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The Change in Cloud Droplet Size
Due to Diffusion

Background

In cloud physics there is an equation, usually called the growth equation, which describes how a cloud droplet grows. Rogers and Yau in their A Short Course in Cloud Physics provide a derivation of this equation. Although overall Rogers and Yau's text is excellent, in the matter of the derivation of the growth equation their presentation is incomplete and points confusing because it misleads the reader. What follows is an attempt to provide a more comprehensive derivation. The presentation of the derivation differs from that of Rogers and Yau's in notation. The generic radial distance variable is denoted by the lower case symbol r and upper case R is used to denote the particular radius of the droplet. This is the opposite of Rogers and Yau. The environmental levels of water vapor density and temperature are identified by a subscript e, in contrast to Rogers and Yau who denote the environmental levels by the absence of a subscript. The values of water vapor density and temperature are denoted here by a subscript R.

The Analysis

Cloud droplets may grow by condensation or diminish by evaporation. Both condensation and evaporation are diffusion processes described by the Fick diffusion equation; i.e., the flux of water vapor passing through a plane M is given by:

M = -D∇ρ

where D is the diffusion coefficient for water vapor and ρ is the density of water vapor in the space surrounding the droplet. The negative sign in the above equation reflects the fact that the mass flow is in the opposite direction from the gradient of water vapor density.

When water vapor diffuses into the droplet it brings heat energy which raises the temperature of the droplet and that heat energy may diffuse out of the droplet into the surrounding moist air. Such diffusion of heat energy obeys the same form of equation as does the diffusion of the water vapor. The heat energy flux Q passing through a plane is given by

Q = -K∇T.

When the net flow of water vapor or heat into or out of an infinitesimal volume is considered they must generate a change in the density and the temperature within that infinitesimal volume; i.e.,

∂ρ/∂t = D∇2ρ
and
c∂T/∂t = K∇2T

Under steady-state conditions (∂ρ/∂t = 0 and ∂T/∂t = 0) the vapor density and temperature fields in the space surrounding the droplet must satisfy the equations:

2ρ = 0
and
2T = 0.

In spherical coordinates under conditions of spherical symmetry the Laplacian ∇2 is:

(1/r2)(∂(r2∂    /∂r)/∂r = 0

where r is the distance from the center of the droplet.

For a droplet of radius R the boundary conditions are:

ρ = ρR and T = TR at r=R
and
ρ -> ρe and T -> Te as r -> ∞

where ρe and Te are the environmental vapor density and temperature, respectively.

The solution to the steady-state diffusion equation reduces to integrating

∂(r2∂ρ/∂r)/∂r = 0

once with respect to r to get

r2∂ρ/∂r = C1

where C1 is a constant.

Dividing by r2 gives the equation

∂ρ/∂r = C1/r2

Integating this with respect to r gives

ρ = -C1/r + C2

The integration constants must be chosen to satisfy the boundary conditions.

ρe = C2
and
ρR = -C1/R + C2

Subtracting the second equation from the first gives

eR) = C1/R
and thus
C1 = R(ρeR)

The solution to the boundary value problem for the steady-state diffusion equation is therefore

ρ(r) = ρe - (R/r)(ρeR)

This equation reflects the fact that if ρe is greater than ρR then ρ is decreasing smoothly as r decreases toward R.

Temperature obeys the same equations as water vapor density so

T(r) = Te - (R/r)(Te-TR)
which can be rearranged to

T(r) = Te + (R/r)(TR-Te).

The latter form reflects the fact that if TR is greater than Te then T increases smoothly as r decreases to R.

At the surface of the droplet

(∂ρ/∂r)r=R = ((ρeR)/R2)R = (ρee)/R
and
(∂T/∂r)r=R = -((TR-Te)/R2)R = -(TR-Te)/R

Therefore the net amounts of water vapor mass and heat energy diffusing into and out of the droplet surface are:

M = (4πR2(-D)(-(ρee))/R = 4πRD(ρeR)
and
Q = (4πR2(-K)(-(TR-Te)/R = 4πRK(TR-Te)

The heat energy supplied to the droplet from the condensation of water vapor into the droplet is:

LM = 4πLD(ρeR)R

For equilibrium this must match the heat energy Q diffused out of the droplet; i.e.,

LM = Q
4πLDR(ρeR) = 4πKR(TR-Te)
and hence
(TR-Te)/(ρRe) = -LD/K

This above equation may be solved for the unknown ρR in terms of the unknown TR as

ρR = ρe - (K/LD)(TR-Te)

The water vapor density at the surface of the droplet, ρR, also obeys the ideal gas law so

ρR = e's(TR)/RVTR

where RV is the gas constant for water vapor and e's(TR) is the saturated vapor pressure at the temperature TR taking into account the Kelvin effect due to the curvature of the droplet.

From the Kelvin Equation and the Raoult effect e's is given by

e's = es[1 + a/R - b/R3]

where es is the vapor pressure of water over an infinite plane.

Combining the two equations for ρR gives

(es/RVTR)[1 + a/R - b/R3] = (ρe - (K/LD)(TR-Te)

Dividing by (es/RVTR) gives:

[1 + a/R - b/R3] = (ρe - (K/LD)(TR-Te)RVTR/es(TR)

The RHS of the above equation is a function of the known parameters and TR; the LHS is a cubic equation in (1/R). Thus for given ρe and Te the solution for R for any value of TR can be found. From the value of TR the value of ρR can be found. It is, of course, the inverse relationships of TR and ρR as functions of R that are desired but that is a mere mathematical transformation.

B.J. Mason's analytical approximation
of the solution to the equation

Let ρVS and eS be the density and vapor pressure of water vapor under saturated conditions. The Clasius-Clapyeron equation is usually expressed in the form:

deS/dT = L/(T(αVSL))

where L is the latent heat of fusion and the α's refer to the specific densities of water vapor and liquid water under conditions of saturation. The specific volumes are just the reciprocals of the densities. Therefore the Clasius-Clapyeron equation can be expressed as:

deS/dT = LρVS/(T(1-ρVSL))

Since ρVS in negligible compared to ρL the above equation reduces to:

deS/dT = LρVS/T

But water vapor satisfies the ideal gas law

eS = ρVSRVT
and so
deS/dT = (dρVS/dT)RVT + ρVSRV

Equating the two expressions for deS/dT and dividing both sides by ρVSRVT gives:

(1/ρVS)(dρVS/dT) = L/(RVT2 - 1/T
or, in differential form
VSVS = (L/RV)(dT/T2) - dT/T

The integration of this latter differential equation from Te to TR yields

ln(ρRSeS)
= (L/RV)[1/Te- 1/TR] - ln(TR/Te)

At this point Mason makes use of an approximation:

If z is small then ln(1+z)=z

Thus if (y/x) is approximately 1

ln(y/x) = ln((x+(y-x))/x)
= ln(1+(y-x)/x) = (y-x)/x

Applying this approximation to the previous equation yields

RSeS)/ρeS
= (L/RV)(TR-Te)/(TeTR) - (TR-Te)/Te
or after factoring
RSeS)/ρeS = [(L/RVTR) - 1](TR-Te)/Te

Since (TR-Te) = L(dm/dt)/(4πKR) the previous equation can be expressed as

RSeS)/ρeS
= [(L/RVTR) - 1]L(dm/dt)/(4πKRTe)

It was previously established that

ρeR = (dm/dt)/(4πD)
or, upon division by ρeS
eR)/ρeS = (dm/dt)/(4πDρeS)

The two equations above can be added together. The addition of the LHS gives:

eR)/ρeS + (ρRSeS)/ρeS
e - ρR + ρRS - ρeS)/ρeS

Under the assumption that ρRRS the above reduces to

e - ρeS)/ρeS = (ρeeS) - 1 = S-1

where S is called the saturation ratio and (S-1) the supersaturation ratio.

The RHS of the two equations which were added together is

[(L/RVTR-1)(L/(4πKRTe))+(1/(4πDRρeS)](dm/dt)

Solving for dm/dt gives

dm/dt
= (4πR(S-1))/[(L/RVTR-1)(L/KTe) + (1/(DρeS)]

Since

m=ρL(4/3)πR3
and thus
dm/dt = ρL4πR2(dR/dt)

It thus follows that

R(dR/dt)
= (S-1)/[(L/RVTR-1)(LρL/KTe) + (ρL/(DρeS)]

The density ρeS can be replaced by eeS/RVTe to obtain the growth equation:

R(dR/dt)
= (S-1)/[(L/RVTR-1)(LρL/KTe) + (ρLRR(dR/dt)
= (S-1)/[(L/RVTR-1)(LρL/KTe) + (ρLRVTe/(DeeS)]

the final modification is to assume TR to be equal to Te. Thus the final version of the growth equation is:

R(dR/dt)
= (S-1)/[(L/RVTe-1)(LρL/KTe) + (ρLRVTe/(DeeS)]

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