# Darcy's Law by niusheng11

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```									    Darcy’s
Law
Philip B. Bedient
Civil and Environmental Engineering
Rice University
Darcy’s Law

• Darcy’s law provides an
accurate description of the flow
of ground water in almost all
hydrogeologic environments.
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  –∆h, and v  1/∆L
Hydraulic Conductivity
• K represents a measure of the ability for flow
through porous media:

•   K is highest for gravels - 0.1 to 1 cm/sec
•   K is high for sands - 10-2 to 10-3 cm/sec
•   K is moderate for silts - 10-4 to 10-5 cm/sec
•   K is lowest for clays - 10-7 to 10-9 cm/sec
Darcy’s Experimental Setup:
Head loss h1 - h2 determines flow rate
Darcy’s Law
• Therefore,

V = – K (∆h/∆L)
and since
Q= VA
•       Q= – KA(dh/dL)
Conditions of Law
• In General, Darcy’s Law holds
for:
1. Saturated flow and unsaturated flow
3. Flow in aquifers and aquitards
4. Flow in homogeneous and
heteogeneous 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
• 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 Velocity &
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.
Darcy Velocity &
Seepage Velocity
• From the Continuity Eqn:
•    Q = A vD = A V Vs
– Where:
Q = flow rate
A = cross-sectional area of
material
AV= area of voids
Vs = seepage velocity
vD= Darcy velocity
Darcy Velocity &
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
Ground Water Flow

• Description of ground water flow is based
on:
1. Darcy’s Law

2. Continuity Equation - describes
conservation of fluid mass
during flow through a porous
medium; results in a partial
differential equation of flow.
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 is 5 km.
Determine the
following:
• a) the rate of flow through the aquifer
• (b) the 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
(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
...to continue

• Seepage Velocity:
Vs = V/n = (0.25) / (0.2) =

• 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 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
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

• 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