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Darcy’s Law and Flow
CIVE 6361
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
x
Vx y
Vy z
Vz 0
Mass In - Mass Out = Change in Storage
Vx x
Vx
Vx
z
y
Derivation of 3-D GW Flow Equation from Darcy’s Law
x
Kx
h
x
y
Ky
h
y
z
Kz
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
x
Kx
h
x
y
Ky
h
y
z
Kz
h
z
Ss
h
t
x
Vx y
Vy z
Vz t
n
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