22-Agric-A2 Soil Physics and Mechanics · December 2019
Question 6 of 6: Confined Aquifer Pumping Well (Thiem Equation)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 04-Agric-A2 Soil Physics & Mechanics,
National Exams December 2019 — a three-hour open-book examination;
any non-communicating calculator is permitted. The cover page states that five (5)
questions constitute a complete exam paper and that only the first five as they appear
in the answer book are marked, that each question is of equal value, and that some
questions require a written answer whose clarity and organization matter for marks. All
six printed questions are worked here, because the set is a study resource rather than a
timed attempt; on exam day a candidate submits only the first five, in order.
Find. Steady-state Qw from the two observation wells (a);
maximum Qw for s = 5 m at the well (b); tracer travel time from r1 at
that flow rate (c).
Confined aquifer with the pumping well and two observation wells; the
red dashed curve is the cone of depression during pumping.
Approach. Apply the Thiem equation between the two observation wells to
find the steady-state flow rate, then extend it between the reference (r2, h2)
observation well and the pumping well itself to cap the discharge at the allowable drawdown,
and finally convert the resulting radial Darcy velocity into a travel time for the tracer.
a) Steady-state discharge from the observation wells. For confined,
axisymmetric radial flow the Thiem equation relates head at any two radii directly to
Qw:
$$\begin{aligned}
Q_w&=\frac{2\pi KB(h_2-h_1)}{\ln(r_2/r_1)}\\
&=\frac{2\pi(10)(20)(57-52)}{\ln(1000/100)}=\frac{6283.2}{2.3026}=\boxed{2729\ \text{m}^3/\text{day}}
\end{aligned}$$
b) Maximum discharge for s = 5 m allowable drawdown. Using rw =
0.15 m and treating the far (r2, h2) observation well as the reference
head, the well head at the allowable drawdown is $h_w=h_2-s=57-5=52\ \text{m}$:
$$\begin{aligned}
Q_{w,max}&=\frac{2\pi KB(h_2-h_w)}{\ln(r_2/r_w)}\\
&=\frac{2\pi(10)(20)(5)}{\ln(1000/0.15)}=\frac{6283.2}{8.806}=\boxed{714\ \text{m}^3/\text{day}}
\end{aligned}$$
(as a check, this is well below the 2729 m³/day of part (a), which is consistent since
the flow observed in part (a) corresponds to a much larger — about 19 m — drawdown
at the well than the 5 m allowed here.)
c) Tracer travel time from r1 to the pumping well. Under the
part (b) flow rate, the radial SEEPAGE (not Darcy) velocity at radius r is
$v(r)=Q_{w,max}/(2\pi rBn)$; integrating $dt=dr/v(r)$ inward from r1 to
rw:
$$\begin{aligned}
t&=\frac{\pi Bn}{Q_{w,max}}\left(r_1^2-r_w^2\right)\\
&=\frac{\pi(20)(0.30)}{714}\left(100^2-0.15^2\right)=0.02638(10{,}000)\\
&=\boxed{264\ \text{days}\ (\approx0.72\ \text{yr})}
\end{aligned}$$
