Question 4 of 5: Layered Landfill-Cap Seepage, Engineered Hydraulic Containment, and Slug/Time-Drawdown Aquifer Tests

Question 4: Layered Landfill-Cap Seepage, Engineered Hydraulic Containment, and Slug/Time-Drawdown Aquifer Tests (equal value)

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. (a) three cap layers as below; ponding depth 3 cm on the topsoil. (b) clay aquitard $L=3$ m, $k=10^{-16}\ \text{m}^2$, piezometer at the top reads 0.5 m of water; target upward Darcy velocity $1\times10^{-3}\ \text{m/yr}$. (c) casing radius $r_c=7.5$ cm, screen radius $r_s=5$ cm, match point $t_1=20$ s at $Tt_1/r_c^2=1$, matched curve $\log\alpha=-6$. (d) $Q=3\ \text{m}^3\text{/min}$, $r=190$ m, drawdown $s_1=3.6$ m at $t_1=4$ h, $s_2=6.0$ m at $t_2=12$ h.

Find. (a) effective vertical $K_v$ and horizontal $K_h$ of the cap; the seepage flow rate through 1 m² of the barrier layer under ponding. (b) the head and pressure needed at the bottom of the aquitard for the target upward velocity, and why an upward gradient is beneficial. (c) $S$ and $T$ from the slug test. (d) $S$ and $T$ from the two-point time-drawdown data.

Approach. Part (a) converts each layer's intrinsic permeability to hydraulic conductivity via $K=k\rho g/\mu$, combines the layers as a thickness-weighted harmonic mean (vertical, series flow) and arithmetic mean (horizontal, parallel flow), then applies the same series-flow logic as Q2(a) to the ponded cap. Part (b) solves Darcy's law for the unknown boundary head that produces the specified upward flux. Part (c) applies the Cooper-Bredehoeft-Papadopulos (CBP) type-curve match directly at the stated match point. Part (d) applies the Cooper-Jacob straight-line method to the two time-drawdown pairs at the same observation well.

1. Part (a) — permeability to hydraulic conductivity. $K=k\rho g/\mu=k\times(1000)(9.81)/0.001=k\times9.81\times10^6$: $$K_{\text{top}}=9.81\times10^{-7}\ \text{m/s},\quad K_{\text{drain}}=1.962\times10^{-3}\ \text{m/s},\quad K_{\text{barrier}}=1.962\times10^{-9}\ \text{m/s}.$$
2. Effective vertical and horizontal conductivity. With $L_{\text{tot}}=0.5+0.4+1.1=2.0$ m: $$K_v=\frac{L_{\text{tot}}}{\sum L_i/K_i}=\boxed{3.56\times10^{-9}\ \text{m/s}},\qquad K_h=\frac{\sum K_iL_i}{L_{\text{tot}}}=\boxed{3.93\times10^{-4}\ \text{m/s}}.$$ $K_v$ is completely dominated by the barrier layer's very low conductivity (series resistance), while $K_h$ is dominated by the drainage layer (parallel/arithmetic average).
3. Seepage flow rate through the barrier under ponding. Ponding adds 3 cm of driving head at the top; the base of the barrier drains freely into the unsaturated waste (zero pressure head there), so the total head drop across the 2.0 m cap is $\Delta H=2.0+0.03=2.03$ m: $$q=\frac{\Delta H}{\sum L_i/K_i}=\frac{2.03}{561{,}000{,}000}=\boxed{3.62\times10^{-9}\ \text{m/s}}\ \ (\approx0.31\ \text{L/day per m}^2).$$ The barrier layer's series resistance so completely dominates the stack that even a saturated cap under active ponding passes a negligible flux — the cap performs as designed.

4. Part (b) — hydraulic conductivity of the clay and target head. $K_{\text{clay}}=k\rho g/\mu=(10^{-16})(9.81\times10^6)=9.81\times10^{-10}\ \text{m/s}=0.0310\ \text{m/yr}$. With the top of the aquitard at $z=3$ m (piezometer reads 0.5 m of water there, $h_{\text{top}}=3.5$ m) and the base at $z=0$, an UPWARD velocity requires $h_{\text{bottom}}\ >\ h_{\text{top}}$: $$h_{\text{bottom}}=h_{\text{top}}+\frac{v\,L}{K_{\text{clay}}}=3.5+\frac{(1\times10^{-3})(3)}{0.0310}=\boxed{3.60\ \text{m}}.$$
5. Pressure at the base of the aquitard. At $z=0$, $\psi_{\text{bottom}}=h_{\text{bottom}}-0=3.60$ m: $$p=\rho_wg\,\psi_{\text{bottom}}=(1000)(9.81)(3.60)=\boxed{35{,}300\ \text{Pa}\ (35.3\ \text{kPa})}.$$ Inducing an upward gradient (aquifer head slightly higher than waste-side head) is a deliberate hydraulic-containment strategy: it forces clean groundwater to migrate up INTO the waste rather than allowing leachate to migrate down into the aquifer, protecting the water supply even if the clay liner develops a defect.

6. Part (c) — CBP transmissivity and storativity. At the match point $Tt_1/r_c^2=1$: $$T=\frac{r_c^2}{t_1}=\frac{(0.075)^2}{20}=\boxed{2.81\times10^{-4}\ \text{m}^2\text{/s}\ (24.3\ \text{m}^2\text{/day})}.$$ The CBP type-curve family is indexed by $\alpha=r_s^2S/r_c^2$; the matched curve has $\log\alpha=-6\Rightarrow\alpha=10^{-6}$: $$S=\alpha\frac{r_c^2}{r_s^2}=10^{-6}\left(\frac{0.075}{0.05}\right)^2=\boxed{2.25\times10^{-6}}.$$

7. Part (d) — Cooper-Jacob straight line between the two time-drawdown points. With $Q=3/60=0.05\ \text{m}^3\text{/s}$, $t_1=4\ \text{h}=14{,}400$ s, $t_2=12\ \text{h}=43{,}200$ s, the drawdown-per-log-cycle slope is: $$\Delta s=\frac{s_2-s_1}{\log_{10}(t_2/t_1)}=\frac{6.0-3.6}{\log_{10}(3)}=5.03\ \text{m per log cycle},\qquad T=\frac{2.303\,Q}{4\pi\,\Delta s}=\boxed{1.82\times10^{-3}\ \text{m}^2\text{/s}\ (157\ \text{m}^2\text{/day})}.$$
8. Storativity from the zero-drawdown time intercept. Extrapolating the line to $s=0$ gives $t_0\approx2770$ s: $$S=\frac{2.25\,T\,t_0}{r^2}=\frac{2.25(1.82\times10^{-3})(2770)}{190^2}=\boxed{3.15\times10^{-4}}.$$