Page 88 - Petrophysics 2E

62 PETROPHYSICS: RESERVOIR ROCK PROPERTIES

as follows:
1
CW (2.11)

where V1 and V2 are the volumes at pressures p1 and p2. The ratio V2/V1
is equivalent to the amount of water expansion as the pressure drops
from p2 to p1.

EXAMPLE

The bottomhole temperature of a gas reservoir is 140°F. Calculate
the amount of water expansion, per unit volume, that will occur when
the pressure is decreased from 4,000 to 3,270 psi.
From Figure 2.13, the estimated compressibility of water at the given
reservoir conditions (i.e., at 4,000 psi) is 2.8 x psi-'.

V2/V1 = [l - (2.8 x 10-6)(3,270 - 4,OOO)I = 1.02

Water compressibility decreases when the water contains hydrocarbon
gases in solution according to the following empirical equation [26,27]:

cm = cw(l.O + 0.0088 x &) (2.12)

where: cm = compressibility of water containing solution gas
(l/kPa or l/psi).
cw = compressibility of water.
Rm = solubility of gas in water, m3 gas/m3 water (ft3/bbl).

Gas Solubility
The solubility of hydrocarbon gases in water at any given pressure does
not change very much as the temperature is increased. The behavior is
similar to compressibility because the solubility decreases slightly as the
temperature is increased from ambient temperature reaching a minimum
solubility at about (66°C) 150°F and then increasing continuously as the
temperature is increased (Figure 2.14). On the other hand, pressure has
a large influence. According to Figure 2.14, the solubility of natural gas
in water at 500 psi and 150°F is about 4.1 ft3/bbl and at 2,000 psi and
150°F the solubility increases to about 11.9 ft3hbl (2.1 m3 gas/m3). The
solubility of gas in water also is influenced by the amount of dissolved
salts. Increasing salinity decreases the solubility of hydrocarbon gases in
water according to the following empirical relationship:

RB = RW[l - X, x (ppm salts)(lO-')I (2.13)