diffusiontoclouddroplet_1

ATMO 551a Fall 08

Diffusion of water vapor onto a cloud droplet

Growth of cloud droplet by diffusion involves two simultaneous processes.1. Water vapor diffuses onto the condensation nucleus/droplet2. Heat released by the latent heat release diffuses outward from the cloud droplet

From Fick’s second law

€

∂n

∂t= Dv∇

2n (1)

Assume steady state

€

∂n

∂t= 0 = Dv∇

2n (2)

In spherical coordinates and assuming, by symmetry, only radial dependence (no angular dependence)

€

∇2n =1

r2

d

drr2 d

drn

⎛

⎝ ⎜

⎞

⎠ ⎟= 0 (3)

€

r2 d

drn

⎛

⎝ ⎜

⎞

⎠ ⎟= C1 (4)

€

d

drn =

C1

r2 (5)

€

dn∫ =C1

r2dr∫ (6)

€

n(∞) − n(r)[ ] = −C1 0 −1

r

⎡ ⎣ ⎢

⎤ ⎦ ⎥=

C1

r

€

n(r) = n(∞) −C1

r(7)

We have the further constraint that at rdrop, n = nr such that

€

n(rdrop ) = n(∞) −C1

rdrop

€

C1 = n(∞) − n(rdrop )[ ]rdrop (8)

such that

€

n(r) = n(∞) −rdrop

rn(∞) − n(rdrop )[ ] (9)

The flux is

€

Fn = −Dv

dn

dr= −Dv

rdrop

r2n(∞) − n(rdrop )[ ]

1 Kursinski 11/12/08

ATMO 551a Fall 08

€

Fn r( ) = −Dv

r

rdrop

rn(∞) − n(rdrop )[ ] = −

Dv

rn(∞) − n(r )[ ] (10)

The total flux at the drop surface where r = rdrop is the flux (density) times the area of the drop:

€

Fn,rdropAdrop = −

Dv

rdrop

n(∞) − n(rdrop )[ ]4πrdrop2 = −Dv n(∞) − n(rdrop )[ ]4πrdrop (11)

The change in the mass of the droplet is this number density flux times the mass per molecule

€

dmdrop

dt= Dv n(∞) − n(rdrop )[ ]4πrdrop mv = Dv ρ v (∞) − ρ v (rdrop )[ ]4πrdrop (12)

where mv is the mass per molecule or per mole depending on the units of n. Clearly, for the droplet to grow, the environmental number density of water molecules,

€

n(∞) , must exceed that of the number density at the surface of the droplet,

€

n(rdrop ) . If on the other hand

€

n(rdrop ) is greater than

€

n(∞) , then the cloud droplet will evaporate.

The latent heat release from the condensation of the water vapor causes the droplet to warm above the ambient temperature and this extra sensible heat diffuses away from the droplet.

€

dQ

dt= 4πrKs T r( ) − T(∞)[ ] (13)

where Ks (= r Cp Ds) is the thermal conductivity of the air in units of kg/m3 J/kg/K m2/s = J/m/s/K.

The change in the temperature of the droplet is

€

4

3πr3ρ LCL

dTr

dt= L

dm

dt−

dQ

dt(14)

If we assume steady state droplet temperature, then

€

Ldm

dt=

dQ

dt= LDv ρ v (∞) − ρ v (rdrop )[ ]4πrdrop = 4πrdropK T rdrop( ) − T(∞)[ ] (15)

€

rv (∞) − ρ v (rdrop )[ ]

T rdrop( ) − T(∞)[ ]=

K

LDv(16)

The water vapor density over the droplet surface is the determined by the saturation vapor pressure over a curved surface such that

€

rvrdrop=

es ' r,Trdrop( )

RvTrdrop

=es r = ∞,Trdrop( )

RvTrdrop

1−b

rdrop3

⎡

⎣ ⎢

⎤

⎦ ⎥exp

a

rdrop

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ (17)

Equations (16) and (17) can be solved simultaneously numerically to determine Tdrop, rvdrop and rdrop versus time. There is an alternative solution shown below.


ATMO 551a Fall 08

Growth of the droplet size: Alternative approximate solution by Mason (1971)

The saturation vapor pressure is related to the water vapor density via the ideal gas law

€

es = ρ vsRvT

€

rvs =es

RvT(18)

The logarithmic derivative is given as

€

dρ vs

ρ vs

=des

es

−dT

T=

L

Rv

dT

T 2−

dT

T(19)

where we have used the Clausius Clapeyron equation (so this equation does NOT involve the droplet surface curvature effects). Integrating (19) from Tdrop to T and assuming T/Tdrop ~ 1 yields

€

dρ vs

ρ vsTdrop

T

∫ = ln ρ vs( )Tdrop

T=

L

Rv

dT

T 2−

dT

T

⎛

⎝ ⎜

⎞

⎠ ⎟

Tdrop

T

∫ = −L

Rv

1

T Tdrop

T

− ln T( )Tdrop

T

€

lnρ vs T[ ]

ρ vs Tdrop[ ]

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟= −

L

Rv

1

T−

1

Tdrop

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟− ln

T

Tdrop

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

€

lnρ vs T[ ]

ρ vs Tdrop[ ]

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟=

L

Rv

T − Tdrop

TdropT

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟− ln 1+

T − Tdrop

Tdrop

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

€

lnρ vs T[ ]

ρ vs Tdrop[ ]

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟=

L

Rv

T − Tdrop

TdropT

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟−

T − Tdrop

Tdrop

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

€

lnρ vs_ env

ρ vs_ drop

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟= Tenv − Tdrop( )

L

RvTdropTenv

−1

Tdrop

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ (20)

€

rvs_ env − ρ vs_ drop

ρ vs_ drop

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟=

Tenv − Tdrop( )

Tenv

L

RvTenv

−1 ⎛

⎝ ⎜

⎞

⎠ ⎟ (21)

Subbing from (13) and then (15)

€

rvs_ env − ρ vs_ drop

ρ vs_ drop

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟= 1−

L

RvTenv

⎛

⎝ ⎜

⎞

⎠ ⎟

1

4πrK sTenv

dQ

dt= 1−

L

RvTenv

⎛

⎝ ⎜

⎞

⎠ ⎟

L

4πrKsTenv

dmdrop

dt(22)

(22) shows the fractional difference between saturation vapor densities of the environment minus that at the droplet for flat surfaces. Note that (22) would actually cause the droplet to evaporate because Tenv < Tdrop.

We also have (12) that includes the actual vapor pressure of the environment which is supersaturated and the actual saturation vapor density over the curved droplet surface. From (12), we can write


ATMO 551a Fall 08

€

rv _ env (∞) − ρ v (rdrop )[ ]

ρ v (rdrop )=

1

Dv 4πrdropρ v (rdrop )

dmdrop

dt(23)

This should be rewritten to include the curvature and solution effects on the droplet surface which means its saturation vapor pressure depends on the size of the droplet according to (9) from the droplet activation notes.

€

es ' r,T( )es r = ∞,T( )

=ρ vs ' r,T( )

ρ vs r = ∞,T( )= 1−

b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp

2σ

rRvρ LT

⎛

⎝ ⎜

⎞

⎠ ⎟= 1−

b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp

a

r

⎛

⎝ ⎜

⎞

⎠ ⎟ (24)

Therefore (23) can be rewritten as

€

rv _ env (∞) − ρ v _ drop (r∞) 1−b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp

a

r

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥

ρ v _ drop (r∞) 1−b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp

a

r

⎛

⎝ ⎜

⎞

⎠ ⎟

=1

Dv 4πrdrop ρ v (rdrop )

dmdrop

dt

€

rv _ env (∞) 1+b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp −

a

r

⎛

⎝ ⎜

⎞

⎠ ⎟− ρ v _ drop (r∞)

⎡

⎣ ⎢

⎤

⎦ ⎥

ρ v _ drop (r∞)=

1

Dv 4πrdrop ρ v (rdrop )

dmdrop

dt(25)

€

rv _ drop (∞) f (r) − ρ v _ drop (r∞)[ ]

ρ v _ drop (r∞)=

1

Dv 4πrdropρ v (rdrop )

dmdrop

dt(26)

where

€

f (r) = 1+b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp −

a

r

⎛

⎝ ⎜

⎞

⎠ ⎟≅ 1−

a

r+

b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥ (27)

Subtracting (22) from (26) and assuming

€

rv _ drop (r∞) = rvs_drop,

€

rv _ env (∞) f (r) − ρ vs_ env[ ]

ρ vs_ drop

=1

Dv 4πrdrop ρ vs_ drop

− 1−L

RvT

⎛

⎝ ⎜

⎞

⎠ ⎟

L

4πrKsT

⎡

⎣ ⎢

⎤

⎦ ⎥dmdrop

dt(28)

€

rv _ env (∞) f (r) − ρ vs_ env[ ]

ρ vs_ drop

=1

Dv 4πrdrop ρ vs_ drop

− 1−L

RvT

⎛

⎝ ⎜

⎞

⎠ ⎟

L

4πrKsT

⎡

⎣ ⎢

⎤

⎦ ⎥ρ L 4πrdrop

2 drdrop

dt(29)

Now we must rearrange (29) to solve for drdrop/dt. So more algebra…

€

rv _ env (∞) f (r) − ρ vs_ env[ ]

ρ vs_ drop

=1

Dvρ vs_ drop

− 1−L

RvT

⎛

⎝ ⎜

⎞

⎠ ⎟

L

KsT

⎡

⎣ ⎢

⎤

⎦ ⎥ρ Lrdrop

drdrop

dt

€

rv _ env (∞) f (r) − ρ vs_ env[ ]

ρ vs_ drop

=ρ LRvT

Dves(rdrop )− 1−

L

RvT

⎛

⎝ ⎜

⎞

⎠ ⎟ρ LL

KsT

⎡

⎣ ⎢

⎤

⎦ ⎥rdrop

drdrop

dt


ATMO 551a Fall 08

€

rdrop

drdrop

dt=

ρ v _ env (∞) f (r) − ρ vs_ env[ ]

ρ vs_ drop

L

RvT−1

⎛

⎝ ⎜

⎞

⎠ ⎟ρ LL

KsT+

ρ LRvT

Dves(rdrop )

⎡

⎣ ⎢

⎤

⎦ ⎥

(30)

We need to do something with the numerator…

€

rdrop

drdrop

dt=

ρ v (∞) f (r) − ρ vs_ env( )

ρ vs_ drop

ρ vs_ env

ρ vs_ env

L

RvT−1

⎛

⎝ ⎜

⎞

⎠ ⎟ρ LL

KsT+

ρ LRvT

Dves(rdrop )

⎡

⎣ ⎢

⎤

⎦ ⎥

=

ρ v _ env (∞) f (r) − ρ vs_ env( )

ρ vs_ env

ρ vs_ env

ρ vs_ drop

L

RvT−1

⎛

⎝ ⎜

⎞

⎠ ⎟ρ LL

KsT+

ρ LRvT

Dves(rdrop )

⎡

⎣ ⎢

⎤

⎦ ⎥

Now we substitute the supersaturation, S = e/es = rv/rvs , and get

€

rdrop

drdrop

dt=

S f (r) −1( )ρ vs_ env

ρ vs_ drop

L

RvT−1

⎛

⎝ ⎜

⎞

⎠ ⎟ρ LL

KsT+

ρ LRvT

Dves(rdrop )

⎡

⎣ ⎢

⎤

⎦ ⎥

(31)

So we now need to consider

€

rvs_ env

ρ vs_ drop which is the saturation vapor density in the environment

divided by the saturation vapor density at the surface of the droplet. There are two issues. First the curvature of the droplet surface means its saturation vapor pressure depends on the size of the droplet according to (9) from the droplet activation notes.

€

es ' r,T( )es r = ∞,T( )

= 1−b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp

2σ

rRvρ LT

⎛

⎝ ⎜

⎞

⎠ ⎟= 1−

b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp

a

r

⎛

⎝ ⎜

⎞

⎠ ⎟ (32)

Second, the temperature at the droplet is higher than that of the environment so the saturation vapor pressure for a flat surface at the droplet is higher than that of the environment.

€

es r = ∞,Tdrop( ) ≅ 1+L Tdrop − Tenv{ }

RvTdropTenv

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥es r = ∞,Tenv( ) > es r = ∞,Tenv( ) (33)

Combining the last two equations

€

es ' rdrop ,Tdrop( ) = es r∞,Tdrop( ) 1−b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp

a

r

⎛

⎝ ⎜

⎞

⎠ ⎟≅ es r∞,Tenv( ) 1+

L Tdrop − Tenv{ }

RvTdropTenv

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥1−

b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp

a

r

⎛

⎝ ⎜

⎞

⎠ ⎟

(34)

Also we are dealing with water vapor densities rather than vapor pressures so there is an additional conversion required

€

rvs_ env

ρ vs_ drop

=evs_ env

evs_ drop

RvTdrop

RvTenv

=evs_ env

evs_ drop

Tdrop

Tenv

(35)


ATMO 551a Fall 08

Combining these last two equations yields

€

rvs_ env

ρ vs_ drop

=evs_ env

evs_ drop

Tdrop

Tenv

=1

1+L Tdrop − Tenv{ }

RvTdropTenv

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥1−

b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp

a

r

⎛

⎝ ⎜

⎞

⎠ ⎟

Tdrop

Tenv (36)

€

rvs_ env

ρ vs_ drop

=Tdrop

Tenv

1−L Tdrop − Tenv{ }

RvTdropTenv

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥1+

b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp −

a

r

⎛

⎝ ⎜

⎞

⎠ ⎟ (37)

€

rvs_ env

ρ vs_ drop

= 1+Tdrop − Tenv

Tenv

−L Tdrop − Tenv{ }

RvTdropTenv

−a

r+

b

r3

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

(38)

€

rvs_ env

ρ vs_ drop

= 1+Tdrop − Tenv

Tenv

1−L

RvTdrop

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟−

a

r+

b

r3

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

(39)

Now let’s try to understand the size of the non-unity terms. From (16) and the definition of K, we can write

€

T rdrop( ) − T(∞)[ ] =LDv

Kρ v (∞) − ρ v (rdrop )[ ] =

LDv

ρCpDs

ρ v (∞) − ρ v (rdrop )[ ] (40)

The two diffusivities, the sensible heat, Ds, and diffusion of water vapor in air, Dv, are similar. The diffusion of sensible heat is the result of the diffusion of N2 and O2 molecules through other N2 and O2 molecules. The diffusion of water vapor is a bit different because it is the diffusion of water vapor molecules through N2 and O2 molecules. We can guess that the collisional crosssection is larger for water molecules which will decrease the diffusivity but the mass of water vapor is smaller which will increase the diffusivity. So we can argue crudely that Ds and Dv, are similar so their ratio is about 1. The supersaturation is perhaps a percent of the saturation value. Near the surface, a rough approximation is rv/ra ~ 1%. So Drv/ra ~ 1e-4. L/Cp ~ 2.5e6/1e3 K = 2.5e3 K. So the temperature difference for near surface conditions is of the order of 2.5e3 K 1e-4 = 0.25 K. At colder temperatures, the temperature difference will be still smaller because less water vapor is available in the air and therefore less latent heating to drive a temperature difference. Therefore the T difference term in (39) is approximately

€

Tdrop − Tenv

Tenv

−L

RvTdrop

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟~ −

0.25

25020 = −0.02

This is comparable to the –a/r term for typical supersaturations. So

€

rvs_ env

ρ vs_ drop

= 1− Δ[ ] (41)

where D can reach a few percent near the surface and is less at colder temperatures at higher altitudes.

Now we return to the other term in the numerator of (31),

€

S f (r) −1( ) .


ATMO 551a Fall 08

€

S f (r) −1( ) =ρ v

ρ vs

1+b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥exp −

a

r

⎛

⎝ ⎜

⎞

⎠ ⎟−1≅

ρ v

ρ vs

1−a

r+

b

r3

⎡ ⎣ ⎢

⎤ ⎦ ⎥−1≅ S −1−

a

r+

b

r3 (42)

Combining (41) and (42), the numerator of (31) becomes

€

S f (r) −1( )ρ vs_ env

ρ vs_ drop

≅ S −1−a

r+

b

r3

⎛

⎝ ⎜

⎞

⎠ ⎟1− Δ[ ] ≅ S −1−

a

r+

b

r3

⎛

⎝ ⎜

⎞

⎠ ⎟ (43)

Plugging (43) into (31) yields when the supersaturations are not too large

€

rdrop

drdrop

dt=

S −1−a

r+

b

r3

⎛

⎝ ⎜

⎞

⎠ ⎟

L

RvT−1

⎛

⎝ ⎜

⎞

⎠ ⎟ρ LL

KsT+

ρ LRvT

Dves(rdrop )

⎡

⎣ ⎢

⎤

⎦ ⎥

(44)

According to Rogers and Yau, this equation estimates the growth rate of cloud droplets quite accurately.

Cloud droplet growth by diffusion

0.1

1

10

100

1000

10 100 1000 10000 100000

time (sec)

Radius (microns)

Series1


diffusiontoclouddroplet_1

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