METR125: Cloud Microphysics
– grow by condensation
Menglin Jin
For advance knowledge: see Wallace and Hobbs, Sections 6.1.1 and 6.4.1
Cold Cloud Processes
Homogeneous Nucleation of Droplets;
Kelvin’s Equation
Cloud Condensation Nuclei.
Warm Clouds.
Growth of Drops by Condensation
Atmospheric Aerosols
Heterogeneous Nucleation of Droplets;
Köhler Curves
Warm Cloud
Processes
Growth of Drops by Collisions.
Courtesy: Steve Platnick, NASA
Ice Nuclei and Ice Crystal in Clouds
Growth of Ice Particles in Clouds
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
• Or: Volume of rain drop = 106 volume of cloud droplet
• 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??
Processes for Cloud Droplet
Growth
•
•
•
•
Natural Condensation -- Condensation commences at RH=100% when air is
cooled. (Would take 24 hours for a droplet to grow to the size of a small raindrop)
Condensation on condensation Nuclei -- Condensation commences with the
presence of Cloud Condensation Nuclei at RH=80%, but droplets only grow rapidly
at RH=100%. (Still takes over 10 hours for droplets to grow to the size of a small
raindrop.)
Collision and Coalescence -- Droplets grow by accretion when they strike other
small droplets. Chief process in the clouds vertical development, but also with
altocumulus and cirrocumulus. Droplet growth is rapid - the greater the volume of the
cloud the more turbulent the flow in that cloud. Best developed in cumulonimbus
clouds.
Ice Crystal (Three Phase) Process -- When RH<100% and water droplets occur
simultaneously with ice crystals (between 0 and -20C) water droplets evaporate, but
ice crystals can continue to grow. Ice crystals resist evaporation (actually
sublimation). Ice crystals can continue to grow as random collisions between water
vapor molecules leaving the evaporating water droplets collide with the ice crystals.
Eventually the water droplets disappear, but the ice crystals grow to such a size that
they fall out through the freezing level and arrive at the ground as small to moderate
sized rain drops.
Processes for Cloud Droplet
Growth
• How does this happen??
• By:
– condensation
– collision/coalescence
– ice-crystal process
today
Video: cloud formation in Tucson
• http://www.youtube.com/watch?v=NiCSk1
zxMEs
Timelapse of Tucson cloud formations
REVIEW
Cloud Droplet Formation
REVIEW
Clouds Formation
Clouds are formed when air containing water vapor
is cooled below a critical temperature called
the dew point and the resulting moisture condenses
into droplets on microscopic dust particles
(condensation nuclei) in the atmosphere.
CLOUDS: A visible mass of liquid/solid water droplets suspended
in the atmosphere above Earth's surface.
http://earthobservatory.nasa.gov/Library/glossary.php3
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 Ttr=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
Another form:
Saturation Vapor Pressure (Clausius-Clapeyron equation)
Another form:
where es is in units of Pa and T is in units of C
to show that as the temperature
increases, es_________
nearly exponentially.
This implies that the atmosphere
is able to hold _____ water
at a higher temperature.
Chow et al., 1988)
Water Cloud Formation
Water clouds form when RH slightly greater than 100% (e.g., 0.3%
supersaturation).
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)
T
unsaturated
q, w, e, T of the mixed air
• See handout
• q= M1
M1+M2
q1 +
M2
M1+M2
q2
W=
e=
T=
Class Participation: Air parcel A: 20g, 10C, 10g/kg
Air Parcel B: 40g, 30C, 20g/kg, What is mixed parcel’s
Important Properties:
The saturation vapor pressure above ice crystals < than
When both Ice Crystal and water
droplet present, which one will
grow?
•
When the air becomes saturated with respect to water,
evaporation = condensation
•
When the air becomes saturated with respect
to water it is supersaturated with respect to ice.
Thus, the ice collects more water molecules
than its loses by sublimation.
Class participation
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
Theory
• Homogeneous (spontaneous) nucleation
(cont.)
– Recall: a system (droplet + environment)
approaches an equilibrium state by reducing
its energy (DE<0) in time
Theory
• Subsaturated conditions (e < es)
If droplet grows (R increases), then DE>0, this won’t happen spontaneously.
Theory
• Subsaturated conditions (e < es)
– Formation of droplets is not favored
– Random collisions of water molecules do
occur, forming very small embryonic droplets
(that evaporate)
– These droplets never grow large enough to
become visible
Theory
• Supersaturated conditions (e > es)
If droplet grows (R increases), then DE can be positive or negative
Theory
• Supersaturated conditions (e > es)
 DE initially increases with increasing R
 DE is a maximum where R = r
 DE decreases with increasing R beyond R = r
Theory
• Supersaturated conditions (e > es)
– Embryonic droplets with R < r tend to
evaporate
– Droplets which grow by chance (collisions)
with R > r will continue to grow spontaneously
by condensation
• They will cause a decrease in the energy (total
energy) of the system
Theory
• Kelvin’s formula can be used to
– calculate the radius r of a droplet which will be
in (unstable) equilibrium with air with a given
water vapor pressure e
– determine the saturation vapor pressure e
over a droplet of specified radius r
Theory
• Kelvin’s formula can be used to
– calculate the radius r of a droplet which will be
in (unstable) equilibrium with air with a given
water vapor pressure e
– determine the saturation vapor pressure e
over a droplet of specified radius r
• r = 0.01 micrometers requires a RH of 112.5%
• r = 1.0 micrometer requires a RH of 100.12%
Discuss: see text book
Diffusion Process
Handout: http://www.shodor.org/os411/courses/411c/module07/unit02/page03.html
• Two phenomena which influence the
growth that occurs by diffusion are the
curvature
solution
_________
effect and the _______
effect.
Droplets, by nature, are _______.
The curvature of a droplet
tends to _______ the concentration of vapor
at the surface of the droplet.
Small droplets have ______ curvature than larger droplets.
In general, given identical atmospheric conditions,
a smaller droplet will have a ________ concentration
of water vapor at its surface than a larger droplet.
Since diffusion is the movement from ______ concentration
to _____ concentrations,
the curvature effect tends to _______ droplet
growth by diffusion.
As a droplet grows, its curvature ______ and becomes
more like a plane surface and the influence of
the curvature effect ________ as well.
Droplets, by nature, are round.
The curvature of a droplet
tends to increase the concentration of vapor
at the surface of the droplet.
Small droplets have more curvature than larger droplets.
In general, given identical atmospheric conditions,
a smaller droplet will have a greater concentration
of water vapor at its surface than a larger droplet.
Since diffusion is the movement from higher concentrations
to lower concentrations,
the curvature effect tends to retard droplet
growth by diffusion.
As a droplet grows, its curvature decreases and becomes
more like a plane surface and the influence of
the curvature effect decreases as well.
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
If the relative humidity decreases,
the droplet will evaporate until it reaches equilibrium.
A droplet that manages to grow to a diameter
of about 20 micrometers will start to grow by
collision and coalescence.
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.
• See Lecture AEROSOLS
Droplet activated
• Find it in text book P214
Water Droplet Growth - Condensation
FYI
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).
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