K VALUES VARY WITHIN SPACE AND WITH
DIRECTION
HETEROGENEITY - describes spatial variation
ANISOTROPY - describes directional variation
HOMOGENEOUS - uniform throughout (K
independent of position)
ISOTROPIC - properties do not vary with direction
LAYERED HETEROGENEITY
Individual layers are homogeneous
K1
K2
K3
1
DISCONTINUOUS HETEROGENEITY
e.g. across a fault
K1
K3
K2
K4
K3
K5
K6
TRENDING HETEROGENEITY
variation in sedimentation patterns
K1
K2
K3
2
RANDOM HETEROGENEITY
small scale variation with a structure that isn't easily tied
to geologic process, although we know it is its basis
we can describe with geostatistics (spatial statistics)
AVERAGING OPTIONS
Weighted Arithmetic
∑K d
∑d
i
i
i
Geometric
N
⎛1
⎞
⎜ (logK1 +logK2 +...+logKN ) ⎟
⎝N
⎠
K1K2 ...KN or 10
⎛1
⎞
⎜ log(K1K2 ...KN ) ⎟
⎝N
⎠
or 10
Above assumes each K represents an equal volume, each could be weighted by the volume represented
Weighted Harmonic
∑d
d
∑K
i
i
i
3
Random distribution of K
*
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*
*
*
*
*
6
1x10-3cm/s
4
1x10-5cm/s
2
1x10-6cm/s
* sample locations
assumed to be
representative of the
proportions
Realize you would only know the values shown
*
*
*
*
*
*
*
*
*
*
*
*
You might have clues based on lithology
*
*
*
*
*
*
*
*
*
*
*
*
4
ANISOTROPY
CONSIDER K IN THE PRINCIPLE DIRECTIONS X, Y, Z
ISOTROPIC - KX = KY = KZ
TRANSVERSELY ISOTROPIC - KX = KY not = KZ
CONSIDER THE RELATIONS OF K IN A VERTICAL CROSS
SECTION
HOMOGENEOUS
ISOTROPIC
HOMOGENEOUS
ANISOTROPIC
HETEROGENEOUS
ANISOTROPIC
HETEROGENEOUS
ISOTROPIC @ each point
z
THERE IS A RELATION BETWEEN
LAYERED HETEROGENEITY & ANISOTROPY
an equivalent K can be calculated
to simplify complex systems thus making it possible
to apply Darcy's Law
dT
d1
K1
Kx1 =Kz1
d2
K2
Kx2 =Kz2
d3
K3
Kx3 =Kz3
d4
Kx4 =Kz4
K4
x
5
Consider Flow Perpendicular to Layering:
Say that K4= K1 < K3 << K2
V = Q/A
Qin = Qout
therefore
Vin = Vout
Δhi = head drop across layer i
Δh = Δh1 + Δh2 + Δh3 + Δh4=htop-hbot
htop
Δh1
K1
Δh2
K2
d3
Δh3
K3
d4
Δh4
d1
d2
dT
Δh
Kx1 =Kz1
Δh=htop-hbot
Kx2 =Kz2
Kx3 =Kz3
Kx4 =Kz4
K4
hbot
By Darcy’s Law:
Rearrange:
V=
K eq Δh
K 1 Δ h 1 K 2 Δh 2
K Δh
=
= ... n n =
d1
d2
dn
dT
Expand head difference term:
Substitute Darcy’s Law for
head differences:
Vd T
Vd T
Vd T
=
==
Vd1 Vd 2
Vd n
Δh
Δh1 + Δh 2 + ... + Δh n
+
+ ... +
K1
K2
Kn
dT
=
Equivalent K for flow
di
perpendicular to layers
∑K
i
K eq =
K eq
d1
d2
dT
Δh
Δh1
Δh2
d3
Δh3
d4
Δh4
K1
K2
K3
K4
6
Consider Flow Parallel to Layering:
Say for example K4= K1 < K3 << K2
V = Q/A
Qin = Qout
therefore
Vin = Vout
dT
d1
K1
d2
K2
d3
K3
d4
hright
hleft
Δh = hleft-hright
K4
By Darcy’s Law:
Δh
d 2 (1)
L
Δh
Q4 = K 4
d 4 (1)
L
Δh
d1 (1)
L
Δh
Q3 = K 3
d 3 (1)
L
Q1 = K 1
Q2 = K 2
1 for unit width into the paper
Total flow is their sum:
dT
Δh
Δh
Δh
Δh
+ K 2d 2
+ K 3d 3
+ K 4d 4
L
L
L
L
d1
K1
d2
K2
d3
K3
d4
hright
hleft
Q T = K 1d 1
K4
7
V=
Q
Q
=
A d T (1)
1 for unit width into the paper
Substitute Eqtn for Q from previous slide
Simplify with a summation
Δh
Δh
Δh
Δh
+ K 2d2
+ K 3d 3
+ K 4d4
L
L
L
L ==
V=
dT
We can divide V by
K1d1
hleft
K eq
V
=
i
dT
V
=
Δh
L
i
i
)Δh
dT L
Cancel the gradients:
∑
=
n
i =1
K id i
dT
d1
K1
d2
K2
d3
K3
d4
hright
gradient to get K:
∑ (K d
K4
Calculate Flow and Heads between boundaries
L1
H1
H3
D K1
K2
H4
L2
H2
H1 = 20cm
H2 = 10cm
K1 = 1cm/sec
K2 = 0.2cm/sec
L1 = 30cm
L2 = 30cm
D = 2cm
Q @ H2 = ??
H3 = ??, H4 = ??
8
Calculate Flow and Heads between boundaries
L1
H1
K1
H3
D
L2
K2
H2
H1 = 20cm
H2 = 10cm
K1 = 1cm/sec
K2 = 0.2cm/sec
L1 = 30cm
L2 = 30cm
D = 2cm
Q @ H2 = ??
H3 = ??
Calculate Equivalent K's of the "Ant Farms"
9
Looking Back
Recall the
options and
when each is
appropriate
AVERAGING OPTIONS
Weighted Arithmetic
∑K d
∑d
i
i
i
Geometric
N
⎛1
⎞
⎜ (logK1 +logK2 +...+logKN ) ⎟
⎝N
⎠
K1K2 ...KN or 10
⎛1
⎞
⎜ log(K1K2 ...KN ) ⎟
⎝N
⎠
or 10
Above assumes each K represents an equal volume, each could be weighted by the volume represented
e.g. each K (or for the middle option log(K)) would be multiplied by di/dmin & then N= sum of di/dmin
Weighted Harmonic
∑d
d
∑K
i
i
i
⎛h −h ⎞
VDarcy = − K ⎜ 2 1 ⎟
⎝ l ⎠
K is equivalent K over the distance l
l is the flow path distance between h2 and h1
gradient is in the direction of flow
when using velocity to calculate volumetric
discharge, area is perpendicular to the gradient
10
Consider a Diagonal Gradient
Given a Keqx and a Keqz
H top left = htl
Δhx = htl -htr = hbl -hbr
iz = Δhz /Lz
18
10
ix = Δhx /Lx
Vx = Keqx ix
Vz = Keqz iz
Δhz = htl -hbl = htr -hbr
20
Lz
H top right = htr
Lx
H bot left = htl- Δhz
Resultant V
H bot right =htr - Δhz
8
Calculate a resultant velocity at the center
now the gradient is different at top and bottom and from side to side
Kx=1x10-5 m/s Ky=1x10-6m/s
42m
32m
26m
26
21m
29
33m
29m
33
32
11
Consider a Gradient Askew to Layering:
The overall velocity is actually composed of individual
velocity vectors in each layer
H top left = htl
Δhx = htl -htr = hbl -hbr
H top right = htr
Lx
dT
d1
ix = Δhx /Lx
K1
iz1 = Δhz1 /dz1
Kx1 =Kz1
d2
ix = Δhx /Lx
K2
iz2 = Δhz2 /dz2
Kx2 =Kz2
d3
ix = Δhx /Lx
K3
iz3 = Δhz3 /dz3
Kx3 =Kz3
d4
ix = Δhx /Lx
iz4 = Δhz4 /dz4
Kx4 =Kz4
K4
H bot left = htl- Δhz
H bot right =htr - Δhz
dT
Δh
di
∑K
Δh
K 1 Δh 1 K 2 Δh 2
K n Δh n K eq Δh
i
=
=
V=
=
= ...
=
d
dT
d1
d2
dn
dT
∑ Ki
i
K i Δh i
Δh
so
=
d
di
∑ Ki
i
To calculate Δhzi:
Δh i =
Δhd i
d
Ki ∑ i
Ki
d1
d2
dT
Δh
Δh1
Δh2
d3
Δh3
d4
Δh4
K1
K2
K3
K4
12
Consider a Gradient Askew to Layering:
The overall velocity is actually composed of individual
velocity vectors in each layer
K4= K1 < K3 << K2
H top left = htl
H top right = htr
Lx
d2
d3
d4
K1
Kx1 =Kz1
K2
Kx2 =Kz2
Kx3 =Kz3
K3
Vx = Keqx ix
Vz = Keqz iz
dT
ix = Δhx /Lx
iz = Δhz /Lz
d1
Kx4 =Kz4
K4
Resultant V
H bot left = htl- Δhz
H bot right =htr - Δhz
Refraction at Layer Interfaces
K 1 tan θ1
=
K 2 tan θ 2
θ1
K4= K1 < K3 << K2
K1
θ2
K1
K2
K2
K2
K3
K3
θ2
θ3
K4
13