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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 .

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darcy s law and flow

Darcy’s Law and Flow

Philip B. Bedient

Civil and Environmental Engineering

Rice University

darcy allows an estimate of
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
  • Darcy’s law provides an accurate description of the flow of ground water in almost all hydrogeologic environments.
darcy s experiment 1856
Darcy’s Experiment (1856):

Flow rate determined by Head loss dh = h1 - h2

darcy s law6
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
darcy s law7
Darcy’s Law

V = – K (∆h/∆L) and since

Q = VA (A = total area)

Q = – KA (dh/dL)

hydraulic conductivity
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
  • 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
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 velocity11
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 & 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.

Av voids

A = total area

darcy seepage velocity13
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 velocity14
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
Equations of Groundwater Flow
  • Description of ground water flow is based on: Darcy’s LawContinuity 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
Derivation of 3-D GW Flow Equation from Darcy’s Law


Mass In - Mass Out = Change in Storage


derivation of 3 d gw flow equation from darcy s law17
Derivation of 3-D GW Flow Equation from Darcy’s Law

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

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

transient saturated flow
Transient Saturated Flow

A change in h will produce change in  and n, replaced

with specific storage Ss = g( + n). Note, is the compressibility of aquifer and B is comp of water,


solutions to gw flow eqns
Solutions to GW Flow Eqns.

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



transient saturated flow20
Transient Saturated Flow

Simplifying by assuming K = constant in all dimensions

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

steady state flow to well
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

example of darcy s law
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.
  • 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
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) / (15x 104 m2) =0.25m/day
  • 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
Limitations of theDarcian 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
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
Confined Aquifer

Confining Layer


30 ft

example 2
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

Constant Head

Falling Head

constant head permeameter
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 sample L = length of sample h = constant head
  • t = time of flow