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22-Agric-B7 Principles of Hydrology · December 2015

Question 6 of 6: Unconfined-Aquifer Well Pumping Rate (Thiem Equation)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2015 — 04-Agric-B7, Principles of Hydrology. Three-hour, open-book exam; any non-communicating calculator is permitted. Format: five questions constitute a complete paper (the first five as they appear in the answer book are marked), each of equal value; most questions require an answer involving calculations. All six questions are solved here as a complete study resource.

Reference texts: Chow, Maidment & Mays, Applied Hydrology — unit hydrographs and convolution, Horton infiltration, flood-frequency analysis, hydrologic routing; Viessman & Lewis, Introduction to Hydrology — hydrologic terminology, detention-pond routing; Todd & Mays, Groundwater Hydrology — Thiem equation for unconfined aquifers, well-test assumptions.

Question 6: Unconfined-Aquifer Well Pumping Rate (Thiem Equation) (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. Fully screened well in an unconfined (surficial) sand & gravel aquifer, steady-state (equilibrium) drawdowns:

QuantityValue
Well diameter0.30 m
Static water table depth below ground2.5 m
Aquifer (sand & gravel) thickness15 m, atop impermeable till
Drawdown at the well, $s_w$6.5 m
Distance to monitoring well, $r_2$100 m
Drawdown at monitoring well, $s_2$0.1 m

Find. (a) The steady pumping rate $Q$. (b) The assumptions required. (c) How the approach and role of evapotranspiration change if the pumped water is returned to the land above the aquifer.

Approach. Apply the Thiem equation for steady, radial flow to a fully penetrating well in an unconfined aquifer, using the well and the monitoring well as the two observation points; because no hydraulic conductivity is given, assume a representative literature value for sand and gravel and flag it explicitly.

  1. Initial saturated thickness and saturated thickness at each observation point. The 15 m of aquifer material sits below a water table 2.5 m below ground, so the initial (pre-pumping) saturated thickness is $$b_0=15-2.5=\boxed{12.5\ \text{m}}$$ Saturated thickness at the well ($r_1=r_w=0.15$ m) and at the monitoring well ($r_2=100$ m): $$h_1=b_0-s_w=12.5-6.5=6.0\ \text{m}\qquad h_2=b_0-s_2=12.5-0.1=12.4\ \text{m}$$
  2. Thiem equation for an unconfined aquifer, with hydraulic conductivity assumed. $$Q=\frac{\pi K\left(h_2^2-h_1^2\right)}{\ln(r_2/r_w)}$$ No hydraulic conductivity is given for this "sand and gravel" aquifer, so a representative literature value is assumed here (flagged below): $K=50$ m/day, within the commonly cited clean sand-and-gravel range of roughly 1–1000 m/day (Freeze & Cherry; Todd & Mays). With $\ln(100/0.15)=6.50$: $$Q=\frac{\pi(50)\left(12.4^2-6.0^2\right)}{6.50}=\frac{\pi(50)(117.8)}{6.50}=\boxed{2845\ \text{m}^3/\text{day}\ (32.9\ \text{L/s})}$$

b. Assumptions required to answer part (a). Beyond the assumed hydraulic conductivity itself, the Thiem solution requires: steady-state (equilibrium) conditions have actually been reached at both the well and the monitoring well before the drawdowns were read; the aquifer is homogeneous and isotropic with uniform thickness over the radius of influence; flow is horizontal and radially symmetric toward the well (the Dupuit assumption — reasonable far from the well, but more approximate here given that the 6.5 m drawdown at the well is over half the initial 12.5 m saturated thickness, so the flow lines near the well are not purely horizontal); there is no recharge boundary, impermeable boundary, or other pumping well within the tested radius that would distort the drawdown pattern; the well is fully screened and fully penetrating (given) with negligible well losses, so the measured well drawdown reflects aquifer response only, not turbulent entrance losses; and the observation (monitoring) well fully reflects the same aquifer at the same depth interval.

c. Effect of returning the pumped water to the land above the aquifer. If the irrigation water percolates back down through the unsaturated zone, the well is no longer simply mining a non-recharging aquifer as the plain Thiem test assumes — it becomes, in effect, its own local recharge source, and the true long-term steady-state analysis must treat that return flow as a distributed (or superposed image-well) source term within the same cone of depression rather than ignore it. This would generally mean less net drawdown for a given pumping rate than the no-recharge Thiem estimate predicts, so continuing to size the well on the plain Thiem answer above would be conservative (an underestimate of sustainable yield) once return flow is accounted for. Evapotranspiration considerations are, however, very important here: not all irrigation water applied at the surface returns to the aquifer — a substantial fraction is consumed by turf transpiration and direct evaporation before it can percolate back down, so the effective recharge is only the applied depth minus crop ET (and any surface runoff losses), not the full pumped volume. The higher the golf course's consumptive ET demand, the smaller the fraction of pumping that actually returns to recharge the aquifer, so a proper long-term water balance for this system needs an estimate of crop/turf ET (e.g. via Penman–Monteith or a similar method) subtracted from the applied irrigation depth before the return-flow recharge term can be quantified.

QuantityValue
Initial saturated thickness, $b_0$12.5 m
Saturated thickness at well, $h_1$6.0 m
Saturated thickness at $r=100$ m, $h_2$12.4 m
Estimated pumping rate, $Q$ (at assumed $K=50$ m/day)2845 m³/day (32.9 L/s)
Check: no hydraulic conductivity is given in the source for this "sand and gravel" aquifer. The assumed $K=50$ m/day is a representative mid-range literature value; $Q$ scales linearly with whatever $K$ a grader intends — e.g. $K=25$ m/day gives $Q\approx16.5$ L/s, and $K=100$ m/day gives $Q\approx65.9$ L/s. The method and the ratio $h_2^2-h_1^2$ are exact; only the absolute rate depends on the assumed $K$.
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