Page 451 - Petrophysics

LINEAR FLOW OF INCOMPRESSIBLE FLUIDS 419

EXAMPLE

A 10-cm long cylindrical core sample was subjected to a laboratory
linear flow test under a pressure differential of 3.4 atm using a fluid
of viscosity 2.5 cp. The diameter of the core is 4 cm. A flow rate of
0.35 cc/sec was obtained. Calculate the permeability of this core sample.

SOLUTION

Figure 7.1 is a schematic representation of the core sample. lJsing
Equation 7.5 the permeability of the core sample is:
0.35 x 2.5 x 10
k= = 0.204 Darcy = 204 mD
12.57 x 3.4

To use Equation 7.6, one must first convert the data to oilfield units.

AP = 3.4(atm) x 14.7 psi/atm = 50 psi
( ::c3) ~- -- (24 D
q= 0.35- (30.148' cE3) (5.215 F)

= 0.19 bbl/D

L=(10cm) --
(30148 c:) = 0'328 ft

The permeability of this core sample using Equation 7.6 is:

k= qPL - 0.19 x 2.5 x 0.328 = 204 mD
-
1.127 x 10sAAP 1.127 x 103 x 0.0135 x 50
Because of the many unit systems employed by the industry, it is very
important that petroleum engineers be able to convert units from one
system to another.
To estimate the pressure at any point in a linear flow system,
Equation 7.4 is integrated between the limits of 0 and x, and p1 and
p, respectively, yielding:

(7.7)

From Equation 7.5, one obtains:
P1 -p2 = (2)L (7.8)