Darcy’s Law and Flow

Philip B. BedientCivil 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

Darcy’s Law

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

Flow in Aquifers

Darcy’s Experiment (1856):

Flow rate determined by Head loss dh = h1 - h2

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 Law

V = – K (∆h/∆L) and since

Q = VA (A = total area)

Q = – KA (dh/dL)

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

Conditions• 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 heterogeneous systems5. Flow in isotropic or anisotropic media6. 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 & 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.

A = total areaAv voids

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 velocityVD = Darcy velocity

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

• 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

€

−∂

∂xρVx( ) −

∂

∂yρVy( ) −

∂

∂zρVz( ) = 0

Mass In - Mass Out = Change in Storage€

ρVx +∂

∂xρVx( )

€

ρVx

z

y

Derivation of 3-D GW Flow Equation from Darcy’s Law

€

∂

∂xρKx

∂h

∂x

⎛

⎝ ⎜

⎞

⎠ ⎟+

∂

∂yρKy

∂h

∂y

⎛

⎝ ⎜

⎞

⎠ ⎟+

∂

∂zρKz

∂h

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

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

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

€

∂ 2h

∂x 2+

∂ 2h

∂y 2+

∂ 2h

∂z2= 0

∇ 2h = 0 called Laplace Eqn.

Transient Saturated Flow

€

∂

∂xKx

∂h

∂x

⎛

⎝ ⎜

⎞

⎠ ⎟+

∂

∂yKy

∂h

∂y

⎛

⎝ ⎜

⎞

⎠ ⎟+

∂

∂zKz

∂h

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟= Ss

∂h

∂t

€

−∂

∂xρVx( ) −

∂

∂yρVy( ) −

∂

∂zρVz( ) =

∂

∂tρn( )

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,

therefore,

Solutions to GW Flow Eqns.

€

∂ 2h

∂x 2+

∂ 2h

∂y 2+

∂ 2h

∂z2= 0

∇ 2h = 0 called Laplace Eqn.

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

h0 h1

Transient Saturated Flow

€

∂

∂x

∂h

∂x

⎛

⎝ ⎜

⎞

⎠ ⎟+

∂

∂y

∂h

∂y

⎛

⎝ ⎜

⎞

⎠ ⎟+

∂

∂z

∂h

∂z

⎛

⎝ ⎜

⎞

⎠ ⎟=

Ss

K

∂h

∂t

Simplifying by assuming K = constant in all dimensions

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

€

∂ 2h

∂x 2+

∂ 2h

∂y 2+

∂ 2h

∂z2=

Ss

K

∂h

∂t

∇ 2h =S

T

∂h

∂t from Jacob, Theis

Steady State Flow to WellSimplifying by assuming K = constant in all dimensions

and assuming that Transmissivity T = Kb and

Q = flow rate to well at point (x,y) yields

€

∂2h

∂x 2+

∂2h

∂y 2= −

Q x, y( )T

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.

Compute:• 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• 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

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.

Limitations of theDarcian Approach1. 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

• 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 Aquifer

30 ft

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 tA = cross-sectional area of the sampleL = length of sampleh = constant head

• t = time of flow