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Darcy's Law

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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
    2. Steady-state and transient 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
  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 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
     • 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
...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.
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




  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
 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.

				
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