# Darcy's Law

```Darcy’s Law
In 1856, Darcy investigated the flow of water through sand filters for water
purification purposes. His experimental apparatus is shown below:
Output manometer
Input manometer
Water
h1-h2
Water in
1
Sand Pack
Flow direction
A
h1
L
h2
Datum
2
Water out
at a rate q
Water
Water
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.
2.
3.
4.
5.
Single-phase flow (only water)
Homogeneous porous medium (sand)
Vertical flow.
Non-reactive fluid (water)
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.:
1
q ∝ h1 − h2
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∝
1
L
Thus we write:
q=C
A
(h1 − h2 )
L
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 ∝ (grain size) = d 2
2
It is inversely proportional to tortuosity; i.e.:
C∝
1
1
=
tortuosity τ
2
and inversely proportional to the specific surface; i.e.:
C∝
1
1
=
specific surface S s
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
γ A
(h − h )
µ L 1 2
Since:
q = vA
Thus we can write:
v=
q
γ (h1 − h2 )
=K
A
µ
L
Now, let us consider the more realistic flow; i.e. the tilted flow for the same sand
pack:
Datum
P2, D2
2
P1, D1
w
Flo
1
io
ect
r
i
D
L
3
n
A
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:
h1 = D1 −
p1
h2 = D2 −
&
γ
p2
γ
Thus:

p  
p 
1
h1 − h2 =  D1 − 1  −  D2 − 2  = ( D1 − D2 ) − ( p1 − p2 )
γ  
γ 
γ

which can be written in a difference form as:
h1 − h2 = ∆D −
1
γ
∆p
Substituting (2) into (1) and rearranging yields:
v=K
γ  ∆D 1 ∆p 
∆D 
K  ∆p
−
=− 
−γ


µ L γ L 
µL
L 
The differential form of Darcy’s equation for single-phase flow is written as follows:
v=−
∂D 
K  ∂p
−γ

∂L 
µ  ∂L
For multi-phase flow, we write Darcy’s equation as follows:
 k   ∂p
∂D 
v = − K r   − γ
∂L 
 µ   ∂L
More conveniently, the compact differential form of Darcy’s equation is written as
follows:
k 
r
v = − K  r  (∇p − γ∇D)
 µ
Where K is the absolute permeability tensor which must be determined
experimentally. It is written as follows:
4
 k xx

K =  k yx
 k zx

k xy
k yy
k zy
k xz 

k yz 
k zz 
Substitute (5) into (4), we obtain:
 k xx
v x 
 kr  
 
v y  = −    k yx
 µ
v 
 k zx
 z

k xy
k yy
k zy
 ∂p
 −γ
k xz   ∂x
  ∂p
k yz   − γ
∂y
k zz   ∂p
 −γ
 ∂z
∂D 
∂x 
∂D 

∂y 
∂D 
∂z 
Solve for velocity components yields:
 k    ∂p
 ∂p
∂D 
∂D 
∂D  
 ∂p
v x = −  r   k xx  − γ

 + k xy  − γ
 + k xz  − γ
 ∂z
∂x 
∂y 
∂z  
 ∂y
 µ    ∂x
 k    ∂p
 ∂p
∂D 
∂D 
∂D  
 ∂p
v y = −  r   k yx  − γ

 + k yy  − γ
 + k yz  − γ
 ∂z
∂x 
∂y 
∂z  
 ∂y
 µ    ∂x
 k    ∂p
 ∂p
∂D 
∂D 
∂D  
 ∂p
v z = − r   k zx  − γ
 + k zy  − γ

 + k zz  − γ
 ∂z
∂x 
∂y 
∂z  
 µ    ∂x
 ∂y
In most practical problems, it is necessary to assume that K is a diagonal tensor; i.e.:
 k xx
K=0

 0
0
k yy
0
0
0

k zz 
and thus velocity components of equation (6) become:
 k   ∂p
∂D 
v x = − k xx  r   − γ

∂x 
 µ   ∂x
 k   ∂p
∂D 
v y = − k yy  r   − γ

∂y 
 µ   ∂y
 k   ∂p
∂D 
v z = − k zz  r   − γ

∂z 
 µ   ∂z
5
Unit Analysis
Since:
KA ∆p
qµ ∆L
⇒
K=
A ∆p
µ ∆L
2
When q in cc/sec, µ in cp, )L in cm, A in cm , and )p in atm, then K is in the units of
Darcy, where:
q=
(1 cm )(1 cp)(1 cm) = (1 cm ) (1 cp)
1 Darcy =
(s) (1 cm )(1 atm) (1 atm) (1 s)
3
2
2
Since
1 cp =
1
1 dyne s
poise =
100
100 cm 2
and
1 atm = 1,013,250
dyne
cm 2
thus we have:
1 cm 2 1 dyne s cm 2
1
1 Darcy = 1000 md =
=
cm 2
2
(100) cm (1,013,250) dyne s 101,325,000
or
1 md =
10 − 9
cm 2 = 986.923x10 −14 cm 2
.
101325
Since:
q=
KA ∆p
µ ∆L
Thus:
bbl
day
(md )( ft 2 )( psi ) (md )( ft )( psi )
=
=
(cp)( ft )
(cp)
2
=
 10 − 9  
lb 
1   1

bbl  144 2 


 
.   2.54 x12   5.6146  
ft 
 101325
 1   1   lb − day 




 47,880   86,400   ft 2 
≈ 0.001127107
6
= 1127106.663x10 − 9
```
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