Page 67 - Fundamentals of Reservoir Engineering

SOME BASIC CONCEPTS IN RESERVOIR ENGINEERING 6

Pressure (psia)
2250 2375 2500 Exploration
Well
5000 2265 2369

Depth GAS
(feet)
GOC : p = p = 2385 GOC 5200'
w
o
5250
TEST RESULTS
OIL at 5250 ft
p = 2402 psia
o
5500 OWC : p = p = 2490 OWC 5500' dp
o
w
dD = 0.35psi/ft
WATER
Fig. 1.3 Pressure regimes in the oil and gas for a typical hydrocarbon accumulation

which assumes a normal hydrostatic pressure regime. Therefore, at the oil-water
contact

p = p w = .45 5500 15 = 2490 (psia)
+
×
o
The linear equation for the oil pressure, above the oil water contact, is then
p o = 0.35D + constant

and since p o = 2490 psia at D = 5500 ft, the constant can be evaluated to give the
equation

+
p = 0.35D 565 (psia ) (1.8)
o
At the gas-oil contact at 5200 ft, the pressure in both fluids must be equal and can be
calculated, using equ. (1.8), to be 2385 psia. The equation of the gas pressure line can
then be determined as

p = 0.08D 1969 (psia ) (1.9)
+
o
Finally, using the latter equation, the gas pressure at the very top of the structure, at
5000 ft, can be calculated as 2369 psia. The pressure lines in the hydrocarbon column
are drawn in the pressure depth diagram, fig. 1.3, from which it can be seen that at the
top of the structure the gas pressure exceeds the normal hydrostatic pressure by
104 psi. Thus in a well drilling through a sealing shale on the very crest of the structure
there will be a sharp pressure kick from 2265 psi to 2369 psia on first penetrating the
reservoir at 5000 ft. The magnitude of the pressure discontinuity on drilling into a
hydrocarbon reservoir depends on the vertical distance between the point of well
penetration and the hydrocarbon water contact and, for a given value of this distance,
will be much greater if the reservoir contains gas alone.