Cloud Microphysics

1. Microphysics of Clouds

Changes in phase are basic to cloud microphysics. The possible changes are: vapor-liquid evaporation, condensation liquid-solid freezing, melting vapor-solid deposition, sublimation The changes from left to right correspond to increasing molecular order. These transitions don’t occur at thermodynamic equilibrium. There is a strong free energy barrier that must be overcome for droplets to form. To form a droplet by condensation of vapor, surface tension must be overcome by a strong gradient of vapor pressure. This means that phase transitions don’t occur at es(∞), or saturation over a plane surface of water. Even if a sample of moist air is cooled adiabatically to the equilibrium saturation point for bulk water, droplets should not be expected to form. In fact, water droplets do begin to condense in pure water vapor only when the relative humidity reaches several hundred percent! This is called Homogeneous Nucleation.

The reason why cloud droplets are observed to form in the atmosphere when ascending air just reaches equilibrium saturation is because the atmosphere con- tains signiﬁcant concentrations of particles of micron and sub-micron size which have an afﬁnity for water. This is called Heterogeneous Nucleation. There are many types of condensation nuclei in the atmosphere. Some become wetted at RH less than 100% and account for haze. Larger condensation nuclei may grow to cloud droplet size. As moist air is cooled adiabatically and RH is close to 100 the hygroscopic CCN begin to serve as centers of condensation. If ascent continues, there is supersaturation by cooling which is depleted by condensation on the nuclei. Supersaturation is when RH exceeds 100%. Air with relative humidity of 101.5% has a supersaturation of 1.5%. It is an important characteristic of the atmosphere that there are always condensation nuclei present and RH rarely exceeds 100%.

1a. Droplet Growth by Condensation i. Homogeneous nucleation (The curvature effect) In homogeneous nucleation, pure vapor condenses to form droplets. The ﬁrst stage is the chance collision of a number of water molecules in the vapor phase to form small embryonic

1 water droplets.Survival of the droplet is determined by a balance between condensation and evaporation. The equilibrium between the droplet and surrounding vapor is described by the Gibbs free energy, but with one modiﬁcation. In addi- tion to expansion work, a spherical droplet performs work in association with its surface tension σ, which has dimensions of energy/area. σdA represents the work to form the incremental area dA of interface between vapor and liquid.

∆G = dmv(dgv − dgw) + σdA (1) The First term represents the decrease in Gibbs Free Energy due to condensation of a mass of vapor into a mass of liquid water. The second term is the work to create a surface area of the droplet. The ﬁrst term can be expressed as:

dmv = −dmw (2)

= −ρwdVw (3)

where Vw is the volume of the droplet. The difference in the chemical poten- tials (this system is not necessarily in chemical equilibrium), can be expressed in terms of the vapor and saturation vapor pressures. For an isothermal, reversible transformation:

dgv = −sdT + αvdp = αvde (4)

dgw = αwde (5)

where e is the pressure. Subtracting gives

d(µv − µw) = (αv − αw)de ≈ αvde (6) Applying the gas law to an individual molecule of water vapor

eαv = RvT (7)

R T (dg − dg ) = v de = R T dlne (8) v w e v

Integrating from the vapor pressure (e) to the saturation pressure es (at which µv = µw) we have

2 e dgv − dgw = RvT ln (9) es Incorporating into Equation (??) and integrating from a reference state of pure vapor (with vapor pressure maintained) we have e ∆G = −VwρwRvT ln + σdA (10) es Forming a spherical droplet of radius r corresponds to the change of Gibbs free energy

2 4 3 e ∆G = 4πr σ − πr RvT ln ρw (11) 3 es The second term is the change in vapor pressure “bulk term”, it is negative because there is a decrease in the Gibbs free energy when condensation occurs. while the ﬁrst term is the surface tension term. For subsaturation e/es < 1, saturation e/es = 1 and supersaturation e/es > 1. The free energy as a function of droplet radius increases monotonically for subsat- urated and saturated conditions, but a system approaches thermodynamic equilibrium by reducing ist free energy so droplet formation is not favored for e/es ≤ 1 - a droplet formed will spontaneously evaporate. For supersaturated conditions, there is a maximum energy at rc and then the energy decreases and condensation is more efﬁcient than evaporation.

r < rc droplet evaporates back to initial state. r > rc droplet grows spontaneously through condensation of vapor and is said to be activated. The critical radius for a given temperature is:

2σ rc = (12) ρwRvT ln(e/es(∞))

e 2σ = (13) es(∞) RvρwT rc This is Kelvin’s Formula. For critical or equilibrium supersaturation: Above rc the droplet grows spontaneously through condensation. c describes an unstable equilibrium - any perturbation will take the droplet away from this state. However, the requirements for supersaturation are very high (much higher

3 than those observed). An embryonic droplet as large as 0.01 µm still requires a supersaturation of 12% to be sustained. Yet, supersaturations exceeding 1% are rarely observed. Cloud formation cannot be explained by homogeneous nucleation. ii. Heterogeneous nucleation (Solution effect) water condenses onto existing particles of atmospheric aerosol termed cloud condensation nuclei CCN. These particles support condensation at supersaturation values well below those required for homogeneous nucleation - primarily because of their size. Hygroscopic particles, like sodium chloride and ammonium sulfate are even more effective. In the presence of moisture NaCl and (NH4)2SO4 absorb vapor and readily dissolve.The resulting solution has a saturation vapor pressure below that of pure water - because es is proportional to the absolute concentration of water molecules on the surface of the droplet. Consequently, a droplet containing dis- solved salt favors condensation more than would a pure water droplet of the same size.

Saturation pressure over solution Droplets. Over a plane water surface, the reduction in vapor pressure due to the presence of non-volatile solute is:

e0 n n −1 = w = 1 + d (14) es(∞) nw + nd nw 0 e is the equilibrium vapor pressure over a solution with nw molecules of water

4 and nd molecules of solute. For nd << nw: We can express nw = mw/Mw (total mass over molecular weight), and mw = 4 3 ( 3 πr ρw − md) and (md is the mass of the solute). In the case of the solute, what really maters is not the number of molecules of solute but the number of molecules formed in the solution (dissociation).While a similar expression may be obtained for nd, except when the solute dissociates into ions, nd = imd/Md where i is the number of ions per molecule i = 2 for NaCl then

e0 im M −1 im M −1 = 1 + d w = 1 + d w (15) 4 3 es(∞) mwMd ( 3 πr ρw − md)Md for small mass of solute we can approximate this expression as:

e0 b −1 b = 1 + 3 ≈ 1 − 3 (16) es(∞) r r where i3m M b = d w (17) 4πρwMd This is Raoult’s Law. iii. Combining the Solution and Curvature effect: Kohler¨ Curves Describe the equilibrium supersaturations for solutions containing speciﬁed amounts of solute.

0 e (r) b a/r = 1 − 3 exp (18) es(∞) r where 2σ a = (19) ρwRvT A good approximation is:

e0(r) a b = 1 + − 3 (20) es(∞) r r Where the ﬁrst term is the curvature term and the second term is the solution term. Using this expression we can obtain Kohler¨ Curves. 0 For a ﬁxed r, e (r)/es(∞) decreases with increasing solute and eventually be-

5 6 come negative (RH < 100%). At r < rt a droplet is in stable equilibrium - perturbed droplet is restored to original state. - haze At r > rt a droplet is in unstable equilibrium - perturbed droplet is restored to original state.- can enlarge into cloud drops

If we plot the variation of the relative humidity adjacent to a solution droplet as a function of its radius, we obtain what is referred to as a Kohler¨ Curves. Several such curves are shown in the ﬁgure. Below a certain droplet radius, the relative humidity adjacent to a solution droplet is less than that which is in equilibrium with a plane surface of pure water at the same temperature (100%). As the droplet increases in size, the solution becomes weaker, the Kelvin curvature effect becomes the dominant inﬂuence, and eventually the relative humidity of the air adjacent to the droplet becomes essentially the same as the adjacent to a pure water droplet of the same size.

If we analyze solution droplets containing 10−19 kg of NaCl. If it were placed in air with a supersaturation 0.4%, condensation would occur on this particle to form a solution droplet, and the droplet would grow along the red curve. As it does so, the supersaturation adjacent to the droplet is less than the ambient supersaturation. Consequently, the droplet will grow over the peak in its Kohler curve and down the right-hand side of this curve to form a fog or cloud droplet. A droplet that has passed over the peak in its Kohler curve and continues to grow is said to be activated.

−19 Now consider a particle of (NH4)2SO4 with mass 10 kg that is placed in the same ambient supersaturation of 0.4%. In this case, condenseation will occur on the particle and it will grow as a solution droplet along its Kohler curve until it reaches point A. At point A the supersaturation adjacent to the droplet is equal to the ambient supersaturation. If the droplet at A should grow slightly, the supersaturation adjacent to it would increase above the ambient supersaturation, and therefore the drop would evaporate back to point A. If the drop at A should evaporate slightly, the supersaturation adjacent to it would decrease below the ambient supersaturationand the droplet would grow by condensation back to A. Hence the droplet at A is in stable equilibrium with the ambient supersaturation. Droplets in this state are said to be unactivated or haze droplets. Haze droplets in the atmosphere can considerably reduce visibility by scattering light.

7 Continental aerosols are the dominant source of CCNs - sea salt is not com- mon in cloud droplets. Cumulus clouds have droplet number densities of order 102cm−3 over maritime regions and 103cm−3 for continental regions. This implies smaller size of the droplets in continental regions because the liquid water content doesn’t differ much.

2. Aerosols

Radiative transfer is modiﬁed by clouds. Clouds represent cooling in the short- wave (SW) energy budget and warming in the longwave (LW) energy budget. Small particulates are produced and removed through a variety of processes, which make the composition, size and distribution of atmospheric aerosols widely vari- able. There is low density of aerosols over the oceans (marine aerosols) with number density n = 103cm−3 and greater over terrestrial areas (continental) with number densities n > 105cm−3 over industrial areas.

Continental Aerosols 1. Crustal species (subtropical deserts like Sahara and Southwestern US) a > 1µm 2. Combustion (industrialized regions) (submicron-scale particles) 3. Carbonaceous components (tropical regions through agricultural burn- ing) Has a bimodal distribution.

Marine Aerosol Are composed primarily of sea salt. Also has a bimodal distribution.

Stratospheric Aerosol Introduced into the stratosphere through penetrative con- vection and volcanic eruptions. Also by gas-to-particle conversions.

Some facts about aerosols... • The larger the size of a particle , and the greater its solubility, the lower the supersaturation at which the particle can serve as a CCN. (≈ .1µm for water insoluble, while ≈ .01µm for water soluble at 1% supersaturation)

• Number density decreases with height by about a factor of 5 between the PBL and the free troposphere.

8 • There is a diurnal variation in CCN concentrations with a minimum at about 6am and maximum about 6pm - apparently due to gas-to-particle conversions that require solar radiation.

• Soil particles and dust serve as CCN but are not really a dominant source.

2a. Size Distribution of Aerosols Aerosols are characterized by its size distribution expressed as the distribution function nd(D) as a function of the diameter (of equi-volume spheres). nd(D)dD is the number of particles per unit volume whose diameters are between D and D + dD. We can also relate to the volume distribution function nv(v)dv for the number of particles with volumes between v and v + dv.

nd(D)dD = nv(v)dv (21) πD3 because v = 6 , dv π πD3 n (D) = n (v) = D2n (22) d v dD 2 v 6 The number of particles per unit volume with diameter smaller than D is

Z D 0 0 N(D) = nd(D )dD (23) 0 dN where nd(D) = dD . It is also convenient to use the logarithmic scale by deﬁning another disctibution function nl(D) where nl(D)d(logD) is the number of particles per unit volume whose diameters are in the interval d(logD) and dN nl(D) = dlogD . Plots of lognl(D) versus logD for aerosol particles can be ﬁt by stright lines over a limited range of diameters. This implies a pwer law depen- dence of nl on D of the form

−β nl(D) = cD (24) β = 3 is often cited a typical over the diameter range from 10−1 to 10µm. Aerosol populations having this form are said to follow a Junge distribution. We can also write the distributions in terms of surface area or volume. Surface area of particles with diameters smaller than D is given by:

9 Z D 02 0 0 S(D) = πD nd(D )dD (25) 0 The volume of particles with diameters smaller than D is:

Z D π 03 0 0 V (D) = D nd(D )dD (26) 0 6 .1 to 1 µm Accumulation mode - collision as a result of Brownian motion. Most important for cloud formation. Larger particles close to5µm are much fewer in number.

10 to 20 µm Coarse particle mode. Combustion sources or high winds.

.01 µm Nucleation mode

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