Generating cloud drops from CCN Wallace & Hobbs (1977)
Cloud Drops and Equilibrium Considera3ons: review We discussed how to compute the equilibrium vapor pressure over a pure water drop, or a solu3on droplet, as a func3on of composi3on (which is in turn dependent on the amount and nature of soluble material in the CCN) and curvature This led us to an equa3on for the Köhler curve, which expresses vapor pressure as a func3on of the droplet size Non- unique rela3onship (above S=1) if generated for a solu3on droplet, except at the cri3cal diameter (s c ) Unique rela3onship if computed for a pure water drop (or insoluble, welable par3cle) How does the drop move along the x- axis (i.e., gain or lose water)? Condensa3on / evapora3on Direc(on is dictated by equilibrium (system wants to move toward equilibrium) But the rate of movement is no longer in the realm of equilibrium thermo, but closer to kine3cs
Droplet Growth By Condensation From our Kohler Curve discussions, we have seen that an activated droplet will grow by vapor condensing on its surface, provided the vapor pressure of the ambient air exceeds the vapor pressure adjacent to (over, or at the surface of) the droplet. For droplets exceeding ~1 µm in radius, the curvature term is ~negligible and hence the droplet behaves nearly as a flat surface of water (no curvature). Droplet Growth --------------- Ambient air is above water saturation Droplet Evaporation ---------- Ambient air is below water saturation Our task here is to derive an expression for the rate that a small droplet grows by vapor condensing on its surface traditionally, the rate is written either in terms of the rate of change of the drop radius or of the drop s water mass
Consider the following situation: r R Droplet of mass m and radius r is embedded within a steady-state field of water vapor To calculate the droplet growth rate by condensation, surround this droplet with an imaginary sphere of radius R, write a mass balance, and use the Fickian Diffusion Law to compute the flux of water vapor through the imaginary spherical surface, towards the droplet s surface. Droplet density Drop mass For steady-state conditions and no storage (accumulation) of water vapor in the region around the droplet, this vapor flux must be equivalent to the growth rate of the droplet. The unsteady-state diffusion of species A (here, water) to the surface of a stationary particle of radius R p is described by c ( ) t = 1 r 2 r r2 J A,r where c(r,t) is the molar concentration of A (moles vol -1 ), and J A,r is the flux of A (units: moles area -1 time -1 ) at any radial position r. (The equation arises from a mass balance in a spherical shell around the particle.) Fick s Law for our problem can be well approximated by J A,r = D V dc dr c t = x J A,x ( ) = x D V c x D V is the diffusivity of vapor in air (Cartesian)
Combining, we get the governing equation, c t = D 2 c V r + 2 2 r c = D V 2 c r We need one initial condition, and two boundary conditions, to solve this. We use c(r,0) = c c(,t) = c c(r p,t) = c s We can find a solution for this full problem (see Chapter 11 of the text by Seinfeld and Pandis), and putting in typical values for the kinds of problems we ll be trying to solve, we find out that the transient is very short. So it is valid to assume we get to a steadystate situation very quickly, and that we can set the time derivative to zero. The steady-state solution is, c(r) c c s c = R p r (*) What happens as r = R p and as r gets very large? The total flow of A (moles time -1 ) toward the particle is J = 4πR p 2 (J A ) r= R p So, take the derivative of the solution (*), plug into the equation for J A,r, and find Maxwellian flux (1877) J = 4πR p D V (c c s ) The transferred moles time -1 increase as R p increases, and are proportional to the difference between the surface concentration and the far-field concentration
Now write a mass (or here, mole) balance on the growing (or evaporating) droplet: ρ l M w d dt 4 3 πr 3 p = J = 4πR p D V (c c s ) This yields an equation for the time rate of change of the drop radius, dr p dt = 1 R p D V M w ρ l (c c s ) Which we could also express in terms of the vapor density (mass vol -1 ) dr p dt = 1 R p D V ρ l (ρ ρ s ) Or as a rate of change of the droplet mass, dm dt = 4πR pd V (ρ w, ρ eq w ) ρ and ρ w, are the same thing; w used to indicate water vapor here; eq means the equilibrium value over the drop The diffusivity of water vapor can be expressed as D V = 0.211 p T 273 1.94 D V in units of cm 2 s -1, T in K, p in atm
For very small droplets, we should modify the droplet growth equation to include noncontinuum effects. We can accomplish this by defining a modified diffusivity, that here includes a water accommodation coefficient αc: DV ʹ = DV D 2πM w 1+ V α C R p RT 1 2 The figure below shows the impact of this correction. For αc=1, the correction is less than 25% for drops larger than 1 µm and less than 5% for those > 5 µm. Recommended values, αc: 0.045 (P & K) 1 (many authors) See recent work by R. Shaw Fig 15.13, Seinfeld and Pandis
Go back to the equations for the rates of change or drop mass or radius: dr p dt = 1 R p D V ρ l (ρ w, ρ eq w ) dm dt = 4πR pd V (ρ w, ρ eq w ) Both rates are proportional to the difference between the environmental vapor pressure and that over the droplet The rate of change of the radius with time is inversely proportional to the radius The radii of the smallest drops grow (or evaporate) the fastest The rate of change of the droplet mass is proportional to the radius The masses of the largest drops increase (or decrease) the fastest Also, there is something a bit strange about these final equations. We derived them initially for a fixed radius R p (using a time-independent steady-state profile around the drop). But now we have developed an equation for how R p changes with time! This treatment implies that the vapor concentration profile near the drop achieves steady state before appreciable growth occurs. Diffusion is much faster than the time rate of change of R p, so we can assume that the adjustment of the vapor profile is very fast compared with the changes brought about by the rate of the drop growth.
The R-squared law arises if one can assume that can easily integrate the equation ρ w,, ρ w eq are constant. Then we dr p dt = 1 R p D V ρ l (ρ w, ρ eq w ) to get R 2 2 p = R p,0 R p R p,0 2 + 2D V ρ l (ρ w, ρ eq w )t =1+ (const) t =1+ τ How does the characteristic time scale with R p,0? What does this mean for growth rates of different-sized drops?
The growth rate is not strictly given by the equations we just derived, since heat is released at the droplet s surface owing to release of latent heat of condensation, L c The heat liberated will cause the droplet to warm, thereby increasing the vapor pressure adjacent to the droplet, reducing the vapor pressure gradient and therefore reducing the growth rate. (ρ ρ s ) = 1 e e V,s R V T T s e V,s with heating, therefore gradient is reduced. The heat can be released either toward the particle or toward the exterior gas phase. As mass transfer continues, the particle surface temperature changes until the rate of heat transfer balances the rate of heat generation (or consumption, if evaporation). The formation of the external T and vapor conc. profiles must be related by a steady-state energy balance, to determine the steady-state surface temperature at all times. The droplet growth rate is thus a function of the rate that heat can be conducted away from the droplet by heat conduction (the energy analogy to vapor diffusion). To derive an expression for the change in temperature, construct a balance between heat released by condensation and conduction of heat away from the droplet: heat conduction term rate that latent heat is released Please note here ρ=ρ and ρ r is at the surface (called ρ s above)
By analogy to diffusion, can be written as, Where is the thermal conductivity of air. is the radial temperature gradient. (-) sign is needed since heat flow in opposite sense to gradient. Integrating (6), For steady-state conditions, Equation (7) reduces to the condition, vapor & thermal fields around droplet satisfy this relationship particle wet bulb temperature
The next step is to combine equation (7) and the drp/dt eqn with the Clausius Clapeyron equation and the Ideal Gas Law to write in terms of the supersaturation of the environment. After much algebra, Assumes pure water, no curvature molecular weight of water k All other variables have been previously defined is the saturation ratio of the environment. All of the Ts in this equation refer to the environmental temperature. This equation is written in the simpler form as, Please also see pages 801-805 of the Seinfeld and Pandis text for a complete, and clear, derivation of this growth equation
For droplets in radius, the curvature and solute effects must be considered, mass of salt We could also write the numerator as S a w exp(kelvin) molecular weight of salt Van t Hoff factor molecular weight of water Environmental saturation ratio saturation ratio over the droplet Condensational growth implies, which means that r 2 r 1 r 0 time 10 hours Condensational growth alone would tend to produce a monodisperse cloud droplet size distribution
growth rate small dm/dt is proportional to r growth rate large Houghton, 1985 ~20s ~40s
Evaporation is described by the same equation as condensation (but opposite sign for the driving force, of course), so we should also expect smaller drops to evaporate faster than larger ones: Distance a drop falls before evaporating, assuming isothermal atmosphere and constant S (Table 7.3, Rogers and Yau) Initial radius (microns) Distance fallen 1 2 µm 3 0.17 mm 10 2.1 cm 30 1.69 m 0.1 mm 208 m 0.15 mm 1.05 km
Rate of growth droplets by condensation on salt nuclei (after Best 1951b) Nuclear mass m Supersaturation=100 (S-1)% Radius (µm) Temperature T=273K 10-14 g 0.05 1. Growth rate (initially) faster for larger initial salt mass (solute effect) 2. For a given salt mass, the growth rate decreases with increasing size. Pressure=900mb 3. Conclude that diffusional growth alone cannot produce precipitation sized drops within reasonable cloud lifetimes. 10-13 g 0.05 Time (s) to grow from initial radius of 0.75µm 10-12 g 0.05 1 2.4 0.15 0.013 2 130 7.0 0.61 5 1000 320 62 10 2700 1800 870 15 5200 4200 2900 20 8500 7400 5900 25 12500 11500 9700 30 17500 16000 14500 35 23000 22000 20000 50 44500 43500 41500 r 0 r 1 r 2 Condensational growth tends to produce uniform size distribution r 0 r 1 r 2
Consider the growth of a population of droplets In clouds many droplets grow at the same time (on activated CCN). Typical concentrations may be a few hundred per cm 3 in maritime clouds and ~500-700 cm -3 for continental clouds. These droplets compete for available H 2 O vapor made available by condensation associated with the rising air parcels. This of course creates a supersaturation which is reduced by condensational growth on the cloud droplets. Time rate of change of supersaturation, Net rate of change = rate of production of supersaturation rate of consumption (by condensation) supersaturation condensational growth rate saturation mixing ratio; updraft velocity where is the droplet concentration, assumed to be
INITIAL RADII α salt mass radii 0 6 60 MONODISPERSE Activated droplets Increasing salt mass non-activated droplets 1) Max supersaturation achieved few tens of meters above cloud base cloud droplet concentration determined 2) drop growth maximum around S max 3) smallest particles become haze droplets 4) Activated droplets become monodisperse in size