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Advection model

If there is a wind, pollutants are moved around after they are emitted (advection). The wind velocity is assumed to be only horizontal and is denoted by $v_x,v_y$. We also assume that it is the same everywhere in the basin at a given time, but that it may vary with time.

How does wind redistribute the pollutants? Suppose it is moving only in the positive $x$ direction with velocity $v_x$. Then over a time $dt$ all the pollutants in a given box move a distance $v_x dt$ in the positive $x$ direction. Since we are keeping our grid fixed in time, this means that some of the pollutants in each box get moved to the neighboring box in the positive $x$ direction. How much is moved? If the distance $v_x dt$ is less than the grid spacing $a$, then each box loses a fraction $v_x dt/a$ of its pollutants to its neighbor in the positive $x$ direction. So we have

\begin{displaymath}
\Delta m(x,y,t) = - (v_x dt/a) m(x,y,t)    \mbox{for $v_x > 0$}
\end{displaymath} (3)

This mass is transferred to the neighbor in the positive $x$ direction, so at the same time we have
\begin{displaymath}
\Delta m(x+a,y,t) = (v_x dt/a) m(x,y,t)    \mbox{for $v_x > 0$}
\end{displaymath} (4)

Note that we require
\begin{displaymath}
v_x dt/a < 1    \mbox{and}    v_y dt/a < 1. \nonumber
\end{displaymath}  

to make sense of the updating rules. However, if these ratios are too small, the advection updating starts producing diffusion as well. Diffusion is introduced because the model assumes that any small bit of mass moved from one cell to the next immediately spreads throughout that cell. It is better to control diffusion more directly through the explicit diffusion step. So if there is a wind at all, we want it to move reasonably big chunks of mass through the advection step to minimize this artifact.

If the wind direction is in the negative $x$ direction (so $v_x$ is negative) the box $x,y$ loses mass to the neighbor in that direction and that neighbor gains mass:

$\displaystyle \Delta m(x,y,t)$ $\textstyle =$ $\displaystyle (v_x dt/a) m(x,y,t)    \mbox{for $v_x < 0$}$  
$\displaystyle \Delta m(x-a,y,t)$ $\textstyle =$ $\displaystyle -(v_x dt/a) m(x,y,t)    \mbox{for $v_x < 0$}$ (5)

Similar considerations apply for the $y$ direction. Contributions to the change in four directions must be added up to get the net change due to advection. (In the limit of small grid spacing $a$ and small time interval $dt$ the change in mass may be written in the form $\partial m/\partial t = -\vec v \cdot \nabla m$.)


next up previous
Next: Diffusion Model Up: Air Pollution Model Previous: Source
Carleton DeTar 2007-12-03