Page 455 - Petrophysics

LINEAR FLOW OF GAS 423

ps = 0.015 cP, L = 12 in. x 2.54 cm/in. = 30.48 cm, and Ap =
(100 psi)/(14.7 psi/atm) = 6.80 atm, the apparent velocityis equal to:

0.245 6.8
v=- x-- - 3.64
0.015 30.48
and the actual velocity is:

3.64
v, = = 21.1 cm/sec
0.24(1 - 0.28)

(b) The mean volumetric flow rate of gas through this sandpack in ft3/D
is obtained from Equation 7.19, where k = 254 mD, L = 1 ft, and
A = 1’~(1/12)~ = 0.0128 ft2:

6.33 x lo3 x 245 x 0.0218 x 100
q= = 225.5 ft3/D
0.015 x 1
or
(
Day
= 225.5- (24 x 60 x 60sec ) (30.483$) = 73.9- cc
::y)
sec
Assuming constant steady flow rate, q, a pressure distribution equation
along a linear sand body also can be derived by combining Boyle’s and
Darcy’s laws, and integrating between p1 and p, and 0 and x. Replacing
L with x in Equation 7.12 and integrating:

(7.22)

From Equation 7.13, one can obtain:

(7.23)

Dividing Equation 7.22 by Equation 7.23 and solving for the variable
pressure p gives:

2x
2
p2 = (p; - P1)- + P1 (7.24)
L
This expression indicates that the pressure decline vs. distance during
steady-state flow of gas through a linear system follows the parabolic
curve. It also indicates that pressure is maintained near the inlet because
of the release of energy stored in the gas, but it is still independent of
fluid and rock properties. Generally, the use of linear steady-state flow of
compressible and incompressible fluids is limited to laboratory testing.