AOSC 620
Cloud Nucleation
Russell Dickerson
2014
Rogers and Yau, Chapt. 6
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
1
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
2
Questions on the Effects of Aerosols
on Clouds and Precipitation
● Why do many people think aerosols inhibit deep convective
cloud formation?
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
3
Opposing Effects of Aerosols on
Clouds and Precipitation
● How do radiative and micro-physical effects of aerosols
compete?
How does suppression of precp change buoyancy?
How does freezing change buoyancy?
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
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Opposing Effects of Aerosols on
Clouds and Precipitation
● How do radiative and micro-physical effects of aerosols
compete?
How does suppression of precp change buoyancy?
negative impact.
How does freezing change buoyancy?
a) if normal precp then freezing enhances
buoyancy.
b) If suppressed precp (too many ccn) then
freezing generates even more buoyancy.
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
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Liquid Water Cloud VDR Yorks et al., 2011
Opposing Effects of aerosols on
Clouds and Precipitation
(Rosenfeld et al., Science 2008)
Radiative Effects:
● Aerosols aloft shield the Earth’s surface from radiation and
stabilize the atmosphere wrt convection and the moisture is
advected away. (Park et al., JGR, 2001; Ramanathan et al., Science, 2001)
● Increased numbers of CCN slow the conversion of droplets
into raindrops and inhibit precipitation, but ingestion of large
particles such as sea salt appears to enhance precip. (Radke et al.,
Science, 1989; Rosenfeld et al., Science, 2002)
● Total water vapor is conserved so suppression of precip here
means more rain there.
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& Z.Q. Li
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The Rain according to Rosenfeld (microphysical effects)
● The extra CCN in hazy air make for more, smaller droplets in the early
stages of a convective cloud.
● The smaller droplets travel higher and more reach colder levels where
they are more likely to release latent heat of freezing and increase
buoyancy – haze means more instability for the same amount of rain.
● Even though aerosols slow the conversion of cloud droplets into rain
drops, convection is eventually invigorated.
● With cold-based clouds (< 0 oC) most of the water is frozen already
and there is no enhancement of precip.
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& Z.Q. Li
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Fig. 2. Evolution of deep convective clouds developing in the pristine (top) and polluted (bottom)
atmosphere
Published by AAAS
D. Rosenfeld et al., Science 321, 1309 -1313 (2008)
Wet (Pseudo-Adiabatic) Parcel Theory (no mixing).
● If all the water in excess of the saturation vapor pressure immediately
condenses and precipitates out, then buoyancy is zero all the way up; this
is the reference for CAPE calculations.
● If all the water is held in the cloud, then buoyancy becomes more
negative with altitude.
● If all the water in excess of the saturation vapor pressure immediately
condenses and freezes at T < – 4oC then buoyancy is enhanced.
● If precip is suppressed until the parcel reaches T = – 4oC then
buoyancy is enhanced further.
The following figure shows an example with the LCL at 960 hPa and
22oC.
Fig. 3. The buoyancy of an unmixed adiabatically raising air parcel
Energy
released
in J kg-1.
no precp.
All precp frozen
suppressed precp.
←Cloud base
Published by AAAS
Who wins – radiation or microphysics?
Particles in the accumulation mode with a diameter around 500
nm are most effective at increasing AOT, but CCN can be
almost any size – it is the number that matters.
Does CCN correlate with AOT?
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& Z.Q. Li
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Fig. 1. Relations between observed aerosol optical thickness at 500 nm and CCN concentrations at
supersaturation of 0.4% from studies where these variables have been measured simultaneously, or
where data from nearby sites at comparable times were available
Published by AAAS
D. Rosenfeld et al., Science 321, 1309 -1313 (2008)
Who wins – radiation or microphysics?
● From this empirical relationship we can estimate the number
of CCN as a function of AOT.
● If the count of CCN is 104 cm-3 then AOT ~ 1.0 and radiation
reaching the Earth’s surface is reduced by an e-folding.
● CAPE reaches a maximum at CCN ~ 1200 cm-3 (AOT ~ 0.25)
; adding more aerosols will inhibit convection.
Bell (GSFC) et al., (JGR, 2008; “Why do tornados and hailstorms
rest on weekends?” 2011) showed a weekday/weekend effect.
Fig. 4. Illustration of the relations between the aerosol microphysical and radiative effects
D. Rosenfeld et al., Science 321, 1309 -1313 (2008)
Published by AAAS
Who wins – radiation or microphysics?
● From this empirical relationship we can estimate the number
of CCN as a function of AOT.
● If the count of CCN is 104 cm-3 then AOT ~ 1.0 and radiation
reaching the Earth’s surface is reduced by an e-folding.
● CAPE reaches a maximum at CCN ~ 1200 cm-3 (AOT ~
0.25) ; adding more aerosols will inhibit convection.
Bell (GSFC) et al., (JGR, 2008) showed a weekday/weekend
effect.
From Rosenfeld and Bell, 2011
Let’s get quantitative; Rogers & Yau, Chapt 6.
Copyright © 2010 R. R. Dickerson & Z.Q. Li
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Phase Change &
Nucleation Process
(inhibited by surface tension)
Condensation
Vapor
Liquid
Evaporation
Deposition
Vapor
Solid
Sublimation
Freezing
Liquid
Solid
Melting
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& Z.Q. Li
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Condensation
• In theory, a cloud droplet may not be
formed until pure water vapor is over
saturated by a few hundreds per cent.
• In nature, super-saturation rate rarely
exceeds a few tenths per cent.
• The reason lies in the presence of plentiful
of water cloud nuclei.
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
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Deposition
• In theory, a cloud droplet may be frozen at a
temperature at 0oC.
• In nature, super-cooled water droplets of
temperature well below the freezing point
are often observed.
• The reason lies in the lack of ice water
cloud nuclei.
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
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The coverage of this lecture
•
•
•
Derivation of equilibrium water vapor pressure
for a small droplet of pure water vs pure bulk
water;
-Homogenous nucleation
Derivation of equilibrium water vapor pressure
for a small droplet of solution water vs pure
water.
-Heterogeneous nucleation
Aerosol and CCN
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
23
Questions to be addressed:
1. How is an embryonic cloud droplet formed
and maintained?
2. Why do cloud droplets have a rather
narrow range in size?
3. How can a cloud exist for certain period of
time?
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& Z.Q. Li
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Homogeneous Nucleation
For a droplet to form by condensation from the vapor, the
surface tension, s, must be overcome by a strong gradient
of vapor pressure.
The Clausius-Claperon equation describes the equilibrium
condition for bulk water and its vapor, which does not
apply to small droplet.
* Surface tension = work required to increase surface area by one unit.
* Store potential energy.
* Volume of liquid tends to assume minimum area-to-volume.
* Small masses  Spherical droplets.
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Take a small particle of radius, r, and divide in half:
Surface tension (s )acting across the plane holds the edges of the sphere together.
Surface tension force per unit length (N/m) is a fundamental property of all liquids
and is relatively high for water.
Force tending to hold the edges together = 2ps r
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For holding together
F1 = 2p s r
For forcing apart
F 2 = p r 2 pi
In equilibrium, F1 = F2
2s
Now pi =
r
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Surface tension causes internal pressure
2s
pi =
r
s = surface tension ( W/A = J m -2 = Nm -1 )
r = radius of droplet (cm)
for H 2 O, s = 0.075 Nm -1
if r = 1m m in radius (10 -4 cm)
p i = 1.5´10+6 dynes/cm 2
= 1.5 Atmospheres
The surface tensions for a solute is lower than that of pure water by
up to one-third, which was attributed to dissolved organics or ions.
Copyright © 2014 R. R. Dickerson
& Z.Q. Li
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Derivation of the Kelvin (1870) Equation
- Curvature effect on saturation
•Surface energy associated with curved surface has
impact on equilibrium vapor pressure and rate of
evaporation.
•Let equilibrium vapor pressure over a flat surface be es.
•And over a curved surface be esr.
•Consider droplet in equilibrium with environment,
temperature = T and vapor pressure = ec
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Specific Gibbs function gives
g = u + pa - Tj
note that g remains constant in an isothermal,
isobaric change of phase.
For the environment
g3 = u 3 + e ca 3 - Tj 3
For the droplet
g 2 = u 2 + (e c + pi )a 2 - Tj 2
Where ec + pi is the total pressure inside the droplet
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
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Note
pi 
2s
r
2s
g 2  U 2  (ec 
) 2  T2
r
Now in equilibriu m g 2  g3
2s
i.e. u 2  (ec 
) 2  T2  u3  ec 3  T3
r
Now conside an isothermal change in phase
But for a droplet of radius r  dr
g  g 2  dg 2  g3  dg3
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& Z.Q. Li
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Remember
dq = du + pdv
2s
dg2 = du2 + da 2 (ec +
)
r
2s
+ a 2 (dec - 2 dr ) - Tdj 2
r
+ j 2 dT
2s
)da 2 = Tdj 2
r
and dT = 0 (isothemal)
But du2 + (ec +
Substituting Tdj 2 into above
2s
dg2 = a 2 (dec - 2 dr)
r
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& Z.Q. Li
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Similar analysis for g 3 gives
dg 3  α3dec
Here
2s
 2(dec  2 dr)   3dec
r
2s
or ( 3   2 )dec   2  2 dr
r
1
But  2 

2s dr
( 3   2 )dec  
 r 2
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In general, specific volume of vapor  3is
much greater th an droplet,  2
Hence formula becomes
2σ dr
α3dec  
 2
 r
and as
RV T
α3 
ec
dec
2s
dr

 2
ec
RvT r
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& Z.Q. Li
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Now integrate equation from a flat plate
(r  , ec  es(T))
To a droplet of radius r (ec  esr )
esr
r
dec
2σ
dr
e ec   RV Tρω  r 2
s

2σ
esr
ln

es(T) RV Tρω r
2σ
es ( r )  es (T) exp (
)
RV Tρω r
Kelvin’s Equation,
R&Y Eq 6.1
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
35
The relative humidity and supersaturation (both with respect
to a plane surface of pure water) for pure water droplets.
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
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An embryonic cloud droplet (molecular cluster) can be formed
by collision of water vapor molecules. Once it exists, it may grow
or decay depending on ambient water vapor pressure.
S = e/es(∞).
e>esr, the droplet tends to grow,
e<esr, the droplet tends to decay.
So, the droplet must be big enough for it to endure.
We will show that the critical radius (S is supersaturation) is:
2s
rc 
Rv  wT ln S
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& Z.Q. Li
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Kelvin Curve
Köhler curve
S* - critical
saturation ratio
r* - critical radius
Haze ←
→ Activated nucleus
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& Z.Q. Li
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Fair Weather Cumulus
Fair weather cumulus
1 pm EST July 7, 2007,
a smoggy day
1.0
0.5
0.0
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If the droplet does not evaporate then the vapor pressure of
the surroundin g environmen t must be supersatur ated
(relative humidity much greater th an 100%)
Not many clouds would form.
Why then do we see clouds?
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& Z.Q. Li
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Köhler Equation - Heterogene ous Nucleation
Most efficient nuclei are hygroscopi c particles –
souble in water. Let us look at growth of a droplet.
Consider a droplet consisting of water and some
dissolved substance.
 ehr as the equilibriu m vapor pressure over
a spherical droplet of this solution.
esr as the equilibriu m vapor pressure over
a spherical droplet of pure water.
 es  equilibriu m vapor pressure over pure plane
water surface.
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we can write
ehr eh ehr
 
es es eh
Where eh  equilibriu m vapor pressure for a plane surface of solution.
Now we have shown that
2σ
esr
 e ρ RV Tr
es
Likewise, for a solution droplet
2σ 
ehr
 e ρRV Tr
eh
σ   surface tension of solution droplet
ρ  density
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We now have to find an equation for
eh
es
•Proportion of surface area occupied by water molecules
n
will be reduced in the approximate ratio
n + n¢
where
n¢ = # of water molecules.
n = # of solute molecules (ions).
•Assumes
(a) Molecules retain their identity.
(b) Cross-sectional area of both molecules are the same.
(c) Molecules are uniformly spread over surface.
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• Qualitativ ely we would expect number of water
molecules escaping surface to go down
– lower equilibriu m vapor pressure.
• eh  es
Raoult’ s law :
eh
n
n 1
*
•
M 
 (1  )
es
n  n
n
M* Mole fraction of water in the solution
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
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we now have
ehr eh ehr
n 1
 
 (1  )  e
es es eh
n
 M* e
2s 
 RV Tr
2s 
 RV Tr
What if molecules of solute dissociate into ions
• n must be replaced by in
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i  # of dissociati ng ions per molecule
in 1
*
• M  (1  )
n
• Note
N M
n o s
ms
n 
No M w
mw
M  mass
m  molecules mass
N o  Avogadr os number
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& Z.Q. Li
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Hence
M s mw
in
i
(
)
n
M w ms
4
3
But M w   r    M s
3
iM s
mw 1
in
 (1  )  (1 

)
3
n
 r 3    M s ms
4
ehr
iM s
mw 1
 (1 

) e
3
es
 r 3    M s ms
4
In a dilute solution M s  M w
  w
s s
Copyright © 2010 R. R. Dickerson
& Z.Q. Li
2s 
 w RV Tr
49
Hence
in 1
in
M  (1  )  1 
n
n
3iM s mw
 1
(
)
3
4 r ms
*
b
 1 3
r
2s
and
Where
ehr
b  w RV Tr
 (1  3 )e
es
r
2s
a
 w RV T
3.3  10 5 o
(
k  cm)
T
3iM s mw
Ms
b
(
)  4.3i
cm 3 mole 1
4w ms
ms
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& Z.Q. Li
50
a
ehr
b
 (1  3 )  e r
es
r
For
T  300 o K
a  1 10  7 cm
r  0.1 10 5  0.1m
a
r
 1 10  2 
a
r
a
e  1
r
Then
ehr
b
a
 (1  3 )(1  )
es
r
r
b
a ab
 1 3   4
r
r
r
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& Z.Q. Li
51
Kelvin Curve
Köhler Curve
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& Z.Q. Li
52
ehr
a b
 1  3
es
r r
See Equation 6.6
Supersatur ation
ehr
Define S  S  1 
1
es
*
a b
  3
r r
r has a critical value given as
1
*
3
b
ds
r *  ( ) 2 where
0
a
dr
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& Z.Q. Li
53
Next lecture will show where
these trends come from.
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& Z.Q. Li
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& Z.Q. Li
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