Question 1 of 5: Concept True/False, Soil Density, Storage Volumes and Two Gradient Problems

Question 1: Concept True/False, Soil Density, Storage Volumes and Two Gradient Problems (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.

Part (a). Each statement is judged against the strict "always true" standard, so a statement that is true only under special circumstances is False.

• I. False — an aquitard is defined by its low permeability; it is the low-K layer that separates or bounds higher-permeability aquifers, not a high-permeability layer.
• II. True — evaporation (from open water and soil) is one of the standard fluxes of the hydrologic cycle, alongside precipitation, infiltration, transpiration and runoff.
• III. False — hydraulic conductivity $K=k\rho g/\mu$ depends on both the porous medium's intrinsic permeability $k$ (grain size/geometry) and the fluid's density and viscosity; the same soil has a different $K$ for water than for oil.
• IV. True — a groundwater divide is the locus from which flow diverges in opposite directions (the water-table crest of a local flow system); like a topographic divide it is itself a flow line, i.e. a streamline, and acts as a no-flow boundary.
• V. False — a slug test is run in a single well: a known volume (slug) is instantaneously added or removed and the water-level recovery in that same well is recorded; no separate pumping well or observation well is needed (that requirement describes a full aquifer pumping test).

Given. (b) Dry bulk density $\rho_d=1.96\ \text{g/cm}^3$, porosity $n=0.35$. (c) Specific yield $S_y=0.07$, aquifer $900\ \text{m}\times2200\ \text{m}$, water-table decline $\Delta h=1.2$ m. (d) Specific storativity $S_s=1.1\times10^{-6}\ \text{m}^{-1}$, thickness $b=50$ m, aquifer $1500\ \text{m}\times2500\ \text{m}$, head decline $\Delta h=2.5$ m. (e) Wells A(0,0), B(0 m east, 250 m north), C(250 m east, 0 m north), heads $h_A=180$, $h_B=195$, $h_C=180$ m.a.s.l. (f) 10 m thick aquitard, top head 162 m, bottom head 175 m.

Find. (b) solid (grain) particle density. (c) volume of water released. (d) volume of water pumped. (e) magnitude and direction of the hydraulic gradient. (f) hydraulic gradient and flow direction across the aquitard.

Approach. (b) uses the standard phase relation between dry bulk density, porosity and grain density. (c)/(d) apply the storage-coefficient definition (specific yield for the unconfined case, storativity $=S_s b$ for the confined case) to the aquifer's plan area. (e) is a three-point problem: fit a planar head surface through A, B, C and read its gradient vector directly from the plane's coefficients. (f) is a one-dimensional Darcy gradient between two stated heads.

1. Part (b) — solid particle density. Dry bulk density is the grain density scaled by the solids' volume fraction, $\rho_d=(1-n)\rho_s$, so $$\rho_s=\frac{\rho_d}{1-n}=\frac{1.96}{1-0.35}=\boxed{3.02\ \text{g/cm}^3}.$$
2. Part (c) — specific-yield volume. The volume released from unconfined storage is the specific yield times the plan area times the water-table decline: $$\Delta V=S_y\,A\,\Delta h=(0.07)(900\times2200)(1.2)=\boxed{166{,}320\ \text{m}^3}.$$
3. Part (d) — confined storativity volume. The storativity is $S=S_sb=(1.1\times10^{-6})(50)=5.5\times10^{-5}$, and the volume pumped from confined storage is $\Delta V=S\,A\,\Delta h$: $$\Delta V=(5.5\times10^{-5})(1500\times2500)(2.5)=\boxed{515.6\ \text{m}^3}.$$
4. Part (e) — fit the head plane. Taking A as the origin (east $=x$, north $=y$), $h(x,y)=a+bx+cy$ must satisfy $h_A=180$ at $(0,0)$, $h_B=195$ at $(0,250)$, and $h_C=180$ at $(250,0)$. The A–C equation gives $250b=h_C-h_A=0\Rightarrow b=0$; the A–B equation gives $250c=h_B-h_A=15\Rightarrow c=0.06$. The gradient vector is $(\partial h/\partial x,\partial h/\partial y)=(b,c)=(0,\,0.06)$.
5. Gradient magnitude and direction. $$|\nabla h|=\sqrt{b^2+c^2}=\sqrt{0^2+0.06^2}=\boxed{0.0600\ (6.00\%)}.$$ Because $b=0$ and $c>0$, the gradient vector points purely in $+y$: head increases due north (this makes sense — A and C share the same 180 m head, so the A–C line is an equipotential, and the gradient must be perpendicular to it, i.e. north–south). Groundwater flows down-gradient, from high head to low head, so the flow direction is due south.
6. Part (f) — aquitard gradient. With $z=0$ taken at the aquitard base, the gradient magnitude is simply the head difference over the thickness: $$i=\frac{h_{\text{bottom}}-h_{\text{top}}}{L}=\frac{175-162}{10}=\boxed{1.30\ (130\%)}.$$ Since $h_{\text{bottom}}(175\ \text{m})>h_{\text{top}}(162\ \text{m})$, head decreases upward, so flow moves from high head to low head — the flow direction is upward through the aquitard.