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CE 491 / 591 · Groundwater Hydrology · Module 3

Darcy's Law and Hydraulic Conductivity

Self-paced · about 5 hours plus the OpenHydroQual exercise · prerequisites: Module 1, and Module 2 (scalar and vector fields, the gradient)

Overview

Darcy's law is to groundwater what Ohm's law is to circuits: a simple proportionality between a flow and the push that drives it. Almost every calculation in the rest of the course, from flow to a well to the spread of a contaminant plume, starts here.

By the end of this module you should be able to:
  1. Compute hydraulic head from elevation and pressure, and use it to decide the direction of flow.
  2. Apply Darcy's law to find discharge, Darcy flux and average pore-water velocity, and estimate travel times.
  3. Distinguish hydraulic conductivity from intrinsic permeability and convert between them for different fluids.
  4. Write Darcy's law in three dimensions for an anisotropic medium.
  5. Compute the equivalent horizontal and vertical conductivity of a layered system, and the refraction of flow lines at a layer boundary.
  6. Estimate K from grain size (Hazen, Kozeny–Carman), from constant- and falling-head permeameter tests, and for fractured rock (cubic law).
  7. Check whether Darcy's law applies, using the REV concept and the Reynolds number.

How to work through it: as in Module 1, each section has an animation, a reading and an interactive tool. Videos tagged required are part of the module; optional ones go deeper. Then do the practice problems, the Brightspace quiz, the OpenHydroQual exercise, and Assignments 3 and 4.

1Darcy's law and hydraulic head

requiredAnimation 1: Darcy's column, hydraulic head, Darcy's law, Darcy flux versus pore velocity, and a worked example.
requiredDane Andersen: Groundwater and Head (4 min). Elevation head, pressure head and piezometers.

Darcy's experiment

In 1856 the French engineer Henry Darcy measured the flow of water through vertical columns of sand while designing filters for the public fountains of Dijon. He found that the discharge \(Q\) is proportional to the cross-sectional area \(A\) of the column and to the difference in head \(\Delta h\) between its ends, and inversely proportional to its length \(L\). With a constant of proportionality \(K\), the hydraulic conductivity:

\[ Q = K A\,\frac{\Delta h}{L} \qquad\text{or, in general,}\qquad Q = -K A\,\frac{dh}{dl} \]

\(dh/dl\) is the hydraulic gradient (dimensionless). The minus sign says flow goes in the direction of decreasing head. \(K\) has units of velocity (m/day, cm/s, ft/day).

Hydraulic head

The head at a point is the elevation to which water rises in a piezometer open at that point. It is the mechanical energy per unit weight of water:

\[ h = z + \frac{p}{\rho g} \qquad (+\,v^2/2g,\ \text{negligible in groundwater}) \]

\(z\) is the elevation above a datum and \(p/\rho g\) the pressure head. Water flows from high head to low head, not necessarily from high pressure to low pressure, nor downhill. In a discharge area groundwater flows upward; under a hill it can flow toward points of higher pressure.

optional Tom Gleeson: hydraulic gradient in nested piezometers (12 min), the vertical direction of flow from piezometers at different depths.

Darcy flux and pore-water velocity

Dividing by the area gives the specific discharge or Darcy flux, the flow per unit of total cross-section:

\[ q = \frac{Q}{A} = -K\,\frac{dh}{dl}, \qquad v = \frac{q}{n_e} \]

Water can only pass through the connected pores, a fraction \(n_e\) (the effective porosity) of the cross-section, so the average pore-water (seepage) velocity \(v\) is larger than \(q\). Use \(q\) for discharges and water budgets; use \(v\) for travel times and for how fast a dissolved contaminant is carried.

Example 3.3. A culvert 5 m long and 0.8 m in diameter has been packed with sand (\(K = 3\) m/day, \(n_e = 0.38\)) by a storm. The culvert is submerged and the water level at one end is 1.6 m higher than at the other. Then \(A = \pi(0.4)^2 = 0.503\) m², \(dh/dl = 1.6/5 = 0.32\), \(Q = 3(0.503)(0.32) = 0.483\) m³/day, \(q = 0.96\) m/day and \(v = 0.96/0.38 = 2.53\) m/day. Water takes about two days to get through.

Virtual Darcy column

Gradient Δh/L
Discharge Q
Darcy flux q
Pore velocity v
Travel time L/v
Reynolds number

Total head, elevation head and pressure head along the column

Predict, then check: tilt the column so the outlet is higher. Does the flow reverse?
No. Flow depends only on \(h_{in} - h_{out}\). Tilting changes how the head is split between elevation and pressure, not the total head, so \(Q\) is unchanged. Water flows uphill whenever the inlet head is higher.
Predict, then check: when does the pressure head go negative?
Where the column rises above the total-head line, \(p/\rho g = h - z < 0\): the water is under suction. That cannot last in a real saturated column; air would enter. Try a steep upward tilt with low heads.
Predict, then check: make K very large with coarse grains. What happens to the Reynolds number?
It rises above 10: flow becomes non-laminar and Darcy's law overpredicts \(Q\). This happens in coarse gravel and karst, rarely in ordinary aquifers.

2Hydraulic conductivity, permeability and Darcy's law in 3-D

requiredGeosearch International: Hydrogeology 101: Introduction to Groundwater Flow (20 min). Darcy's law, conductivity and gradients, from the same series as Module 1.
optionalOn The Rocks: Hydraulic Conductivity, Permeability, and Darcy's Law (21 min).

Intrinsic permeability

Hubbert (1956) noted that \(K\) depends on both the medium and the fluid. Flow is faster for a denser fluid (larger driving weight \(\rho g\)), slower for a more viscous one (\(\mu\)), and proportional to the square of the mean grain diameter \(d\):

\[ K = \frac{C d^2 \rho g}{\mu} = \frac{k\,\rho g}{\mu}, \qquad k = C d^2 \]

\(k\) is the intrinsic permeability, a property of the porous medium alone, with units of area (m², cm²). Petroleum engineers use the darcy: 1 darcy \(= 9.87\times10^{-9}\) cm² \(= 9.87\times10^{-13}\) m². For water at 20 °C, \(\rho g/\mu \approx 9.8\times10^{6}\) m⁻¹s⁻¹, so 1 darcy corresponds to \(K \approx 10^{-5}\) m/s \(\approx 0.84\) m/day.

Example 3.4. How much more conductive is a sand to PCE (tetrachloroethylene: \(\rho = 1.63\) g/cm³, \(\mu = 9.0\times10^{-4}\) N·s/m²) than to water (\(\rho = 1.0\) g/cm³, \(\mu = 1.0\times10^{-3}\) N·s/m²)? With the same \(k\), \(\dfrac{K_{PCE}}{K_w} = \dfrac{\rho_{PCE}}{\rho_w}\dfrac{\mu_w}{\mu_{PCE}} = 1.63 \times \dfrac{1.0}{0.9} = 1.81\). Dense, low-viscosity solvents move through aquifers faster than water under the same gradient.

Typical values

\(K\) varies over more orders of magnitude than almost any other physical property used in engineering (Fetter, 2001):

MaterialK (cm/s)K (m/day)
Clay10⁻⁹ – 10⁻⁶10⁻⁶ – 10⁻³
Silt, sandy silt, clayey sand, till10⁻⁶ – 10⁻⁴10⁻³ – 0.1
Silty sand, fine sand10⁻⁵ – 10⁻³0.01 – 1
Well-sorted sand, glacial outwash10⁻³ – 10⁻¹1 – 100
Well-sorted gravel10⁻² – 110 – 1000

Homogeneity, isotropy and Darcy's law in three dimensions

A medium is homogeneous if \(K\) is the same at every point, otherwise heterogeneous. It is isotropic if \(K\) at a point is the same in every direction, otherwise anisotropic. Flow in an aquifer is three-dimensional. When the coordinate axes are aligned with the principal directions of \(K\):

\[ q_x = -K_x \frac{\partial h}{\partial x}, \qquad q_y = -K_y \frac{\partial h}{\partial y}, \qquad q_z = -K_z \frac{\partial h}{\partial z} \]

In vector form \(\mathbf{q} = -\mathbf{K}\,\nabla h\) with \(\mathbf{K} = \mathrm{diag}(K_x, K_y, K_z)\), and the magnitude of the flux is \(|\mathbf{q}| = \sqrt{q_x^2+q_y^2+q_z^2}\). This is where the gradient from Module 2 comes in: in an isotropic medium \(\mathbf{q}\) points straight down the head gradient, perpendicular to the contours of equal head. In an anisotropic medium it is turned toward the direction of larger \(K\).

Example 3.5 (simplified). In an unconfined sand, the horizontal head gradient is \(\partial h/\partial x = -0.0011\) and \(K_x = 8\) ft/day, so \(q_x = 0.0088\) ft/day. Recharge of 1.25 ft/yr \(= 0.00342\) ft/day crosses the water table downward. If piezometers show \(\partial h/\partial z = 0.03\) (head increasing upward, flow downward), then \(K_z = 0.00342/0.03 = 0.114\) ft/day: about 70 times smaller than \(K_x\). The total flux is \(\sqrt{0.0088^2 + 0.00342^2} = 0.0094\) ft/day, dipping \(\tan^{-1}(0.00342/0.0088) = 21°\) below horizontal. You will do the full three-piezometer version in Assignment 3.

3Layered systems and the refraction of flow lines

requiredAnimation 2: equivalent K for flow parallel to and across layers, the head profile, and refraction at a layer boundary.
optionalGroundwater Modeling: Darcy's Law, Layered Systems (16 min). A whiteboard derivation of both formulas.

Equivalent hydraulic conductivity

Layering is the most common heterogeneity, because sediments are deposited in successive beds. We replace the stack of layers (thicknesses \(b_i\), conductivities \(K_i\)) by one equivalent, homogeneous but anisotropic medium.

Flow parallel to the layers. Every layer has the same gradient \(dh/dl\), so \(Q_i = K_i b_i\,(dh/dl)\) per unit width, and the flows add: \(\sum K_i b_i\,(dh/dl) = K_h\left(\sum b_i\right)(dh/dl)\).

Flow across the layers. The same flux \(q\) passes through every layer, so the head drops add: \(\Delta h = \sum \Delta h_i = q\sum b_i/K_i = q \sum b_i / K_v\).

\[ K_h = \frac{\sum_i K_i b_i}{\sum_i b_i} \quad\text{(arithmetic mean)}, \qquad K_v = \frac{\sum_i b_i}{\sum_i b_i/K_i} \quad\text{(harmonic mean)}, \qquad K_h \ge K_v \]

The most permeable layer controls horizontal flow; the least permeable controls vertical flow. A single thin clay layer can make a thick sand nearly impermeable to vertical flow.

Layered-system calculator

Edit the layers, top to bottom. K in m/day, b in m.

Total thickness
Kh (arithmetic)
Kv (harmonic)
Anisotropy Kh/Kv

Left: horizontal flux in each layer (same gradient). Right: head profile for vertical flow (same q).

Refraction of flow lines

Where flow crosses a boundary from \(K_1\) to \(K_2\), two conditions hold. The flux normal to the boundary is continuous (mass conservation), \(K_1\,\partial h/\partial n|_1 = K_2\,\partial h/\partial n|_2\). And the head is continuous along the boundary, so its tangential derivative is the same on both sides. Dividing one by the other, with angles measured from the normal:

\[ \frac{\tan\theta_1}{\tan\theta_2} = \frac{K_1}{K_2} \]

Example 3.7. With \(K_1 = 10^{-6}\) cm/s, \(K_2 = 10^{-3}\) cm/s and \(\theta_1 = 5°\): \(\tan\theta_2 = \tan 5° \times 1000 = 87.5\), so \(\theta_2 = 89.3°\). Flow entering the aquifer turns almost parallel to it. This is why flow is nearly horizontal in aquifers and nearly vertical in aquitards.

K₂/K₁
Refracted angle θ₂

A flow line crossing the boundary (K in m/day)

4Estimating and measuring K

requiredAnimation 3: permeameters, grain-size formulas, the cubic law for fractures, and when Darcy's law applies.
requiredSoil Mechanics and Engineering Geology: Falling head permeability test, procedure and data analysis (8 min). The real lab test behind the virtual one below.

optional Constant head permeability test, procedure and data analysis (7 min), same channel.

Answering almost any quantitative groundwater question requires \(K\). There are three ways to get it: empirical formulas from grain size, laboratory tests on samples, and field tests (pumping and slug tests, covered with well hydraulics). Field tests sample the largest volume and are usually the most representative.

Grain-size formulas

\[ \text{Hazen:}\quad K = C\,d_{10}^{\,2} \qquad\qquad \text{Kozeny–Carman:}\quad K = \frac{\rho g}{\mu}\,\frac{d_m^{\,2}}{180}\,\frac{n^3}{(1-n)^2} \]

In Hazen's formula \(K\) is in cm/s and \(d_{10}\) in cm, with \(C\) between about 40 (very fine, poorly sorted sand) and 150 (coarse, well-sorted sand). It is meant for sands with \(d_{10}\) between about 0.1 and 3 mm. Kozeny–Carman uses a representative grain size \(d_m\) (often \(d_{50}\)) and the porosity, in any consistent units.

Example 3.8. A sand has \(d_{10} = 0.1\) mm, \(d_{50} = 0.24\) mm and \(n = 0.38\). Hazen with \(C = 100\)–\(150\): \(K = C(0.01)^2 = 0.010\)–\(0.015\) cm/s. Kozeny–Carman with \(d_{50}\): \(k = \dfrac{(2.4\times10^{-4})^2}{180}\,\dfrac{0.38^3}{0.62^2} = 4.57\times10^{-11}\) m², \(K = 9.81\times10^{6}\,k = 4.5\times10^{-4}\) m/s \(= 0.045\) cm/s. A factor of three or four between formulas is typical.

Permeameters

Constant head (sands, gravels): keep the head difference \(h\) across a sample of length \(L\) and area \(A\) fixed, and collect a volume \(V\) in time \(t\). Darcy's law gives directly

\[ K = \frac{V L}{A\,t\,h} \]

Falling head (silts, clays, where flow is too slow to collect): the head in a narrow tube of diameter \(d_t\) above a sample of diameter \(d_c\) falls as water drains through. The water leaving the tube equals the flow through the sample, \(-\frac{\pi d_t^2}{4}\frac{dh}{dt} = K\,\frac{\pi d_c^2}{4}\,\frac{h}{L}\). Separating variables and integrating from \(h_0\) at \(t = 0\):

\[ K = \frac{d_t^{\,2}\,L}{d_c^{\,2}\,t}\,\ln\frac{h_0}{h} \]

Example 3.9. A sandy-silt sample 30 cm long and 10 cm in diameter, with an expected \(K \approx 10^{-5}\) cm/s, is tested with an 8 mm burette. The time for the head to halve is \(t = \dfrac{d_t^2 L}{d_c^2 K}\ln 2 = \dfrac{(0.8)^2(30)}{(10)^2(10^{-5})}(0.693) \approx 13{,}000\) s \(\approx 3.7\) hours.

Virtual falling-head permeameter

Each sample has an unknown K. Run the test, read the head at a few times, then compute K from the data, either from two readings or from the slope of ln(h₀/h) against t.

Sample length L
Sample diameter dc
Burette diameter dt
Initial head h₀
Elapsed time
Current head h
Your K =cm/s

Your readings

t (s)h (cm)ln(h₀/h)

Flow in fractured rock: the cubic law

Seepage under dams, into tunnels and through crystalline rock happens mostly in fractures. If the fractures are idealized as smooth parallel plates with aperture \(b\), laminar flow between them gives, per fracture of width \(w\),

\[ Q = \frac{\rho g\, b^3 w}{12\,\mu}\,\frac{dh}{dl} \qquad\Longrightarrow\qquad K_{eq} = \frac{N \rho g\, b^3}{12\,\mu} \]

where \(N\) is the number of fractures per unit distance normal to them. Doubling the aperture multiplies the flow by eight. Real fractures are rough and variable, so the cubic law gives an order of magnitude.

Example 3.10. A granite has one fracture every 2 m (\(N = 0.5\) m⁻¹), aperture 0.3 mm: \(K = \dfrac{0.5(1000)(9.81)(3\times10^{-4})^3}{12(1.0\times10^{-3})} = 1.10\times10^{-5}\) m/s, about 1 m/day: the same as a fine sand, from rock that is almost entirely solid.

When does Darcy's law apply?

Scale. The representative elementary volume (REV) is the smallest volume over which a measured property is representative of the whole. Below the REV, porosity jumps between 0 (a grain) and 1 (a pore), and \(K\) is undefined. Darcy's law applies at the REV scale and above.

Laminar flow. Darcy's law assumes viscous forces dominate inertia. With the Reynolds number based on the Darcy flux and the mean grain diameter,

\[ Re = \frac{\rho\, q\, d}{\mu} \qquad\text{Darcy's law is valid for } Re \lesssim 1 \text{ to } 10 \]

That covers nearly all natural groundwater flow. Exceptions are karst conduits, very coarse gravels, and the immediate vicinity of large pumping wells.

5Practice (ungraded)

New numbers every time. Aim for two correct attempts in a row on each type before taking the quiz. Scientific notation can be typed as 2.5e-4.

A · Darcy's law: discharge, flux, velocity, travel time

B · Hydraulic head and the direction of flow

C · Intrinsic permeability and other fluids

D · Darcy's law in 2-D with anisotropy

E · Equivalent K of a layered system

F · Refraction at a layer boundary

G · Permeameter tests

H · Hazen and Kozeny–Carman

I · Fractured rock (cubic law) and the Reynolds number

Concept checks

1. Piezometer A is screened 10 m deep with a water level 2.0 m below the ground; piezometer B, beside it, is screened 30 m deep with a water level 1.5 m below the ground. Which way is the vertical flow?

2. A sand aquifer 20 m thick contains a continuous clay layer 0.5 m thick. Which statement is correct?

3. A dye tracer travels 30 m between two wells in 10 days. A student uses 3 m/day as the Darcy flux \(q\) to compute the discharge through the aquifer. What is wrong?

6Graded quiz

7OpenHydroQual exercise: Darcy's column in a computer

You will run a numerical version of Darcy's experiment, a sand column split into ten cells between two reservoirs, check that the model reproduces Darcy's law, and then put a less permeable layer in the column and compare the model with the series formula for \(K_v\).

Open the exercise instructions →