Page 186 - Introduction to Petroleum Engineering

RESISTIVITY LOGS 173

1000000
F(0.35, 1.14)
100 000 F(0.35, 2.52)
F(4.78, 1.14)
Formation factor (F(a,m)) 1000
F(4.78, 2.52)
10000

100

10

1
0.00 0.10 0.20 0.30 0.40 0.50
Porosity (fraction)

FIGuRE 9.7 Sketch of formation resistivity factor for sands.
from 1.14 to 2.52. Tortuosity factor a depends on pore geometry. Formation
resistivity factor F(a,m) is plotted as a function of porosity for sands in Figure 9.7.
In the case of sands, the empirical parameters m and a vary from 1.14 to 2.52 and
from 0.35 to 4.78 (Bassiouni, 1994), respectively.
Archie’s equation is an empirical relationship for the formation resistivity R of a
t
porous medium that is partially saturated by an electrically conducting wetting phase
with saturation S . Archie’s equation is
w
R = R S − w n (9.12)
0
t
where the empirical parameter n is the saturation exponent. If the porous medium is
completely saturated by the wetting phase so that S = 1, then R = R from Archie’s
t
w
0
equation. Archie’s equation for wetting‐phase saturation
 FR  1/ n
S =  w  (9.13)
w
 R t 
is obtained by combining Equations 9.10 through 9.12. The wetting‐phase saturation
is often water saturation.

Example 9.5 Resistivity Log
A consolidated sandstone has tortuosity factor = 0.81, cementation factor = 2,
and formation factor = 90. What is the porosity of the rock?

answer
m
−
/m
1
Solve Fa m( , ) = aφ for φ = (aF ) to get φ = (.08190/ ) / 12 = .095
/
0