4. Continuous Point Sources of Pollutant

Consider a point source at $x=y=z=0$ emitting pollutant continuously in time. The pollutant will be carried down stream by the wind and will disperse by turbulent diffusion. Suppose that the wind has speed $u$ in the $x$ direction. The situation is sketched in Fig. 8.

Fig. 8. A plan view of a plume of pollutant from a continuous point source S.

Consider a slice of air 1 m thick moving in the $x$-direction, extending to infinity in the $y$ and $z$ directions and moving with the mean wind $u$. The time taken for the slice to pass a fixed point is $1/u$ seconds. If the source emits $q$ kgs$^{-1}$ of pollutant, then the amount in the slice is $Q=q/u$. The pollutant diffuses in the $x$, $y$ and $z$ directions. But since the source is continuous, about the same amount of pollutant diffuses into the sheet in the $x$-direction as diffuses out through the opposite side. Diffusion in the $x$-direction therefore has negligible effect. The diffusion problem reduces to that of 2-dimensional diffusion in the $y$, $z$ plane but in a frame of reference moving in the $x$-direction with speed $u$. The 2-dimensional diffusion solution Eq. (22) therefore applies, but with $Q=q/u$. In other words

\begin{displaymath} C(y,z,x)={q\over 2\pi\sigma_y\sigma_zu}{\rm exp}\left ( -{y^2\over 2\sigma_y^2}-{z^2\over 2\sigma_z^2}\right ).\eqno{(47)} \end{displaymath}

This is the Gaussian plume equation which is the basis of much dispersion modelling. Note that following the flow, $t=x/u$ so that we expect $\sigma_y$ and $\sigma_z$ to behave like

\begin{displaymath} \sigma_y=\sqrt{2\varepsilon _y{x\over u}},\qquad \sigma_z=\sqrt{2\varepsilon _z{x\over u}} \eqno{(48)} \end{displaymath}

In fact this is not the case because $\varepsilon _y$ and $\varepsilon _z$ are not constant.

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