Question 1 of 6: Unconfined Transmissivity, Soil Index Properties and a Landfill Aquitard

Question 1: Unconfined Transmissivity, Soil Index Properties and a Landfill Aquitard (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.

Part (a). In an unconfined (water-table) aquifer the transmissivity $T=Kb$ is the product of the hydraulic conductivity and the saturated thickness $b$, and unlike a confined aquifer — where $b$ is fixed by the overlying and underlying confining beds — the saturated thickness of an unconfined aquifer is literally the position of the free water table above the base of the aquifer. Any process that moves the water table (seasonal recharge and evapotranspiration, a nearby pumping well's drawdown, a river-stage change) changes $b$ directly, so $T$ rises and falls with it even though the intrinsic conductivity $K$ of the material stays essentially constant. This is the physical reason the Theis solution is not exact for unconfined aquifers — it assumes constant $T$ — and why unconfined drawdown data are conventionally corrected (Jacob's correction, $s'=s-s^2/2b$) before matching to a confined-aquifer type curve.

Given. Soil sample: moist mass $M_t=1025$ g, total volume $V_t=660\ \text{cm}^3$, oven-dried solids mass $M_s=952$ g, grain density $\rho_s=2.62\ \text{g/cm}^3$ (water density $1.00\ \text{g/cm}^3$). Landfill aquitard: thickness $L=3$ m, intrinsic permeability $k=10^{-17}\ \text{m}^2$, piezometer at the top reads a 0.5 m water column, target upward Darcy velocity $v=1\times10^{-3}\ \text{m/yr}$.

Find. (b) dry bulk density, porosity, void ratio, saturation and moisture content of the sample. (c) the total head and pressure needed at the bottom of the aquitard to drive a $1\times10^{-3}$ m/yr upward Darcy velocity through it, and why an upward gradient is a favourable design choice.

Approach. Part (b) is a standard phase-relation reduction (mass/volume of solids, water and voids) from the four measured quantities. Part (c) first converts the given intrinsic permeability to a hydraulic conductivity ($K=k\rho g/\mu$), then applies Darcy's law between the base and the top of the aquitard to solve for the base's total head, and finally converts that head to a pressure.

1. Part (b) — volume of solids and voids. The oven-dried solids occupy $V_s=M_s/\rho_s$, and whatever total volume is left over is void space: $$\begin{aligned} V_s &=\frac{952\ \text{g}}{2.62\ \text{g/cm}^3}=363.4\ \text{cm}^3 \\ V_v &=V_t-V_s=660-363.4=296.6\ \text{cm}^3 \end{aligned}$$
2. Dry bulk density, porosity and void ratio. These follow directly from $V_s$ and $V_v$: $$\begin{aligned} \rho_d &=\frac{M_s}{V_t}=\frac{952}{660}=\boxed{1.44\ \text{g/cm}^3} \\ n &=\frac{V_v}{V_t}=\frac{296.6}{660}=\boxed{0.450\ (45.0\%)} \\ e &=\frac{V_v}{V_s}=\frac{296.6}{363.4}=\boxed{0.816} \end{aligned}$$
3. Saturation and moisture content. The pore water mass is $M_w=M_t-M_s=1025-952=73$ g, which at unit water density occupies $V_w=73\ \text{cm}^3$: $$\begin{aligned} S &=\frac{V_w}{V_v}=\frac{73}{296.6}=\boxed{0.246\ (24.6\%)} \\ w &=\frac{M_w}{M_s}=\frac{73}{952}=\boxed{0.0767\ (7.67\%)} \end{aligned}$$ That completes part (b) — a sample that is roughly one-quarter saturated, consistent with it having been described only as "moist," not submerged.
4. Part (c) — convert intrinsic permeability to hydraulic conductivity. The two are related by the fluid properties, $K=k\rho_w g/\mu_w$: $$K=\frac{(10^{-17}\ \text{m}^2)(1000\ \text{kg/m}^3)(9.81\ \text{m/s}^2)}{0.001\ \text{kg/m-sec}}=\boxed{9.81\times10^{-11}\ \text{m/s}},$$ a value squarely in the clay range, consistent with the material description.
5. Total head at the top of the aquitard. Taking the base of the aquitard as datum $z=0$ (so the top is at $z=3$ m), the piezometer's 0.5 m water column is the pressure head there, so $$h_{\text{top}}=z_{\text{top}}+\psi_{\text{top}}=3+0.5=3.5\ \text{m}.$$
6. Required head at the base. Darcy's law over the aquitard, $v=K(h_{\text{bottom}}-h_{\text{top}})/L$, is solved for $h_{\text{bottom}}$ after converting $v$ to consistent units ($1\times10^{-3}\ \text{m/yr}=3.17\times10^{-11}\ \text{m/s}$): $$h_{\text{bottom}}=h_{\text{top}}+\frac{vL}{K}=3.5+\frac{(3.17\times10^{-11})(3)}{9.81\times10^{-11}}=\boxed{4.47\ \text{m}}.$$ Because $h_{\text{bottom}}$ works out greater than $h_{\text{top}}$, the gradient does indeed drive flow upward, as required.
7. Pressure at the base. The pressure head there is $\psi_{\text{bottom}}=h_{\text{bottom}}-z_{\text{bottom}}=4.47-0=4.47$ m, so $$P_{\text{bottom}}=\rho_w g\,\psi_{\text{bottom}}=(1000)(9.81)(4.47)=\boxed{43{,}800\ \text{Pa}\ (\approx 43.8\ \text{kPa})}.$$

Inducing the upward gradient is a good idea because it establishes hydraulic containment: with the head higher beneath the aquitard than above it, the net advective flow through the clay is from the aquifer up toward the landfill, not from the landfill down into the aquifer. Any leachate that reached the base of the waste cell would then be opposed by the upward flow instead of being carried downward into the potable aquifer — the aquitard becomes a one-way hydraulic barrier rather than merely a low-permeability delay. This is why regulated landfill designs frequently specify an underdrain or a pumped leachate-collection sump beneath (or at the base of) the liner: keeping the aquifer-side head elevated relative to the waste side is an active, verifiable substitute for relying on the clay's permeability alone.