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

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.