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Darcy’s Law
Philip B. BedientCivil and Environmental Engineering
Rice University
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Darcy’s Law
• Darcy’s law provides an accurate description of the flow of ground water in almost all hydrogeologic environments.
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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
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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
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Darcy’s Experimental Setup:
Head loss h1 - h2 determines flow rate
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Darcy’s Law• Therefore,
V = – K (∆h/∆L) and since Q= VA
• Q= – KA(dh/dL)
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Conditions of Law
• In General, Darcy’s Law holds for:
1. Saturated flow and unsaturated flow2. Steady-state and transient flow3. Flow in aquifers and aquitards4. Flow in homogeneous and heteogeneous systems5. Flow in isotropic or anisotropic media6. Flow in rocks and granular media
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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 ….
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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.
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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.
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Darcy Velocity & Seepage Velocity
• From the Continuity Eqn:
• Q = A vD = AV Vs
– Where:Q = flow rate
A = cross-sectional area of material
AV= area
of voids Vs
= seepage velocityvD= Darcy velocity
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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
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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.
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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.
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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
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…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
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...to continue
• 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.
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Limitations of theDarcian Approach1. 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)
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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.
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Confined Aquifer
Confining Layer
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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
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Permeameters
Constant Head Falling Head
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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 tA = cross-sectional area of the sampleL = length of sampleh = constant head
• t = time of flow
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Darcy’s Law
Darcy’s Law can be used to compute flow rate in almost any aquifer system where heads and areas are known from wells.