Unconfined Aquifer — Well Discharge and Drawdown: 18-Env-A3 Geotechnical and Hydrogeological Engineering, May 2015, Q4

Question 4: Unconfined Aquifer — Well Discharge and Drawdown (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.

Given.

Given data
Quantity	Symbol	Value
Well radius	$r_w$	30 cm diameter = 0.15 m
Aquifer thickness (ground surface to base)	—	10 m
Static water table depth below ground	—	2 m ($h_0=10-2=8$ m saturated thickness)
Hydraulic conductivity	$K$	75 m/day
Radius of influence	$R_0$	300 m
Max. allowable drawdown in well	$s_{w,max}$	4 m

Find. (a) maximum discharge $Q_{max}$ (m³/day) for 4 m of allowable drawdown in the well; (b) drawdown at $r=5$ m from the well when $Q=1.5$ L/s.

Fig. Q4 — unconfined radial flow to a fully-penetrating well; the Dupuit–Thiem parabola falls from the static water table ($h_0=8$ m above the aquifer base) toward the pumped well.

Approach. Model the drawdown as steady, radially-symmetric Dupuit flow in an unconfined aquifer (Thiem's equation for $h^2$), anchored at $h=h_0$ at the radius of influence $R_0$.

Part (a) — maximum discharge. With $h_w=h_0-s_{w,max}=8-4=4\ \text{m}$, the unconfined Thiem equation gives $$Q_{max}=\frac{\pi K\left(h_0^2-h_w^2\right)}{\ln(R_0/r_w)}=\frac{\pi\times75\times(8^2-4^2)}{\ln(300/0.15)}=\frac{\pi\times75\times48}{\ln(2000)}=\frac{11310}{7.601}=\boxed{1488\ \text{m}^3/\text{day}}.$$
Part (b) — drawdown at r = 5 m for Q = 1.5 L/s. Convert $Q=1.5\ \text{L/s}=1.5\times86400/1000=129.6\ \text{m}^3/\text{day}$, then solve the same Thiem relation for the head at $r=5\ \text{m}$, anchored at $h_0$ at $R_0$: $$h_r^2=h_0^2-\frac{Q\ln(R_0/r)}{\pi K}=64-\frac{129.6\times\ln(300/5)}{\pi\times75}=64-2.252=61.75,$$ $$h_r=7.858\ \text{m},\qquad s_r=h_0-h_r=8-7.858=\boxed{0.142\ \text{m}}.$$

Check: the 129.6 m³/day pumping rate used in part (b) is far below the 1488 m³/day maximum from part (a), so the small 0.14 m drawdown at 5 m is consistent — the well is being pumped well within its capacity.

Quantity	Value
(a) Saturated thickness at static level, $h_0$	8.0 m
(a) Head in well at max. drawdown, $h_w$	4.0 m
(a) Maximum discharge, $Q_{max}$	1488 m³/day
(b) Head at r = 5 m, $h_r$	7.86 m
(b) Drawdown at r = 5 m	0.142 m

Concept & Theory

Topic. Steady radial (Dupuit–Thiem) flow to a well in an unconfined aquifer.

Key relations. $Q=\dfrac{\pi K(h_1^2-h_2^2)}{\ln(r_1/r_2)}$ between any two radii where the head is known or sought; equivalently $h(r)^2=h_0^2-\dfrac{Q\ln(R_0/r)}{\pi K}$ anchored at the radius of influence.

Why this works. The Dupuit assumption (near-horizontal flow, negligible vertical head gradient) reduces the true two-dimensional unconfined flow problem to a one-dimensional radial ODE in $h^2$ whose solution is logarithmic in $r$ — this is why the same equation can be anchored between any two radii (well-to-$r$, $r$-to-$R_0$, or well-to-$R_0$) and still give an internally consistent head profile.

Common pitfall. Using the CONFINED Thiem equation (linear in $h$, not $h^2$) for an unconfined aquifer, or forgetting to convert the pumping rate's units (L/s to m³/day) before combining with $K$ in m/day.

Source. Freeze & Cherry, Groundwater (1979), Ch. 8 — wells and well hydraulics; Das, Principles of Geotechnical Engineering (9th ed.), Ch. 7.

Question 4 of 6: Unconfined Aquifer — Well Discharge and Drawdown

Question 4: Unconfined Aquifer — Well Discharge and Drawdown (20 marks)