Groundwater Contamination Plume and Monitoring Strategy
A shallow unconfined sandy aquifer (hydraulic conductivity 8 m/day, hydraulic gradient 0.004, porosity 0.30) underlies a site where a solvent spill occurred. A monitoring well 50 m downgradient shows a concentration of 40 µg/L; a well 120 m downgradient shows 3 µg/L. The regulatory limit is 5 µg/L. A nearby supply well is 300 m downgradient. The aquifer is 6 m thick and the water table is 3 m below ground. A consultant proposes monitored natural attenuation with no active remediation.
1. Estimate the seepage velocity and the time for the plume front to reach the supply well, and assess whether natural attenuation is a defensible sole remedy.
Read the worked answer
Darcy velocity: v = K·i = 8 × 0.004 = 0.032 m/day.
Seepage (pore-water) velocity: v_s = v/n = 0.032/0.30 = 0.107 m/day ≈ 39 m/year.
Travel time to the supply well: the plume has already reached 120 m; remaining distance = 300 − 120 = 180 m. Time = 180/0.107 ≈ 1,680 days ≈ 4.6 years.
Judgement: natural attenuation as the sole remedy is not defensible without demonstrating that attenuation outpaces transport. The plume is already above the limit at 50 m (40 µg/L) and still above at 120 m (3 µg/L is below 5, but only just). The 4.6-year arrival time at a supply well is short relative to regulatory review cycles. A defensible NAA case requires: (a) a demonstrated, sustained downward concentration trend at multiple wells over several years, (b) evidence the attenuation rate exceeds the transport rate, and (c) a contingency plan. None of that is shown here.
2. Use the two well concentrations to estimate whether attenuation is occurring fast enough to protect the supply well, and state what additional data you would require.
Read the worked answer
Assume first-order decay along the flow path: C = C0·e^(−k·t), with t = distance/v_s.
Between 50 m and 120 m: Δx = 70 m, travel time = 70/0.107 ≈ 654 days.
40 → 3 µg/L: ln(40/3) = ln(13.3) = 2.59 = k × 654 → k ≈ 0.00396 /day (half-life ≈ 175 days).
Extrapolate to 300 m (another 180 m, t ≈ 1,680 days): C = 3 × e^(−0.00396 × 1680) = 3 × e^(−6.65) ≈ 3 × 0.0013 ≈ 0.004 µg/L.
On this basis attenuation appears sufficient - predicted concentration at the supply well is far below 5 µg/L.
But this is fragile: it rests on two points only, assumes steady-state plug flow with no dispersion or retardation changes, and ignores that the 3 µg/L well is already near the limit. Additional data required: (1) a time series at each well (is the trend sustained or a single snapshot?), (2) at least one intermediate well to confirm the decay is monotonic, (3) aquifer heterogeneity and any preferential pathways, (4) the solvent's degradation products (which may be more mobile or toxic than the parent), and (5) the supply well's pumping, which will distort the flow field and pull the plume faster than the natural gradient predicts.