Page 490 - Petrophysics

458 PETROPHYSICS: RESERVOIR ROCK PROPERTIES

Thus, Equation 7.1 13 becomes:

2
aq, + bqm fc = 0
This is, of course, a quadratic equation with two solutions: one negative
and one positive. Inasmuch as the negative value has no physical mean-
ing, the solution is:
1
qm = -[-b + (b2 - 4a~)'.~]
2a
For k = 225 mD and 9 = 20% the value of p from Equation 7.116a is
3 x lo7 ft-l. Inlaboratoryunits [g, = 1,013,420(g - cm)/(atm/cm2)(s2)],
p is equal to:

3 x 10' = 0.97 atm-sec2/g
= (2.54)( 12)( 1,013,420)

The cross-sectional area is equal to:

4
The other variables in laboratory units for qm in g/s are:

M = 29 g/g-mole.
pg = 0.02 CP.
R = 82.06 cm3-atm/(g-mole) (OK).
T = 297.2"K.
L=5cm.
p: - pi = 6.802 - 1= 45.28 atm2.
k = 0.225 Darcy.

The values of the constants a, b, and c are

0.97
a= = 15.44
(0.02)(3.14)
1
b=-- - 4.44
0.225
(29)(3.14)(45.28)
C= = -0.85
(2)( 1)(82.06)(297.2)(0.02)(5)

Thus, the equation describing the mass flow rate through the core is:

15.449; + 4.44qm - 0.85 = 0