Question 7 of 7: Evaporative Loss Time for a Gasoline Spill in Sandy Loam Soil

Question B-2: Evaporative Loss Time for a Gasoline Spill in Sandy Loam Soil (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.

Find. (a) the number of days for all of the spilled gasoline to evaporate at the given flux; (b) measures to speed up evaporation.

Approach. The evaporative flux applies over the surface area of the contaminated footprint, so that area must first be recovered from the mass balance: the total spilled mass (from volume and gasoline density) equals the measured soil concentration times the mass of dry soil in the contaminated volume (footprint area × depth × dry bulk density). Once the area is known, the flux directly gives a mass-loss rate, and total mass divided by that rate gives the time to evaporate everything.

1. Total gasoline mass spilled. $V=5{,}000\ \text{L}=5.0\ \text{m}^3$; $m=V\rho_g=5.0\times726=\boxed{3{,}630\ \text{kg}}$ ($3.63\times10^6$ g).
2. Contaminated footprint area (from the mass balance). The same total mass, expressed via the soil concentration over the contaminated volume $A\,z$: $m=C_{soil}\,\rho_b\,A\,z$, so $A=\dfrac{m}{C_{soil}\,\rho_b\,z}=\dfrac{3.63\times10^{9}\ \text{mg}}{3200\times1400\times2.0}=\dfrac{3.63\times10^{9}}{8.96\times10^{6}}=\boxed{405\ \text{m}^2}$.
3. Mass-loss rate from the evaporative flux. $A=405\ \text{m}^2=4.051\times10^{6}\ \text{cm}^2$; rate $=J\,A=3\times10^{-6}\times4.051\times10^{6}=\boxed{12.15\ \text{g/s}}$ ($\approx$1,050 kg/day).
4. Time to evaporate the full spilled mass. $t=\dfrac{m}{J\,A}=\dfrac{3.63\times10^{6}\ \text{g}}{12.15\ \text{g/s}}=2.987\times10^{5}\ \text{s}=\boxed{\approx3.5\ \text{days}}$ ($\approx$83 hours).

Part (b) — speeding up evaporation. Since evaporative loss is directly proportional to exposed surface area and to the vapour-concentration gradient driving diffusion away from the soil, the effective measures all act on one of those two levers. Increase surface area: excavate and spread the contaminated soil into a thin layer (land-farming/aeration windrow) instead of leaving it at 2.0 m depth, and periodically till/turn the pile to continuously expose fresh, still-wet material rather than letting a dry, low-permeability crust form at the surface and choke off further loss. Increase the concentration gradient/airflow: apply forced ventilation or vacuum-enhanced soil vapour extraction (SVE) to actively pull air through the soil and continuously remove vapour before the boundary layer saturates, which is far faster than relying on passive atmospheric diffusion. Raise temperature: gasoline's vapour pressure rises sharply with temperature, so warming the soil (solar cover/greenhouse enclosure over a land-farm cell, or heated-air injection for an SVE system) directly increases the evaporation rate. Manage soil moisture: if a shallow or perched water table is present, lowering it (or improving drainage) increases the air-filled pore space available for vapour movement, since water-saturated pores block vapour-phase transport entirely.