I found a formula for the dry deposition speed of particles and an electric circuit analogy to go with it. I would like some clarification on this figure:


I'm mostly interested in the top right circuit. Here, $R_a$ is the aerodynamic resistance, $R_{cp}$ is an (effective) canopy resistance and $V_g$ is the gravitational settling velocity of the particles. I don't understand why there is a $V_g^{-1}$ in parallel with the other resistors and I'm not sure where the $R_a R_{cp} V_g$ comes from either. Some hints would be greatly appreciated. I got this figure from Hicks et al. 1987. The final deposition velocity would then be

$$V_d = (R_a + R_{cp} + R_a R_{cp}V_g)^{-1} + V_g$$

which can easily be derived from the circuit above.


1 Answer 1


I'm not sure if you've gone one citation deeper into the report,

Hicks et al (1985) On the Use of Monitored Air Concentrations to Infer Dry Deposition, NOAA Technical Memorandum ERLO ARLZ141 (https://repository.library.noaa.gov/view/noaa/19701)

where there's more information in Appendix C on the origin of those diagrams.

They make the argument that the turbulent and gravitational (settling) deposition fluxes are separable and additive such that the total deposition velocity is,

$$v_p = v_d + v_g = -F_p / C$$

where $F_p$ and $C$ are the particle flux and concentration, respectively. They then note that,

This simple procedure introduces a conceptual problem, however, since obviously both $v_d$ and $C$ are functions of height and $v_g$ is not.

at which point they assume two equations for $F_p$ (Eqns. C5 and C6), one for the turbulent layer and one for the canopy layer:

$$ F_p = - \left[C(z) - C(z_0)\right] \,r_a^{-1} - C(z) \, v_g\\ F_p = -C(z_0) \, r_{cp}^{-1} - C(z_0)\, v_g $$

These can be combined to eliminate $C(z_0)$, and the result combined with the $v_p$ equation above and rearranged to yield Eq C7,

$$v_p = (r_a + r_{cp} + r_a r_{cp} v_g)^{-1} + v_g$$

The circuit diagram you show in the top right is an expression of this equation, which shows how the traditional circuit diagram can be modified to account for turbulent transfer in the presence of gravitational settling. They note, however, that,

In practice, the triple-product term in Equation (C.7) is unlikely to be of great significance unless $v_g$ is relatively large. Detailed consideration of the ramifications is not warranted, since great uncertainty surrounds the quantity $r_{cp}$ for the case of particles.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.