Maximum ground-level concentration from a stack

Derivation

The ground-level centreline concentration is

$$C = \frac{Q}{\pi u \sigma_y \sigma_z}\exp\left(-\frac{H^2}{2\sigma_z^2}\right)$$

Let $\sigma_y = \sigma_z / k$ where $k = \sigma_z/\sigma_y$ is constant. Then

$$C = \frac{kQ}{\pi u}\,\frac{1}{\sigma_z^2}\exp\left(-\frac{H^2}{2\sigma_z^2}\right)$$

Differentiating with respect to $\sigma_z$ and setting $dC/d\sigma_z = 0$ (let $s=\sigma_z$):

$$\frac{d}{ds}\left[s^{-2}\exp\left(-\frac{H^2}{2s^2}\right)\right] = \exp\left(-\frac{H^2}{2s^2}\right)\left[-2s^{-3} + s^{-2}\cdot\frac{H^2}{s^3}\right] = 0$$ $$\Rightarrow \frac{H^2}{s^2} = 2 \Rightarrow \sigma_z = \frac{H}{\sqrt{2}}$$

Substituting back, $\exp(-1) = 1/e$ and $\sigma_z^2 = H^2/2$:

$$C_{max} = \frac{kQ}{\pi u}\cdot\frac{2}{H^2}\cdot\frac{1}{e} = \frac{2Q}{\pi e\, u H^2}\left(\frac{\sigma_z}{\sigma_y}\right)$$

The maximum occurs at the distance $x$ where $\sigma_z = H/\sqrt{2}$. It falls as $1/H^2$, so raising the stack height is very effective.

Numerical

$Q = 100$ g/s, $H = 80$ m, $u = 4$ m/s, $k = 0.6$.

$$\sigma_z = \frac{80}{\sqrt 2} = 56.6\ \text{m}$$ $$C_{max} = \frac{2\times100}{\pi\times 2.718\times 4\times 80^2}\times0.6 = \frac{200\times0.6}{218\,620} = 5.49\times10^{-4}\ \text{g/m}^3$$

Answer: $C_{max} \approx 5.5\times10^{-4}$ g/m$^3$ = 549 $\mu$g/m$^3$, at the distance where $\sigma_z = 56.6$ m.