<p>After some more work I got it working, this might not be the shortest solution possible. There are no further assumptions needed, just pure algebra.
The solution is a bit tedious to reach, so it's not too surprising that this is usually not included in the literature.</p>

<p>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:
$$
\frac{1}{r_t} = \frac{1}{r_a + r_b + r_a r_b v_g } + v_g
$$</p>

CC BY-SA 3.0
2016-08-11T10:34:15.903Z
2016-08-11T10:34:15.903Z
metnix
<p>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. </p>

<p>How is $(1)$ derived?</p>

<p>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 <a href="https://books.google.se/books?hl=sv&amp;lr=&amp;id=n_RmCgAAQBAJ&amp;oi=fnd&amp;pg=PR23&amp;ots=gQM6KLscgL&amp;sig=UAVnf4v1jDmb_9Jd0cooP8BL_lM&amp;redir_esc=y#v=onepage&amp;q&amp;f=false">this book by Seinfeld and Pandis</a> (ISBN: 9781118947401), chapter 19.</p>

<p>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.</p>

<p>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:
$$
\frac{C_3}{r_t}
= \frac{C_3-C_2}{r_A} + v_g C_3
= \frac{C_2}{r_B} + v_g C_2
\hspace{10mm} (3)
$$</p>

<p>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:
$$
v_d = \frac{1}{r_t}
= \frac{1}{r_A+r_B+r_A r_B v_g} + \frac{v_g}{r_A(\frac{1}{r_B}+v_g) +1}
$$</p>

<p>So either my algebra is a bit rusty or there are additional assumption involved...</p>

<meteorology><models>
How to derive the (dry) deposition velocity for particles in the atmosphere
metnix
# How to derive the (dry) deposition velocity for particles in the atmosphere

(Asked by: username_0 on 2016-08-11)

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](https://books.google.se/books?hl=sv&lr=&id=n_RmCgAAQBAJ&oi=fnd&pg=PR23&ots=gQM6KLscgL&sig=UAVnf4v1jDmb_9Jd0cooP8BL_lM&redir_esc=y#v=onepage&q&f=false) (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...

**Tags:** meteorology, models


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## 1 answers

### Answer by username_0 on 2016-08-11. Score: 4 | Accepted answer

After some more work I got it working, this might not be the shortest solution possible. There are no further assumptions needed, just pure algebra.
The solution is a bit tedious to reach, so it's not too surprising that this is usually not included in the literature.

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:
$$
\frac{1}{r\_t} = \frac{1}{r\_a + r\_b + r\_a r\_b v\_g } + v\_g
$$


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