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18-Env-A3 Geotechnical and Hydrogeological Engineering · December 2015

Question 4 of 6: Anisotropic Unconfined Aquifer — Well Drawdown

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

Notes on this paper

National Exams — December 2015 — 04-Env-A3 / Geotechnical & Hydrogeological Engineering. 3 hours duration; open book exam, any non-communicating calculator permitted. The first five questions as they appear in the answer book are marked (20 marks each, 100 marks total); all six are solved below for completeness.

Reference texts. Braja M. Das, Principles of Geotechnical Engineering (9th ed.) — weight–volume relations, compaction, lateral earth pressure and retaining-wall stability chapters; Craig & Knappett, Craig's Soil Mechanics (8th ed.) — seepage/flow-net theory and Rankine earth-pressure cross-reference; Freeze & Cherry, Groundwater (1979) — Darcy's law, anisotropic layered media and the Dupuit–Thiem equation for radial flow to a well.

Question 4: Anisotropic Unconfined Aquifer — Well 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
QuantitySymbolValue
Sand layer thickness / conductivity$d_1,K_1$4 m, 0.05 cm/s
Gravel layer thickness / conductivity$d_2,K_2$4 m, 0.20 cm/s
Far-field reference head above base$h_R$ at $R$6 m at $R=5000$ m
Pumping rate$Q$50,000 L/day $=50\ \text{m}^3/\text{day}$

Find. (a) equivalent $K_x$ (horizontal) and $K_y$ (vertical); (b) drawdown at $r=10$ m and $r=100$ m.

Sandy soil, K1 = 0.05 cm/s (4 m) Gravel, K2 = 0.20 cm/s (4 m) Impermeable base Q = 50,000 L/day static level, h(R=5 km) = 6 m r = 10 m r = 100 m (schematic — not to scale) 8 m Figure Q4 — Anisotropic unconfined aquifer, fully-penetrating pumping well
Two-layer anisotropic unconfined aquifer with a fully-penetrating pumping well; static head referenced 5 km away.

Approach. Combine the two layers into equivalent horizontal/vertical conductivities (arithmetic mean for flow parallel to layering, harmonic mean for flow across it), then apply the Dupuit–Thiem unconfined radial-flow equation anchored at the given far-field reference head to get drawdown at each radius.

  1. Equivalent horizontal conductivity $K_x$ (flow parallel to the layering — each layer contributes in proportion to its own thickness, like resistors in parallel): $$K_x=\frac{K_1d_1+K_2d_2}{d_1+d_2}=\frac{0.05(4)+0.20(4)}{8}=\frac{1.00}{8}=\boxed{0.125\ \text{cm/s}}\ (=108\ \text{m/day}).$$
  2. Equivalent vertical conductivity $K_y$ (flow across the layering must pass through both in series — harmonic mean): $$K_y=\frac{d_1+d_2}{d_1/K_1+d_2/K_2}=\frac{8}{4/0.05+4/0.20}=\frac{8}{100}=\boxed{0.080\ \text{cm/s}}.$$ (Note $K_y
  3. Dupuit–Thiem unconfined radial flow, anchored at the given far-field condition $h_R=6$ m at $R=5000$ m, using only $K_x$ (horizontal radial flow governs for a fully-penetrating well): $$h(r)^2=h_R^2-\frac{Q}{\pi K_x}\ln\!\frac{R}{r}.$$
  4. Drawdown at $r=10$ m: $$h(10)^2=36-\frac{50}{\pi(108)}\ln\!\frac{5000}{10}=36-0.1474(6.215)=35.08,$$ $$h(10)=5.923\ \text{m},\qquad s(10)=h_R-h(10)=\boxed{0.077\ \text{m}\ (7.7\ \text{cm})}.$$
  5. Drawdown at $r=100$ m: $$h(100)^2=36-0.1474\ln\!\frac{5000}{100}=36-0.1474(3.912)=35.42,$$ $$h(100)=5.952\ \text{m},\qquad s(100)=\boxed{0.048\ \text{m}\ (4.8\ \text{cm})}.$$
QuantityValue
(a) Equivalent horizontal conductivity, $K_x$0.125 cm/s (108 m/day)
(a) Equivalent vertical conductivity, $K_y$0.080 cm/s
(b) Drawdown at $r=10$ m0.077 m
(b) Drawdown at $r=100$ m0.048 m