16.1 Transport Across a Membrane

Consider the two reservoirs (indicated with subscripts 1 and 2) of Fig. 16.1, having equal volume, V, and size, L, separated by a porous membrane of section S = V/L, porosity ε, and thickness δ. The two reservoirs contain a well-mixed A–B mixture; initially, the molar fractions of A in the two reservoirs are x A10 and x A20 < x A10. Then, as time progresses, A will diffuse across the membrane from left to right, while B will diffuse in the opposite direction. We want to determine how the difference between the molar fractions of A in the two reservoirs, x A1−x A2, changes with time, so that the diffusion coefficient D of A into B within the membrane can be determine.

Fig. 16.1
figure 1

Material transport across a membrane

Full size image

This problem is almost identical to the one that was solved in Sect. 9.4.

First of all, we observe that the total moles of A must be conserved. Therefore, as for long times the system must be at chemical equilibrium, with uniform concentration x Af , assuming that the total concentration, c, is constant we obtain:

$$x_{A1} + x_{A2} = x_{A1,0} + x_{A2,0} = 2x_{Af} \Rightarrow x_{Af} = \left( {x_{A1,0} + x_{A2,0} } \right)/2$$
(16.1.1)

where x A1,0 and x A2,0 are the initial compositions in the two reservoirs and we have neglected the number of moles within the membrane. Assuming that x A  = O(1), this latter assumption means that the wetted volume V w occupied by the fluid mixture within the membrane is much smaller than the volume V of the reservoir. Therefore, since the ratio between V w and the membrane total volume, Sδ, is the porosity ε (see Eq. 5.4.3), this assumption becomes:

$$\frac{\varepsilon \delta }{L} \ll 1.$$
(16.1.2)

Now, in order to find the transient solution of the problem, we must write a separate molar balance of the A species for each reservoir separately, obtaining,

$$\frac{d}{dt}\left( {cx_{A1} V} \right) = - J_{A1} S\varepsilon ,$$
(16.1.3a)
$$\frac{d}{dt}\left( {cx_{A2} V} \right) = - J_{A2} S\varepsilon ,$$
(16.1.3b)

where J A1 and J A2 are the molar fluxes leaving the reservoirs 1 and 2, respectively. Here, we have assumed that the porous membrane is isotropic, so that its porosity is also equal to the ratio between the wetted area and the total cross sectional area (see Eq. 4.4.6), and therefore the effective cross section area of the membrane is \(S\varepsilon\).

Now, subjected to a posteriori verification, suppose that the hypothesis of Quasi Steady State (QSS) can be applied. That means that the fluxes J A1 and J A2 correspond to the steady fluxes of A induced by a given, constant composition x A1 and x A2 in the two reservoirs. In other words, if x A1 or x A2 change abruptly, the material flux of A will adjust to the new condition, reaching a new steady state value within a characteristic relaxation time τ ss . Whenever the QSS approximation is valid, τ ss is so short, compared to the time τ characterizing the variation of x A , that the relaxation process can be considered to be instantaneous. Therefore, since at steady state there cannot be any accumulation of A within the membrane, the flux of A leaving 1 is equal to the flux of A entering 2, i.e.,

$$J_{A1} = - J_{A2}.$$
(16.1.4)

In addition, in the general constitutive relation (14.3.9),

$$J_{A} = x_{A} (J_{A} + J_{B} ) - cD - \frac{{dx_{A} }}{dz},$$

the convective term is zero, because, as the total molar concentration c = c A  + c B is constant, for each mole of A moving from right to left there must be a mole of B moving in the opposite direction, so that the fluxes of A and B are equimolar and countercurrent,

$$J_{A} = - J_{B}.$$
(16.1.5)

Concluding, as the flux J A (and J B as well) is diffusive, in the QSS approximation the concentration profile is linear between x A1 and x A2, that is:

$$J_{A} = \frac{cD}{\delta }\left( {x_{A1} - x_{A2} } \right).$$
(16.1.6)

Let us see when the QSS approximation can be applied. As we saw, the QSS hypothesis is applicable when the characteristic time, τ eq , describing the temporal variation of x A1 and x A2 (in fact, x A1 and x A2 are the boundary values of the concentration profile x A within the membrane) is much larger than the time τ ss that takes the flux J A to reach its stationary value (16.1.6). Thus, as x A1 and x A2 change, tending in a time τ eq , to their common equilibrium value, x Af  = x A1 = x A2, the flux J A will adapt to the changing boundary conditions within a time, τ ss , that is much smaller than τ eq . From Eqs. (16.1.3a, b) we derive the following estimate,

$$\tau_{eq} \approx \frac{cV}{{J_{A} S\varepsilon }} \approx \frac{L\delta }{D\varepsilon },$$
(16.1.7)

with L = V/S, where we have assumed that x A  = O(1), and we have substituted Eq. (16.1.5) for the flux J A . In addition, as in every diffusive process, the time necessary to reach steady state is τ ss  ≈ δ 2/D. Therefore, the QSS hypothesis is valid when:

$$\tau_{ss} \ll \tau_{eq} \Rightarrow (\delta \varepsilon )/L \ll 1.$$
(16.1.8)

This condition is identical to Eq. (16.1.2), indicating that that QSS approximation means assuming that the wetted volume of the membrane is much smaller than the volumes of the reservoirs. It is satisfied when the membrane thickness, δ, is much smaller than the size of the reservoir, L, or/and when the membrane porosity, ε, is very small.

Now, let us find the time-dependent solution of the problem. Substituting Eq. (16.1.5) into (16.1.1) and (16.1.2) we obtain:

$$L\frac{{dx_{A1} }}{dt} = - \frac{\varepsilon D}{\delta }\left( {x_{A1} - x_{A2} } \right),$$
(16.1.9a)
$$L\frac{{dx_{A2} }}{dt} = \frac{\varepsilon D}{\delta }\left( {x_{A1} - x_{A2} } \right),$$
(16.1.9b)

to be solved with initial conditions x A1(0) = x A10 and x A2(0) = x A20. Summing these two equations we find again the condition (16.1.1). Instead, subtracting them from one another and integrating between t = 0 and t, we obtain:

$${ \ln }\frac{{{\Delta} x_{0} }}{{{\Delta} x}} = \frac{2D\varepsilon }{\delta L}t = 2\frac{t}{{\tau_{eq} }} \Leftrightarrow {\Delta} x = {\Delta} x_{0} e^{{ - 2t/\tau_{eq} }} ,$$
(16.1.10)

where Δx = x A1 − x A2, while τ eq is defined in (16.1.7). This solution confirms that τ eq is indeed the characteristic time that the system needs to reach its equilibrium state, with uniform composition. Finally, substituting Eq. (16.1.10) into (16.1.9a) and integrating, we find:

$$x_{A1} = x_{A10} - \frac{{\varepsilon D\tau_{eq}\Delta x_{0} }}{2\delta L}\left( {1 - e^{{ - 2t/\tau_{eq} }} } \right).$$
(16.1.11)

Starting from Eq. (16.1.10), we see that the diffusion coefficient D can be determined by measuring the concentration at two times, t = 0 and t, i.e.,

$$D = \frac{\delta L}{2\varepsilon t}\text{ln}\frac{{\Delta x_{0} }}{{\Delta x}}.$$
(16.1.12)

16.2 Evaporation of a Liquid from a Reservoir

This problem is the natural continuation of the steady diffusion across a stagnant film of Sect. 15.1, where we studied the sublimation of a solid species A, occupying the region 0 < z < z 1 of a reservoir, with its vapor crossing a stagnant gas, B, laying in the region z 1 < z < z 2, with Δz = z 2 − z 1 (see Fig. 16.2). Obviously, the problem could also consist of the evaporation of a liquid from a container where it is initially confined. The molar fraction x A of the vapor A at the interface z = z 1 is fixed at its saturation value, \(x_{A1} = x_{A}^{sat} ,\) from the condition of local equilibrium, while at the upper boundary z = z 2 it is kept constant at x A2. Unlike the case of Sect. 15.1, however, here the height z 1 is not constant, as A progressively sublimates (or evaporates), thus emptying the reservoir. We intend to determine the characteristic emptying time, assuming that initially z 1 = z 1,i.

Fig. 16.2
figure 2

Evaporation of a liquid from a reservoir

Full size image

Consider the material balance within the volume indicated in Fig. 16.2,

$$\frac{d}{dt}\int\limits_{V(t)} {c_{A} dV} = - \oint\limits_{S(t)} {J_{Ar} dS,}$$
(16.2.1)

where J Ar is the flux of A exiting the solid volume V, relative to the velocity w = dz 1/dt of the interface z = z 1. Therefore, since J A  = c A v A is the flux of A referred to a fixed reference frame, we have:

$$J_{Ar} = C_{A} \left( {v_{A} - w} \right) = J_{A} - C_{A} \frac{{dz_{1} }}{dt},$$
(16.2.2)

so that:

$$C_{A}^{S} \frac{dV}{dt} = - {J_{Ar}}^{S} \left( {z_{1} } \right)S = - {J_{Ar}}^{V} \left( {z_{1} } \right)S,$$
(16.2.3)

where S is the area of the interface. Here, \(J_{Ar}^{S} \left( {z_{1} } \right)\) and \(J_{Ar}^{V} \left( {z_{1} } \right)\) are, respectively, the flux of A on the solid side, \(z = z_{1}^{ - } ,\) and that on the vapor side, \(z = z_{1}^{ + } ,\) of the interface, and we have considered that, from the simple material balance illustrated in Fig. 16.3, by continuity the two fluxes are equal to each other. Therefore, considering that V = Sz 1, we finally obtain:

$${c_{A}}^{S} \frac{{dz_{1} }}{dt} = - J_{A1} + c_{A1} \frac{{dz_{1} }}{dt},$$
(16.2.4)

where \({c_{A}}^{S}\) is the concentration of A in the solid phase, while c A1 = cx A1 and J A1 are the concentration and the flux of A in the vapor phase at the interface z = z 1, respectively.

Fig. 16.3
figure 3

Continuity of the material flux at the interface

Full size image

Now, subjected to a posteriori verification, let us assume that the QSS approximation can be applied. That means that the characteristic relaxation time, τ ss  ≈ (Δz)2/D, that is needed to reach steady state, is much shorter than the emptying time, τ, that is the time characterizing the interface motion. Consequently, the diffusion speed, \(v_{A} \approx D/\Delta z,\) is much larger than the speed of the interface, \(w = dz_{1} /dt \approx \varDelta z/\tau\). Therefore:

  1. 1.

    As \(dz_{1} /dt \ll v_{A} ,\) from Eq. (16.2.2) we see that the relative flux can be approximated as the absolute flux, i.e. J Ar  = J A , so that the last term in Eq. (16.2.4) can be neglected, showing that \(c_{A1} \ll {c_{A}}^{s} .\)

  2. 2.

    The flux J A1 can be expressed as the steady state expression (15.1.7), i.e.,

$$J_{A1} = \frac{{cD\left( {x_{A1} - x_{A2} } \right)}}{{\left( {z_{2} - z_{1} } \right)\left( {x_{B} } \right)_{{\text{ln}}} }}.$$
(16.2.5)

Substituting Eq. (16.2.5) into (16.2.4) yields:

$$\frac{{dz_{1} (t)}}{dt} = - \frac{W}{{z_{2} - z_{1} \left( t \right)}},\quad \text{with}\quad W\text{ = }\frac{{cD\left( {x_{A1} - x_{A2} } \right)}}{{{c_{A}}^{S} \left( {x_{B} } \right)_{{\text{ln}}} }},$$
(16.2.6)

and integrating between t = 0 and t, with z 1(t = 0) = z 10, we obtain the final result:

$$\left( {z_{2} - z_{1} } \right)^{2} - \left( {z_{2} - z_{10} } \right)^{2} = 2Wt.$$
(16.2.7)

In particular, we find the emptying time, τ, defined as z 1(τ) = 0,

$$\tau = \frac{{{z_{2}}^{2} - (z_{2} - z_{10} )^{2} }}{2W}.$$
(16.2.8)

Note that, as it always happens in diffusive processes, time is proportional to the square of a length.

Now, let us see when the QSS approximation can be applied:

$$\tau_{ss} \ll \tau \Rightarrow \frac{{\left( {{\Delta} z} \right)^{2} }}{D} \ll \frac{{\left( {{\Delta} z} \right)^{2} }}{W} \approx \frac{{\left( {{\Delta} z} \right)^{2} c_{A}^{S} }}{{Dc_{A1} }} \Rightarrow c_{A1} \ll c_{A}^{S} ,$$
(16.2.9)

where we have considered that \(\tau \approx \left( {\varDelta z} \right)^{2} /W\) and \(W \approx c_{A1} /c_{A}^{S} ,\) with \(\left( {x_{B} } \right)_{{\text{ln}}} = O\left( 1 \right).\) We could obtain the same result by imposing that the diffusion speed, \(v_{A} \approx D/\Delta z,\) is much larger than the speed of the interface, \(w = dz_{1} /dt \approx W/\Delta z\) (see Eq. 16.2.6).

The condition \(c_{A1} \ll c_{A}^{S}\) is generally verified. For example, in the case of the evaporation of water, \(c_{A}^{S} = 55\,{\text{mole/lt,}}\) while, assuming that the vapor behaves as an ideal gas, at ambient conditions \(c_{A1} \cong 10^{ - 3} \,{\text{mole/lt,}}\) so that \(c_{A1} /c_{A}^{S} \approx 10^{ - 4} .\) Finally, note that in the dilute case, \(x_{A} \ll 1,\) we obtain the same result (16.2.7), with x B  = 1.

16.3 Slow Combustion of a Coal Particle

Consider a spherical particle of coal, having radius R and concentration c S C of carbon (species C), immersed in air, containing oxygen (species A) with molar fraction \({ x_{A}}^{\infty }\). Oxygen diffuses towards the particle surface, where it reacts with carbon to produce carbon dioxide (species B), with a first-order heterogeneous chemical reaction, with reaction rate k M ,Footnote 1

$${\text{C}} + {\text{O}}_{2} \to {\text{CO}}_{2} \quad J_{A} \left( {r = R} \right) = - k_{M} c_{A} \left( {r = R} \right).$$
(16.3.1)

Knowing that initially the particle radius is R 0, we want to calculate the time needed for the complete combustion.

A material balance of the moles of carbon within the particle volume yields:

$$\frac{d}{dt}\left( {{c_{C}}^{S} \frac{4}{3}\pi R^{3} } \right) = - J_{Br} \left( {4\pi R^{2} } \right) = J_{Ar} \left( {4\pi R^{2} } \right),$$
(16.3.2)

where J Br and J Ar are, respectively, the flux of carbon dioxide and of oxygen at the gas-solid interface, relative to the speed dR/dt of the interface,Footnote 2 and we have considered that for each mole of C consumed in the volume there is one mole of B leaving and one mole of A arriving at the interface.

Now we assume that the QSS hypothesis is satisfied. As we saw in the previous section, that amounts to assuming that the interface velocity, dR/dt, is negligible with respect to the diffusion speed, so that the relative material flux, J Ar , is approximately equal to the absolute material flux, J A , obtaining:

$${c_{C}}^{S} \frac{dR}{dt} = J_{A} \left( {r = R} \right).$$
(16.3.3)

The material flux J A satisfies the constitutive relation,

$$J_{A} = x_{A} \left( {J_{A} + J_{B} + J_{N} } \right) - cD\frac{{dx_{A} }}{dr}$$
(16.3.4)

where the subscript N indicates the nitrogen of the air which, being inert, is stagnant and therefore J N  = 0; in addition, as we saw, the flux of oxygen, J A , and the flux of carbon dioxide, J B , are countercurrent and equimolar, so that J A  = −J B , and therefore the oxygen flux is purely diffusive,

$$J_{A} = - cD\frac{{dx_{A} }}{dr}.$$
(16.3.5)

Note that the diffusion coefficient D is an effective parameter, accounting for the fact that A diffuses in a multicomponent mixture. This approximation must be verified case by case, but in general it is always valid in the dilute case, for \(x_{A} \ll 1,\) when D indicates the diffusivity of oxygen in nitrogen.

Now, at steady state (congruent with the QSS approximation), imposing that \(J_{A} r^{ 2}\) is constant, we obtain the usual spherically symmetric solution,

$$- \frac{{dx_{A} }}{dr}r^{2} = B \Rightarrow x_{A} \left( r \right) = A + \frac{B}{r},$$
(16.3.6)

where A and B are constant to be determined imposing that the following boundary conditions are satisfied,

$$x_{A} = {x_{A}}^{\infty } \quad \text{as}\;r \to \infty ,$$
(16.3.7)
$$J_{A} = - cD\frac{{dx_{A} }}{dr} = - k{}_{M}c_{A} \quad \text{at}\;r = R.$$
(16.3.8)

Thus:

$$x_{A} \left( r \right) = {x_{A}}^{\infty } \left[ {1 - \left( {\frac{{k_{M} R^{2} }}{{D + k_{M} R}}} \right)\frac{1}{r}} \right]\quad \text{and}\;J_{A} \left( {r = R} \right) = - \frac{{cDk_{M} {x_{A}}^{\infty } }}{{D + k_{M} R}}.$$
(16.3.9)

Substituting this results into Eq. (16.3.3), we obtain:

$$\frac{dR}{dt} = \frac{{J_{A} \left( {r = R} \right)}}{{{c_{C}}^{S} }} = - \frac{{Dk_{M} \left( {{c_{A}}^{\infty } /{c_{C}}^{S} } \right)}}{{D + k_{M} R}},$$
(16.3.10)

where \({c_{A}}^{\infty } = {cx_{A}}^{\infty }.\) Finally, integrating with R(t = 0) = R 0 and \(R\left( {t = \tau } \right) = 0,\) we find:

$$\tau = \frac{{R_{0}^{2} }}{2D}\frac{{c_{C}^{S} }}{{c_{A}^{\infty } }}\left( {1 + \frac{1}{{Bi_{M} }}} \right) = \tau_{diff} + \tau_{react} ,\quad {\text{with}}\;\tau_{diff} = \frac{{c_{C}^{S} }}{{c_{A}^{\infty } }}\frac{{R_{0}^{2} }}{2D}\quad {\text{and}}\;\tau_{react} = \frac{{c_{C}^{S} }}{{c_{A}^{\infty } }}\frac{{R_{0} }}{{k_{M} }},$$
(16.3.11)

where \(Bi_{M} = k_{M} R_{0} /2D = \tau_{diff} /\tau_{react}\) is the material Biot number, expressing the ratio between the material flux at the wall, due to the heterogeneous chemical reaction, and the diffusive flux occurring at r > R. The material Biot number is analogous to its thermal counterpart, defined in Eq. (9.2.7). Note that the kinetics of the process is controlled by the slowest of the two processes, that is diffusion when \(\tau_{diff} \gg \tau_{react} ,\) and reaction when \(\tau_{diff} \gg \tau_{react}\).

The QSS hypothesis is verified assuming that \(\tau \gg \tau_{ss} \approx R_{0}^{2} /D,\) which means also assuming that the interface velocity, \(dR/dt \approx R_{0} /\tau ,\) is much smaller than the diffusion speed, D/R 0, obtaining,

$$\left( {1 + \frac{1}{{Bi_{M0} }}} \right)\left( {\frac{{c_{C}^{S} }}{{c_{A}^{\infty } }}} \right) \gg 1.$$
(16.3.12)

This condition is always satisfied, provided that \({c_{A}}^{\infty } \ll {c_{C}}^{S} ,\) which is normally true in normal conditions, as we saw in the previous section.

Now, it is instructive to consider the two limit cases, when \(Bi_{M} \gg 1\) and \(Bi_{M} \ll 1\).

  1. (a)

    \(Bi_{M} \gg 1.\)

In this case, reaction is much faster than diffusion, meaning that the oxygen A-molecules react as soon as they reach the particle surface. Therefore, \(k_{M} \to \infty ,\) and the boundary condition (16.3.8) reduces to x A (r = R) = 0, so that the solution (16.3.9) becomes:

$$x_{A} \left( r \right) = {x_{A}}^{\infty } \left( {1 - \frac{R}{r}} \right);\quad J_{A} \left( {r = R} \right) = - \frac{{{Dc_{A}}^{\infty } }}{R},$$
(16.3.13)

and:

$${R_{0}}^{2} - R^{2} = 2\frac{{{c_{A}}^{\infty } }}{{{c_{C}}^{S} }}Dt.$$
(16.3.14)

Therefore, the characteristic time needed to consume a particle of initial radius R 0 is:

$$\tau = \tau_{diff} = \frac{{R_{0}^{2} }}{2D}\frac{{c_{C}^{S} }}{{c_{A}^{\infty } }},$$
(16.3.15)

and the QSS hypothesis is satisfied provided that \({c_{A}}^{\infty } \ll {c_{C}}^{S}\). As expected, diffusion here is the controlling process, since it is much slower than reaction, with \(\tau_{diff} \gg t_{react}.\)

  1. (b)
    $$Bi_{M} \ll 1.$$

In this case, diffusion is much faster than reaction, meaning that the oxygen A-molecules diffuse very rapidly towards the particle surface, where they slowly react. Therefore, \(k_{M} \to 0\), and the boundary condition (16.3.8) reduces to \(J_{A} \left( {r = R} \right) = 0,\) so that the solution (16.3.9) becomes:

$$x_{A} \left( r \right) = {x_{A}}^{\infty } .$$
(16.3.16)

Substituting this result into Eqs. (16.3.1) and (16.3.3), we obtain: \(dR/dt = - k_{M} c_{A}^{\infty } /{c_{C}}^{S}\), and thus:

$$R_{0} - R = \frac{{{c_{A}}^{\infty } }}{{{c_{C}}^{S} }}k_{M} t.$$
(16.3.17)

Therefore, the characteristic time needed to consume a particle of initial radius R 0 is:

$$\tau = \tau_{react} = \frac{{R_{0} }}{{k_{M} }}\frac{{c_{C}^{S} }}{{c_{A}^{\infty } }},$$
(16.3.18)

and the QSS hypothesis is satisfied provided that \(Bi_{M} {c_{A}}^{\infty } \ll {c_{C}}^{S} .\) This condition is satisfied even more easily than in the previous case. As expected, the controlling process here is reaction, since it is much slower than diffusion, with \(\tau_{diff} \ll \tau_{react}\).

16.4 Unsteady Evaporation

Consider a liquid A that evaporates into a vapor B in a tube of infinite length, assuming that the liquid level is kept at position z = 0 and that B is insoluble in A. In addition, the system is kept at constant pressure and temperature and the vapors A and B form an ideal gas mixture, so that the concentration c is constant in the gas phase. Therefore, as A starts to evaporate, it pushes B forward, along the tube z-axis.

The continuity equations for A and B in the semi-infinite region \(z \ge 0\) are:

$$\frac{{\partial c_{A} }}{\partial t} + \frac{{\partial J_{A} }}{\partial z} = 0,$$
(16.4.1)
$$\frac{{\partial c_{B} }}{\partial t} + \frac{{\partial J_{B} }}{\partial z} = 0,$$
(16.4.2)

where J A and J B are the material fluxes, satisfying the following boundary and initial conditions for x A (and similar ones for x B ),

$$x_{A} \left( {z,t = 0} \right) = 0;\quad x_{A} \left( {z = 0,t} \right) = x_{A0} ;\quad x_{A} \left( {z \to \infty ,t} \right) = 0.$$
(16.4.3)

Here, we have assumed that at time t = 0, B fills the whole region, while at the interface z = 0 the A vapor is at local equilibrium with its liquid, so that \(x_{A0} = x_{A}^{sat} .\)

Now, adding Eqs. (16.4.1) and (16.4.2), and considering that c = c A  + c B is constant, we conclude that J A  + J B is independent of z and therefore it is a function of time alone. Therefore, since J B (z = 0, t) = 0, as B is insoluble into liquid A, we see that,

$$J_{A} \left( {z,t} \right) + J_{B} \left( {z,t} \right) = J_{A} \left( {0,t} \right) \equiv J_{A0} \left( t \right).$$
(16.4.4)

Obviously, J A0 decreases with time and is larger at shorter times (it is infinite at t = 0). In addition, since J A is zero as \(z \to \infty ,\) then \(J_{A0} \left( t \right) = J_{B} \left( {z \to \infty ,t} \right),\) indicating that, as expected, A pushes B along the tube axis, z.

The material flux J A is given by the constitutive relation,

$$J_{A} \left( {z,t} \right) = x_{A} \left( {J_{A} + J_{B} } \right) - cD\frac{{\partial x_{A} }}{\partial z} = x_{A} \left( {z,t} \right)J_{A0} \left( t \right) - cD\frac{{\partial x_{A} \left( {z,t} \right)}}{\partial z}.$$
(16.4.5)

At z = 0, we obtain:

$$J_{A0} \left( t \right) = - \frac{cD}{{1 - x_{A0} }}\left( {\frac{{\partial x_{A} }}{\partial z}} \right)_{z = o} .$$
(16.4.6)

At this point, substituting Eq. (16.4.6) into (16.4.5) and then applying Eq. (16.4.1), we obtain a partial differential equation (of second order in z and first order in t) to be solved with two boundary conditions and one initial condition [see Eqs. (16.4.3)].

In the dilute case, with \(x_{A0} \ll 1,\) we can neglect the convective contribution of the material transport [i.e., the first term on the RHS of Eq. (16.4.5)] and therefore, as we have seen in Sect. 7.4, we obtain the typical self-similar solution,

$$\frac{{x_{A} \left( {z,t} \right)}}{{x_{A0} }} = f\left( \eta \right) = erfc\left( \eta \right) = \left[ {1 - erf\left( \eta \right)} \right],\quad \text{with}\;\eta = \frac{z}{{\sqrt {4Dt} }},$$
(16.4.7)

where erf indicates the error function (7.4.9), represented in Fig. 7.6. Substituting this result into Eq. (16.4.6) we find the flux at the interface, \(J_{A0} \left( t \right) = cx_{A0} \sqrt {D/\pi t}\) [see Eq. (7.4.10)], that is,

$$\left( {J_{A0} } \right)_{diff} = cD\frac{{x_{A0} - x_{A\infty } }}{{\partial_{diff} }},\quad \text{where}\;\partial_{diff} = \sqrt {\pi Dt.}$$
(16.4.8)

Here, x A∞  = 0, while \(\delta_{diff}\) is the diffusion length, that is a typical distance covered by an A-molecule within a time t.

In the general, non-dilute case, we again look for a self-similar solution f(η), expressed in terms of the dimensionless variable η defined in Eq. (16.4.7), Then, the following equation is obtained:

$$\frac{{d^{2} f}}{{d\eta^{2} }} + 2\left( {\eta - \phi } \right)\frac{df}{d\eta } = 0,$$
(16.4.9)

where ϕ is a constant, still to be determined, due to the contribution of convection,

$$\phi = - \frac{1}{2}\frac{{x_{A0} }}{{1 - x_{A0} }}\frac{df}{d\eta }\left( 0 \right).$$
(16.4.10)

Naturally, when ϕ = 0, we find the error function solution of the purely diffusive, or dilute, case. As for the conditions to be imposed, following the discussion of Sect. 7.4, here we note again that the most stringent condition in finding self-similar solutions is that Eq. (16.4.9) can only satisfy 2 conditions, while our initial problem imposes the 3 conditions (16.4.3). In our case, however, these three relations collapse into the following two conditions,

$$f\left( 0 \right) = 1\quad \text{and}\quad f\left( \infty \right) = 0,$$
(16.4.11)

and so, from Eq. (16.4.9) we obtain:

$$f\left( \eta \right) = \frac{{1 - erf\left( {\eta - \phi } \right)}}{1 + erf\left( \phi \right)}.$$
(16.4.12)

In addition, substituting Eq. (16.4.12) into (16.4.10), we can determine ϕ implicitly as:

$$x_{A0} = \frac{2\phi }{{2\phi - f^{\prime}\left( 0 \right)}} = \left\{ {1 - \frac{{f^{\prime}\left( 0 \right)}}{2\phi }} \right\}^{ - 1} = \left\{ {1 + \left[ {\sqrt \pi \left( {1 + erf\left( \phi \right)} \right)\phi \exp \left( {\phi^{2} } \right)} \right]^{ - 1} } \right\}^{ - 1} ,$$
(16.4.13)

where we have considered that \(f^{\prime}\left( 0 \right) = - 2\exp \left( { - \phi^{2} } \right)/\sqrt \pi \left( {1 + erf\phi } \right).\) Physically, ϕ is a function of x A0 and represents a dimensionless molar velocity.

Now, let us calculate the material flux (16.4.6) of A at the interface,

$$J_{A0} = - \frac{{cDx_{A0} }}{{1 - x_{A0} }}\left( {\frac{\partial f}{\partial z}} \right)_{z = 0} = - \frac{{cDx_{A0} }}{{1 - x_{A0} }}\frac{1}{{\sqrt {4Dt} }}\frac{df}{d\eta }\left( 0 \right) = c\phi \sqrt {\frac{D}{t}.}$$
(16.4.14)

Finally, we can determine the Sherwood number, that is the ratio between the flux J A0 and its purely diffusive part, i.e.,

$$Sh = \frac{{J_{A0} }}{{\left( {J_{A0} } \right)_{diff} }} = \frac{{\sqrt {\pi \varphi } }}{{x_{A0} }}.$$
(16.4.15)

x A0

ϕ

Sh

0

0

1

0.25

0.156

1.11

0.50

0.358

1.27

0.75

0.662

1.56

1

∞

∞

As we see from the Table above, the convective contribution starts to be relevant only when the molar fraction of A at the interface is almost unity. Finally, the flux diverges at x A0 = 1, as we have already seen at the end of Sect. 15.1.

In analogy with Eq. (16.4.8), the result (16.4.14) can be written as:

$$J_{A0} = cD\frac{{x_{A0} - x_{A\infty } }}{{\delta_{diff} }},\quad \text{where}\;\delta_{diff} = \frac{{\sqrt {\pi Dt} }}{Sh}.$$
(16.4.16)

Here we see that in the presence of convection, with Sh > 1, the diffusion length \(\delta_{diff}\) decreases, as one would expect, since the concentration profile at the interface becomes steeper.

16.5 Problems

  1. 16.1

    Derive Eq. (16.2.1) by applying the integral Reynolds transport theorem (6.2.7).

  2. 16.2

    A little drop of water (species A) of radius R i is suspended at time t = 0 in an atmosphere of dry air (species B), that is, with molar fraction of water vapor, x ∞  = 0. As the drop evaporates, at the liquid-vapor interface the local equilibrium condition determines the molar fraction of water vapor, x 0. Using the QSS approximation, determine the evaporation time.

  3. 16.3

    Model the combustion of methane (A) immersed in an atmosphere of nitrogen and oxygen (B), according to the reaction 2O2 + CH4 → CO2 + 2H2. Assume that the reaction is very rapid, A and B are dilute and the problem can be schematized as being plane, with the following conditions: (a) at z < 0 there is only A; (b) initially, at z > 0, there is only B. Determine the expression of the methane flux.

  4. 16.4

    A spherical catalyst of radius R is immersed in a gas mixture containing a dilute species A, with concentration c A , with which it reacts instantaneously. Determine the flux of A at the surface of the catalyst.