Note little curvature over 20 microns

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Note little curvature over 20 microns
Large droplets have less curvature than small
droplets. It is easier for molecules to
escape from the smaller droplets.
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Smaller droplets require supersaturation to
Remain in equilibrium.
es(r)/ es(8) exp(2?/rRv ?LT)
rc 2?/( Rv ?T ln S)
Adding in the solute effect,
es(r)/ es(8) 1 a/r b/r3
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r (3b/a)1/2
S 1 (4a3/27b) 1/2
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Eq. 7.7   Eq. 7.8   Eq. 7.18
    Heat Diffusion   Vapor
Diffusion     Eq. 7.19   Eq. 7.20    
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Example 1 How large will a droplet with initial
radius of 1µm be in 5 minutes if the growth
constant ? 0.2 µm2s-1 ?
Using
r2 r02 2(0.2 µm2s-1)(300s) r2 1µm2
120µm2 r 11 µm
Example 2 How large will a droplet with initial
radiusof 1µm be in 5 minutes if the growth
parameter ?1 100 µm2s-1 in an environment of
0.1 supersaturation? ? (S-1)?1 (.001)(100
µm2s-1) 0.1µm2s-1 r2 r02 2(0.1
µm2s-1)(300s) r2 1µm2 60µm2 r 7.8 µm
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Example 3 A cloud droplet grows from 5 µmto 20
µm in 15 minutes at 10C and 600 hPa. How much
supersaturation was required? r(t)2 r02 2
(S-1)(?1) t ?1 102.05 112 µm2s-1 400 µm2
25 µm2 2(S-1)(112 µm2s-1 )(900s) S-1 375 µm2
/ (900s)(112 µm2s-1 ) .0037 S-1 0.37 (S
1.0037, R.H. 100.37)
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