Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2018 / 4 / a

$S=e/e_s^l$ is the ambient saturation ratio over a flat liquid-water surface; $S-1$ is the liquid atmospheric supersaturation. Curvature increases the surface equilibrium vapour pressure, producing the Kelvin penalty $A/r$, while dissolved solute lowers it and gives the opposing Raoult contribution $B/r^3$. Thus the dilute Köhler equilibrium curve is $S_{\mathrm{eq}}-1=A/r-B/r^3$, and the numerator is the atmospheric supersaturation available to drive growth beyond surface equilibrium.

$F_d^l$ describes the resistance of vapour diffusion through the surrounding air. $F_k^l$ describes the thermal-conduction resistance to removing latent heat released by condensation: a warmed droplet surface has a higher equilibrium vapour pressure and hence a weaker mass-transfer drive. Common continuum approximations are $F_d^l\simeq R_{v}T/(D_v{e}_s^l)$ and $F_k^l\simeq L_v[k_{\mathrm{air}}T{]}^{-1}[L_v/(R_{v}T)-1]$, showing the roles of diffusivity, conductivity, saturation pressure and latent heat. Both resistances have the supplied units $\mathrm{s}\mathrm{m}\mathrm{k}{\mathrm{g}}^{-1}$ and add for this coupled transport problem.

For a spherical droplet of constant density, $m=4\pi \rho r^3/3$. Multiplying the radius equation by $4\pi \rho r$ givesAt sufficiently large radius, curvature and solute corrections are negligible, soThis limit also requires sustained positive atmospheric supersaturation. It need not describe the earliest submicrometre stage or an unactivated haze particle.