Page 111 - Fundamentals of Reservoir Engineering

PVT ANALYSIS FOR OIL 50

The shapes of the B o and R s curves below the bubble point, shown in fig. 2.5(a) and
(b), are easily explained. As the pressure declines below p b, more and more gas is
liberated from the saturated oil and thus R s, which represents the amount of gas
dissolved in a stb at the current reservoir pressure, continually decreases. Similarly,
since each reservoir volume of oil contains a smaller amount of dissolved gas as the
pressure declines, one stb of oil will be obtained from progressively smaller volumes of
reservoir oil and B o steadily declines with the pressure.

EXERCISE 2.1 UNDERGROUND WITHDRAWAL

The oil and gas rates, measured at a particular time during the producing life of a
reservoir are, x stb oil/day and y scf gas/day.

1) What is the corresponding underground withdrawal rate in reservoir barrels/day.

2) If the average reservoir pressure at the time the above measurements are made
is 2400 psia, calculate the daily underground withdrawal corresponding to an oil
production of 2500 stb/day and a gas rate of 2.125 MMscf/day. Use the PVT
relationships shown in figs. 2.5(a) − (c), which are also listed in table 2.4.

3) If the density of the oil at standard conditions is 52.8 lb/cu.ft and the gas gravity is
0.67 (air = 1) calculate the oil pressure gradient in the reservoir at 2400 psia.

EXERCISE 2.1 SOLUTION

1) The instantaneous or producing gas oil ratio is R = y/x scf/stb. If, at the time the

surface rates are measured, the average reservoir pressure is known, then B o, R s
and B g can be determined from the PVT relationships at that particular pressure.

The daily volume of oil plus dissolved gas produced from the reservoir is then
y
xB o rb, and the liberated gas volume removed daily is x( − R ) B g rb. Thus the
x s
total underground withdrawal is

y
x(B + ( − R )B ) rb / day (2.4)
o
s
g
x
2) At a reservoir pressure of 2400 psia, the PVT parameters obtained from table 2.4
are:

B o = 1.1822 rb/stb; R s = 352 scf/stb and B g = .0012 rb/ scf

Therefore, evaluating equ. (2.4) for x = 2500 stb/d and y = 2.125 MMscf/d gives a
total underground withdrawal rate of

2500 (1.1822 + (850 − 352) × .0012) = 4450 rb/d

3) The liquid oil gradient in the reservoir can be calculated by applying mass
conservation, as demonstrated in exercise 1.1 for the calculation of the gas
gradient. In the present case the mass balance is