Ice in the Atmosphere
W+H 6.5; S+P Ch. 17
• Start with some terminology
– Warm clouds = T > 0 ºC (= 273.15 K)
– Cold clouds = T < 0 ºC
• Cold clouds may or may not contain ice!
• Liquid drops below 0 ºC -- supercooled
– crystals and drops = mixed phase
– Ice only = glaciated
Zone of ice clouds
(Using global mean profile)
z
G = 6.5 K/km
Cold and
glaciated
~8 km
-40 ºC
Cold zone
2 km
0 ºC
Warm zone
T
15 ºC
Frequency of occurrence
How does it get to be this way?
• Clouds forming in cold conditions are ice clouds
• Clouds forming in warm conditions, then transported to cold conditions can
contain supercooled water
State of cold clouds vs. temp
% Containing some ice
Marine
(clean)
Continental
Freezing requires nucleation
To create an ice crystal, a surface must
be created. This requires a
“nucleation” event, just as formation of
liquid drops in saturated air requires
nucleation.
Homogeneous freezing of liquid water
occurs at ~-40 ºC
If an aerosol in the air or the liquid drop
has a crystal matrix similar to ice, it can
cause heterogeneous freezing between
0 ºC and -40 ºC.
AgI is a classic heterogeneous nucleus
Method of Ice Nucleation
• Homogeneous Freezing of liquid water
drops – No solids in supercooled drop
• Heterogeneous Nucleation – Freezing
mediated by an ice nucleus (IN)
– Types of Heterogeneous Ice Nucleation
• Freezing Nucleus – Impurity in a drop
• Contact Nucleus – IN in air contacts a drop + starts
freezing
• Deposition Nucleus – Supersaturation in air
w/respect to ice deposits on an aerosol particle
Clausius Clapyron & Phase Diagram
General Equation:
dps (T ) L M W pS

dT
RT 2
L = LS for sublimation
L = LV for vaporization
Since LS > LV, supercooled H2O is
always supersaturated with respect
to ice
When ambient vapor pressure is above both curves, vapor condenses to both ice
and supercooled drops. When ambient vapor falls between the two curves, drops
evaporate, and crystals grow.
Note: Some IN work better when liquid deposits first, and then freezes:
Parameterization of IN Concentrations
For CCN, we used power law with respect to supersaturation
 S 
N ( s)  C 

 1% 
k
where C and k are a best fit to observations, or perhaps based on a powerlaw aerosol size distribution
For IN, we base things on freezing temperature instead of supersaturation.
and for which NIN is measured in particles per liter (i.e. 10-3 cm-3) using:
N IN (T )  expaT1  T 
OR
N IN (T )  expaT1  T   bsi 
The warmest freezing temperatures (-4 ºC) tend to be primary organics (leaf
debris, plankton, etc). Certain types of mica dust are good in the -10 to -20
ºC range.
The Freezing Process
• Latent heat is released during droplet freezing.
– The droplet is warmer than surroundings during freezing
– Pronounced for supercooled drops suddenly freezing.
• Lf = 3.3x105 J/kg;
• Cpw = 4.2x103 J/kg/K
– Lf/Cpw = 78 ºC!
• This means a supercooled droplet would have to be -78 ºC at the
beginning of the freezing process for it to completely freeze after
nucleation.
• If surface of drop freezes prior to interior, it may explode
(ice multiplication)… freezing is an expansive process
– This may explain the observation that observed ice crystal
densities far exceed the observed IN concentrations
– Another explanation is the potential that H2SO4 crystals may
survive convection and act as IN
– A third is that fragile ice crystals may break up.
Growth of ice crystals
• Growth by deposition:
– Produces a number of crystalline forms, based on the
crystal’s history of temperature and supersaturation.
• Plate-like vs. Column-like
• http://www.its.caltech.edu/~atomic/snowcrystals/primer/prime
r.htm
• Growth by riming
– Ice crystals collide with supercooled drops when
falling, and get coated with rapidly frozen drops.
Growth by deposition in mixed phase clouds
Consider NC ice crystals of mean diameter DpC, and ND
supercooled drops of mean diameter DpD
dM
 M w 2Dv D p (nv  neq )
Each grow by diffusion:
dt
The net impact of deposition to the particles on the ambient vapor
concentration is:
dnv
  N C 2Dv D pC (nv  neq,ice )  N D 2Dv D pD (nv  neq,liq )
dt
Suppose now that we assert steady-state conditions –
dnv
 0   N C 2Dv D pC (nv  neq,ice )  N D 2Dv D pD (nv  neq,liq )
dt
N D DpD (neq,liq  neq,ice )
dMC
This yields
 MW 2Dv DpC
dt
NC DpC  N D DpD
NC DpC (neq,ice )  N D DpD (neq,liq )
nv 
NC DpC  N D DpD
NC DpC (neq,liq  neq,ice )
dMD
 MW 2Dv DpD
dt
NC DpC  N D DpD
Growth by deposition – crystal habits
• Key controlling factors: T, sliq
Guide to
snowflakes
Kenneth Librecht
http://www.its.caltech.edu/~at
omic/snowcrystals/class/class
.htm