Chapter 2. Examples of Differential (1-D) Balances

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Chapter 2. Examples of Differential (1-D) Balances

Learning Objectives

After completing this chapter, you will be able to:

Develop differential models for problems in mass transfer where the concentration varies along only one spatial dimension.
Derive differential equations for the species concentration by use of the conservation law followed by Fick’s law in Cartesian coordinates.
Derive the model equations in cylindrical and spherical coordinates and understand the differences arising due to geometry.
State the equation in vector form and be able to use it in any coordinate system.
Solve some simpler problems, especially those involving mass transfer with reaction, and examine the property of the solution.

This chapter illustrates the development of differential models by examining simple examples where the transport occurs in only one spatial direction. This simplifies both the model and the mathematics of deriving the solution. By focusing on such simple problems, we build our understanding of the model development. In Chapter 5, we will generalize this understanding to 3-D problems where we derive the complete differential equation for the concentration field and where the particular cases studied here can be obtained as a simplification of the general model. However, the model development is better understood by studying particular cases individually, as is done in this chapter.

To begin, we consider problems posed in Cartesian geometry. Here the concentration is assumed to be a function of only one coordinate, say x. We start off with a simple diffusion problem, followed by diffusion with reaction, transient diffusion, and diffusion with convection. Model development is illustrated in a step-by-step manner, followed by solution of some illustrative problems.

Problems posed in cylindrical and spherical coordinates where the concentration varies only in the radial direction are examined next. One distinguishing feature in contrast to the case involving Cartesian coordinates is that the area for diffusion varies as a function of the radial coordinates; hence the resulting equations appear to be slightly different from those in Cartesian coordinates. Illustrative problems are then solved for problems posed in these geometries. This chapter also indicates that the equations for all three geometries can be unified by using the divergence and the Laplacian operator from vector calculus. This approach provides the vector representation of the model, which is general and useful in any chosen coordinate system.

2.1 Cartesian Coordinates

The first set of problems we will cover are those concerned with mass transfer in a slab of finite thickness in the x-direction with rather large height and width in the y- and z-directions. This makes the transport one-dimensional. First we examine a problem with no generation, no accumulation, and no externally imposed flow across the slab. We can progressively include these terms in the analysis.

2.1.1 Steady State Diffusion across a Slab

Steady state diffusion across a slab is the simplest problem in mass transport analysis. It is so simple that a linear profile across the slab is often taken for granted in steady state analysis. Nevertheless, the steps and the assumptions involved in arriving the result should be examined and the problem should be analyzed step by step since the approach is similar to that used in other cases and in other coordinate systems.

An illustrative problem statement is as follows: A membrane or a medium of thickness L has concentrations C_A₀ on one side and C_AL on the other side. This can, for example, be accomplished by circulating a gas containing the solute A at the appropriate concentration levels on both sides of the membrane. The diffusion takes place along the x-direction only and the concentration profile in the slab and the quantity of species transported across the slab are to be computed. The schematic of the problem is shown in Figure 2.1.

Diagrammatic illustration of the steady-state diffusion across a slab, with respect to the problem statement.

Figure 2.1 Schematic of steady state transport across a slab. The shaded section is the differential control volume used in the analysis.

We start with the conservation law, which reduces for this case as follows since there is no generation nor accumulation:

in – out = 0

Input at x is equal to the flux times the area evaluated at x and can be represented as N_AxA. Here A is the area of cross-section in the direction perpendicular to diffusion (perpendicular to x here). Similarly, the output is N_AxA but evaluated at x + Δx.

The conservation principle therefore gives

\begin{matrix} {[N_{A x} A]}_{x} - {[N_{A x} A]}_{x + Δ x = 0} & (2.1) \end{matrix}

$\begin{matrix} {[N_{A x} A]}_{x} - {[N_{A x} A]}_{x + Δ x = 0} & (2.1) \end{matrix}$

If one divides the expression by Δx and takes the limit as Δx tends to zero, the following differential equation for the flux is obtained:

\begin{matrix} \frac{d}{d x} (N_{A x} A) = 0 & (2.2) \end{matrix}

$\begin{matrix} \frac{d}{d x} (N_{A x} A) = 0 & (2.2) \end{matrix}$

Note: This expression is general and holds for all three geometries examined in this chapter when there is no generation due to chemical reaction and a steady state exists.

The areas for diffusion at x and x + Δx are the same and hence this term cancels out.

\begin{matrix} \frac{d}{d x} (N_{A x}) = 0 & (2.3) \end{matrix}

$\begin{matrix} \frac{d}{d x} (N_{A x}) = 0 & (2.3) \end{matrix}$

So far we have dealt with the combined flux to keep the analysis general. In addition, the area term did not matter and canceled out. For some other cases (a cylinder and a sphere, for example), the area is a function of the position. These cases are examined in Sections 2.2 and 2.3.

Although the problem examined here is a diffusion-dominated process and there is no forced flow, convective flow can arise due to diffusion itself. The effect of such convection is examined in Chapters 5 and 6. For the time being, let us assume that the convection flux is negligible and the combined flux has only the contribution resulting from the diffusion flux. With this assumption we have

N_Ax ≈ J_Ax

which is sometimes referred to as a low flux approximation. In other words, any convective flux (resulting from the diffusion process itself or due to any superimposed external flow) is neglected, and the conservation model reduces to

\begin{matrix} N_{A x} \approx J_{A x} & (2.4) \end{matrix}

$\begin{matrix} N_{A x} \approx J_{A x} & (2.4) \end{matrix}$

Note: The effect of diffusion-induced convection (examined in Chapter 6) is sometimes referred to the high flux model. The effect of a superimposed (external) flow in the direction of diffusion (the so-called blowing problem) is discussed in Section 2.1.4.

This is as far as we can proceed just by use of the conservation law alone. A model for diffusion flux is needed, and we use Fick’s law as the model:

J_{A x} = - D_{A} \frac{d C_{A}}{d x}

$J_{A x} = - D_{A} \frac{d C_{A}}{d x}$

where D_A is the diffusion coefficient of A in the medium or the membrane. Substituting Fick’s law into the flux conservation equation, Equation 2.4, the following equation is obtained for the concentration profile:

\begin{matrix} \frac{d}{d x} (D_{A} \frac{d C_{A}}{d x}) = 0 & (2.5) \end{matrix}

$\begin{matrix} \frac{d}{d x} (D_{A} \frac{d C_{A}}{d x}) = 0 & (2.5) \end{matrix}$

We assume now that the diffusion coefficient is not a function of concentration, which turns out to be good assumption for many cases (e.g., diffusion of gases such as oxygen in polymeric membranes). This permits us to take D_A out of the derivative sign. It cancels out as well, leading to the following expression:

\frac{d^{2} C_{A}}{{d x}^{2}} = 0

$\frac{d^{2} C_{A}}{{d x}^{2}} = 0$

Integrating twice gives us a general expression for the concentration profile:

C_A = B₁ + B₂x

Here B₁ and B₂ are two integration constants that are evaluated to satisfy the boundary conditions imposed at the two ends.

Using the values of C_A₀ at x = 0 and C_AL at x = L, we can find the integration constants and eliminate them from the previous equation. The final expression for the concentration profile is

\begin{matrix} C_{A} = C_{A 0} - (\frac{C_{A 0} - C_{A L}}{L}) x & (2.6) \end{matrix}

$\begin{matrix} C_{A} = C_{A 0} - (\frac{C_{A 0} - C_{A L}}{L}) x & (2.6) \end{matrix}$

The concentration profile is simply a straight line connecting the concentration values at the two ends. The flux from the slope of this straight line is therefore

\begin{matrix} J_{A x} = D_{A} (\frac{C_{A 0} - C_{A L}}{L}) & (2.7) \end{matrix}

$\begin{matrix} J_{A x} = D_{A} (\frac{C_{A 0} - C_{A L}}{L}) & (2.7) \end{matrix}$

The rate at which the solute is transported across the media, ${\dot{M}}_{A}$ ${\dot{M}}_{A}$ , can now be calculated as J_Ax times A and is equal to

{\dot{M}}_{A} = A J_{A x} = A D_{A} (\frac{C_{A 0} - C_{A L}}{L})

${\dot{M}}_{A} = A J_{A x} = A D_{A} (\frac{C_{A 0} - C_{A L}}{L})$

We now have both the concentration profile and the quantity transported across the slab, and the solution is complete. Let us apply the results to examine a simple problem dealing with membrane transport, which is shown in Example 2.1.

Example 2.1 Helium Leakage Rate across a Glass Wall

Light gases such as helium (He) has reasonable solubility in materials such as glass, so there may be leakage of this gas from a helium-containing glass shell that is used for a helium–cadmium laser in a copying machine. Calculate the helium leakage rate across a glass shell of thickness 1.5 mm with pure helium on one side and air containing negligible helium on the other side. Assume that the glass shell is at a constant temperature of 115 °C and the pressure inside the shell is 460 Pa at the given instant of time. (Problem adapted from Mills, 1995.)

The diffusion coefficient of He (m²/s) in the solid glass is reported as (Mills, 1995)

D_A = 1.40 × 10^–8 exp(–3280/T)

The solubility parameter has to be known as well and is given in a part of the solution.

Solution

Equation 2.7 is applicable for the flux. However, the concentrations of He are in the glass near the wall, not on the gas near the wall. Hence the concentration jump condition (introduced in Section 1.4) must to be used. We need to know the solubility relationship, which is the additional data needed to solve the problem.

The solubility parameter reported by Mills (1995) is

C_As = KC_AG

where K is reported as

K = 3.0 × 10^–5 T – 0.0012

For the given conditions, T = 115 + 273 = 388 kelvin, so K equals 0.0104.

On the gas side we have pure helium, whose concentration is calculated as

C_{A G} (x = 0) = \frac{P}{R_{g} T} = \frac{460}{8.314 \times 388} = 0.1426 mol / m^{3}

$C_{A G} (x = 0) = \frac{P}{R_{g} T} = \frac{460}{8.314 \times 388} = 0.1426 mol / m^{3}$

Using the K value we find C_A₀ = 0.0104 × 0.1426 = 1.15 × 10³ mol/m³, which is the concentration in the glass shell on the helium-exposed side.

On the air side, the helium concentration in air is zero. The glass-side concentration of helium C_AL is also equal to zero.

The diffusion coefficient has a value of 2.98 × 10^–12 m²/s at 388 K using the equation and the data given earlier.

The flux can be calculated as

\begin{matrix} J_{A x} & = & \frac{D_{A}}{L} (C_{A 0} - C_{A L}) = \frac{(2.98 \times 10^{- 12})}{(1.5 \times 10^{- 3})} (1.15 \times 10^{3} - 0) \\ = & 2.92 \times 10^{- 12} {mol/m}^{2} s \end{matrix}

$\begin{matrix} J_{A x} & = & \frac{D_{A}}{L} (C_{A 0} - C_{A L}) = \frac{(2.98 \times 10^{- 12})}{(1.5 \times 10^{- 3})} (1.15 \times 10^{3} - 0) \\ = & 2.92 \times 10^{- 12} {mol/m}^{2} s \end{matrix}$

which is the leakage rate per unit surface area of the glass.

Note that thermodynamic relation was needed in addition to the diffusion flux expression to account for the concentration jump at the gas–glass interface. A similar approach is used to model the transport rate in gas separation membranes as shown in Section 27.1.

2.1.2 Steady State Diffusion with Reaction in a Slab

Now we extend the model by assuming that the diffusing solute undergoes a chemical reaction in the slab. The problem statement is as follows: Consider a material in the form of a long slab with a chemical reaction taking place at a rate of R_A per unit volume of the slab. An example would be a porous catalyst cast in the form of a long slab. A second example would be oxygen diffusing into a pool of liquid containing some biomass or a liquid-phase reactant that consumes the diffusing oxygen. Many other examples can be given.

Assume that the mass transfer of species A is only in the x-direction. In this section, we develop a differential equation for the concentration variation of A in the system as a function of x and use this in the flux expression to obtain the rate of mass transfer into the slab.

We proceed in a similar manner as in the earlier transfer problem, but include the chemical reaction term in the conservation law. This law should now read as follows:

in – out + net generation = 0

The input and output terms are the same as before and the additional term due to generation is equal to R_A(A Δx). Hence

[N_AxA]_x – [N_AxA]_x+Δx + R_AA Δx = 0

Further simplification is done by taking the diffusion flux alone (N_Ax ≈ J_Ax)—that is, by ignoring any effects due to diffusion-induced convection. Also A is constant and does not vary with x. We next divide through by Δx and take the limit as Δx → 0 to obtain a differential equation for the variation of the diffusion flux in the system:

\begin{matrix} \frac{d}{d x} (J_{A x}) = R_{A} & (2.8) \end{matrix}

$\begin{matrix} \frac{d}{d x} (J_{A x}) = R_{A} & (2.8) \end{matrix}$

This is the basic model for the system based on the conservation principle. The model can be completed if a constitutive (transport) law for diffusion of a species A in a porous solid is available or developed. A Fick’s law type of model is now used here to complete the problem:

J_{A x} = - D_{e A} \frac{d C_{A}}{d x}

$J_{A x} = - D_{e A} \frac{d C_{A}}{d x}$

where D_eA is the (effective) diffusion coefficient of A in the porous medium.

Note: The rate of diffusion is modified due to the porous matrix. Specifically, a pore structure–dependent effective diffusivity is used in these cases in lieu of fluid-phase diffusivity. This topic is discussed in further detail in Section 7.6. We use the simplified notation D_A for the effective diffusivity D_eA in the remainder of this chapter.

Using a Fick’s law type of model for pore diffusion as shown in the mass conservation equation and assuming constant diffusivity (Equation 2.8), we get

\begin{matrix} \begin{matrix} D_{A} \frac{d^{2} C_{A}}{d x^{2}} = - R_{A} \end{matrix} & (2.9) \end{matrix}

$\begin{matrix} \begin{matrix} D_{A} \frac{d^{2} C_{A}}{d x^{2}} = - R_{A} \end{matrix} & (2.9) \end{matrix}$

This equation is known as the diffusion-reaction equation and is widely applied in the analysis of transport in catalysts. Note that the second derivative term on the left side of the equation is the x-dependent part of the Laplacian equation. Hence the equation can be compacted as follows:

\begin{matrix} D_{A} \nabla^{2} C_{A} + R_{A} = 0 & (2.10) \end{matrix}

$\begin{matrix} D_{A} \nabla^{2} C_{A} + R_{A} = 0 & (2.10) \end{matrix}$

Further information is needed on the kinetics of the reaction. In general, R_A is a function of the concentration of species A and temperature. It is also often a function of the concentration of other species in the system (e.g., bimolecular reactions, reactions with strong product inhibition). These dependencies are given by a kinetic model for the process. In Examples 2.2 and 2.3, we examine the (general) solution to the problem for simple forms for R_A— namely, for a zero-order reaction and a first-order reaction.

Example 2.2 Diffusion with Zero-Order Reaction

Derive a general expression for the concentration profile for diffusion accompanied by a zero-order reaction in a slab geometry.

Solution

For a zero-order reaction, the rate is given as

R_A = –k₀

where k₀ is the volumetric rate of consumption of reactant A (units of mol/m³ s). Equation 2.9 now takes the following form:

\begin{matrix} D_{A} \frac{d^{2} C_{A}}{d x^{2}} = k_{0} & (2.11) \end{matrix}

$\begin{matrix} D_{A} \frac{d^{2} C_{A}}{d x^{2}} = k_{0} & (2.11) \end{matrix}$

The resulting differential equation can be integrated twice to obtain

C_{A} (x) = \frac{k_{0}}{2 D_{A}} x^{2} + B_{1} x + B_{2}

$C_{A} (x) = \frac{k_{0}}{2 D_{A}} x^{2} + B_{1} x + B_{2}$

which is the general solution for the concentration profile. The constants of integration B₁ and B₂ can be found if the boundary conditions are specified. The boundary conditions are problem specific and depend on what is happening at the two ends of the system.

A common application of this model is in trickling bed filters used in waste water treatment. It is also used for modeling oxygen transport in tissues, where the oxygen metabolism follows a zero-order rate law for most cases.

Example 2.3 shows the application to a first-order reaction.

Example 2.3 Diffusion with First-Order Reaction

Derive a general expression for the concentration profile for diffusion accompanied by a first-order reaction in a porous catalyst in a slab geometry.

Solution

In this case the rate is given as

R_A = –k₁C_A

where k₁ is the rate constant for a first-order reaction in the units of s^–1. Equation 2.9 now takes the following form:

\begin{matrix} D_{A} \frac{d^{2} C_{A}}{d x^{2}} - k_{1} C_{A} = 0 & (2.12) \end{matrix}

$\begin{matrix} D_{A} \frac{d^{2} C_{A}}{d x^{2}} - k_{1} C_{A} = 0 & (2.12) \end{matrix}$

This equation can be solved by treating it as a linear second-order differential equation. The general solution can be expressed in two equivalent forms:

\begin{matrix} C_{A} (x) = A_{1} cosh (x \sqrt{k_{1} / D_{A}}) + A_{2} sinh (x / \sqrt{k_{1} / D_{A}}) & (2.13) \end{matrix}

$\begin{matrix} C_{A} (x) = A_{1} cosh (x \sqrt{k_{1} / D_{A}}) + A_{2} sinh (x / \sqrt{k_{1} / D_{A}}) & (2.13) \end{matrix}$

with A₁ and A₂ being integration constants or

C_{A} (x) = B_{1} exp (x \sqrt{k_{1} / D_{A}}) + B_{2} exp (- x / \sqrt{k_{1} / D_{A}})

$C_{A} (x) = B_{1} exp (x \sqrt{k_{1} / D_{A}}) + B_{2} exp (- x / \sqrt{k_{1} / D_{A}})$

where B₁ and B₂ now represent the integration constants.

Note: The B₁ and B₂ will be a linear combination of A₁ and A₂, and vice versa.

A common application is in constructing a diffusion model for a catalyst pore—an application that is discussed in Chapter 18. At this point, the model development and the nature of the resulting differential equation should be noted. The general solution with two integration constants is also noteworthy since it can be applied to many different boundary conditions. Various types of boundary conditions of common occurrence are summarized in Chapter 5.

In Example 2.4, we apply Equation 2.13 to a specific problem. We also illustrate the use of two common types of boundary conditions.

Example 2.4 Oxygen Profile in a Pool of Liquid

A pool of liquid is 10 cm deep and a gas A dissolves and reacts in the liquid. The solubility of the oxygen is such that the interfacial concentration is equal to 2 mol/m³ in the liquid and the diffusivity of the dissolved gas is 2 × 10^–9 m²/s.

Assume a first-order reaction with a rate constant of 10^–6 s^–1 in the liquid. Find the concentration profile in the liquid. Express the solution in dimensionless form. Also find the flux of oxygen into the pool of liquid.

Solution

The schematic of the problem and the anticipated concentration profile of oxygen in the pool of liquid are shown in Figure 2.2. Equation 2.13 is applicable for the concentration profile. Also, Equation 2.13 (cosh and sinh form) is more convenient here (rather than the exponential form), and the constants of integration must be evaluated using the boundary conditions. You will find that taking x = 0 at the bottom will simplify the math in the cosh and sinh form of the solution. The location x = L (10 cm here) is then the position of the gas–liquid interface. We need to evaluate the two constants by specifying appropriate boundary conditions.

Figure 2.2 Schematic of oxygen diffusion with reaction in a pool of liquid and the expected nature of the concentration profile.

Since no oxygen diffuses into the bottom of the tank, we take dC_A/dx = 0 at x = 0, the bottom of the tank. This type of boundary condition is generally known the boundary condition of the second kind, also known as the Neumann boundary condition.

The procedure to get one of the constants of integration is as follows. First, differentiate Equation 2.13 to get the derivative of the concentration:

\frac{d C_{A}}{d x} = A_{1} [\sqrt{k_{1} / D_{A}}] sinh (x \sqrt{k_{1} / D_{A}}) + A_{2} [\sqrt{k_{1} / D_{A}}] cosh (x / \sqrt{k_{1} / D_{A}})

$\frac{d C_{A}}{d x} = A_{1} [\sqrt{k_{1} / D_{A}}] sinh (x \sqrt{k_{1} / D_{A}}) + A_{2} [\sqrt{k_{1} / D_{A}}] cosh (x / \sqrt{k_{1} / D_{A}})$

Then set this expression equal to zero at x = 0. We find A₂ = 0.

The solution now reduces to

C_{A} (x) = A_{1} cosh (x \sqrt{k_{1} / D_{A}})

$C_{A} (x) = A_{1} cosh (x \sqrt{k_{1} / D_{A}})$

Now use the boundary condition at x = L to get A₁. The concentration is set equal to C_As, the interfacial concentration of dissolved oxygen at x = L. This is known as the boundary condition of the first kind, also referred to as the Dirichlet condition. Using this information and some minor algebra yields

A_{1} = \frac{C_{A s}}{cosh (L / \sqrt{k_{1} / D_{A}})}

$A_{1} = \frac{C_{A s}}{cosh (L / \sqrt{k_{1} / D_{A}})}$

Both constants are evaluated and the solution for the concentration profile is completed. Backsubstituting of A₁ leads to the following result:

\begin{matrix} \frac{C_{A}}{C_{A} s} = \frac{cosh (x \sqrt{k_{1} / D_{A}})}{cosh (L \sqrt{k_{1} / D_{A}})} & (2.14) \end{matrix}

$\begin{matrix} \frac{C_{A}}{C_{A} s} = \frac{cosh (x \sqrt{k_{1} / D_{A}})}{cosh (L \sqrt{k_{1} / D_{A}})} & (2.14) \end{matrix}$

The dimensionless version of the solution requires the use of dimensionless concentration c_A defined as C_A/C_As and a dimensionless distance parameter, ξ = x/L. In addition, a dimensionless parameter called the Thiele modulus for diffusion-reaction systems is needed, which is defined here:

φ = L \sqrt{k_{1} / D_{A}}

$φ = L \sqrt{k_{1} / D_{A}}$

With these definitions, Equation 2.14 can be written in dimensionless form:

\begin{matrix} c_{A} = \frac{cosh (φ ξ)}{cosh φ} & (2.15) \end{matrix}

$\begin{matrix} c_{A} = \frac{cosh (φ ξ)}{cosh φ} & (2.15) \end{matrix}$

The use of dimensionless quantities is a common practice in representation and solution to engineering models, since it makes the mathematical manipulations far easier and allows the final results to take a more compact form.

The rate of absorption of oxygen is calculated using Fick’s law at the gas–liquid interface:

{\dot{M}}_{A} = A D_{A} {(\frac{d C_{A}}{d x})}_{x = L}

${\dot{M}}_{A} = A D_{A} {(\frac{d C_{A}}{d x})}_{x = L}$

where A is the exposed surface area for oxygen transfer.

Note the plus sign used in Fick’s law for the flux; it is appropriate because we are looking at the flux in a direction of decreasing x. Substituting for the derivative of the concentration leads to

\begin{matrix} {\dot{M}}_{A} = A \sqrt{D_{A} k_{1}} tanh (L \sqrt{k_{1} / D_{A}}) \cdot C_{A s} & (2.16) \end{matrix}

$\begin{matrix} {\dot{M}}_{A} = A \sqrt{D_{A} k_{1}} tanh (L \sqrt{k_{1} / D_{A}}) \cdot C_{A s} & (2.16) \end{matrix}$

This result has wide applicability in reaction engineering.

Substituting the values, the following results are obtained:

\begin{matrix} oxygen concentration at the bottom & = & 1 / cosh (L \sqrt{k_{1} / D_{A}}) \\ = & 0.4227 {mol/m}^{3} \end{matrix}

$\begin{matrix} oxygen concentration at the bottom & = & 1 / cosh (L \sqrt{k_{1} / D_{A}}) \\ = & 0.4227 {mol/m}^{3} \end{matrix}$

oxygen transfer rate per unit area using Equation 2.16 = 8.74 × 10^–8 mol/m² s

2.1.3 Transient Diffusion in a Slab

Consider the problem examined in Section 2.1.2, but now assume the slab is initially at a uniform concentration of C_Ai. At t > 0, one end of the slab (say, x = 0) is exposed to a different concentration of C_A₀. The other end is still at C_Ai. The problem is how to predict the temporal evolution of the concentration profile in the system and how use this information to get the instantaneous rate of mass transfer. Here we derive the governing differential equation for the transient diffusion process.

The conservation law should now include the accumulation term:

input – output = accumulation

The input and output terms are the same as before and the additional term due to accumulation is equal to the time derivative of moles of A in the control volume, C_AA Δx. Hence:

{[N_{A x} A]}_{x} - {[N_{A x} A]}_{x + Δ x} \cdot = \frac{\partial}{\partial t} (C_{A} A Δ x)

${[N_{A x} A]}_{x} - {[N_{A x} A]}_{x + Δ x} \cdot = \frac{\partial}{\partial t} (C_{A} A Δ x)$

The final equation with the approximation of N_A = J_A and the use of Fick’s law for J_A leads to the following partial differential equation for the transient problem:

\begin{matrix} \begin{matrix} D_{A} \frac{\partial^{2} C_{A}}{\partial x^{2}} = \frac{\partial C_{A}}{\partial t} \end{matrix} & (2.17) \end{matrix}

$\begin{matrix} \begin{matrix} D_{A} \frac{\partial^{2} C_{A}}{\partial x^{2}} = \frac{\partial C_{A}}{\partial t} \end{matrix} & (2.17) \end{matrix}$

This equation is sometimes referred to as Fick’s second law, but in reality is a combination of the conservation law and Fick’s first law. Constant diffusivity is also implied. The equation needs two boundary conditions and one initial condition since it is now a partial differential equation (PDE). The solution method is described in detail in Section 8.2, but Example 2.5 shows an illustrative result. The example also shows how to represent the problem in dimensionless form.

Example 2.5 Temporal Evolution of a Concentration Profile in a Slab

A membrane, cast as a slab, has zero concentration at time zero. Suddenly, after time zero, one surface is exposed to a gas such that the concentration in the membrane at this point is C_A₀. The other side of the membrane is kept at zero concentration.

State the governing equation and the boundary and initial conditions to be used. Also sketch the anticipated temporal evolution of the concentration profile in slab.

Solution

The governing equation is the transient model given by Equation 2.17, Fick’s second law. Before stating the required conditions it is useful to write the equation in a dimensionless form. This can be done by using the following variables:

\begin{matrix} dimensionless concentration c_{A} = \frac{C_{A}}{C_{A 0}} \\ dimensionless distance ξ = \frac{x}{L} \end{matrix}

$\begin{matrix} dimensionless concentration c_{A} = \frac{C_{A}}{C_{A 0}} \\ dimensionless distance ξ = \frac{x}{L} \end{matrix}$

Notice that L²/D_A has the same units as time, so we can use it to scale the actual time. Thus we can define a dimensionless time as

dimensionless time τ = \frac{t}{(L^{2} / D_{A})} = \frac{t D_{A}}{L^{2}}

$dimensionless time τ = \frac{t}{(L^{2} / D_{A})} = \frac{t D_{A}}{L^{2}}$

The differential equation reduces to a simpler form with these variables:

\begin{matrix} \frac{\partial^{2} c_{A}}{\partial ξ^{2}} = \frac{\partial c_{A}}{\partial τ} & (2.18) \end{matrix}

$\begin{matrix} \frac{\partial^{2} c_{A}}{\partial ξ^{2}} = \frac{\partial c_{A}}{\partial τ} & (2.18) \end{matrix}$

Note that there are no free parameters, which is a characteristic of a well-scaled problem. You should verify the dimensionless version starting from the original version using the chain rule.

The boundary conditions to be used are as follows: At x = 0, C_A = C_A₀, which in dimensionless form is c_A(ξ = 0) = 1.

At x = L, C_A = 0, which in dimensionless form is c_A(ξ = 1) = 0

Both of these relations hold for all values of τ > 0.

The initial condition is at τ = 0, c_A = 0 for all ξ in the domain to 1, since the slab is maintained at zero concentration at the begining of the process.

The problem specification is now complete. This problem can be solved analytically by a modified method of separation of variables or by numerical methods (e.g., MATLAB PDEPE solver). These methods are elaborated in a later chapter. At this point you should focus on problem formulation and dimensionless representation. The computed result using MATLAB PDEPE solver is shown in Figure 2.3. Note that for large time, a linear profile is obtained that is the steady state profile for diffusion across a slab.

Figure 2.3 Evolution of concentration profile for transient diffusion across a slab.

L²/D_A is the characteristic time scale for diffusion. In other words, it is the time scale for any disturbance to be smoothed out by diffusion, and one can expect that a steady state profile will be reached around this time. This parameter is called the diffusion time. In this particular case, the steady state is reached around 0.3 × L²/D_A.

2.1.4 Diffusion with Convection

We now examine a simple problem where the convective flow is superimposed on diffusion. Specifically, the case involving steady state and no chemical reaction is analyzed. Consider again the problem discussed in Section 2.1.1, but with the addition of a superimposed flow across the system with a velocity of v_x. This information is used to derive the differential equation and to find the flux across the system. The problem is illustrated in Figure 2.4.

Diagrammatic illustration of steady-state diffusion with convection, related to the problem statement.

The medium of flow is represented using two vertical lines, the left marked 0 and the right marked L. The concentration at 0 is C subscriptA0 and the concentration at L is C subscript AL. The flow is shown to be from left to right, with the thickness expressed by variable x. The velocity of flow at x is V subscriptx. A graph on the right shows the profiles of flow with and without convection. The horizontal axis represents the medium thickness ranging from x equals 0 to x equals L. The vertical axis represents the concentration C subscript A. A solid, decreasing, concave down curve is labeled "v subscriptx greater than 0" and a dashed, decreasing, concave up curve is labeled "v subscriptx lesser than 0." An expression above the diagrams reads (D subscript A times second order derivative of C subscript A with respect to x) minus ( v subscriptx times derivative of C subscript A with respect to x) equals 0.

Figure 2.4 Schematic of steady state diffusion with superimposed convection in a Cartesian geometry. The profile will be a straight line for no convection. The dashed line is that for flow (convection) in the opposite direction to the concentration difference.

The application of the conservation law shows that the combined flux is the same at x (in) and at x + Δx (out), since this is steady state and no reaction is taking place. Therefore Equation 2.3 applies here as well; it is repeated here for ease of reference:

\frac{d}{d x} (N_{A x}) = 0

$\frac{d}{d x} (N_{A x}) = 0$

The combined flux has now contribution from both diffusion and convection:

N_{A x} = - D_{A} \frac{d C_{A}}{d x} + υ_{x} C_{A}

$N_{A x} = - D_{A} \frac{d C_{A}}{d x} + υ_{x} C_{A}$

where v_x is the blowing or transpiration velocity. Substituting into the flux equation and assuming constant D_A and v_x, we obtain the following second-order differential equation:

\begin{matrix} D_{A} \frac{d^{2} C_{A}}{d x^{2}} - υ_{x} \frac{d C_{A}}{d x} = 0 & (2.19) \end{matrix}

$\begin{matrix} D_{A} \frac{d^{2} C_{A}}{d x^{2}} - υ_{x} \frac{d C_{A}}{d x} = 0 & (2.19) \end{matrix}$

This is the simplified version of the general 3-D convection-diffusion equation that will be discussed in more detail in Chapter 5. This general equation is stated here to help you better appreciate the vector representation of mass transfer models:

D_A∇²C_A – v · ∇C_A = 0

The general solution to Equation 2.19 can be written as follows:

\begin{matrix} C_{A} = A_{1} + A_{2} exp (x υ_{x} / D_{A}) & (2.20) \end{matrix}

$\begin{matrix} C_{A} = A_{1} + A_{2} exp (x υ_{x} / D_{A}) & (2.20) \end{matrix}$

where A₁ and A₂ are two integration constants. (You may wish to verify this by direct substitution.)

The boundary conditions of the first kind are useful for this case since the concentrations are specified at the ends. At x = 0, the concentration is C_A₀; at x = L, the concentration is C_AL. Using these conditions, we can obtain the constants of integration and complete the solution. The details are left as an exercise. In this scenario, the profiles are no longer linear, in contrast to the result obtained in the pure diffusion case.

More useful than the concentration profile is the (combined) flux across the system. This can be calculated from the concentration profile as

N_{A x} = - D_{A} {(\frac{d C_{A}}{d x})}_{x = 0} + υ_{x} C_{A 0}

$N_{A x} = - D_{A} {(\frac{d C_{A}}{d x})}_{x = 0} + υ_{x} C_{A 0}$

or equivalently as

N_{A x} = - D_{A} {(\frac{d C_{A}}{d x})}_{x = L} + υ_{x} C_{A L}

$N_{A x} = - D_{A} {(\frac{d C_{A}}{d x})}_{x = L} + υ_{x} C_{A L}$

The result after some standard algebraic manipulations is as follows (which you should verify):

\begin{matrix} N_{A x} = υ_{x} (C_{A 0} - C_{A L}) \frac{exp (υ_{x} L / D_{A})}{exp (υ_{x} L / D_{A}) - 1} & (2.21) \end{matrix}

$\begin{matrix} N_{A x} = υ_{x} (C_{A 0} - C_{A L}) \frac{exp (υ_{x} L / D_{A})}{exp (υ_{x} L / D_{A}) - 1} & (2.21) \end{matrix}$

The expression can also be written as

\begin{matrix} N_{A x} = \frac{D_{A}}{L} (C_{A 0} - C_{A L}) [\frac{(υ_{x} L / D_{A}) exp (υ_{x} L / D_{A})}{exp (υ_{x} L / D_{A}) - 1}] & (2.22) \end{matrix}

$\begin{matrix} N_{A x} = \frac{D_{A}}{L} (C_{A 0} - C_{A L}) [\frac{(υ_{x} L / D_{A}) exp (υ_{x} L / D_{A})}{exp (υ_{x} L / D_{A}) - 1}] & (2.22) \end{matrix}$

where the bracketed term on the right side can be identified as the factor by which the blowing enhances the rate of mass transfer. This is illustrated in Example 2.6.

Example 2.6 Transport Rate in Presence of Convection

A porous plug 10 cm long has hydrogen at a partial pressure of 0.1 atm on one side and air at 1 atm on the other side at 293 K. Find the transport rate of hydrogen if the system is stagnant, if there is a superimposed gas flow at a rate of 0.1 cm/s in the same direction as that for diffusion, and if the flow is in the opposite direction to diffusion.

Solution

The parameter v_xL/D_A is the key dimensionless group in Equation 2.22 showing the effect of the blowing velocity. This parameter is known as the Peclet number:

Peclet number = \frac{υ_{x} L}{D_{A}} = \frac{Convection rate}{Diffusion rate}

$Peclet number = \frac{υ_{x} L}{D_{A}} = \frac{Convection rate}{Diffusion rate}$

It can be interpreted as the ratio of the convection rate to the diffusion rate. For no blowing, the pure diffusion model applies:

N_{A x} = \frac{D_{A}}{L} (C_{A 0} - C_{A L})

$N_{A x} = \frac{D_{A}}{L} (C_{A 0} - C_{A L})$

Equation 2.22 indicates that the pure diffusion flux should be corrected by the following factor:

Correction for blowing = \frac{(υ_{x} L / D_{A}) exp (υ_{x} L / D_{A})}{exp (υ_{x} L / D_{A}) - 1}

$Correction for blowing = \frac{(υ_{x} L / D_{A}) exp (υ_{x} L / D_{A})}{exp (υ_{x} L / D_{A}) - 1}$

The correction factor is more compactly expressed using the Peclet number as

\begin{matrix} Correction for blowing = \frac{P e exp (P e)}{exp (P e) - 1} & (2.23) \end{matrix}

$\begin{matrix} Correction for blowing = \frac{P e exp (P e)}{exp (P e) - 1} & (2.23) \end{matrix}$

For numerical calculations, the value of the diffusion coefficient of hydrogen in air is needed. We will use a value of 0.7 cm²/s here. Hence the Peclet number is Pe = 0.1 * 10/0.7 = 1.4286 and the correction factor for blowing calculated using Equation 2.23 is 1.8788.

The concentration of hydrogen at one end C_A₀ is calculated as p_h/R_gT , or 4.03 mol/m³.

Flux in the absence of blowing is D_A(C_A₀ – 0)/L, which is found to be 0.0028 mol/m² s. The flux in the presence of blowing is 0.0028 × 1.8788 = 0.0053 mol/m² s.

If the flow is in the opposite direction, v_x is negative (suction condition) and hence Pe = –1.4286. The correction factor is now 0.45 (less than 1), so the flux is 0.0013 mol/m² s. Flux is reduced due to the blowing counter to the direction of diffusion.

We have now examined four prototypical problems in Cartesian coordinates: (1) diffusion alone, (2) diffusion balanced by reaction, (3) transient diffusion with accumulation, and (4) diffusion with superimposed flow. We now study similar problems posed in cylindrical coordinates but again with transport taking place only in the radial direction.

2.2 Cylindrical Coordinates

In this section, we repeat the analysis for the case of a cylindrical geometry to show some nuances that arise primarily due to the nature of the coordinate system. Specifically, we look at problems where the concentration is a function of only the radial position. The analysis is similar to that for a slab geometry, but the point to bear in mind is that the area over which diffusion takes place is a function of the radial position. We repeat some preliminary steps in the slab analysis and show the differences for clarity.

Steady state diffusion is considered first, followed by diffusion with reaction and then transient diffusion.

2.2.1 Steady State Radial Diffusion

The differential control volume, shown in Figure 2.5, is a ring element located at r and r + Δr. We start with the conservation law for this control volume, which reduces for the case of no generation and accumulation to

Diagrammatic representation of steady-state radial diffusion with respect to the problem statement.

Figure 2.5 Differential control volume used in the analysis of radial diffusion in a cylinder.

in – out = 0

Input at r is equal to the flux times the area evaluated at r and can be represented as N_ArA. Similarly, the output is N_ArA but is evaluated at r + Δr. The conservation principle therefore gives

\begin{matrix} {[N_{A r} A]}_{r} - {[N_{A r} A]}_{r + Δ r} \cdot = 0 & (2.24) \end{matrix}

$\begin{matrix} {[N_{A r} A]}_{r} - {[N_{A r} A]}_{r + Δ r} \cdot = 0 & (2.24) \end{matrix}$

Next, we note that the area across which diffusion is taking place is equal to 2πrL. (This is different at r and r + Δr.) The conservation equation with the area A term substituted is

\begin{matrix} {[2 π r L N_{A r}]}_{r} - {[2 π r L N_{A r}]}_{r + Δ r} = 0 & (2.25) \end{matrix}

$\begin{matrix} {[2 π r L N_{A r}]}_{r} - {[2 π r L N_{A r}]}_{r + Δ r} = 0 & (2.25) \end{matrix}$

Dividing by Δr and taking the limit leads to

\begin{matrix} \frac{d}{d r} (r N_{A r}) = 0 & (2.26) \end{matrix}

$\begin{matrix} \frac{d}{d r} (r N_{A r}) = 0 & (2.26) \end{matrix}$

The combined flux is taken as diffusion flux for further analysis. Using Fick’s law to close the problem for J_Ar

J_{A r} = - D_{A} \frac{d C_{A}}{d r}

$J_{A r} = - D_{A} \frac{d C_{A}}{d r}$

leads to

\begin{matrix} \frac{d}{d r} (D_{A} r \frac{d C_{A}}{d r}) = 0 & (2.27) \end{matrix}

$\begin{matrix} \frac{d}{d r} (D_{A} r \frac{d C_{A}}{d r}) = 0 & (2.27) \end{matrix}$

For a constant diffusivity case, the following differential equation is obtained:

\begin{matrix} \frac{d}{d r} (r \frac{d C_{A}}{d r}) = 0 & (2.28) \end{matrix}

$\begin{matrix} \frac{d}{d r} (r \frac{d C_{A}}{d r}) = 0 & (2.28) \end{matrix}$

This suggests that

r \frac{d C_{A}}{d r} = constant, say A_{1}

$r \frac{d C_{A}}{d r} = constant, say A_{1}$

Now a second integration can be done and a general expression for the concentration profile can be obtained:

C_A = A₁ ln r + A₂

The constants A₁ and A₂ can be found by fitting the boundary conditions.

Diffusion across a Cylindrical Shell

Consider the diffusion across a cylindrical shell with radius of R_i at the inner surface and R_o at the outer surface. If we set the concentrations as C_Ai and C_Ao at these points, we can evaluate the constants and obtain the following equation for the concentration profile in this shell:

\begin{matrix} C_{A} (r) - C_{A o} = (C_{A i} - C_{A o}) \frac{ln (r / R_{o})}{ln (R_{i} / R_{o})} & (2.29) \end{matrix}

$\begin{matrix} C_{A} (r) - C_{A o} = (C_{A i} - C_{A o}) \frac{ln (r / R_{o})}{ln (R_{i} / R_{o})} & (2.29) \end{matrix}$

The concentration profile is now logarithmic in r, rather than a linear profile across the shell.

The rate of transport across the system can be computed as

{\dot{M}}_{A} = - (2 π R_{i} L) D_{A} {(\frac{d C_{A}}{d r})}_{r = R_{i}}

${\dot{M}}_{A} = - (2 π R_{i} L) D_{A} {(\frac{d C_{A}}{d r})}_{r = R_{i}}$

or as

{\dot{M}}_{A} = - (2 π R_{o} L) D_{A} {(\frac{d C_{A}}{d r})}_{r = R_{o}}

${\dot{M}}_{A} = - (2 π R_{o} L) D_{A} {(\frac{d C_{A}}{d r})}_{r = R_{o}}$

Both expressions will lead to the same result:

{\dot{M}}_{A} = \frac{2 π R_{o} L (C_{A i} - C_{A o})}{ln (R_{o} / R_{i})}

${\dot{M}}_{A} = \frac{2 π R_{o} L (C_{A i} - C_{A o})}{ln (R_{o} / R_{i})}$

Mean Area for Diffusion

The expression for moles transported is often written in a format similar to that used in the Cartesian case:

{\dot{M}}_{A} = A_{m} \frac{D_{A}}{t} (C_{A i} - C_{A o})

${\dot{M}}_{A} = A_{m} \frac{D_{A}}{t} (C_{A i} - C_{A o})$

where t is the thickness for diffusion (equal to R_o – R_i here) and A_m is the mean area for diffusion. Comparing the two expressions for ${\dot{M}}_{A}$ ${\dot{M}}_{A}$ , we find the mean area for diffusion is the logarithmic mean area of the inner and outer cylinder:

A_{m} = 2 π L \frac{(R_{o} - R_{i})}{ln (R_{o} / R_{i})}

$A_{m} = 2 π L \frac{(R_{o} - R_{i})}{ln (R_{o} / R_{i})}$

2.2.2 Steady State Mass Transfer with Reaction

In our next scenario, we include the effect of reaction. This problem is representative of diffusion with reaction in a catalyst, which now takes the shape of a long cylinder or a concentric annular cylinder. We follow the method shown in Section 2.1.2, with the only addition being the reaction term, which is equal to control volume times the rate of reaction per unit volume, or (2πLr Δr)R_A. The final equation with the same assumption of constant D_A can be derived as

\begin{matrix} \begin{matrix} \frac{D_{A}}{r} \frac{d}{d r} (r \frac{d C_{A}}{d r}) = - R_{A} \end{matrix} & (2.30) \end{matrix}

$\begin{matrix} \begin{matrix} \frac{D_{A}}{r} \frac{d}{d r} (r \frac{d C_{A}}{d r}) = - R_{A} \end{matrix} & (2.30) \end{matrix}$

Note that the differential term on the left side of the equation is the r-dependent part of the Laplacian operator. Also note the following two forms for this part:

Laplacian, cylinder, r -only = \frac{1}{r} \frac{d}{d r} (r \frac{d C_{A}}{d r})

$Laplacian, cylinder, r -only = \frac{1}{r} \frac{d}{d r} (r \frac{d C_{A}}{d r})$

or equivalently

Laplacian, cylinder, r -only = \frac{d^{2} C_{A}}{d r^{2}} + \frac{1}{r} \frac{d C_{A}}{d r}

$Laplacian, cylinder, r -only = \frac{d^{2} C_{A}}{d r^{2}} + \frac{1}{r} \frac{d C_{A}}{d r}$

We now examine the solution for simple cases for R_A. However, we also note that in general a numerical solution is needed for reactions with nonlinear dependency on concentration.

Zero-Order Reaction

For a zero-order reaction R_A = –k₀, the governing equation is

\begin{matrix} \frac{D_{A}}{r} \frac{d}{d r} (r \frac{d C_{A}}{d r}) = k_{0} & (2.31) \end{matrix}

$\begin{matrix} \frac{D_{A}}{r} \frac{d}{d r} (r \frac{d C_{A}}{d r}) = k_{0} & (2.31) \end{matrix}$

This model holds for both a solid cylinder and an annular ring cylinder. The general solution is obtained easily by performing integration twice. The final solution is, however, different due to the change in the location where the boundary conditions are applied.

The geometry considered further here is a solid cylinder with a radius R. The domain of the solution is therefore 0 < r < R.

The boundary condition of constant surface concentration C_As is imposed at r = R, while the condition dC_A/dr = 0 at r = 0 is imposed at the center. The final result for the concentration profile is as follows (which you should verify by doing some minor algebra):

C_{A} = C_{A s} - \frac{k_{0}}{4 D_{A}} (R^{2} - r^{2})

$C_{A} = C_{A s} - \frac{k_{0}}{4 D_{A}} (R^{2} - r^{2})$

The final solution for a concentric cylinder is somewhat different since the domain of the solution is R_i < r < R_o; it is left as an exercise. An application of this geometry is in oxygen transport in tissues, a problem of importance in biomedical engineering that is discussed in Section 23.2.

First-Order Reaction

The solution for a first-order case where the rate is a linear function of concentration is somewhat complicated. The rate is given as

R_A = –k₁C_A

Use of this rate form in Equation 2.30 leads to

\begin{matrix} \frac{D_{A}}{r} \frac{d}{d r} (r \frac{d C_{A}}{d r}) = k_{1} C_{A} & (2.32) \end{matrix}

$\begin{matrix} \frac{D_{A}}{r} \frac{d}{d r} (r \frac{d C_{A}}{d r}) = k_{1} C_{A} & (2.32) \end{matrix}$

The solution of this differential equation is stated in terms of the Bessel functions. The general solution is

\begin{matrix} C_{A} (r) = A_{1} I_{0} (r \sqrt{k_{1} / D_{A}}) + A_{2} K_{0} (r \sqrt{k_{1} / D_{A}}) & (2.33) \end{matrix}

$\begin{matrix} C_{A} (r) = A_{1} I_{0} (r \sqrt{k_{1} / D_{A}}) + A_{2} K_{0} (r \sqrt{k_{1} / D_{A}}) & (2.33) \end{matrix}$

The I₀ and K₀ components are the modified Bessel functions of the first and second kind, respectively.

This solution may be applied to diffusion with reaction in a catalyst having the shape of a long solid cylinder or long concentric cylinder (discussed further in Section 23.2). The constants of integration depend both on the geometry (solid or concentric cylinder) and on the imposed boundary conditions at the end points of the geometry. The solid cylinder case is described in Example 2.7.

Example 2.7 Diffusion with First-Order Reaction in a Long Cylinder

A porous catalyst is cast in the form of a long cylinder and is exposed to a gas concentration of C_As at the surface. State the boundary conditions and derive an expression for the concentration profile.

Solution

The general solution shown in Equation 2.33, applies but the concentration should be finite at r = 0. Note that this can be taken as one of the boundary conditions. Since the K₀ function goes to infinity at r = 0, we have to set A₂ = 0 so that the concentration does not becomes unbounded. The solution reduces to

C_{A} (r) = A_{1} I_{0} (r \sqrt{k_{1} / D_{A}})

$C_{A} (r) = A_{1} I_{0} (r \sqrt{k_{1} / D_{A}})$

The second boundary condition at r = R is applied. Here C_A is set as C_As, the prescribed value. A₁ can be estimated and substituted back to get the final result for the concentration profile. The result is compactly represented in dimensionless form as

\begin{matrix} c_{A} = \frac{I_{0} (φ ξ)}{I_{0} (φ)} & (2.34) \end{matrix}

$\begin{matrix} c_{A} = \frac{I_{0} (φ ξ)}{I_{0} (φ)} & (2.34) \end{matrix}$

where c_A is a scaled or dimensionless concentration C_A/C_As, and ø is a dimensionless regrouping of the variables. You should verify that ø is defined as

φ = R \sqrt{k_{1} / D_{e A}}

$φ = R \sqrt{k_{1} / D_{e A}}$

This dimensionless parameter is referred to as the Thiele modulus. Finally, ξ = r/R is the dimensionless radial coordinate.

2.2.3 Transient Diffusion in a Cylinder

In the case of transient diffusion in a cylinder, we balance the in – out term with the accumulation. Details are not shown but you should be able to derive the following PDE for transient diffusion in a long cylinder with diffusion in only the radial direction:

\begin{matrix} \begin{matrix} \frac{D_{A}}{r} \frac{\partial}{\partial r} (r \frac{\partial C_{A}}{\partial r}) = \frac{\partial C_{A}}{\partial t} \end{matrix} & (2.35) \end{matrix}

$\begin{matrix} \begin{matrix} \frac{D_{A}}{r} \frac{\partial}{\partial r} (r \frac{\partial C_{A}}{\partial r}) = \frac{\partial C_{A}}{\partial t} \end{matrix} & (2.35) \end{matrix}$

Solution methods for such PDEs are examined in detail in Chapter 8.

2.3 Spherical Coordinates

In this section we show examples of problems posed in spherical coordinates. Diffusion in a spherical shell, diffusion with reaction in a porous spherical catalyst, and transient diffusion from a sphere are the three examples explored here, as they are prototypical problems encountered in many applications.

2.3.1 Steady State Diffusion across a Spherical Shell

The control volume in spherical coordinates is a spherical shell contained between r and r + Δr, as shown in Figure 2.6. The key point to note is that the area for diffusion is 4πr² at any location r. Similarly, the volume contained in the differential control volume is equal to 4πr² Δr. With these modifications the governing differential equations can be set up for various cases, following a similar procedure as the two earlier cases.

Diagrammatic representation of a 1-dimensional spherical control volume inside a sphere.

Figure 2.6 Control volume for 1-D differential analysis in a sphere.

For the diffusion-only case, we obtain

\frac{d}{d r} (r^{2} \frac{d C_{A}}{d r}) = 0

$\frac{d}{d r} (r^{2} \frac{d C_{A}}{d r}) = 0$

This suggests that

r^{2} \frac{d C_{A}}{d r} = constant, say A_{1}

$r^{2} \frac{d C_{A}}{d r} = constant, say A_{1}$

Now a second integration can be done and a general expression for the concentration profile can be obtained:

C_{A} = - \frac{A_{1}}{r} + A_{2}

$C_{A} = - \frac{A_{1}}{r} + A_{2}$

This concentration profile is no longer linear. The constants A₁ and A₂ can be found by fitting the boundary conditions at r = R_i and at r = R_o. Further details are straightforward and not shown.

The rate of transport across the shell is of more interest and can be computed as

{\dot{M}}_{A} = - (4 π R_{i}^{2}) D_{A} {(\frac{d C_{A}}{d r})}_{r = R_{i}}

${\dot{M}}_{A} = - (4 π R_{i}^{2}) D_{A} {(\frac{d C_{A}}{d r})}_{r = R_{i}}$

Here Fick’s law is used at r = R_i and multiplied by the area for diffusion at this radius. Alternatively, we can use Fick’s law at r = R_o:

{\dot{M}}_{A} = - (4 π R_{o}^{2}) D_{A} {(\frac{d C_{A}}{d r})}_{r = R_{o}}

${\dot{M}}_{A} = - (4 π R_{o}^{2}) D_{A} {(\frac{d C_{A}}{d r})}_{r = R_{o}}$

Both expressions will lead to the same result. The final result can be put in a form similar to the Cartesian geometry solution, using a mean area parameter for the diffusion area. The result is

{\dot{M}}_{A} = A_{m} \frac{D_{A}}{t} C_{A i} - C_{A o})

${\dot{M}}_{A} = A_{m} \frac{D_{A}}{t} C_{A i} - C_{A o})$

where t is the diffusion length equal to R_o – R_i and A_m is the area for diffusion. The latter can be shown to be equal to the “geometric” mean area of the inner and outer shell:

A_m = 4πR_iR_o

2.3.2 Diffusion and Reaction

In this section, a chemical reaction is assumed to take place together with diffusion. Again we consider concentration to vary as a function of r only. A shell balance approach leads to the following equation for diffusion with reaction in a spherical geometry:

\begin{matrix} \begin{matrix} \frac{D_{A}}{r^{2}} \frac{d}{d r} (r^{2} \frac{d C_{A}}{d r}) = - R_{A} \end{matrix} & (2.36) \end{matrix}

$\begin{matrix} \begin{matrix} \frac{D_{A}}{r^{2}} \frac{d}{d r} (r^{2} \frac{d C_{A}}{d r}) = - R_{A} \end{matrix} & (2.36) \end{matrix}$

Again, the leading differential term on the left side of Equation 2.36 is found to be the Laplacian operator

Laplacian, sphere, r -only = \frac{1}{r^{2}} \frac{d}{d r} (r^{2} \frac{d C_{A}}{d r})

$Laplacian, sphere, r -only = \frac{1}{r^{2}} \frac{d}{d r} (r^{2} \frac{d C_{A}}{d r})$

or equivalently

Laplacian, sphere, r -only = \frac{d^{2} C_{A}}{d r^{2}} + \frac{2}{r} \frac{d C_{A}}{d r}

$Laplacian, sphere, r -only = \frac{d^{2} C_{A}}{d r^{2}} + \frac{2}{r} \frac{d C_{A}}{d r}$

We find the diffusion-reaction problem for all the geometries can be generalized as

\begin{matrix} \begin{matrix} D_{A} \nabla^{2} C_{A} + R_{A} = 0 \end{matrix} & (2.37) \end{matrix}

$\begin{matrix} \begin{matrix} D_{A} \nabla^{2} C_{A} + R_{A} = 0 \end{matrix} & (2.37) \end{matrix}$

A constant value for the diffusion coefficient is implied in this model.

Details and the implications of the solution for calculation of the rate of reaction in a porous catalyst are examined in Chapter 18.

For now, it is useful to present the final solution for a spherical catalyst that is exposed to a surface concentration C_As. Solutions are presented in dimensionless form, with their detailed derivation left as exercises.

Zero-Order Reaction

The solution for a zero-order reaction in dimensionless form is

\begin{matrix} c_{A} = 1 - \frac{φ_{0}^{2}}{6} (1 - ξ^{2}) & (2.38) \end{matrix}

$\begin{matrix} c_{A} = 1 - \frac{φ_{0}^{2}}{6} (1 - ξ^{2}) & (2.38) \end{matrix}$

where ξ = r/R and $φ_{0} = R / \sqrt{(k_{0} / (D_{A} C_{A s})}$ $φ_{0} = R / \sqrt{(k_{0} / (D_{A} C_{A s})}$ .

The solution should be used only if $φ_{0} < \sqrt{6}$ $φ_{0} < \sqrt{6}$ . If $φ_{0} > \sqrt{6}$ $φ_{0} > \sqrt{6}$ , a concentration-depleted layer (a region of negative concentrations, which is unrealistic) develops near ξ = 0, the center of the sphere. A modified analysis is needed in such a case. Section 18.2.3 takes this issue up in more detail.

First-Order Reaction

The solution for a first-order reaction can be shown to be

\begin{matrix} c_{A} = \frac{sinh (φ ξ)}{ξ sinh (φ)} & (2.39) \end{matrix}

$\begin{matrix} c_{A} = \frac{sinh (φ ξ)}{ξ sinh (φ)} & (2.39) \end{matrix}$

where ξ = r/R and $φ = R \sqrt{(k_{1} / D_{A})}$ $φ = R \sqrt{(k_{1} / D_{A})}$ .

2.3.3 Transient Diffusion in Spherical Coordinates

Transient diffusion in a sphere in considered here briefly, as it has important applications in many areas. For example, this case may be used to find the rate of drug release from a capsule. A shell balance approach leads to the following partial differential equation for transient diffusion in a spherical geometry:

\begin{matrix} \frac{D_{A}}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial C_{A}}{\partial r}) = \frac{\partial C_{A}}{\partial t} & (2.40) \end{matrix}

$\begin{matrix} \frac{D_{A}}{r^{2}} \frac{\partial}{\partial r} (r^{2} \frac{\partial C_{A}}{\partial r}) = \frac{\partial C_{A}}{\partial t} & (2.40) \end{matrix}$

Here we illustrate the case in which the sphere has a uniform concentration C_Ai, which provides the initial condition. The sphere is exposed to a constant surface concentration C_A₀ at r = R after time zero. This provides one of the boundary conditions. The concentration is symmetric at r = 0 and hence dC_A/dr = 0 at this point, which provides the second boundary condition. The problem formulation is complete and analytical solutions or numerical solutions are used to solve for the temporal evolution of the concentration profile and the flux at the surface. These details are examined in Chapter 8.

For illustrative purposes, a sketch of dimensionless concentration versus position is shown in Figure 2.7. It is intended to help you get a qualitative feel for the result.

Figure 2.7 Evolution of the concentration profile for transient diffusion across a sphere. The final concentration should be 0. We reach almost 10% of the final value at time τ = 0.3.

Figure 2.7 uses dimensionless variables—specifically, dimensionless distance, ξ = r/R; dimensionless time, τ = tD_A/R²; and dimensionless concentration (more of a ratio of concentration differences), defined as:

c_{A} = \frac{C_{A} - C_{A 0}}{C_{A i} - C_{A 0}}

$c_{A} = \frac{C_{A} - C_{A 0}}{C_{A i} - C_{A 0}}$

This concentration is actually a dimensionless concentration difference. Thus the initial value of c_A is 1, and the final value is equal to the surface concentration of 0. Key points to note from the solution shown in Figure 2.7 are the following: At the initial time, the concentration drop is confined to the region near the surface, so the initial profile is rather steep. A nearly steady state profile is achieved at τ = 0.3. In terms of actual time, the value is about 0.3R²/D_A. The final steady state (dimensionless) concentration should be 0.

Summary

Diffusion in a slab of finite thickness with the ends held at different concentrations is the simplest problem in mass transfer. For constant diffusivity, the concentration profile is linear and the flux can be calculated from the slope of this linear profile.
An important application of slab diffusion is to calculate the transport rate of a gaseous species across a membrane. Note that the solubility of the gas in the membrane is also needed to solve such a problem, because the concentrations given in Section 2.1 are in the membrane phase, rather than in the external fluid phase.
The concentration profile for diffusion in long cylindrical shell is not a linear function of r, unlike the case in slab geometry. Instead, this profile is a logarithmic function. The difference arises due to the geometry or curvature effect—that is, due to the area of diffusion changing in the direction of diffusion. The curvature effect can be neglected for thin shells and the problem may be approximated as though a slab model applies.
Likewise, the concentration profile for diffusion across a spherical shell is an inverse function of r due to the curvature effect.
Diffusion with reaction is a well-studied and important problem in reaction engineering. The governing equation is the diffusion reaction equation given by Equation 2.9 for a slab, Equation 2.30 for a cylinder, and Equation 2.36 for a sphere. The equations can be compacted using vector notation and with the Laplacian operator of the concentration. The vector form applies to three-dimensional problems in any coordinate system.
The solution of the equation for diffusion with reaction for a zero-order reaction is the quadratic function for a slab. It can be shown to be a combination of a quadratic function and a logarithmic function in cylindrical coordinates (with r-variation only), but is a combination of r and 1/r in spherical coordinates.
Diffusion with a first-order reaction is amenable to analytical solution for the three common geometries considered in this chapter. Simple solutions in terms of exponential or hyperbolic functions are obtained for a slab. By comparison, the solutions in cylindrical coordinates are stated in terms of modified Bessel functions, while those for spherical coordinates are stated in terms of spherical Bessel functions.
Transient diffusion is governed by a PDE (partial differential equation). An initial condition is also required. Results are usually shown using a dimensionless time. The definitions of time constant and dimensionless time are worth remembering.
Convection imposed in the same direction as diffusion is one of the simplest problems in convective mass transfer. The effect of convection can be represented by a correction factor (blowing factor), which depends on a dimensionless number called the Peclet number.

Review Questions

2.1 Equation 2.6 shows that the concentration profile for diffusion in a slab is linear. State the assumptions that have gone into this result.

2.2 When can the concentration profile for diffusion in a slab geometry be nonlinear?

2.3 State the differential equation for concentration in a slab geometry for diffusion with a second-order reaction.

2.4 State the two commonly observed boundary conditions in diffusion-reaction problems. State which condition applies when.

2.5 What is the definition of the Thiele modulus for a first-order reaction and what is its significance?

2.6 How is the Thiele modulus defined for a zero-order reaction?

2.7 What is the definition of diffusion time and what is its significance?

2.8 What is the definition of the Peclet number and what is its significance?

2.9 Define blowing factor and state its use.

2.10 A catalyst is cast in the shape of a long concentric cylinder. A first-order reaction is taking place here. What is the general solution for the concentration profile?

2.11 State the solution to diffusion with a first-order reaction in a spherical catalyst particle.

Problems

2.1 Membrane for oxygen separation. Data for gas permeation in polymers have been reported by Seader et al. (2010). Table 2.1 shows these data for low-density polyethylene. Note that both diffusivity and solubility are needed to calculate the rate of diffusion.

Table 2.1 Diffusivity and Solubility Data for Gases in Low-Density Polyethylene

Gases	H₂	O₂	N₂	CH₄
D × 10¹⁰ m²/s	0.474	0.46	0.32	0.193
H × 10⁶m³ STP/m³ · Pa	1.58	0.472	0.228	1.13

The solubility parameter H here is defined as the concentration in the solid (membrane) divided by the partial pressure of the gas at equilibrium. The concentration in the solid is defined as m³ gas in STP perm³ solid (q_A) rather than in mol/m³ Thus the Henry’s law constant is defined as q_A = H_Ap_A.

Find the rate of oxygen and nitrogen transport rate (in mol/m² s) for the previously described membrane when it is exposed to air with 150 psia and 78 °F on one side and atmospheric pressure on the other side. If a rate ratio of at least 5 is needed for good oxygen separation, is this a good membrane to enrich oxygen from air? The membrane thickness is 5 μ m.

2.2 Slab diffusion with diffusivity varying with concentration. A membrane is set up such that the diffusion coefficient varies as a function of concentration according to the relation

D = D₀(1 + αC_A)

The membrane is of thickness L with the concentrations of C_A₀ and C_AL as the end point values. Find an expression for the concentration profile and the flux across the system. Compare this expression with the flux value using the flux based on a constant value of the diffusion coefficient based on the average concentration—that is, the diffusivity based on (C_A₀ + C_AL)/2.

2.3 Hydrogen leakage from a steel tank. A spherical steel tank of 1 L capacity and a 2-mm wall thickness is used to store hydrogen at 673 K. The initial pressure in the tank is 5 bar, and the hydrogen partial pressure outside the tank is zero. Hydrogen has finite solubility in steel and leaks out of the tank. Find the rate of leakage at the start of the process where the pressure in the tank is 5 bar.

The solubility and diffusivity of hydrogen are functions of temperature and given by the following relations:

C_{A} = 3.74 \times 10^{3} exp (- 3950 / T) \sqrt{p A}

$C_{A} = 3.74 \times 10^{3} exp (- 3950 / T) \sqrt{p A}$

where C_A is the concentration of dissolved hydrogen, mol/m³, in equilibrium with a pressure of p_A, with pressure unit in bars.

D = 1.65 × 10^–6 exp(–4630/T)

with T in K and D in m²/s.

Also calculate the leakage rate if the tank pressure is 10 bar.

Note that the solubility relation is nonlinear. This is often observed for gases such as hydrogen, which dissociates at the metal surface and is present as H-atoms in the solid matrix rather than as H₂ molecules in the gas phase. Hence doubling the pressure will not double the transport rate.

2.4 Mean area for diffusion in a cylindrical shell. Consider the diffusion across a shell of long cylinder of inner radius R_i and outer radius R₀. Often the results are expressed in the form similar to that for a slab:

{\dot{M}}_{A} = A_{m} D_{A} \frac{C_{A i} - C_{A o}}{t}

${\dot{M}}_{A} = A_{m} D_{A} \frac{C_{A i} - C_{A o}}{t}$

where t is the thickness, which is equal to R₀ – R_i

The area A_m is some representative average area for mass transfer. Verify that the area should be log mean of the inside area 2πR_iL and the outer surface area 2πR_oL.

Also show that if the membrane thickness is small, either of the areas (inner or outer) can be used as an engineering approximation. The slab model is sufficient for such cases and the actual logarithmic profile is close to a linear profile.

2.5 Mean area for diffusion in a spherical shell. Repeat problem 4 for diffusion across a spherical shell. Show that the area should be the geometric mean of the inside area $4 π R_{i}^{2}$ $4 π R_{i}^{2}$ and the outer surface area $4 π R_{o}^{2}$ $4 π R_{o}^{2}$ .

2.6 General convection-diffusion equation. Write out in detail the convection-diffusion equation for a 3-D problem in Cartesian coordinates. Show that Equation 2.19 will be obtained after making suitable simplifications.

Verify the general solution given by Equation 2.20 for the 1-D case shown in the text. Find the solution for a unit concentration at x = 0 and zero concentration at x = L.

2.7 Convection augmentation of mass transfer in a slab. Plot the augmentation factor due to convection (the blowing parameter) as a function of the Peclet number. Find the limiting values of this factor for large values of Pe. Plot the factor if the velocity is opposite to the direction of diffusion. What would be flux under large velocity values?

2.8 Convection effects in a cylinder. Consider a porous membrane cast as the shell of a long cylinder of inner radius R_i and outer radius R₀. Concentrations are maintained at different values C_Ai and C_Ao at each of these surfaces, and diffusion is taking place. What is the number of moles transported across the system? This provides the base value in the absence of convection.

Now to increase the transport rate, a gas at a flow rate of Q is forced across the system in the radial direction, and convection is also taking place in addition to diffusion. Set up the model for this system and derive the governing differential equation. Solve the equation and find the augmentation in the rate of mass transfer due to the flow.

2.9 Effectiveness factor of a catalyst. A catalyst is cast in the form a thin slab of thickness 2L. Species A diffuses and undergoes a first-order reaction. Set up a 1-D differential model. Show that the solution can be expressed in dimensionless form using a Thiele parameter, ø, defined as

φ = L \sqrt{k_{1} / D_{e A}}

$φ = L \sqrt{k_{1} / D_{e A}}$

Verify the following solution:

c_{A} = \frac{cosh (φ ξ)}{cosh (φ)}

$c_{A} = \frac{cosh (φ ξ)}{cosh (φ)}$

where c_A is C_A/C_As and ξ = x/L.

The concentration variation along position causes the rate of reaction to vary along with the position. For design purposes, an average rate of reaction, $- {\bar{R}}_{A}$ $- {\bar{R}}_{A}$ is often interest. This is defined as

- {\bar{R}}_{A} = \frac{1}{L} \int_{0}^{L} k_{1} C_{A} d x

$- {\bar{R}}_{A} = \frac{1}{L} \int_{0}^{L} k_{1} C_{A} d x$

Derive an expression for this concentration variation.

The average reaction rate is often scaled by the maximum reaction rate and the ratio is called the effectiveness factor of the catalyst. It is defined as

η = \frac{{\bar{R}}_{A}}{- k_{1} C_{A s}}

$η = \frac{{\bar{R}}_{A}}{- k_{1} C_{A s}}$

Show that the following expression can be derived for the effectiveness factor:

η = \frac{tanh φ}{φ}

$η = \frac{tanh φ}{φ}$

Show that this simplifies to η = 1/ø for ø > 3.

Calculate the effectiveness factor for a catalyst slab with thickness L = 3 mm having an effective diffusion coefficient of 2 × 10^–5 m²/s. The rate constant has a value of 0.2 s^–1.

2.10 Concentration profiles in a cylindrical catalyst. A porous catalyst is used for CO oxidation and the process is modeled as a first-order reaction with a rate constant of 0.2 s^–1. The effective diffusion coefficient for pore diffusion was estimated as 4 × 10^–6 m²/s. The catalyst is exposed to a bulk gas at 600 K and 1 bar pressure with a CO mole fraction of 2%. External mass transfer coefficient is assumed to be not rate limiting, so a type I condition can be used. The catalyst is in the form of a long cylinder of radius 5 mm.

What is the solution for the concentration profile? Write the solution for concentration in terms of a Thiele parameter defined as $R \sqrt{k_{1} / D_{e A}}$ $R \sqrt{k_{1} / D_{e A}}$ . Plot the CO concentration within the catalyst as a function of radial position. Find the concentration value at the center of the catalyst.

2.11 Zero-order reaction in catalyst of three shapes. Derive an expression for the concentration profile for a zero-order reaction taking place in (a) a rectangular slab catalyst, (b) a long cylinder, and (b) a sphere.

2.12 Transport of oxygen in a tissue. Krogh (1919), in a Nobel Prize–winning paper, modeled oxygen transport in tissues by assuming the tissue to be an annular cylinder where a blood vessel of radius R_i is surrounded by a tissue region with an outer radius of R_o.

Oxygen diffuses into the tissues and undergoes a zero-order reaction. What is the governing differential equation? State the boundary conditions under the following assumptions:

r = R_i, the inner radius, is in contact with blood and the oxygen concentration at this point is fixed as the concentration in the blood.

Oxygen does not diffuse past r = R_o, the outer radius of the tissue. Solve the concentration profile in the tissue. If the oxygen concentration at the outer edge of tissue is nearly zero, what is the concentration in the blood?

2.13 First-order reaction in a spherical catalyst. A spherical catalyst is 6 mm in diameter and is maintained at a concentration of 20, 000 g · mol/m³ at the surface. The effective diffusion coefficient is 0.02 cm²/s and the rate constant is 0.2 s^–1. What is the center concentration?

An average rate of reaction can be defined as

〈 R_{A} 〉 = \frac{1}{V_{p}} \int_{0}^{R} 4 π r^{2} k_{1} C_{A} d r

$〈 R_{A} 〉 = \frac{1}{V_{p}} \int_{0}^{R} 4 π r^{2} k_{1} C_{A} d r$

where V_p is the volume of the sphere equal to 4πR³/3. What is the average rate of reaction?

The ratio of the average rate of reaction to the value based on the surface concentration is called the effectiveness factor of the catalyst. Derive an expression for the effectiveness factor. Calculate the numerical value for the given data.

2.14 Effect of particle size. In problem 13, if the catalyst size is reduced by half what would be the effectiveness factor? If an effectiveness factor of 0.9 is needed, what should be the diameter of the catalyst?

2.15 Drug release from a capsule. A drug capsule is a 0.6-cm-diameter sphere and has an active component that diffuses out the capsule. If the diffusion coefficient is 3 × 10^–6 cm²/s, estimate the approximate time for which the active component will be effective.

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Table of Contents for Chapter 2. Examples of Differential (1-D) Balances

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Chapter 2. Examples of Differential (1-D) Balances

2.1 Cartesian Coordinates

2.1.1 Steady State Diffusion across a Slab

2.1.2 Steady State Diffusion with Reaction in a Slab

2.1.3 Transient Diffusion in a Slab

2.1.4 Diffusion with Convection

2.2 Cylindrical Coordinates

2.2.1 Steady State Radial Diffusion

Diffusion across a Cylindrical Shell

Mean Area for Diffusion

2.2.2 Steady State Mass Transfer with Reaction

Zero-Order Reaction

First-Order Reaction

2.2.3 Transient Diffusion in a Cylinder

2.3 Spherical Coordinates

2.3.1 Steady State Diffusion across a Spherical Shell

2.3.2 Diffusion and Reaction

Zero-Order Reaction

First-Order Reaction

2.3.3 Transient Diffusion in Spherical Coordinates

Summary

Review Questions

Problems

Table of Contents for
Chapter 2. Examples of Differential (1-D) Balances