darcy’s law and flowlibvolume3.xyz/civil/btech/semester6/groundwater... · 2014-12-30 ·...
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Darcy’s
Law and Flow
Philip B. Bedient
Civil and Environmental Engineering
Rice University
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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
• Darcy’s law provides an accurate description of the flow of ground water in almost all hydrogeologic environments.
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Flow in Aquifers
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Darcy’s Experiment (1856):
Flow rate determined by Head loss dh = h1 - h2
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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 α – ∆h, and V α 1/∆L
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Darcy’s Law
V = – K (∆h/∆L)
and since
Q = VA (A = total area)
Q = – KA (dh/dL)
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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
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Conditions
• 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
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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 Darcy’s 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.
A = total area Av voids
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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
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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 Groundwater Flow
• Description of ground water flow is based on: Darcy’s Law Continuity 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
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Derivation of 3-D GW Flow
Equation from Darcy’s Law
−∂
∂xρV x( )−
∂∂y
ρV y( )−∂
∂zρV z( ) = 0
Mass In - Mass Out = Change in Storage
ρVx +∂
∂xρVx( ) ρVx
z
y
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Derivation of 3-D GW Flow
Equation from Darcy’s Law
∂
∂xρK x
∂h∂x
+
∂∂y
ρK y
∂h∂y
+
∂∂z
ρK z
∂h∂z
= 0
Replace Vx, Vy, and Vz with Darcy using Kx, Ky, and Kz
Divide out constant ρ, and assume Kx= Ky= Kz = K
∂ 2h
∂x 2+
∂ 2h
∂y 2+
∂ 2h
∂z 2= 0
∇ 2h = 0 called Laplace Eqn.
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Transient Saturated Flow
∂
∂xK x
∂h∂x
+
∂∂y
K y
∂h∂y
+
∂∂z
K z
∂h∂z
= S s
∂h∂t
−∂
∂xρVx( ) −
∂∂y
ρVy( )−∂
∂zρV z( ) =
∂∂t
ρn( )
A change in h will produce change in ρ and n, replaced with specific storage Ss = ρg(α + nΒ). Note, α is the compressibility of aquifer and B is comp of water,
therefore,
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Solutions to GW Flow Eqns.
∂ 2h
∂x 2+
∂ 2h
∂y 2+
∂ 2h
∂z 2= 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
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Transient Saturated Flow
∂
∂x∂h∂x
+
∂∂y
∂h∂y
+
∂∂z
∂h∂z
=
S sK
∂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
∂z 2=S sK
∂h∂ t
∇ 2h =S
T
∂h∂t
from Jacob, Theis
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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
∂2h
∂x 2+
∂2h
∂y 2= −
Q x, y( )T
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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.
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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
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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 the
Darcian 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)
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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.
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Confined Aquifer
Confining Layer Aquifer
30 ft
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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 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