METR125: Cloud Microphysics
– Nucleation of water vapor
condensation
Wallace and Hobbs, Sections 6.1 and
6.4
Menglin Jin
Prof. Jin and Prof. Wallace, AMS
Courtesy: Steve Platnick, NASA
Warm Cloud
Processes
Cold Cloud Processes
Rain Drops, Cloud Droplets, and
CCN
• relative sizes of rain drops, cloud drops, and CCN:
– raindrops - 2000 μm = 2 mm
• fall at a speed of 4-5 ms-1
– cloud drops - 20 μm = 0.02 mm
• remain suspended in the air
– CCN - 0.2 μm = 0.0002 mm
• remain suspended in the air
• To get a droplet (20 μm) to grow to raindrop size (2000μm) it must
increase in size by a factor of 100 (two orders of magnitude):
– 2000μm/20μm = 100
• this occurs in about 30 minutes in a thunderstorm!!!
• this is like a 150 lb person growing in size to 15,000 lbs in half an
hour!!!
• Q: How does this happen??
Video: cloud formation in Tucson
• http://www.youtube.com/watch?v=NiCSk1
zxMEs
Timelapse of Tucson cloud formations
Q: How and why do clouds form on some
regions and not on others????
Q: Why does the atmosphere sometimes produce
stratus clouds (thin layered) while other times we get cumulus,
or cumulonimbus clouds to form??
The answer is largely related to the concept of
atmospheric stability.....
Cloud Development - stable environment
Consider this simple situation of a marble in the bottom of a bowl:
if you push the marble up the side of the bowl, it will fall back down
to the bottom, to it's original position
Stable air (parcel) - vertical motion is inhibited
if clouds form, they will be shallow, layered clouds like stratus
Cloud Development - unstable environment
If the marble is on the top of the bowl
and you give it a little push, it rolls off the bowl....
does NOT come back to its original position
This is an unstable situation
Unstable air (parcel) - vertical motion occurs
•commonly produces cumulus, cumulonimbus clouds
So, the question becomes, how does one determine the stability of the atmosphere?
Assessing Atmospheric Stability
– To determine whether or not a parcel will rise
or sink in the atmosphere, one must compare
the parcels temperature (Tp) with that of the
environment (Te) at some altitude:
– if Tp > Te what will the parcel do?
It will rise since it is less dense than the air
– if Tp = Te what will the parcel do?
It will remain at that location
– if Tp < Te what will the parcel do?
IT WILL SINK SINCE DENSER THAN THE AIR
Assessing Atmospheric Stability
to assess stability, what two pieces
of information do we need ?
We need to know
•the vertical temperature profile and
• temperature of the parcel of the air
Determining Air Parcel Temperature:
Rising air parcels and adiabatic cooling
• consider a rising parcel of air -->>
•As the parcel rises,
it will adiabatically expand and cool
•
adiabatic - a process where the parcel temperature
changes due to an expansion or compression,
no heat is added or taken away from the parcel
the parcel expands since the lower pressure outside allows the air molecules to push out on the parcel walls
since it takes energy for the parcel molecules to "push out" on the parcel walls, they use up some of their
internal energy in the process.
therefore, the parcel also cools since temperature is proportional to molecular internal energy
Determining Air Parcel Temperature:
Rising air parcels and adiabatic cooling
• consider a sinking parcel of air -->>
As the parcel sinks, it will adiabatically compress
and warm
• the parcel compresses since it is moving into a
region of higher pressure
• due to the parcel compression, the air molecules
gain internal energy
• hence, the mean temperature of the parcel
increases
Dry adiabatic lapse rate
• As a parcel of air rises, it cools, but at
what rate???
•rate of temperature change with height is
called the lapse rate
•units of lapse rate are °C km-1
First case: an unsaturated parcel of air
unsaturated parcels cool at a rate of 10°C km-1
- this is called the dry-adiabatic lapse rate
What will be the parcel's temperature be at 1 km?
Answer: 30 degrees Celsius
Dry adiabatic lapse rate, cont
• What will be the parcel's temperature be at
2 km?
Answer: 20C
Moist Adiabatic Lapse Rate
For a saturated parcel of air, i.e.,
when its T=Td, then it cools at
the moist adiabatic lapse rate = 6°C km-1
What will be the parcel's temperature be at 3 km?
ANSWER: 14C
Moist Adiabatic Lapse Rate
• What will be the parcel's temperature be at
4 km?
Answer: 8C
Moist Adiabatic Lapse Rate
Why does the parcel cool at a
slower rate (6 °C km-1) when it is saturated
than at 10°C km-1 when it is unsaturated??
Answer: Because latent heat is released into the air parcel as vapor condenses into water
The latent heat counteract the adiabatic cooling.
Dry versus Moist-Adiabatic
Process
• the moist adiabatic lapse rate is less than the dry
adiabatic lapse rate because as vapor condenses into
water (or water freezes into ice) for a saturated parcel,
latent heat is released into the parcel, mitigating the
adiabatic cooling -->
Absolute Stability
Absolute Stability
•
•
consider the diagram to the right, notice that:
Tup < Tsp < Te
– where Tup is the temperature of an unsaturated parcel = 0°C
– Tsp = temperature of a saturated parcel = 10°C
– Te = environmental temperature = 20°C.
•
•
•
•
generally, notice that Te is always larger than Tsp and Tup at any level
Hence, an unsaturated or saturated parcel will always be cooler than the
environment and will sink back down to the ground
This is an example of absolute stability
The condition for absolute stability is:
– Gd>Gm>Ge
•
•
•
•
Gd is the dry adiabatic lapse rate (10°C km-1)
Gm is the moist adiabatic lapse rate (6°C km-1)
Ge is the environmental lapse rate (variable - 0°C km-1 in this case)
Q: How would you characterize the stability of an inversion layer?
Stability of Inversion Layers
How would you characterize the stability of an inversion layer?
They are absolutely stable
see diagram to the right -->>
note that the absolute stability criteria:
Ge<Gm<Gd
Atmospheric Instability and
Cloud Development
What determines the base (bottom) of a cloud??
Q: What determines the height to which the cloud
will grow??
let's use the previous example of a rising air parcel -->>
Q: On this diagram, where is cloud base?
Q: On this diagram, where is cloud top?
LFC – level of free convection where in atmosphere
The environment temperature decreases faster
Than the moist adiabatic lapse rate
LCL –Listed Convection Level
Atmospheric Instability and
Cloud Development
•
•
•
•
•
Q: On this diagram, where is cloud base?
A: Where the parcel reaches saturation - 2 km
Q: On this diagram, where is cloud top?
A: Where the parcel will no longer be able to rise - 9 km
Here, Tp = Te - this is often referred to as the equilibrium level (EL)
Example Microphysical Measurements in
Marine Sc Clouds (ASTEX field campaign, near Azores, 1992)
Data from U.
Washington C-131
aircraft
PHYS 622 - Clouds, spring ‘04, lect.2, Platnick
Cloud Droplet Formation
S. Platnick notes
Water Cloud Formation
Water clouds form when RH slightly greater than 100% (e.g., 0.3%
supersaturation). This is a result of a subset of the atmospheric aerosol
serving as nucleation sites (to be discussed later).
Common ways for exceed saturation:
1. Mixing of air masses (warm moist with cool air)
2. Cooling via parcel expansion (adiabatic)
3. Radiative cooling (e.g. ground fog, can lead to process 2)
PHYS 622 - Clouds, spring ‘04, lect. 2, Platnick
Concepts
es(T)
(T1,e1)
e
Radiative
Cooling
saturated
Mixing
(T2,e2)
unsaturated
T
PHYS 622 - Clouds, spring ‘04, lect. 2, Platnick
q, w, e, T of the mixed air
• See textbook and notes
• q= M1
M1+M2
W=
e=
T=
q1 +
M2
M1+M2
q2
Saturation Vapor Pressure (Clausius-Clapeyron equation)
At equilibrium, evaporation and condensation have the
same rate, and the air above the liquid is saturated
with water vapor; the partial pressure of water vapor, or
the Saturation Vapor Pressure (es) is:

es (T)  es Ttr  e
Air and
water vapor
T
T Water
L 1 1
( 
)
R v T Ttr
Where Ts=triple point temperature (273.16K), L is the latent heat of
vaporization (2.5106 J/kg), es(Ttr) = 611Pa (or 6.11 mb). Rv is the
gas constant for water vapor (461.5 J-kg1-K1).
specific
Platnick
Saturation Vapor Pressure
An approximation for the saturation vapor pressure
(Rogers & Yau):

e s (T )  Ae
Over liquid water:
L = latent heat of vaporization/condensation,
A=2.53 x 108 kPa, B = 5.42 x 103 K.
Over ice:
L = latent heat of sublimation,
A=3.41 x 109 kPa, B = 6.13 x
103 K.
Platnick
B
T
Rain Drops, Cloud Droplets, and
CCN
Processes for Cloud Droplet
Growth
• How does this happen??
• By:
– condensation
– collision/coalescence
– ice-crystal process
today
Water Droplet Growth
Condensation & Collision
•
Condensational growth: diffusion of vapor to droplet
•
Collisional growth: collision and coalescence (accretion, coagulation)
between droplets
PHYS 622 - Clouds, spring ‘04, lect.4, Platnick
Water Droplet Growth - Condensation
Flux of vapor to droplet (schematic shows “net flux” of vapor towards
droplet, i.e., droplet grows)
Need to consider:
1.
Vapor flux due to gradient between saturation vapor pressure at droplet
surface and environment (at ∞).
2.
Effect of Latent heat effecting droplet saturation vapor pressure
(equilibrium temperature accounting for heat flux away from droplet).
PHYS 622 - Clouds, spring ‘04, lect.4, Platnick
Cloud Droplet Growth by
Condensation
• Consider pure water in equilibrium with air
above it
C-C equation to calculate es
Growth by Condensation Cloud
Droplet
Consider pure water in equilibrium with air above it:
• then the RH = 100%
• evaporation = condensation
• vapor pressure (e) = saturation vapor pressure (es)
• if evaporation > condensation, water is _________
• if evaporation < condensation, water is ________
• Now, a droplet surface is not flat, instead, it has
curvature.....
Q: how does curvature affect the evaporation/condensation
process??
Flat versus Curved Water
Surfaces
Flat versus Curved Water
Surfaces: curvature effect
•
•
•
•
•
•
•
•
more energy is required to maintain the "curvature" of the drop
therefore, the water molecules on the surface of the drop have more energy
therefore, they evaporate more readily that from the flat water surface
(compare the length of the red arrows)
therefore: evaporation rate off curved surface > evaporation rate off of flat
surface
since air above both surfaces is saturated, then
evaporation rate = condensation rate
therefore, condensation rate onto droplet > condensation rate onto flat water
surface
therefore, esdrop > esflat
therefore:
– if RHflat = 100%, then RHdrop > 100%
– the air surrounding the drop must be supersaturated!!
•
This is called the curvature effect
Kelvin’s Equation
Curvature Effect
Curvature effect -->
•notice that for the droplet to be in equilibrium
(evaporation off drop = condensation onto drop),
the environment must be supersaturated
•also notice that the curvature effect
is larger for smaller drops
this makes sense since smaller drops
have more curvature that larger drops
Discuss: see text book
Class activity-Curvature Effect
• Q: what will happen to
a drop 1.9 μm in size
that is in a cloud
where the RH is
100.05%?
• Q: what will happen to
a drop 1.9 μm in size
that is in a cloud
where the RH is
100.15%?
Class activity-Curvature Effect
• Q: what will happen to
a drop 1.9 μm in size
that is in a cloud
where the RH is
100.05%?
• Q: what will happen to
a drop 1.9 μm in size
that is in a cloud
where the RH is
100.15%?
QUESTIONS FOR THOUGHT:
1. At what relative humidity will pure water
droplets of the following sizes grow by
condensation:
a. 10 microns
b. 4 microns
c. 1 micron
2. Explain why very small cloud droplets of
pure water evaporate even when the
relative humidity is 100%.
Solution Droplets
Note that the previous discussion
is valid for a pure water drop
• if a droplet is comprised of a
solution - it can be in
equilibrium with the
environment at a much lower
RH -->
• this explains the formation of
haze
• This process of condensation
will grow drops , but not to
precipitation sizes (~ 2 mm)
Q: So, if a droplet grows to some
size by condensation, how can
it continue to grow to
precipitation size???
QUESTION FOR THOUGHT:
• Haze particles can form when the relative
humidity is less than 100%. Are these
haze particles pure water droplets or
solution droplets? Why?
Köhler Curve
How is amount of solution
change water droplet formation?
Give example.
How different solution change
water droplet formation?
Give example.
Extra reading about curvature and
solutions effects!!!
• http://www.shodor.org/os411/courses/411c/
module07/unit02/page03.html
Droplet activated
• Find it in text book P214
Water Droplet Growth - Condensation
Growth slows down with increasing droplet size:
large droplets :
G s 
dr
~ env
dt
r
Since large droplets grow slower, there is a narrowing of the size distribution
with time.
R&Y, p. 111
PHYS 622 - Clouds, spring ‘04, lect.4, Platnick
rdry= 0.1
0.22
0.48 m
Assumes supersaturation=0.05%, p=900 hPa, T=273K
Water Droplet Growth - Condensation
Diffusional growth summary:
• Accounted for vapor and thermal fluxes to/away from droplet.
• Growth slows down as droplets get larger, size distribution narrows.
• Initial nucleated droplet size distribution depends on CCN spectrum & ds/dt
seen by air parcel.
• Inefficient mechanism for generating large precipitation sized cloud drops
(requires hours). Condensation does not account for precipitation
(collision/coalescence is the needed for “warm” clouds - to be discussed).