Landfill Leachate Collection Through a Clay Liner: 18-Env-A3 Geotechnical and Hydrogeological Engineering, December 2018, Q6

Question 6: Landfill Leachate Collection Through a Clay Liner (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.

Fig. Q6 — landfill leachate collection system: 0.5 m ponded leachate over a 1 m clay liner, gravel collection layer with collector pipes below, bottom clay liner protecting groundwater.

Given.

Given data
Quantity	Symbol	Value
Ponded leachate depth	—	0.5 m
Top liner thickness	$L$	1 m
Liner hydraulic conductivity	$K$	2.1×10⁻⁸ cm/s
Landfill base area	$A$	5000 m²
Liner porosity	$n$	10%

Find. (a) annual volume of leachate collected through the top liner; (b) time for leachate to penetrate the 1 m top liner.

Approach. The gravel layer beneath the liner drains freely (zero pressure head at its top), so the full ponding depth plus the liner thickness acts as the driving head over the liner's own thickness; apply Darcy's law for the flow rate and the seepage (interstitial) velocity for the penetration time.

Hydraulic gradient across the liner. Head loss = ponding depth + liner thickness (free-draining gravel below): $$i=\frac{0.5+1.0}{1.0}=1.5.$$
Part (a) — annual collected volume. Darcy flux, converting $K$ to m/s first ($2.1\times10^{-8}\ \text{cm/s}=2.1\times10^{-10}\ \text{m/s}$): $$q=Ki=2.1\times10^{-10}\times1.5=3.15\times10^{-10}\ \text{m/s}.$$ Multiplying by area and seconds per year ($365\times86{,}400=3.154\times10^7\ \text{s}$): $$Q_{yr}=q\,A\,(3.154\times10^7)=3.15\times10^{-10}\times5000\times3.154\times10^7=\boxed{49.7\ \text{m}^3/\text{year}}.$$
Part (b) — seepage velocity through the liner. The interstitial (true) velocity governs how fast leachate actually migrates through the pore space: $$v_s=\frac{q}{n}=\frac{3.15\times10^{-10}}{0.10}=3.15\times10^{-9}\ \text{m/s}.$$
Penetration time. $$t=\frac{L}{v_s}=\frac{1.0}{3.15\times10^{-9}}=3.17\times10^{8}\ \text{s} = \boxed{10.1\ \text{years}}.$$

Check: the annual volume (a) uses the Darcy (superficial) flux over the full liner area, while the penetration time (b) needs the faster, porosity-corrected seepage velocity through the actual pore channels — conflating the two would understate the travel speed by the factor $1/n=10\times$.

Quantity	Value
Hydraulic gradient across liner	1.5
(a) Annual leachate volume collected	49.7 m³/year
(b) Seepage velocity through liner	3.15×10⁻⁹ m/s
(b) Time to penetrate the liner	10.1 years

Concept & Theory

Topic. Darcy flow and pore-scale travel time through a compacted-clay barrier liner.

Key relations. $i=\Delta H/L$; $q=Ki$ (Darcy flux); $v_s=q/n$ (seepage velocity); $t=L/v_s$.

Why this works. The free-draining gravel layer below the liner holds essentially zero pressure head at the liner's base, so the full ponding depth adds directly to the liner thickness as driving head — a textbook one-layer Darcy problem. Because the liner's actual flow channels are only the pore space (fraction $n$ of the bulk volume), the fluid travels through it faster than the bulk (Darcy) flux alone would suggest, by the factor $1/n$.

Common pitfall. Using the Darcy (superficial) velocity directly for the breakthrough-time calculation instead of dividing by porosity — this understates the true migration speed by a factor of $1/n$ and can give a dangerously over-optimistic liner-integrity estimate.

Source. Das, Principles of Geotechnical Engineering (9th ed.), Ch. 7 (Darcy's law, seepage velocity); Freeze & Cherry, Groundwater (1979), Ch. 2 (porosity and seepage velocity).

Question 6 of 6: Landfill Leachate Collection Through a Clay Liner

Question 6: Landfill Leachate Collection Through a Clay Liner (20 marks)