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  (sc)  

•  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

Droplet density

Drop mass

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.

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 Rp is described by

where c(r,t) is the molar concentration of A (moles vol-1), and JA,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 €

∂c∂t

= −1r2

∂∂r

r2JA ,r( )

€

JA ,r = −DVdcdr

DV is the diffusivity of vapor in air

€

∂c∂t

= −∂∂x

JA ,x( ) =∂∂x

DV∂c∂x

⎛

⎝ ⎜

⎞

⎠ ⎟ (Cartesian)

Combining, we get the governing equation,

We need one initial condition, and two boundary conditions, to solve this. We use

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 steady-state situation very quickly, and that we can set the time derivative to zero. The steady-state solution is,

(*)

The total flow of A (moles time-1) toward the particle is

So, take the derivative of the solution (*), plug into the equation for JA,r, and find

€

∂c∂t

= DV∂ 2c∂r2

+2r∂c∂r

⎛

⎝ ⎜

⎞

⎠ ⎟ = DV∇

2c

€

c(r,0) = c∞c(∞,t) = c∞c(Rp,t) = cs

€

c(r) − c∞cs − c∞

=Rp

r

€

J = 4πRp2 (JA )r=Rp

€

J = 4πRpDV (c∞ − cs)

What happens as r = Rp and as r gets very large?

The transferred moles time-1 increase as Rp increases, and are proportional to the difference between the surface concentration and the far-field concentration “Maxwellian flux” (1877)

Now write a mass (or here, mole) balance on the growing (or evaporating) droplet:

This yields an equation for the time rate of change of the drop radius,

Which we could also express in terms of the vapor density (mass vol-1)

Or as a rate of change of the droplet mass,

The diffusivity of water vapor can be expressed as

€

ρlMw

ddt

43πRp

3⎛

⎝ ⎜

⎞

⎠ ⎟ = J = 4πRpDV (c∞ − cs)

€

dRp

dt=1Rp

DVMw

ρl(c∞ − cs)

€

dRp

dt=1Rp

DV

ρl(ρ∞ − ρs)

€

dmdt

= 4πRpDV (ρw,∞ − ρweq )

ρ∞ and ρw,∞ are the same thing; w used to indicate water vapor here; eq means the equilibrium value over the drop

€

DV =0.211p

T273⎛

⎝ ⎜

⎞

⎠ ⎟ 1.94

DV in units of cm2 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:

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.

Fig 15.13, Seinfeld and Pandis

€

DVʹ′ =

DV

1+DV

αCRp

2πMw

RT⎛

⎝ ⎜

⎞

⎠ ⎟

12

Recommended values, αc:

0.045 (P & K) 1 (many authors)

See recent work by R. Shaw

Go back to the equations for the rates of change or drop mass or radius:

€

dRp

dt=1Rp

DV

ρl(ρw,∞ − ρw

eq )

€

dmdt

= 4πRpDV (ρw,∞ − ρweq )

Both rates are proportional to the difference between the environmental vapor pressure and that

over the droplet

The rate of change of the droplet mass is proportional to the radius

The masses of the largest drops increase (or decrease) the fastest

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

Also, there is something a bit strange about these final equations. We derived them initially for a fixed radius Rp (using a time-independent steady-state profile around the drop). But now we have developed an equation for how Rp 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 Rp, 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 are constant. Then we can easily integrate the equation

to get

€

dRp

dt=1Rp

DV

ρl(ρw,∞ − ρw

eq )€

ρw,∞ , ρweq

€

Rp2 = Rp,0

2 +2DV

ρl(ρw,∞ − ρw

eq )t

Rp

Rp,0

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

2

=1+ (const) × t =1+ τ

How does the characteristic time scale with Rp,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, Lc

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.

eV,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

€

(ρ∞ − ρs) =1RV

e∞T∞−eV ,sTs

⎛

⎝ ⎜

⎞

⎠ ⎟

Please note here ρ=ρ∞ and ρr is at the surface (called ρs 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,

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,

molecular weight of water

All other variables have been previously defined

Please also see pages 801-805 of the Seinfeld and Pandis text for a complete, and clear, derivation of this growth equation

k

Assumes pure water, no curvature

For droplets in radius, the curvature and solute effects must be considered,

mass of salt

molecular weight of salt

Van’t Hoff factor

molecular weight of water

Condensational growth implies,

which means that

r2

r1

r0

time 10 hours

Condensational growth alone would tend to produce a monodisperse cloud droplet size distribution

We could also write the numerator as S∞ – awexp(Kelvin)

Environmental saturation ratio

saturation ratio over the droplet

growth rate small

growth rate large

~20s ~40s

dm/dt is proportional to r

Houghton, 1985

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

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:

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.

3.  Conclude that diffusional growth alone cannot produce precipitation sized drops within reasonable cloud lifetimes.

Condensational growth tends to produce uniform size distribution

Rate of growth droplets by condensation on salt nuclei (after Best 1951b)

Temperature T=273K Pressure=900mb

Nuclear mass m Supersaturation=100

(S-1)%

10-14 g 0.05

10-13 g 0.05

10-12 g 0.05

Radius (µm) Time (s) to grow from initial radius of 0.75µm

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

r0 r1 r2

r0 r1 r2

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 cm3 in maritime clouds and ~500-700 cm-3 for continental clouds. These droplets compete for available H2O 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,

supersaturation

condensational growth rate

saturation mixing ratio; updraft velocity

where is the droplet concentration, assumed to be

Net rate of change = rate of production of supersaturation – rate of consumption (by condensation)

cloud droplet concentration determined

2) drop growth maximum around Smax

3) smallest particles become haze droplets

4) Activated droplets become monodisperse in size

INITIAL RADII α salt mass radii

1) Max supersaturation achieved few tens of meters above cloud base

Increasing salt mass

non-activated droplets

Activated droplets

MONODISPERSE 0 6 60