# 11.7: Can we relate this turbulent flux to a molecular flux?

$F_{m o l e c u l e s}=-D_{v} \frac{\partial n}{\partial r}$

where $$F_{\text {molecules}}$$ is the molecular flux (S) units of molecules $$m^{-2} s^{-1}$$, $$\partial n / \partial r$$ is the change in concentration nn (SI units of molecules $$m^{-3}$$) as a function of radial distance rr from the drop (SI units of m), and Dv is the molecular diffusion coefficient $$\left(S | \text { units of } m^{2} s^{-1}\right) .$$ When $$n$$ increases with $$r,$$ then the flux is negative, which means that the flux is toward the drop, in the negative rr direction.

Molecular diffusion, by the way, is very slow at transferring molecules from one place to another in the troposphere. By solving the equations of motion for a simple case, we find that the characteristic time to travel a distance L by molecular diffusion is:

$\tau=\frac{L^{2}}{D_{v}}$

By molecular diffusion, how long would it take water vapor molecules to move from Earth's surface to the top of the planetary boundary layer, 1 km away? A typical value for $$D_{V}$$ is $$2 \times 10^{-5} \mathrm{m}^{2} \mathrm{s}^{-1}$$
$F_{\text {heat}}=\overline{w^{\prime} \theta^{\prime}}=-K \frac{\partial \bar{\theta}}{\partial z}$