Ambaum M., Thermal Physics of the Atmosphere

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146 CH 7 CLOUD DROPS

where we have introduced a reduced vapour density 



. (7.76)

This density difference then leads to the total mass ﬂux,

I F

= 4Dr



(RH − 1). (7.77)

As can be seen, Eq. 7.77 is the same as Eq. 7.69 with the far-ﬁeld saturated

vapour density replaced by the reduced vapour density 

It now becomes clear that latent heating slows the growth of the of the

drop because for all 

we have



<

. (7.78)

Physically this makes sense: a warmer drop reduces the supersaturation with

respect to the far ﬁeld and thus reduces the vapour diffusion towards the

drop. There are two limiting cases of interest. If the heat conduction coefﬁ-

cient K is very large, then any excess heat is immediately diffused away from

the drop. We thus expect the heating effect to be small. Indeed, for K →∞

we have 

→∞and we ﬁnd



→ 

. (7.79)

So the mass ﬂux in Eq. 7.77 reduces to Eq. 7.69. The other limiting case is

for a small heat conduction coefﬁcient. In this case the latent heat cannot

be transported away from the drop and we expect a strong reduction of the

supersaturation. Indeed, for K → 0 we have 

→ 0 and we ﬁnd



→ 

→ 0, (7.80)

so the mass ﬂux to the drop vanishes. Realistic values of the diffusion and

conduction coefﬁcients give a value of 

of about 7 × 10

−3

kg m

−3

and it

is typically similar or somewhat larger than 

. The heating effect therefore

reduces the growth by a factor of two to four compared to the case without

heating.

The reduction becomes larger for higher temperatures. In fact, the reduced

vapour density is a much weaker function of temperature than the saturated

vapour density, see Figure 7.9. As can be seen in this ﬁgure, the reduced

density is very nearly a linear function of the temperature. The approximation



= (2.80 + 0.11 T(

◦

C)) × 10

−3

kg m

−3

(7.81)

is accurate to within 4% between temperatures of –5

◦

C and 50

◦

7.5 DROPLET GROWTH 147

100

0 10 20 30 40

T (ºC)

vapour density (gkg

−1

)

FIGURE 7.9 Saturated vapour density 

and reduced vapour density 

as a function of

temperature T.

Now we have an expression for the mass-ﬂux, we can calculate the change

in droplet radius r

. We have

= 

= 4

(7.82)

with 

the density of the liquid and V the droplet volume, V = (4/3)r

. Sub-

stituting the expression for the mass ﬂux, we ﬁnd

= D



(RH − 1). (7.83)

The right-hand side of Eq. 7.83 is a constant at given temperature and relative

humidity, so we can integrate the equation to ﬁnd:

I r

(t) =



(0) + 2D t with D = D



(RH − 1). (7.84)

The droplet radius grows as the square root of time. When the relative hu-

midity is smaller than 100%, this equation describes how the droplet radius

reduces in time through evaporation.

Square root change of scale with time is typical of phenomena that are

driven by diffusive processes. For example, the typical size of a cloud of

tracer released by a local source grows with the square root of time and heat

penetrates a conductor to a depth that grows with the square root of time.

An interesting property of Eq. 7.84 is that the radii of two droplets that

start growing from different initial radii become more and more similar. From

148 CH 7 CLOUD DROPS

FIGURE 7.10 Growth of droplet radius through vapour diffusion. The distribution of

droplet radii becomes ever narrower.

Eq. 7.84 we have

(t) − r

(t) = r

(0) − r

(0), (7.85)

with r

and r

the radii of two different droplets growing in the same back-

ground. This is equivalent to

(t) − r

(t) =

(0) − r

(0)

(t) + r

(t)

. (7.86)

Because both r

and r

will grow with time, their radii will get closer with

time. This effect is illustrated in Figure 7.10.

Note that with changing droplet radius, the mass ﬂux F

is not constant

in time. We then need to ask ourselves whether the steady state assumption

leading to Eq. 7.61 is accurate. Any density ﬂuctuations in the vapour ﬁeld

around the drop will adjust to a steady state through a diffusion process with

diffusion coefﬁcient D. The boundary condition change through the changing

droplet radius occurs with an effective diffusion coefﬁcient D . Because

D  D, (7.87)

the vapour ﬁeld adjusts to a steady state on a much faster time scale than that

of the changing boundary (the relevant time scales are r

/D and r

/D). We

can therefore safely assume that the vapour ﬂux is constant for any given

droplet radius during the growth process.

Let us now put some numbers in Eq. 7.84. A typical supersaturation at

activation is about 2% or less. At 10

◦

C, the reduced vapour density 

about 4 × 10

−3

kg m

−3

. The time taken to grow from a small droplet to a

typical rain drop of radius about 1 mm is several days. Even for growth to

modest droplet size of several tens of micrometres we ﬁnd times of several

hours. The typical time scale for rain to form in a convective cloud is very

much shorter, perhaps half an hour or less. Diffusive growth cannot be the

complete story to go from activation to raindrop.

The next stage in droplet growth is described by the process of collision and

coalescence. In its simplest form, drops at different radii will fall at different

7.5 DROPLET GROWTH 149

terminal velocities. This then allows the drops to collide and possibly coa-

lesce. This is a more efﬁcient mechanism for droplet growth than diffusional

growth.

The terminal velocity of a drop increases with increasing size. Now consider

the following idealized case: A relatively large drop of radius r

is falling

through an environment with a number of smaller drops with a relative speed

of V . The large drop would sweep out a volume of V r

per unit time. The

smaller drops represent a total liquid water density, 

. We assume that a

fraction E of the small drops in the volume swept out by the large drop would

collide and coalesce with the large drop. The mass increase of the large drop

would then be

= E V r



. (7.88)

The effective volume swept out by the drop is somewhat larger than this

because its effective radius has to be increased with the radius of the small

droplets. The fraction E is called the collection efﬁciency and it is a strong

function of the radius of the large drop and the radius of the smaller drops. The

collection efﬁciency is typically a few per cent if the small droplets are much

smaller than the larger droplets. This is because the air ﬂow around the large

drop drags most small droplets around the large drop and thus prevents them

from colliding with the large drop. For small droplets larger than about 20 ␮m

the collection efﬁciency is typically 90% or higher.

As in Eq. 7.82, this mass increase is related to a radius increase and we can

rewrite the above equation as

= E

V 

4

. (7.89)

This is not a closed equation because the terminal velocity V is a function of

the radius of the drop. For small drops (typical cloud drops with radii smaller

than about 30 ␮m), the terminal velocity is low and the Reynolds number of

the ﬂow around the drop is small. This is the viscous Stokes regime, where

the terminal velocity is proportional to the square of the radius. For large

drops (larger than about a millimetre), the terminal velocity is large and the

Reynolds number of the ﬂow is large. This is the turbulent regime where the

terminal velocity is proportional to the square root of the droplet radius.

In the intermediate regime, an empirical match is found for the terminal

In the Stokes regime, the terminal velocity is a function of the effective droplet buoy-

ancy g



, the viscosity , and the droplet radius r

. The only combination of these variables

with the dimension of speed is (g



) r

, which is, up to a constant factor, the Stokes

formula. For high Reynolds numbers, the terminal velocity cannot be a function of the

viscosity. The only possible combination of buoyancy and droplet radius with the dimen-

sion of speed is

√



. The effective buoyancy is g



= g (

− )/. Note that the effective

droplet inertia in the denominator is given by the density of the displaced air, .

150 CH 7 CLOUD DROPS

velocity. In its simplest form, a terminal velocity proportional to the radius is

chosen. Approximate formulae for droplet terminal velocity are

V = 1.2 × 10

−1

if r

∼

30 ␮m, (7.90a)

V = 250

√

−1

if r

∼

1mm, (7.90b)

V = 8 × 10

−1

otherwise. (7.90c)

These fall speed equations can be combined with Eq. 7.89 to ﬁnd the growth

of drops by collisions and coalescence.

This theory has many approximations. The collection efﬁciency is not a

constant but a complex function of the sizes of the two colliding drops as

well as the relative speed of the colliding drops and of the turbulence in the

surrounding air. Also, to achieve realistic droplet growths from this model,

we need to assume a whole spectrum of initial droplet sizes. The largest of

the initial droplets will then grow at the expense of the smaller drops because

this process corresponds to the highest collection efﬁciencies.

We have only discussed the formation and growth of water droplets. So

we have been mainly looking at processes that occur in warm clouds, that is,

clouds below the freezing level. However, many clouds are above the freezing

level and are therefore likely to be at least partly made up of ice. Cold cloud

physics has a more empirical character as the ice processes are so complex.

Ice crystal activation is similar to droplet activation. Homogeneous nucle-

ation does not usually occur; we need a substrate for the ice crystal to nucleate

on. The most obvious choice would be for water droplets to freeze. It turns

out that for small drops this happens when the water droplet is severely su-

percooled. All drops would be frozen typically at around T<−40

◦

C. Droplets

can also freeze at higher temperatures, especially if they contain a solid nu-

cleus. However, many clouds contain substantial amounts of supercooled

water. Analogous to heterogeneous nucleation of water drops, ice crystals

can form on suitable substrates. Wind-blown clay-dust is a typical cause of nu-

cleation. Ice nuclei become more efﬁcient at lower temperatures. Indeed, the

effective ice nucleus concentration increases exponentially with decreasing

temperatures.

When ice crystals are activated they can grow through diffusional growth,

just like water drops. In fact we ﬁnd an analogue of Eq. 7.63 for ice crystals,

/C = 4D ((∞) − (r

)). (7.91)

Here the radius of the droplet in Eq. 7.63 is replaced by a so-called capacitance

C, which takes into account the shape of the ice crystal.

The capacitance

of a sphere is r. The word ‘capacitance’ has its origin in electrostatics: the

calculation for diffusional ﬂux around a crystal turns out to be analogous to

See Westbrook, C. D. et al. (2008) J. Atmos. Sci. 65, 206–219.

7.5 DROPLET GROWTH 151

the calculation of the electric ﬁeld around a capacitor of a certain shape with

capacitance C.

It is important to note that the growth rate of the crystal is now propor-

tional to the supersaturation over ice rather than over water. The saturated

vapour pressure over ice is lower than that over water, which means that the

supersaturation with respect to ice is higher than the supersaturation with re-

spect to water. It may then happen that in clouds where ice crystals and water

droplets are present, the ice crystals grow faster than the water droplets, per-

haps up to the point where the water droplets become subsaturated because

the ice crystals will have scavenged the water vapour. At this point the ice

crystals will grow at the expense of the water drops. This process is called the

Bergeron–Findeisen process. Hole clouds are thought to be spectacular mani-

festations of the Bergeron–Findeisen process: an aeroplane ﬂying through a

supercooled water cloud can initiate ice crystal growth at the expense of the

water drops. The cloud is left with a hole ﬁlled with ice crystals.

The droplet can grow further by the collision of ice crystals with super-

cooled drops, a process usually called accrection, or the clumping together

of different ice crystals, a process usually called aggregation. The collection

efﬁciencies for these processes are hard to determine as they depend strongly

on crystal shape and temperature.

PROBLEMS

7.1. For a drop which grows by condensation, calculate a typical temperature

difference between the drop and the far ﬁeld.

7.2. Wilson cloud chamber. Take a cylinder with some water and its satu-

rated vapour. Now expand the cylinder rapidly. The adiabatic expansion

is so fast the vapour does not immediately adjust to its new equilib-

rium. Using the Clausius–Clapeyron equation in its approximate form,

show that the relative humidity after rapid expansion from volume V

to volume V

satisﬁes

RH =





exp







− 1



For water vapour, the speciﬁc gas constant is R

= 461Jkg

−1

and

the speciﬁc heat capacity at constant volume is c

= 1410 J kg

−1

Hence show that an expansion by a factor of 1.3 is sufﬁcient to achieve

a relative humidity of 400%. How does this result change when the

cylinder is ﬁlled with air at standard pressure and temperature, some

water, and its saturated vapour?

For further reading on cloud physics, see Mason, B. J. (1971) The physics of clouds, 2nd

edn. Oxford University Press, Oxford; Rogers R. R. & Yau, M. K. (1989) A short course in

cloud physics, 3rd edn. Butterworth–Heinemann; Pruppacher, H. R. & Klett, J. D. (1997)

Microphysics of clouds and precipitation, 2nd edn. Kluwer, Dordrecht.

152 CH 7 CLOUD DROPS

7.3. Supersaturation in updrafts. How far do we need to lift an initially sat-

urated air parcel to achieve a supersaturation of 1%? What is a typical

updraft velocity needed to prevent a cloud-drop from falling from the

air?

7.4. Evaporation of raindrops. Assume that raindrops have a terminal ve-

locity of V = Cr

with r

the drop radius and C = 8 × 10

−1

. Show

that, if we ignore the temperature dependence in the reduced density



, the radius of a raindrop falling through an environment with relative

humidity RH changes with height z as

(z) = r

) +

3D

C

(RH − 1) (z

− z),

with z

the initial height. For a cloud base of 1 km, an environmental

temperature of 280 K, and relative humidity RH = 70%, how large does

the initial raindrop need to be to reach the ground? Estimate the error

made by assuming a constant 

. The effective diffusion coefﬁcient for

a falling drop will be increased because of ventilation. How will this

change the answer?

Mixtures and solutions

Variations in the composition of substances have thermodynamic implica-

tions. This contrasts with our previous emphasis on so-called simple sub-

stances, substances where the composition does not vary during the processes

considered. Here we will mainly concentrate on how the presence of solutes

in water affects the properties of the water. Although of considerable interest

to atmospheric science, we will not consider the thermodynamic aspects of

chemical reactions.

8.1 CHEMICAL POTENTIALS

Consider a substance which is made up of N different composites with the

i-th composite having mass M

. So the total mass M is

M =



. (8.1)

The Gibbs function G of such a composite substance will be a function of the

pressure p, temperature T and the composition,

G = G(p, T, M

,...). (8.2)

The differential of G can be written as

dG =



∂G

∂p



T,M

dp +



∂G

∂T



p,M

dT +





∂G

∂M



p,T

= V dp − S dT +



, (8.3)

where the g

are deﬁned by the partial differentiation of the Gibbs function

with respect to M

I g



∂G

∂M



T,p

. (8.4)

153

Thermal Physics of the Atmosphere Maarten H. P. Ambaum

154 CH 8 MIXTURES AND SOLUTIONS

The g

are called chemical potentials; the chemical potentials are intensive

quantities that are functions of the pressure, temperature, and the composi-

tion of the substance.

Note that the chemical potentials are not functions of the absolute values

of the M

but only of the ratios M

: any uniform scaling of the masses M

by some factor does not change the deﬁnition of the chemical potentials, as

the Gibbs function G itself also scales with this factor. Such a uniform scaling

of the masses can be written as

= M

d˛, (8.5)

where d˛ = dM

is the same for all composites. So for constant composi-

tion we ﬁnd that at ﬁxed temperature and pressure

dG =



d˛. (8.6)

Because the g

do not change in this process, this equation can be integrated

to ﬁnd

G =



. (8.7)

It now becomes clear that the chemical potentials g

are the multi-component

analogues of the speciﬁc Gibbs function for simple substances. The chemical

potentials g

are not the same as the speciﬁc Gibbs functions for the pure com-

ponents, as the chemical potentials in general will depend on the composition

of the substance.

The analogy between chemical potentials and speciﬁc Gibbs functions also

applies to the equilibrium between different phases. As shown in Chapter 5

this equilibrium follows from dG = 0 at constant pressure and temperature.

For simple substances this leads to the result that the speciﬁc Gibbs functions

of the substance in the two phases have to be the same, see Eq. 5.4. This can

be generalized to multi-component substances.

The phase equilibrium at constant pressure and temperature is still deﬁned

by dG = 0, where G is the Gibbs function for the total system containing the

two phases. We can now consider the variation in the Gibbs function due

to the transfer of component i from one phase to another. For the sake of

argument, consider a system made up of a liquid and its vapour, and the

evaporation of a mass ıM

of component i from the liquid to the vapour. At

constant pressure and temperature the change in Gibbs function ıG is then

given by

ıG =



i,v

− g

i,l

) ıM

. (8.8)

As ıG has to vanish for any variation in the system, we conclude that for all

components the chemical potentials have to be the same between the two

8.2 IDEAL GAS MIXTURES AND IDEAL SOLUTIONS 155

phases in equilibrium,

I g

i,v

= g

i,l

. (8.9)

This is the multi-component generalization of Eq. 5.4. Note that on adding a

component we also add another equilibrium relation for the chemical poten-

tial so the equilibrium state remains well deﬁned.

The reliance on the use of the Gibbs function G to deﬁne chemical potentials

and phase equilibria is only apparent; the choice is one of convenience, not

necessity. The situation is analogous to that in the derivation of the Clausius–

Clapeyron equation in Section 5.1. For example, consider the enthalpy H of

a substance made up of several components. Because H = G +TS we can use

Eq. 8.3 to write for the differential of the enthalpy

dH = dG + T dS + S dT = V dp + T dS +



. (8.10)

So we can also deﬁne the chemical potentials g

as partial derivatives of the

enthalpy g

= (∂H/∂M

)

S,p

. Analogous arguments work for the other thermo-

dynamic potentials and they lead to four equivalent deﬁnitions of the chem-

ical potentials:



∂U

∂M



S,V



∂H

∂M



S,p



∂F

∂M



T,V



∂G

∂M



T,p

. (8.11)

In contrast, the arguments leading to Eq. 8.7 cannot be straightforwardly

translated to the other thermodynamic potentials: only the Gibbs function

has two intensive variables as natural variables (T and p); the other potentials

have at least one extensive natural variable, which would change on a uniform

scaling of the masses of all constituents. So although we can use any of the

four standard thermodynamic potentials to deﬁne the chemical potential, the

chemical potential is the multi-component generalization of the speciﬁc Gibbs

function.

8.2 IDEAL GAS MIXTURES AND IDEAL SOLUTIONS

For ideal gas mixtures we can write down explicit expressions for the chem-

ical potentials. In order to calculate the chemical potentials in an ideal gas

mixture we will ﬁrst calculate its total Gibbs function. We will take it that for

an ideal gas mixture the individual components contribute independently to

the extensive variables as each component, by deﬁnition of the ideal gas, is

not inﬂuenced by the presence of the other components.

For an ideal gas mixture the total internal energy U is given by

U =



. (8.12)