Transcript Darcy's Law - Rice University

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
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.
2.
3.
4.
Saturated flow and unsaturated flow
Steady-state and transient flow
Flow in aquifers and aquitards
Flow in homogeneous and
heterogeneous systems
5. Flow in isotropic or anisotropic media
6. 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.
Av voids
A = total area
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 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 = V D / n
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
Derivation of 3-D GW Flow
Equation from Darcy’s Law
z

Vx  Vx 
x
Vx

y
Mass In - Mass Out =Change in Storage




Vx   Vy  Vz   0
x
y
z
Derivation of 3-D GW Flow
Equation from Darcy’s Law
Replace Vx, Vy, and Vz with Darcy using Kx, Ky, and Kz
 
h   
h   
h 
K x

 0
K y
 K z
x 
x  y 
y  z 
z 
Divide out constant , and assume Kx= Ky= Kz = K
 2h  2h  2h
 2  2 0
2
x
y
z
 2 h  0 called Laplace Eqn.
Transient Saturated Flow





Vx   Vy  Vz   n
x
y
z
t
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,
  h    h    h 
h
K x

 Ss
K y
 K z
x  x  y  y  z  z 
t
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.
 2h  2h  2h
 2  2 0
2
x
y
z
 2 h  0 called Laplace Eqn.
h0

h1
Transient Saturated Flow
Simplifying by assuming K = constant in all dimensions
And assuming that S = Ssb, and that T = Kb yields
 h   h   h  Ss h
 
   
x x  y y  z z  K t

 2 h  2 h  2 h Ss h
 2  2 
2
x
y
z
K t
S h
2
 h
from Jacob,T heis
T t
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
Qx, y 
2 h 2 h
 2 
2
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 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)
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
30 ft
Aquifer
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 t
A = cross-sectional area of the sample
L = length of sample
h = constant head
•
t = time of flow