Fundamentals of mass diffusion

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1 Errata

• Slide 11 has an incorrect species subscript in the mass-fraction denominator. Use

  𝑤𝐴=𝑥𝐴𝑀𝐴𝑥𝐴𝑀𝐴+𝑥𝐵𝑀𝐵$$w_A=\frac{x_AM_A}{x_AM_A+x_BM_B}$$

• Slide 13 retains species subscripts on the summed mixture fluxes. Use the total fluxes

  𝑛=∑𝑖𝜌𝑖𝑣𝑖,𝑁=∑𝑖𝑐𝑖𝑣𝑖$$n=\sum_i\rho_iv_i,\qquad N=\sum_i c_iv_i$$

  The molar-flux description should say molar flow rate per unit area.

• Slide 14 mislabels the second velocity heading. Use “Molar-average velocity” with

  𝑣∗=∑𝑖𝑐𝑖𝑣𝑖𝑐$$v^*=\frac{\sum_i c_iv_i}{c}$$

2 Reading list

• Bird et al. (2006), chapters 17 and 18.
• Geankoplis (2003), chapter 6.
• Welty et al. (2001), chapter 24.

Use consistent kmol units with 𝑅𝑢 =8314 Pa m3 kmol−1 K−1$R_u=8314\ \mathrm{Pa\,m^3\,kmol^{-1}\,K^{-1}}$.

3 Choose a steady-diffusion model

Given conditions	Model and checks
Equal and opposite molar fluxes	Set total molar flux to zero and use Eq. 8.
Carrier B has zero net flux	Retain bulk molar motion and use the stagnant-carrier logarithmic relation, Eq. 9.
Radial transfer through a cylinder or sphere	Write an area-weighted shell balance; conserve molar rate rather than radial flux.
A liquid boundary moves slowly	Use a pseudo-steady gas profile plus the interface inventory balance, Eq. 11.
Concentration changes through time inside a body	Move to the transient-diffusion topic; a steady film model alone is insufficient.

4 Fick’s law of diffusion

Molecular diffusion

Species transport relative to the bulk mixture caused by molecular motion and a thermodynamic driving force. In the simple Fickian models used here, the driving force is a concentration gradient.

When two gases with different compositions meet, random molecular motion mixes them even if there is no imposed flow. Across a plane in the gas, molecules of A arrive from both sides. More arrive from the side with the greater concentration of A, so their random motion produces a net transfer towards the lower concentration.

Fick’s law relates that net transfer to the concentration gradient. For a binary mixture at constant total molar concentration, the molar diffusive flux relative to the mixture’s molar-average velocity is

𝐽𝐴=−𝐷𝐴𝐵𝑑𝑐𝐴𝑑𝑧$$J_A=-D_{AB}\frac{dc_A}{dz}$$(1)

𝐽𝐴$J_A$ is in kmol m−2 s−1$\mathrm{kmol\,m^{-2}\,s^{-1}}$, 𝑐𝐴$c_A$ in kmol m−3$\mathrm{kmol\,m^{-3}}$, 𝑧$z$ in m and 𝐷𝐴𝐵$D_{AB}$ in m2 s−1$\mathrm{m^2\,s^{-1}}$. The gradient has units kmol m−4$\mathrm{kmol\,m^{-4}}$, so the dimensions close. The minus sign predicts transport down the gradient. The specified reference frame matters. A moving mixture can also transport A past the tube wall, so the flux measured by a fixed observer may differ from 𝐽𝐴$J_A$.

At constant total molar concentration, 𝑐𝐴$c_A$ decreases as 𝑧$z$ increases and 𝐷𝐴𝐵 >0$D_{AB}>0$. What is the direction of the molar diffusive flux 𝐽𝐴$J_A$?

• It is negative because the gradient is negative.
• It is zero unless the bulk fluid moves.
• It points in positive 𝑧$z$, down the concentration gradient. ✓
• It cannot be determined from the gradient.

Explanation

Fick’s law gives

𝐽𝐴=−𝐷𝐴𝐵𝑑𝑐𝐴𝑑𝑧$$J_A=-D_{AB}\frac{dc_A}{dz}$$

The concentration gradient is negative, so the diffusive flux is positive. A moves toward increasing position, from higher to lower concentration. Bulk motion can make the total flux different from this diffusive flux.

4.1 Examples: estimating diffusion times

Diffusion smooths a concentration difference over a distance 𝐿$L$ on the time scale

𝑡diff∼𝐿2𝐷𝐴𝐵$$t_{\mathrm{diff}}\sim\frac{L^2}{D_{AB}}$$

The estimate follows by comparing the accumulation term 𝑐𝐴/𝑡$c_A/t$ with the diffusion term 𝐷𝐴𝐵𝑐𝐴/𝐿2$D_{AB}c_A/L^2$ in the transient species balance. It estimates how quickly a profile changes, rather than the exact time to reach a specified concentration.

For 𝐿 =1 mm =10−3 m$L=1\ \mathrm{mm}=10^{-3}\ \mathrm m$, the estimate gives

Gas:𝑡diff∼(10−3)210−5=0.1 sLiquid:𝑡diff∼(10−3)210−9=1000 sSolid:𝑡diff∼(10−3)210−14=108 s$$\begin{aligned} \text{Gas:}\qquad t_{\mathrm{diff}}&\sim\frac{(10^{-3})^2}{10^{-5}}=0.1\ \mathrm s\\ \text{Liquid:}\qquad t_{\mathrm{diff}}&\sim\frac{(10^{-3})^2}{10^{-9}}=1000\ \mathrm s\\ \text{Solid:}\qquad t_{\mathrm{diff}}&\sim\frac{(10^{-3})^2}{10^{-14}}=10^8\ \mathrm s \end{aligned}$$

The values of 𝐷$D$ are illustrative. Diffusivity depends on the species pair, phase and temperature. Over the same distance, molecular diffusion can mix a gas in a fraction of a second while taking years in a solid. Doubling the distance multiplies the estimated time by four.

For modest dilute-gas extrapolations, kinetic theory gives

𝐷1𝐷2=𝑃2𝑃1(𝑇1𝑇2)3/2$$\frac{D_1}{D_2}=\frac{P_2}{P_1}\left(\frac{T_1}{T_2}\right)^{3/2}$$(2)

Use absolute temperatures and matching pressure units. Doubling temperature at fixed pressure predicts a factor 2.83 with this estimate, compared with 3.36 using a 𝑇1.75$T^{1.75}$ fit. Measured data take precedence.

For hydrogen in nickel, 𝐷1 =1.2 ×10−12$D_1=1.2\times10^{-12}$ at 𝑇1 =358 K$T_1=358\ \mathrm K$ and 𝐷2 =10−11 m2 s−1$D_2=10^{-11}\ \mathrm{m^{2}\,s^{-1}}$ at 𝑇2 =438 K$T_2=438\ \mathrm K$. Solid-state diffusion is often represented by the Arrhenius relation,

𝐷=𝐷0exp⁡(−𝐸𝑎𝑅𝑢𝑇)$$D=D_0\exp\left(-\frac{E_a}{R_uT}\right)$$

Taking logarithms at the two temperatures eliminates the unknown prefactor 𝐷0$D_0$. The activation energy is therefore

𝐸𝑎=𝑅𝑢ln⁡(𝐷2𝐷1)1𝑇1−1𝑇2=8.314ln⁡[10−111.2×10−12]1358−1438=3.46×104 J mol−1$$E_a=\frac{R_u\ln\left(\frac{D_2}{D_1}\right)}{\frac{1}{T_1}-\frac{1}{T_2}} =\frac{8.314\ln\left[\frac{10^{-11}}{1.2\times10^{-12}}\right]}{\frac{1}{358}-\frac{1}{438}} =3.46\times10^4\ \mathrm{J\,mol^{-1}}$$

Thus the two supplied values imply 𝐸𝑎 ≃35 kJ mol−1$E_a\simeq35\ \mathrm{kJ\,mol^{-1}}$. The gas constant here is on a molar basis, 𝑅𝑢 =8.314 J mol−1 K−1$R_u=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$.

For a porous medium using

𝐷eff=𝐷𝜖𝜏,$$D_{\mathrm{eff}}=\frac{D\epsilon}{\tau},$$

𝜖 =0.4$\epsilon=0.4$ and 𝜏 =3$\tau=3$ give 𝐷eff =0.133𝐷$D_{\mathrm{eff}}=0.133D$.

Diffusivity

Diffusivity, 𝐷𝐴𝐵$D_{AB}$, is the material property relating the molar diffusive flux of A to its concentration gradient in the binary Fickian model at constant total concentration, 𝐽𝐴 = −𝐷𝐴𝐵𝑑𝑐𝐴/𝑑𝑧$J_A=-D_{AB}dc_A/dz$. Its SI unit is m2 s−1$\mathrm{m^2\,s^{-1}}$. It depends on the species pair, temperature, pressure and composition.

A species has constant diffusivity 𝐷 =10−9 m2 s−1$D=10^{-9}\ \mathrm{m^2\,s^{-1}}$. What is the diffusion time scale across 1$1$ mm?

• 1000$1000$ s. ✓
• 1$1$ s.
• 10−3$10^{-3}$ s.
• 106$10^6$ s.

Explanation

First convert the diffusion distance:

𝐿=1 mm=10−3 m$$L=1\ \mathrm{mm}=10^{-3}\ \mathrm m$$

Then estimate the diffusion time:

𝑡diff∼𝐿2𝐷=(10−3)210−9=1000 s$$t_{\mathrm{diff}}\sim\frac{L^2}{D}=\frac{(10^{-3})^2}{10^{-9}}=1000\ \mathrm s$$

5 Units of concentration terms

Mass fraction

The mass of a species divided by the total mixture mass. It is dimensionless, and all species mass fractions sum to one.

Mole fraction

The molar amount of a species divided by the total molar amount of the mixture. It is dimensionless, and all species mole fractions sum to one.

Let 𝜌𝑖$\rho_i$ be species mass concentration in kg m−3$\mathrm{kg\,m^{-3}}$, 𝑐𝑖$c_i$ molar concentration in kmol m−3$\mathrm{kmol\,m^{-3}}$, 𝑀𝑖$M_i$ molecular weight in kg kmol⁻¹, and 𝑤𝑖$w_i$ the dimensionless mass fraction. The source also writes 𝜔𝑖$\omega_i$ for the same fraction. Use 𝑦𝑖$y_i$ for gas mole fraction and 𝑥𝑖$x_i$ for liquid mole fraction; a general mixture mole fraction below is denoted 𝜒𝑖$\chi_i$. The source shell balance uses 𝑥𝑖$x_i$ even for gas, which denotes the same quantity as our 𝑦𝑖$y_i$.

𝜌𝑖=𝑐𝑖𝑀𝑖$$\rho_i=c_iM_i$$

The total mass and molar concentrations are sums over the species,

𝜌=∑𝑖𝜌𝑖$$\rho=\sum_i\rho_i$$ 𝑐=∑𝑖𝑐𝑖$$c=\sum_i c_i$$

The mass fraction and mole fraction are the corresponding species-to-total ratios,

𝑤𝑖=𝜌𝑖𝜌$$w_i=\frac{\rho_i}{\rho}$$ 𝜒𝑖=𝑐𝑖𝑐$$\chi_i=\frac{c_i}{c}$$

Both sets of fractions sum to one,

∑𝑖𝑤𝑖=∑𝑖𝜒𝑖=1$$\sum_iw_i=\sum_i\chi_i=1$$(3)

For an ideal gas 𝑐𝑖 =𝑝𝑖/(𝑅𝑢𝑇)$c_i=p_i/(R_uT)$, 𝑐 =𝑃/(𝑅𝑢𝑇)$c=P/(R_uT)$ and 𝑦𝑖 =𝑝𝑖/𝑃$y_i=p_i/P$. Here 𝑝𝑖$p_i$ is partial pressure and 𝑃$P$ total pressure in Pa; 𝑅𝑢 =8314 Pa m3 kmol−1 K−1$R_u=8314\ \mathrm{Pa\,m^3\,kmol^{-1}\,K^{-1}}$. For a liquid, 𝑐 =𝜌/¯¯¯¯¯𝑀$c=\rho/\overline M$, where ¯¯¯¯¯𝑀 =∑𝑖𝜒𝑖𝑀𝑖$\overline M=\sum_i\chi_i M_i$ is the mixture molecular weight, not the solute’s molecular weight. Conversion between mass and mole fractions follows

𝜒𝐴=𝑤𝐴𝑀𝐴𝑤𝐴𝑀𝐴+𝑤𝐵𝑀𝐵𝑤𝐴=𝜒𝐴𝑀𝐴𝜒𝐴𝑀𝐴+𝜒𝐵𝑀𝐵$$\begin{gathered} \chi_A=\frac{\frac{w_A}{M_A}}{\frac{w_A}{M_A}+\frac{w_B}{M_B}}\\ w_A=\frac{\chi_A M_A}{\chi_A M_A+\chi_B M_B} \end{gathered}$$(4)

In a binary mixture 𝜒𝐵 =1 −𝜒𝐴$\chi_B=1-\chi_A$, so the gradients are opposite. At constant 𝑐$c$, this identity and 𝐽𝐴 +𝐽𝐵 =0$J_A+J_B=0$ require the same mutual diffusivity for the pair. They do not imply that laboratory-frame total molar flux is zero.

A binary mixture contains equal moles of A and B, but A has the greater molar mass. Which statement is true?

• Both mass fractions must be one half.
• A has the smaller mass fraction.
• Mole fractions cannot be found without density.
• Both mole fractions are one half, but A has the greater mass fraction. ✓

Explanation

Mole fraction counts moles, while mass fraction weights those moles by molar mass. Equal molar amounts therefore give equal mole fractions and unequal mass fractions when molar masses differ.

6 Component velocity in a mixture

A species velocity 𝑣𝑖$v_i$ in m s⁻¹ is a local average of many random molecular velocities, leaving a small drift. It is not a molecule’s thermal speed. A fixed observer measures mass flux 𝑛𝑖 =𝜌𝑖𝑣𝑖$n_i=\rho_iv_i$ in kg m−2 s−1$\mathrm{kg\,m^{-2}\,s^{-1}}$ or molar flux 𝑁𝑖 =𝑐𝑖𝑣𝑖$N_i=c_iv_i$ in kmol m−2 s−1$\mathrm{kmol\,m^{-2}\,s^{-1}}$. Their sums 𝑛 =∑𝑖𝑛𝑖$n=\sum_i n_i$ and 𝑁 =∑𝑖𝑁𝑖$N=\sum_iN_i$ describe the mixture.

The two mixture averages weight different quantities:

𝑣=∑𝑖𝜌𝑖𝑣𝑖𝜌$$v=\frac{\sum_i\rho_i v_i}{\rho}$$

𝑣∗=∑𝑖𝑐𝑖𝑣𝑖𝑐$$v^*=\frac{\sum_i c_i v_i}{c}$$(5)

𝑣$v$ is the mass-average velocity used in overall mass continuity; 𝑣∗$v^*$ is the molar-average velocity natural for molar diffusion. In an equal-mole hydrogen/nitrogen mixture with 𝑣𝐵 = −𝑣𝐴$v_B=-v_A$ and molecular weights 2 and 28, 𝑣∗ =0$v^*=0$ but

𝑣=(2−28)𝑣𝐴30=−0.867𝑣𝐴.$$v=\frac{(2-28)v_A}{30}=-0.867v_A.$$

These averages coincide when molecular weights are equal.

A fixed observer sees 𝑣𝑖$v_i$, a mass-average observer sees 𝑣𝑖 −𝑣$v_i-v$, and a molar-average observer sees 𝑣𝑖 −𝑣∗$v_i-v^*$. A diffusion velocity measures the species drift relative to the chosen mixture average. It can be positive, negative or zero. Weighted sums of diffusion velocities vanish by construction.

7 Bulk flux and diffusion flux

Diffusive flux

Species transport rate per unit area relative to a specified mixture-average velocity. Its definition must identify both the mass or molar basis and the reference frame.

Total molar flux

Molar transport rate of a species per unit area past a fixed observer. It includes transport by bulk motion and diffusion, and has units such as mol m−2 s−1$\mathrm{mol\,m^{-2}\,s^{-1}}$.

By Eq. 5, 𝑐𝑣∗ =𝑁𝐴 +𝑁𝐵$cv^*=N_A+N_B$. Expanding 𝐽𝐴 =𝑐𝐴(𝑣𝐴 −𝑣∗)$J_A=c_A(v_A-v^*)$ gives

𝐽𝐴=𝑁𝐴−𝜒𝐴(𝑁𝐴+𝑁𝐵),$$J_A=N_A-\chi_A(N_A+N_B),$$

so

𝑁𝐴=𝜒𝐴(𝑁𝐴+𝑁𝐵)+𝐽𝐴$$N_A=\chi_A(N_A+N_B)+J_A$$(6)

Figure 1 shows how transport by bulk motion and diffusion add to give the total molar flux in Eq. 6. For stagnant B, the bulk contribution is the fraction 𝜒𝐴$\chi_A$ of 𝑁𝐴$N_A$. For arbitrary 𝑁𝐵$N_B$, use the full sum 𝑁𝐴 +𝑁𝐵$N_A+N_B$ to determine the bulk contribution. At dilute A the stagnant-carrier correction is small because A makes up only a small fraction of the moving mixture. At higher concentrations, it must be included.

Figure 1: Total molar flux is the sum of transport by bulk motion and diffusion, shown here in the same direction. Dashed guides align the component arrows with the total.

In a binary mixture, the molar-frame diffusive fluxes satisfy 𝐽𝐴 +𝐽𝐵 =0$J_A+J_B=0$. What follows about the total molar flux 𝑁𝐴 +𝑁𝐵$N_A+N_B$ past a fixed observer?

• It must always be zero.
• It can be nonzero because bulk motion carries moles past the observer. ✓
• It must equal the total mass flux numerically.
• It must be negative.

Explanation

The diffusive fluxes sum to zero in the molar-average frame. Past a fixed observer, bulk motion also transports material:

𝑁𝐴+𝑁𝐵=𝑐𝑣∗$$N_A+N_B=cv^*$$

The total molar flux can therefore be nonzero. It is zero only if the molar-average velocity is zero, as in equimolar counter-diffusion.

8 Diffusion flux

Substituting Eq. 1 into Eq. 6 gives

𝑁𝐴=𝑐𝐴𝑐(𝑁𝐴+𝑁𝐵)−𝐷𝐴𝐵𝑑𝑐𝐴𝑑𝑧𝑁𝐴=𝜒𝐴(𝑁𝐴+𝑁𝐵)−𝑐𝐷𝐴𝐵𝑑𝜒𝐴𝑑𝑧$$\begin{gathered} N_A=\frac{c_A}{c}(N_A+N_B)-D_{AB}\frac{dc_A}{dz}\\ N_A=\chi_A(N_A+N_B)-cD_{AB}\frac{d\chi_A}{dz} \end{gathered}$$(7)

The product derivative

𝑑𝑐𝐴𝑑𝑧=𝑐 𝑑𝜒𝐴𝑑𝑧+𝜒𝐴 𝑑𝑐𝑑𝑧$$\frac{dc_A}{dz}=c\,\frac{d\chi_A}{dz}+\chi_A\,\frac{dc}{dz}$$

explains the constant-𝑐$c$ restriction. In an isothermal isobaric ideal gas it is exact. A liquid may need a composition-dependent model or a justified average concentration. Positive flux means motion in positive 𝑧$z$; a negative result reports the direction rather than an arithmetic failure.

9 Diffusion flux: two limiting cases

Equimolar counter-diffusion imposes 𝑁𝐴 +𝑁𝐵 =0$N_A+N_B=0$. Burning solid carbon to only carbon dioxide consumes one oxygen molecule for one carbon dioxide molecule produced, so this is a useful gas-film model. Formation of CO changes the ratio and invalidates the equimolar assumption. Binary distillation with equal molar latent heats is another approximate example.

A crossing stagnant B imposes 𝑁𝐵 =0$N_B=0$. Water absorbed into concentrated acid crosses the interface while insoluble carrier air does not. Air can still diffuse towards the interface: bulk motion cancels its diffusion exactly. This is zero net flux, not zero microscopic motion. Other stoichiometric ratios or imposed flows must be inserted into Eq. 7 explicitly.

10 Equimolar counter-diffusion of gases A and B

Equimolar counter-diffusion

Binary transport with equal and opposite total molar fluxes, 𝑁𝐴 = −𝑁𝐵$N_A=-N_B$. The net mixture molar flux is zero.

Take steady, one-dimensional, nonreacting transfer across 0 ≤𝑧 ≤𝐿$0\le z\le L$ with constant 𝐷$D$, 𝑇$T$ and 𝑃$P$. Conservation makes 𝑁𝐴$N_A$ constant. Since 𝑐𝐴 =𝑝𝐴/(𝑅𝑢𝑇)$c_A=p_A/(R_uT)$,

𝑁𝐴=−𝐷𝐴𝐵𝑅𝑢𝑇𝑑𝑝𝐴𝑑𝑧𝑁𝐴𝐿=−𝐷𝐴𝐵𝑅𝑢𝑇∫𝑝𝐴2𝑝𝐴1𝑑𝑝𝐴𝑁𝐴=𝐷𝐴𝐵(𝑝𝐴1−𝑝𝐴2)𝑅𝑢𝑇𝐿$$\begin{gathered} N_A=-\frac{D_{AB}}{R_uT}\frac{dp_A}{dz}\\ N_AL=-\frac{D_{AB}}{R_uT}\int_{p_{A1}}^{p_{A2}}dp_A\\ N_A=\frac{D_{AB}(p_{A1}-p_{A2})}{R_uTL} \end{gathered}$$(8)

Partial pressures and concentrations are linear, with 𝑝𝐴 +𝑝𝐵 =𝑃$p_A+p_B=P$. The pressure-based resistance is 𝑅𝑢𝑇𝐿/𝐷$R_uTL/D$, with units Pa m2 s kmol−1$\mathrm{Pa\,m^2\,s\,kmol^{-1}}$, while concentration-based resistance 𝐿/𝐷$L/D$ has units s m⁻¹. Use the same driving-force basis when adding resistances in series.

11 A diffusing through stagnant, non-diffusing B

Stagnant carrier

A species with zero total flux past the chosen fixed observer, such as 𝑁𝐵 =0$N_B=0$. Its diffusive and bulk contributions can still be nonzero and cancel each other.

Now set 𝑁𝐵 =0$N_B=0$ in Eq. 7. Rearranging and separating gives

𝑁𝐴(1−𝑦𝐴)=−𝑐𝐷𝐴𝐵𝑑𝑦𝐴𝑑𝑧𝑁𝐴∫𝐿0𝑑𝑧=−𝑐𝐷𝐴𝐵∫𝑦𝐴2𝑦𝐴1𝑑𝑦𝐴1−𝑦𝐴𝑁𝐴=𝑐𝐷𝐴𝐵𝐿ln⁡1−𝑦𝐴21−𝑦𝐴1$$\begin{gathered} N_A(1-y_A)=-cD_{AB}\frac{dy_A}{dz}\\ N_A\int_0^L dz=-cD_{AB}\int_{y_{A1}}^{y_{A2}}\frac{dy_A}{1-y_A}\\ N_A=\frac{cD_{AB}}{L}\ln\frac{1-y_{A2}}{1-y_{A1}} \end{gathered}$$(9)

Equivalently

𝑁𝐴=𝐷𝐴𝐵𝑃𝑅𝑢𝑇𝐿ln⁡(𝑃−𝑝𝐴2𝑃−𝑝𝐴1).$$N_A=\frac{D_{AB}P}{R_uTL}\ln\left(\frac{P-p_{A2}}{P-p_{A1}}\right).$$

When 𝑦𝐴1 >𝑦𝐴2$y_{A1}>y_{A2}$, the logarithm is positive. The local multiplier 1/(1 −𝑦𝐴)$1/(1-y_A)$ is 2 at 𝑦𝐴 =0.5$y_A=0.5$ and 10 at 0.9. At low concentration the logarithm approaches 𝑦𝐴1 −𝑦𝐴2$y_{A1}-y_{A2}$ and Eq. 8 returns.

Define the log-mean stagnant-species partial pressure and mole fraction:

𝑝𝐵𝑀=𝑝𝐵2−𝑝𝐵1ln⁡(𝑝𝐵2𝑝𝐵1),𝑦𝐵𝑀=𝑝𝐵𝑀𝑃𝑁𝐴=𝐷𝐴𝐵(𝑝𝐴1−𝑝𝐴2)𝑅𝑢𝑇𝐿𝑃𝑝𝐵𝑀𝑁𝐴=𝑐𝐷𝐴𝐵(𝑦𝐴1−𝑦𝐴2)𝐿𝑦𝐵𝑀$$\begin{gathered} p_{BM}=\frac{p_{B2}-p_{B1}}{\ln\left(\frac{p_{B2}}{p_{B1}}\right)},\quad y_{BM}=\frac{p_{BM}}{P}\\ N_A=\frac{D_{AB}(p_{A1}-p_{A2})}{R_uTL}\frac{P}{p_{BM}}\\ N_A=\frac{cD_{AB}(y_{A1}-y_{A2})}{Ly_{BM}} \end{gathered}$$(10)

Stefan-flow enhancement factor

For A diffusing through stagnant B in the stated binary gas film,

𝐹𝑆=1𝑦𝐵𝑀=𝑃𝑝𝐵𝑀$$F_S=\frac{1}{y_{BM}}=\frac{P}{p_{BM}}$$

multiplies the equimolar-diffusion flux based on the same endpoint concentration difference. It accounts for bulk molar motion. The local multiplier 1/(1 −𝑦𝐴)$1/(1-y_A)$ is distinct from this integrated factor.

If the endpoints coincide, take the common value as the log mean. The exact enhancement is 𝑃/𝑝𝐵𝑀$P/p_{BM}$, not an arbitrarily chosen local fraction. The logarithm follows from integrating 1/(1 −𝑦𝐴)$1/(1-y_A)$ and the profile is curved. Its gradient magnitude increases towards the low-A end because

∣𝑑𝑦𝐴𝑑𝑧∣=𝑁𝐴(1−𝑦𝐴)𝑐𝐷.$$\left|\frac{dy_A}{dz}\right|=\frac{N_A(1-y_A)}{cD}.$$

The smaller fraction of A at that end means that a steeper gradient is needed to maintain the same total flux.

As A becomes dilute, 𝐹𝑆$F_S$ approaches one; when B’s log-mean fraction is 0.5, 𝐹𝑆 =2$F_S=2$. This correction needs the stagnant-B model. A change in reaction stoichiometry or imposed bulk flow changes the multiplier.

A diffuses through stagnant B with 𝑁𝐵 =0$N_B=0$. What does stagnant mean here?

• B molecules have no random motion.
• The mixture molar-average velocity must be zero.
• B has zero total molar flux, while its bulk and diffusive contributions can cancel. ✓
• The concentration gradient of B must be zero.

Explanation

Stagnant B means that no net amount of B crosses a fixed surface. B can still diffuse in one direction while bulk motion carries it in the opposite direction. These contributions cancel. It does not mean that B molecules stop moving or that the mixture has zero bulk velocity.

11.1 Pseudo-steady-state analysis with a moving boundary

Let 𝐿(𝑡)$L(t)$ be the distance from a falling pure-liquid surface to a fixed swept tube mouth. The surface equilibrium composition and the mouth composition remain constant. The gas adjusts over 𝐿2/𝐷$L^2/D$ while the liquid retreats much more slowly, so approximate the gas profile as steady at each instantaneous length. Liquid molar density is 𝑐𝐿 =𝜌𝐿/𝑀𝐴$c_L=\rho_L/M_A$, with 𝜌𝐿$\rho_L$ a mass density, not a molar density.

𝑐𝐿𝑑𝐿𝑑𝑡=𝑐𝐷𝐴𝐵Λ𝐿,Λ=ln⁡1−𝑦𝐴21−𝑦𝐴1∫𝑡0𝑑𝑡=𝑐𝐿𝑐𝐷𝐴𝐵Λ∫𝐿𝑓𝐿0𝐿 𝑑𝐿𝑡=𝑐𝐿(𝐿2𝑓−𝐿20)2𝑐𝐷𝐴𝐵Λ$$\begin{gathered} c_L\frac{dL}{dt}=\frac{cD_{AB}\Lambda}{L},\quad \Lambda=\ln\frac{1-y_{A2}}{1-y_{A1}}\\ \int_0^t dt=\frac{c_L}{cD_{AB}\Lambda}\int_{L_0}^{L_f}L\,dL\\ t=\frac{c_L(L_f^2-L_0^2)}{2cD_{AB}\Lambda} \end{gathered}$$(11)

Since

Λ=𝑦𝐴1−𝑦𝐴2𝑦𝐵𝑀,$$\Lambda=\frac{y_{A1}-y_{A2}}{y_{BM}},$$

this also reproduces the source’s log-mean form. Only 𝐿$L$ changes during integration. Plotting 𝐿2$L^2$ against time is the Stefan-tube method of estimating diffusivity. Its slope is 2𝑐𝐷Λ/𝑐𝐿$2cD\Lambda/c_L$. Maintaining temperature, a dry mouth and negligible convection is necessary.

Figure 2: A Stefan tube with a liquid surface below a dry swept mouth. Vapour moves upwards through air with zero net air flux.

12 Classroom example 1: water evaporation

Water evaporation

Water in the bottom of a narrow metal tube is held at a constant temperature of 293 K. The total pressure of air (assumed dry) is 1.01325 ×105 Pa$1.01325\times10^5\ \mathrm{Pa}$ and the temperature is 293 K. Water evaporates and diffuses through the air in the tube, and the diffusion path 𝑧2 −𝑧1$z_2-z_1$ is 0.1524 m. Calculate the rate of evaporation at steady state in kgmol/(s ⋅m2)$\mathrm{kgmol/(s\cdot m^2)}$. The diffusivity of water vapour at 293 K and 1 atm pressure is 0.250 ×10−4 m2/s$0.250\times10^{-4}\ \mathrm{m^2/s}$. Assume that the system is isothermal. The vapour pressure of water at 1 atm and 293 K is 17.535 mmHg.

Additional note: Moving boundaries or pseudo-steady state diffusion. How long would it take for the water level to decrease by 0.1 m?

Solution

Figure 3: Water vapour diffuses through the gas above the liquid to the tube mouth. As the liquid surface falls, the gas diffusion distance increases.

The evaporation path is the gas space between the water surface and the tube mouth, as shown in Figure 2. Water vapour moves upwards. Air has zero net flux across the liquid interface.

Data

Quantity	Value
Initial gas diffusion length	𝐿0 =0.1524 m$L_0=0.1524\ \mathrm m$
Temperature	𝑇 =293 K$T=293\ \mathrm K$
Pressure	𝑃 =101325 Pa$P=101325\ \mathrm{Pa}$
Diffusivity	𝐷𝐴𝐵 =2.50 ×10−5 m2 s−1$D_{AB}=2.50\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Surface vapour pressure	𝑝𝐴,𝑠 =17.535 mmHg$p_{A,s}=17.535\ \mathrm{mmHg}$
Liquid density for the moving-boundary calculation	𝜌𝐿 =1000 kg m−3$\rho_L=1000\ \mathrm{kg\,m^{-3}}$
Water molar mass	𝑀𝐴 =18 kg kmol−1$M_A=18\ \mathrm{kg\,kmol^{-1}}$
Final gas diffusion length	𝐿𝑓 =0.1524 +0.1000 =0.2524 m$L_f=0.1524+0.1000=0.2524\ \mathrm m$

The source unit kgmol is equivalent to kmol. Liquid density and molar mass are supplementary inputs needed to convert liquid-depth loss to an amount evaporated.

Assumptions

• Unidirectional diffusion through the gas space.
• Steady gas profile for the initial flux calculation.
• Pseudo-steady gas profile as the liquid surface recedes.
• Ideal gas behaviour.
• Constant temperature, pressure and diffusivity.
• Vapour-liquid equilibrium at the water surface.
• Zero net air flux through the liquid surface.
• Negligible natural convection.

Calculations

At the liquid surface, equilibrium fixes the vapour partial pressure. At the mouth, the dry air gives 𝑝𝐴2 =0$p_{A2}=0$. Converting the surface vapour pressure to Pa gives

𝑝𝐴1=17.535760(101325)=2337.8 Pa$$p_{A1}=\frac{17.535}{760}(101325)=2337.8\ \mathrm{Pa}$$

The surface mole fraction and total gas concentration are

𝑦𝐴1=𝑝𝐴1𝑃=2337.8101325=0.0230724$$y_{A1}=\frac{p_{A1}}P=\frac{2337.8}{101325}=0.0230724$$𝑐=𝑃𝑅𝑢𝑇=1013258314(293)=0.0415948 kmol m−3$$c=\frac{P}{R_uT}=\frac{101325}{8314(293)}=0.0415948\ \mathrm{kmol\,m^{-3}}$$

Because air has zero net flux, the appropriate governing relation is the stagnant-carrier result, Eq. 9,

𝑁𝐴=𝑐𝐷𝐴𝐵𝐿0ln⁡1−𝑦𝐴21−𝑦𝐴1$$N_A=\frac{cD_{AB}}{L_0}\ln\frac{1-y_{A2}}{1-y_{A1}}$$

Substituting the boundary fractions and the initial gas length gives

𝑁𝐴=0.0415948(2.50×10−5)0.1524ln⁡11−0.0230724=1.593×10−7 kmol m−2 s−1$$N_A=\frac{0.0415948(2.50\times10^{-5})}{0.1524} \ln\frac{1}{1-0.0230724} =1.593\times10^{-7}\ \mathrm{kmol\,m^{-2}\,s^{-1}}$$(12)

Multiplying by molar mass converts the molar flux to mass flux,

𝑛𝐴=𝑀𝐴𝑁𝐴=18(1.593×10−7)=2.867×10−6 kg m−2 s−1$$n_A=M_AN_A=18(1.593\times10^{-7})=2.867\times10^{-6}\ \mathrm{kg\,m^{-2}\,s^{-1}}$$

The initial rate is about 0.248 kg m−2$0.248\ \mathrm{kg\,m^{-2}}$ per day. The stagnant-carrier flux is only 1.17 percent above the equimolar estimate because the water vapour is dilute.

For the falling liquid level, the gas length increases and the evaporation flux decreases. Define the constant logarithmic driving force,

Λ=ln⁡1−𝑦𝐴21−𝑦𝐴1=ln⁡11−0.0230724=0.0233427$$\Lambda=\ln\frac{1-y_{A2}}{1-y_{A1}}=\ln\frac{1}{1-0.0230724}=0.0233427$$

The liquid molar density is

𝑐𝐿=𝜌𝐿𝑀𝐴=100018=55.5556 kmol m−3$$c_L=\frac{\rho_L}{M_A}=\frac{1000}{18}=55.5556\ \mathrm{kmol\,m^{-3}}$$

A decrease in liquid depth 𝑑𝐿$dL$ removes 𝑐𝐿𝑑𝐿$c_LdL$ kmol per unit tube area. Equating that loss to the instantaneous evaporation flux gives

𝑐𝐿𝑑𝐿𝑑𝑡=𝑐𝐷𝐴𝐵Λ𝐿$$c_L\frac{dL}{dt}=\frac{cD_{AB}\Lambda}{L}$$

Separating variables and integrating between the initial and final levels gives the moving-boundary relation, Eq. 11,

𝑡=𝑐𝐿(𝐿2𝑓−𝐿20)2𝑐𝐷𝐴𝐵Λ$$t=\frac{c_L(L_f^2-L_0^2)}{2cD_{AB}\Lambda}$$

Substitute the consistent kmol-based concentrations,

𝑡=55.5556(0.25242−0.15242)2(0.0415948)(2.50×10−5)(0.0233427)=4.632×107 s=536 days$$t=\frac{55.5556(0.2524^2-0.1524^2)} {2(0.0415948)(2.50\times10^{-5})(0.0233427)} =4.632\times10^7\ \mathrm s=536\ \mathrm{days}$$(13)

As a check, removing 0.1 m$0.1\ \mathrm m$ of water requires 5.556 kmol m−2$5.556\ \mathrm{kmol\,m^{-2}}$. Even if the initial flux stayed constant, evaporation would take

5.5561.593×10−7=3.49×107 s.$$\frac{5.556}{1.593\times10^{-7}}=3.49\times10^7\ \mathrm s.$$

The falling level lengthens the diffusion path, so the calculated time must be greater than that lower bound.

The gas diffusion time is 𝐿20/𝐷𝐴𝐵 ≃929 s$L_0^2/D_{AB}\simeq929\ \mathrm s$, much shorter than the liquid retreat time. This supports the pseudo-steady approximation. Maintaining the stipulated isothermal, convection-free conditions for 536 days would be difficult in an experiment.

Answer

The initial evaporation flux is 𝑁𝐴 =1.59 ×10−7 kgmol m−2 s−1$N_A=1.59\times10^{-7}\ \mathrm{kgmol\,m^{-2}\,s^{-1}}$, directed towards the tube mouth. The corresponding mass flux is 2.87 ×10−6 kg m−2 s−1$2.87\times10^{-6}\ \mathrm{kg\,m^{-2}\,s^{-1}}$.

Under the pseudo-steady moving-boundary model, the water level takes 4.63 ×107 s$4.63\times10^7\ \mathrm s$, or about 536 days, to fall by 0.1 m$0.1\ \mathrm m$.

13 Classroom example 2: butanol in water

Butanol in water

Calculate the rate of diffusion of butanol at 20 °C$20\ \degree\mathrm{C}$ under unidirectional steady state conditions through a 0.1 cm thick film of water when the concentrations of butanol at the opposite sides of the film are, 10% and 4% butanol by weight, respectively.

𝐷𝐴𝐵$D_{AB}$ (butanol in water solution)=5.9 ×10−6 cm2/s$5.9\times10^{-6}\ \mathrm{cm^2/s}$. The densities of 10% and 4% butanol solutions at 20 °C$20\ \degree\mathrm{C}$ may be taken as 0.971 and 0.992 g/cc, respectively. Molecular weight of Butanol is 74, and that of water is 18.

Solution

Data

Quantity	Value
Film thickness	𝐿 =0.1 cm$L=0.1\ \mathrm{cm}$
Diffusivity	𝐷𝐴𝐵 =5.9 ×10−6 cm2 s−1$D_{AB}=5.9\times10^{-6}\ \mathrm{cm^2\,s^{-1}}$
Butanol mass fraction at side 1	𝑤𝐴1 =0.10$w_{A1}=0.10$
Density at side 1	𝜌1 =0.971 g cm−3$\rho_1=0.971\ \mathrm{g\,cm^{-3}}$
Butanol mass fraction at side 2	𝑤𝐴2 =0.04$w_{A2}=0.04$
Density at side 2	𝜌2 =0.992 g cm−3$\rho_2=0.992\ \mathrm{g\,cm^{-3}}$
Butanol molar mass	𝑀𝐴 =74 g mol−1$M_A=74\ \mathrm{g\,mol^{-1}}$
Water molar mass	𝑀𝐵 =18 g mol−1$M_B=18\ \mathrm{g\,mol^{-1}}$

Assumptions

• Steady state.
• Unidirectional diffusion across the film.
• Isothermal conditions at 20 °C$20\ \degree\mathrm{C}$.
• No chemical reaction.
• Zero net water flux.
• Constant diffusivity.
• Total molar concentration approximated by its arithmetic mean across the film.

Calculations

Let A be butanol and B be water, with positive flux from the 10 percent side to the 4 percent side. The stagnant-water model uses mole fractions, so first convert the supplied mass fractions. On a 100 g$100\ \mathrm g$ basis, side 1 contains 10/74$10/74$ mol butanol and 90/18$90/18$ mol water. Hence

𝑥𝐴1=10741074+9018=0.0263158$$x_{A1}=\frac{\frac{10}{74}}{\frac{10}{74}+\frac{90}{18}}=0.0263158$$

For side 2, the same conversion gives

𝑥𝐴2=474474+9618=0.0100334$$x_{A2}=\frac{\frac{4}{74}}{\frac{4}{74}+\frac{96}{18}}=0.0100334$$

The total molar concentration equals the total moles divided by the solution volume. Since a 100 g$100\ \mathrm g$ sample occupies 100/𝜌𝑖 cm3$100/\rho_i\ \mathrm{cm^3}$,

𝑐1=1074+90181000.971=0.0498622 mol cm−3$$c_1=\frac{\frac{10}{74}+\frac{90}{18}}{\frac{100}{0.971}}=0.0498622\ \mathrm{mol\,cm^{-3}}$$𝑐2=474+96181000.992=0.0534429 mol cm−3$$c_2=\frac{\frac{4}{74}+\frac{96}{18}}{\frac{100}{0.992}}=0.0534429\ \mathrm{mol\,cm^{-3}}$$

Their arithmetic mean is

¯¯¯¯𝑐=𝑐1+𝑐22=0.0498622+0.05344292=0.0516525 mol cm−3$$\overline c=\frac{c_1+c_2}{2}=\frac{0.0498622+0.0534429}{2}=0.0516525\ \mathrm{mol\,cm^{-3}}$$(14)

The total molar concentration changes by about 7 percent across the film. Treating it as uniform is an approximation for this liquid mixture. Constant temperature and pressure alone do not make a liquid’s molar concentration constant.

With zero net water flux, the governing total-flux relation is

𝑁𝐴=−¯¯¯¯𝑐𝐷𝐴𝐵1−𝑥𝐴𝑑𝑥𝐴𝑑𝑧$$N_A=-\frac{\overline cD_{AB}}{1-x_A}\frac{dx_A}{dz}$$

Integrating over the film gives the liquid version of Eq. 9,

𝑁𝐴=¯¯¯¯𝑐𝐷𝐴𝐵𝐿ln⁡1−𝑥𝐴21−𝑥𝐴1$$N_A=\frac{\overline cD_{AB}}{L}\ln\frac{1-x_{A2}}{1-x_{A1}}$$

Substituting in centimetre and mol units gives

𝑁𝐴=0.0516525(5.9×10−6)0.1ln⁡1−0.01003341−0.0263158=5.054×10−8 mol cm−2 s−1=5.054×10−7 kmol m−2 s−1$$\begin{aligned} N_A&=\frac{0.0516525(5.9\times10^{-6})}{0.1} \ln\frac{1-0.0100334}{1-0.0263158}\\ &=5.054\times10^{-8}\ \mathrm{mol\,cm^{-2}\,s^{-1}}\\ &=5.054\times10^{-7}\ \mathrm{kmol\,m^{-2}\,s^{-1}} \end{aligned}$$(15)

The mass flux follows by multiplying by the molar mass of butanol,

𝑛𝐴=𝑀𝐴𝑁𝐴=74(5.054×10−8)=3.740×10−6 g cm−2 s−1$$n_A=M_AN_A=74(5.054\times10^{-8})=3.740\times10^{-6}\ \mathrm{g\,cm^{-2}\,s^{-1}}$$

As a check, neglecting the stagnant-water correction gives

𝑁𝐴,eq=¯¯¯¯𝑐𝐷𝐴𝐵(𝑥𝐴1−𝑥𝐴2)𝐿=4.962×10−8 mol cm−2 s−1.$$N_{A,\mathrm{eq}}=\frac{\overline cD_{AB}(x_{A1}-x_{A2})}{L}=4.962\times10^{-8}\ \mathrm{mol\,cm^{-2}\,s^{-1}}.$$

The full result is 1.85 percent larger, consistent with dilute butanol.

The flux is about 3.23 kg m−2$3.23\ \mathrm{kg\,m^{-2}}$ per day. Comparing it with the water-evaporation example requires the concentration difference and film thickness as well as the diffusivity.

Answer

Butanol diffuses from the 10 percent side to the 4 percent side at 𝑁𝐴 =5.05 ×10−7 kmol m−2 s−1$N_A=5.05\times10^{-7}\ \mathrm{kmol\,m^{-2}\,s^{-1}}$, equivalent to a mass flux of 3.74 ×10−5 kg m−2 s−1$3.74\times10^{-5}\ \mathrm{kg\,m^{-2}\,s^{-1}}$.

14 Steady-state mass transport: shell balance approach

14.1 Species balance and homogeneous reaction

At steady state, species inflow minus outflow plus generation is zero. Total 𝑁𝐴$N_A$ already includes convection and diffusion. For a plane shell of area 𝐴𝑠 =Δ𝑥Δ𝑦$A_s=\Delta x\Delta y$ and thickness Δ𝑧$\Delta z$,

𝐴𝑠[𝑁𝐴(𝑧)−𝑁𝐴(𝑧+Δ𝑧)]+𝐴𝑠Δ𝑧𝑅𝐴=0𝑑𝑁𝐴𝑑𝑧=𝑅𝐴$$\begin{gathered} A_s\left[N_A(z)-N_A(z+\Delta z)\right]+A_s\Delta z R_A=0\\ \frac{dN_A}{dz}=R_A \end{gathered}$$(16)

𝑅𝐴$R_A$ is net production in kmol m−3 s−1$\mathrm{kmol\,m^{-3}\,s^{-1}}$, negative for consumption. A homogeneous 𝑛$n$th-order rate may be written 𝑅𝐴 = ±𝑘′′′𝑛𝑐𝑛𝐴$R_A=\pm k_n'''c_A^n$, with 𝑘′′′𝑛$k_n'''$ units (kmol m−3)1−n s−1$\mathrm{(kmol\,m^{-3})^{1-n}\,s^{-1}}$. Multiplying by volume produces kmol s⁻¹, while multiplying 𝑁𝐴$N_A$ by face area produces the same units. A heterogeneous reaction acts at a boundary, so it sets a surface-flux condition instead of a volumetric source. A porous catalyst can distribute internal reacting area throughout its modelled volume, requiring an effective volumetric rate and diffusivity.

14.2 Setting interface conditions

Use the equilibrium relation appropriate to the example, such as Raoult’s law 𝑝𝐴 =𝑥𝐴,𝐿𝑝sat𝐴$p_A=x_{A,L}p_A^{\mathrm{sat}}$ or Henry’s law 𝑝𝐴 =𝐻𝑥𝐴,𝐿$p_A=H x_{A,L}$, with 𝐻$H$ in Pa for this convention. An impermeable face has 𝑁𝐴 =0$N_A=0$; a reacting face has a flux condition from its stoichiometry.

15 Case study 1: diffusion through stagnant gas film

15.1 Balance and constitutive law

Consider the same straight tube at an instant, with interface 𝑧1$z_1$ and mouth 𝑧2$z_2$. B cannot cross the liquid surface; steadiness without reaction then requires 𝑁𝐵 =0$N_B=0$ throughout. A’s balance is Eq. 16 with 𝑅𝐴 =0$R_A=0$, so 𝑑𝑁𝐴/𝑑𝑧 =0$dN_A/dz=0$. Every mole leaving the liquid reaches the mouth. Substitute Eq. 7:

𝑁𝐴=−𝑐𝐷𝐴𝐵1−𝑦𝐴𝑑𝑦𝐴𝑑𝑧𝑑𝑑𝑧[𝑐𝐷𝐴𝐵1−𝑦𝐴𝑑𝑦𝐴𝑑𝑧]=0$$\begin{gathered} N_A=-\frac{cD_{AB}}{1-y_A}\frac{dy_A}{dz}\\ \frac{d}{dz}\left[\frac{cD_{AB}}{1-y_A}\frac{dy_A}{dz}\right]=0 \end{gathered}$$(17)

The isothermal isobaric gas has constant 𝑐$c$, and constant diffusivity is assumed. Remove these constants before integrating. If properties vary strongly they must stay inside the derivative.

15.2 Two integrations and boundary conditions

Integrating Eq. 17 once and then again gives

11−𝑦𝐴𝑑𝑦𝐴𝑑𝑧=𝐶1−ln⁡(1−𝑦𝐴)=𝐶1𝑧+𝐶2$$\begin{gathered} \frac{1}{1-y_A}\frac{dy_A}{dz}=C_1\\ -\ln(1-y_A)=C_1z+C_2 \end{gathered}$$(18)

𝐶1$C_1$ has units m−1$\mathrm{m^{-1}}$ and 𝐶2$C_2$ is dimensionless. At 𝑧1$z_1$ and 𝑧2$z_2$, impose 𝑦𝐴1$y_{A1}$ and 𝑦𝐴2$y_{A2}$. Subtraction yields

Λ=ln⁡1−𝑦𝐴21−𝑦𝐴1𝐿=𝑧2−𝑧1𝐶1=−Λ𝐿𝐶2=−ln⁡(1−𝑦𝐴1)−𝐶1𝑧1$$\begin{gathered} \Lambda=\ln\frac{1-y_{A2}}{1-y_{A1}}\\ L=z_2-z_1\\ C_1=-\frac{\Lambda}{L}\\ C_2=-\ln(1-y_{A1})-C_1z_1 \end{gathered}$$(19)

Write the profile using a dimensionless distance measured from the liquid surface:

𝜂=𝑧−𝑧1𝐿1−𝑦𝐴=(1−𝑦𝐴1)(1−𝑦𝐴21−𝑦𝐴1)𝜂𝑁𝐴=−𝑐𝐷𝐴𝐵𝐶1=𝑐𝐷𝐴𝐵𝐿Λ$$\begin{gathered} \eta=\frac{z-z_1}{L}\\ 1-y_A=(1-y_{A1})\left(\frac{1-y_{A2}}{1-y_{A1}}\right)^\eta\\ N_A=-cD_{AB}C_1=\frac{cD_{AB}}{L}\Lambda \end{gathered}$$(20)

The endpoints give the specified values, and differentiating the profile reproduces Eq. 9 at every position. This is the benefit of shell balance: the profile is now known, rather than only a resistance formula. Dilute A makes it approximately linear.

16 Case study 2: steady-state diffusion from a suspended droplet

Take a spherical liquid droplet of radius 𝑎$a$ surrounded by a gas film ending at radius 𝑏$b$. Radial coordinate 𝑟$r$ points outward. Assume quasi-steady spherical symmetry, constant 𝑐$c$ and 𝐷$D$, no reaction or external flow, interface gas fraction 𝑦𝐴1$y_{A1}$ and outer fraction 𝑦𝐴2$y_{A2}$, and insoluble B. At a pure-liquid interface Raoult’s law sets the surface vapour fraction at the droplet temperature. A shrinking droplet is only instantaneously steady; neglect thermal cooling or include it in a coupled model.

Figure 4: Spherical diffusion from a liquid droplet of radius a to the outer edge of a gas film at radius b.

A shell has different inner and outer areas. Conservation gives

4𝜋[𝑟2𝑁𝐴(𝑟)−(𝑟+Δ𝑟)2𝑁𝐴(𝑟+Δ𝑟)]=0𝑑𝑑𝑟(𝑟2𝑁𝐴)=0,𝑟2𝑁𝐴=𝐶𝑁𝑁𝐴=−𝑐𝐷𝐴𝐵1−𝑦𝐴𝑑𝑦𝐴𝑑𝑟$$\begin{gathered} 4\pi\left[r^2N_A(r)-(r+\Delta r)^2N_A(r+\Delta r)\right]=0\\ \frac{d}{dr}(r^2N_A)=0,\qquad r^2N_A=C_N\\ N_A=-\frac{cD_{AB}}{1-y_A}\frac{dy_A}{dr} \end{gathered}$$(21)

Thus the flux falls as 1/𝑟2$1/r^2$ while total molar rate stays fixed. Substituting and integrating gives

𝑟21−𝑦𝐴𝑑𝑦𝐴𝑑𝑟=𝐶1−ln⁡(1−𝑦𝐴)=−𝐶1𝑟+𝐶2𝐶1=Λ1𝑏−1𝑎𝐶2=−ln⁡(1−𝑦𝐴1)+𝐶1𝑎$$\begin{gathered} \frac{r^2}{1-y_A}\frac{dy_A}{dr}=C_1\\ -\ln(1-y_A)=-\frac{C_1}{r}+C_2\\ C_1=\frac{\Lambda}{\frac{1}{b}-\frac{1}{a}}\\ C_2=-\ln(1-y_{A1})+\frac{C_1}{a} \end{gathered}$$(22)

Here 𝐶1$C_1$ has units m, unlike the plane integration constant. Subtract the surface equation to avoid dimensioned exponents:

ln⁡1−𝑦𝐴(𝑟)1−𝑦𝐴1=Λ1𝑎−1𝑟1𝑎−1𝑏𝑁𝐴(𝑎)=𝑐𝐷𝐴𝐵Λ𝑎2(1𝑎−1𝑏)˙𝑛𝐴=4𝜋𝑎2𝑁𝐴(𝑎)=4𝜋𝑐𝐷𝐴𝐵Λ1𝑎−1𝑏$$\begin{gathered} \ln\frac{1-y_A(r)}{1-y_{A1}} =\Lambda\frac{\frac{1}{a}-\frac{1}{r}}{\frac{1}{a}-\frac{1}{b}}\\ N_A(a)=\frac{cD_{AB}\Lambda}{a^2\left(\frac{1}{a}-\frac{1}{b}\right)}\\ \dot n_A=4\pi a^2N_A(a)=\frac{4\pi cD_{AB}\Lambda}{\frac{1}{a}-\frac{1}{b}} \end{gathered}$$(23)

˙𝑛𝐴$\dot n_A$ is a total molar rate in kmol s⁻¹; 𝑛𝑖$n_i$ earlier was a mass flux, so the overdot and A label matter. The source calls this rate 𝑊𝐴$W_A$. Both boundary fractions are recovered by Eq. 23. For 𝑏 →∞$b\to\infty$, 𝑁𝐴(𝑎) =𝑐𝐷Λ/𝑎$N_A(a)=cD\Lambda/a$ and ˙𝑛𝐴 =4𝜋𝑎𝑐𝐷Λ$\dot n_A=4\pi acD\Lambda$.

Sherwood number

The dimensionless mass transfer coefficient Sh =𝑘𝑐𝐿/𝐷𝐴𝐵$\mathrm{Sh}=k_cL/D_{AB}$, comparing surface transfer with diffusion across the fluid length 𝐿$L$. The coefficient 𝑘𝑐$k_c$ must use a concentration difference on the same basis as the species diffusivity. It is the mass counterpart of Nusselt number.

Large Sh means a steep interfacial concentration gradient relative to the reference gradient across 𝐿$L$. Values of order one mean that reference diffusion scale is comparable. A stagnant sphere gives Sh two with diameter as 𝐿$L$, as derived below, rather than a universal value of one. For 𝑘𝑐 =0.01 m s−1$k_c=0.01\ \mathrm{m\,s^{-1}}$, 𝐿 =0.02 m$L=0.02\ \mathrm m$ and 𝐷𝐴𝐵 =2 ×10−5 m2 s−1$D_{AB}=2\times10^{-5}\ \mathrm{m^2\,s^{-1}}$, Sh =10$\mathrm{Sh}=10$. Using radius instead halves it; use the length required by the correlation. For a local coefficient use the matching local length; for a surface average use the average coefficient and the specified overall length. The convective topic derives these local and average forms.

For dilute A, Λ ≃𝑦𝐴1 −𝑦𝐴2$\Lambda\simeq y_{A1}-y_{A2}$. Compare the surface flux with 𝑁𝐴 =𝑘𝑐(𝑐𝐴1 −𝑐𝐴2)$N_A=k_c(c_{A1}-c_{A2})$ to obtain

𝑘𝑐=𝐷𝐴𝐵𝑎,𝑑𝑝=2𝑎Sh=𝑘𝑐𝑑𝑝𝐷𝐴𝐵=2$$\begin{gathered} k_c=\frac{D_{AB}}{a},\quad d_p=2a\\ \mathrm{Sh}=\frac{k_cd_p}{D_{AB}}=2 \end{gathered}$$(24)

The leading two in sphere correlations is this diffusion limit, analogous to the heat-transfer Nu =2$\mathrm{Nu}=2$ limit. At concentrated A, Stefan flow increases the coefficient based on the simple concentration difference, so Sh equals two only for the stated dilute limit. Combining ˙𝑛𝐴$\dot n_A$ with the liquid balance

−(𝜌𝐿𝑀𝐴)4𝜋𝑎2 𝑑𝑎𝑑𝑡=˙𝑛𝐴$$-\left(\frac{\rho_L}{M_A}\right)4\pi a^2\,\frac{da}{dt}=\dot n_A$$

gives

𝑑(𝑎2)𝑑𝑡=−2𝑀𝐴𝑐𝐷Λ𝜌𝐿$$\frac{d(a^2)}{dt}=-\frac{2M_AcD\Lambda}{\rho_L}$$

in the infinite-film constant-property approximation. Diameter squared then decreases linearly. Real droplets can cool, encounter flow or change properties; those effects change the model. For a flowing gas, a suitable mass-transfer correlation accounts for the changed concentration field.

In steady spherically symmetric diffusion with no volumetric reaction, which quantity stays constant as radius increases?

• 4𝜋𝑟2𝑁𝐴𝑟$4\pi r^2N_{Ar}$, the total molar rate. ✓
• 𝑁𝐴𝑟$N_{Ar}$, the local flux.
• 𝑟𝑁𝐴𝑟$rN_{Ar}$ in every spherical problem.
• The concentration 𝑐𝐴$c_A$.

Explanation

The same molar rate crosses every sphere surrounding the droplet:

˙𝑛𝐴=4𝜋𝑟2𝑁𝐴𝑟=constant$$\dot n_A=4\pi r^2N_{Ar}=\text{constant}$$

Thus,

𝑁𝐴𝑟=˙𝑛𝐴4𝜋𝑟2$$N_{Ar}=\frac{\dot n_A}{4\pi r^2}$$

The area increases with the square of radius, so the local flux decreases as the inverse square. The total rate stays constant.

17 Additional examples

Compare equimolar diffusion and a stagnant carrier

A binary ideal gas is at 300 K and 100 kPa. Across a 5.0 mm plane film, the mole fraction of A falls from 0.30 to zero. Use 𝐷𝐴𝐵 =2.0 ×10−5$D_{AB}=2.0\times10^{-5}$ m² s⁻¹ and 𝑅 =8.314$R=8.314$ J mol⁻¹ K⁻¹. Compare the steady molar flux of A for equimolar counter-diffusion and for zero net flux of B. Take positive flux from the high-A face to the low-A face. Explain the difference and determine the flux ratio if the high-face mole fraction is reduced to 0.010, with all other data unchanged.

Solution

Data

Quantity	Value
Temperature	𝑇 =300 K$T=300\ \mathrm K$
Pressure	𝑃 =100000 Pa$P=100000\ \mathrm{Pa}$
Film thickness	𝐿 =0.0050 m$L=0.0050\ \mathrm m$
Diffusivity	𝐷𝐴𝐵 =2.0 ×10−5 m2 s−1$D_{AB}=2.0\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Gas constant	𝑅 =8.314 J mol−1 K−1$R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Side 1 mole fraction in the first comparison	𝑦𝐴1 =0.30$y_{A1}=0.30$
Side 2 mole fraction	𝑦𝐴2 =0$y_{A2}=0$
Side 1 mole fraction in the second comparison	𝑦𝐴1 =0.010$y_{A1}=0.010$

Assumptions

• Steady state.
• One-dimensional diffusion across a plane film.
• Ideal gas behaviour.
• Constant temperature, pressure and diffusivity.
• No reaction in the film.

Calculations

Positive flux points from the high-A face towards the low-A face. The total concentration is

𝑐=𝑃𝑅𝑇=1000008.314(300)=40.09 mol m−3$$c=\frac{P}{RT}=\frac{100000}{8.314(300)}=40.09\ \mathrm{mol\,m^{-3}}$$

For equimolar counter-diffusion, 𝑁𝐴 +𝑁𝐵 =0$N_A+N_B=0$. As in Eq. 8, the bulk contribution vanishes and the governing relation is

𝑁𝐴=−𝑐𝐷𝐴𝐵𝑑𝑦𝐴𝑑𝑧.$$N_A=-cD_{AB}\frac{dy_A}{dz}.$$

Integrating between the film faces gives

𝑁𝐴,eq=𝑐𝐷𝐴𝐵𝐿(𝑦𝐴1−𝑦𝐴2)=40.09(2.0×10−5)0.0050(0.30−0)=0.04811 mol m−2 s−1$$\begin{aligned} N_{A,\mathrm{eq}}&=\frac{cD_{AB}}L(y_{A1}-y_{A2})\\ &=\frac{40.09(2.0\times10^{-5})}{0.0050}(0.30-0)\\ &=0.04811\ \mathrm{mol\,m^{-2}\,s^{-1}} \end{aligned}$$

For zero net flux of B, the total flux satisfies

𝑁𝐴=−𝑐𝐷𝐴𝐵1−𝑦𝐴𝑑𝑦𝐴𝑑𝑧.$$N_A=-\frac{cD_{AB}}{1-y_A}\frac{dy_A}{dz}.$$

Integrating this relation gives

𝑁𝐴,stag=𝑐𝐷𝐴𝐵𝐿ln⁡1−𝑦𝐴21−𝑦𝐴1=40.09(2.0×10−5)0.0050ln⁡10.70=0.05720 mol m−2 s−1$$\begin{aligned} N_{A,\mathrm{stag}}&=\frac{cD_{AB}}L\ln\frac{1-y_{A2}}{1-y_{A1}}\\ &=\frac{40.09(2.0\times10^{-5})}{0.0050}\ln\frac{1}{0.70}\\ &=0.05720\ \mathrm{mol\,m^{-2}\,s^{-1}} \end{aligned}$$

A enters the film without a compensating net flux of B, producing molar flow in A’s direction. That bulk flow transports additional A. B still diffuses towards the high-A side, but its bulk contribution cancels its diffusion.

With 𝑦𝐴2 =0$y_{A2}=0$, dividing the two fluxes gives

𝑁𝐴,stag𝑁𝐴,eq=−ln⁡(1−𝑦𝐴1)𝑦𝐴1$$\frac{N_{A,\mathrm{stag}}}{N_{A,\mathrm{eq}}} =\frac{-\ln(1-y_{A1})}{y_{A1}}$$

The common concentration, diffusivity and film thickness cancel. The ratios at the two supplied compositions are

[𝑁𝐴,stag𝑁𝐴,eq]𝑦𝐴1=0.30=−ln⁡(0.70)0.30=1.189$$\left[\frac{N_{A,\mathrm{stag}}}{N_{A,\mathrm{eq}}}\right]_{y_{A1}=0.30} =\frac{-\ln(0.70)}{0.30}=1.189$$[𝑁𝐴,stag𝑁𝐴,eq]𝑦𝐴1=0.010=−ln⁡(0.990)0.010=1.0050$$\left[\frac{N_{A,\mathrm{stag}}}{N_{A,\mathrm{eq}}}\right]_{y_{A1}=0.010} =\frac{-\ln(0.990)}{0.010}=1.0050$$

As a dilute-limit check,

−ln⁡(1−𝑦)=𝑦+𝑦22+⋯ ,$$-\ln(1-y)=y+\frac{y^2}{2}+\cdots,$$

so the ratio approaches one as 𝑦 →0$y\to0$. The two models then give nearly the same flux.

Answer

At 𝑦𝐴1 =0.30$y_{A1}=0.30$, the equimolar flux is 0.0481 mol m−2 s−1$0.0481\ \mathrm{mol\,m^{-2}\,s^{-1}}$ and the stagnant-carrier flux is 0.0572 mol m−2 s−1$0.0572\ \mathrm{mol\,m^{-2}\,s^{-1}}$. The latter is 18.9 percent larger. At 𝑦𝐴1 =0.010$y_{A1}=0.010$, their ratio falls to 1.005.

18 What you should now be able to calculate

Convert concentration and flux bases, distinguish equimolar from stagnant-carrier diffusion, and calculate plane or radial rates and slow interface motion.

Practice: Workshop 8 for evaporation and shrinking droplets.

19 Report an error

Email Ranjeet Utikar with the title Fundamentals of mass diffusion, the URL /mass-transport/diffusion-fundamentals/, and the equation or passage to check.

References

Bird, R. Byron, Warren E. Stewart, and Edwin N. Lightfoot. 2006. Transport Phenomena. Revised 2nd. John Wiley & Sons. https://www.wiley-vch.de/en/areas-interest/engineering/transport-phenomena-978-0-470-11539-8.

Geankoplis, Christie J. 2003. Transport Processes and Separation Process Principles. 4th ed. Prentice Hall.

Shah, Milinkumar. n.d. “CHEN3001-TP: Mass Transport 1.” Curtin University. Unpublished.

Welty, James R., Charles E. Wicks, Robert E. Wilson, and Gregory L. Rorrer. 2001. Fundamentals of Momentum, Heat, and Mass Transfer. 4th ed. John Wiley & Sons.