# 7.4: Liquid Droplet Growth by Diffusion

• • Contributed by Roland Stull
• Professor & Director, Geophysical Disaster Computational Fluid Dynamics Center at University of British Columbia

In a supersaturated environment, condensation onto a growing droplet removes water vapor from the adjacent air (Fig. 7.11). This lowers the humidity near the droplet, creating a humidity gradient down which water vapor can diffuse. Figure 7.11 Humidity gradients in supersaturated environment near growing droplets. S% is supersaturation percentage, x is distance. The nucleus for both droplets is salt of mass 10–16 g. Background supersaturation is 1%, the small droplet has radius 0.2 µm, and the large drop 1.0 µm. Equilibrium supersaturation adjacent to the droplets is taken from Fig. 7.7b. Nearest neighbor droplets are roughly 1000 µm distant.

Diffusion is the process where individual water-vapor molecules meander through air via Brownian motion (i.e., random walk). The net direction of diffusion is always down the humidity gradient toward drier air.

The diffusive moisture flux F in kgwater·m–2·s–1 is

\ \begin{align} \mathbb{F}=-D \cdot \frac{\Delta \rho_{v}}{\Delta x}\tag{7.19}\end{align}

where x is distance, D is diffusivity, and ρv is absolute humidity (water-vapor density) in kgwater m–3. To rewrite this in kinematic form, divide by dry-air density ρair to give:

\ \begin{align} F=-D \cdot \frac{\Delta r}{\Delta x}\tag{7.20}\end{align}

where r is water-vapor mixing ratio, and kinematic flux F has units of mixing ratio times velocity [(kgwater kgair–1)·(m s–1)]. Larger gradients cause larger fluxes, which cause droplets to grow faster.

The mixing ratio gradient is ∆r/∆x . This is related to the supersaturation gradient ∆S/∆x by:

\ \begin{align} \frac{\Delta r}{\Delta x}=\frac{\Delta\left[(1+S) \cdot r_{S}\right]}{\Delta x} \approx r_{S} \cdot \frac{\Delta S}{\Delta x}\tag{7.21}\end{align}

where rs is the saturation mixing ratio over a flat water surface. For this expression, the supersaturation fraction is defined by eq. (7.4), and supersaturation percentage by eq. (7.2).

Sample Application

Given a supersaturation gradient of 1% per 2 µm near a droplet in a cloud at 4 km altitude where saturated mixing ratio is 1.5 g kg–1 and the diffusivity is 2x10–5 m2·s–1. Find (a) the mixing-ratio gradient, (b) the kinematic moisture flux, and (c) the dynamic moisture flux.

Given: ∆S%/∆x = 1% / 2 µm, rs = 0.0015 kgwater/kgair

D = 2x10–5 m2·s–1.

Find: (a) ∆r/∆x = ? (kgwater/kgair)/m , (b) F = ?

(kgwater/kgair)·(m s–1) , (c) F = ? kgwater·m–2·s–1.

First, convert units in the saturation gradient:

∆S/∆x = [1%/2 µm]·(106µm/1m)/100% = 5000 m–1.

Next, use eq. (7.21):

∆r/∆x = (0.0015kg/kg)·(5000m–1) = 7.5 (kgwater/kgair)·m–1.

Then use eq. (7.20):

F = –(2x10–5 m2·s–1)·(7.5 (kg/kg)·m–1) = 1.5x10–4 (kgwater/kgair)·(m s–1).

Finally, use the definition of kinematic flux from the Solar & IR Radiation chapter: F = F / ρair . As a first guess for density, assume a standard atmosphere at z = 2 km, and use Table 1-5 in Chapter 1 to estimate ρair ≈ 0.82 kgair m–3 . Thus: F = (ρair)·(F)

= (0.82kgair m–3)·[–1.5x10–4(kgwater/kgair)·(m s–1)] =1.23x10–4 kgwater·m–2·s–1.

Check: Units OK. Physics OK.

Exposition: Diffusive fluxes are very small, which is why it can take a couple hours for droplets to grow to their maximum drop radius. The negative sign means water vapor flows from high to low humidity.

Larger drops create a more gentle gradient than smaller drops. The humidity profile is given by:

\ \begin{align} S \approx S_{\infty}+\frac{R}{|x|} \cdot\left(S_{R}-S_{\infty}\right)\tag{7.22}\end{align}

where S is supersaturation fraction at distance x from the center of the drop, S is background supersaturation at a large distance from the droplet, SR is equilibrium supersaturation adjacent to the drop, and R is droplet radius. Eq. (7.22) was solved on a spreadsheet to create Fig. 7.11.

The diffusivity D is approximately

\ \begin{align} D=c \cdot \frac{P_{o}}{P} \cdot\left(\frac{T}{T_{o}}\right)^{1.94}\tag{7.23}\end{align}

where c = 2.11x10–5 m2·s–1 is an empirical constant, Po = 101.3 kPa, and To = 273.15 K. This molecular diffusivity for moisture is similar to the thermal conductivity for heat, discussed in the Heat Budgets chapter.

During droplet growth, not only must water vapor diffuse through the air toward the droplet, but heat must conduct away from the drop. This is the latent heat that was released during condensation. Without conduction of heat away from the drop, it would become warm enough to prevent further condensation, and would stop growing.

Droplet radius R increases with the square-root of time t, as governed by the combined effects of water diffusivity and heat conductivity:

\ \begin{align} R \approx c_{4} \cdot\left(D \cdot S_{\infty} \cdot t\right)^{1 / 2}\tag{7.24}\end{align}

where S is the background supersaturation fraction far from the drop. Also, dimensionless constant c4 is:

\ \begin{align} c_{4}=\left(2 \cdot r_{\infty} \cdot \rho_{a i r} / \rho_{l i q . w a t e r}\right)^{1 / 2}\tag{7.25}\end{align}

where background mixing ratio is r. If the supersaturation in air surrounding droplets is great enough, the smaller droplets grow by diffusion faster than larger droplets, because of the greater humidity gradients near the smaller drops (Fig. 7.11 and eq. 7.22). Thus, the small droplets will tend to catch up to the larger droplets.

The result is a drop size distribution that tends to become monodisperse, where most of the drops have approximately the same radius. Also, eq. (7.24) suggests that time periods of many days would be necessary to grow rain-size drops by diffusion alone. But real raindrops form in much less time (tens of minutes), and are known to have a wide range of sizes. Hence, diffusion cannot be the only physical process contributing to rain formation.

Sample Application

Find the water vapor diffusivity at P = 100 kPa and T = –10°C.

Given: P = 100 kPa and T = 263 K.

Find: D = ? m2·s–1.

Use eq. (7.23):

\begin{aligned} D &=\left(2.11 \times 10^{-5} \mathrm{m}^{2} \mathrm{s}^{-1}\right)\left(\frac{101.3 \mathrm{kPa}}{100 \mathrm{kPa}}\right)\left(\frac{263 \mathrm{K}}{273.15 \mathrm{K}}\right)^{1.94} =1.99 \times 10^{-5} \mathrm{m}^{2} \cdot \mathrm{s}^{-1} \end{aligned}

Check: Units OK. Physics OK.

Exposition: Such a small diffusivity means that a large gradient is needed to drive the vapor flux.

Sample Application

Find and plot drop radius vs. time for diffusive growth, for the same conditions as the previous Sample Application. Assume 1% supersaturation.

Given: P = 100 kPa, T = 263 K, D = 2x10–5 m2·s–1.

Find: R(µm) vs. t

First get ρair from the ideal gas law:

\begin{aligned} \rho_{a i r} &=\frac{P}{\Re_{d} \cdot T}=\frac{100 \mathrm{kPa}}{\left(0.287 \mathrm{kPa} \cdot \mathrm{K}^{-1} \cdot \mathrm{m}^{3} \cdot \mathrm{kg}^{-1}\right) \cdot(263 \mathrm{K})} =1.325 \mathrm{kg} \cdot \mathrm{m}^{-3} \end{aligned}

rs ≈ 1.8 g kg–1 = 0.0018 kg kg–1 from thermo diagram in Ch. 5. But supersaturation S = 1% = 0.01 = [ r / rs ] – 1.

Thus r = [ 1 + S ]·rs = 1.01·(0.0018 kg kg–1) = 0.00182 kg kg–1.

Using this in eq. (7.25): c4 = [2·(0.00182 kgw/kga)·(1.325 kga·m–3)/(1000 kgw·m–3)]1/2

c4 = 0.0022 (dimensionless)

Finally solve eq. (7.24) on a spreadsheet:

R ≈ 0.0022 · [(2x10–5 m2 s–1) · (0.01) ·t ] 1/2 Check: Units OK. Physics OK.

Exposition: The droplet radius increases with the square root of time — fast initially and slower later. After 3 hours, it has a size on the borderline between cloud and rain drops. Thus, it is virtually impossible to grow full-size rain drops solely by diffusion.