Page 140 - Fundamentals of Reservoir Engineering

MATERIAL BALANCE APPLIED TO OIL RESERVOIRS 79

(B − B oi )
o
c =
o
B ∆ p
oi
and substituting this in equ. (3.15) gives

(c S + c )
NB = NB oi c o w wc f ∆ p (3.16)
o
p
1S wc
−
Since there are only two fluids in the reservoir, oil and connate water, then the sum of
the fluid saturations must be 100% of the pore volume, or
S o + S wc = 1

and substituting the latter in equ. (3.16) gives the reduced form of material balance as

cS + c S + c
NB = NB oi o o w wc f ∆ p (3.17)
p
o
1S wc
−
or

NB = NB c ∆ p (3.18)
p
e
oi
o
in which
1
c = (c S + c S + c ) (3.19)
e
1S o o w wc f
−
wc
is the effective, saturation-weighted compressibility of the reservoir system. Since the
saturations are conventionally expressed as fractions of the pore volume, dividing by
1 − S wc expresses them as fractions of the hydrocarbon pore volume.
Thus the compressibility, as defined in equ. (3.19),must be used in conjunction with the
hydrocarbon pore volume. Equation (3.18) illustrates how the material balance can be
reduced to nothing more than the basic definition of compressibility, equ. (1.12), in
which N pB o = dV, the reservoir production expressed as an underground withdrawal,
and NB oi = V the initial hydrocarbon pore volume.

EXERCISE 3.1 SOLUTION GAS DRIVE; UNDERSATURATED OIL RESERVOIR

Determine the fractional oil recovery, during depletion down to bubble point pressure,
for the reservoir whose PVT parameters are listed in table 2.4 and for which
-6
= 3.0 × 10 / psi S wc = .20
c w
-6
= 8.6 × 10 / psi
c f
EXERCISE 3.1 SOLUTION

The data required from table 2.4 are

= 4000 psi = 1.2417 rb/stb
p i B oi