Darcy’s law
Groundwater Hydraulics
Daene C. McKinney
Outline
•
•
•
•
•
Darcy’s Law
Hydraulic Conductivity
Heterogeneity and Anisotropy
Refraction of Streamlines
Generalized Darcy’s Law
Darcy
http://biosystems.okstate.edu/Darcy/English/index.htm
Darcy’s Experiments
• Discharge is
Proportional to
hL
P1/g
– Area
– Head difference
• Coefficient of
proportionality is
v
h1
Q
z1
QA
L
Sand
column
Datum
plane
K = hydraulic conductivity
h1  h2
h2
,A
– Length
P2/g
Ar
ea
Inversely proportional to
L
z2
Q
Q = -KA
h2 - h1
L
Q = -KA
Dh
L
Darcy’s Data
35
Flow, Q (l/min)
Set 1, Series 1
30
Set 1, Series 2
25
Set 1, Series 3
Set 1, Series 4
20
Set 2
15
10
5
0
0
5
10
Gradient (m/m)
15
20
Hydraulic Conductivity
• Has dimensions of velocity [L/T]
• A combined property of the medium and the fluid
• Ease with which fluid moves through the medium
k
ρ
µ
g
= cd2
=
=
=
intrinsic permeability
density
dynamic viscosity
specific weight
Porous medium property
Fluid properties
Hydraulic Conductivity
Groundwater Velocity
• q - Specific discharge
Discharge from a unit crosssection area of aquifer
formation normal to the
direction of flow.
• v - Average velocity
Average velocity of fluid
flowing per unit crosssectional area where flow is
ONLY in pores.
q
Q
A
v
q


Q
A
Example
h1 = 12m
K = 1x10-5 m/s
 = 0.3
Find q, Q, and v
h2 = 12m
/”
10m
Flow
Porous medium
5m
L = 100m
dh = (h2 - h1) = (10 m – 12 m) = -2 m
J = dh/dx = (-2 m)/100 m = -0.02 m/m
q = -KJ = -(1x10-5 m/s) x (-0.02 m/m) = 2x10-7 m/s
Q = qA = (2x10-7 m/s) x 50 m2 = 1x10-5 m3/s
v = q/ = 2x10-7 m/s / 0.3 = 6.6x10-7 m/s
Hydraulic Gradient
Gradient vector points in the direction of greatest rate of increase of h
Specific discharge vector points in the opposite direction of h
Well Pumping in an Aquifer
Hydraulic gradient
y
Circular hydraulic
head contours
Dh
K, conductivity,
Is constant
q
Specific discharge
x
Well, Q
h1
h2
h3
h1 < h2 < h3
Aquifer (plan view)
Validity of Darcy’s Law
• We ignored kinetic energy (low velocity)
• We assumed laminar flow
• We can calculate a Reynolds Number for the flow
rqd10
NR =
m
q = Specific discharge
d10 = effective grain size diameter
• Darcy’s Law is valid for NR < 1 (maybe up to 10)
Specific Discharge vs Head Gradient
Re = 10
Re = 1
Experiment
shows this
Darcy Law
predicts this
a
tan-1(a)=
q
(1/K)
Estimating Conductivity
Kozeny – Carman Equation
• Kozeny used bundle of capillary tubes model to derive an
expression for permeability in terms of a constant (c) and
the grain size (d)
3

 2


d
k  cd 
 180(1   ) 2 


2
Kozeny – Carman eq.
• So how do we get the parameters we need for this
equation?
Measuring Conductivity
Permeameter Lab Measurements
• Darcy’s Law is useless unless we can measure the
parameters
• Set up a flow pattern such that
– We can derive a solution
– We can produce the flow pattern experimentally
• Hydraulic Conductivity is measured in the lab with a
permeameter
– Steady or unsteady 1-D flow
– Small cylindrical sample of medium
Measuring Conductivity
Constant Head Permeameter
• Flow is steady
• Sample: Right circular cylinder
– Length, L
– Area, A
Continuous Flow
head difference
• Constant head difference (h) is
applied across the sample
producing a flow rate Q
flow
• Darcy’s Law
Q = KA
b
L
Overflow
A
Outflow
Q
Sample
Measuring Conductivity
Falling Head Permeameter
• Flow rate in the tube must equal that in the column
2
Qtube = prtube
dh
dt
2
Qcolumn = prcolumn
K
Initial head
h
L
Final head
æ rtube ö æ L ö dh
ç
÷ ç ÷ = dt
r
è column ø è K ø h
2
flow
Outflow
Q
Sample
Heterogeneity and Anisotropy
• Homogeneous
– Properties same at every
point
• Heterogeneous
– Properties different at every
point
• Isotropic
– Properties same in every
direction
• Anisotropic
– Properties different in different
directions
• Often results from stratification
during sedimentation
K horizontal  K vertical
www.usgs.gov
Example
•
•
•
•
a = ???, b = 4.673x10-10 m2/N, g = 9798 N/m3,
S = 6.8x10-4, b = 50 m,  = 0.25,
Saquifer = gabb  ???
Swater = gbb
• % storage attributable to water expansion
•
• %storage attributable to aquifer expansion
•
Layered Porous Media
(Flow Parallel to Layers)
Piezometric surface
Dh
h1
h2
datum
3
𝑏𝑖 𝐾𝑖 = 𝑏𝐾
b1
K1
Q1
b b2
K2
Q2
b3
K3
Q3
𝑖=1
W
Q
Layered Porous Media
(Flow Perpendicular to Layers)
Piezometric surface
Dh1
Dh2 Dh
Dh3
K1
b
K2
K3
Q
Q
3
𝑖=1
𝐿𝑖
𝐿
=
𝐾𝑖 𝐾
L1
L2
L
L3
Example
Flow Q
• Find average K
Flow Q
• Find average K
Example
Anisotrpoic Porous Media
• General relationship between specific
discharge and hydraulic gradient
K is symmetric, i.e., Kij = Kji.
Principal Directions
• Often we can align the
coordinate axes in the
principal directions of
layering
• Horizontal conductivity
often order of
magnitude larger than
vertical conductivity
K xx = K yy = K Horiz >> K zz = KVert
éq x ù éK xx
ê ú ê
êq y ú = -ê 0
êë qz úû êë 0
0
K yy
0
¶h
¶x
¶h
qy = -K yy
¶y
¶h
qz = -K zz
¶z
qx = -K xx
é¶h ù
ê ú
0 ùê¶x ú
ú ¶h
0 úê ú
ê ¶y ú
ú
K zz ûê¶h ú
êë ¶z úû
Flow between 2 adjacent flow lines
𝑑ℎ
𝑞=𝐾
𝑑𝑚
𝑑𝑠
For the squares of the flow net
𝑑𝑠 = 𝑑𝑚
so
𝑞 = 𝐾𝑑ℎ
For entire flow net, total head
loss h is divided into n squares
𝑑ℎ =
ℎ
𝑛
If flow is divided into m channels
by flow lines
𝑚ℎ
𝑄 = 𝑚𝑞 = 𝐾
𝑛
KU/KL = 1/50
Flow lines are
perpendicular
to water table
contours
Flow lines are
parallel to
impermeable
boundaries
KU/KL = 50
Contour Map of Groundwater Levels
• Contours of
groundwater level
(equipotential lines)
and Flowlines
(perpendicular to
equipotiential lines)
indicate areas of
recharge and discharge
Groundwater Flow Direction
• Water level
measurements from
three wells can be used
to determine
groundwater flow
Head Gradient, J
direction
Groundwater
Contours
hi
h1(x1,y1)
hi > hj > hk
hj
hk
h3(x3,y3)
z
y
Groundwater
Flow, Q
h2(x2,y2)
x
Groundwater Flow Direction
Head gradient =
Magnitude of head gradient =
Angle of head gradient =
Groundwater Flow Direction
Head Gradient, J
h1(x1,y1)
h3(x3,y3)
z
Equation of a plane in 2D
3 points can be used to
define a plane
y
Groundwater
Flow, Q
h2(x2,y2)
x
Set of linear equations can be solved for a,
b and c given (xi, hi, i=1, 2, 3)
Groundwater Flow Direction
Negative of head gradient in x direction
Negative of head gradient in y direction
Magnitude of head gradient
Direction of flow
Example
Find:
y
Well 2
(200 m, 340 m)
55.11 m
Magnitude of head gradient
Direction of flow
Well 1
(0 m,0 m)
57.79 m
x
Well 3
(190 m, -150 m)
52.80 m
Refraction of Streamlines
• Vertical component of
velocity must be the same
on both sides of interface
qy1 = qy2
y
K1
q1
q1 cos q1 = q2 sin q 2
• Head continuity along
interface
h1 = h2 @ y = 0
• So
K1 tan q1
=
K 2 tan q 2
K2
K2  K1
1
Upper Formation
2
x
q2
Lower Formation
Consider a leaky confined aquifer
with 4.5 m/d horizontal hydraulic
conductivity is overlain by an
aquitard with 0.052 m/d vertical
hydraulic conductivity. If the flow in
the aquitard is in the downward
direction and makes an angle of 5o
with the vertical, determine q2.
𝐾1 𝑡𝑎𝑛𝜃1
=
𝐾2 𝑡𝑎𝑛𝜃2
0.052 𝑚/𝑑 𝑡𝑎𝑛 5𝑜
=
4.5 𝑚/𝑑
𝑡𝑎𝑛𝜃2
𝜃2 = 82.5𝑜
Summary
• Properties – Aquifer Storage
• Darcy’s Law
–
–
–
–
Darcy’s Experiment
Specific Discharge
Average Velocity
Validity of Darcy’s Law
• Hydraulic Conductivity
–
–
–
–
Permeability
Kozeny-Carman Equation
Constant Head Permeameter
Falling Head Permeameter
• Heterogeneity and Anisotropy
– Layered Porous Media
• Refraction of Streamlines
• Generalized Darcy’s Law
Darcy’s Law
Examples
Example
•
•
•
•
•
•
•
•
a = ???, b = 4.673x10-10 m2/N, g = 9798 N/m3,
S = 6.8x10-4, b = 50 m,  = 0.25,
Saquifer = gabb  ???
Swater = gbb = (9798 N/m3)(4.673x10-10 m2/N)(0.25)(50 m)
= 5.72x10-5
percent of storage coefficient attributable to water expansion
= Swater /S = 5.72x10-5 /6.8x10-4 *100 = 8.4%
percent of storage coefficient attributable to aquifer
expansion
• = Saquifer /S = 1 – (Swater /S ) = 91.6%
Example
Flow Q
K h,A =
K1z1 + K 2 z2 (2.3 m / d)(15 m) + (12.8 m / d)(15 m)
=
= 7.55 m / d
z1 + z2
(15 m) + (15 m)
Flow Q
K v,A =
Example
z1 + z2
(15 m) + (15 m)
=
= 3.90 m / d
z1 z2
15 m
15 m
+
+
K1 K 2 2.3 m / d 12.8 m / d
Example
Well 2
(200, 340)
55.11 m
x
Well 1
(0,0)
57.79 m
 = -5.3 deg
Well 3
(190, -150)
52.80 m