Confined-Aquifer Water Supply and a Three-Point Flow-Net Problem: 18-Geol-A2 Hydrogeology, May 2014, Q3

Question 3: Confined-Aquifer Water Supply and a Three-Point Flow-Net Problem (equal value)

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. (a) Well spacing $L=1.2$ km, depth to water 20 m (well 1) and 26.5 m (well 2), $K=10^{-3}$ cm/s, aquifer thickness $b=22$ m, aquifer width $W=5$ km, daily demand 1000 m³/day. (b) Piezometer geometry (A 3200 m south of B; C 2000 m from both A and B, due east) with surface elevations and depths to water as tabulated; aquifer $K=10^{-6}$ m/s, thickness $b=12$ m.

Piezometer	Surface elev.	Depth to water	Water-table elevation $h$
A	532 m	25 m	507 m
B	521 m	46 m	475 m
C	506 m	6 m	500 m

Find. (a) whether the aquifer's natural throughflow can meet the 1000 m³/day demand. (b) the hydraulic gradient, flow direction and discharge per unit width of the aquifer.

Approach. Part (a) treats the two static water levels as defining the regional hydraulic gradient and applies Darcy's law across the aquifer's full cross-section (thickness × width) to get a volumetric throughflow to compare against demand. Part (b) is the classical three-point problem: fit the three water-table elevations to a locally planar head surface, from which the gradient magnitude and direction (and hence an equipotential line and the flow direction, perpendicular to it) follow directly, then Darcy's law over the given thickness gives the discharge per unit width.

Part (a) — regional hydraulic gradient. The two static water levels 1.2 km apart give the natural gradient driving flow through the aquifer: $$i=\frac{26.5-20}{1200}=\boxed{5.42\times10^{-3}}.$$
Aquifer throughflow. With $K=10^{-3}\ \text{cm/s}=1\times10^{-5}\ \text{m/s}$ and a flow cross-section $A=b\,W=(22)(5000)=110{,}000\ \text{m}^2$, Darcy's law gives the total natural discharge through the aquifer: $$Q=KiA=(1\times10^{-5})(5.42\times10^{-3})(110{,}000)=5.96\times10^{-3}\ \text{m}^3/\text{s}=\boxed{515\ \text{m}^3/\text{day}}.$$ Since 515 m³/day is less than the required 1000 m³/day, the aquifer's natural regional throughflow alone cannot meet the subdivision's demand — the aquifer is not suitable as characterized, and a larger well field or an alternative source would be needed.
Part (b) — locate piezometer C and fit the head plane. Placing $B=(0,0)$ and $A=(0,-3200)$ (A due south of B), C lies on the perpendicular bisector of AB at distance $\sqrt{2000^2-1600^2}=1200$ m east of it, giving $C=(1200,-1600)$. Fitting the water-table elevations to a plane $h(x,y)=a+bx+cy$ through the three points (x = east, y = north, metres) gives $$h(x,y)=475+0.0075\,x-0.0100\,y,$$ so the head gradient vector is $\nabla h=(0.0075,-0.0100)$, pointing toward higher head (east-southeast).
Hydraulic gradient and flow direction. The gradient magnitude is the length of $\nabla h$, and groundwater flows the opposite way, from high head toward low head: $$i=|\nabla h|=\sqrt{0.0075^2+0.0100^2}=\boxed{0.0125},$$ with flow directed opposite to $\nabla h$, i.e. toward the north-northwest (bearing approximately N37°W, toward piezometer B, the lowest of the three heads). The 500 m equipotential line runs through C and through point D on segment AB (found by linear interpolation, 700 m from A, at elevation 500 m) — shown as the dashed line in the figure — and the flow arrow is perpendicular to it, as drawn.
Discharge per unit width. Darcy's law applied over the aquifer thickness (per metre of width normal to flow) gives $$q=K\,i\,b=(10^{-6})(0.0125)(12)=\boxed{1.50\times10^{-7}\ \text{m}^2/\text{s}}\ \left(\approx 1.30\times10^{-2}\ \text{m}^3/\text{day per metre of width}\right).$$

Figure 2 — Three-point problem: triangle A-B-C in map view, the interpolated 500 m equipotential D-C (dashed), and the groundwater-flow direction (perpendicular to it, toward the lowest head at B).

Quantity	Result
(a) Regional gradient / natural throughflow	5.42×10⁻³ / 515 m³/day
(a) Suitability vs. 1000 m³/day demand	Not suitable (throughflow < demand)
(b) Hydraulic gradient	0.0125
(b) Flow direction	≈ N37°W (toward B)
(b) Discharge per unit width	1.50×10⁻&sup7; m²/s

Concept & Theory

Topic. Darcy's law as an aquifer-capacity check, and the three-point method for determining hydraulic gradient and flow direction from scattered piezometer data.

Key relations. $Q=KiA$; three-point plane fit $h(x,y)=a+bx+cy$; gradient $i=|\nabla h|$; flow direction $=-\nabla h/|\nabla h|$; unit-width discharge $q=Kib$.

Why this works. A locally uniform aquifer produces a water table that is very nearly planar over distances of a few kilometres, so three independent head measurements are exactly enough to fix that plane (three points determine a plane in 3-D). The plane's steepest-descent direction is the true flow direction, and its magnitude is the hydraulic gradient — identical information to constructing equipotential contours graphically, just solved algebraically instead of by hand-drawn interpolation.

Common pitfall. In part (a), a common error is comparing the daily requirement against the wrong quantity — e.g. a single well's expected yield — rather than the aquifer's total natural cross-sectional throughflow implied by the given width. In part (b), students often average the three heads or eyeball a flow direction instead of solving the plane fit, which is unreliable whenever (as here) the three piezometers are not symmetrically spaced; and it is easy to compute gradient direction but forget to check that it points toward, not away from, the lowest head.

Source. Freeze & Cherry, Groundwater, Ch. 2 (Darcy's law, aquifer discharge) and Ch. 6 (regional flow-net construction, the three-point problem).

Question 3 of 6: Confined-Aquifer Water Supply and a Three-Point Flow-Net Problem

Question 3: Confined-Aquifer Water Supply and a Three-Point Flow-Net Problem (equal value)