1. GEF2200 Stordal - based on Durkee 10/11/2015 Relative sizes of cloud droplets and raindrops; r is the radius in micrometers, n the number per liter of air, and v the terminal fall speed in centimeters per second. The circumference of the circles are drawn approximately to scale, but the black dot representing a typical CCN is twenty-five times larger than it should be relative to the other circles. Adapted from Adv. in Geophys. 5, 244 (1958). Fig 6.18 W&H

2. GEF2200 Stordal - based on Durkee 10/11/2015 Formation of Cloud Droplets Why do cloud droplets form almost immediately upon reaching supersaturation? In air containing water vapor above the saturation pressure, can chance collisions form a stable droplet of pure water?  is the surface tension (energy/area or force/length) Smaller drops require higher e s for equilibrium

3. GEF2200 Stordal - based on Durkee 10/11/2015

4. Growth depends on the difference between e s (r) and e e < e s (r) decay(vapor moves away from the drop) e > e s (r) growth(vapor moves toward the drop) When the radius is such that e = e s (r) the droplet is just large enough to be stable: where S = e/e s (  ) is the saturation ratio (Eq. 6.5 W&H) Statistical thermodynamic calculations show that S must be 300-600% for one homogeneous nucleation event per cm 3 per second in the natural atmosphere. Since S rarely exceeds 1-2%, homogeneous nucleation is never consistently achieved.

5. GEF2200 Stordal - based on Durkee 10/11/2015 The relative humidity and supersaturation (both with respect to a plane surface of pure water) with which pure water droplets are in (unstable) equilibrium at 5ºC. Fig. 6.2 W&H

6. GEF2200 Stordal - based on Durkee 10/11/2015 Curvature effect Increased r, decreases equilibrium/saturation vapor pressure over the drop (fewer molecules required outside the drop at equilibrium) To attain this new equilibrium, vapor molecules will want to enter the drop at a higher rate than they leave (growth) But this positive feedback can’t get started at typical atmospheric saturation ratios. - +- - + + Adding solute, decreases equilibrium vapor pressure over the drop since fewer liquid molecules are available to escape (fewer molecules required outside at equilibrium ) Solution effect To attain this new equilibrium, vapor molecules will want enter the drop at a higher rate than they leave (growth) (note: saturation vapor pressure is the vapor pressure required for equilibrium) Positive feedback:

7. GEF2200 Stordal - based on Durkee 10/11/2015 Nucleation of droplets requires a particle (condensation nucleus). Hygroscopic nuclei are soluble in water and decrease e s (r) significantly. hygroscopic hydrophobic r e s (r) + - + + + + + + + + + + + + - - - - -- - - - - - - With non-water molecules on the surface, the equilibrium (equal transfer across the interface) occurs at lower pressure M = mass of solute C = 3imv/4prLms ~Eq. 6.6 W&H

8. GEF2200 Stordal - based on Durkee 10/11/2015 haze “activated” Now for a solution droplet (compared to a pure water plane surface) the equilibrium vapor pressure is increased due to curvature effects and decreased due to solution effects: Köhler curve (where a=2  /  L R v T, and b=3im v M/4  L m s ) Which term dominates below 100% RH? Why does the Köhler curve approach 1.0 for large r? Growth does not continue without bound since drops start to compete for vapor ~Fig. 6.3 W&H

9. GEF2200 Stordal - based on Durkee 10/11/2015 r*, S* as dry particle diameter(or mass) increases r*, S* as dry solute molecular weight increases N c as S increases From Seinfeld and Pandis ~Fig. 6.3 W&H

10. GEF2200 Stordal - based on Durkee 10/11/2015 S = S max OD model of CCN activation Guibert et al. 2003

11. GEF2200 Stordal - based on Durkee 10/11/2015 Activity Spectrum = number of activated particles at some supersaturation S and below N c = C s k (where s=(S-1)x100%) Marine: C=150 k=0.6 Continental: C=1500 k=1.1 maritime:C=30-300 cm -3 ; k=0.3-1.0 continentalC=300-3000 cm -3 ; k=0.2-2.0 ~Fig. 6.5 W&H

12. GEF2200 Stordal - based on Durkee 10/11/2015 =Fig. 6.5 W&H

13. GEF2200 Stordal - based on Durkee 10/11/2015 Supersaturation is controlled by updraft velocity so... Marine: C=150 k=0.6 Continental: C=1500 k=1.1

14. GEF2200 Stordal - based on Durkee 10/11/2015 Marine: C=150 k=0.6 Continental: C=1500 k=1.1 And the maximum supersaturation becomes…

15. GEF2200 Stordal - based on Durkee 10/11/2015 NaCl nuclei with N c = (650cm -3 ) s 0.7 Why is maximum supersaturation higher for 2m/s updraft velocity? Why is final droplet concentration greater for 2m/s updraft velocity? Why is average radius greater for 0.5m/s updraft velocity? Why is final deviation of radius greater for 0.5m/s updraft velocity? Why is LWC greater for 0.5m/s updraft velocity?

16. GEF2200 Stordal - based on Durkee 10/11/2015

17. Note that x 0 is not R+r since drops will deflect as they approach due to aerodynamic forces x Collision Efficiency x, the separation between the drop centers, or impact parameter, has a maximum value of R+r R r The collision efficiency then is the fraction of the drops that collide compared to those that could collide: If x 0 is the maximum impact parameter for a given r and R that will result in a collision. Why do small r/R have low efficiencies? Why does efficiency decrease beyond r/R~0.6? How could efficiency exceed 1.0 (near r/R~1)?

18. GEF2200 Stordal - based on Durkee 10/11/2015 Growth of all drops in the distribution - stochastic coalescence Observations: initial single mode evolves to two modes by about 20 minutes r f describes first mode and r g describes the second mode