Page 270 - Introduction to Petroleum Engineering

OIL RESERVOIR MATERIAL BALANCE 257
can be compared with volumetric estimates of original volumes. The two methods
are independent and provide a comparison between a geologic method that
depends on static data and an engineering method that depends on dynamic, time‐
dependent data.
Different material balance methods are used for oil reservoirs and gas reservoirs.
The material balance method for oil reservoirs is discussed here, and the material
balance method for gas reservoirs is discussed in the next section.

13.4.1 Undersaturated Oil Reservoir with Water Influx
We illustrate the material balance method for analyzing an oil reservoir by deriving
the material balance equation for an undersaturated oil reservoir with water influx.
Water influx is water flowing into the reservoir from a natural source such as an
aquifer. We simplify the derivation by assuming the effect of compressibility is
negligible. In this case, the decrease in oil volume V at reservoir conditions that
o
occurs when oil is produced must be matched by an increase in water volume V at
w
reservoir conditions so that
V w V o (13.8)
The change in oil volume at reservoir conditions is

V NB N N B (13.9)
o oi p o
where N is OOIP (STB) and NB is initial OIP at reservoir conditions (RB) for initial
oi
oil formation volume factor (FVF) B (RB/STB). N is oil produced (STB), so
oi
p
(N − N )B is remaining OIP at reservoir conditions.
p
o
The change in water volume at reservoir conditions is
V W W W B WB W WB (13.10)
w e p w w e p w
where W is OWIP (STB), W is water influx (STB), W is cumulative water
p
e
production (STB), and B is water FVF (RB/STB) which is often approximated as
w
1.0 RB/STB. Water influx refers to water entering the system from a natural source
such as an aquifer or by water injection.
The material balance equation for the oil–water system is obtained by substituting
Equations 13.9 and 13.10 into Equation 13.8 to give
W W B NB N N B (13.11)
e p w oi p o
This equation is one equation for the two unknowns N and W . Solving
e
Equation 13.11 for N gives

NB W W B
N p o p e w (13.12)
B o B oi