Page 80 - Well Logging and Formation Evaluation

70 Well Logging and Formation Evaluation

This plot was made assuming an m* value of 2.0. From the plot it was
possible to determine: R w = 0.05ohmm; f c = 0.13; and C = 0.01. Using
an assumed value of n*, it is then possible to calculate Waxman-Smits
saturations in the hydrocarbon leg.
If SCAL (special core analysis) data are available, it is possible to
derive m* and n* from the experiments as follows. In a conventional
cementation-exponent (m) measurement, the formation factor F is plotted
against f on a log-log scale. In Archie’s model, the following would be
true:

F = ( R R w ) = f - m (5.1.6)
o

where R o is the resistivity of the 100% water-saturated rock. Hence
log(F) =-m*log(f) and the gradient of the line yields m. Since for
the Waxman-Smits equation it is clearly not the case that F =f -m * , a cor-
rection must be made. Let:

F* = (1 + R BQ v * ) F = f - m* (5.1.7)
w
Now, if F* is plotted against f on a log-log scale, it is indeed the case that
the gradient yields m*.
Having derived m*, it should then be used to rederive C wa from R t and
f. This will then lead to revised values of R w, f c , and C. This may in turn
lead to a revised value of m*. Usually a couple of iterations are sufficient
to get m* to converge to a value that fits both the C wa vs. 1/f plot and the
F* vs. f plot. Since F* will exceed F by a larger amount at low porosi-
ties than at high porosities, m* will always be greater than m. Typically,
if an m value of 2.0 is measured, the value of m* will be around 2.2. A
similar procedure is followed for n*. Archie’s model assumes that:

I = ( R R o ) = S w - n (5.1.8)
t
Hence, if log(I) is plotted against log(S w ), the gradient should yield n.
For Waxman-Smits, it is necessary to derive I*, which is given by

n
-
)
I* = (1 + R BQ S w * ) I (1 + R BQ v = S w *. (5.1.9)
w
v
w
Plotting log(I*) versus log(S w ) yields the corrected saturation exponent
n*. As with m*, n* will be found to be lower than n, typically by
about 0.2.