Maximum Ground-Level Concentration

Rather than asking what the concentration is at a chosen distance, this asks the question a permit actually turns on: how bad does it ever get? The answer is a closed form, $C_{max} = 2Q\sigma_z/(e\pi u H^2 \sigma_y)$. For 100 g/s in a 5 m/s wind from an effective height of 100 m with a spread ratio of 0.6, $e\pi = 8.5397$, so the denominator is $8.5397 \times 5 \times 10000 = 426987$, and $C_{max} = (0.2/426987) \times 0.6 = 2.81\times10^{-7}$ kg/m³, or 281 micrograms per cubic metre.

The e in that denominator is not decoration, and finding where it comes from is the most satisfying derivation in this whole section. Substitute $\sigma_y = \sigma_z/r$ into the ground-level equation and differentiate with respect to $\sigma_z^2$, holding the ratio r fixed. The maximum falls at $\sigma_z = H/\sqrt2$. Put that back into the exponential and the argument becomes $-H^2/(2 \cdot H^2/2) = -1$, so the exponential is exactly $e^{-1}$ at the peak, and $1/e$ migrates to the denominator. Everything else is bookkeeping. It also hands you the location for free: the maximum sits wherever the vertical spread has grown to $H/\sqrt2$. For a 100 m plume that is $\sigma_z = 70.7$ m, and with the coefficients $a = 0.113, b = 0.911$ that occurs at $x = (70.7/0.113)^{1/0.911} = 1170$ m downwind.

The $1/H^2$ is the whole economic argument of stack design. Doubling effective height quarters the peak. But height and wind are coupled in a way that makes the worst case non-obvious: $C_{max}$ falls as $1/u$ directly, while plume rise also falls as $1/u$, so H shrinks with wind and $1/H^2$ grows. When rise dominates the effective height, the two effects combine to make $C_{max}$ grow roughly in proportion to u; when the physical stack dominates, it falls as $1/u$. Somewhere between sits a critical wind speed at which the peak is worst, and for a tall buoyant source it usually lands between 3 and 8 m/s. This is why a screening study sweeps wind speed rather than assuming that the calmest hour is the worst one.

Two limitations before this is trusted. The spread ratio $\sigma_z/\sigma_y$ is treated as constant with distance, which is what makes the closed form possible at all, and it genuinely is not: it drifts with distance and varies from about 0.5 to 1.0 in neutral air and much lower in stable air, so a plausible range of ratios moves the answer by a factor of two on its own. And the derivation assumes the plume is still free to grow vertically, so it is invalid once $\sigma_z$ approaches the mixing height, which for a tall stack under a low inversion can happen before the peak is ever reached. As with the point-concentration form, this is a screening tool for comparing designs and sizing a first guess, not a substitute for running the hours.