ice in the atmosphere w+h 6.5; s+p ch. 17 start with some terminology –warm clouds = t > 0 ºc...

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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

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Page 1: 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

Ice in the AtmosphereW+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

Page 2: 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

Zone of ice clouds(Using global mean profile)

T

z

15 ºC

2 km

Warm zone

= 6.5 K/km

~8 kmCold and glaciated

Cold zone

0 ºC

-40 ºC

Page 3: 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

Frequency of occurrence

% Containing some iceState of cold clouds vs. temp

Continental

Marine (clean)

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

Page 4: 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

Freezing requires nucleationTo 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

Page 5: 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

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

Page 6: 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

Clausius Clapyron & Phase Diagram

2

)(

RT

pML

dT

Tdp SWs

General Equation:

L = LS for sublimationL = 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:

Page 7: 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
Page 8: 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

Parameterization of IN Concentrations

For CCN, we used power law with respect to supersaturationk

SCsN

%1

)(

where C and k are a best fit to observations, or perhaps based on a power-law 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:

TTaTN IN 1exp)(

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.

iIN bsTTaTN 1exp)(OR

Page 9: 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

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.

Page 10: 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
Page 11: 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

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.

Page 12: 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

Growth by deposition in mixed phase clouds

Consider NC ice crystals of mean diameter DpC, and ND supercooled drops of mean diameter DpD

Each grow by diffusion:

The net impact of deposition to the particles on the ambient vapor concentration is:

Suppose now that we assert steady-state conditions –

This yields

)(2 eqvpvw nnDDMdt

dM

)(2)(2 ,, liqeqvpDvDiceeqvpCvCv nnDDNnnDDNdt

dn

)(2)(20 ,, liqeqvpDvDiceeqvpCvCv nnDDNnnDDNdt

dn

pDDpCC

liqeqpDDiceeqpCCv DNDN

nDNnDNn

)()( ,, pDDpCC

iceeqliqeqpDDpCvW

C

DNDN

nnDNDDM

dt

dM

)(

2 ,,

pDDpCC

iceeqliqeqpCCpDvW

D

DNDN

nnDNDDM

dt

dM

)(

2 ,,

Page 13: 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

Growth by deposition – crystal habits

• Key controlling factors: T, sliq

Page 15: 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