Given. Three compacted lifts, 2 m each (6 m total refuse depth), separated by 250 mm clay interlayers, 1 m clay cap at 4% slope; annual precipitation $P = 900\ \text{mm/yr}$, 67% lost to evapotranspiration; waste as-delivered density $= 300\ \text{kg/m}^3$, as-delivered moisture content $= 25\%$ by weight; compacted density $= 600\ \text{kg/m}^3$; maximum (field-capacity) moisture content of compacted refuse $= 30\%$ by volume.
Find. 15.1 The annual quantity of leachate generated (once the refuse reaches field capacity); 15.2 The time before the full 6 m of refuse (all three lifts) saturates and leachate begins to flow.
Field-capacity water-balance method: infiltration less runoff and evapotranspiration percolates into the refuse; leachate begins once the refuse's moisture content reaches the given 30%-by-volume maximum.
Approach. Apply the field-capacity water-balance method: (a) compute net percolation into the refuse from a climate water balance on the clay cap, which becomes the steady-state leachate rate once the refuse below reaches its maximum (field-capacity) moisture content; (b) convert the as-delivered moisture content to a compacted, by-volume basis, compare it against the given 30% maximum, and divide the resulting storage deficit by the percolation rate to get the time to saturation.
Check: a runoff coefficient of 20% of precipitation is assumed for the 4% clay-cap slope. The 6 m total refuse depth (three 2 m lifts) is treated as one continuous moisture-storage reservoir for the water balance — the thin 250 mm clay interlayers between lifts are field-practice daily/intermediate cover rather than a full engineered barrier, and are assumed not to prevent moisture continuity between lifts; if they do act as partial capillary barriers, the true time to saturation would be somewhat longer than estimated here. Moisture content by weight (25%) is assumed unchanged by compaction (no water lost or gained during compaction).
Net percolation through the cap (climate water balance).
$$\text{Percolation} = P - RO - ET = 900 - (0.20\times900) - (0.67\times900) = 900 - 180 - 603 = \boxed{117\ \text{mm/yr}}$$
This is the steady-state leachate generation rate once the refuse below reaches its maximum moisture content — equivalently $1{,}170\ \text{m}^3/\text{ha/yr}$.
Initial moisture content of the compacted refuse, by volume. The as-delivered moisture content (25% by weight) is assumed retained through compaction; converting to a volume basis at the compacted density:
$$MC_{vol} = w \times \dfrac{\rho_{compacted}}{\rho_{water}} = 0.25 \times \dfrac{600}{1000} = 0.15 = 15\%\ \text{by volume}$$
Additional water storage capacity before saturation.
$$\Delta MC = FC - MC_{vol} = 0.30 - 0.15 = 0.15 \ \ \Rightarrow \ \ \Delta S = \Delta MC \times \text{depth} = 0.15 \times 6\ \text{m} = 0.90\ \text{m} = 900\ \text{mm}$$
Time to saturation.
$$t = \dfrac{\Delta S}{\text{Percolation rate}} = \dfrac{900\ \text{mm}}{117\ \text{mm/yr}} = \boxed{7.7\ \text{yr}\ (\approx 92\ \text{months})}$$
Quantity
Value
15.1 Annual leachate quantity (once saturated)
117 mm/yr ≈ 1,170 m³/ha/yr
15.2 Time to saturation of the refuse (6 m, 3 lifts)
