\text{}\mathrm{\nabla}\text{}C}_{\text{A}}& \left(\text{146}\right)\end{array}$
D_{AB} is the diffusivity of A in B $\left(\frac{{\text{m}}^{2}}{\text{s}}\right)$. Combining Equations (145) and (146), we obtain an expression for the molar flux of A in terms of concentration for constant total concentration
Molar flux equation
$\begin{array}{c}\hline {\mathbf{\text{W}}}_{\text{A}}={D}_{\text{AB}}\text{}\mathrm{\nabla}\text{}{C}_{\text{A}}+{C}_{\text{A}}\mathbf{\text{U}}\\ \hline\end{array}\begin{array}{c}\left(\text{147}\right)\end{array}$
In one dimension, that is, z, the molar flux term is
$\begin{array}{cc}{{\mathbf{\text{W}}}_{\text{A}z}=D}_{\text{AB}}\frac{{dC}_{\text{A}}}{dz}+{C}_{\text{A}}{\mathbf{\text{U}}}_{\text{Z}}& \left(\text{148a}\right)\end{array}$
Where U_{z} is the axial velocity, in radial coordinates with no variations in the angular (θ) direction
$\begin{array}{cc}{{\mathbf{\text{W}}}_{\text{A}r}=D}_{\text{AB}}\frac{{dC}_{\text{A}}}{dz}+{C}_{\text{A}}{\mathbf{\text{U}}}_{r}& \left(\text{148b}\right)\end{array}$
Where U_{r} is the fluid’s radial velocity
Although many systems involve more than two components, the diffusion of each species can be treated as if it were diffusing through another single species rather than through a mixture by defining an effective diffusivity.
Now the task is to evaluate the bulkflow term.
We now consider A diffusing in B. Substituting Equation (146) into Equation (144) we obtain
$\begin{array}{cc}{\mathbf{\text{W}}}_{\text{A}}={D}_{\text{AB}}\text{}\text{}\nabla {C}_{\text{A}}+{y}_{\text{A}}({\mathbf{\text{W}}}_{\text{A}}+{\mathbf{\text{W}}}_{\text{B}})& \left(\text{149}\right)\end{array}$
Before going to Section 14.2.2 it would be helpful to evaluate the bulk flow term (y_{A} (W_{A} + W_{B})) for five limiting situations. These situations are given by Equations 1410 through 1413 in Table 141 on page 745.
TABLE 141 EVALUATING W_{A} FOR SPECIES A DIFFUSING IN SPECIES B

^{1} C. N. Satterfield, Mass Transfer in Heterogeneous Catalysis, Cambridge: MIT Press, 1970, pp. 41–42, discusses Knudsen flow in catalysis and gives the expression for calculating D_{K}.
When accounting for diffusional effects, the molar flow rate of species A, F_{A}, in a specific direction z, is the product of molar flux in that direction, W_{A}_{z}, and the crosssectional area normal to the direction of flow, A_{c}
F_{A}_{z} = A_{c}W_{A}_{z}
In terms of concentration, the flux is
The molar flow rate is
$\begin{array}{cc}{F}_{\text{A}z}={W}_{\text{A}z}\text{}{\text{A}}_{C}=[{D}_{\text{AB}}\frac{{dC}_{\text{A}}}{dz}+{{C}_{\text{A}}U}_{z}]{\text{A}}_{C}& \left(\text{1414}\right)\end{array}$
Similar expressions follow for W_{A}_{x} and W_{A}_{y}. Substituting for the flux W_{A}_{x}, W_{A}_{y}, and W_{A}_{z} into Equation (141), we obtain
Flow, diffusion, and reaction
This form is used in COMSOL Multiphysics.
$\begin{array}{c}\hline {D}_{\text{AB}}[\frac{{{\partial}^{2}C}_{\text{A}}}{{\partial x}^{2}}+\frac{{{\partial}^{2}C}_{\text{A}}}{{\partial y}^{2}}+\frac{{{\partial}^{2}C}_{\text{A}}}{{\partial z}^{2}}]{U}_{x}\frac{{\partial C}_{\text{A}}}{\partial x}{U}_{y}\frac{{\partial C}_{\text{A}}}{\partial y}{U}_{z}\frac{{\partial C}_{\text{A}}}{\partial z}+{r}_{\text{A}}=\frac{{\partial C}_{\text{A}}}{\partial t}\\ \hline\end{array}\begin{array}{c}\left(\text{1415}\right)\end{array}$
In terms of axial and radial coordinates with no angular variation and no radial velocity U_{r} we have
$\begin{array}{cc}{D}_{\text{AB}}\left[\frac{\partial}{\partial r}\left(\frac{{rdC}_{\text{A}}}{\partial r}\right)+\frac{{{\partial}^{2}C}_{\text{A}}}{{\partial z}^{2}}\right]{U}_{\text{z}}\frac{{\partial C}_{\text{A}}}{\partial z}+{r}_{\text{A}}=\frac{{\partial C}_{\text{A}}}{\partial t}& \left(\text{1416}\right)\end{array}$
Equation (1415) and (1416) are in a userfriendly form to apply to the PDE solver, COMSOL. For one dimension at steady state, Equation (1415) reduces to
$\begin{array}{c}\hline \begin{array}{c}{D}_{\text{A}\text{B}}\frac{{d}^{2}{C}_{\text{A}}}{d{z}^{2}}{U}_{\text{z}}\frac{d{C}_{\text{A}}}{dz}+{r}_{\text{A}}=0\end{array}\\ \hline\end{array}\begin{array}{c}\left(\text{1417}\right)\end{array}$
In order to solve Equation (1417) we need to specify the boundary conditions. In this chapter we will consider some of the simple boundary conditions, and in Chapter 18 we will consider the more complicated boundary conditions, such as the Danckwerts’ boundary conditions.
We will now use this form of the molar flow rate in our mole balance in the z direction of a tubular flow reactor
$\frac{{dF}_{\text{A}}}{dV}=\frac{d\left({\text{A}}_{c}{W}_{\text{A}z}\right)}{d\left({\text{A}}_{c}z\right)}=\frac{{dW}_{\text{A}z}}{dz}={r}_{\text{A}}$
However, we first have to discuss the boundary conditions in solving this equation.
The most common boundary conditions are presented in Table 142.
TABLE 142 TYPES OF BOUNDARY CONDITIONS

Before closing this brief discussion on masstransfer fundamentals, further mention should be made of the diffusion coefficient. Equations for predicting gas diffusivities are given by Fuller and are also given in Perry’s Handbook.^{2,3} The orders of magnitude of the diffusivities for gases, liquids, and solids and the manner in which they vary with temperature and pressure are given in Table 143.^{4} We note that the Knudsen, liquid, and solid diffusivities are independent of total pressure.
^{2} E. N. Fuller, P. D. Schettler, and J. C. Giddings, Ind. Eng. Chem., 58(5), 19 (1966).
^{3} R. H. Perry and D. W. Green, Chemical Engineer’s Handbook, 7th ed. New York: McGrawHill, 1999.
^{4} To estimate liquid diffusivities for binary systems, see K. A. Reddy and L. K. Doraiswamy, Ind. Eng. Chem. Fund., 6, 77 (1967).
TABLE 143 DIFFUSIVITY RELATIONSHIPS FOR GASES, LIQUIDS, AND SOLIDS

Order of Magnitude 



Phase  cm^{2}/s  m^{2}/s  Temperature and Pressure Dependences^{a} 
Gas  
Bulk  10^{−1}  10^{−5}  ${D}_{\text{AB}}({T}_{2},{P}_{2})={D}_{\text{AB}}({T}_{1},{P}_{1})\frac{{P}_{1}}{{P}_{2}}{\left(\frac{{T}_{2}}{{T}_{1}}\right)}^{1.75}$ 
Knudsen  10^{−2}  10^{−6}  ${{D}_{\text{A}}\left({T}_{2}\right)={D}_{\text{A}}\left({T}_{1}\right)\left(\frac{{T}_{2}}{{T}_{1}}\right)}^{1/2}$ 
Liquid  10^{−5}  10^{−9}  ${D}_{\text{AB}}\left({T}_{2}\right)={D}_{\text{AB}}\left({T}_{1}\right)\frac{{\mu}_{1}}{{\mu}_{2}}\left(\frac{{T}_{2}}{{T}_{1}}\right)$ 
Solid  10^{−9}  10^{−13}  ${D}_{\text{AB}}\left({T}_{2}\right)={D}_{\text{AB}}\left({T}_{1}\right)\text{exp}\left[\frac{{E}_{D}}{R}\left(\frac{{T}_{2}{T}_{1}}{{T}_{1}{T}_{2}}\right)\right]$ 
^{a}μ_{1}, μ_{2}, liquid viscosities at temperatures T_{1} and T_{2}, respectively; E_{D}, diffusion activation energy.
It is important to know the magnitude and the T and P dependence of the diffusivity.
We now consider the situation where species A does not react as it diffuses through a hypothetical stagnant film to a surface where it does react. In Chapter 15 we consider the case when species A does react as it diffuses through a stagnant film. Table 144 provides an algorithm for both of these situations.
TABLE 144 STEPS IN MODELING CHEMICAL SYSTEMS WITH DIFFUSION AND REACTION

^{a}In some instances it may be easier to integrate the resulting differential equation in Step 4 before substituting for WA.
The purpose of presenting algorithms (e.g., Table 144) to solve reaction engineering problems is to give the readers a starting point or framework with which to work if they were to get stuck. It is expected that once readers are familiar and comfortable using the algorithm/framework, they will be able to move in and out of the framework as they develop creative solutions to nonstandard chemical reaction engineering problems.
Use Table 144 to Move In ⇄ Out of the algorithm (Steps 1 → 10) to generate creative solutions.
We will first discuss the diffusion of reactants from the bulk fluid to the external surface of a particle that is either solid or liquid. Here our attention will focus on the flow past a single particle as shown in Figure 142(a) and its corresponding boundary layer shown in Figure 142(b). The particle can be either a liquid droplet, a catalyst pellet or a combustible solid grain. The reaction takes place only on the external surface and not in the fluid surrounding it. The fluid velocity in the vicinity of the spherical pellet will vary with position around the sphere. The hydrodynamic boundary layer is usually defined as the distance from a solid object to where the fluid velocity is 99% of the bulk velocity, U_{0}. Similarly, the mass transfer boundarylayer thickness, =, is defined as the distance from a solid object to where the concentration of the diffusing species reaches 99% of the bulk concentration.
A reasonable representation of the concentration profile for a reactant A diffusing to the external surface is shown in Figure 142. As illustrated, the change in concentration of A from C_{A}_{b} to C_{A}_{s} takes place in a very narrow fluid layer next to the surface of the sphere. Nearly all of the resistance to mass transfer is found in this layer.
The concept of a hypothetical stagnant film within which all the resistance to external mass transfer exists
A useful way of modeling diffusive transport is to treat the fluid layer next to a solid boundary as a hypothetical stagnant film of thickness δ, which we cannot measure. We say that all the resistance to mass transfer is found (i.e., lumped) within this hypothetical stagnant film of thickness δ, and the properties (i.e., concentration, temperature) of the fluid at the outer edge of the film are identical to those of the bulk fluid. This model can readily be used to solve the differential equation for diffusion through a stagnant film. The dashed line in Figure 142(b) represents the concentration profile predicted by the hypothetical stagnant film model, while the solid line gives the actual profile. If the film thickness is much smaller than the radius of the pellet (which is usually the case), curvature effects can be neglected. As a result, only the onedimensional diffusion equation must be solved, as shown in Figure 143.
We are going to carry out a mole balance on species A diffusing through the fluid between z = z and z = z + Δz at steady state for the unit crosssectional area, A_{c}
$\begin{array}{ccccccc}\text{In}& & \text{Out}& +& \text{Generation}& =& \text{Accumulation}\\ {W}_{\text{A}z}{}_{z}& & {W}_{\text{A}z}{}_{\text{z+}\mathrm{\Delta}\text{z}}& +& 0& =& 0\end{array}$
dividing by Δz and taking the limit as Δz → 0
$\frac{{dW}_{\text{A}z}}{dz}=0$
For diffusion through a stagnant film at dilute concentrations
$\begin{array}{cc}{J}_{\text{A}}\text{}\gg \text{}{y}_{\text{A}}({W}_{\text{A}}+{W}_{\text{B}})& \left(\text{1423}\right)\end{array}$
or for EMCD, we have using Fick’s first law
$\begin{array}{cc}{W}_{\text{A}z}={D}_{\text{AB}}\frac{{dC}_{\text{A}}}{dz}& \left(\text{1424}\right)\end{array}$
Substituting for W_{A}_{z} and dividing by D_{AB} we have
$\frac{{{d}^{2}C}_{\text{A}}}{{dz}^{2}}=0$
Integrating twice to get C_{A} = K_{1}z + K_{2}, using the boundary conditions at
we obtain the concentration profile
To find the flux to the surface we substitute Equation (1425) into Equation (1424) to obtain
At steady state the flux of A to the surface will be equal to the rate of reaction of A on the surface. We also note that another example of diffusion through a stagnant film as applied to transdermal drug delivery is given in the Chapter 14 Expanded Material on the CRE Web site.
We now interpret the ratio (D_{AB}/δ ) in Equation (1426).
While the boundarylayer thickness will vary around the sphere, we will take it to have a mean film thickness δ. The ratio of the diffusivity D_{AB} to the film thickness δ is the mass transfer coefficient, k_{c}, that is,
$\begin{array}{c}\hline {k}_{c}=\frac{{D}_{\text{AB}}}{\delta}\\ \hline\end{array}\begin{array}{c}\left(\text{1427}\right)\end{array}$
Combining Equations (1426) and (1427), we obtain the average molar flux from the bulk fluid to the surface
$\begin{array}{c}\hline {W}_{\text{A}z}={k}_{c}({C}_{\text{A}b}{C}_{\text{As}})\\ \hline\end{array}\begin{array}{c}\left(\text{1428}\right)\end{array}$
The mass transfer coefficient
In this stagnant film model, we consider all the resistance to mass transfer to be lumped into the thickness δ. The reciprocal of the mass transfer coefficient can be thought of as this resistance
$\begin{array}{c}\hline {W}_{\text{A}z}=\text{Flux}=\frac{\text{Driving force}}{\text{Resistance}}=\frac{{C}_{\text{A}b}{C}_{\text{As}}}{(1/{k}_{c})}\\ \hline\end{array}\begin{array}{c}\left(\text{1429}\right)\end{array}$
Molar flux of A to the surface
How Do I find the Mass Transfer Coefficient? The mass transfer coefficient is found either by experimentation or from correlations analogous to what one finds for a heat transfer coefficient. These correlations are usually in the form of the Sherwood number, Sh, as a function of the Reynolds number, Re, and the Schmidt number, Sc, that is,
$\begin{array}{cc}\text{Sh}=\text{f}\left(\text{Re, Sc}\right)& \left(\text{1430}\right)\end{array}$
where
Sherwood
$\begin{array}{cc}\text{Sh}=\frac{{k}_{c}\u2033L\u2033}{{D}_{\text{AB}}}& \left(\text{1431}\right)\end{array}$
Schmidt
$\begin{array}{cc}\text{Sc}=\frac{\upsilon}{{D}_{\text{AB}}}=\frac{\mu /\rho}{{D}_{\text{AB}}}& \left(14\text{32}\right)\end{array}$
Reynolds
$\begin{array}{cc}\text{Re}=\frac{U\u2033L\u2033}{\upsilon}=\frac{U\u2033L\u2033\rho}{\mu}& \left(\text{1433}\right)\end{array}$
where
″L″ is the characteristic length (m) (e.g., d_{P}, diameter of the particle)
υ = Kinematic viscosity (m^{2}/s) = μ/ρ
μ = viscosity (kg/m · s)
ρ = density (kg/m^{3})
U = free stream velocity (m/s)
D_{AB} = diffusivity (m^{2}/s)
As an example, the mass transfer coefficient for flow around a single spherical particle can be found from the Frössling correlation.^{†}
^{†}N. Frössling, Gerlands Beitr. Geophys., 52, 170 (1938).
$\begin{array}{cc}\text{Sh}=2+0.6{\text{Re}}^{1/2}{\text{Sc}}^{1/3}& \left(\text{1434}\right)\end{array}$
For turbulent flow, the number 2 in this equation can be neglected with respect to the second term and the resulting correlation is shown in Table 145.^{‡}
^{‡} Just out… An article in August 2019 issue of the AIChE Journal expands this correlation. Y. Wang and J G. Brasseur, “Enhancement of mass transfer from particles by local shearrate and correlations with application to drug dissolution,” AIChE J., 65 (8), (August 2019). Equation (1434) is derived for dissolution in an infinite fluid while Professor Wang’s article AIChE J. discusses dissolution in a confined domain. The corrections are particularly important at low Reynolds number and this article guides you through these corrections. However, the nomenclature is different. For example, the Sherwood number is given as Sh = R/δ, so it might help if you referred to Professor Wang’s earlier article 2012 Y. Yang, et.al., Mol. Pharm., 9, 1052 (2012).
The correlation given for low Re is reported as
Sh = Sh_{0} + 0.0177 Re^{0.46} Sc^{0.68}
where Sh_{0} is a function of the Schmidt number Sc. For example, when
5 < Sc < 100
then
Sh_{0} = 1.2 + Sc^{0.82}
Other values of Sh_{0} are given in Professor Wang’s article.
After calculating the numerical value of Sh, given the parameters to calculate Re and Sc, the mass transfer coefficient can be calculated
$\begin{array}{cc}\hline \begin{array}{cc}{k}_{\text{}c}=\frac{\text{Sh}\text{}{D}_{\text{AB}}}{{d}_{p}}& \end{array}& \\ \hline\end{array}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{cc}\left(\text{m/s}\right)& \end{array}\end{array}\end{array}\left(\text{1435}\right)\end{array}$
Correlation for geometries other than a single spherical particle are given in Table 145.
TABLE 145 MASS TRANSFER CORRELATIONS
Turbulent flow, mass transfer to pipe wall 
Sh = .332 (Re)^{1/2} (Sc)^{1/3} 
Mass transfer to a single sphere 
Sh = 2 = 0.6 Re^{1/2} Sc^{1/3} 
Mass transfer in fluidized beds 
${\varphi J}_{\text{D}}=\frac{0.765}{{\text{Re}}^{82}}+\frac{0.365}{{\text{Re}}^{0.386}}$ 
Mass transfer to packed beds 
ϕJ_{D} = 0.453 Re^{0.453} 
${J}_{\text{D}}=\frac{\text{Sh}}{{\text{ReSc}}^{1/3}}$ 
What if I cannot find the mass transfer correlation for my situation or geometry? In this case, see whether a correlation for the heat transfer coefficient exists and then go to (http://www.umich.edu/~elements/6e/14chap/obj.html#/additionalmaterials/) to learn how to turn that heat transfer correlation into a mass transfer correlation.
The Sherwood, Reynolds, and Schmidt numbers are used in forced convection mass transfer correlations.
Mass transfer to single particles is important in catalytic reactions and in dust explosions. In this section we consider two limiting cases of diffusion and reaction on a catalyst particle.^{5} In the first case, the reaction is so rapid that the rate of diffusion of the reactant to the surface limits the reaction rate. In the second case, the reaction is so slow that virtually no concentration gradient exists in the gas phase (i.e., rapid diffusion with respect to surface reaction).
^{5} A comprehensive list of correlations for mass transfer to particles is given by G. A. Hughmark, Ind. Eng. Chem. Fund., 19(2), 198 (1980).
To easily show the limitations of mass transfer and reaction rate we will consider first order kinetics. For the case of either the burning of a combustible dust particle or for reaction on a catalyst surface at high temperatures, the rate law is taken as apparent first order.
Figure 144 shows the mass transfer flux of A to the surface, W_{A}, the reaction on the surface ${r}_{\text{A}s}^{\u2033}$, and the mass transfer flux of B away from the surface. In the examples discussed here species A can be thought of as oxygen and B as the combustion products, for example, CO_{2}. The reaction on the surface is taken as apparent first order.
$\begin{array}{cc}{r}_{\text{A}s}^{\u2033}={{k}_{r}C}_{\text{As}}& \left(\text{1436}\right)\end{array}$
Using boundary conditions 2b. and 2c. in Table 142, we obtain
The concentration at the surface, C_{A}_{s}, is not as easily measured as is the bulk concentration, C_{A}. Consequently, we need to eliminate C_{A}_{s} from the equation for the flux and rate of reaction. Solving Equation (1438) for C_{A}_{s} yields
and the rate of reaction on the surface becomes
Molar flux of A to the surface is equal to the rate of consumption of A on the surface.
$\begin{array}{c}\hline {W}_{\text{A}}={r}_{\text{A}s}^{\u2033}=\frac{{{k}_{c}{k}_{r}C}_{\text{A}}}{{k}_{r}+{k}_{c}}\\ \hline\end{array}\begin{array}{c}\left(\text{1440}\right)\end{array}$
One will often find the flux to or from the surface written in terms of an effective transport coefficient k_{eff}
where
We will now consider the extremes of rapid and slow reaction at the particle surface.
Rapid Reaction. We first consider how the overall rate of reaction may be increased when the rate of mass transfer to the surface limits the overall rate of reaction. Under these circumstances, the specific reactionrate constant is much greater than the mass transfer coefficient
${k}_{r}\text{\u226b}\text{}{k}_{c}$
and
k_{eff} = k_{c}
$\begin{array}{cc}{W}_{\text{A}}={r}_{\text{As}}^{\u2033}=\frac{{k}_{c}{C}_{\text{A}}}{1+{k}_{c}/{k}_{r}}\approx {k}_{c}{C}_{\text{A}}& \left(\text{1443}\right)\end{array}$
To increase the rate of reaction per unit surface area of a solid sphere, one must increase C_{A} and/or k_{c}. In this gasphase catalytic reaction example, and for most liquids, the Schmidt number is sufficiently large that the number 2 in Equation (1434) is negligible with respect to the second term when the Reynolds number is greater than 25. As a result, Equation (1434) gives
It is important to know how the mass transfer coefficient varies with fluid velocity, particle size, and physical properties.
$\begin{array}{rr}\begin{array}{c}{k}_{c}=0.6\text{}\left(\frac{{D}_{\text{AB}}}{{d}_{p}}\right)\text{}{\text{Re}}^{1/2}{\text{Sc}}^{1/3}\\ \begin{array}{cc}\text{}\text{}=0.6\text{}\left(\frac{{D}_{\text{AB}}}{{d}_{p}}\right){\left(\frac{U{d}_{p}}{v}\right)}^{1/2}{\left(\frac{v}{{D}_{\text{AB}}}\right)}^{1/3}& \end{array}\end{array}\hfill & \left(\text{1444}\right)\end{array}$
$\begin{array}{cc}\begin{array}{c}\begin{array}{c}{k}_{c}=0.6\times \frac{{D}_{\text{AB}}^{2/3}}{{v}^{1/6}}\times \frac{{U}^{1/2}}{{d}_{p}^{1/2}}\end{array}\hfill \\ {k}_{c}=0.6\times \left(\text{Term 1}\right)\times \left(\text{Term 2}\right)\end{array}\hfill & \left(\text{1445}\right)\end{array}$
Mass Transfer Limited
Term 1 is a function of the physical properties D_{AB} and ν, which depend on temperature and pressure only. The diffusivity always increases with increasing temperature for both gas and liquid systems. However, the kinematic viscosity ν increases with temperature (ν ∝ T^{3/2}) for gases and decreases exponentially with temperature for liquids. Term 2 is a function of flow conditions and particle size. Consequently, to increase k_{c} and thus the overall rate of reaction per unit surface area, one may either decrease the particle size or increase the velocity of the fluid flowing past the particle. For this particular case of flow past a single sphere, we see that if the velocity is doubled, the mass transfer coefficient and consequently the rate of reaction is increased by a factor of
(U_{2}/U_{1})^{0.5} = 2^{0.5} = 1.41 or 41%
Reaction Rate Limited
Slow Reaction. Here, the specific reactionrate constant is small with respect to the mass transfer coefficient
$\begin{array}{c}{k}_{r}\text{\u226a}{k}_{c}\\ \begin{array}{cc}{W}_{\text{A}}={r}_{\text{A}S}^{\u2033}=\frac{{{k}_{r}C}_{\text{A}}}{1+\text{}{k}_{r}/{k}_{c}}\approx {{k}_{r}C}_{\text{A}}& \left(\text{1446}\right)\end{array}\end{array}$
Mass transfer effects are not important when the reaction rate is limiting.
The specific reaction rate is independent of the velocity of fluid and for the solid sphere considered here, independent of particle size. However, for porous catalyst pellets, k_{r} may depend on particle size for certain situations, as shown in Chapter 15.
Figure 145 shows the variation in reaction rate with Term 2 in Equation (1445), the ratio of velocity to particle size. At low velocities, the mass transfer boundarylayer thickness is large and diffusion limits the reaction. As the velocity past the sphere is increased, the boundarylayer thickness decreases, and the mass transfer across the boundary layer no longer limits the rate of reaction. One also notes that for a given (i.e., fixed) velocity, reactionlimiting conditions can be achieved by using very small particles. However, the smaller the particle size, the greater the pressure drop in a packed bed. When one is obtaining reactionrate data in the laboratory, one must operate at sufficiently high velocities or sufficiently small particle sizes to ensure that the reaction is not mass transferlimited when collecting data.
When collecting ratelaw data, operate in the reactionlimited region.
Before we analyze Equation (1440) in more detail, we will describe examples involving a dust particle fire of granular carbon in air and a catalyst particle in a liquid slurry. In both cases we will use k_{r} >> k_{s} so that $\u2013{r}_{\text{As}}^{\u2033}={W}_{\text{A}}={k}_{c}{C}_{\text{A}}$.
A schematic of the transfer of oxygen to the particle is shown in Figure 144. The gasphase diffusivity in the flame (1000 K) surrounding the particle is taken as 10^{–4} m^{2}/s and the particle diameter is 50 μm (5 × 10^{–5} m). The concentration of oxygen in the bulk air (21% at 1 atm and 298 K) is 8.58 mol/m^{3}. The reaction is virtually instantaneous so that the concentration of oxygen on the surface of the particle is zero, that is, C_{A}_{s} = 0. The heat of reaction is ΔH_{Rx} = –93.5 kJ/mol carbon.
Calculate molar flux, W_{A} (mol/m^{2}/s) and molar flow ${\dot{m}}_{\text{A}}$ (mol/s) to the particle. For a dust cloud density of 200 g/m^{3}, calculate the heat generated per particle ${\dot{q}}_{p}$, and the heat generated per volume of cloud.
Solution
The flux to the surface is
$\begin{array}{cc}{W}_{\text{A}}={k}_{c}[{C}_{\text{A}\infty}{C}_{\text{As}}]\cong {k}_{c}{C}_{\text{A}\infty}& \left(\text{E141.1}\right)\end{array}$
For particles that are sufficiently small such that they are in Stoke’s flow, (i.e., it follows the fluid) the fluid velocity relative to the particle is zero, (i.e., U ≌ 0). For a single particle in Stoke’s flow, the Sherwood number for a spherical particle
Sh = 2 + 0.61 (Re^{1/2} Sc^{1/3}) ≅ 2
reduces to
$\begin{array}{cc}\text{Sh}=\frac{{{k}_{c}d}_{p}}{{D}_{\text{AB}}}=2& \left(\text{E141.2}\right)\end{array}$
The corresponding molar flux of oxygen to the surface where it reacts virtually instantaneously (C_{A}_{s} = 0) at the surface
$\begin{array}{cc}\begin{array}{c}\begin{array}{c}\begin{array}{c}{k}_{c}=\frac{\text{Sh}}{{d}_{p}/{D}_{\text{AB}}}=\frac{2{D}_{\text{AB}}}{{d}_{p}}=\frac{2({10}^{4}{\text{m}}^{2}/\text{s})}{5\times {10}^{5}\text{m}}=4\text{m/s}\\ {W}_{\text{A}}={k}_{c}\left({C}_{\text{A}\infty}\right)=2\frac{{D}_{\text{AB}}}{{d}_{p}}{C}_{\text{A}\infty}\hfill \end{array}\end{array}\\ {W}_{\text{A}}=\frac{2({10}^{4}{\text{m}}^{2}/\text{s})}{{50\times 10}^{6}\text{m}}\left(8.58\frac{\text{mol}}{{\text{m}}^{3}}\right)={34.3\text{mol/m}}^{2}\cdot \text{s}\end{array}\hfill & \left(\text{E141.3}\right)\end{array}$
The molar flux (mol/m^{2}/s) of oxygen to one particle is
$\begin{array}{c}\hline {W}_{\text{A}}={r}_{\text{A}}^{\u2033}=34.3{\text{mol/m}}^{2}\cdot \text{s}\\ \hline\end{array}$
The molar flow (mol/s) of oxygen, $\dot{{m}_{\text{A}}}$, to one particle with surface area A_{P} is
${{\dot{m}}_{\text{A}}={{\text{A}}_{P}W}_{\text{A}}={\pi d}_{p}^{2}\text{}{\text{W}}_{\text{A}}=\pi {(50\times {10}^{6}\text{m})}^{2}34.3\text{mol/m}}^{2}\cdot \text{s}$
$\begin{array}{c}\hline {{\dot{m}}_{\text{A}}=2.7\times 10}^{7}\text{mol/s}\\ \hline\end{array}$
Heat Effects
Now let’s estimate the heat generated for this burning particle and then for the dust cloud with n_{p}, the number of particles per unit volume. Figure 8.40 of Ogle^{†} gives the minimum dust explosion concentration, in air, Ccloud, as approximately 200 g/m^{3}. The mass of one 50 μm carbon dust particle is
^{†} Ibid, page 480.
$\begin{array}{lll}{m}_{P}={\rho}_{c}V={\rho}_{c}\frac{{\pi d}_{p}^{3}}{6}& =& \left({{2.226\times 10}^{6}\text{g/m}}^{3}\right)\frac{\pi}{6}{\left({5\times 10}^{5}\right)}^{3}{\text{m}}^{3}\\ & =& {\begin{array}{c}14.5\times 10\end{array}}^{8}\text{g/particle}\end{array}$
The corresponding particle concentration is
${n}_{P}=\frac{{C}_{\text{cloud}}}{{m}_{p}}=\frac{{200\text{g/m}}^{3}}{{14.5\times 10}^{8}\text{g/particle}}$
$\begin{array}{c}\hline {{{n}_{p}=1.38\times 10}^{9}\text{particles/m}}^{3}\\ \hline\end{array}$
The heat generated by a single particle ${\dot{q}}_{P}$ (kJ/particle) is just the molar flux of O_{2} (i.e., ${r}_{\text{A}s}^{\u2033}={W}_{\text{A}}$) times the surface area of particle times the heat of reaction, that is,
${\dot{q}}_{P}=({r}_{\text{A}s}^{\u2033}{\text{A}}_{P})({\mathrm{\Delta}H}_{\text{Rx}})$
The heat generated per unit volume of dust cloud, $\dot{{Q}_{gd}}$, with a concentration of dust particles, n_{P}, is
${\dot{Q}}_{gd}={{n}_{P}\dot{q}}_{P}=\left({r}_{\text{A}s}^{\u2033}{n}_{P}{\text{A}}_{P}\right)\left({\mathrm{\Delta}H}_{\text{Rx}}\right)$
The rate of heat generated per particle, ${\dot{q}}_{P}$, times the particle concentration, n_{P}, gives the heat generated per unit volume of dust as
${\dot{Q}}_{gd}={{n}_{P}\dot{q}}_{P}={{n}_{P}\dot{q}}_{P}={{n}_{P}\dot{m}}_{\text{A}}\left({\mathrm{\Delta}H}_{\text{Rx}}\right)=\frac{{1.38\times 10}^{9}}{{\text{m}}^{3}}\times 2.7\times {10}^{7}\frac{\text{mol}}{\text{s}}\times \left(\text{93.5}\frac{\text{kJ}}{\text{mol}}\right)$
$\begin{array}{c}\hline {Q}_{gd}=34586\text{kJ/s}\cdot {\text{m}}^{3}\\ \hline\end{array}$
Analysis: We calculated the molar flux as well as the mass flow of oxygen to the surface of a burning particle. The molar flux of O_{2} to the surface is equal to the rate of reaction per unit surface area. Knowing the reaction rate, the dust density, and the heat of reaction we calculated the heat generated per particle and per unit volume of dust cloud.
We will continue our discussion on single dust particles when we discuss the shrinking core model. However, let’s first make a comparison with a larger particle, 0.1 cm, suspended in a body of flowing liquid.
Calculate the molar flux, W_{A}_{r}, of reactant A to a single catalyst pellet 1 cm in diameter suspended in a large body of liquid B. The reactant is present in dilute concentrations, and the reaction is considered to take place instantaneously at the external pellet surface (i.e., C_{A}_{s} ≃ 0). The bulk concentration of the reactant A is 1.0 M, and the freestream liquid velocity past the sphere is 0.1 m/s. The kinematic viscosity (i.e., $\frac{\mu}{\rho}$) is 0.5 centistoke (cS; 1 centistoke = 10^{6} m^{2}/s), and the liquid diffusivity o A in B is D_{AB} = 10^{–10} m^{2}/s at 300 K.
If the surface reaction is rapid, then diffusion limits the overall rate.
Solution
For dilute concentrations of the solute, the radial flux is
$\begin{array}{cc}{W}_{\text{A}r}={k}_{c}({C}_{\text{}\text{A}b}{C}_{\text{As}})& \left(\text{E142.1}\right)\end{array}$
Because reaction is assumed to occur instantaneously on the external surface of the pellet, C_{A}_{s} = 0. Also, C_{A}_{b} is given as 1 mol/dm^{3}.
$\begin{array}{cc}{W}_{\text{A}r}={k}_{c}{C}_{\text{}\text{A}b}& \left(\text{E142.2}\right)\end{array}$
The mass transfer coefficient for single spheres is calculated from the Frössling correlation
$\begin{array}{cc}\begin{array}{ccc}\text{Sh}& =& \frac{{{k}_{c}d}_{p}}{{D}_{\text{AB}}}=2+{{0.6\text{Re}}^{1/2}\text{Sc}}^{1/3}\\ \text{Re}& =& \frac{{\rho d}_{p}U}{\mu}=\frac{{d}_{p}U}{v}=\frac{\left(0.01\text{m}\right)\left(\text{0.1 m/s}\right)}{{0.5\times 10}^{6}{\text{m}}^{2}/\text{s}}=2000\end{array}& \left(\text{1434}\right)\end{array}$
Liquid Phase
Re = 2000
Sc = 5000
Sh = 460
k_{c} = 4.6 × 10^{6} m/s
$\text{Sc}=\frac{v}{{D}_{\text{AB}}}=\frac{{{5\times 10}^{7}\text{m}}^{2}/\text{s}}{{10}^{\text{10}}{\text{m}}^{2/\text{S}}}=5000$
Substituting these values into Equation (1434) gives us
$\begin{array}{cc}{{\text{Sh}=2+0.6\left(\text{2000}\right)}^{0.5}\left(\text{5000}\right)}^{1/3}=460.7& \left(\text{E142.3}\right)\end{array}$
$\begin{array}{c}\begin{array}{cc}{k}_{c}=\frac{{D}_{\text{AB}}}{{d}_{p}}\text{Sh}=\frac{{10}^{10}{\text{m}}^{2}/\text{s}}{0.01\text{m}}\times \text{460.7}=4.61\times {10}^{6}\text{m/s}& \left(\text{E142.4}\right)\end{array}\\ {\begin{array}{c}{\begin{array}{c}{C}_{\text{A}b}=1.0\text{mol/dm}\end{array}}^{3}={10}^{3}\text{mol/m}\end{array}}^{3}\end{array}$
Substituting for k_{c} and C_{A}_{b} in Equation (E142.2), the molar flux to the surface is
W_{A}_{r} = (4.61 × 10^{−6}) m/s (10^{3} − 0) mol/m^{3} = 4.61 × 10^{−3} mol/m^{2} · s
Because ${W}_{\text{A}r}={r}_{\text{A}S}^{\u2033}$, this rate is also the rate of reaction per unit surface area of catalyst.
$\begin{array}{c}\hline {{{{r}_{\text{A}S}^{\u2033}=0.0046\text{mol/m}}^{2}\cdot \text{s}=\text{4.6}\times 10}^{5}\text{mol/dm}}^{2}\cdot \text{s}\\ \hline\end{array}$
The mass transfer flow to the particle is
$\begin{array}{ccc}\hfill {\dot{m}}_{\text{A}}={\pi d}_{p}^{2}({r}_{\text{A}}^{\u2033})& =& \pi {\left(\text{0.01m}\right)}^{2}({\text{0.0046 mol/m}}^{2}\cdot \text{s})\hfill \\ \hfill {\dot{m}}_{\text{A}}& =& \text{1.45}\times {10}^{6}\text{mol/s}\hfill \end{array}$
Analysis: In this example we calculated the rate of reaction on the external surface of a catalyst pellet in a liquid reactant when external mass transfer was limiting the reaction rate. To determine the rate of reaction, we used correlations to calculate the mass transfer coefficient and then used k_{c} to calculate the flux to the surface, which in turn was equal to the rate of surface reaction.
Many situations arise in heterogeneous reactions where a gasphase species reacts with a species contained in an inert solid matrix or solid combustible dust particles. The burning of a combustible dust is shown in Figure 146(a). The removal of carbon from catalyst particles that have been deactivated by fouling is shown in Figure 146(b) (see Section 10.7.1). The catalyst regeneration process to reactivate the catalyst by burning off the carbon is discussed in the expanded material on the CRE Web site.
The shrinking core model is used to describe situations in which solid particles are being consumed either by dissolution or reaction and, as a result, the amount of the material being consumed is “shrinking.” For example, to design the time release of drugs into the body’s system, one must focus on the rate of dissolution of capsules and solid pills in the stomach. We now apply the shrinking core model to dust explosions where solid organic particles such as sugar are burned, to the formation of an ash layer around a burning coal particle, to catalyst regeneration. In this section we focus primarily on dust explosions and catalyst regeneration, and leave other applications such as drug delivery as exercises at the end of the chapter.
To illustrate the principles of the shrinking core model, we shall consider the burning of carbon dust particle. As the carbon continues to burn the radius of dust particles shrink from R_{0} initially to R at time t as shown in Figure 146(a).
As shown in Figure 147, oxygen diffuses from the bulk gas at R_{∞} to the radius R, where it reacts with carbon to form carbon dioxide, which then diffuses out from the particle. The reaction
C + O_{2} → CO_{2}
at the solid surface is very rapid, so the rate of oxygen diffusion to the surface controls the rate of carbon removal from the core. Although the core of
Oxygen must diffuse through the porous pellet matrix until it reaches the unreacted carbon core.
carbon is shrinking with time (an unsteadystate process), we assume the concentration profiles at any instant in time to be the steadystate profiles over the distance (R_{∞} – R) when R_{∞} is some larger distance from the surface of the particle, for example, R_{∞} ≈ ∞. This assumption is known as the quasisteady state assumption (QSSA).
QSSA Use steadystate profiles
To study how the radius of unreacted carbon changes with time, we must first find the rate of diffusion of oxygen to the carbon surface. Next, we perform a mole balance on the elemental carbon and equate the rate of consumption of the carbon particle to the rate of diffusion of oxygen to the gas–carbon interface.
In applying a differential oxygen mole balance over the increment Δr located somewhere between R_{∞} and R, we recognize that O_{2} does not react in this region and reacts only when it reaches the solid carbon interface located at r = R. We shall let species A represent O_{2}.
Step 1: The mole balance on O_{2} (i.e., A) between r and r + Δr is
$\begin{array}{cccccc}\left[\begin{array}{c}\text{Rate}\\ \text{in}\end{array}\right]& & \left[\begin{array}{c}\text{Rate}\\ \text{out}\end{array}\right]& +& \left[\begin{array}{c}\text{Rate}\text{of}\\ \text{generation}\end{array}\right]& \begin{array}{cc}\begin{array}{c}=\end{array}& \left[\begin{array}{c}\text{Rate}\text{of}\\ \text{accumulation}\end{array}\right]\end{array}\\ {\begin{array}{c}{\begin{array}{c}{W}_{\text{A}r}4\pi r\end{array}}^{2}\end{array}}_{r}& & {\begin{array}{c}{\begin{array}{c}{W}_{\text{A}r}4\pi r\end{array}}^{2}\end{array}}_{r+\mathrm{\Delta}r}& +& 0& \begin{array}{cc}\begin{array}{c}=\end{array}& 0\end{array}\end{array}$
Dividing through by –4πΔr and taking the limit gives
Mole balance on oxygen
$\begin{array}{cc}\stackrel{\mathbf{\text{lim}}}{\mathrm{\Delta}r\to 0}\frac{{{{{W}_{\text{A}r}r}^{2}}_{r+\mathrm{\Delta}r}{W}_{\text{A}r}{r}^{2}}_{r}}{\mathrm{\Delta}r}=\frac{d\left({{W}_{\text{A}r}r}^{2}\right)}{dr}=0& \left(\text{1447}\right)\end{array}$
Step 2: The constitutive equation for constant total molar concentration becomes
$\begin{array}{cc}{W}_{\text{A}r}={D}_{\text{AB}}\frac{{dC}_{\text{A}}}{dr}& \left(\text{1448}\right)\end{array}$
Combining Equations (1417) and (1448) and dividing by (–D_{AB}) gives
$\begin{array}{c}\hline \frac{d}{dr}\left({r}^{2}\frac{{dC}_{\text{A}}}{dr}\right)=0\\ \hline\end{array}\begin{array}{c}\left(\text{1449}\right)\end{array}$
Step 3: Boundary conditions for burning of a dust particle At a large distance from the dust particle r ~ ∞ then C_{A} = C_{A}_{∞} At the dust particle’s surface, r = R then C_{A} = 0
Step 4: Integrating twice yields
${r}^{2}\frac{{dc}_{\text{A}}}{dr}={K}_{1}$
${C}_{\text{A}}=\frac{{K}_{1}}{r}+{K}_{2}$
Using the boundary conditions for a spherical dust particle to eliminate K_{1} and K_{2}, the concentration profiles is
$\begin{array}{cc}{C}_{\text{A}}={C}_{\text{A}\infty}(1\frac{R}{r})& \left(\text{1450}\right)\end{array}$
A schematic representation of the profile of O_{2} is shown in Figure 148 at a time when the inner core has receded to a radius R. The zero on the r axis corresponds to the center of the sphere
Step 5: The molar flux of O_{2} to the gas–carbon interface for a dust particle is
$\begin{array}{cc}{W}_{\text{A}r}={D}_{\text{AB}}\frac{{dC}_{\text{A}}}{dr}=\frac{{D}_{\text{AB}}{C}_{\text{A}\infty}}{R}& \left(\text{1451}\right)\end{array}$
Step 6: We now carry out an overall balance on elemental carbon. Elemental carbon does not enter or leave the particle.
$\begin{array}{cc}\begin{array}{ccccccc}\left[\begin{array}{c}\text{Rate}\\ \text{in}\end{array}\right]& & \left[\begin{array}{c}\text{Rate}\\ \text{out}\end{array}\right]& +& \left[\begin{array}{c}\text{Rate of}\\ \text{generation}\end{array}\right]& =& \left[\begin{array}{c}\text{Rate of}\\ \text{accumulation}\end{array}\right]\\ 0& & 0& +& {\begin{array}{c}{r}_{\text{A}}^{\u2033}\cdot 4\pi R\end{array}}^{2}& =& \frac{d\left({\frac{4}{3}\pi R}^{3}{\rho}_{c}{\varphi}_{c}\right)}{dt}\end{array}& \left(\text{1452}\right)\end{array}$
Where ρ_{c} is the molar density of the solid carbon, simplifying gives
$\begin{array}{cc}\frac{dR}{dt}=\frac{{r\text{}\text{}}_{\text{A}}^{\u2033}}{{{\rho}_{c}\varphi}_{c}}& \left(\text{1453}\right)\end{array}$
In terms of particle diameter, d_{P},
$\begin{array}{cc}\frac{{dd}_{p}}{dt}=\frac{2{r\text{}}_{\text{A}}^{\u2033}}{{{\rho}_{c}\varphi}_{c}}& \left(14\text{54}\right)\end{array}$
Step 7: The rate of disappearance of carbon is equal to the flux of O_{2} to the gas–carbon interface:
A. Surface reaction limiting
$\begin{array}{cc}{r}_{\text{A}}^{\u2033}={k}_{r}{C}_{\text{A}S}& \left(\text{1455}\right)\end{array}$
$\begin{array}{cc}\frac{ddp}{dt}=\frac{2{r}_{\text{A}}^{\u2033}}{{{\rho}_{c}\varphi}_{c}}=\frac{{2k}_{c}{C}_{\text{A}\infty}}{{{\rho}_{c}\varphi}_{c}}& \left(\text{1456}\right)\end{array}$
The initial conditions are
$\begin{array}{c}\hline {d}_{p0}^{2}{d}_{p}^{2}=\frac{{2k}_{r}}{{{\rho}_{c}\varphi}_{c}}{C}_{\text{A}s}t\\ \hline\end{array}\begin{array}{c}\left(\text{1458}\right)\end{array}$
B. Diffusion Limiting
$\begin{array}{c}{r}_{\text{A}}^{\u2033}={W}_{\text{A}}={k}_{c}{C}_{\text{A}\infty}\\ \frac{ddp}{dt}=\frac{2{r}_{\text{A}}^{\u2033}}{{{\rho}_{c}\varphi}_{c}}=2\left(\frac{{2D}_{\text{AB}}}{{d}_{p}}\right)\frac{{C}_{\text{A}\infty}}{{{\rho}_{c}\varphi}_{c}}=\frac{{4D}_{\text{AB}}{C}_{\text{A}\infty}}{{{\rho}_{c}\varphi}_{c}}\frac{1}{{d}_{p}}\end{array}$
Again using the boundary conditions in Equation (1457), we obtain
$\begin{array}{c}\hline {d}_{p0}^{2}{d}_{p}^{2}=\frac{{8D}_{\text{AB}}{C}_{\text{A}\infty}}{{{\rho}_{c}\varphi}_{c}}t={K}_{S}t\\ \hline\end{array}\begin{array}{c}\left(\text{1459}\right)\end{array}$
${K}_{S}=\frac{{8D}_{\text{AB}}{C}_{\text{A}\infty}}{{{\rho}_{c}\varphi}_{c}}$
where K_{S} is the burning rate constant, s^{1}.^{†}
^{†}R. A. Ogle Dust Explosion Dynamics, Amsterdam: Elsevier, 2017.
The time for complete combustion (d_{P} = 0) of a particle of diameter d_{p}_{0} is
$\begin{array}{c}\hline {t}_{c}=\frac{{d}_{p0}^{2}}{{K}_{\text{S}}}=\frac{{d}_{p0}^{2}{{\rho}_{c}\varphi}_{c}}{{{8D}_{\text{AB}}C}_{\text{A}\infty}}\\ \hline\end{array}\begin{array}{c}\left(\text{1460}\right)\end{array}$
There is a 100 micron carbon dust particle in air at 1 atm. Calculate K_{S} and the time for the particle to burn completely (d_{p} = 0). The average value of the diffusivity near the burning particle is 10^{–4} m^{2}/s and the density of the carbon particle is 2.26 × 10^{6} g/m^{3}.
Solution
The concentration of gas at 1 atm and 273 K is 0.046 mol/dm^{3}. The corresponding concentration of O_{2} at the outside the boundary layer correcting for the temperature at 298 K, that is, C_{A}_{∞} is
$\begin{array}{cc}{C}_{{O}_{2}}=0.21& {C}_{t}=\left(0.21\right)\left(0.0446\frac{\text{mol}}{{\text{dm}}^{3}}\right)\frac{273}{298}{\left(\frac{10\text{dm}}{\text{m}}\right)}^{3}=8.58\frac{\text{mol}}{{\text{m}}^{3}}\end{array}$
The molar density of the solid carbon particle is
$\begin{array}{ccc}{\rho}_{c}& =& \frac{2.26\times {10}^{6}{\text{g/m}}^{3}}{12\text{g/mol}}=188333\frac{\text{mol}}{{\text{m}}^{3}}\hfill \\ {\varphi}_{c}& =& 1\hfill \\ {K}_{\text{S}}=\frac{{8D}_{\text{AB}}{C}_{\text{A}\infty}}{{\rho}_{c}}& =& \frac{8({10}^{4}{\text{m}}^{2}/\text{S})\left(\frac{8.58\text{mol}}{{\text{m}}^{3}}\right)}{{0.1883\cdot 10}^{6}{\text{mol/m}}^{3}}={365\times 10}^{10}{\text{m}}^{2}/\text{s}\hfill \end{array}$
The units used by R. A. Ogle in his book Dust Explosion Dynamic, Elsevier 2017, are μm and ms, in which case K_{S} becomes
${K}_{S}=365\times {10}^{10}{\text{m}}^{2}/\text{s}{\left(\frac{{10}^{6}\mu \text{m}}{1\text{m}}\right)}^{2}\left(\frac{1\text{s}}{{10}^{3}\text{ms}}\right)=36.5\text{}\mu {\text{m}}^{2}\text{/ms}$
This value of K_{S} is about an order of magnitude smaller than a burning liquid droplet. Calculate, t_{c}, the time for the dust particle to completely burn
$\begin{array}{c}\hline {t}_{c}=\frac{{d}_{p0}^{2}}{{K}_{S}}=\frac{{\left(100\text{}\mu \text{m}\right)}^{2}}{36.5\text{}{\mu \text{m}}^{2}/\text{ms}}=274\text{ms}\\ \hline\end{array}$
The time for a blink of an eye is approximately 400 ms.
Analysis: This analysis shows how to calculate time to consume a 100 micron carbon particle. The time is quite short, leading to a dust explosion. Be sure to view the Chemical Safety Board (CSB) video on the CRE Web site (http://www.csb.gov/imperialsugarcompanydustexplosionandfire/).
A number of industrial reactions are potentially mass transferlimited because they may be carried out at high temperatures without the occurrence of undesirable side reactions. In mass transferdominated reactions, the surface reaction is so rapid that the rate of transfer of reactant from the bulk gas or liquid phase to the surface limits the overall rate of reaction. Consequently, mass transferlimited reactions respond quite differently to changes in temperature and flow conditions than do the ratelimited reactions discussed in previous chapters. In this section the basic equations describing the variation of conversion with the various PBR design parameters (catalyst weight, flow conditions) will be developed. To achieve this goal, we begin by carrying out a mole balance on the following generic mass transferlimited reaction
$\begin{array}{cc}\text{A}+\frac{b}{a}\text{B}\to \frac{c}{a}\text{}C+\frac{d}{a}\text{D}& \left(\text{1461}\right)\end{array}$
carried out in a packedbed reactor (Figure 149). A steadystate mole balance on reactant A in the reactor segment between z and z + Δz is
$\begin{array}{cc}\begin{array}{ccccccc}\left[\begin{array}{c}\text{Molar}\\ \text{rate in}\end{array}\right]& & \left[\begin{array}{c}\text{Molar}\\ \text{rate out}\end{array}\right]& +& \left[\begin{array}{c}\text{Molar rate of}\\ \text{generation}\end{array}\right]& =& \left[\begin{array}{c}\begin{array}{c}\text{Molar rate of}\\ \text{accumulation}\end{array}\end{array}\right]\\ {F}_{\text{A}zz}& & {F}_{\text{A}zz+\mathrm{\Delta}z}& +& {r}_{\text{A}}^{\u2033}{\text{a}}_{c}\left({\text{A}}_{c}{\mathrm{\Delta}}_{z}\right)& =& 0\end{array}& \left(\text{1462}\right)\end{array}$
where ${r}_{\text{A}}^{\u2033}$ = rate of generation of A per unit of catalytic surface area, mol/s · m^{2}
a_{c} = external surface area of catalyst per volume of catalytic bed, m^{2}/m^{3}
${\text{A}}_{c}=\frac{\text{Volume of solid}}{\text{Volume of bed}}\times \frac{\text{Surface area}}{\text{Volume of solid}}=(1\varphi )[{\pi d}_{p}^{2}/({\pi d}_{p}^{3}/6)]$
= 6(1 = ϕ)/d_{p} for packed beds, m^{2}/m^{3}
ϕ = porosity of the bed (i.e., void fraction)^{6}
^{6} In the nomenclature for Chapter 4, for the Ergun equation for pressure drop.
d_{p} = particle diameter, m
A_{c} = crosssectional area of tube containing the catalyst, m^{2}
Dividing Equation (1462) by A_{c}Δz and taking the limit as Δz → 0, we have
$\begin{array}{cc}\frac{1}{{\text{A}}_{c}}\left(\frac{{dF}_{\text{A}z}}{dz}\right)+{r}_{\text{A}}^{\u2033}{\text{A}}_{c}=0& \left(\text{1463}\right)\end{array}$
We now need to express F_{A}_{z} and ${r}_{\text{A}}^{\u2033}$ in terms of concentration.
The molar flow rate of A in the axial direction is
$\begin{array}{cc}{F}_{\text{A}z}={\text{A}}_{c}{W}_{\text{A}z}=({J}_{\text{A}z}+{B}_{\text{A}z}){\text{A}}_{c}& \left(\text{1464}\right)\end{array}$
Axial diffusion is neglected.
In almost all situations involving flow in packedbed reactors, the amount of material transported by diffusion or dispersion in the axial direction is negligible compared with that transported by convection (i.e., bulk flow)
${{J}_{\text{A}z}\text{}\ll \text{}B}_{\text{A}z}$
(In Chapter 18 we consider the case when dispersive effects (e.g., diffusion) must be taken into account.) Neglecting dispersion, Equation (1414) becomes
$\begin{array}{cc}{F}_{\text{A}z}={\text{A}}_{c}{W}_{\text{A}z}={\text{A}}_{c}{B}_{\text{A}z}={UC}_{\text{A}}{\text{A}}_{c}& \left(\text{1465}\right)\end{array}$
where U is the superficial molar average velocity through the bed (m/s). Substituting for F_{A}_{z} in Equation (1463) gives us
$\begin{array}{cc}\frac{d\left({C}_{\text{A}}U\right)}{dz}+{r}_{\text{A}}^{\u2033}{a}_{c}=0& \left(\text{1466}\right)\end{array}$
For the case of constant superficial velocity U
Differential equation describing flow and reaction in a packed bed
$\begin{array}{c}\hline U\frac{{dC}_{\text{A}}}{dz}+{r}_{\text{A}}^{\u2033}{a}_{c}=0\\ \hline\end{array}\begin{array}{c}\left(\text{1467}\right)\end{array}$
For reactions at steady state, the molar flux of A to the particle surface, W_{A}_{r} (mol/m^{2} · s) (see Figure 1410), is equal to the rate of disappearance of A on the surface ${r}_{\text{A}}^{\u2033}$ (mol/m^{2} · s); that is
$\begin{array}{cc}{r}_{\text{A}}^{\u2033}={W}_{\text{A}r}& \left(\text{1468}\right)\end{array}$
From Section 14.4, the boundary condition at the external surface is
$\begin{array}{cc}{r}_{\text{A}}^{\u2033}={W}_{\text{A}r}={k}_{c}({C}_{\text{A}}{C}_{\text{A}s})& \left(14\text{69}\right)\end{array}$
where k_{c} = mass transfer coefficient = (D_{AB}/δ), (m/s)
C_{A} = bulk concentration of A (mol/m^{3})
C_{A}_{s} = concentration of A at the catalytic surface (mol/m^{3})
Substituting for ${r}_{\text{A}}^{\u2033}$ in Equation (1467), we have
$\begin{array}{cc}U\frac{{dC}_{\text{A}}}{dz}{k}_{c}{a}_{c}({C}_{\text{A}}{C}_{\text{A}s})=0& \left(\text{1470}\right)\end{array}$
In reactions that are completely mass transferlimited, it is not necessary to know the rate law.
In most mass transferlimited reactions, the surface concentration is negligible with respect to the bulk concentration (i.e., ${C}_{\text{A}}\text{}\gg \text{}{C}_{\text{A}S}$)
$\begin{array}{cc}U\frac{{dC}_{\text{A}}}{dz}={k}_{c}{a}_{c}{C}_{\text{A}}& \left(\text{1471}\right)\end{array}$
Integrating with the limit, at z = 0, C_{A} = C_{A0}
$\begin{array}{c}\hline \frac{{C}_{\text{A}}}{{C}_{\text{A0}}}=\text{exp}(\frac{{k}_{c}{a}_{c}}{U}z)\\ \hline\end{array}\begin{array}{c}\left(14\text{72}\right)\end{array}$
The corresponding variation of reaction rate along the length of the reactor is
$\begin{array}{cc}{r}_{\text{A}}^{\u2033}={k}_{c}{C}_{\text{A0}}\text{exp}(\frac{{k}_{c}{a}_{c}}{U}z)& \left(\text{1473}\right)\end{array}$
The concentration and conversion profiles down a reactor of length L are shown in Figure 1411.
To determine the reactor length L necessary to achieve a conversion X, we combine the definition of conversion
$\begin{array}{cc}X=\frac{{C}_{\text{A0}}{C}_{\text{AL}}}{{C}_{\text{A0}}}& \left(\text{1474}\right)\end{array}$
Reactor concentration profile for a mass transferlimited reaction
with the evaluation of Equation (1472) at z = L to obtain
$\begin{array}{c}\hline \text{ln}\frac{1}{1X}=\frac{{k}_{c}{a}_{c}}{U}L\\ \hline\end{array}\begin{array}{c}\left(\text{1475}\right)\end{array}$
Robert is an engineer who is always worried (which is a Jofostanian trait). He always thinks something bad will happen if we change an operating condition such as flow rate or temperature, or an equipment parameter such as particle size. Robert’s motto is “If it ain’t broke, don’t fix it.” We can help Robert be a little more adventuresome by analyzing how the important parameters vary as we change operating conditions in order to predict the outcome of such a change. We first look at Equation (1475) and see that conversion depends on the parameters k_{c}, a_{c}, U, and L. We now examine how each of these parameters will change as we change operating conditions. We first consider the effects of temperature and flow rate on conversion.
To learn the effect of flow rate on conversion, we need to know how flow rate affects the mass transfer coefficient. That is, we must determine the correlation for the mass transfer coefficient for the particular geometry and flow field. For flow through a packed bed, the correlation given by Thoenes and Kramers for 0.25 < ϕ < 0.5, 40 < Re′ < 4000, and 1 < Sc < 4000 is^{7}
^{7} D. Thoenes, Jr. and H. Kramers, Chem. Eng. Sci., 8, 271 (1958).
Thoenes–Kramers correlation for flow through packed beds
$\begin{array}{c}\hline {{\text{Sh\u2032}=1.0\left(\text{Re\u2032}\right)}^{1/2\text{}}\text{Sc}}^{1/3}\\ \hline\end{array}\begin{array}{c}\left(\text{1476}\right)\end{array}$
$\begin{array}{c}\hline \left[\frac{{k}_{c}{d}_{p}}{{D}_{\text{AB}}}\left(\frac{\varphi}{1\varphi}\right)\frac{1}{\gamma}\right]={\left[\frac{U{d}_{p}\rho}{\mu (1\varphi )\gamma}\right]}^{1/2}{\left(\frac{\mu}{\rho {D}_{\text{AB}}}\right)}^{1/3}\\ \hline\end{array}\begin{array}{c}\left(14\text{77}\right)\end{array}$
where $\text{Re\u2032}=\frac{\text{Re}}{(1\varphi )\gamma}$
$\text{Sh\u2032}=\frac{\text{Sh}\varphi}{(1\varphi )\gamma}$
d_{p} = particle diameter (equivalent diameter of a sphere of the same volume), m
= [(6/π) (volume of pellet)]^{1/3}, m
ϕ = void fraction (porosity) of packed bed
γ = shape factor (external surface area divided by ${\pi d}_{p}^{2}$)
and
U, ρ, μ, v, and D_{AB} are as previously defined.
For constant fluid properties and particle diameter
$\begin{array}{cc}{{k}_{c}\propto \text{}U}^{1/2}& \left(\text{1478}\right)\end{array}$
We see that the mass transfer coefficient increases with the square root of the superficial velocity through the bed. Therefore, for a fixed concentration, C_{A}, such as that found in a differential reactor, the rate of reaction should vary with U^{1/2}
For diffusionlimited reactions, reaction rate depends on particle size and fluid velocity.
${{r}_{\text{A}}^{\u2033}\text{}\propto \text{}{k}_{c}{C}_{\text{A}}\propto \text{}U}^{1/2}$
However, if the gas velocity is continually increased, a point is reached where the reaction becomes reaction rate–limited and, consequently, is independent of the superficial gas velocity, as shown in Figure 145.
Many mass transfer correlations in the literature are often reported in terms of the Colburn J factor (i.e., J_{D}), which is a function of the Reynolds number. The relationship between J_{D} and the numbers we have been discussing is
$\begin{array}{c}\hline {J}_{\text{D}}=\frac{\text{Sh}}{{\text{Sc}}^{1/3}\text{Re}}\\ \hline\end{array}\begin{array}{c}\left(\text{1479}\right)\end{array}$
Figure 1412 shows data from a number of investigations for the J factor as a function of the Reynolds number for a wide range of particle shapes and gasflow conditions. Note: There are serious deviations from the Colburn analogy when the concentration gradient and temperature gradient are coupled, as shown by Venkatesan and Fogler.^{8}
^{8} R. Venkatesan and H. S. Fogler, AIChE J., 50, 1623 (July 2004).
Colburn J factor
A correlation for flow through packed beds in terms of the Colburn J factor
Dwidevi and Upadhyay review a number of mass transfer correlations for both fixed and fluidized beds and arrive at the following correlation, which is valid for both gases (Re > 10) and liquids (Re > 0.01) in either fixed or fluidized beds:^{9}
^{9} P. N. Dwidevi and S. N. Upadhyay, Ind. Eng. Chem. Process Des. Dev., 16, 157 (1977).
$\begin{array}{c}\hline {\varphi J}_{\text{D}}=\frac{0.765}{{\text{Re}}^{0.82}}+\frac{0.365}{{\text{Re}}^{0.386}}\\ \hline\end{array}\begin{array}{c}\left(\text{1480}\right)\end{array}$
For nonspherical particles, the equivalent diameter used in the Reynolds and Sherwood numbers is ${d}_{p}=\sqrt{{\text{A}}_{p}/\pi}=0.564\text{}\sqrt{{\text{A}}_{p}}$, where A_{p} is the external surface area of the pellet.
To obtain correlations for mass transfer coefficients for a variety of systems and geometries, see either D. Kunii and O. Levenspiel, Fluidization Engineering, 2nd ed. ButterworthHeinemann, 1991, Chap. 7, or W. L. McCabe, J. C. Smith, and P. Harriott, Unit Operations in Chemical Engineering, 6th ed. New York: McGrawHill, 2000. For other correlations for packed beds with different packing arrangements, see I. ColquhounLee and J. Stepanek, Chemical Engineer, 108 (Feb. 1974).
Hydrazine has been studied extensively for use in monopropellant thrusters for space flights of long duration. Thrusters are used for altitude control of communication satellites. Here, the decomposition of hydrazine over a packed bed of aluminasupported iridium catalyst is of interest.^{10} In a proposed study, a 2% hydrazine in 98% helium mixture is to be passed over a packed bed of cylindrical particles 0.25 cm in diameter and 0.5 cm in length at a gasphase velocity of 150 m/s and a temperature of 450 K. The kinematic viscosity of helium at this temperature is 4.94 × 10^{–5} m^{2}/s. The hydrazine decomposition reaction is believed to be externally mass transferlimited under these conditions. If the packed bed is 0.05 m in length, what conversion can be expected? Assume isothermal operation.
^{10} O. I. Smith and W. C. Solomon, Ind. Eng. Chem. Fund., 21, 374.
Actual case history and current application
Additional information:
D_{AB} = 0.69 × 10^{–4} m^{2}/s at 298 K
Bed porosity: 40%
Bed fluidicity: 95.7%
Solution
The following solution is detailed and a bit tedious, but it is important to know the details of how a mass transfer coefficient is calculated.
Rearranging Equation (1464) gives us
$\begin{array}{c}\hline {X=1e}^{({k}_{c}{a}_{c}/U)L}\\ \hline\end{array}\begin{array}{c}\left(\text{E144.1}\right)\end{array}$
(a) Using the Thoenes–Kramers correlation to calculate the mass transfer coefficient, k_{c}
First we find the volumeaverage particle diameter
$\begin{array}{cc}{{d}_{p}={\left(\frac{6V}{\pi}\right)}^{1/3}=\left(6\frac{{\pi D}^{2}}{4}\frac{L}{\pi}\right)}^{1/3}& \left(\text{E144.2}\right)\end{array}$
Surface area per volume of bed
$\begin{array}{cc}{a}_{c}=6\left(\frac{10.4}{{d}_{p}}\right)=6\left(\frac{10.4}{3.61\times {10}^{3}\text{m}}\right)={998\text{m}}^{2}{/\text{m}}^{3}& \left(\text{E144.3}\right)\end{array}$
Tedious reading and calculations, but we gotta know how to do the nitty–gritty.
Mass transfer coefficient
$\text{Re}=\frac{{d}_{p}U}{v}=\frac{(3.61\times {10}^{3}\text{m})\left(150\text{m/s}\right)}{{4.94\times 10}^{5}{\text{m}}^{2/5}}=10942$
For cylindrical pellets
$\begin{array}{c}\begin{array}{cc}\gamma =\frac{{2\pi rL}_{p}+{2\pi r}^{2}}{{\pi d}_{p}^{2}}=\frac{{\left(2\right)(0.005/2)\left(0.005\right)+\left(2\right)(0.0025/2)}^{2}}{({3.61\times {10}^{3})}^{2}}=1.20& \left(\text{E144.4}\right)\end{array}\\ \text{Re\u2032}=\text{}\frac{\text{Re}}{(1\varphi )\gamma}=\frac{10942}{\left(0.6\right)\left(1.2\right)}=15173\end{array}$
Correcting the diffusivity to 450 K using Table 143 gives us
$\begin{array}{cc}\begin{array}{l}\begin{array}{cc}\begin{array}{cc}{{D}_{\text{AB}}\left(\text{450 K}\right)={D}_{\text{AB}}\left(298\text{K}\right)\times \left(\frac{450}{298}\right)}^{1.75}& =\end{array}& ({\text{0.69}\times {10}^{4}\text{m}}^{2}/\text{s})\left(2.06\right)\end{array}\\ {D}_{\text{AB}}\left(\text{450 K}\right)=1.42\times {10}^{5}{\text{m}}^{2/\text{s}}\end{array}& \left(\text{E144.5}\right)\end{array}$
$\text{Sc}=\frac{v}{{D}_{\text{AB}}}=\frac{{{4.94\times 10}^{5}\text{m}}^{2}/\text{s}}{{{1.42\times 10}^{4}\text{m}}^{2}/\text{s}}=0.35$
Representative values
Gas Phase
Re′ = 15173
Sc = 0.35
Sh′ = 86.66
k_{c} = 6.15 m/s
Substituting Re′ and Sc into Equation (1465) yields
$\begin{array}{cc}{{\text{Sh\u2032}=\left(15173.92\right)}^{1/2}\left(0.35\right)}^{1/3}=\left(123.18\right)\left(0.70\right)=86.6& \left(\text{E144.6}\right)\end{array}$
${k}_{c}=\frac{{D}_{\text{AB}}(1\varphi )}{{d}_{p}\varphi}\gamma \left(\text{Sh\u2032}\right)=\left(\frac{1.42{\times {10}^{4}\text{m}}^{2}/\text{s}}{{3.61\times 10}^{3}\text{m}}\right)\left(\frac{10.4}{0.4}\right)\times \left(1.2\right)\left(86.66\right)$
$\begin{array}{c}\hline {k}_{c}=6.15\text{m/s}\\ \hline\end{array}\begin{array}{c}\left(\text{E144.7}\right)\end{array}$
The conversion is
$\begin{array}{l}\begin{array}{cc}X=1\text{exp}[\left(6.15\text{m/s}\right)\left(\frac{998{\text{m}}^{2}/{\text{m}}^{3}}{150\text{m/s}}\right)\left(0.05\text{m}\right)]& \left(\text{E144.8}\right)\end{array}\\ =10.13=0.87\end{array}$
We find 87% conversion.
(b) Colburn J_{D} factor to calculate k_{c}. To find k_{c}, we first calculate the surfaceareaaverage particle diameter.
For cylindrical pellets, the external surface area is
$\begin{array}{cc}\text{A}={\pi dL}_{p}+2\pi \left(\frac{{d}^{2}}{4}\right)& \left(\text{E144.9}\right)\end{array}$
$\begin{array}{ccc}{d}_{p}& =& \sqrt{\frac{\text{A}}{\pi}}=\sqrt{\frac{\pi d{L}_{p}+2\pi ({d}^{2}/4)}{\pi}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\text{E144.10}\right)\hfill \\ & =& \sqrt{\left(0.0025\right)\left(0.005\right)+\frac{{\left(0.0025\right)}^{2}}{2}}=3.95\times {10}^{3}\text{m}\hfill \\ {a}_{c}& =& \frac{6(1\phi )}{{d}_{p}}=910.74{\text{m}}^{2}/{\text{m}}^{3}\hfill \\ \text{Re}& =& \frac{{d}_{p}U}{v}=\frac{(3.95\times {10}^{3}\text{m})\left(150\text{m/s}\right)}{4.94\times {10}^{5}{\text{m}}^{2}/\text{s}}\hfill \\ & =& 11996.04\hfill \end{array}$
Typical values
Gas Phase
Re = 11996
J_{D} = 0.025
Sc = 0.35
Sh = 212
k_{c} = 7.63 m/s
$\begin{array}{cc}\varphi {J}_{\text{D}}=\frac{0.765}{{\text{Re}}^{0.82}}+\frac{0.365}{{\text{Re}}^{0.386}}& \left(\text{1479}\right)\end{array}$
$\begin{array}{l}\begin{array}{cc}=\frac{0.765}{{\left(11996\right)}^{0.82}}+\frac{0,365}{{\left(11996\right)}^{0.386}}=3.5\times {10}^{4}+9.7\times {10}^{3}& \left(\text{E144.11}\right)\end{array}\\ =0.010\end{array}$
$\begin{array}{cc}{J}_{\text{D}}=\frac{0.010}{0.4}=0.025& \left(\text{E144.12}\right)\end{array}$
$\begin{array}{llll}\text{Sh}& =& {\text{Sc}}^{1/3}\text{Re}\left({\text{J}}_{\text{D}}\right)& \left(\text{E144.13}\right)\\ & =& {\begin{array}{c}\left(\text{0.35}\right)\end{array}}^{1/3}\left(11996\right)\left(0.025\right)=212& \\ {k}_{c}& =& \frac{{D}_{\text{AB}}}{{d}_{p}}\text{}\text{Sh}=\frac{1.42\times {10}^{4}}{3.95\times {10}^{3}}\left(212\right)=7.63\text{m/s}& \end{array}$
$\begin{array}{cc}\begin{array}{ccc}\begin{array}{l}\text{Then}X\\ \end{array}& =& 1\text{exp}\left[\left(7.63\text{m/s}\right)\left(\frac{910{\text{m}}^{2}/{\text{m}}^{3}}{150\text{m/s}}\right)\left(0.05\text{m}\right)\right]\hfill \\ & \simeq & 0.9\hfill \\ & & \end{array}& \left(\text{E144.14}\right)\end{array}$
Fluidicity?? Red herring!
If there were such a thing as the bed fluidicity, given in the problem statement, it would be a useless piece of information. Make sure that you know what information you need to solve problems, and go after it. Do not let additional data confuse you or lead you astray with useless information or facts that represent someone else’s bias, and which are probably not well founded.
As we have stressed many times, one of the most important skills of an engineer is to be able to predict the effects of changes of system variables on the operation of a process. The engineer needs to determine these effects quickly through approximate but reasonably close calculations, which are sometimes referred to as “backoftheenvelope calculations.”^{11} This type of calculation is used to answer such questions as “What will happen if I decrease the particle size?” “What if I triple the flow rate through the reactor?”
^{11} Prof. J. D. Goddard, University of Michigan, 1963–1976. Professor Emeritus at the University of California, San Diego.
J. D. Goddard’s
Back of the Envelope
To help answer these questions, we recall Equation (1445) and our discussion on page 754. There, we showed the mass transfer coefficient for a packed bed was related to the product of two terms: Term 1 was dependent on the physical properties and Term 2 was dependent on the system properties. Rewriting Equation (1445) as
$\begin{array}{c}\hline {k}_{c}\propto \left(\frac{{D}_{\text{AB}}^{2/3}}{{v}^{1/6}}\right)\left(\frac{{U}^{1/2}}{{d}_{p}^{1/2}}\right)\\ \hline\end{array}\begin{array}{c}\left(14\text{81}\right)\end{array}$
one observes from this equation that the mass transfer coefficient increases as the particle size decreases. The use of sufficiently small particles offers another technique to escape from the mass transferlimited regime into the reactionratelimited regime.
Find out how the mass transfer coefficient varies with changes in physical properties and system properties.
A mass transferlimited reaction is being carried out in two reactors of equal volume and packing, connected in series as shown in Figure E145.1. Currently, 86.5% conversion is being achieved with this arrangement. It is suggested that the reactors be separated and the flow rate be divided equally among each of the two reactors (Figure E145.2) to decrease the pressure drop and hence the pumping requirements. In terms of achieving a higher conversion, Robert is wondering if this is a good idea.
Reactors in series versus reactors in parallel
Solution
For the series arrangement we were given, X_{1} = 0.865, and for the parallel arrangement, the conversion is unknown, that is, X_{2} = ? As a first approximation, we neglect the effects of small changes in temperature and pressure on mass transfer.
We recall Equation (1475), which gives conversion as a function of reactor length. For a mass transferlimited reaction
$\begin{array}{cc}\text{ln}\frac{1}{1X}=\frac{{k}_{c}{a}_{c}}{U}L& \left(\text{1475}\right)\end{array}$
For case 1, the undivided system
$\begin{array}{cccc}\left(\begin{array}{c}\text{ln}\frac{1}{1{X}_{1}}\end{array}\right)& =& \frac{{k}_{c1}{a}_{c}}{{U}_{1}}{L}_{1}& \text{}\left(\text{E1451}\right)\\ {X}_{1}& =& 0.865& \underset{\xaf}{\begin{array}{c}\text{Answer 1}\end{array}}\end{array}$
For case 2, the divided system
$\begin{array}{cc}\begin{array}{c}\left(\text{ln}\frac{1}{1{X}_{2}}\right)=\frac{{k}_{c2}{a}_{c}}{{U}_{2}}{L}_{2}\\ {X}_{2}=?\end{array}& \left(\text{E145.2}\right)\end{array}$
We now take the ratio of case 2 (divided system) to case 1 (undivided system)
$\begin{array}{cc}\frac{\text{ln}\frac{1}{1{X}_{2}}}{\text{ln}\frac{1}{1{X}_{1}}}=\frac{{k}_{c2}}{{k}_{c1}}\left(\frac{{L}_{2}}{{L}_{1}}\right)\frac{{U}_{1}}{{U}_{2}}& \left(\text{E145.3}\right)\end{array}$
The surface area per unit volume a_{c} is the same for both systems.
From the conditions of the problem statement we know that
$\begin{array}{lll}{L}_{2}& =& \frac{1}{2}{L}_{1},{U}_{2}=\frac{1}{2}{U}_{1},\text{and}{X}_{1}=0.865\\ {X}_{2}& =& ?\end{array}$
However, we must also consider the effect of the division on the mass transfer coefficient. From Equation (1481) we know that
k_{c} ∝ U^{1/2}
$\begin{array}{cc}{\frac{{k}_{c2}}{{k}_{c1}}=\left(\frac{{U}_{2}}{{U}_{1}}\right)}^{1/2}& \left(\text{E145.4}\right)\end{array}$
Multiplying by the ratio of superficial velocities yields
$\begin{array}{cc}{\frac{{U}_{1}}{{U}_{2}}\left(\frac{{k}_{c2}}{{k}_{c1}}\right)=\left(\frac{{U}_{1}}{{U}_{2}}\right)}^{1/2}& \text{}\left(\text{E145.5}\right)\end{array}$
$\begin{array}{cc}\hline \begin{array}{ccc}\text{ln}\frac{1}{1{X}_{2}}& =& {\begin{array}{c}\left(\text{ln}\frac{1}{1{X}_{1}}\right)\frac{{L}_{2}}{{L}_{1}}\left(\frac{{U}_{1}}{{U}_{2}}\right)\end{array}}^{1/2}\hfill \end{array}& \\ \hline\end{array}$
$\begin{array}{cc}\begin{array}{ccc}& =& \left(\text{ln}\frac{1}{10.865}\right)\left[\frac{\frac{1}{2}{L}_{1}}{{L}_{1}}{\left(\frac{{U}_{1}}{\frac{1}{2}{U}_{1}}\right)}^{1/2}\right]\hfill \\ & =& 2.00\left(\frac{1}{2}\right)\sqrt{2}=1.414\hfill \end{array}& \left(\text{E145.6}\right)\end{array}$
Solving for X_{2} gives us
X_{2} = 0.76 Answer 2
Analysis: Consequently, we see that although the divided arrangement will have the advantage of a smaller pressure drop across the bed, it is a bad idea in terms of conversion. Recall that the series arrangement gave X_{1} = 0.865; therefore (X_{2} < X_{1}). Bad idea!! But every chemical engineering student in Jofostan knew that! Recall that if the reaction were reaction rate–limited, both arrangements would give the same conversion.
Bad idea!! Robert was right to worry.
The same reaction as that in Example 145 is being carried out in the same two reactors in series. A new engineer suggests that the rate of reaction could be increased by a factor of 2^{10} by increasing the reaction temperature from 400°C to 500°C, reasoning that the reaction rate doubles for every 10 = C increase in temperature. Another engineer arrives on the scene and berates the new engineer with quotations from Chapter 3 concerning this rule of thumb. She points out that it is valid only for a specific activation energy within a specific temperature range. She then suggests that he go ahead with the proposed temperature increase but should only expect an increase on the order of 2^{3} or 2^{4}. What do you think? Who is correct?
Robert worries if this temperature increase will be worth the trouble.
Solution
Because almost all surface reaction rates increase more rapidly with temperature than do diffusion rates, increasing the temperature will only increase the degree to which the reaction is mass transferlimited.
We now consider the following two cases:
$\begin{array}{cc}\text{Case 1:}T=400\xb0\text{C}& X=0.865\hfill \\ \text{Case 2:}T=500\xb0\text{C}& X=?\hfill \end{array}$
Taking the ratio of case 2 to case 1 and noting that the reactor length is the same for both cases (L_{1} = L_{2}), we obtain
$\begin{array}{cc}\frac{\text{ln}\frac{1}{1{X}_{2}}}{\text{In}\frac{1}{1{X}_{1}}}=\frac{{k}_{c2}}{{k}_{c1}}\left(\frac{{L}_{2}}{{L}_{1}}\right)\frac{{U}_{1}}{{U}_{2}}=\frac{{k}_{c2}}{{k}_{c1}}\left(\frac{{U}_{1}}{{U}_{2}}\right)& \left(\text{E146.1}\right)\end{array}$
The molar feed rate F_{T}_{0} remains unchanged
$\begin{array}{cc}{F}_{\text{T0}}={\upsilon}_{01}\left(\frac{{P}_{01}}{{RT}_{01}}\right)={\upsilon}_{02}\left(\frac{{P}_{02}}{{RT}_{02}}\right)& \left(\text{E146.2}\right)\end{array}$
the pressure remains constant so
$\frac{{\upsilon}_{01}}{{T}_{1}}=\frac{{\upsilon}_{02}}{{T}_{2}}$
Because υ = A_{c}U, the superficial velocity at temperature T_{2} is
$\begin{array}{cc}{U}_{2}=\frac{{T}_{2}}{{T}_{1}}{U}_{1}& \left(\text{E146.3}\right)\end{array}$
J. D. G.
We now wish to learn the dependence of the mass transfer coefficient on temperature
$\begin{array}{cc}{k}_{c}\propto \left(\frac{{U}^{1/2}}{{d}_{p}^{1/2}}\right)\left(\frac{{D}_{\text{AB}}^{2/3}}{{v}^{1/6}}\right)& \left(\text{E146.4}\right)\end{array}$
Taking the ratio of case 2 to case 1 and realizing that the particle diameter is the same for both cases gives us
$\begin{array}{cc}{\frac{{k}_{c2}}{{k}_{c1}}={\left(\frac{{U}_{2}}{{U}_{1}}\right)}^{1/2}{\left(\frac{{D}_{\text{AB2}}}{{D}_{\text{AB1}}}\right)}^{2/3}\left(\frac{{v}_{1}}{{v}_{2}}\right)}^{1/6}& \left(\text{E146.5}\right)\end{array}$
The temperature dependence of the gasphase diffusivity is (from Table 143)
$\begin{array}{cc}{D}_{\text{AB}}\propto \text{}{T}^{1.75}& \left(\text{E146.6}\right)\end{array}$
For most gases, viscosity increases with increasing temperature according to the relation
μ ∝ T^{1/2}
From the ideal gas law
ρ ∝ T^{1}
Then
$\begin{array}{cc}v=\frac{\mu}{\rho}{\propto T}^{3/2}& \left(\text{E146.7}\right)\end{array}$
$\begin{array}{c}\begin{array}{c}\begin{array}{ccc}\begin{array}{cc}{\frac{\text{ln}\frac{1}{1{X}_{2}}}{\text{ln}\frac{1}{1{X}_{1}}}}^{}& =\end{array}& \frac{{U}_{1}}{{U}_{2}}\left(\frac{{k}_{c2}}{{k}_{c1}}\right)={\left(\frac{{U}_{1}}{{U}_{2}}\right)}^{1/2}{\left(\frac{{D}_{\text{AB2}}}{{D}_{\text{AB1}}}\right)}^{2/3}(\frac{{\upsilon}_{1}}{{v}_{2}}{)}^{1/6}& \text{}\left(\text{E146.8}\right)\end{array}\end{array}\end{array}$
$\begin{array}{c}{\begin{array}{cc}=& {\begin{array}{c}\left(\frac{{T}_{1}}{{T}_{2}}\right)\end{array}}^{1/2}{\left[{\left(\frac{{T}_{2}}{{T}_{1}}\right)}^{1.75}\right]}^{2/3}\left[{\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{3/2}\right]\end{array}}^{1/6}\end{array}$
$\begin{array}{cc}\begin{array}{ccc}\frac{{U}_{1}{k}_{c2}}{{U}_{2}{k}_{c1}}& =& {\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{1/2}{\left(\frac{{T}_{2}}{{T}_{1}}\right)}^{7/6}{\left(\frac{{T}_{1}}{{T}_{2}}\right)}^{1/4}=(\frac{{T}_{2}}{{T}_{1}}{)}^{5/12}\hfill \\ & =& {\begin{array}{c}\left(\frac{773}{673}\right)\end{array}}^{5/12}=1.059\hfill \end{array}& \left(\text{E146.9}\right)\end{array}$
It’s really important to know how to do this type of analysis.
Rearranging Equation (E146.1) in the form
$\begin{array}{cc}\begin{array}{c}\begin{array}{c}\text{ln}\frac{1}{1{X}_{2}}=\frac{{k}_{c2}{U}_{1}}{{k}_{c1}{U}_{2}}\text{ln}\frac{1}{1X1}\\ \text{ln}\frac{1}{1{X}_{2}}=1.059\left(\text{ln}\frac{1}{10.685}\right)=1.059\left(2\right)\end{array}\hfill \\ \begin{array}{cc}\begin{array}{c}{X}_{2}=0.88\end{array}\underset{\xaf}{\text{Answer}}& \end{array}\end{array}\hfill & \left(\text{E146.10}\right)\end{array}$
Analysis: Consequently, we see that increasing the temperature from 400°C to 500°C increases the conversion by only 1.7%, that is, X = 0.865 compared to X = 0.88. Bad idea! Bad, bad idea! Both engineers would have benefited from a more thorough study of this chapter.
Bad idea!! Robert was right to worry.
For a packed catalyst bed, the temperaturedependence part of the mass transfer coefficient for a gasphase reaction can be written as
$\begin{array}{cc}\begin{array}{cc}{k}_{c}\propto \text{}{U}^{1/2}& ({D}_{\text{AB}}^{2/3}/{v}^{1/6})\end{array}& \left(\text{1482}\right)\end{array}$
$\begin{array}{cc}{k}_{c}\text{}\propto \text{}{U}^{1/2}{T}^{11/12}& \left(\text{1483}\right)\end{array}$
Taking the ratio of U to U_{0} and T to T_{0} and k_{c}_{0} (T_{0}, U_{0}) the mass transfer coefficient at any other T and U can be found from the Equation
$\begin{array}{cc}{{{k}_{c}={k}_{c0}\left(\frac{U}{{U}_{0}}\right)}^{1/2}\left(\frac{T}{{T}_{0}}\right)}^{11/12}& \left(\text{1484}\right)\end{array}$
One can use Equation (1481) when studying mass transferlimited gasphase reaction.
Depending on how one fixes or changes the molar feed rate, F_{T}_{0}, U may also depend on the feed temperature.
As an engineer, it is extremely important that you reason out the effects of changing conditions, as illustrated in the preceding two examples.
Important concept
Let’s revisit a packedbed reactor shown in Figure E144.1 in Example 144. We want to put the algorithm in a form where one can use Wolfram or Python to vary all the physical parameters and operating variables in order to study a firstorder gasphase reaction in a PBR. The reaction will be carried out isothermally and pressure drop is neglected.
Study the conversion and reaction rate profile for a reaction that is partially diffusion limited by first plotting the conversion as a function of catalyst weight up to a value W = 100 kg. Next use Wolfram to vary all the parameters and describe what you find.
Solution
Mole Balance
$\begin{array}{cc}\frac{dX}{dW}={r}_{\text{A}}^{\prime}/{F}_{\text{A0}}& \left(\text{E147.1}\right)\end{array}$
Rate Law
2A. First order, ${r}_{\text{A}}^{\u2033}={k}_{r}{C}_{\text{As}}$
$\begin{array}{cc}{r}_{\text{A}}^{\prime}={r}_{\text{A}}^{\u2033}*{a}_{c}^{\prime}& \left(\text{E147.2}\right)\end{array}$
At steady state the molar flux to the surface is equal to the rate of reaction on the surface
$\begin{array}{cc}{W}_{\text{A}}={k}_{c}({C}_{\text{A0}}{C}_{\text{As}})={k}_{r}{C}_{\text{As}}={r}_{\text{A}}^{\u2033}=\frac{{k}_{c}{k}_{r}}{{k}_{r}+{k}_{c}}{C}_{\text{A}}={kC}_{\text{A}}& \left(\text{E147.3}\right)\end{array}$
where
k_{r} = reactionrate constant on the surface (m/s)
k_{c} = mass transfer coefficient (m/s)
$k=\frac{{k}_{c}{k}_{r}}{{k}_{r}+{k}_{c}}$ (m/s) overall transfer coefficient
${a}_{c}^{\prime}$ = external surface area of the catalyst per unit mass of catalyst (m^{2}/kgcat)
2B. Using the Thoenes and Kramer’s correlation, Equations (1476) and (1477)
$\begin{array}{cc}\text{Sh\u2032}=1.0{\left(\text{Re\u2032}\right)}^{1/2}{\text{Sc}}^{1/3}& \left(\text{1476}\right)\end{array}$
$\begin{array}{cc}{\left[\frac{{k}_{c}{d}_{p}}{{D}_{\text{AB}}}\left(\frac{\varphi}{1\varphi}\right)\frac{1}{\gamma}\right]=\left[\frac{{Ud}_{p}\rho}{\mu (1\varphi )\gamma}\right]\left(\frac{\mu}{{\rho D}_{\text{AB}}}\right)}^{1/3}& \left(\text{1477}\right)\end{array}$
Correlation for k_{c} is
$\begin{array}{cc}{k}_{c}=\frac{\text{Sh\u2032}\text{}{D}_{\text{AB}}}{{d}_{p}}\frac{(1\varphi )\gamma}{\varphi}& \left(\text{E147.4}\right)\end{array}$
$\begin{array}{cc}{{\text{Sh\u2032}=1.0\left(\text{Re\u2032}\right)}^{1/2}\text{Sc}}^{1/3}& \left(\text{E147.5}\right)\end{array}$
$\begin{array}{cc}\text{Re\u2032}=\frac{\text{Re}}{(1\varphi )\gamma}& \left(\text{E147.6}\right)\end{array}$
$\begin{array}{cc}\text{Re}=\frac{{d}_{p}U}{v}& \left(\text{E147.7}\right)\end{array}$
$\begin{array}{cc}\text{Sc}=\frac{v}{{D}_{\text{AB}}}& \left(\text{E147.8}\right)\end{array}$
$\begin{array}{cc}v=\frac{\mu}{\rho}& \left(\text{E147.9}\right)\end{array}$
$\begin{array}{cc}{a}_{c}^{\prime}=6\frac{(1\varphi )}{{d}_{p}^{*}{\rho}_{c}}& \left(\text{E147.10}\right)\end{array}$
Stoichiometry with P = P_{0} and T = T_{0}
$\begin{array}{cc}{C}_{\text{A}}={C}_{\text{A0}}(1X)& \left(\text{E147.11}\right)\end{array}$
Combine Polymath
We will now combine Equations (E147.1) through (E147.11) into Polymath, Wolfram, or Python to vary the parameters. The Polymath program is shown in Table E147.2 and the Wolfram solution in Figure E147.1.
Evaluate
The nominal values of the parameter to be varied are shown in Table E147.1.
TABLE E147.1 NOMINAL PARAMETER VALUES
Parameters 
Liquids 
Gases 

υ 
10^{–6} m^{2}/s 
10^{–5} m^{2}/s 
ρ 
1000 kg/m^{3} 
1 kg/m^{3} 
D_{AB} 
10^{–8} m^{2}/s 
10^{–5} m^{2}/s 
μ 
10^{–3} kg/m/s 
10^{–5} kg/m/s 
U 
1 m/s 
100 m/s 
d_{p} 
10^{–3} m 
10^{–2} m 
k_{r} 
0.5 m/s 
5 m/s 
A_{c} 
5*10^{–2} m^{2} 
5*10^{–2} m^{2} 
γ 
1 
1 
ϕ 
0.4 
0.4 
ρ_{c} 
3000 kg/m^{3} 
3000 kg/m^{3} 
C_{A0} 
1 mol/m^{3} 
0.01 mol/m^{3} (Ca 1 atm) 
Plot the conversion as a function of catalyst weight.
Analysis: In this example we modeled a partially diffusion limited reaction in a PBR. The rest of the example is now up to you, the reader. Go to the LEP 147 and vary a number of physical parameters and operating variables, and write a set of observations and conclusions.
Dust Explosion at Imperial Sugar Company
According to the Chemical Safety Board, the yearly average number of dust explosions has been 5 per year over the last 5 years. In Examples 141 and 143, we discussed and calculated the rate of mass transfer of oxygen to small carbon dust particles allowing the particle to burn and possibly causing a dust explosion. Dust explosions can produce deadly consequences. On February 7, 2008, combustible fine sugar was ignited at the Imperial Sugar Company causing an explosion. Because the fine granulated sugar has high surface area, it has the potential to rapidly burn and explode. Prior to the explosion the dust collection system was not working properly; as a result fine sugar that had spilled over on the floor and collected on the ceiling and piping became airborne. When the belt transport system was enclosed, the airborne sugar dust found an ignition source and ignited. The explosion was fueled by the large accumulation of airborne sugar dust throughout the packaging plant. The fire and explosion caused 14 deaths and 38 injuries, including lifethreatening burns.
(https://www.csb.gov/imperialsugarcompanydustexplosionandfire/)
As chemical reaction engineers, it is important that we understand this accident, why it happened, and how it could have been prevented to ensure similar accidents may be prevented. In order to become familiar with a strategy for accident awareness and prevention, view the Chemical Safety Board video on the explosion and determine whether all the important concepts to prevent and mitigate the incident were captured in the BowTie diagram below.
Two other excellent videos on dust explosions are (https://www.chemengonline.com/combustibledustfiresexplosionsrecentdatalessonslearned/) and (https://www.aiche.org/academy/webinars/dustexplosion). The BowTie diagram for the sugar dust plan explosion is shown in Figure 1413.
The molar flux of A in a binary mixture of A and B is
$\begin{array}{cc}{\mathbf{\text{W}}}_{\text{A}}={D}_{\text{AB}}\text{}\mathrm{\nabla}{C}_{\text{A}}+{y}_{\text{A}}({\mathbf{\text{W}}}_{\text{A}}+{\mathbf{\text{W}}}_{\text{B}})& \left(\text{S141}\right)\end{array}$
For equimolar counterdiffusion (EMCD) or for dilute concentration of the solute
$\begin{array}{cc}{\mathbf{\text{W}}}_{\text{A}}={\mathbf{\text{J}}}_{\text{A}}={D}_{\text{AB}}\text{}\mathrm{\nabla}{\text{C}}_{\text{A}}& \left(\text{S142}\right)\end{array}$
For diffusion through a stagnant gas
$\begin{array}{cc}{\mathbf{\text{W}}}_{\text{A}}={cD}_{\text{AB}}\text{}\mathrm{\nabla}\text{ln}(1{y}_{\text{A}})& \left(\text{S143}\right)\end{array}$
For negligible diffusion
$\begin{array}{cc}{\mathbf{\text{W}}}_{\text{A}}={y}_{\text{A}}\text{}\text{}\mathbf{\text{W}}={y}_{\text{A}}({\mathbf{\text{W}}}_{\text{A}}+{\mathbf{\text{W}}}_{\text{B}})={\mathbf{\text{C}}}_{\text{A}}\text{U}& \left(\text{S144}\right)\end{array}$
The rate of mass transfer from the bulk fluid to a boundary at concentration C_{A}_{s} is
$\begin{array}{cc}{W}_{\text{A}}={k}_{c}({C}_{\text{A}b}{C}_{\text{As}})& \left(\text{S145}\right)\end{array}$
where k_{c} is the mass transfer coefficient.
The Sherwood and Schmidt numbers are, respectively,
If a heat transfer correlation exists for a given system and geometry, the mass transfer correlation may be found by replacing the Nusselt number by the Sherwood number and the Prandtl number by the Schmidt number in the existing heat transfer correlation.
Increasing the gasphase velocity and decreasing the particle size will increase the overall rate of reaction for reactions that are externally mass transferlimited.
The conversion for externally mass transferlimited reactions can be found from the equation
$\begin{array}{cc}\text{ln}\frac{1}{1X}=\frac{{k}_{c}{a}_{c}}{U}L& \left(\text{S148}\right)\end{array}$
Backoftheenvelope calculations should be carried out to determine the magnitude and direction that changes in process variables will have on conversion. What if . . .?
(http://www.umich.edu/~elements/6e/14chap/obj.html#/)
• Additional Material on the Web site
Transdermal Drug Delivery Example
Transdermal drug delivery schematic.
The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.
A = • B = ▪ C = ♦ D = ♦♦
Q141_{A} QBR (Question Before Reading). Under what circumstances will the conversion predicted by mass transfer limitations be greater than that for surface reaction limitations?
Q142 Read over the problems at the end of this chapter. Make up an original problem that uses the concepts presented in this chapter. See Problem P51_{A} for the guidelines. To obtain a solution:
Make up your data and reaction.
Use a real reaction and real data.
The journals listed at the end of Chapter 1 may be useful for part (b).
Q143 (Sergeant Ambercromby). Capt. Apollo is piloting a shuttlecraft on his way to space station Klingon. Just as he is about to maneuver to dock his craft using the hydrazine system discussed in Example 142, the shuttle craft’s thrusters do not respond properly and it crashes into the station, killing Capt. Apollo (Star Wars 7 (fall 2015)). An investigation reveals that Lt. Darkside prepared the packed beds used to maneuver the shuttle and Lt. Data prepared the hydrazine–helium gas mixture. Foul play is suspected and Sgt. Ambercromby arrives on the scene to investigate.
What are the first three questions he asks?
Make a list of possible explanations for the crash, supporting each one by an equation or reason.
Q144 How would your answers change if the temperature was increased by 50°C, the particle diameter was doubled, and fluid velocity was cut in half? Assume properties of water can be used for this system.
Q145 How would your answers change in Example 143 if you had a 50–50 mixture of hydrazine and helium? If you increase d_{p} by a factor of 5?
Q146_{A} After viewing the video, what was new to you with regard to dust explosions? Could you list three things one should do or have in place to prevent dust explosions?
Q147 Go to the LearnChemE screencast link for Chapter 14 (http://www.learncheme.com/screencasts/kineticsreactordesign). View one or more of the screencast 5 to 6minute videos and write a twosentence evaluation of what you learned.
Q148 AWFOS–S14 View the CSB video to list two things in which dust explosions are different from those involving flammable liquids.
P141_{B}
Example 141: Mass Transfer of Oxygen to a Burning Carbon Particle
Wolfram and Python
Vary each slider to find the parameter to which the flux W_{A}_{r} is most sensitive.
What happens when the liquid diffusity and viscosity are both increased simultaneously?
Vary the velocity as shown in Figure 145 and describe its effect on the flux.
Write a set of conclusions based on the above experiments.
Example 142: Rapid LiquidPhase Reaction on the Surface of a Catalyst Wolfram
and Python
Vary each of the parameters and tell which one the mass transfer coefficient (k_{c}) is most sensitive.
Change D_{AB} and U simultaneously both up and down and describe what you find.
Write a set of three conclusions from your experiments (i) and (ii).
Example 143: Combustion Time for a Single Particle
Wolfram and Python
Compare the burning time for a 100 μm porous carbon particle with a 10% solid with t_{c} in Example 143.
Vary all the parameters you can think of that would give a burning time of 500 ms.
Write a set of conclusions from your Wolfram experiments (i) and (ii).
Example 144: Mass Transfer Effects in Maneuvering a Space Satellite. What if you were asked for representative values for Re, Sc, Sh, and k_{c} for both liquid and gasphase systems for a velocity of 10 cm/s and a pipe diameter of 5 cm (or a packedbed diameter of 0.2 cm)? What numbers would you give?
Example 145: The Case of Divide and Be Conquered. How would your answers change if the reaction were carried out in the liquid phase where kinematic viscosity, υ, varied as $v\left({T}_{2}\right)=v\left({T}_{1}\right)\text{exp}[\text{400 K}(\frac{1}{{T}_{1}}\frac{1}{{T}_{2}})]?$
Example 147: Flow, Diffusion, and Reaction in a Packed Bed
Wolfram and Python
Vary k_{r} and U and view X, W_{A}, and k as a function of W.
Vary D_{AB} and d_{p} and view k as a function of W.
Vary the ration (D_{AB}/υ) and describe what you find.
Vary the parameters by a factor of 10 above and below their nominal values and describe what you find (e.g., what were the most sensitive and least sensitive parameters? Hint: See Equation (1481).
Write a set of conclusions based on all your experiments.
P142_{B} Assume the minimum respiration rate of a chipmunk is 1.5 micromoles of O_{2}/min. The corresponding volumetric rate of gas intake is 0.05 dm^{3}/min at STP.
What is the deepest a chipmunk can burrow a 3cm diameter hole beneath the surface in Ann Arbor, Michigan? D_{AB} = 1.8 × 10^{–5} m^{2}/s
In Boulder, Colorado?
How would your answers to (a) and (b) change in the dead of winter when T = 0°F?
Critique and extend this problem (e.g., CO_{2} poisoning). Thanks to Professor Robert Kabel at Pennsylvania State University.
Hint: Review derivations and equations for W_{A} and W_{B} to see how they can be applied to this problem.
P143_{B} Pure oxygen is being absorbed by xylene in a catalyzed reaction in the experimental apparatus sketched in Figure P143_{B}. Under constant conditions of temperature and liquid composition, the following data were obtained:
Rate of Uptake of O_{2} (mL/h) for System Pressure (absolute) 


Stirrer Speed (rpm) 
1.2 atm 
1.6 atm 
2.0 atm 
3.0 atm 
400 
15 
31 
75 
152 
800 
20 
59 
102 
205 
1200 
21 
62 
105 
208 
1600 
21 
61 
106 
207 
No gaseous products were formed by the chemical reaction. What would you conclude about the relative importance of liquidphase diffusion and about the order of the kinetics of this reaction? (California Professional Engineers Exam)
P144_{B} Derive an equation for the time necessary to completely burn a 100 μm carbon particle as a function of D_{0}. Also calculate the burning rate constant. Use the K_{S} values of the parameter values, for example, C_{A}_{∞}, given in Example 143.
diffusion controlled with D_{AB} = 10^{–4} m^{2}/s
reaction controlled with k_{r} = 0.01 m/s
combined reaction and diffusion controlled
Additional Information:
ρ_{c} = 188 mol/m^{3}
P145_{C} In a divingchamber experiment, a human subject breathed a mixture of O_{2} and He while small areas of his skin were exposed to nitrogen gas. After some time, the exposed areas became blotchy, with small blisters forming on the skin. Model the skin as consisting of two adjacent layers, one of thickness δ_{1} and the other of thickness δ_{2}. If counterdiffusion of He out through the skin occurs at the same time as N_{2} diffuses into the skin, at what point in the skin layers is the sum of the partial pressures a maximum? If the saturation partial pressure for the sum of the gases is 101 kPa, can the blisters be a result of the sum of the gas partial pressures exceeding the saturation partial pressure and the gas coming out of the solution (i.e., the skin)?
Before answering any of these questions, derive the concentration profiles for N_{2} and He in the skin layers.
Diffusivity of He and N_{2} in the inner skin layer = 5 × 10^{−7} cm^{2}/s and 1.5 × 10^{−7} cm^{2}/s, respectively.
Diffusivity of He and N_{2} in the outer skin layer = 10^{−5} cm^{2}/s and 3.3 × 10^{4} cm^{2}/s, respectively.
External Skin Boundary Partial Pressure 
Internal Skin Boundary Partial Pressure 


N_{2}  101 kPa 
0 
He  0 
81 kPa 
δ_{1}  20 μm 
Stratum corneum 
δ_{2}  80 μm 
Epidermis 
Hint: See Transdermal Drug Delivery in Expanded Material on the CRE Web site.
P146_{B} The decomposition of cyclohexane to benzene and hydrogen is mass transferlimited at high temperatures. The reaction is carried out in a 5cmID pipe 20 m in length packed with cylindrical pellets 0.5 cm in diameter and 0.5 cm in length. The pellets are coated with the catalyst only on the outside. The bed porosity is 40%. The entering volumetric flow rate is 60 dm^{3}/min.
Calculate the number of pipes necessary to achieve 99.9% conversion of cyclohexane from an entering gas stream of 5% cyclohexane and 95% H_{2} at 2 atm and 500°C.
Plot conversion as a function of pipe length.
How much would your answer change if the pellet diameter and length were each cut in half?
How would your answer to part (a) change if the feed were pure cyclohexane?
What do you believe is the point of this problem? Is the focus really green CRE? How so?
P147_{C} Lead titanate, PbTiO_{3}, is a material having remarkable ferroelectric, pyroelectric, and piezoelectric properties [ J. Elec. Chem. Soc., 135, 3137 (1988)]. A thin film of PbTiO_{3} was deposited in a CVD reactor. The deposition rate is given below as a function of a temperature and flow rate over the film.
Flow Rate (SCCM) 
Temperature (°C) 
Deposition Rate (mg/cm^{2} · h) 

500 
650 
0.2 
750 
0.8 

800 
1.2 

600 
650 
0.35 
750 
1.0 

800 
1.5 

750 
650 
0.53 
750 
1.45 

800 
2.0 

1000 
650 
0.55 
750 
1.5 

800 
2.0 
What are all the things, qualitative and quantative, that you can learn from these data?
P148_{B} OEQ (Old Exam Question). A plant is removing a trace of Cl_{2} from a wastegas stream by passing it over a solid granulm absorbent in a tubular packed bed (Figure P148_{B}). At present, 63.2% removal is being acomplished, but it is believed that greater removal could be achieved if the flow rate were increased by a factor of 4, the particle diameter were decreased by a factor of 3, and the packed tube length increased by 50%. What percentage of chlorine would be removed under the proposed scheme? (The chlorine transferring to the absorbent is removed completely by a virtually instantaneous chemical reaction.) (Ans: 98%)
P149_{B} OEQ (Old Exam Question). In a certain chemical plant, a reversible fluidphase isomerization
$\text{A}\underleftarrow{\to}\text{}\text{B}$
is carried out over a solid catalyst in a tubular packedbed reactor. If the reaction is so rapid that mass transfer between the catalyst surface and the bulk fluid is ratelimiting, show that the kinetics are described in terms of the bulk concentrations C_{A} and C_{B} by
${r}_{\text{A}}^{\u2033}=\frac{{k}_{\text{B}}[{C}_{\text{A}}(1/K){C}_{\text{B}}]}{1/K+{k}_{\text{B}}/{k}_{\text{A}}}$
where ${r}_{\text{A}}^{\prime}$ = moles of A reacting per unit area catalyst per
k_{A}, k_{B} = transfer coefficients for A and B
K = equilibrium constant
It is desired to double the capacity of the existing plant by processing twice the feed of reactant A while maintaining the same fractional conversion of A to B in the reactor. How much larger a reactor, in terms of catalyst weight, would be required if all other operating variables are held constant? You may use the Thoenes–Kramers correlation for mass transfer coefficients in a packed bed.
P1410_{B} The irreversible gasphase reaction
$\text{A}\stackrel{\text{cat}}{\to}\text{B}$
is carried out adiabatically over a packed bed of solid catalyst particles. The reaction is first order in the concentration of A on the catalyst surface
${r}_{\text{As}}^{\prime}={k\prime C}_{\text{As}}$
The feed consists of 50% (mole) A and 50% inerts, and enters the bed at a temperature of 300 K. The entering volumetric flow rate is 10 dm^{3}/s (i.e., 10000 cm^{3}/s). The relationship between the Sherwood number and the Reynolds number is
Sh = 100 Re^{1/2}
As a first approximation, one may neglect pressure drop. The entering concentration of A is 1.0 M. Calculate the catalyst weight necessary to achieve 60% conversion of A for
isothermal operation.
adiabatic operation.
Additional information:
Kinematic viscosity: μ/ρ = 0.02 cm^{2}/s
Particle diameter: d_{p} = 0.1 cm
Superficial velocity: U = 10 cm/s
Catalyst surface area/mass of catalyst bed: a = 60 cm^{2}/gcat
Diffusivity of A: D_{e} = 10^{−2} cm^{2}/s
Heat of reaction: $\mathrm{\Delta}{H}_{\text{Rx}}^{\circ}$= −10000 cal/g mol A
Heat capacities: C_{p}_{A} = C_{p}_{B} = 25 cal/g mol · K, C_{pS} (solvent) = 75 cal/g mol · K
k′ (300 K) = 0.01 cm^{3}/s · gcat with E = 4000 cal/mol
P1411_{B} Transdermal Drug Delivery. See photo on page 781. The principles of steadystate diffusion have been used in a number of drugdelivery systems. Specifically, medicated patches are commonly attached to the skin to deliver drugs for nicotine withdrawal, birth control, and motion sickness, to name a few. The U.S. transdermal drugdelivery is a multibillion dollar market. Equations similar to Equation (1424) have been used to model the release, diffusion, and absorption of the drug from the patch into the body. The figure shown in the Additional Material and on page 781 shows a drugdelivery vehicle (patch) along with the concentration gradient in the epidermis and dermis skin layers.
Use a shell balance to show
$\frac{{dW}_{\text{A}z}}{dz}=0$
Show the concentration profile in the epidermis layer
$\frac{{\text{C}}_{\text{A0}}{C}_{\text{A}}}{{{C}_{\text{A}}}_{0}{C}_{\text{A1}}}=\frac{z}{{\delta}_{1}}$
Show the concentration profile in the dermis layer
$\frac{{C}_{\text{A}}}{{C}_{\text{A1}}}=\frac{{\delta}_{2}z}{{{\delta}_{2}}^{}{\delta}_{1}}$
Equate the fluxes using ${\text{W}}_{\text{A1}}={D}_{\text{A1}}\frac{{dC}_{\text{A}}}{dz}$ and ${\text{W}}_{\text{A2}}={D}_{\text{A2}}\frac{{dC}_{\text{A}}}{dz}$ and to show
${C}_{\text{A1}}=\frac{{D}_{\text{A1}}\frac{{C}_{\text{A0}}}{{\delta}_{1}}}{\frac{{D}_{\text{A2}}}{{\delta}_{2}{\delta}_{1}}+\frac{{D}_{\text{A1}}}{{\delta}_{1}}}$
What are the concentration profiles in the dermis and epidermis layers?
Show the flux in the dermis layer is
${W}_{\text{A}z}=\frac{{D}_{\text{A2}}}{{\delta}_{2}{\delta}_{1}}\left[\frac{{D}_{\text{A1}}\frac{{C}_{\text{A0}}}{{\delta}_{1}}}{\frac{{D}_{\text{A2}}}{{\delta}_{2}{\delta}_{1}}+\frac{{D}_{\text{A1}}}{{\delta}_{1}}}\right]$
What is the flux in the epidermis layer?
P1412_{D}(Estimating glacial ages) The following oxygen18 data were obtained from soil samples taken at different depths in Ontario, Canada. Assuming that all the ^{18}O was laid down during the last glacial age and that the transport of ^{18}O to the surface takes place by molecular diffusion, estimate the number of years since the last glacial age from the following data. Independent measurements give the diffusivity of ^{18}O in soil as 2.64 = 10^{−10} m^{2}/s.
(surface) 


Depth (m)  0 
3 
6 
9 
12 
18 
^{18}O Conc. Ratio (C/C_{0})  0 
0.35 
0.65 
0.83 
0.94 
1.0 
C_{0} is the concentration of ^{18}O at 25 m. Hint: A knowledge of error function solutions may or may not be helpful. (Ans: t = 5,616 years)
P1413_{B} OEQ (Old Exam Question). A spherical particle is dissolving in a liquid. The rate of dissolution is first order in the solvent concentration, C. Assuming that the solvent is in excess, show that the following conversiontime relationships hold.
RateLimiting Regime 
Conversion–Time Relationship 

Surface reaction  ${1(1X)}^{1/3}=\frac{\alpha t}{{D}_{i}}$ 
Mass transfer  $\frac{{D}_{i}}{2D*}[1{(1X)}^{2/3}]=\frac{\alpha t}{{D}_{i}}$ 
Mixed  ${[1(1X)]}^{1/3}+\frac{{D}_{i}}{2D*}\left[1{(1X)}^{2/3}\right]=\frac{\alpha t}{{D}_{i}}$ 
P1414_{C} Derive the diffusion and reaction equation in spherical coordinates to describe the dissolution of a drug in the form of a spherical pellet. Plot the drug concentration as a function distance r and time t. Also plot the flux and particle diameter as a function of time.
P1415_{B} A powder is to be completely dissolved in an aqueous solution in a large, wellmixed tank. An acid must be added to the solution to render the spherical particle soluble. The particles are sufficiently small that they are unaffected by fluid velocity effects in the tank. For the case of excess acid, C_{0} = 2 M, derive an equation for the diameter of the particle as a function of time when
Mass transfer limits the dissolution: –W_{A} = k_{c}C_{A0}
Reaction limits the dissolution:${r}_{\text{A}}^{\u2033}={k}_{r}^{\text{A}}{C}_{\text{A0}}$
What is the time for complete dissolution in each case?
Now assume that the acid is not in excess and that mass transfer is limiting the dissolution. One mole of acid is required to dissolve 1 mol of solid. The molar concentration of acid is 0.1 M, the tank is 100 L, and 9.8 mol of solid is added to the tank at time t = 0. Derive an expression for the radius of the particles as a function of time and calculate the time for the particles to dissolve completely.
How could you make the powder dissolve faster? Slower?
Additional information:
$\begin{array}{c}\begin{array}{cc}{{D}_{e}={10}^{10}\text{m}}^{2}/\text{s,}& {\begin{array}{c}k=10\end{array}}^{18}/s\end{array}\\ \text{initial diameter}={10}^{5}\text{m}\end{array}$
P1416_{B} (Pills) An antibiotic drug is contained in a solid inner core and is surrounded by an outer coating that makes it palatable. The outer coating and the drug are dissolved at different rates in the stomach, owing to their differences in equilibrium solubilities.
If D_{2} = 4 mm and D_{1} = 3 mm, calculate the time necessary for the pill to dissolve completely.
Assuming firstorder kinetics (k_{A} = 10 h^{–1}) for the absorption of the dissolved drug (i.e., in solution in the stomach) into the bloodstream, plot the concentration in grams of the drug in the blood per gram of body weight as a function of time when the following three pills are taken simultaneously:
$\begin{array}{ccc}\text{Pill 1:}& {D}_{2}=5\text{mm,}& {D}_{1}=3\text{mm}\\ \text{Pill 2:}& {D}_{2}=4\text{mm,}& {D}_{1}=3\text{mm}\\ \text{Pill 3:}& {D}_{2}=3.5\text{mm,}& {D}_{1}=3\text{mm}\\ & & \\ & & \end{array}$
Discuss how you would maintain the drug level in the blood at a constant level using differentsize pills?
How could you arrange a distribution of pill sizes so that the concentration in the blood was constant over a period (e.g., 3 hr) of time?
Additional information:
Amount of drug in inner core = 500 mg
Solubility of outer layer at stomach conditions = 1.0 mg/cm^{3}
Solubility of inner layer at stomach conditions = 0.4 mg/cm^{3}
Volume of fluid in stomach = 1.2 L
Typical body weight = 75 kg
Sh = 2, D_{AB} = 6 × 10^{–4} cm^{2}/min
P1417_{B} If disposal of industrial liquid wastes by incineration is to be a feasible process, it is important that the toxic chemicals be completely decomposed into harmless substances. One study carried out concerned the atomization and burning of a liquid stream of “principal” organic hazardous constituents (POHCs) [Environ. Prog., 8, 152 (1989)]. The following data give the burning droplet diameter as a function of time (both diameter and time are given in arbitrary units):
Time  20 
40 
50 
70 
90 
110 
Diameter  9.7 
8.8 
8.4 
7.1 
5.6 
4.0 
What can you learn from these data?
The fundamentals of diffusional mass transfer may or may not be found in
R. B. BIRD, W. E. STEWART, and E. N. LIGHTFOOT, Transport Phenomena, 2nd ed. New York: Wiley, 2002, Chaps. 17 and 18.
S. Collins, Mockingjay (The Final Book of the Hunger Games). New York: Scholastic, 2014.
E. L. CUSSLER, Diffusion Mass Transfer in Fluid Systems, 3rd ed. New York: Cambridge University Press, 2009.
C. J. GEANKOPLIS, Transport Processes and Unit Operations. Upper Saddle River, NJ: Prentice Hall, 2003.
V. G. LEVICH, Physiochemical Hydrodynamics. Upper Saddle River, NJ: Prentice Hall, 1962, Chaps. 1 and 4.
Experimental values of the diffusivity can be found in a number of sources, two of which are
R. H. PERRY, D. W. GREEN, and J. O. MALONEY, Chemical Engineers’ Handbook, 8th ed. New York: McGrawHill, 2007.
T. K. SHERWOOD, R. L. PIGFORD, and C. R. WILKE, Mass Transfer. New York: McGrawHill, 1975.
A number of correlations for the mass transfer coefficient can be found in
A. L. LYDERSEN, Mass Transfer in Engineering Practice. New York: WileyInterscience, 1983, Chap. 1.
W. L. MCCABE, J. C. SMITH, and P. HARRIOTT, Unit Operations of Chemical Engineering, 6th ed. New York: McGrawHill, 2000, Chap. 17.