Question 4 of 6: Unconfined Well Hydraulics and Seepage Through an Earthen Dam

Question 4: Unconfined Well Hydraulics and Seepage Through an Earthen Dam (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) Well radius $r_w=0.10$ m, discharge $Q=5$ L/s, steady water level at the well $h_w=50$ m, radius of influence $R=200$ m, initial (undisturbed) aquifer depth $H_0=55$ m. (b) Dam base length $L=123$ m, $K=1.33$ m/day, reservoir depth $h_1=18.5$ m, tailwater depth $h_2=4.6$ m, recharge (infiltration) rate $w=0.007$ m/day.

Find. (a) the aquifer's hydraulic conductivity and the steady-state water level 80 m from the pumping well. (b) the head at $x=80$ m from the reservoir face.

Approach. Part (a) applies the Thiem equation for steady radial flow to an unconfined well (the Dupuit form, in $h^2$) between the well and the radius of influence to back out $K$, then re-applies it between the well and the observation point. Part (b) uses the Dupuit-Forchheimer solution for one-dimensional unconfined flow through an earthen dam with areal recharge, which gives $h^2(x)$ as a parabola satisfying the two reservoir/tailwater boundary conditions.

1. Part (a) — hydraulic conductivity from the Thiem (Dupuit) equation. For steady unconfined radial flow, $Q=\pi K\left(H_0^2-h_w^2\right)/\ln(R/r_w)$; solving for $K$: $$K=\frac{Q\ln(R/r_w)}{\pi\left(H_0^2-h_w^2\right)}=\frac{(0.005)\ln(200/0.10)}{\pi\left(55^2-50^2\right)}=\frac{(0.005)(7.601)}{\pi(525)}=\boxed{2.30\times10^{-5}\ \text{m/s}}\ (\approx 1.99\ \text{m/day}).$$
2. Water level 80 m from the well. The same Thiem equation, applied between the well ($r_w$, $h_w$) and the observation point ($r=80$ m), gives $$h_{80}^2=h_w^2+\frac{Q\ln(80/r_w)}{\pi K}=2500+\frac{(0.005)\ln(800)}{\pi(2.304\times10^{-5})}=2500+461.6=2961.6,$$ $$h_{80}=\sqrt{2961.6}=\boxed{54.4\ \text{m}},$$ comfortably between the well's 50 m and the undisturbed 55 m, as the monotonically flattening Thiem cone requires.
3. Part (b) — Dupuit-Forchheimer seepage with recharge. With $x$ measured from the reservoir face, unconfined flow through the dam with uniform areal recharge $w$ satisfies $d^2(h^2)/dx^2=-2w/K$, whose solution meeting $h(0)=h_1$ and $h(L)=h_2$ is $$h^2(x)=h_1^2-\frac{h_1^2-h_2^2}{L}\,x+\frac{w}{K}\,x(L-x).$$
4. Evaluate at $x=80$ m. Substituting the given values ($h_1^2=342.25$, $h_2^2=21.16$, $w/K=0.007/1.33=5.263\times10^{-3}$): $$h^2(80)=342.25-\left(\frac{321.09}{123}\right)(80)+(5.263\times10^{-3})(80)(43)=342.25-208.8+18.1=151.5,$$ $$h(80)=\sqrt{151.5}=\boxed{12.3\ \text{m}}.$$ The recharge term adds roughly 0.8 m to what a straight linear-in-$h^2$ interpolation would give (11.5 m), consistent with infiltration mounding the water table above the no-recharge seepage line.