# How to derive the (dry) deposition velocity for particles in the atmosphere

In atmospheric dispersion modelling the deposition velocity of particulate matter is often described by: $$v_d = \frac{1}{r_A+r_B+r_A r_B v_g}+v_g \hspace{10mm}(1)$$ where $r_A$ is the aerodynamic resistance (describing turbulent fluxes in the surface layer), $r_B$ is the viscous resistance (molecular diffusion in the quasi-laminar sublayer) and $v_g$ is the terminal fall velocity of the particles.

How is $(1)$ derived?

It is used in numerous models but I haven't found any references explaining how the expression is derived. The closest I've found is in this book by Seinfeld and Pandis (ISBN: 9781118947401), chapter 19.

There, it is explained that $(1)$ is derived from the following system of equations describing the total flux through an assumed constant flux layer: $$F = \frac{C_3-C_2}{r_A} + v_g C_3 = \frac{C_2+C_1}{r_B} + v_g C_2 = \frac{C_3-C_1}{r_t} \hspace{10mm} (2)$$ where $C_3$ is the particle concentration at the top of the surface layer, $C_2$ is the concentration between the surface layer and the viscous sublayer and $C_1$ is the concentration of still airborne particles at the surface.

At this stage it is assumed all particles close to the surface will adhere to the surface so that $C_1$ is practically 0. Then $(2)$ can be simplified to: $$\frac{C_3}{r_t} = \frac{C_3-C_2}{r_A} + v_g C_3 = \frac{C_2}{r_B} + v_g C_2 \hspace{10mm} (3)$$

By elimination of $C_3$ and $C_2$ from $(3)$ one is supposed to end up with $(1)$, since $v_d=r_t^{-1}$. I've tried doing this but end up with this instead: $$v_d = \frac{1}{r_t} = \frac{1}{r_A+r_B+r_A r_B v_g} + \frac{v_g}{r_A(\frac{1}{r_B}+v_g) +1}$$

So either my algebra is a bit rusty or there are additional assumption involved...

Start by splitting $(3)$ into two separate equations $$\begin{cases} \frac{C_3}{r_t}=\frac{C_3-C_2}{r_a}+v_g C_3\hspace{10mm}(4)\\ \frac{C_3}{r_t}=\frac{C_2}{r_b}+v_g C_2\hspace{17mm}(5) \end{cases}$$ Get $C_3$ by multiplying $(5)$ with $r_t$ and breaking out $C_2$: $$C_3 = C_2 r_t (\frac{1}{r_b} + v_g)\hspace{10mm}(6)$$ Divide $(4)$ by $C_3$: $$\frac{1}{r_t}=\frac{1-C_2/C_3}{r_a}+v_g\hspace{10mm}(7)$$ Insert $(6)$ in $(7)$: $$\frac{1}{r_t} =\frac{1-\frac{C_2}{C_2r_t(1/r_t+v_g)}}{r_a}+v_g =\frac{ \frac{1}{r_b}+v_g-\frac{1}{r_t} }{ r_a(\frac{1}{r_b}+v_g)}+v_g \hspace{10mm}$$ Move $r_t$ to LHS, i.e. add $\left(r_t \cdot r_a(\frac{1}{r_b}+v_g)\right)^{-1}$ to both sides: $$\frac{1}{r_t}\left(1 + \frac{1}{r_a(\frac{1}{r_b}+v_g)}\right) = \frac{1}{r_a}+v_g$$ Divide by $\left(1+\frac{1}{r_a(\frac{1}{r_b}+v_g)} \right)$ on both sides: $$\frac{1}{r_t} =\frac{1/r_a+v_g}{\left(1+\frac{1}{r_a(\frac{1}{r_b}+v_g)} \right)}$$ Multiply numerator and denominator (RHS) by $r_a$ and simplify: $$= \frac{1+v_g r_a}{\left(r_a+\frac{1}{(\frac{1}{r_b}+v_g)} \right)} = \frac{1+v_g r_a}{\left(r_a\frac{(\frac{1}{r_b}+v_g)}{(\frac{1}{r_b}+v_g)}+\frac{1}{(\frac{1}{r_b}+v_g)} \right)} = \frac{(1+v_gr_a)(\frac{1}{r_b}+v_g)}{r_a(\frac{1}{r_b}+v_g)+1}$$ Re-organize the numerator and split it: $$=\frac{ \frac{1}{r_b}+v_g+v_g\frac{r_a}{r_b} +v_g^2 r_a }{r_a(\frac{1}{r_b}+v_g)+1} =\frac{ \frac{1}{r_b}+v_g\left(r_a(\frac{1}{r_b} +v_g)+1\right) }{r_a(\frac{1}{r_b}+v_g)+1} =\frac{ \frac{1}{r_b}}{r_a(\frac{1}{r_b}+v_g)+1} + v_g$$ If we now move $r_b$ in the first term on the RHS to the denominator we end up with the final expression: $$\frac{1}{r_t} = \frac{1}{r_a + r_b + r_a r_b v_g } + v_g$$