Question 4 of 6: Downwind Vapor Dispersion — ERPG-1 Evacuation Zone

Question 4: Downwind Vapor Dispersion — ERPG-1 Evacuation Zone (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Continuous ground-level vapour source, mass emission rate $\dot Q=20$ lb/min; wind speed $u=3.4$ mph; stability Class F, rural; evacuation threshold $C^*=12.5\ \text{mg/m}^3$ (ERPG-1).

Find. The ground area downwind where the vapour concentration exceeds the ERPG-1 threshold.

Approach. Model the release as a continuous, ground-level Gaussian plume (no elevation, no dense-gas correction as instructed). The threshold is already given in mass-per-volume units, so the calculation stays entirely in mass units (no molecular weight/ppm conversion is needed). Solve for the maximum downwind distance $x_{max}$ at which the plume CENTRELINE concentration equals the threshold using the stability-Class-F rural dispersion-coefficient correlations, then integrate the crosswind isopleth width from the source out to $x_{max}$ to get the evacuation area.

1. Convert to consistent SI-mass units. $$\dot Q=20\ \tfrac{lb}{min}\times453.6\ \tfrac{g}{lb}\times\tfrac{1}{60\,min/s}=151.2\ \text{g/s}=151197\ \text{mg/s}$$ $$u=3.4\ \text{mph}\times0.4470\ \tfrac{m/s}{mph}=1.52\ \text{m/s}$$
2. Ground-level continuous-plume centreline concentration. For a ground-level source (already includes the ground-reflection term), the downwind centreline concentration is $$C(x,0,0)=\frac{\dot Q}{\pi\,u\,\sigma_y(x)\,\sigma_z(x)}$$ with the Class-F rural Pasquill–Gifford correlations (Crowl & Louvar Table 5-2, $x$ in metres): $$\sigma_y(x)=\frac{0.04x}{\sqrt{1+0.0001x}}\qquad\sigma_z(x)=\frac{0.016x}{\sqrt{1+0.0003x}}$$
3. Solve for the maximum downwind distance. $C(x,0,0)$ decreases monotonically with $x$ once $\sigma_y\sigma_z$ grows faster than any near-field peculiarity, so a numerical (bisection) solve of $C(x_{max},0,0)=C^*$ gives: $$\sigma_y(x_{max})=86.4\ \text{m}\qquad\sigma_z(x_{max})=29.3\ \text{m}$$ $$\boxed{x_{max}\approx2405\ \text{m}\ (\approx2.41\ \text{km})}$$
4. Crosswind isopleth width and area. At any $x < x_{max}$ the plume crosses the threshold at a crosswind offset $$y_{1/2}(x)=\sigma_y(x)\sqrt{2\ln\!\left(\frac{C(x,0,0)}{C^*}\right)}$$ Numerically integrating the full isopleth width $2y_{1/2}(x)$ from the source out to $x_{max}$ gives the evacuation footprint: $$\boxed{A=\int_0^{x_{max}}2y_{1/2}(x)\,dx\approx255848\ \text{m}^2\ (\approx0.256\ \text{km}^2)}$$ with a maximum crosswind half-width of $72.0$ m (full width $\approx144.0$ m) near the source.

The evacuation zone is a long, narrow lens stretching roughly 2.4 km downwind of the release but only about 144 m across at its widest — a direct consequence of Class-F "stable" conditions, which suppress vertical and lateral mixing far more than a neutral or unstable atmosphere would, so the plume travels a long way before it dilutes below the ERPG-1 threshold rather than spreading out and diluting quickly nearby.