Darcy’s

Law and Flow

Philip B. Bedient

Civil and Environmental Engineering

Rice University

Darcy allows an estimate of:

• the velocity or flow rate moving within the aquifer

• the average time of travel from the head of the

aquifer to a point located downstream

Darcy’s Law

• Darcy’s law provides an accurate description of the flow of ground water in almost all hydrogeologic environments.

Flow in Aquifers

Darcy’s Experiment (1856):

Flow rate determined by Head loss dh = h1 - h2

Darcy’s Law

• Henri Darcy established empirically that the

flux of water through a permeable formation

is proportional to the distance between top

and bottom of the soil column.

• The constant of proportionality is called the

hydraulic conductivity (K).

• V = Q/A, V α – ∆h, and V α 1/∆L

Darcy’s Law

V = – K (∆h/∆L)

and since

Q = VA (A = total area)

Q = – KA (dh/dL)

Hydraulic Conductivity

• K represents a measure of the ability for flow

through porous media:

• Gravels - 0.1 to 1 cm/sec

• Sands - 10-2 to 10-3 cm/sec

• Silts - 10-4 to 10-5 cm/sec

• Clays - 10-7 to 10-9 cm/sec

Conditions

• Darcy’s Law holds for:

1. Saturated flow and unsaturated flow

2. Steady-state and transient flow

3. Flow in aquifers and aquitards

4. Flow in homogeneous and

heterogeneous systems

5. Flow in isotropic or anisotropic media

6. Flow in rocks and granular media

Darcy Velocity

• V is the specific discharge (�Darcy velocity).

• (–) indicates that V occurs in the direction of

the decreasing head.

• Specific discharge has units of velocity.

• The specific discharge is a macroscopic

concept, and is easily measured. It should be

noted that Darcy’s velocity is different ?.

Darcy Velocity

• ...from the microscopic velocities

associated with the actual paths if

individual particles of water as they wind

their way through the grains of sand.

• The microscopic velocities are real, but

are probably impossible to measure.

Darcy & Seepage Velocity

• Darcy velocity is a fictitious velocity since it assumes that flow occurs across the entire cross-section of the soil sample. Flow actually takes place only through interconnected pore channels.

A = total area Av voids

Darcy & Seepage Velocity

• From the Continuity Eqn:

• Q = A vD = AV Vs

– Where:

Q = flow rate

A = total cross-sectional area of

material

AV = area of voids

Vs = seepage velocity

VD = Darcy velocity

Darcy & Seepage Velocity

• Therefore: VS = VD ( A/AV)

• Multiplying both sides by the length of the

medium (L)

VS = VD ( AL / AVL ) = VD ( VT / VV )

• Where:

VT = total volume

VV = void volume

• By Definition, Vv / VT = n, the soil porosity

• Thus VS = VD / n

Equations of Groundwater Flow

• Description of ground water flow is based on: Darcy’s Law Continuity Equation - describes conservation of fluid mass during flow through a porous medium; results in a partial differential equation of flow.

• Laplace’s Eqn - most important in math

Derivation of 3-D GW Flow

Equation from Darcy’s Law

−∂

∂xρV x( )−

∂∂y

ρV y( )−∂

∂zρV z( ) = 0

Mass In - Mass Out = Change in Storage

ρVx +∂

∂xρVx( ) ρVx

z

y

Derivation of 3-D GW Flow

Equation from Darcy’s Law

∂

∂xρK x

∂h∂x

+

∂∂y

ρK y

∂h∂y

+

∂∂z

ρK z

∂h∂z

= 0

Replace Vx, Vy, and Vz with Darcy using Kx, Ky, and Kz

Divide out constant ρ, and assume Kx= Ky= Kz = K

∂ 2h

∂x 2+

∂ 2h

∂y 2+

∂ 2h

∂z 2= 0

∇ 2h = 0 called Laplace Eqn.

Transient Saturated Flow

∂

∂xK x

∂h∂x

+

∂∂y

K y

∂h∂y

+

∂∂z

K z

∂h∂z

= S s

∂h∂t

−∂

∂xρVx( ) −

∂∂y

ρVy( )−∂

∂zρV z( ) =

∂∂t

ρn( )

A change in h will produce change in ρ and n, replaced with specific storage Ss = ρg(α + nΒ). Note, α is the compressibility of aquifer and B is comp of water,

therefore,

Solutions to GW Flow Eqns.

∂ 2h

∂x 2+

∂ 2h

∂y 2+

∂ 2h

∂z 2= 0

∇ 2h = 0 called Laplace Eqn.

Solutions for only a few simple problems can be obtained directly - generally need to apply numerical methods to address complex boundary conditions.

h0 h1

Transient Saturated Flow

∂

∂x∂h∂x

+

∂∂y

∂h∂y

+

∂∂z

∂h∂z

=

S sK

∂h∂t

Simplifying by assuming K = constant in all dimensions

And assuming that S = Ssb, and that T = Kb yields

∂ 2h

∂x 2+

∂ 2h

∂y 2+

∂ 2h

∂z 2=S sK

∂h∂ t

∇ 2h =S

T

∂h∂t

from Jacob, Theis

Steady State Flow to Well

Simplifying by assuming K = constant in all dimensions

and assuming that Transmissivity T = Kb and

Q = flow rate to well at point (x,y) yields

∂2h

∂x 2+

∂2h

∂y 2= −

Q x, y( )T

Example of Darcy’s Law

• A confined aquifer has a source of recharge.

• K for the aquifer is 50 m/day, and n is 0.2.

• The piezometric head in two wells 1000 m

apart is 55 m and 50 m respectively, from a

common datum.

• The average thickness of the aquifer is 30

m, and the average width of aquifer is 5 km.

Compute:

• a) the rate of flow through the aquifer

• (b) the average time of travel from the head of the

aquifer to a point 4 km downstream

• *assume no dispersion or diffusion

The solution

• Cross-Sectional area=

30(5)(1000) = 15 x 104

m2

• Hydraulic gradient =

(55-50)/1000 = 5 x 10-3

• Rate of Flow for K = 50 m/day

Q = (50 m/day) (75 x 101

m2)

= 37,500 m3/day

• Darcy Velocity:

V = Q/A = (37,500m3/day) / (15

x 104

m2) = 0.25m/day

And • Seepage Velocity:

Vs = V/n = (0.25) / (0.2) =

1.25 m/day (about 4.1 ft/day)

• Time to travel 4 km downstream:

T = 4(1000m) / (1.25m/day) =

3200 days or 8.77 years

• This example shows that water moves

very slowly underground.

Limitations of the

Darcian Approach

1. For Reynold’s Number, Re, > 10 or where the flow is turbulent, as in the immediate vicinity of pumped wells.

2. Where water flows through extremely fine-grained

materials (colloidal clay)

Darcy’s Law:

Example 2

• A channel runs almost parallel to a river, and they

are 2000 ft apart.

• The water level in the river is at an elevation of 120

ft and 110ft in the channel.

• A pervious formation averaging 30 ft thick and with

K of 0.25 ft/hr joins them.

• Determine the rate of seepage or flow from the

river to the channel.

Confined Aquifer

Confining Layer Aquifer

30 ft

Example 2

• Consider a 1-ft length of river (and channel).

Q = KA [(h1 – h2) / L]

• Where:

A = (30 x 1) = 30 ft2

K = (0.25 ft/hr) (24 hr/day) = 6 ft/day

• Therefore,

Q = [6 (30) (120 – 110)] / 2000

= 0.9 ft3/day/ft length = 0.9 ft2/day

Permeameters

Constant Head Falling Head

Constant head Permeameter

• Apply Darcy’s Law to find K: V/t = Q = KA(h/L) or: K = (VL) / (Ath)

• Where: V = volume flowing in time t A = cross-sectional area of the sample L = length of sample h = constant head

• t = time of flow