Page 125 - Petrophysics

98 PETROPHYSICS: RESERVOIR ROCK PROPERTIES

initial oil-in-place (N in bbl) by this method is:

(3.5)

where:

As = surface area of the reservoir, acres.
h = thickness of the formation, ft.
@ = porosity, fraction.
Soi = initial oil saturation, fraction.

Equation 3.5 gives the volume of oil contained in the porous rock at
reservoir conditions of pressure and temperature. However, the surface
or “stock tank” oil as finally sold by the producer is different from
the liquid volume that existed underground. The difference is due to
the changes in the oil properties as the pressure is decreased from
high underground pressure and temperature to surface pressure and
temperature. This reduction in p and T causes some of the volatile
components to come out of solution (evaporate), causing the liquid
volume to shrink. This reduction in volume is expressed by the oil
formation volume factor, Bo. Thus, the stock tank oil initially in
place is:

(3.6)

where Boi is in reservoir barrels per stock tank barrel or bbVSTB. In
this equation, Soi is replaced by (1 - Si,), where Si, is the irreducible
or connate water saturation. This implies that no free gas is present in
the pore space. Because no petroleum reservoir is homogenous, the
factors A,, h, @, and Si,, must be averaged. The constant 7,758 becomes
10,000 if A,, h are expressed in hectares (1 hectare = 10,000 m2) and m,
respectively, and N in m3.

EXAMPLE

Calculate the initial oil-in-place (N) of an oil reservoir ifA = 1,600 acres,
h = 32 ft, @ = 22%, Si, = 20%, and Boi = 1.23 bbl/STB.

SOLUTION

Using Equation 3.6, we have:

N = 7,758(1,600)(32)(0.22)(1 - 0.20)/1.23 = 56.8 x lo6 STB