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

$S=e/e_s^l$ 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.

Use the large-radius condensational-growth approximation from part (a), with fixed ambient atmospheric supersaturation and transport resistances. It givesTaking $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$, $S-1=0.01$, and negligible initial $r_0^2$ givesThe squared-radius law makes condensation relatively fast for reaching cloud-droplet sizes but far too slow for producing millimetre rain within a typical evolving cloud. Collision–coalescence, or ice growth followed by aggregation/riming and melting, bypasses this bottleneck. In reality atmospheric supersaturation is shared by many particles and changes with ascent and vapour depletion, so even the constant-$1\%$ estimate is idealized.

These are growth times after activation, with negligible activation delay. The statement $r_0\ll 1\mu \mathrm{m}$ alone does not ensure that a seed grows: for pure water with $B=0$, a seed with $r_0<A/(S-1)$ has a negative growth rate and shrinks. For soluble seeds the Köhler barrier depends on $A,B$ and the environmental atmospheric supersaturation. The literal full-model time iswhich cannot be assigned a unique initial-stage numerical value without those microphysical parameters. The two boxed estimates are the intended continuum condensation scales for activated droplets, rather than a claim that every arbitrarily small seed reaches those sizes.

In the quasisteady diffusion approximation, the vapour-density disturbance outside a small stationary particle satisfies Laplace's equation, with a specified equilibrium value on its surface and a uniform far-field value. This has the same mathematical boundary-value problem as the electrostatic potential outside a conductor held at fixed potential. The integrated normal gradient is proportional to the conductor's geometrical capacitance.

Here $C$ is a length: a sphere of radius $r$ has $C=r$. The corresponding electrical capacitance in SI units would be $4\pi {ϵ}_{0}C$, not $C$ farads. Replacing the spherical diffusive conductance $4\pi r$ by $4\pi C$ accounts for the crystal's non-spherical geometry. Coupling the analogous heat-conduction problem to vapour diffusion supplies the ice-specific thermal and diffusional resistances.

The approximation assumes continuum diffusion, equilibration of the surface vapour concentration, an approximately isothermal particle, a slowly changing shape, weak perturbations and a far field sufficiently distant relative to the crystal size. It neglects ventilation by falling motion, interactions with neighbouring particles and facet-dependent attachment kinetics. Shape affects diffusive growth through capacitance, not merely through projected collecting area as in a collision calculation.

The target liquid-water mass isThe thin-plate ice-growth time uses the prescribed ice mass law, which therefore requires $r_{i\ast }=(m_{\ast }/\alpha {)}^{1/3}=45.31\mu \mathrm{m}$, not the same radius as the water droplet. With $C=2r_i/\pi$ and $F_i=F_k^i+F_d^i$,Hence, for negligible seed mass and constant conditions,This is about $8.24$ times faster than the liquid estimate to the same mass. Both laws have $dm/dt\propto m^{1/3}$ under the given shape assumptions. Ice benefits here from five times the atmospheric supersaturation and a larger geometrical capacitance for the same mass, partly offset by its slightly greater transport resistance. The mass coefficient $\alpha$ encodes the assumed plate shape; replacing it by bulk ice density would change the supplied model.

In a mixed-phase cloud, saturation vapour pressure over ice is lower than over supercooled liquid water. Air can therefore be subsaturated with respect to liquid yet supersaturated with respect to ice: droplets evaporate while ice crystals grow, transferring water through the vapour phase. This is the Wegener–Bergeron–Findeisen mechanism. Subsequent aggregation, riming and melting can produce atmospheric precipitation much faster than condensation alone.

The large difference is not a universal fixed ice/liquid time ratio. Atmospheric supersaturation evolves, latent heating increases thermal resistance, crystal habits and capacitance change, and surface attachment or ventilation may matter. Ice nucleation and the availability of freezing nuclei also precede the calculated growth. The boxed time describes the specified already-present crystals under maintained conditions, not a complete prediction of the time until rain begins.