π (r 1 + r 2 ) 2 πr 2 1 v T1 v T2 v T1

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1 How do we get rain? So far we ve discussed droplet growth by vapor diffusion, but this is not the process that by itself is primarily responsible for precipitation in warm clouds. The primary production mechanism for growth of precipitation sized droplets is collision-coalescence. In this process droplets collide due to differential settling velocities, and coalesce. If we say the big collector drop is r 1 and the smaller drops it is collecting have radius r 2 and r 1 r 2 then the rate of collection depends on 1. The collision cross-sectional area is 2. The relative collection velocity is π (r 1 + r 2 ) 2 πr 2 1 v T1 v T2 v T1 3. The coalescence efficiency (i.e. the probability that two colliding droplets stick), which is approximately unity if r 1 > 10r 2 4. The collision efficiency E (r 1, r 2 )(i.e. the probability that two droplets that are within the sweep area of the collector drop actually collide. 5. The LWC of the collected droplets Therefore the growth rate of the collector drop is dm 1 dt = πr 2 1E (r 1, r 2 )v T1 LWC 2 Notice that we could have gotten this equation from the non-equilibrium thermodynamic solution we derived earlier = V u ρ which, since the volume swept out by a falling droplet is V = A x, and ρ = ρ/ x, it could be re-written as = Av T ρ or, = Av T LWC and noting that the effective cross-section here is A = Eπr1 2 1 = πr 2 1Ev T LWC Now, noting that dm dt = dm dr dr dt 1

2 and dm dr = 4πρr2 we can show that dr 1 dt = E (r 1, r 2 )v T1 LWC 2 4ρ l Note that r 1 > 20 µm appears to be a critical initial size for initiation of the collision-coalescence process. If droplet are unable to grow to this size in sufficient concentrations by vapor diffusion, then typically rain does not form. Figure 1: Growth rate by collision compared to diffusion. Terminal velocity For very particles smaller than about 40 µm radius, we can use a solution that can be derived from first principles, where the drag force is proportional to radius 6πµv T r = 4 3 πρr3 g v T = 2ρr2 g 9µ = k 1r 2 2

3 where k cm 1 s 1. However this assumes Stokes drag. However, for rain drops with radius 0.6 mm <r < 2 mm the drag force is in fact proportional to r 2. The solution then for v T is v T k 2 r 1/2 where k cm 1 s 1. In the intermediate range 40 µm <r < 0.6 mm v T k 3 r where k s 1 Terminal velocities as a function of droplet size are given in the table below Figure 2: Table of terminal fall speed Collision Efficiency The computation of the collision efficiency from first principles is a very difficult problem owing to the complexities of the flow field around a sphere falling through a viscous medium. Rogers and Yau state For any size of collector drop, the collision efficiency is small for small values of r 2 /r 1. The collected droplets are then small, have little inertia, and are easily deflected by the flow around the collector drop. The inertia of the droplets increases with r 2 /r 1 accounting for an increase in collision efficiency up to a radius ratio of about 0.6. Two counteracting effects come into play as r 2 /r 1 increases beyond this value. Because the difference in the size of the drops is getting smaller, the relative velocity between the drops is reduced, prolonging the time of interaction. The flow fields interact strongly, and the time can be sufficient for the droplet to be deflected around the drop without collision. On the other hand there is a possibility for a trailing droplet to be attracted 3

4 into the wake of a drop falling close by at nearly the same speed. This effect can lead to wake capture and to collision efficiencies that exced unity for values of r 2 /r 1 1. Figure 3: Collision efficiency (Rogers and Yau) There is also a coalescence efficiency. Big drops can bounce of each other. Really, though it is the collision efficiency that is of most interest for collision-coalescence, because the concentrations of these bouncy droplets are so low. Example Imagine a cumulus cloud with and LWC of 0.4 g m 3 1 km thick with a monodisperse droplet mode at 10 µm radius and a second mode with concentrations of 1/litre at 100 µm radius at cloud top. The updraft velocity in the cloud is 0.05 m/s. 1. Derive an expression for the size of the drizzle drops as a function of distance as they descend 4

5 through the cloud dr dt = LWCv TE (r 1, r 2 ) 4ρ l This distance the droplet falls from cloud top is z = v T t, so dr dt /dz dt = dr dz = LWCv TE (r 1, r 2 ) (v T W)4ρ l Thus, H 0 LWCdz = 4ρ l rh r 0 v T W v T E dr 1 Okay. This integrand on the RHS is a bit nasty, but if we assume that in general W v T (which maybe isn t so good), integrating we get the radius in meters as 2. What is the droplet size at cloud base? Just substitue z = 1000m to get r = 180µm 3. What is the precipitation flux in mm/hr Flux = LWC precip v Tprecip r = z Flux = 4/3πρr 3 1 N k 3r Flux = 4/3π (1000) ( ) 3 ( ) ( ) with appropriate conversions, I get 4.4 mm/hr. 4. In the absence of updrafts generating new cloud, what is the time scale for depletion of all the liquid water in the cloud? loss rate = ϕ = LWC = LWC 0 exp ( ϕt) precip rate cloud water depth = precip rate ρ l cloud depth LWC = (1000)( ) 10/hr This leads to a time scale of about 6 min. Well this is interesting, because it indicates that the cloud exhausts itself of water 10 times an hour. How then is it possible for us to see a cloud for longer than only a few minutes? Another interesting conclusion is that ϕ is the same as the loss rate of the cloud forming aerosol from the entire boundary layer, or cloud condensation nuclei. How is this the case? 5