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22-Agric-A2 Soil Physics and Mechanics · May 2014

Question 5 of 7: Hydraulic Conductivity and Well Drawdown from an Unconfined Aquifer Pumping Test

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 May 2014 — 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 seven 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.

Reference texts. B.M. Das, Principles of Geotechnical Engineering, 9th ed. (weight-volume relationships, permeability, effective stress, shear strength, particle-size classification, flow to wells); R.F. Craig, Craig's Soil Mechanics, 9th ed. (effective stress, seepage, shear strength).

Question 5: Hydraulic Conductivity and Well Drawdown from an Unconfined Aquifer Pumping Test (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.

QuantityValue
Pumping well diameter1.0 m (radius rw = 0.5 m)
Initial saturated thickness, b05.0 m
Steady pumping rate, Q50,000 L/day = 50 m³/day
Observation well 1r1 = 20 m, drawdown s1 = 0.001 m
Observation well 2r2 = 5 m, drawdown s2 = 0.021 m

Find. Aquifer hydraulic conductivity K (a); the assumptions used (b); and the drawdown expected right at the pumping well (c).

Approach. For steady radial flow to a fully penetrating well in an unconfined aquifer, the Thiem–Dupuit equation gives $Q = \pi K(h_1^2-h_2^2)/\ln(r_1/r_2)$; solve it for K using the two observation wells, then extrapolate the same $h^2$-vs-$\ln r$ straight-line trend inward to the pumping-well radius to get the drawdown there.

  1. Saturated thickness at each observation well. $$h_1 = b_0 - s_1 = 5.0 - 0.001 = 4.999\ \text{m}, \qquad h_2 = b_0 - s_2 = 5.0 - 0.021 = 4.979\ \text{m}$$
  2. a) Hydraulic conductivity (Thiem–Dupuit). Solving the unconfined radial-flow equation for K, with $Q = 50\ \text{m}^3/\text{day}$, $$K = \frac{Q\ln(r_1/r_2)}{\pi\left(h_1^2-h_2^2\right)} = \frac{50\ln(20/5)}{\pi\left(4.999^2-4.979^2\right)} = \frac{50(1.386)}{\pi(0.1996)} = \boxed{110.6\ \text{m/day}\ (1.28\times10^{-3}\ \text{m/s},\ 0.128\ \text{cm/s})}$$
  3. c) Drawdown at the pumping well. The same K (and hence the same slope of $h^2$ against $\ln r$) extends from observation well 2 in to the pumping-well face at $r_w = 0.5\ \text{m}$: $$h_w^2 = h_2^2 - \frac{Q}{\pi K}\ln\!\left(\frac{r_2}{r_w}\right) = 4.979^2 - \frac{50}{\pi(110.6)}\ln\!\left(\frac{5}{0.5}\right) = 24.790 - 0.332 = 24.459\ \text{m}^2$$ $$h_w = 4.946\ \text{m} \ \Rightarrow\ s_w = b_0 - h_w = 5.0 - 4.946 = \boxed{0.054\ \text{m}\ (\approx 54\ \text{mm})}$$

b) Assumptions. The Dupuit–Forchheimer approximation (flow is essentially horizontal with vertical equipotentials, so the small vertical flow components very near the well are ignored); the aquifer is homogeneous, isotropic and of effectively infinite horizontal extent with no boundaries (recharge or barrier) within the radius tested; steady state has genuinely been reached at both observation wells (drawdown no longer changing with time); there is no vertical leakage into or out of the aquifer; and the two observation wells, like the pumping well, fully penetrate the aquifer's saturated thickness so the radial-flow geometry is uniform with depth. Because the observed drawdowns are tiny relative to b₀ (0.001–0.021 m against 5.0 m), the additional small-drawdown assumption behind the linearised Dupuit form is very comfortably satisfied here.

QuantityValue
Hydraulic conductivity, K110.6 m/day = 1.28×10-3 m/s = 0.128 cm/s
Drawdown at the pumping well, sw≈0.054 m (54 mm)