Darcy law details

1. 1
Darcy’s Law
In 1856, Darcy investigated the flow of water through sand filters for water
purification purposes. His experimental apparatus is shown below:
Sand
Pack
A
L
Water
Water
Water
h1
h2
h1-h2
Datum
Input manometer
Output manometer
Water in
Water out
at a rate q
Flow
direction
1
2
Where q is the volume flow rate of water downward through the cylindrical sand
pack. The sand pack has a length L and a cross-sectional area A. h1 is the height
above a datum of water in a manometer located at the input face. h2 is the height
above a datum of water in a manometer located at the output face. The following
assumptions are implicit in Darcy’s experiment:
1. Single-phase flow (only water)
2. Homogeneous porous medium (sand)
3. Vertical flow.
4. Non-reactive fluid (water)
5. Single geometry.
From this experiment, Darcy concluded the following points:
The volume flow rate is directly proportional to the difference of water level in the
two manometers; i.e.:

2. 2
q h h
∝ −
1 2
The volume flow rate is directly proportional to the cross-sectional area of the sand
pack; i.e.:
q A
∝
The volume flow rate is inversely proportional to the length of the sand pack; i.e.:
q
L
∝
1
Thus we write:
( )
q C
A
L
h h
= −
1 2
Where:
A is the cross-sectional area of the sand pack
L is the length of the sand pack
h1 is the height above a datum of water in a manometer located at the input face
h2 is the height above a datum of water in a manometer located at the output face
C is the proportionality constant which depends on the rock and fluid properties. For
the fluid effect, C is directly proportional to the fluid specific weight; i.e.
C ∝ γ
and inversely proportional to the fluid viscosity; i.e.:
C ∝
1
µ
Thus:
C ∝
γ
µ
For the rock effect, C is directly proportional to the square of grain size; i.e.:
( )
C d
∝ =
grain size
2 2
It is inversely proportional to tortuosity; i.e.:
C ∝ =
1
tortuosity
1
τ

3. 3
Flow Direction
2
1
A
L
Datum
P2, D2
P1, D1
and inversely proportional to the specific surface; i.e.:
C
Ss
∝ =
1
specific surface
1
where Ss is given by:
Ss =
Interstitial surface area
bulk volume
Combine the above un-measurable rock properties into one property, call it
permeability, and denote it by K, we get:
( )
q K
A
L
h h
= −
γ
µ 1 2
Since:
q vA
=
Thus we can write:
( )
v
q
A
K
h h
L
= =
−
γ
µ
1 2
Now, let us consider the more realistic flow; i.e. the tilted flow for the same sand
pack:

4. 4
Note that the fluid flows from point 1 to point 2, which means that the pressure at
point 1 is higher than the pressure at point 2. Since:
h D
p
h D
p
1 1
1
2 2
2
= − = −
γ γ
&
Thus:
( ) ( )
h h D
p
D
p
D D p p
1 2 1
1
2
2
1 2 1 2
1
− = −





 − −





 = − − −
γ γ γ
which can be written in a difference form as:
h h D p
1 2
1
− = −
∆ ∆
γ
Substituting (2) into (1) and rearranging yields:
v K
D
L
p
L
K p
L
D
L
= −





 = − −






γ
µ γ µ
γ
∆ ∆ ∆ ∆
1
The differential form of Darcy’s equation for single-phase flow is written as follows:
v
K p
L
D
L
= − −






µ
∂
∂
γ
∂
∂
For multi-phase flow, we write Darcy’s equation as follows:
v K
k p
L
D
L
r
= −





 −






µ
∂
∂
γ
∂
∂
More conveniently, the compact differential form of Darcy’s equation is written as
follows:
( )
r
v K
k
p D
r
= −





 ∇ − ∇
µ
γ
Where K is the absolute permeability tensor which must be determined
experimentally. It is written as follows:

5. 5
K
k k k
k k k
k k k
xx xy xz
yx yy yz
zx zy zz
=










Substitute (5) into (4), we obtain:
v
v
v
k
k k k
k k k
k k k
p
x
D
x
p
y
D
y
p
z
D
z
x
y
z
r
xx xy xz
yx yy yz
zx zy zz










= −
















−
−
−


















µ
∂
∂
γ
∂
∂
∂
∂
γ
∂
∂
∂
∂
γ
∂
∂
Solve for velocity components yields:
v
k
k
p
x
D
x
k
p
y
D
y
k
p
z
D
z
x
r
xx xy xz
= −





 −





 + −





 + −












µ
∂
∂
γ
∂
∂
∂
∂
γ
∂
∂
∂
∂
γ
∂
∂
v
k
k
p
x
D
x
k
p
y
D
y
k
p
z
D
z
y
r
yx yy yz
= −





 −





 + −





 + −












µ
∂
∂
γ
∂
∂
∂
∂
γ
∂
∂
∂
∂
γ
∂
∂
v
k
k
p
x
D
x
k
p
y
D
y
k
p
z
D
z
z
r
zx zy zz
= −





 −





 + −





 + −












µ
∂
∂
γ
∂
∂
∂
∂
γ
∂
∂
∂
∂
γ
∂
∂
In most practical problems, it is necessary to assume that K is a diagonal tensor; i.e.:
K
k
k
k
xx
yy
zz
=










0 0
0 0
0 0
and thus velocity components of equation (6) become:
v k
k p
x
D
x
x xx
r
= −





 −






µ
∂
∂
γ
∂
∂
v k
k p
y
D
y
y yy
r
= −





 −






µ
∂
∂
γ
∂
∂
v k
k p
z
D
z
z zz
r
= −





 −






µ
∂
∂
γ
∂
∂

6. 6
Unit Analysis
Since:
q
KA p
L
K
q
A
L
p
= ⇒ =
µ
µ
∆
∆
∆
∆
When q in cc/sec, µ in cp, )L in cm, A in cm2
, and )p in atm, then K is in the units of
Darcy, where:
( )( )( )
( ) ( )( )
( ) ( )
( ) ( )
1 Darcy
1 cm 1 cp 1 cm
s 1 cm 1 atm
1 cm 1 cp
1 atm 1 s
3
2
2
= =
Since
1 cp =
1
100
poise =
1
100
dyne s
cm2
and
1 atm = 1,013,250
dyne
cm2
thus we have:
( ) ( )
1 Darcy md =
1 cm 1 dyne s cm
100 cm 1,013,250 dyne s
cm
2 2
2
2
= =
1000
1
101325 000
, ,
or
1
10
101325
986923 10
9
14
md = cm cm
2 2
−
−
=
.
. x
Since:
q
KA p
L
=
µ
∆
∆
Thus:
( )( )( )
( )( )
( )( )( )
( )
bbl
day
md ft psi
cp ft
md ft psi
cp
x
bbl
lb
ft
lb day
ft
x
= =
=




































−






=
≈
−
−
2
9 2
2
2
9
10
101325
1
254 12
1
56146
144
1
47 880
1
86 400
1127106 663 10
0001127107
. . .
, ,
.
.