Page 129 - Introduction to Petroleum Engineering

ACTIVITIES 115

Example 6.6 Gas in Place
A well is draining a gas–water reservoir. The drainage area of the well is 160
acres and has a net thickness of 20 ft. Initial properties are 15% porosity, 70% gas
saturation, and gas FVF of 0.0016 RB/SCF. What was the original gas in place
in the drainage area?

Answer
OGIP is
7758φ Ah S 7758 . )(160acres )(20ft )070
(015
.
×
G = ggi = =16310 9 SCF
.
F
.
B 0 0016RB/SCF
gi
6.5.3 Recovery Factor and Estimated Ultimate Recovery
RF is the fraction of OHIP that can be produced from a reservoir. EUR is calculated
from RF and OHIP as

GRV××φ NTG×S
×
EUR = OHIP RF = hi × RF (6.3)
B hi
where EUR is estimated ultimate recovery (standard conditions), OHIP is original
hydrocarbon in place (standard conditions), RF is recovery factor (fraction) to
economic limit, GRV is gross rock volume (reservoir conditions), φ is average
porosity in net pay (reservoir conditions), NTG is net‐to‐gross ratio (reservoir condi-
tions), S is initial average hydrocarbon saturation in net pay (reservoir conditions),
hi
and B is initial hydrocarbon formation volume factor in a consistent set of units.
hi
Formation volume factor is the ratio of reservoir volume to surface volume. The
product GRV×× NTG is reservoir pore volume. The volume of fluid produced
φ
from a reservoir is the product of RF and original fluid in place. For example, a 30%
oil RF means that 30% of the OOIP can be produced. It also implies that 70% of
OOIP will remain in the reservoir. EUR is the volume of fluid produced at a specified
economic limit.

6.6 ACTIVITIES

6.6.1 Further Reading
For more information about geology, see Selley and Sonnenberg (2015), Hyne
(2012), Bjørlykke (2010), Reynolds et al. (2008), and Gluyas and Swarbrick
(2004).