Post on 13-Jan-2016
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
GEF2200 Stordal - based on Durkee 04/21/23
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()
GEF2200 Stordal - based on Durkee 04/21/23
GEF2200 Stordal - based on Durkee 04/21/23
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
GEF2200 Stordal - based on Durkee 04/21/23
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
GEF2200 Stordal - based on Durkee 04/21/23
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:
GEF2200 Stordal - based on Durkee 04/21/23
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
GEF2200 Stordal - based on Durkee 04/21/23
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
GEF2200 Stordal - based on Durkee 04/21/23
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
GEF2200 Stordal - based on Durkee 04/21/23
S = Smax
OD model of CCN activation
dtdrQwQ
dtdS
.2.1
)//( 3rBrASdtdrr
Guibert et al. 2003
GEF2200 Stordal - based on Durkee 04/21/23
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
GEF2200 Stordal - based on Durkee 04/21/23
=Fig. 6.5
W&H
GEF2200 Stordal - based on Durkee 04/21/23
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)
GEF2200 Stordal - based on Durkee 04/21/23
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)
GEF2200 Stordal - based on Durkee 04/21/23
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?
GEF2200 Stordal - based on Durkee 04/21/23
GEF2200 Stordal - based on Durkee 04/21/23
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)?
GEF2200 Stordal - based on Durkee 04/21/23
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