Darcy

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Darcys Law

• Philip B. Bedient
• Civil and Environmental Engineering
• Rice University

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Darcys Law

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

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Darcys 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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Darcys Experimental Setup
Head loss h1 - h2 determines flow rate
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Darcys Law

• Therefore, V K (?h/?L)
  and since Q VA
• Q KA(dh/dL)

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Conditions

• In General, Darcys 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

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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
  Darcys 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 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 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 velocity vD Darcy velocity

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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

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Equations of Ground Water Flow

• Description of ground water flow is based
  on 1. Darcys 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 Darcys 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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Calculate

• 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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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.

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Limitations of theDarcian Approach

• 1. For Reynolds Number, Re, gt 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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Darcys LawExample 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 Darcys 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

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Darcys Law
Darcys Law can be used to compute flow rate in
almost any aquifer system where heads and areas
are known from wells.