Effective Hydraulic Conducitivity of Layered Media
Effective Conductivity Parallel to Layers
h1
h1
h
h2
h
h2
Q1
b1
K1
Q2
QT
K2
Q3
Ke
b2
K3
b3
x
Layered media
Equivalent homogenous media
The volmetric flow through each layer
(y is the width of the area through which flow occurs)
Q1 = -y b1 K1 (h/x)
Q2 = -y b2 K2 (h/x)
Q3 =-y b3 K3 (h/x)
(1a)
(1b)
(1c)
QT = -y bT Ke (h/x)
(2)
Assume that the flows through the layered media and the equivalent media are equal
Q1 + Q2 + Q3 = QT
(3)
Substituting (1) and (2) into (3)
-y bT Ke (h/x) = -y b1 K1 (h/x) - y b2 K2 (h/x) - y b3 K3 (h/x)
cancelling and dividing by bT gives the effective hydraulic conductivity.
Ke =
1
{b1 K1 + b2 K2 + b3 K3 }
bT
(4)
This is simpler when written in terms of transmissivity, T=Kb
Te = T1 + T2 + T3
bT
(5)
Effective Hydraulic Conducitivity of Layered Media
Effective Conductivity Normal to Layers
h1
h1
h2
h14
h14
h4
h4
h3
q
K1
b1
K2
b2
q
K3
Ke
bT
b3
Layered media
Equivalent homogeneous media
According to Darcy’s Law, the flux through each layer is
h  h1
h
b
q1   K1 2
  K1 12 ;
or
h12  q1 1
b1
b1
K1
h3  h2
h
  K2 23 ;
b2
b2
h  h3
h
q 3   K3 4
  K3 34 ;
b3
b3
and
h  h1
h
q e   Ke 4
  Ke 14 ;
bT
bT
q 2   K2
or
or
or
b2
K2
b
h34  q 3 3
K3
h23  q 2
h14  q e
bT
Ke
where h23 is shorthand for the head drop across layer 2, etc. Assuming that the head
drop across the equivalent media is the same as the total drop across the layers
b
b
b
b
h14  h12  h23  h34 ;
or q e T  q1 1  q 2 2  q 3 3
Ke
K1
K2
K3
But the flux through each layer is the same, and they are equal to the flux through the
equivalent layer, so
n
bT
bT
Ke =
or for n layers Ke = 
 b1
b2
b 
i=1  bi 
 3
 
 
K2
K3 
 K1
 Ki 