GEF2200 Stordal - based on Durkee 10/11/2015 Relative sizes of cloud droplets and raindrops; r is...

GEF2200 Stordal - based on Durkee 04/21/23

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


Formation of Cloud DropletsFormation 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 es for equilibrium

rres(r) es()


Growth depends on the difference between es(r) and e

e < es(r) decay (vapor moves away from the drop)e > es(r) growth (vapor moves toward the drop)

When the radius is such that e = es(r) the droplet is just large enough to be stable:

where S = e/es() 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 cm3 per second in the natural atmosphere.

Since S rarely exceeds 1-2%, homogeneous nucleation is never consistently achieved.

1( )s

e a

e r


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


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 thevapor pressure required for equilibrium)

Positive feedback:


Nucleation of droplets requires a particle (condensation nucleus).

Hygroscopic nuclei are soluble in water and decrease es(r) significantly.

hygroscopic hydrophobic

rres(r)+

-

+

++

+

+

+

+

+

+

+

+

+

-

-

-

-

--

-

-

-

-

-

-

With non-water molecules on the surface, the equilibrium (equal transfer across the interface) occurs at lower pressure

M = mass of soluteC = 3imv/4prLms

~Eq. 6.6 W&H


hazehaze

““activated”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/LRvT, and b=3imvM/4Lms)

Which term dominates below 100% RH?

Why does the Köhler curve approach 1.0 for large r?

Growth does not continuewithout bound since dropsstart to compete for vapor

~Fig. 6.3 W&H


r* , S* as dry particle diameter(or mass) increases

r* , S* as dry solute molecular weight increases

Nc as S increases

From Seinfeld and Pandis

~Fig. 6.3

W&H


S = Smax

OD model of CCN activation

dtdrQwQ

dtdS

.2.1

)//( 3rBrASdtdrr

Guibert et al. 2003


Activity Spectrum = number of activated particles at some supersaturation S and below Nc = C sk

(where s=(S-1)x100%)

Marine: C=150 k=0.6

Continental: C=1500

k=1.110

100

1000

10000

0 0.5 1 1.5 2

Supersaturation (%)

Act

ivit

y Spect

rum

(cm

-3)

N(marine)

N(continental)

maritime: C=30-300 cm-3; k=0.3-1.0continental C=300-3000 cm-3; k=0.2-2.0

~Fig. 6.5

W&H


=Fig. 6.5

W&H


Supersaturation is controlled by updraft velocity so...

Marine: C=150 k=0.6

Continental: C=1500

k=1.10

100

200

300

400

500

600

700

800

0 50 100 150 200

Vertical Velocity (cm/s)

Num

ber

of

Act

ivate

d D

rople

ts (

cm-3

)

N(marine)

N(continental)


Marine: C=150 k=0.6

Continental: C=1500

k=1.1

And the maximum supersaturation becomes…

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 50 100 150 200

Vertical Velocity (cm/s)

Maxi

mum

Supers

atu

rati

on (

%)

N(marine)

N(continental)


NaCl nuclei with Nc = (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?


Note that x0 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 x0 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)?


Growth of all drops in the distribution - stochastic coalescence

Observations:• initial single mode evolves to two modes by about 20 minutes

• rf describes first mode and rg describes the second mode

GEF2200 Stordal - based on Durkee 10/11/2015 Relative sizes of cloud droplets and raindrops; r is...

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