Chapter 3 – Permeability
3.2 Darcy’s Law
In 1856, Darcy investigated the flow of water through sand filters for water
purification.
His experimental apparatus is shown in Figure 3.11.
By empirical
observation
Figure 3.11 Schematic of Darcy’s experiment on flow of water through sand.
Darcy noticed that fluid flow was directly proportional to the hydraulic gradient, resulting in
the following equation,
h h
q  KA 2 1
l
(3.3)
where q represents the volumetric flow rate of water downward through the cylindrical sand
pack of cross-sectional area A and height l, h1 and h2 are the hydraulic head above the
standard datum of the water in the manometer located at the input and output ports
respectively, and K is a constant of proportionality found to be characteristic of the rock
media. It was observed that for any angle of the sandpack, a given constant flow rate would
result in the same h; i.e., Eq. (3.3) is independent of the direction of flow.
Fluid movement is due to the difference in potential energy. The two components of
the potential energy are fluid pressure and elevation or gravity. By defining a relationship
3.12
Chapter 3 – Permeability
between hydraulic head and pressure, it is possible to determine pressure at any point in the
flow path. The total potential function per unit mass can be written as,

p

 gz  gh
(3.4)
where g is the acceleration of gravity, p is pressure and  is the density of the fluid and z is
the elevation of a given point in the system. Substituting Eq. (3.4) into Eq. (3.3) and writing
in differential form, we obtain,
q

K d p
A   gz 
g dl  

(3.5)
The parameter, K is known as the hydraulic conductivity. A sequence of experiments
maintained at constant fluid potential gradient investigated the effect of different properties
on the flow rate. The resulting expression for the hydraulic conductivity is,
K
2
cd g

(3.6)
where d is grain diameter,  is the fluid viscosity, and c is a constant of proportionality.
Substituting into Eq. (3.5) we get,
q
kA  dp
dz 
 g 

  dl
dl 
(3.7)
where k is the permeability of the porous media. Equation (3.7) is known as Darcy’s Law. It
is applicable for steady state, laminar flow of an incompressible fluid, for a homogeneous
and isotropic porous media.
The negative sign in the previous equation is a result of the pressure and distance both
measured positive in the same direction (see Fig. 3.12). Therefore, the gradient within the
bracket must be negative to move the flow in the same direction as the direction from high to
lower potential. For example, take the simplified case of horizontal flow (z = 0), then
since p1 > p2 the gradient must be 
 p  p1 
dp
.
  2


dL
L


3.13
Chapter 3 – Permeability
P1
P2
q
0
L
Figure 3.12 Schematic of sign convention for potential gradient in horizontal flow
For flow at any angle, , we can substitute sin  for the elevation gradient; i.e., sin  
dz
.
dL
The sign convention is referenced such that positive is upward flow as shown in Figure 3.13.
Flow upward
(+) sin 90 = 1
dL

dz
Flow downward
sin -90 = -1
Figure 3.13 Schematic of sign convention for any flow angle
Darcy’s Law (Eq. 3.7), written for any angle of flow becomes,
q
kA  dp

 g sin  

  dl

(3.8)
The dimensions of permeability can be determined by substituting for the appropriate
cgs units found in table 3.1.
2 
2
cc k (cm )  dyne / cm 


sec
poise 
cm


3
2
 2
L
k ( L )  m 

 k  L 
T
(m / LT )  2 2 
 
L T 
3.14
Chapter 3 – Permeability
Units
cgs
Darcy
Field
q – flow rate
cc/s
cc/s
Bpd
A – cross-sectional area
cm2
cm2
ft2
L – length
cm
cm
Ft
fluid viscosity
poise
centipoise
Cp
P – pressure
dyne/cm2
atm
Psia
density
gm/cc
gm/cc
lbm/ft3
g – gravity
cm/s2
cm/s2
ft/s2
k - permeability
cm2
Darcy
Md
Table 3.1Various units used in petroleum engineering
The unit for permeability in the cgs system is 0.987 x 10-8 cm2. From a practical standpoint
this measure is too small; therefore we define 1 darcy = 0.987 x 10-12 m2  (m)2. That is, a
permeability of 1.0 Darcy is defined such that a single-phase fluid of 1.0 cp flows at a rate of
1.0 cc/sec/cm2 under a pressure of 1 atm/cm. In Darcy units, Eq. (3.7) becomes,
kA  dp
g sin  
q

6
  dl
1.013x10 

(3.9)
Example 3.2
Consider fresh water with a viscosity of 1.0 cp and density of 1.0 gm/cc, injected at a rate of
10 cc/sec through a sandpack with dimensions illustrated below.
After steady state is
achieved the pressure readings are recorded. Determine the permeability for this sandpack?
Note: Barometric pressure = 14.7 psi.
Pin=10psig
d = 2 cm
Pout=0psig
L = 5 cm
Solution
Since the core is horizontal the elevation term is zero; therefore Eq. (3.8) is reduced to,
3.15
Chapter 3 – Permeability
kA dP
 dL
2  0  14.7 10  14.7 
k (1) 

14.7
14.7 

10  
(1)(5)
k  10.8 darcies
q
Conversion to field units results in the following expression for Darcy’s Law,
q  1.127 x10
 3 kA  dp

 0.4335 sin  

  dl

where  is the specific gravity of the fluid.
3.16
(3.10)