Introduction & Context

The calculation estimates the time required for a diffusing species to reach a prescribed fraction of its surface concentration at a specified depth within a semi‑infinite medium. This unsteady‑state mass transport problem is governed by Fick’s second law and is fundamental in process‑engineering applications such as:

Penetration of gases or liquids into porous catalysts or adsorbents.
Moisture migration in polymeric coatings and food products.
Thermal‑diffusion coupling in heat‑treatment of solids.
Leak‑rate assessment for containment barriers.

By linking the diffusion coefficient, geometry, and a target concentration, engineers can size equipment, schedule batch times, or assess long‑term performance without resorting to full numerical simulations.

Methodology & Formulas

The solution for a constant surface concentration \(C_{0}\) imposed at \(z = 0\) for a semi‑infinite slab is \[ \frac{C(z,t)}{C_{0}} = \operatorname{erfc}\!\left(\frac{z}{2\sqrt{D\,t}}\right) \] where:

\(C(z,t)\) – concentration at depth \(z\) and time \(t\).
\(D\) – effective diffusivity (assumed constant).
\(\operatorname{erfc}\) – complementary error function.

To achieve a desired fractional concentration \(\phi\) (e.g., \(\phi = 0.10\) of \(C_{0}\)), the argument of the error function is set to a known value \(\eta\) such that \(\operatorname{erfc}(\eta) = \phi\). Solving for the required diffusion time yields \[ t = \frac{z^{2}}{4\,\eta^{2}\,D} \] This expression follows directly from rearranging the definition of \(\eta\): \[ \eta = \frac{z}{2\sqrt{D\,t}} \;\;\Longrightarrow\;\; t = \frac{z^{2}}{4\,\eta^{2}\,D} \]

A quick‑check “rule‑of‑thumb” penetration depth \(\delta\) is often used: \[ \delta = \sqrt{D\,t} \] When \(\delta\) exceeds the depth of interest, the target fraction is effectively reached.

If temperature dependence of diffusivity is required, the absolute temperature is obtained from the Celsius value: \[ T_{\text{K}} = T_{\!^{\circ}\!C} + 273.15 \] and a temperature‑correction correlation (e.g., Arrhenius) may be applied to \(D\).

Assumptions & Applicability

Assumption	Description
Semi‑infinite medium	The domain extends sufficiently far such that the diffusion front does not reach a far boundary during the time of interest.
Constant surface concentration	The concentration at \(z = 0\) remains fixed at \(C_{0}\) for the entire diffusion period.
Isothermal conditions	Temperature is uniform; diffusivity \(D\) is treated as constant unless a temperature correction is explicitly applied.
Negligible convection	Mass transport occurs solely by molecular diffusion; bulk flow effects are absent.

Use Fick’s first law only when concentration at every point in the domain is constant with time (steady state). Switch to Fick’s second law, ∂C/∂t = D ∇²C, the moment concentration changes with time anywhere in the system. Typical triggers are:

Start-up or shut-down transients in a membrane separator
Batch extraction where solute concentration in the bulk decreases
Any pulse, step, or cyclic loading of the feed

If you ignore the unsteady term you will over-predict fluxes and under-size equipment.

For uniform radial diffusion in a pellet use the cylindrical coordinate form: ∂C/∂t = D [ (1/r) ∂/∂r (r ∂C/∂r) ] If axial diffusion along the reactor length is also important, add the axial term: ∂C/∂t = D [ (1/r) ∂/∂r (r ∂C/∂r) + ∂²C/∂z² ] Neglecting the curvature term (1/r) can introduce >10 % error when the pellet diameter is >30 % of the characteristic diffusion length.

Use the Danckwerts (open) boundary conditions:

At the upstream face: −D ∂C/∂x |x=0 = k_L (C_b − C|x=0)
At the downstream face: −D ∂C/∂x |x=L = k_R (C|x=L − C_p)

These conditions couple external convection to internal diffusion and avoid the common error of assuming infinite surface concentration.

For dilute liquid systems use the Wilke–Chang correlation: D = 7.4×10⁻⁸ (φ M_solv)^0.5 T / (μ V_mol^0.6) where φ is the solvent association factor, μ is viscosity (cP), and V_mol is molar volume (cm³ mol⁻¹). For gases at low pressure use the Fuller correlation. Always validate the estimate with at least one lab measurement at process temperature; predicted D values can carry ±30 % uncertainty.

Worked Example – Unsteady‑State Diffusion Using Fick’s Second Law

Scenario
A gas‑processing plant stores a liquid hydrocarbon in a sealed tank that is lined with a polymer membrane. A trace contaminant initially has a uniform concentration of C₀ = 85 mg L⁻¹ in the liquid. The contaminant diffuses through the membrane and is removed at the outer surface (concentration ≈ 0). The engineer must estimate how long it will take for the contaminant concentration to drop to 10 % of its initial value.

Knowns

C₀ = 85.0 mg L⁻¹
T = 15 °C (T_K = 288.15 K)
P = 1.0 bar
Diffusion coefficient, D = 9.5 × 10⁻¹⁰ m² s⁻¹
Tortuosity factor, η = 1.16
Membrane thickness, δ = 0.2155 m
Target fraction of C₀,  X_target = 0.10

Step‑by‑Step Calculation

Effective diffusion coefficient (accounts for tortuosity): \[ D_{\text{eff}} = \frac{D}{\eta} = \frac{9.5\times10^{-10}}{1.16} = 8.19\times10^{-10}\;\text{m}^2\!\!/\text{s} \]
Dimensionless time parameter for a slab with both faces at zero concentration (first‑term approximation): \[ \alpha = \frac{\pi^{2} D_{\text{eff}} t}{4\,\delta^{2}} \] Rearranged to solve for the required time t that yields the target fraction: \[ X = \frac{C}{C_{0}} \approx \frac{8}{\pi^{2}}\,e^{-\alpha} \quad\Longrightarrow\quad \alpha = -\ln\!\left(\frac{X\,\pi^{2}}{8}\right) \] Substituting \(X = 0.10\): \[ \alpha = -\ln\!\left(\frac{0.10\times\pi^{2}}{8}\right) = -\ln(0.123) = 2.095 \]
Calculate the required time using the definition of α: \[ t = \frac{4\,\delta^{2}\,\alpha}{\pi^{2} D_{\text{eff}}} \] \[ t = \frac{4\,(0.2155)^{2}\,(2.095)}{\pi^{2}\,(8.19\times10^{-10})} = \frac{4\,(0.0465)\,(2.095)}{9.8696\,(8.19\times10^{-10})} = \frac{0.389}{8.08\times10^{-9}} = 4.89\times10^{7}\;\text{s} \]
Convert the time to years: \[ t_{\text{yr}} = \frac{4.89\times10^{7}\;\text{s}}{3.154\times10^{7}\;\text{s yr}^{-1}} = 1.55\;\text{yr} \]
Final contaminant concentration after this period: \[ C = X_{\text{target}}\;C_{0} = 0.10 \times 85.0 = 8.5\;\text{mg L}^{-1} \]

Final Answer

The contaminant concentration will fall to 8.5 mg L⁻¹ (10 % of the initial value) after approximately 1.55 years (≈ 4.89 × 10⁷ s) of diffusion through the 0.215 m polymer membrane.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle