Suppose we partition the available excess water equally between all hydrometeors (for example, for all liquid water droplets). In this way, we can estimate the average radius R for each droplet

due only to condensation (i.e., before collisions between droplets allow some to merge and grow into larger drops): R = 3 Pair TE 4π Pwater n where excess-water mixing ratio re is in kgwater kgair¹, p is density, and n is the number density of hydrometeors (the count of hydrometeors per cubic meter of air). Typical values are R = 2 to 50 μm, which is small compared to the 1000 μm separation between droplets, and is too small to be precipitation. This is an important consideration. Namely, even if we ignore the slowness of the diffusion process, the hydrometeors stop growing by condensation or deposition before they become precipitation. The reason is that there are too many hydrometeors, all competing for water molecules, thus limiting each to grow only a little. Within a cloud, suppose air density is 1 kg m-³/nkgair¹, p is density, and n is the number density of hydrometeors (the count of hydrometeors per cubic meter of air). Typical values are R = 2 to 50 μm, which is small compared to the 1000 μm separation between droplets, and is too small to be precipitation. This is an important consideration. Namely, even if we ignore the slowness of the diffusion process, the hydrometeors stop growing by condensation or deposition before they become precipitation. The reason is that there are too many hydrometeors, all competing for water molecules, thus limiting each to grow only a little. Within a cloud, suppose air density is 1 kg m-³ and the excess water mixing ratio is 2 g kg-¹. Find the final drop radius using the above equation to find the final drop radius for hydrometeor counts of 108 m³ in units of microns. (Pair=1 kg m-³, Pwater =1000 kg m-³) Watch the units, especially with re. Answer: