Question 4 of 6: Downwind LFL Distance for an LPG Leak

Question 4: Downwind LFL Distance for an LPG Leak (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.

Downwind fire-hazard (LFL) distance

Given. Liquid LPG leak rate $Q_L=0.23\ \text{m}^3/\text{s}$; liquid density $\rho_L=450\ \text{kg/m}^3$; vapour density $\rho_v=1.8\ \text{kg/m}^3$; LFL $=3\%$ by volume; wind speed $u=11\ \text{m/s}$; ground-level release, rural, Pasquill–Gifford stability Class F (stable); leak duration 5 minutes (undetected).

Find. The downwind distance beyond which the ground-level centreline vapour concentration falls below the LFL.

Approach. Convert the liquid leak rate to a total mass emission rate (worst case: the released liquid fully flashes/evaporates to vapour), convert the LFL to an equivalent mass concentration using the given vapour density, then solve the steady-state ground-level Gaussian plume equation for the downwind distance $x$ where the centreline concentration equals that value, using the Pasquill–Gifford Class F rural dispersion-coefficient correlations.

1. Total mass emission rate. Treating the entire leaking liquid as fully vapourising (a conservative worst case for a fire-hazard assessment): $$\dot m=Q_L\,\rho_L=0.23\times450=\boxed{103.5\ \text{kg/s}}$$
2. LFL as a mass concentration. At the low dilutions relevant to a flammability boundary, the mixture's molar volume is close to that of air, so the partial mass concentration is well approximated by the volume fraction times the pure vapour density: $$C_{LFL}=y_{LFL}\,\rho_v=0.03\times1.8=\boxed{0.054\ \text{kg/m}^3}$$
3. Ground-level, centreline Gaussian plume equation. For a continuous ground-level release with no reflection needed (source already at grade), rural Class F: $$C(x,0,0)=\frac{\dot m}{\pi\,u\,\sigma_y(x)\,\sigma_z(x)}$$ with the Pasquill–Gifford Table 5-2 rural correlations for Class F: $$\sigma_y(x)=\frac{0.04x}{\sqrt{1+0.0001x}},\qquad \sigma_z(x)=\frac{0.016x}{\sqrt{1+0.0003x}}\quad(x,\sigma\ \text{in m})$$
4. Solve for $x$ by bisection. Setting $C(x,0,0)=C_{LFL}=0.054\ \text{kg/m}^3$ and solving numerically for $x$ (the equation is transcendental in $x$ through $\sigma_y,\sigma_z$): $$x_{LFL}\approx\boxed{303\ \text{m}}$$ at which point $\sigma_y\approx11.9\ \text{m}$, $\sigma_z\approx4.6\ \text{m}$.
5. Check against the leak duration. In 5 minutes at 11 m/s the leading edge of the plume has travelled $$x_{travel}=u\,t=11\times(5\times60)=3300\ \text{m}$$ well beyond $x_{LFL}=303\ \text{m}$, so the plume has fully established a quasi-steady concentration profile out to the LFL boundary well within the 5-minute undetected window — the leak duration does not limit the hazard distance found above.

A leak of this size, dispersing under stable (Class F) conditions where the plume stays narrow and concentrated rather than spreading and diluting quickly, creates a flammable zone extending roughly 300 m downwind — large enough to reach well beyond the tanker's own footprint and into surrounding roadway/facility areas, which is exactly why stable, low-mixing conditions are the worst case for a ground-level flammable-vapour release even though they are often associated with calmer weather.