Page 489 - Petrophysics

TURBULENT FLOW OF GAS 457

C(kifi)2 (35.6 x 0.40)2 + (155 x 0.60)2
=
= 8851.77

kifi = k = 35.6 x 0.40 + 155 x 0.60

= 107.24mD

From Equation 7.1 16d, the average non-Darcy flow coefficient is:

- 93.2 x 10l2
= 9.82 x 10'ft-I
= 8851.77 x 107.24
2. The arithmetic average porosity of the two layers is (0.166 +

0.138)/2 = 0.152, and the average permeability is 107.24 mD.
Substituting these values into Equations 7.116a and b, we find,
respectively:

4.11 x 10"'
= ( 107.24)4/3 = 8 x lo7 ft-'
4.8s x io4
P= 0.1525.5 x Jm
= 14.8 x 10' ft-I

The value of p obtained from Equation 7.116a compares well with the
lab-derived p.

EXAMPLE

A consolidated sand core 2cm in diameter and 5cm long has a
permeability of 225 mD and a porosity of 20%. Air at 75°F is injected into
this core. The inlet pressure is 100 psia and the outlet pressure 14.7 psia.
The viscosity of air is 0.02 cP, and the compressibility is assumed to be
equal to 1 .O. Calculate the mass flow rate.

SOLUTION

The mass flow rate for air can be calculated from Equation 7.1 13. Let