15
PERMEABILITY ESTIMATION
A continuing challenge for hydrogeologists (a holy grail
of sorts) is parameter estimation. If we know all of the
fundamental properties of some porous material, then we
should be able to accurately estimate the hydraulic
properties, right? This could be called the problem of
forward parameter estimation. For the parameter of
hydraulic conductivity, several researchers used capillary
tube analogs to get values of permeability.
The Hazen method started with an analogy to flow
through glass beads, which gave:
K = C (d10 )
where
2
K is the hydraulic conductivity (cm/s),
C is a coefficient based on grain size and sorting,
d10 is the size of particle which only 10% are
smaller than.
very fine sand
clean, medium sand
coarse sand, poor sorting
clean, coarse sand
40-80
80-120
80-120
120-150
16
K = C (d10 )
j
Shepard found that as the texture of the sediment was
more mixed and rough, the exponent (j) got smaller.
Glass beads were about 2, but mixed, rough sand was
about 1.5. Notice that the equation is, therefore,
dimensionally incorrect. Extreme care must be used
when selecting the coefficient. (Dimensionally
incorrect equations are usually evil).
The Hazen method is somewhat subjective, since the
coefficient is up to the judgement of the user.
The Kozeny-Carmen equation (F&C) tryed to eliminate
some of the guesswork with the coefficient, and
suggested that it was a function of the porosity (accounts
for packing):
 ρg  η3  d m2 


K =  
2 
 µ  (1 − η)  180 
where
dm is a “representative” grain size.
17
This immediately suggests that the porosity is a measure of
(or a function of) the packing and shape. But the equation
is still subjective: what is dm?
The Fair-Hatch method tries to eliminate even more
subjectivity by explicitly accounting for the various pore
sizes:
3

 ρg  η 
1

K =  
2
2 
µ
−
η
(
1
)
 
  θ
Pi 

m
∑
 100 d i 
where
m = packing factor (= 5 by experiment)
θ = 6 for spherical grains
θ = 7.7 for angular grains
Pi = percentage (by weight) of grains held
between sieves
di = diameter of these grains (= geometric mean
of adjacent sieves
18
sieve
size (mm)
10
5
% held
10
10
geometric mean
size (mm)
10 × ? =
5 × 10 = 7.1
1× 5 = 2.24
1.0
25
0.5
35
0.5 × 1.0 = 0.71
0.1
10
.1× .5 = 0.22
0.01
10
.01× .1 = 0.032
Pi
10
10
35
25 10 10
=
+
+
+
∑ d 0.032 0.22 0.71 2.24 + 7.1 + 10 ?
i
19
PERMEAMETERS
Constant Head
Q = Kπr 2 ( h2 − h1 ) / L
QL
K= 2
πr ( h2 − h1 )
r
h2
Q
h1
datum
(z = 0)
20
PERMEAMETERS
Falling Head
Qin = − πr 2
dh
= Qout = Kπr 2 ht / L
dt
ht
t
dh
K
∫h ht = − L ∫0 dt
0
dh
Kh
=− t
dt
L
ln( h0 / ht ) = Kt / L
L
K = ln( h0 / ht )
t
r
Qin
slope=K
ln(h0/ht)
ht
ho
0
drain
Qout
0
t/L
21
SPECIFIC YIELD (Sy)
Specific yield is defined as the ratio of the volume of
water released from a saturated sample due to gravity to
the total sample volume:
Sy =
Vgravity drained water
V
This number is always less than the porosity of a
sample, since all of the water is cannot be drained by
gravity. Some water is held by the grains by surface
tension. The Specific retention (Sr). The specific
retention is defined as the volume of water retained
after gravity drainage divided by total rock (sample)
volume
Sr =
Vwater held against gravity
V
This suggests that a sample of porous material with
smaller grains will different have Sr than coarse material.
THOUGHT EXPERIMENT: will fine-grained or coarsegrained spheres have greater Sr? Why?
22
Obviously, the amount of water drained +retained from
a saturated sample is equal to the total pore space, so:
η = S y + Sr
AQUIFER CHARACTERISTICS AND DEFINITIONS
An aquifer is a geologic unit that can store and
transmit usable amounts of water. This definition is
very subjective. “Usable amounts” will be different
for a single family dwelling or a cotton farmer.
A confining layer, or aquitard is a geologic unit with
relatively low K (typically <10-7 m/s). Once again, this
is a relative number. A better definition includes the
context of the adjacent aquifer: An aquitard is a unit
with K at least 100 times lower than the adjacent
aquifer. Further, the confining layer is only appropriate
if it confines some water to an aquifer either above or
below it. Often, a confining layer is not laterally
continuous and is called a leaky confining layer.
23
The water table is the elevation at which the pore
water is at zero gauge pressure, or at atmospheric
pressure. Above the water table, the water is held is
tension by attraction to soil grains. Below the water
table, the water is under pressure greater than
atmospheric pressure.
The porous medium is saturated for some distance
(usually small) above the water table. This zone that is
saturated and above the water table is called the
capillary zone, or the capillary fringe. In gravel, this
may be millimeters thick. In clay, it may be meters
thick.
Above the capillary fringe, the pores become more and
more filled with air. The zone above the capillary
zone with some air in the pores is the vadose zone.
The vadose zone can be zero to hundreds of meters
thick.
Infiltrating water sometimes encounters fine-grained
material and “piles up,” creating a saturated (and
positive pore pressure) zone above the regional water
table. This local zone, which is underlain by more
vadose zone, is called a perched water table.
24
perched zone
water table aquifer
confining layer
confined aquifer
25
ANISOTROPY AND HETEROGENEITY
Most sediments are deposited in a preferential way:
flat objects are laid down flat. This leads to material
that has a different effective K in different directions.
In a similar manner, if aquifer permeability is derived
from joints or faults, the pattern of the stress field is
imposed upon the aquifer. If K differs with direction,
the material is anisotropic:
isotropic
anisotropic
length of arrow indicate ease of flow: permeability
26
If the material is different from one location to the next,
it is called heterogeneous (whether or not it is isotropic
or not)
heterogeneous
homogeneous
THOUGHT EXPERIMENT: Does the scale of
observation sometimes make heterogeneous material
appear anisotropic? What is the fundamental difference?
27
ENERGY AND GROUNDWATER DRIVING FORCES
When we looked at Darcy’s Law (so we could talk about a
thing called hydraulic conductivity), I made up a quantity
called “hydraulic head.” It seems to dictate the rate and
direction of water movement. But surely water is like
anything else on earth, and follows strict Laws of thermodynamics. Can we derive hydraulic head?
Start with the fact that thermodynamic Laws apply.
Chiefly:
1)
2)
3)
energy and mass cannot be created or
destroyed,
energy transformations involving friction are
irreversible, and
water will move from areas of high energy to
low energy, i.e. energy must be lost as water
travels.
What are the common types of energy:
•
•
•
•
•
potential (elevation in a gravity field)
kinetic (water velocity)
elastic (pressure)
chemical
heat
28
In order to keep track of energy, we need to define an
object that has energy. This introduces 2 fundamentally
different “bookkeeping” methods, LaGrangian and
Eulerian.
•Lagrangian follows a distinct moving body (particle)
and keeps track of the losses or gains to that particle as it
moves. For example, the energy transformations taking
place in a speeding bullet are best handled by a
Lagrangian framework
•Eulerian defines an appropriate-sized control volume
and tracks things entering, leaving and being stored in
the control volume.
Using the Eulerian method in the bullet example, we
would define a parcel of air. Nothing much would
happen for a while, then suddenly the energy would
sharply increase as the bullet entered a control volume.
Soon after, the energy would decrease as the bullet left.
In order to fully keep track, we’d have to define control
volumes all the way from the gun to the target.
On the other hand, if we used the Lagrangian method for,
say, the weather, we’d have to keep track of an infinite
number of particles. Better to define blocks of the
atmosphere and track energy entering and leaving.
29
FLUID MECHANICS EXAMPLES - LaGrangian
Imagine a sphere of oil (or gas, etc.) in water:
ρw
ρo
y
r
x
Sum the forces acting on the sphere (buoyancy
[gravity] and drag):
Buoyant force: Fb = g(ρw-ρo)V = 4πr3g(ρw-ρo)/3
Drag Force: Fd = 6πrµv = 6πrµdy/dt (Stoke’s Law)
∑ F = ma = F − F
b
d
4 3 d2y 4 3
dy
r
g
r
ρo πr
=
π
(
ρ
−
ρ
)
−
6
π
µ
o
w
dt 2 3
3
dt
30
which is a linear, 2nd-order ordinary differential
equation (with a stock solution that can be fished out of
a D.E. book)
B   e − At − 1 
y =  t + 
 
A   A 
r=1
y
r=0.5
t
To get the terminal velocity, set the acceleration (d2y/dt2)
to zero:
dy
4 3
πr g (ρo − ρ w ) = 6πrµ
dt
3
2 2
dy
r g ( ρo − ρ w ) =
9µ
dt
31
For problems of groundwater flow, we wish to use an
Eulerian framework. To do so, we do not want to have
to always keep track of the volume or the mass of the
control volume. This means that we need a “normalized”
quantity, i.e. energy per unit mass. Now if my control
volume is twice as big (enclosing twice the mass), the
energy is twice as much, but the normalized energy
(energy per unit mass) remains the same. Now two
different people can communicate about the energy.
Energy is the potential to do work, and has the same
dimensions, or units, as work:
ML2
W = E = FD = maD = 2
T
So our energy per unit mass is
E maD
L2
=
= aD = 2
m
m
T
Since we are always on the earth (and gravity is
constant) we can simplify even further by using
energy per unit weight (where a = g):
E
mgD
=
=D=L
weight
mg
32
This is the most basic normalization we can use when g =
constant.
When we looked at Darcy’s Law, I suggested that
elevation and pressure were the only important energies.
We skipped over some other forms of energy that might be
important, such as velocity. Chemical potential can also
be important in some applications (i.e. osmosis through
clays, etc.). We will only look at the big 3.
Finally, we all know that mechanical energy is a relative
thing. Think of the potential energy of a ball on the table.
It can only go to the floor, so the maximum available
energy is limited relative to the floor. If the table is next
to the stairwell (and the door is open), the ball has more
potential energy. So energy must be measured relative to
a reference state.
Let’s move a blob of water from one state to another and
look at the energy differences:
z=z1, P=P1, V=V1
z=0, P=P0, V=0 (REFERENCE STATE)
33
z1
Elevation:
Velocity:
Elastic:
E z = mg ∫ dz = mgz1
0
Ev = m ∫
v1
0
mv12
vdv =
2
dP
P0 ρ
E p = m∫
P1
i.e. how much bigger
does the volume get
(how much wok on the
surrounding can the
volume do) for a given
pressure change?
The sum represents the total mechanical energy
relative to the initial reference state. We can also
normalize to energy per unit mass at this stage:
P1 dP
M .E.
v12
= gz1 + + ∫
P0 ρ
unit mass
2
In frictionless flow (say, laminar flow in a big pipe), this
energy remains constant (energy is conserved). The
equation is then known as Bernoulli’s equation.
34
MOST of the time, in groundwater flow, v is small, so v2
is VERY small. This is not always the case. Darcy’s
Law clearly neglects the velocity term, which is
applicable when the elevation and pressure terms are
much larger than the velocity term, leaving Hubbert’s
Force Potential:
P1 dP
M .E .
*
= φ = gz1 + ∫
P0 ρ
unit mass
THOUGHT EXPERIMENT: 1) Why is it clear that
Darcy’s Law ignores the velocity term? 2) Can you list
3 instances that a hydrogeologist might encounter
where this isn’t valid?
In order to evaluate the integral, we need to know how
the density of the fluid (or gas) depends upon the
pressure. Fluid and gasses are very different in this
respect! Ideal gas density varies about linearly with
pressure (Ideal Gas Law), while fluids are pretty much
uncompressible. We’ll only concentrate of water at this
point. For water:
ρ = ρ0 exp[β( P − P0 )]
35
The exponential can be expanded:
2
3
x
x
ex = 1+ x +
+ + ...
2! 3!
For small arguments (x<<1), the squared and higher
terms are small, leaving
ex ≈ 1+ x
So ρ ≈ ρ0 [1 + β( P − P0 )]
For water, a relatively incompressible fluid, the value
of β is small enough, and pressures are small enough
that the second term in the parentheses is very small
compared to 1. This is not always the case. For
almost all groundwater applications, we can generalize
that
β ( P − P0 ) << 1 So ρ ≈ ρ0 = cons tan t
P1 − P0
dP
φ = gz1 + ∫
= gz1 +
P0 ρ
ρ
*
P1
36
We’re getting close to the whole Darcy thing... Since the
reference state is sort of arbitrary, why not choose
something simple. We will always talk about gage
pressure, i.e. Patm = 0. Now if we always measure water’s
gage pressure, the reference pressure P0=Patm=0. Now
the energy per unit mass at some location is
φ* = gz1 +
P1
P
= gz +
ρ
ρ
Converting to Mechanical Energy per unit weight
(divide everything by the acceleration of gravity):
φ*
P
=φ=h= z+
g
ρg
This is hydraulic head, or the potentiometric head, or
sometimes the piezometric head.
•
•
•
•
•
z = elevation above some datum,
P = gauge fluid pressure,
ρ = fluid density (a constant here),
g = gravitational acceleration,
kinetic energy is neglected (v2/2g = 0)
37
A fair question might be: Why does fluid flow from high
energy (head) to low energy? This is answered in reverse.
Energy is lost as a fluid moves. The lost energy is due to
the fluid’s viscosity, or internal friction. The drag that the
solid grains imparts to the water is transmitted throughout
the fluid. The viscosity is a measure of the friction of fluid
upon itself.
The lost energy turns into heat, which dissipates
throughout the fluid. This heat cannot be transformed
back into energy (2nd Law of thermodynamics), so water
always flows from high to low energy.
SUMMARY
Forces acting on groundwater:
driving forces:
1) changes in pressure
2) gravity
resisting forces: 1) friction between fluid and solid
2) internal (viscous) fluid friction
driving force
Q
dh
k ρg dh
= q = −K
=−
A
dx
µ dx
resisting force
38
Other forms of Darcy’s Law:
k ρg dh
q=−
µ dx
k ρg dh
Q=−
A
µ
dx
where q is called the specific discharge. The units are
Q/A=L3/(L2T)=L/T. This suggests a water velocity,
but it really isn’t. Its a flow rate per unit area, not
velocity. The unit area is composed of solid and
liquid, but only the liquid part is moving. If we really
want to know the velocity of the fluid, we need to
divide the flow rate (Q) by the cross-sectional area of
the water:
pipe
what is the ratio of
fluid are to the total
are across the sandfilled pipe shown
here?
Awater
=η
Apipe
39
So the average pore water water velocity is
Q
Q
q
1 k ρg dh
v=
=
= =−
Awater ηA η
η µ dx
or simply
q
K dh
v= =−
η
η dx
CONTRARY TO MANY BOOKS we will always
call q the specific discharge, and v the pore velocity.
Some books call v the discharge, some call q the
Darcy velocity. Both are bad. Also, some simple
algebraic manipulation gives other forms
kρg dh
q=−
µ dx
P
h=
+z
ρg
q=−

kρg d  P

+ z 
µ dx  ρg

dz 
k  dP
+ ρg 
q=− 
µ  dx
dx 
THOUGHT EXPERIMENT: what is dz/dx in a horizontal
column? In a vertical column?
40
We have already explored how a WATER TABLE,
or unconfined aquifer stores water: It drains and
fills according to the specific yield (Sy). Imagine a
1m x 1m column of sediments 10 m high. It is
initially filled with sand with Sy=0.10, a porosity of
0.3 and water up to 9m.
If I drain some water so that
the water level drops to 8m,
how much did I drain? The
answer is Sy times the volume
evacuated, or:
9m
8m
Vwater = S y ( ∆h ) × A
1m
So another way of defining Sy is the
volume of water drained, per unit area,
from an unconfined aquifer for a given
change in head. This is really
important, because Darcy’s Law dictates
spatial change in head. The specific
yield tells us how the head will change
in one spot if some water is removes, so
we can use this to predict head change
from one time to the next.
41
What about for a confined aquifer,
where the sediments do not actually drain? Any water
stored in such an aquifer must result in expansion of the
aquifer, or the water, or both. First, lets look at stress
and pressure within an aquifer. What is stress? It is
force applied to a solid per unit area. What is pressure?
Force per unit area applied by a fluid.
All this overburden
has weight (force),
therefore exerts
stress
... and must be
supported by the rock
and fluid beneath
confining
layer
b
what are the forces acting on a horizontal plane in the
aquifer?
σT
σe
P
42
If the plane is not moving, then the forces per unit
area (stresses and pressure) are equal:
σT = P + σe
where
σT is the total stress(F/A = M/(LT2)),
P is the water pressure (F/A = M/(LT2)),
σe is defined as the effective stress M/(LT2).
Taking a differential form:
dσT = dP + dσe
If we assume that the overburden (total stress) doesn’t
change much, then any change in pressure causes an
opposite change in the effective stress. In other words,
a drop in water pressure causes the solid grains to
support more weight:
dP = −dσe
43
Another way to think of this is that an increase in
pressure causes the grains to spread apart somewhat
Increase fluid
pressure
Thought experiment: which configuration has the most
cohesion? How should this affect the probability of
earthquakes? In reverse - what should happen in wells
after an earthquake?
We defined water compressibility earlier as the change in
volume (per unit volume) with a given change in pressure:
β=
− 1 dVw
Vw dP
(1)
The negative sign indicates that an increase in pressure
results in a decrease in the volume of water. We can
similarly define the compressibility of the aquifer material:
44
− 1 dVT
α=
VT dσ e
where
(2)
α is aquifer compressibility,
VT is the total volume of a sample, and
σe is the effective stress.
Hereafter, for
simplicity, we must
assume that a is a
simple constant, i.e. the
graph of VT versus σe
is a single line.
20kg
20kg
20kg
VT
dVT
σe
VT
45
− 1 dVT
α=
VT dσ e
(2)
Remember that the porosity is related to the total volume
and the water volume:
Vw
η=
VT
and
VT = Vw + Vs
Vw
V s = VT − V w = VT (1 − ) = (1 − η )VT
VT
Using all of these relationships, we can figure out how
much water is produced by a given change in pressure.
Since we are looking at a fixed plane (z=constant) a
change is pressure is equivalent to a change in head.
Starting with the water produced by the expansion of
water:
dVw = − βVw dP
dVw = − βηVT dP
We need to know the volume of water produced in a
unit volume, i.e. VT = 1
dVw = − βηdP
46
We will assume that the volume of the soil solids is
relatively constant, i.e. the stress changes cause deformation
of grains and re-packing, but not structural(phase?) changes
in the solids. So Vs is a constant. Also, since we are
tracking the amount of water taken out of the matrix, the
reduction in VT is a gain in Vw (i.e. the sign is changed)
VT = −Vw + cons tan t
dVT = − dVw
Recall that dσe = -dP, so the definition of aquifer
compressibility gives:
1 dVT
α=
VT dσ e
− 1 dVw
α=
VT dP
αdP = − dVw
The total amount of water that is derived from a unit of
confined aquifer is the sum
dVw = −βη dP − αdP
47
Now relate the pressure change to a change in head at a
constant elevation:
h = P/(ρg)+z
ρgdh = dp
dVw = −(βη + α )ρgdh
Finally, we wish to be able to compare this amount
of water to the amount derived from an unconfined
aquifer. One way to define Sy was the amount
drained from a unit volume for a unit decline in head
(see the figure showing drainage from a 10m column
of soil). So let dh = -1 (i.e. lowering the head from
9m to 8m in 1m3 of soil. This is a new coefficient
similar to specific yield, but for confined aquifers. It
is called the specific storage (Ss).
S s = (βη + α)ρg
48
But it is very different from specific yield. The
dimensions are 1/L. Why is this? When we measured Sy,
we knew how much of the aquifer thickness was
contributing to the measurement (the upper 1m in the first
figure). But the whole aquifer is contributing to the
specific storage. A confined aquifer twice as thick can
store twice as much water by elastic expansion. An
unconfined aquifer can store the same amount per rise in
head (it just “lays on top”) no matter how far down the
bottom is. In order to relate the storage properties of a
confined aquifer to an unconfined aquifer, simply multiply
the specific storage times the aquifer thickness. this is a
new parameter called Storativity (S).
S = Ssb
This is a dimensionless parameter. Values are typically
on the order of 10-5 to 10-3. Where does this come from?
material type
clay
sand
gravel
jointed rock
water (β)
α (m2/N)_
10-8 - 10-6
10-9 - 10-7
10-10 - 10-8
10-10 - 10-8
4.4×10-10
49
S s = (βη + α)ρg
S s = (4 ×10 −10
1000kg 10m
ms 2
× 0 .3 + α )
kg
m3
s2
≈ (10 −10 + α)10 4 m −1
≈ α ×10 4 m −1
≈ 10 −3 to 10 −6 m −1
Let’s say a typical aquifer has a compressibility of 10-8
m2/N. How thick must a confined aquifer be in order to
have the same storage ability as an unconfined aquifer?
Assume Sy = 10% (i.e. 0.1).
Ss ≈ 0.0001 m-1. Ss x b = 0.1 b = 1000m.
50
When we looked at Darcy’s Law, we assumed the
flow was in a single direction. Surely flow can go
just about anywhere. How do we account for this?
We must define the flow in terms of vectors, which
have components in one or more dimensions or
directions.
qx
y
|q|
qy
θ
j
i
x
We can represent the q vector in terms of the components
in the x and y (and z if we wish) directions:
r
r
r
q = qx i + q y j
where
qx and qy are the magnitudes of flow in the x
and y directions, and
i and j are unit vectors in the x and y directions
51
Throughout this class, vectors will be denoted by a
number of things, including an arrow, a boldface, an
underline, and maybe even a squiggle.
θ=
qy
qx
and
q = q x2 + q y2
Now Darcy’s Law must be written in terms of the
gradients in each direction:
∂h
qx = − K
∂x
∂h
qy = −K
∂y
∂h
qz = − K
∂z
where ∂ h/ ∂x is the partial derivative of head WRT x.
Partial derivatives are the same old derivatives except
that any variation that is not in the x-direction is ignored
(i.e. is a constant).
52
For example, look at the function
f ( x, y ) = 3x 2 − 12 y + 2
∂
f ( x, y ) = 6 x
∂x
∂
f ( x, y) = −12
∂y
Very often, vertical flow is assumed negligible, that is
only a function of x and y. If the head field is planar,
then
h( x, y ) = Ax + By + C
∂h
qx = − K
= − KA
∂x
∂h
qy = −K
= − KB
∂y
53
What about flow in anisotropic media? Surely the
hydraulic conductivity can be different in the x and y
directions, yes? Let’s just use 2-D for now:
qy
Ky
y
Kx
qx
x
∂h
∂x
∂h
qy = −K y
∂y
qx = − K x
So what is the relative
discharge and effective
hydraulic conductivity for
some direction other than x or
y?
∂h
qs = − K s
∂s
qx
y
qs
θ
x
∂h
q
=
q
cos(
θ
)
=
−
K
s
x
qy x
∂x
∂h
q y = q s sin( θ) = − K y
∂y
54
The Chain Rule (from calculus i or ii)
∂h ∂h dx ∂h dy
+
=
∂s ∂x ds ∂y ds
ds
θ
dy
where
dx/ds = cos(θ) and
dy/ds = sin(θ)
dx
∂h ∂h
∂h
=
cos(θ) + sin( θ)
∂s ∂x
∂y
(1)
From the last page
q
∂h
=− s
Ks
∂s
q x qs cos(θ)
∂h
=−
−
Kx
Kx
∂x
qy
qs sin(θ)
∂h
=−
=−
∂y
Ky
Ky
(2)
55
Inserting the various relations from (2) into (1)
−
qs
q cos(θ)
q sin(θ)
cos(θ) − s
sin(θ)
=− s
Ks
Kx
Ky
1 cos 2 (θ) sin 2 (θ)
=
+
Ks
Kx
Ky
A review of polar coordinates:
y
r
θ
x
x=rcos(θ)
y=rsin(θ)
x2=r2cos2(θ)
y2=r2sin2(θ)
r2 = x2 + y2
r2
x2 y2
=
+
Ks Kx K y
56
This is the equation of an ellipse with axes √Kx and √Ky
Ky
Ks
Kx
All of this assumes that the axes of the conductivity
ellipse are aligned with the x and y axes that I
defined. What is the difference? In a nutshell, if
they are not aligned, then a gradient in the x
direction will cause some flow in the y direction:
K ellipse aligned with
coordinate axes
y
K ellipse NOT aligned
with coordinate axes
qx
y
gradient
gradient
x
x
57
ADVANCED TOPIC
In this case, there is some flow in the y direction from
the gradient in the x direction. (Convince yourself that
there would also be some flow in the x direction from a
pure y-direction gradient). So a complete description is
∂h
∂h
q x = − K xx
− K xy
∂x
∂y
∂h
∂h
q y = − K yy
− K yx
∂y
∂x
where Kij means “hydraulic conductivity in the i
direction from a gradient applied in the j direction.”
Substitute x and y in every combination for i and j.
Note that when the K ellipse is aligned with x and y, the
Kxy and Kyx are simply zero, and Kxx is called Kx
(Kyy is called Ky) It is generally easy to align the
ellipse and your axes. If not, the 2-D Darcy’s Law can
be given in matrix form:
q x 
 K xx
  = −
q y 
 K yx
 ∂h 
K xy   ∂x 
 
K yy   ∂h 
 ∂y 
58
STILL ADVANCED TOPIC
The hydraulic conductivity is a TENSOR, which is a
fancy way of describing any quantity that depends on
direction. A vector is a first-rank tensor, since the
components of a vector can be classified by one
subscript. K is a second-rank tensor because we need
two subscripts. The gradient is a vector sometimes
given the shorthand notation ∇h. This leads to the
really shorthand notations:
r
q = − K ⋅ ∇h
q = − K ⋅ ∇h
which are equivalent to
 K xx
q x 

 
q y  = −  K yx
q 
 K zx
 z

K xy
K yy
K zy
 ∂h 
 
K xz   ∂x 
  ∂h 
K yz   
∂y 


K zz  ∂h
 
 ∂z 
59
And if we are fortunate enough to align the principle
directions of the K ellipse with x, y, and z, then the
off-diagonals are zero, leaving:
K x
q x 
 
0
=
−
q
 y

q 
 0
 z
0
Ky
0
 ∂h 
 
0   ∂x 
 ∂h 

0  
∂y 

K z  ∂h
 
 ∂z 
60
CONSERVATION OF FLUID MASS
ρq z + dz
∂ (ρq z )
∂z
dx
dz
ρq x
ρq x + dx
z
∂ (ρ q x )
∂x
ρq y
y
x
dy
ρqz
where ρ is the density of water. Where does the funky
dx
∂ (ρ q x )
∂x
a + ∆x
df
dx
come from?
some function f(x)
a
a
slope = df/dx
∆x
61
We wish to sum the mass inflows and outflows into a
differential (Eulerian) control volume. We can assign any
coordinate system to do this. The standard 3-D Cartesian
(x,y,z) is as good as any, although certain problems are
much more conveniently described by other coordinates.
Radial coordinates are an obvious example that are
perfectly suited for flow towards a well.
The mass total inflow from the left side is
ρ Q x = ρ q x dy ⋅ dz
The total mass outflow is the original flow plus the
deviation (like on the last figure), or:
∂ρ q x 

ρQ x + ∆ ρQ x =  ρ q x +
dx  dy ⋅ dz
∂x


Inflow - Outflow =
∂ρq x 

dx  dy ⋅ dz =
ρq x dy ⋅ dz −  ρq x +
∂x


∂ρ q x
dx ⋅ dy ⋅ dz
−
∂x
62
Similarly, in the y-direction, the inflow into the front face is
ρq y dx ⋅ dz
The total mass outflow out the top is:
∂ρ q y 

 ρq y +
dy  dx ⋅ dz
∂y


Inflow - Outflow in the y-direction =
∂ρq y 

ρq y dx ⋅ dz −  ρq y +
dy  dx ⋅ dz =
∂y


∂ρq y
−
dy ⋅ dx ⋅ dz
∂y
A similar procedure in the z direction (top and bottom
faces) gives:
Inflow - Outflow in the y-direction =
∂ρ q z
−
dz ⋅ dx ⋅ dy
∂z
63
Now the sum of all the inflows-outflows gives
∂ρq y
∂ρ q z
∂ρq x
dx ⋅ dy ⋅ dz −
dy ⋅ dx ⋅ dz −
dz ⋅ dx ⋅ dy
−
∂y
∂z
∂x
What are the units?
d M / L3 ⋅ L / T
d
∂ρq z
dz ⋅ dx ⋅ dy =
⋅L⋅L⋅L=
∂z
L
M
T
This represents the RATE at which water mass is being
packed into the little volume. We know how to represent
this in a equation - it’s just
∂
∂M w ∂
∂
= (ρVw ) = (ρVw ) = dx ⋅ dy ⋅ dz (ρη )
∂t
∂t
∂t
∂t
So the sum of the inflows-outflows is equal to the time rate
of mass change:
64
∂ρq y
∂ρq x
∂ρ q z
dx ⋅ dy ⋅ dz −
dy ⋅ dx ⋅ dz −
dz ⋅ dx ⋅ dy
−
∂y
∂z
∂x
∂ρη
= dx ⋅ dy ⋅ dz
∂t
Canceling like terms gives:
∂ρq x ∂ρq y ∂ρq z ∂ρη
=
−
−
−
∂y
∂z
∂t
∂x
Now there’s an easy (shortcut) way to continue and
a long way. I’ll start with the shortcut way and then
do a full development. We remember the derivation
of the specific storage (Ss) of aquifer material. It is a
direct measure of the ability of aquifer material to
store water. More precisely, it relates the change in
water volume stored for a given change in head per
unit volume (dxdydz). If the density of water is
relatively constant, then the change in water mass
stored is the density times Ss. So the original
d(Mw)/dt is simply
∂M w ∂ρVw ∂ρS s h
∂ρh
=
=
= Ss
∂t
∂t
∂t
∂t
65
Now the RHS reads
∂ρh
∂ρq x ∂ρq y ∂ρq z
−
= Ss
−
−
∂y
∂z
∂t
∂x
This is not an equation that does us any good. Why?
One equation and 2 unknowns (q and h). But we have
another equation that relates head and q that we can
substitute into the above - Darcy’s Law:
∂h
qx = − K x
∂x
∂h
qy = −K y
∂y
∂h
qz = − K z
∂z
∂ 
∂h  ∂ 
∂h  ∂ 
∂h 
∂ρh
 +  ρK z
 ρK x  +  ρK y
 = Ss
∂x 
∂x  ∂y 
∂y  ∂z 
∂z 
∂t
This is the continuity equation for groundwater flow. You
notice that is describes the single dependent variable (h) as a
function of independent variables (space, time) and
measurable parameters (K,Ss,ρ).
66
A number of simplifications can be made if the
parameters do not vary significantly in space and/or time.
For example, if the density of groundwater does not
change much in space, then it is a constant, and can
move outside of the derivatives:
∂ 
∂h 
∂ 
∂h 
∂  ∂h 
∂h
 + ρ  K z
ρ  K x  + ρ  K y
=
ρ
S

s
∂x 
∂x 
∂y 
∂y 
∂z 
∂z 
∂t
∂ 
∂h  ∂ 
∂h  ∂  ∂h 
∂h


+
+
K
K
K
=
S
 x 
 z

y
s


∂x 
∂x  ∂y 
∂y  ∂z 
∂z 
∂t
This is the usual form that is presented. If the hydraulic
conductivity is isotropic: Kx = Ky = Kz = K
∂  ∂h  ∂  ∂h  ∂  ∂h 
∂h


+
+
K
K
K
=
S




s


∂x  ∂x  ∂y  ∂y  ∂z  ∂z 
∂t
If the hydraulic conductivity does not vary much in space:
∂  ∂h 
∂  ∂h 
∂  ∂h 
∂h
K x   + K y   + K z   = S s
∂x  ∂x 
∂y  ∂y 
∂z  ∂z 
∂t
∂ 2h
∂ 2h
∂ 2h
∂h
K x 2 + K y 2 + K z 2 = Ss
∂x
∂y
∂z
∂t
67
∂ 2h
∂ 2h
∂ 2h
∂h
K x 2 + K y 2 + K z 2 = Ss
∂x
∂y
∂z
∂t
If K is homogeneous AND isotropic
∂ 2h
∂ 2h
∂ 2h
∂h
K 2 + K 2 + K 2 = Ss
∂x
∂y
∂z
∂t
∂ 2 h ∂ 2 h ∂ 2 h S s ∂h
+ 2+ 2 =
2
∂x
∂y
∂z
K ∂t
If the vertical flow is negligible, then the thickness of
the differential unit is (b). Two things happen. First,
dz = b, so the change storage is best described by the
Storativity S = bSs. Multiplying both sides by b:
∂ 
∂h  ∂ 
∂h 
∂h
 K x b  +  K y b  = bS s
∂x 
∂x  ∂y 
∂y 
∂t
This suggests that a new parameter Kb is useful to keep
track of. Indeed, it is called the transmissivity (T) and is
widely used when 2-D flow is a good assumption (or
when K is not kept track of in the vertical dimension). If
the same sorts of regularity are found, (T is homogeneous,
isotropic, then:
68
∂  ∂h  ∂  ∂h 
∂h


T
+
T
=
S
 x 
y
∂x  ∂x  ∂y  ∂y 
∂t
∂ 2 h ∂ 2 h S ∂h
+ 2 =
2
∂x
∂y
T ∂t
Poisson Equation
This is an extremely useful form of the equation. It
means that if flow is essentially 2-D (i.e. within an aquifer
sandwiched between confining units), then I just need to
measure 2 aquifer parameters. If I know T (the
transmissivity) and S (the Storativity), I have an equation
that tells me how fast the heads change at all points in the
aquifer.
Steady-State
Lets say that an aquifer has seen approximately the same
inputs and outputs for a very long time. Then the
aquifer has been approaching equilibrium. Another
phrase for equilibrium is steady state. The inputs and
outputs roughly match, and the overall picture doesn’t
change much as we watch. That means that the time
derivative is very small. Let’s set it to zero and see what
we get:
69
∂ 2h ∂ 2h
+ 2 =0
2
∂x
∂y
LaPlace Equation
Example: Does the LaPlace Equation predict the
results of Darcy’s experiment?
When Darcy ran his tests, he limited the flow to one
direction. Call it the x-direction. The top and sidewalls
of his sand column ensured that no flow occurred in the
other directions. Another way of saying this is that
there were no head gradients in the y and z directions.
So the Laplace equation becomes
d 2h
=0
2
dx
A solution to this equation starts with:
dh
=C
dx
h2
x2
h1
x1
∫ dh = ∫ Cdx
70
h2 − h1
Q
=C =
x 2 − x1
K
Example: You are monitoring a confined aquifer 10 m
thick with isotropic K of 0.5 m/day and porosity of 0.2.
The top of well #1 is 2000 m MSL. The top of well #2 is
at elevation 2010 m MSL. You measure the depth to
water in #1 as 31 m TOC and #2 as 42 m TOC. One day
you detect a strong pulse of PCE in well #1. How long
before this pulse reaches the drinking water factory at #2,
which is 100m away at well #2?
duds&
suds
#1
#2
bedrock
club
soda
Inc.
71
Answer: average pore water velocity = v
q
K ∆h
v= =−
η
η ∆x
h1 = elevation of water in well #1 (=aquifer
elevation+pressure head) = 2000-31 = 1969 m.
h2 = 2010-42 = 1968 m.
1m
∆h
=
= 0.01
∆x 100 m
K
10m / d
v = 0.01 =
0.01 = 5m / d
η
0.2
The total distance is 100m, so the PCE will take an
average time of 20 days to reach the downstream well.
72
Example: A 2-D aquifer is anisotropic. The K in the xdirection is 0.1 m/d, and the K in the y-direction is 1.0
m/d. You put in 3 wells and measure the heads:
h=1,102 m
100m
h=1,100 m
h=1,105 m
100m
y
x
Which way is the gradient? Which way does water flow?
∂h 1,105 − 1,100
=
= 0.05
∂x
100 − 0
∂h 1,102 − 1,100
=
= 0.02
∂y
100 − 0
∂h
= 0.02
∂y
r
J = −∇h
∇h
∂h
= 0.05
∂x
angle of gradient to x-direction = tan-1(0.05/0.02) = 22°
73
∂h
qx = − K x
= KxJx
∂x
∂h
qy = −K y
= KyJy
∂x
∂h
qx = − K x
= −0.005m / d
∂x
∂h
qy = −K y
= −0.02 m / d
∂x
q x = − 0.005 m / d
r
J
q y = −0.02m / d
∂h
= 0.02
∂y
∇h
∂h
= 0.05
∂x
| q |= (−0.0052 + 0.02 2 )1/ 2
= 0.0206m / d
Angle of flow relative to x-direction =tan-1(0.02/0.005)=76°
74
Example: a confined aquifer pinches out slowly in the xdirection. The aquifer thickness (b) at well #1 is 100m. A
kilometer away, the thickness is 10m. Head at well #1 is
1350 m MSL. The aquifer material has a K of 1 m/d.
Recharge to the aquifer is estimated at 50 m2/yr for every
meter along the mountain front. What is the head along
the x-direction?
Q=50 m3/yr per m
100m
h=1350m
K=1m/d
x
1000 m
m2
⋅w
Qin = 50
yr
dh
dh
m2
= − Kbw
= Qin = 50
⋅w
Q = − KA
dx
dx
yr
dh
m2
− Kb
= 50
dx
yr
90 m
x
b = 100 m −
1000 m
10m
75
dh
m2
− K (100m − 0.09 x ) = 50
dx
yr
m2
50
dx
yr
dh =
m 365d (100m − 0.09 x)
−1
d yr
h( x)
x
dl
dh = −0.137 m ∫
∫
(100 m − 0.09l )
x =0
1350 m
x
1
ln(100 m − 0.09 ⋅ l )
1350 m − h( x) = 0.137 m
− 0.09
0
h( x) = 1350 m +
0.137 m 100 m − 0.09 ⋅ x
)
ln(
− 0.09
100 m
1350.5
1350
1349.5
1349
h(m)
1348.5
1348
1347.5
1347
1346.5
1346
0
200
400
600
800
1000
1200
x(m)
Thought Experiment: What happens at x = 111.11...?
76
Steady-state flow in an Unconfined Aquifer
Flow in an unconfined aquifer is somewhat like the
example we just went through. In order for water to flow
in an unconfined aquifer, the head must be lower in the
direction of flow. If the head is lower, then the saturated
thickness is less, i.e. the effective aquifer thickness
decreases in the direction of flow.
0
h=
+z=z
ρg
z
z=0
flow
P
h=
+z
ρg
impermeable
77
Darcy’s Law for an unconfined aquifer:
Q = − KA
dh
dx
A is the area through which the water flows = height x width
A = h x w so:
Q = − Kwh
dh
dx
divide both sides by the aquifer width (perpendicular to
flow):
Q
dh
′
≡ Q = − Kh
w
dx
where Q’ is defined as the flow rate per unit width. This
is called Darcy’s Law for unconfined flow. How is it
different? It takes into account the aquifer thinning.
Let’s calculate how the head changes from one well to
the next (from the previous figure).
− Q′
dh
=h
dx
h
78
L
h2
′
−Q
dx = ∫ hdh
∫
K 0
h1
where x = 0 at the upstream well (#1) and x=L at the
downstream well (#2). The water level (head) in well
1 is labeled h1. The water level (head) in well 2 is
labeled h2.
2 h2
h
− Q′x
=
K 0
2
L
h1
− Q′L h12 − h22
=
K
2
− K h12 − h22
Q′ =
2
L
This is called the Dupuit-Forchheimer Discharge
Formula. It looks alot like Darcy’s Law, but depends on
the square of the heads.
79
The Dupuit Assumption(s)
We have ignored something along the way here
(without reading ahead, what is it?).
We have assumed that flow is horizontal, i.e. there is
no vertical flow. How good an assumption is this?
Take a look at an exaggerated blowup of the water
table:
q
h=z
q
ds
dz
dx
80
ds
dz
θ
dx
where s is a flowline along the water table. On the water
table, P = 0, so h = z
dz
dh
qs = − K
= −K
ds
ds
and
dz
= sin(θ)
ds
qs = − K sin(θ)
All along, we’ve been saying (or we would like to be
able to accurately say) that
qs = − K
dh
dz
= −K
= − K tan(θ)
dx
dx
when is sin(θ) approximately equal to tan(θ)?
81
θ (deg)
sin(θ)
tan(θ)
%difference
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0.01745
0.03489
0.05233
0.06975
0.08715
0.10452
0.12186
0.13917
0.15643
0.17364
0.19080
0.20791
0.22495
0.24192
0.25881
0.01745
0.03492
0.05240
0.06992
0.08748
0.10510
0.12278
0.14054
0.15838
0.17632
0.19438
0.21255
0.23086
0.24932
0.26794
0.0152
0.0609
0.1371
0.2436
0.3805
0.5478
0.7454
0.9732
1.2312
1.5192
1.8373
2.1853
2.5630
2.9704
3.4074
This looks like a pretty good assumption up to some
very steep water tables. So the Dupuit assumption is:
• the hydraulic gradient = slope of the water table, i.e.
dz/dx = dz/ds or dh/dx = dh/ds
This also means that flow is horizontal.