Page 543 - Petrophysics

5 10 PETROPHYSICS: RESERVOIR ROCK PROPERTIES

flushed zone. Using Eq. 8.10b:

The formation resistivity factor F and the water saturation in the
invaded zone Sxo are estimated from equations 8.8~ and 8.12,
respectively, assuming z Z 1 and CX = 0.2:
1
F= = 31.2
0.14l.'5

Using Equation 8.13 the porosity partitioning coefficient is:

0.19
=
V= (L - L) 0.24
O.ld(0.22 - 0.738) 95 9.72
This value indicates that fractures contribute 24 percent of the total
pore space.
(2) We now can estimate the matrix porosity and fracture porosity from
Equations 8.14 and 8.15:

@m = @t(l - V) = 0.14(1 - 0.24) = 0.106

(3) Assuming the water saturation in the fractures is equal to the water
saturation in the matrix, the initial oil in place in the matrix and
fractures are calculated from Equations 8.2 and 8.3, respectively:

(7,758)(3,000)(52)(0.11)(1 - 0.22)(1 - 0.034)
Nom =
1.25
= 80,186,700 STB

(7,758)(3,000)(52)(0.034)(1 - 0.22)
Nof = 1.25 = 25,540,565 STB

The total oil in place in this naturally fractured reservoir is:

Not = 80,186,700 + 25,540,565 = 105.7~10~ STB

This total oil volume is correct, assuming the porosity partitioning
coefficient is the same in the entire reservoir. This is highly unlikely