Darcy Velocity, Seepage Velocity, Permeameter K, and Storage Changes: 18-Geol-A2 Hydrogeology, December 2016, Q1

Question 1: Darcy Velocity, Seepage Velocity, Permeameter K, and Storage Changes (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)–(b) $K=10^{-6}\ \text{m/s}$, hydraulic gradient $i = 100\ \text{ft/mile}$, effective porosity $n_e = 0.1$. (c) constant-head permeameter, cross-section $A=225\ \text{cm}^2$, sample length $L=25\ \text{cm}$, applied head $h=15\ \text{cm}$, discharge $Q_v = 50\ \text{cm}^3$ in $t=456\ \text{s}$; fluid properties from the cover page ($\mu=0.001\ \text{kg/(m}\cdot\text{s)}$, $\rho=1000\ \text{kg/m}^3$, $g=9.81\ \text{m/s}^2$). (d) unconfined aquifer, specific yield (storativity) $S_y=0.23$, area $A=3.2\times10^{8}\ \text{m}^2$, water-table drop $\Delta h=1.5\ \text{m}$. (e) confined aquifer, $\Delta V = 200{,}000\ \text{m}^3$, $A=3.2\times10^{8}\ \text{m}^2$, storativity $S=0.0005$.

Sub-part	Datum	Value
a	K	$10^{-6}$ m/s
a	Gradient	100 ft/mile
b	Effective porosity $n_e$	0.1
c	A, L, h, Q, t	225 cm², 25 cm, 15 cm, 50 cm³, 456 s
d	$S_y$, A, $\Delta h$	0.23, $3.2\times10^8$ m², 1.5 m
e	$\Delta V$, A, S	200,000 m³, $3.2\times10^8$ m², 0.0005

Find. (a) Darcy (specific-discharge) velocity; (b) average linear seepage velocity; (c) hydraulic conductivity in cm/s and intrinsic permeability; (d) volume of water released from unconfined storage; (e) head decline corresponding to a given confined-storage release.

[Figure not reproduced: Figure 1 (redrawn). Constant-head permeameter: outflow head $h$ drives flow along length $L$ through cross-section $A$; $Q$ is measured at the outlet over time $t$. See the official exam paper.]

Approach. Apply Darcy's law $v=Ki$ for the specific discharge, divide by porosity for the seepage velocity, use the permeameter form $K=QL/(Aht)$ for the lab test, convert $K$ to intrinsic permeability with $k=K\mu/(\rho g)$, and use $\Delta V = S\,A\,\Delta h$ (rearranged for whichever unknown is asked) for the two storage sub-parts.

(a) Convert the gradient and apply Darcy's law. $100\ \text{ft/mile} = \dfrac{100\times0.3048\ \text{m}}{1609.344\ \text{m}} = 0.01894$ (dimensionless): $$v_{Darcy}=Ki=(10^{-6}\ \text{m/s})(0.01894)=\boxed{1.89\times10^{-8}\ \text{m/s}}.$$
(b) Divide by effective porosity. Seepage (average linear) velocity is the Darcy velocity spread over the pores actually carrying flow: $$v_{seep}=\frac{v_{Darcy}}{n_e}=\frac{1.89\times10^{-8}}{0.1}=\boxed{1.89\times10^{-7}\ \text{m/s}}.$$
(c) Permeameter K, then intrinsic permeability. For a constant-head test, $K = QL/(Aht)$: $$K=\frac{(50\ \text{cm}^3)(25\ \text{cm})}{(225\ \text{cm}^2)(15\ \text{cm})(456\ \text{s})}=\boxed{8.12\times10^{-4}\ \text{cm/s}}.$$ Converting to m/s ($8.12\times10^{-6}\ \text{m/s}$) and applying $k=K\mu/(\rho g)$: $$k=\frac{(8.12\times10^{-6})(0.001)}{(1000)(9.81)}=\boxed{8.28\times10^{-13}\ \text{m}^2}\;(\approx 0.84\ \text{Darcy}).$$
(d) Unconfined storage release. An unconfined aquifer releases water essentially by gravity drainage, so its storativity is the specific yield $S_y$; the volume released is $\Delta V = S_y A\,\Delta h$: $$\Delta V=(0.23)(3.2\times10^{8}\ \text{m}^2)(1.5\ \text{m})=\boxed{1.104\times10^{8}\ \text{m}^3}.$$
(e) Confined head decline. A confined aquifer releases water elastically, $\Delta V = S A\,\Delta h$, so solving for $\Delta h$: $$\Delta h=\frac{\Delta V}{S\,A}=\frac{200{,}000}{(0.0005)(3.2\times10^{8})}=\boxed{1.25\ \text{m}}.$$

Sub-part	Result
(a) Darcy velocity	$1.89\times10^{-8}$ m/s
(b) Seepage velocity	$1.89\times10^{-7}$ m/s
(c) Hydraulic conductivity	$8.12\times10^{-4}$ cm/s
(c) Intrinsic permeability	$8.28\times10^{-13}$ m² (≈ 0.84 Darcy)
(d) Water lost, unconfined storage	$1.104\times10^{8}$ m³
(e) Head decline, confined storage	1.25 m

Concept & Theory

Topic. Darcy's law and the two forms of aquifer storage (specific yield vs. storativity).

Key relations. Darcy's law $v=Ki$ (bulk/specific discharge) and $v_{seep}=v/n_e$ (true average pore velocity); constant-head permeameter $K=QL/(Aht)$; intrinsic permeability $k=K\mu/(\rho g)$; storage-release volume $\Delta V = S\,A\,\Delta h$, with $S=S_y\sim 0.01$–0.3 for unconfined (gravity drainage) aquifers and $S\sim10^{-5}$–$10^{-3}$ for confined (elastic) aquifers.

Why this works. $v_{Darcy}$ is a bulk flux through the *whole* cross-section (voids + grains); dividing by $n_e$ recovers the faster true velocity of water threading through the connected pores, which is what governs contaminant travel time. The permeameter equation is just Darcy's law solved for $K$ using the measured $Q$, $A$, $h/L$ (the gradient) and $t$. The intrinsic permeability $k$ strips out the fluid properties, leaving a property of the porous medium alone — useful for comparing across fluids (e.g., oil vs. water). The storage relations both say "volume released = storage coefficient × area × head change," but the *mechanism* differs by three to four orders of magnitude: unconfined storage is dominated by pores physically draining, confined storage by the small elastic compressibility of water and matrix.

Common pitfall. Using $S$ (confined storativity, dimensionless but tiny) where $S_y$ (specific yield) is needed, or vice-versa — this is exactly the trap Q1(d)/(e) sets up side-by-side. Also, forgetting unit conversions when a gradient or length is given in imperial units (ft/mile here) before combining with an SI $K$.

Source. Freeze & Cherry, Groundwater, Ch. 2 (Darcy's law, permeameters, intrinsic permeability) and Ch. 2.11/8 (storativity and specific yield).

Question 1 of 5: Darcy Velocity, Seepage Velocity, Permeameter K, and Storage Changes

Question 1: Darcy Velocity, Seepage Velocity, Permeameter K, and Storage Changes (equal value)