Reference ID: MET-A02C | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
The Kelvin equation quantifies how the solubility of a small crystal increases as its radius of curvature decreases. In process engineering this is critical for:
Design of anti-solvent or cooling crystallizers where nuclei < 1 µm appear
Predicting Ostwald-ripening rates in suspensions stored or transported in pipelines
Setting supersaturation targets to avoid unwanted primary nucleation during seeding
Estimating the minimum stable crystal size in milling or sonication circuits
Because the equilibrium solubility \(C_{\text{r}}\) rises exponentially when the particle shrinks, undersaturated solutions can re-dissolve fines while larger grains grow—a phenomenon exploited in industrial ripening and dissolution-recrystallization purification steps.
Methodology & Formulas
Convert practical inputs to SI base units:
\[T\ [\text{K}] = T\ [\text{°C}] + 273.15\]
\[V_{\text{m}}\ [\text{m}^{3}\text{mol}^{-1}] = V_{\text{m}}\ [\text{cm}^{3}\text{mol}^{-1}] \times 10^{-6}\]
\[r\ [\text{m}] = r\ [\text{µm}] \times 10^{-6}\]
Compute the natural logarithm of the solubility ratio:
\[\ln\left(\frac{C_{\text{r}}}{C_{\infty}}\right) = \frac{2\ \gamma\ V_{\text{m}}}{r\ R\ T}\]
where
\(\gamma\) = solid–solution interfacial tension (N m⁻¹)
\(R\) = 8.314 J mol⁻¹ K⁻¹
\(C_{\infty}\) = bulk solubility of an infinitely large crystal (same units as \(C_{\text{r}}\))
Enforce physical limits:
Parameter
Lower bound
Upper bound
Remark
r
10 nm
—
Below 10 nm the continuum–thermodynamic derivation loses validity
\(\gamma\)
0.02 N m⁻¹
0.05 N m⁻¹
Typical range for ice–water at 0 °C
T
273.15 K
373.15 K
Absolute zero and the normal boiling point of water
The code guards against numerical failure by clamping the exponent to a minimum of 1 × 10⁻⁹ and issues warnings when the radius or interfacial tension lies outside the accepted engineering range.
The Kelvin equation relates the equilibrium vapor pressure (or solubility) of a substance to the curvature of its surface. For a spherical crystal of radius r, the equation is:
ln(Sr / S∞) = (2γVm) / (rRT),
where Sr is the solubility of the small crystal, S∞ is the solubility of a flat surface, γ is the surface tension, Vm is the molar volume, R is the gas constant, and T is temperature. Because the right‑hand side is positive for a convex surface,
small crystals are more soluble than bulk crystals. This effect becomes significant when r is on the order of nanometers to a few micrometers, which is common in high‑purity or pharmaceutical processes.
Step‑by‑step application:
Determine the average radius of the crystals in the batch (e.g., from microscopy or laser diffraction).
Obtain the surface tension (γ) of the solute–solvent system from literature or experimental measurement.
Calculate the molar volume (Vm) of the solute in the solid phase.
Insert the values into the Kelvin equation to compute the relative solubility increase (Sr / S∞).
Adjust the process parameters (temperature, supersaturation, stirring) to target a crystal size that yields the desired solubility.
Tip: Use the Kelvin equation as a correction factor when modeling nucleation and growth kinetics in simulation tools.
Key assumptions and limitations:
The crystal is assumed to be spherical and homogeneous; real crystals may have facets or irregular shapes.
Surface tension γ is treated as a constant, but it can vary with crystal orientation and impurities.
The equation neglects elastic strain and electrostatic interactions that can be significant for very small particles.
It assumes equilibrium conditions; during rapid crystallization, kinetic effects may dominate.
Temperature dependence of γ and Vm is often ignored, which can introduce errors at high temperature ranges.
Conclusion: Use the Kelvin equation as a qualitative guide or for small corrections, but validate predictions with experimental data.
Practical scale‑up tips:
Ensure consistent crystal size distribution by controlling nucleation and growth rates; use seeding or anti‑nucleation agents.
Measure actual surface tension under process conditions, as it can change with concentration and temperature.
Account for temperature gradients in large vessels; local supersaturation may differ from the bulk value.
Use in‑situ monitoring (e.g., laser diffraction, dynamic light scattering) to track crystal size evolution and update the Kelvin correction in real time.
Combine the Kelvin equation with thermodynamic solubility data to avoid over‑ or under‑estimating the effect on product purity.
Result: By integrating these considerations, process engineers can more accurately predict and control solubility changes due to crystal curvature during scale‑up.
Worked Example: Kelvin Equation for a 1 µm Ice Crystal
A process engineer is evaluating the solubility of a microscopic ice crystal in supercooled water at 0 °C. The goal is to estimate how the curvature of a 1 µm radius crystal raises its equilibrium concentration relative to a flat interface.
Universal gas constant, R = 8.314 J mol⁻¹ K⁻¹
Temperature, T = 0 °C = 273.15 K
Solid–liquid interfacial tension, γ = 0.032 N m⁻¹
Molar volume of ice, Vm = 1.96 × 10⁻⁵ m³ mol⁻¹
Crystal radius, r = 1.0 µm = 1.0 × 10⁻⁶ m
Bulk (flat‑surface) solubility, C∞ = 1.0 × 10⁻⁹ g L⁻¹
Write the Kelvin equation for the ratio of the curved‑surface concentration Cr to the flat‑surface concentration C∞:
\[
\ln\!\left(\frac{C_{r}}{C_{\infty}}\right)=\frac{2\gamma V_{m}}{R\,T\,r}
\]
Calculate the numerator \(2\gamma V_{m}\):
\[
2\gamma V_{m}=2(0.032)(1.96\times10^{-5})=1.254\times10^{-6}
\]
Calculate the denominator \(R\,T\,r\):
\[
R\,T\,r=8.314(273.15)(1.0\times10^{-6})=2.273\times10^{-3}
\]
Evaluate the logarithmic term:
\[
\ln\!\left(\frac{C_{r}}{C_{\infty}}\right)=\frac{1.254\times10^{-6}}{2.273\times10^{-3}}=5.52\times10^{-4}
\]
Exponentiate to obtain the concentration ratio:
\[
\frac{C_{r}}{C_{\infty}}=\exp(5.52\times10^{-4})\approx1.001
\]
Calculate the curved‑surface solubility:
\[
C_{r}=C_{\infty}\times1.001=1.0\times10^{-9}\times1.001=1.001\times10^{-9}\;\text{g L}^{-1}
\]
Final Answer: The equilibrium solubility of a 1 µm ice crystal at 0 °C is approximately 1.001 × 10⁻⁹ g L⁻¹, a 0.1 % increase over the flat‑surface value.
"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
