Skip to main content
Geosciences LibreTexts

11.10: Can we relate this turbulent flux to a molecular flux?

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

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

    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:



    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 \(2 \times 10^{-5} \mathrm{m}^{2} \mathrm{s}^{-1}\)

    Molecular diffusion cannot transport anything fast enough for the atmosphere except on small scales of a centimeter or less. However, on the spatial and temporal scales of the planetary boundary layer, eddies are quite effective at moving heat, molecules, and momentum. In the last section, we saw that turbulence tends to move heat from heights where the air is warmer to heights where the air is cooler. Eddy "diffusion" shares this characteristic with molecular diffusion.

    We can write the heat flux in the same way that we write the molecular flux:

    \[F_{\text {heat}}=\overline{w^{\prime} \theta^{\prime}}=-K \frac{\partial \bar{\theta}}{\partial z}\]

    where K is the eddy diffusion coefficient. Since K is always positive, this equation makes it clear that the flux of any quantity goes from where there is more of that quantity to where there is less of that quantity.

    11.10: Can we relate this turbulent flux to a molecular flux? is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?