Core-Sample Soil Phase Relations, Anisotropic Darcy Velocity, and Falling-Head Permeameter: 18-Geol-A2 Hydrogeology, May 2018, Q1

Question 1: Core-Sample Soil Phase Relations, Anisotropic Darcy Velocity, and Falling-Head Permeameter (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) Core sample $D=5.5$ cm, $L=10$ cm, saturated mass $M_t=510$ g, oven-dry mass $M_s=460$ g, grain (solid) density $\rho_s=2.65\ \text{g/cm}^3$. (b) Conductivity tensor $K_{xx}=10^{-3}$, $K_{xy}=5.0\times10^{-4}$, $K_{yy}=10^{-4}$ cm/sec, gradients $dh/dx=0.01$, $dh/dy=-0.005$. (c) Falling-head tube diameter 1.8 cm, sample-tube diameter 12.5 cm, flow length $L=15$ cm, head falls from $h_0=6.0$ cm to $h_1=0.50$ cm over $t=550$ min at a constant test temperature of 25°C.

Find. (a) bulk density, void ratio, gravimetric and volumetric water content, porosity, and saturation percentage. (b) Darcy velocities $q_x$, $q_y$, and the magnitude/direction of the resultant velocity vector. (c) hydraulic conductivity $K$ and intrinsic permeability $k$ of the soil at the stated test temperature.

Approach. Part (a) is a standard four-measurement soil phase-relation reduction from total, solid and pore volumes. Part (b) applies the anisotropic (tensor) form of Darcy's law, $q_i=-K_{ij}\,\partial h/\partial x_j$, component by component, then combines the two components into a magnitude and angle. Part (c) uses the falling-head permeameter equation for $K$ (a purely geometric result) and then converts to intrinsic permeability using water properties evaluated AT the stated 25°C test temperature, not the paper's blanket 1000 kg/m³/0.001 kg/m-sec default (which applies only "unless otherwise specified" — here it is).

Part (a) — solid and void volume. Total sample volume $V_t=\pi(D/2)^2L=\pi(2.75)^2(10)=237.58\ \text{cm}^3$. The oven-dried solids occupy $V_s=M_s/\rho_s=460/2.65=173.58\ \text{cm}^3$, leaving void volume $V_v=V_t-V_s=237.58-173.58=\boxed{64.00\ \text{cm}^3}$.
Porosity and void ratio. $$n=\frac{V_v}{V_t}=\frac{64.00}{237.58}=\boxed{0.269\ (26.9\%)},\qquad e=\frac{V_v}{V_s}=\frac{64.00}{173.58}=\boxed{0.369}.$$
Bulk density and water content. The pore-water mass is $M_w=M_t-M_s=510-460=50$ g, so with water density 1 g/cm³, $V_w=50\ \text{cm}^3$. In hydrogeology (Freeze & Cherry) "bulk density" normally means the DRY bulk density $\rho_b=M_s/V_t$; the moist (as-received) value is given alongside: $$\rho_b=\frac{M_s}{V_t}=\frac{460}{237.58}=\boxed{1.94\ \text{g/cm}^3},\qquad \rho_{\text{moist}}=\frac{M_t}{V_t}=\frac{510}{237.58}=\boxed{2.15\ \text{g/cm}^3},\qquad w=\frac{M_w}{M_s}=\frac{50}{460}=\boxed{0.109\ (10.9\%)},\qquad \theta=\frac{V_w}{V_t}=\frac{50}{237.58}=\boxed{0.211\ (21.1\%)}.$$
Saturation percentage. $$S_r=\frac{V_w}{V_v}=\frac{50}{64.00}=\boxed{78.1\%}.$$

Check: the sample is described as "completely water saturated" before drying, which would imply $S_r=100\%$, but the mass data given only support $S_r=78.1\%$. This is a genuine mismatch between the wording and the numbers (not a computational choice) — the question explicitly asks the candidate to "determine" saturation percentage, so the computed 78.1% (consistent with entrapped air not fully displaced during the saturation attempt) is reported rather than the assumed 100%.

Part (b) — anisotropic Darcy velocity components. With the tensor form $q_x=-(K_{xx}\,dh/dx+K_{xy}\,dh/dy)$ and $q_y=-(K_{xy}\,dh/dx+K_{yy}\,dh/dy)$: $$q_x=-\big[(10^{-3})(0.01)+(5.0\times10^{-4})(-0.005)\big]=7.5\times10^{-6}\ \text{cm/sec (in the }-x\text{ sense, i.e. }-7.5\times10^{-6}\ \text{cm/sec)},$$ $$q_y=-\big[(5.0\times10^{-4})(0.01)+(10^{-4})(-0.005)\big]=-4.5\times10^{-6}\ \text{cm/sec}.$$ Converting to SI: $q_x=\boxed{-7.50\times10^{-8}\ \text{m/s}}$, $q_y=\boxed{-4.50\times10^{-8}\ \text{m/s}}$ (both components negative — flow is toward $-x,-y$).
Magnitude and direction. $$|q|=\sqrt{q_x^2+q_y^2}=\sqrt{(7.50\times10^{-8})^2+(4.50\times10^{-8})^2}=\boxed{8.75\times10^{-8}\ \text{m/s}}.$$ The angle from the $+x$ axis is $\theta=\operatorname{atan2}(q_y,q_x)=\boxed{-149.0^{\circ}}$ (i.e. $149.0^{\circ}$ measured into the third quadrant, equivalent to a compass bearing of about S59°W if $+x$=east and $+y$=north).

Darcy velocity vector plotted on the local $x$-$y$ axes: both components negative, so the resultant points into the third quadrant at 149.0° measured clockwise from the $+x$ axis ($\theta=-149.0^{\circ}$).

Part (c) — hydraulic conductivity from the falling-head test. Standpipe area $a=\pi(1.8/2)^2=2.545\ \text{cm}^2$, sample area $A=\pi(12.5/2)^2=122.72\ \text{cm}^2$, $t=550\times60=33{,}000$ s: $$K=\frac{aL}{At}\ln\!\left(\frac{h_0}{h_1}\right)=\frac{(2.545)(15)}{(122.72)(33{,}000)}\ln\!\left(\frac{6.0}{0.50}\right)=\boxed{2.34\times10^{-5}\ \text{cm/s}\ (2.34\times10^{-7}\ \text{m/s})}.$$
Intrinsic permeability at the test temperature. Using water properties AT 25°C ($\mu_{25}=0.890\times10^{-3}\ \text{kg/m-s}$, $\rho_{25}=997.0\ \text{kg/m}^3$), not the paper's 1000 kg/m³/0.001 kg/m-sec default: $$k=\frac{K\mu}{\rho g}=\frac{(2.34\times10^{-7})(0.890\times10^{-3})}{(997.0)(9.81)}=\boxed{2.13\times10^{-14}\ \text{m}^2}.$$ At $K\approx2.3\times10^{-7}\ \text{m/s}$ the material falls in Freeze & Cherry's silt–to–very-fine-sand range, not the clean-sand range the exam's cover story might suggest.

Quantity	Result
(a) Void ratio $e$	0.369
(a) Porosity $n$	26.9%
(a) Bulk density, dry $ ho_b$ (moist)	1.94 g/cm³ (2.15 g/cm³)
(a) Water content, gravimetric $w$	10.9%
(a) Water content, volumetric $\theta$	21.1%
(a) Saturation $S_r$	78.1%
(b) $q_x$, $q_y$	−7.50×10⁻⁸ m/s, −4.50×10⁻⁸ m/s
(b) $\|q\|$, direction	8.75×10⁻⁸ m/s at −149.0° from +x, i.e. 149.0° clockwise (≈S59°W)
(c) Hydraulic conductivity $K$ (at 25°C)	2.34×10⁻⁷ m/s (2.34×10⁻⁵ cm/s)
(c) Intrinsic permeability $k$	2.13×10⁻¹⁴ m²

Concept & Theory

Topic. Soil phase relations, the anisotropic (tensor) form of Darcy's law, and temperature-dependent conversion between hydraulic conductivity and intrinsic permeability.

Key relations. $n=V_v/V_t$, $e=V_v/V_s$, $w=M_w/M_s$, $\theta=V_w/V_t$, $S_r=V_w/V_v$; $q_i=-K_{ij}\,\partial h/\partial x_j$ for a symmetric 2-D conductivity tensor; $K=\dfrac{aL}{At}\ln(h_0/h_1)$; $k=K\mu/(\rho g)$.

Why this works. When hydraulic conductivity varies with direction (layering, fracturing, sedimentary fabric), a single scalar $K$ cannot describe flow that is not aligned with the head gradient — the full tensor is needed, and its off-diagonal terms $K_{xy}=K_{yx}$ couple the $x$- and $y$-gradients into each velocity component. The falling-head test isolates $K$ purely from geometry and the time it takes the head ratio to decay, so $K$ is exactly what was measured at the test temperature; converting to intrinsic permeability is a genuinely fluid-dependent step and must use the fluid properties at that same temperature.

Common pitfall. Treating the conductivity "tensor" as if only the diagonal terms mattered (dropping $K_{xy}$) silently rotates the computed velocity vector away from its true direction; and using the exam's default 20°C-ish viscosity/density instead of the stated 25°C values shifts $k$ by roughly 10–15%.

Source. Freeze & Cherry, Groundwater, Ch. 2.11 (soil phase relations, effective stress), Ch. 2.6 (anisotropic conductivity tensors) and Ch. 2.2 (permeameters, intrinsic permeability).

Question 1 of 5: Core-Sample Soil Phase Relations, Anisotropic Darcy Velocity, and Falling-Head Permeameter

Question 1: Core-Sample Soil Phase Relations, Anisotropic Darcy Velocity, and Falling-Head Permeameter (equal value)