Page 171 - Fundamentals of Reservoir Engineering

DARCY'S LAW AND APPLICATIONS 109

EXERCISE 4.1 SOLUTION

1) For linear, horizontal flow of an incompressible fluid

2
k(D) A(cm ) dp
q(cc / sec) =− (atm / cm) (Darcyunits)
µ (cp) dl
and

2
2
2
k(cm ) A(cm ) dp (dyne / cm )
q(cc / sec) =− (Absolute cgs units)
µ (poise) dl (cm)
The former equation can be converted from Darcy to cgs, absolute units by balancing
both sides of the resulting equation, as follows

2 D 2 atm
2
k(cm ) 2 A(cm ) (dyne / cm ) dyne / cm 2

q(cc / sec) =− cm cp dp (cm)

dl
µ (poise)
poise
and evaluating the conversion factors

2 D
k(cm ) 2 A
2
q =− cm dp (dyne / cm ) 1 6
µ (poise) [100 ] dl 1.0133 10
×
or

2 D
k(cm ) 2 A
q =− cm dp 1 8

µ dl 1.0133 10
×
But the numerical constant in this equation must be unity, therefore

D 8
×
2 = 1.0133 10
cm
2
-12
-8
2
so that 1 Darcy ≈ 10 cm = 10 metre .
It is proposed that the industry will eventually convert to Sl (Système Internationale)
absolute units, (table 4.1), in which case the basic unit of permeability will be the
2
metre . Because this is such an impracticably large unit, it has been tentatively
6
2
2
suggested that a practical unit, the micrometre (µm ), be "allowable" within the new
system. Since
2
-12
1 µm = 10 m 2
then

1 Darcy ≈ 1 µm 2