SPWLA TWENTI-SEVEN ANNUAL LOGGING SYMPOSIUM, JUNE 9-13,
1986Universidad Nacional Autnoma de Mxico.Facultad de
Ingeniera.Divisin de Ciencias de la Tierra.
Caracterizacin Esttica de Yacimientos.
RASMUS II: A summary of the effects of various pore geometries
and their wettabilities on measured and in-situ values of
entrapment and saturation index exponents
Profesor: Fis. Gustavo Mendoza Romero.
lvarez Snchez Ogilvie. Grupo: 04Semestre 2015-2
English A SUMMARY OF THE EFFECTS OF VARIOUS PORE GEOMETRIES AND
THEIR WETTABILITIES ON MEASURED AND IN-SITU VALUES OF ENTRAPMENT
AND SATURATION INDEX EXPONENTSJohn C. Rasmus Schlumberger of
CanadaCalgary, Alberta
ABSTRACTA number of papers written concerning a m, and n derived
from lab measurements concentrate on reporting the observed values
especially if they appear to be out of the normal ranges expected.
A clear understanding of the physical aspects of the rock
responsible for the observed measurements is not often presented
mathematically or visually. Some papers do attempt to quantify the
physical measurements observed. It is the purpose of this paper to
summarize these findings and also contribute to the quantification
of the effects of fractures and vugs on lab measured as well as
in-situ entrapment and index of saturation exponents. These
quantifications will allow a petrophysicist to better integrate lab
and log measurements with the geological realities of the
reservoir.INDUCTIONEntrapment exponent is widely accepted as a
measurement of the tortuosity of the pore geometry to current flow.
In reality, this only applies to the tortuosity of the
intergranular porosity. Water-filled fractures and vugs both
represent less tortuous current paths, yet fractures appear to
decrease entrapment exponent dramatically while vugs appear to
increase the entrapment exponent. This paradox is understood and
quantified by mathematically modelling the current paths through
these respective pore geometries.It is shown, irrespective of
saturation exponent, that the distribution of hydrocarbons in vug
porosity does not greatly influence the overall measured
resistivity, whereas the fluid distribution in a fracture system
has a pronounced effect on the measured resistivity. The in-situ
fluid distribution in these pore systems may differ from that found
in the lab. As a consequence, the entrapment exponent found in the
lab may or may not be applicable for in-situ measurements. Several
researchers have proved experimentally that saturation exponents
are a function of rock wettability. In addition, Wardlaw has shown
how a combination of large and small capillaries in conjunction
with wettability will affect the resulting fluid distribution in
the various capillaries. This paper will demonstrate mathematically
how these different capillary sizes in series and parallel with one
another will influence the desaturation of a core and the
subsequent calculation of saturation exponent. The effect of oil
properties, mineralogy, diagenesis and other factors governing
wettability will not be discussed. The presence of both fractures
and vugs will tend to perturb the lab-measured value of saturation
exponent. When the in-situ fluid distribution and wettability of
these nonintergranular pore systems is taken into account, the
saturation exponent to be used in log analysis may or may not be
the lab-derived value.DEFINITIONS AND SOME HISTORYThe familiar
Archi equation is generally written as
Where:
We will assume that the effect of clay conductance has been
properly corrected for since this paper will be concentrating on
vug and fracture pore geometries, which are normally associated
with carbones. Measured quantities for use in the above equation
include Rt, , and Rw. Variables either assumed or quantified by lab
measurements include a, m, and n. we will discuss how these
variables are quantified and applied in this paper.Resistivity
FactorWinn, in his paper The Fundamentals of Quantitative Analysis
of Electric Logs, derived the expression for resistivity factor
from a consideration of a unit cube of one meter. Appendis A
reproduces the derivation. His equation takes the form of
Where
The resistivity factor is shown to be a rock property that is a
function of the porosity () and the tortuosity ( of the pore
system. Notice that if the porosity is equal to 1.0 (no rock
present), es equal to and the measured resistivity Ro is simply
equal to Rw. The resistivity factor for a fracture or open volume
of water is then 1/However, the resistivity factor for a mixture of
rock and water has been under debate since Archie fist proposed the
relationship in his 1942 paper. Archie found that the tortuosity of
the pore system ( was proportional to the reciprocal of porosity
(1/) and proposed the relationship
Where m was introduced as a entrapment exponent. Notice that m
is simply a parameter that has allowed the tortuosity of the pore
system to be correlated with the porosity, a measurable quantity.
Unconsolidated, noncemented sands were found to have an m of 1.3,
whereas highly cemented rock had an m of 2.2. This relationship
also correctly predicts that resistivity factor = 1.0 when
=1.0.Winsaner et al. found from lab measurements in 1952 that many
sandstone were best approximated by the relationship:
Notice that this equation no longer satisfies the requirement
that resistivity factor=1.o when =1.0.Wyllie. In 1953 concluded
from measurements of variously cemented glass beads that the
equation should take the form:
Where C is defined as a constant controlled by the porosity of
the unconsolidated matrix prior to cementation.One must be certain
that clay effects are correctly accounted for when considering
resistivity factor relationships. Waxman and Thomas state that
nonclay-corrected resistivity factors yielded values of entrapment
exponent as low as 1.4, whereas clay-corrected resistivity factors
yielded exponents within the range of 1.8 to 2.1. Clavier, Coates,
and Dumanoir found the entrapment exponent to be independent of
shaliness once the bound layer volume was accounted for and to
range from 1.89 to 2.13 in the sample studied.The above equations
and many others have been proposed in the literature and
demonstrate the difficulty in correlating the tortuosity of a pore
system (Lw1/Lt) to only the porosity. It is not the purpose of this
paper to propose a new relationship for resistivity factor or
entrapment exponent, but to illustrate and quantify how vug and
fracture pore volumes influence measured resistivities. These
nonintergranular pore systems produce vastly different resistivity
to porosity relationships which cannot be adequately described by a
single parameters such as entrapment exponent.
Resistivity IndexThe derivation of resistivity index as given by
Winn and Amyx is shown in appendix B. The equation takes the
form:
Where
The quantity Lw1 was mentioned as a nonmeasurable quantity. It
follows that Lw2 is equally as elusive. One can see that when the
index of water saturation is 1.0, Lw2 equals Lw1 and. what happens
to Lw2 at partial index water saturations has been under as much
debate as the term Lw1.Early researchers concentrate on reporting
resistivities of rock as a function of the index of water
saturation. These included Wyckoff in 1936, Jakosky in 1937, Martin
in 1938, and Leverett in 1938. Archie in 1942 compiled and
correlated this experimental data and suggested a best fit was
approximated by the equation
Where n was introduced as a saturation exponent. Notice that the
quantity has been equated with 1/Sw. Archie determined that a
saturation exponent of 2.0 fit that data sufficiently well.Morse et
al. in 1947 recogni in 1947 recognized the effect of oil
wettability on the measured saturation exponent. Artificially
consolidated sands containing water and air(believed to be water
wet) exhibited a saturation exponent of .82, while the same
material with water and oil mixtures (believed to be oil wet) had
saturation exponents of 2.51.Dunlap et al. in 1949 found that the
saturation exponent varied from 1.11 to 2.24 in Cotton Valley and
Strawn sandstone cores. The Ottawa silica(Unconsolidated) cores had
saturation exponents ranging from 1.69 to 2.90. they also noted
that n depended upon the order in whicli the naptha and brine were
flowed through one particular core. There was also debate as to
whether flowing an emulsion through the core or desaturating the
core by capillary pressure was the best method of determining
n.Rust in 1952 found the saturation exponent from a Woodbine sand
outcrop to range from 2.31 to 2.40, the Planulina sand to be 1.78
and the Clear Fork sand to be 2.70. Also illustrated in his paper
was a noticeable decrease of the resistivity index (low n) at low
saturations which was attributed to a clay effect. Some of the work
concerning limestone was done in 1953 by Whiting, who found the
saturation exponent to vary from 1.52 to 2.56 depending on whether
the core was dynamically or statically desaturated. Keller in 1953
performed experiments on the Bradford third sand in which he
deliberately altered the wettability of the sand. A value of
saturation exponent of 1.50 was found in the water-wet case,
whereas a value of 11.7 was found in the oil-wet case.Morgan and
Pirson in 1969 used mixtures of water-wet ond oil-wet and oil-wet
glass beads to determine the effect of fractional wettability on
the saturation exponent. Their data showed a nearly linear
relationship of increasing saturation exponent to the fraction of
oil-wet beads in the mixture.These early researchers neglected the
effect of clay conductance on their meansurements. Waxman and
Thomas in 1974 reported how the saturation exponent went from 1.3
to 2.0 once clay effects were properly accounted for. It would be
impossible to tell how much of the earlier research on both
saturation and cementation exponents was perturbed by unaccounted
for clay effects.However, Rosepiler reported in 1981 that the
saturation exponent in the Cotton Valley sand was found to be 1.36
even after the Waxman-Smits clay corrections were applied.
Diederix in 1982 demostrated how glass beads with a rough
surface texture exhibited a much lower saturation exponent than
smooth surfaced glass beads. The rough surface was responsible for
retaining a relatively thick capillary waterlayer even at low water
saturations, there by providing a electrical current path that was
less sensitive to water saturation. This resulted in a saturation
exponent which decreased as the water saturation decreased. This
profile had been observed on core samples from the Rotliegend
sandstone which was coated by illite and kaolinite clays. The
measured resistivities had been corrected for clay effects
utilizing the Waxman-Smiths technique. The recomputed log data
using the lab-derived saturation exponents agreed quite well with
the capillary pressure data.Swanson in 1985 illustrated how
microporosity in cherts and clays can also cause the saturation
exponent to decrease as the water saturation decreases. A
Waxman-Smits clay correction was inadequate in predicting this
behavior.The above motioned variations found in saturation
exponents demonstrate the difficulty in equating the quantity with
. There has been little pubshed literature reporting measured
saturation exponents in vuggy of fractured rocks. This paper will
not add to the existing literature in this respect, but will show
mathematically how vug and fracture pore volumes influence measured
resistivities at partial saturation. The resulting resistivity to
saturation relationships are difficult to characterize by the
traditional parameter n, saturation exponent.MATHEMATICAL MODELING
OF VUG PORE GEOMETRIES. Sen et al. adapted the Maxwell-Garnett
mathematical relationship to model the dielectric properties of
mixtures of rock grains and water. Kenyon and Rasmus have used
these expressions to model low frequency conductivity and high
frequency dielectric measurements in oomoldic rocks. Oomoldic pore
geometries are a specific subset of rocks containing secondary
porosity. Secondary porosity will be referred to in this paper as
that porosity which is significantly larger than the
intercrystalline or intergranular porosity. The secondary porosity
in oomoldic rocks is spherical in shape and disseminate throughout
the rock in a somewhat homogeneous manner. In contrast, the
secondary porosity in vuggy rocks is generally irregular in shape
and heterogeneous ously disseminated. Nevertheless, the
Maxwell-Garnett relationship will be used to model the low
frequency (induction or laterolog ) current response in both shape
of secondary porosity. It has been found that the current responds
mainly to the intergranular porosity so the shape or placement of
the secondary porosity is not important. Kenyon and Rasmus give the
low frequency conductivity response of a mixture of water filled
spherical pores imbedded in a host material as
Where
In order to model the effect of hydrocarbon placement, the
equation must be enhanced in the following manner:1.-The
conductivity of the host material will be modelled by the Archie
relationship:
Where
2.- The conductivity of the spherical inclusion will be modelled
by the conductivity of an open volume
Where
Modeling the Effect of Vugs on Factor of ResistivityEqs. 8, 9
and 10 can be used to model the factor of resistivity in rocks
containing secondary porosity by setting the saturation of the vugs
equal to 1.0 and the saturation of the intergranular porosity to
1.0. Various combinations of values for and can be substituted in
the equations and the conductivity of the mixture () solved for.
This conductivity is converted to resistivity or factor of
resistivity and plotted against total () porosity.
One other possible saturation combination was also considered.
Ward law has shown how the imbibition of water into vuggy rocks is
controlled by the overall wettability of the intergranular and
vugular pores. If both pore system are oil wet , water will
preferentially reside in the larger capillaries (vugs). If both
pore system are water wet , oil will preferentially reside in the
larger capillaries (vugs). It is quite possible that within the
transition zone of a water-wet vuggy reservoir, the vugs will
contain oil while the intergranular porosity is fully saturated
with connate water. For this reason, the total conductivity en Eq.
8 was computed with the term in Eq. 10 set equal to zero while ,
and were allowed to vary. Although this condition would not occur
in the lab, it was studied mathematically to determine the
difference in the resistivity response for the two extreme cases of
vug saturation.
Modeling the Effect of Vugs on Resistivity IndexEqs. 8,9 and 10
were used to computer resistivity for various values of for both
cases of vug saturation mentioned above. First, was set equal to
zero for all values of intergranular saturation to simulate the
fluid distribution obtained when desaturating a water-wet vuggy
rock. The second case where was set equal to 1.0 for all values of
intergranular saturation was done to simulate the desaturation of
an oil-wet vuggy rock. The resistivities obtained from Eq. 8 at the
various values of intergranular saturations () for both cases of
vug saturation were then divided by the corresponding resistivity
ot full saturation (Ro). Only the value of Ro for the case when and
was retained for the calculation of resistivity index. This
resistivity index was then plotted as a function of total water
saturation. This made it possible to study the effect of various
desaturation schemes on resistivity index and saturation
exponent.
MATHEMATICAL MODELING OF FRACTURE PORE GEOMETRIES.Rasmus has
derived the resistivity relationship for a fracture volume imbedded
in a rock containing intergranular porosity. The equation is given
as:
Where
Eq. 11 was derived for the case where a fracture volume is
placed in parallel within a volume of rock containing intergranular
porosity. This equation does not account for the fact that a
measuring device`s current will not be perfectly focused in front
of a fracture and therefore not be exactly represented by the
theoretically derived equation. Nevertheless, it has shown to be an
accurate representation of the overall current response in
naturally fractured reservoirs.The tortuosity of a fracture or
fracture system, (Tf), is difficult to gauge. For simplicity it
will be assumed to be unity for this study. This implies that the
current path through a fracture is the same length as the total
length of the unit cube usedin the derivation of resistivity in
appendix A.The index of saturation exponent of the fracture (nf)
was taken equal to 1.0 since this is the exponent for an open
volume.
Modeling the Effect of Fractures on Resistivity Factor.In order
to study the effects of fracture volumes on resistivity factor, and
were set equal to 1.0 in Eq. 11. The resulting expression is
then:
The volumes of and mig were varied and resistivity factor
computed. This was then plotted as a function of total pore volume
() to study the effect of fracture volumes on formation factor and
cementation exponent.Modeling the Effect of Fractures on
Resistivity Index.Eq. 11 can also be used to compute Rt for given
volumes of and and varying values of and . These resistivities are
used in conjunction with Ro computed with Eq. 12 to calculate
resistivity index. This was then plotted versus the total water
saturation index to study the effects of different desaturation
schemes on the saturation exponent index.
PRESENTATION OF DATAFig. 1 illustrates the effect of vug
porosity on formation factor. The case for and =1.0 are both
presented. The spine shown on the graph represents the case when
and . The ribs emanating from the spine and going to the right
represent the addition of vug porosity to a particular volume of
intergranular porosity. This value of intergranular porosity is
represented by the intersection of the spine and rib.Fig. 2
illustrate the effect of fracture porosity on resistivity factor.
Here the saturation of the fracture porosity was set equal to 1.0
for all cases. The spine represents the case where =0.0 and =2.0.
the ribs emanating from the spine and pointing down represent the
addition of fracture porosity to a given volume of intergranular
porosity.Fig. 3 illustrates the effect of vug porosity on the
resistivity index for an intergranular porosity of 0.10. The spine
in this figure represents the case when only intergranular porosity
is present and its saturation exponent is 2.0. The ribs to the left
of the spine represent the case where the rock is taken to be water
wet . The vug in this case will then be full of oil , regardless of
the intergranular saturation. The points along the ribs represent
various vug porosities. The ribs to the right of the spine
represent the case where the rock is taken to be oil wet. The vugs
in this case are taken to be water filled regardless of the
intergranular saturation.Fig. 4a illustrates the effect on
resistivity index of 0.01 volumes of fracture porosity contained
within an intergranular pore volume of 0.05. The spine represents
the case when only the intergranular porosity is present and its
saturation exponent is 2.0. The ribs emanating from the spine and
pointing down represent the effect that the fracture volume at
various saturations has on the measured resistivity index. Fig. 4b
is the same as Fig. 4a except it represents the case of 0.01 pore
volumes of fracture porosity contained within an intergranular pore
volume of 0.20.
INTERPRETATION OF DATA CONCERNING RESISTIVITY FACTOR AND
ENTRAPMENT EXPONENT.It is important to remember that Eq. 8 has been
used to describe the total resistivity in terms of the individual
contributions of intergranular and vug pore volumes. The same
should be stated regarding Eq. 11 where the total resistivity was
computed for the various combinations of intergranular and fracture
pore volumes are not differentiated. The task becomes one of trying
to explain the total resistivity response in terms of the total
(inter granular + vug or intergranular + fracture) porosity. The
following discussions concerning the interpretation of the data
will examine the difficulties in trying to do this, and where
applicable, describe techniques to circumvent this
problem.Interpretation of the Effect of Vugs on Resistivity Factor
In order to illustrate the resistivity response to vug porosity
let`s consider the case where the intergranular porosity is 0.10
and the vug porosity is 0.20. this would be an abnormal percentage
of secondary porosity in a vuggy carbonate but within the upper
limit of secondary porosity seen in oomildic carbonates.It would be
convenient to express the resistivity in terms of bulk volume water
(BVW) or resistivity porosity in discussing this graph. Archie`s
equation rearranged and solved for bulk volume water is given
as:
When n and m are set equal to 2.0, the bulk volume water is
proportional to the square root of true resistivity. The spine on
Fig. 1 can now be thought of as the solution of Eq. 13 for water
porosity for a given value of true resistivity. (Well set Rw equal
to 1.0 for simplicity).If the resistivity porosity equal to the
total porosity. The water saturation index is 100% . When the
resistivity porosity(BVW) is less than total porosity we say the
zone has a volume of hydrocarbons in place which is equal to
-BVW.Let`s apply the resistivity porosity concept to Fig. 1. When
only intergranular porosity is present in the amount of 0.10, the
resistivity would read 100 ohms. Adding 0.20 pore volumes of
water-bearing secondary porosity causes the porosity measurement to
read 0.30 and the resistivity to decrease to a value of 58 ohms. A
value of 58 ohms on the resistivity log is shown on the spine to
correspond to a resistivity porosity of 0.13 pore volumes (assuming
m=n=2.0) if we had some measurement of the intergranular porosity
(such as sonic) we would say the intergranular saturation index was
0.13/0.10, or 130.0%. If we had only total porosity to work with we
would say the zone has a saturation of 43.3%(0.13/0.30). In
reality, both total and intergranular saturations are 100.00%. In
other words the resistivity log has responded to 0.03 pore volumes
of vug porosity out of 0.20 or only (0.03/0.20) 15.0% of the vug
porosity even though it was filled with water!Let`s now consider
the case where the vug is taken to be full of oil and electrically
isolated. The resistivity would now read 137 ohms. This corresponds
to a resistivity porosity of 0.085. this would give us an
intergranular saturation of (0.085/0.10) 85.0%. If we had only
total porosity to work with we would have a saturation of
(0.085/0.30) 28.3%. For comparison, remember the actual
intergranular saturation is 100% and the total saturation is
33.3%.In summary, the resistivity measurement gives us a range from
85% to 130% for the intergranular saturation when it was actually
100%, a value of total saturation of 43.3% when it was actually
100%, and a value of total saturation of 28.3% when it was actually
33.3%.This leads us to the conclusion that the resistivity
measurement is responding mainly to the intergranular pore water
volume regardless of the fluid saturation of the vugs. It follows
that the entrapment exponent should be left at 2.0 (or the
intergranular porosity entrapment exponent), and the resulting BVW
compared to intergranular porosity to determine its saturation.
What is in the vugs could possibly be deduced from knowledge of the
saturation of the intergranular porosity and capillary pressure
considerations.One could comment `l thought using a higher
entrapment exponent in vuggy rocks accounted for this resistivity
response! Unfortunately, it`s not quite that easy. If you have an
intergranular porosity measurement, we have seen that a entrapment
exponent of 2.0 works fairly well.Even if you could somehow measure
the intergranular entrapment exponent, notice that it would change
depending if the vug were filled with water (with m=2.0 we computed
130.0% Swig so m would drop to 1.76 to make Swig=100%) or filled
with hydrocarbons (with m=2.0 we computed Swig=85.0% so m would now
be 2.14 to make Swig=100%).A greater variance in m occurs if you
work with total porosity. With water in the vugs an m of 3.37 would
be needed to compute 100% Swt. With oil in the vugs an m of 2.14
would be needed to match the actual total saturation of 33.3%.
Remember that we are working with fully saturated intergranular
porosities so saturation exponents do nor come into play yet.In the
lab, a value of m is determined by drawing a best fit line(or spine
in our case) through the range of porosities measured. A Lab
measurement of porosity will include the secondary porosity in
addition to the intergranular porosity. The volume of secondary
porosity will not remain either a constant volume or a constant
percentage throughout the range of porosities encountered in a
reservoir. This makes it difficult to draw a best fit line through
the lab date. The result is entrapment exponent that do not reflect
the true tortuosity of the pore system and a values other than 1.0.
Essentially m could be, and probably is, different for each sample
plotted. Another complicating factor is the fact that in the lab
the vugs will always be filled with water, whereas in the reservoir
this will not always be the case. The above calculations
illustrated the errors in working with total porosity and m to get
total water saturations. What is needed is a lab measurement of
intergranular porosity so that an intergranular entrapment exponent
can be used with intergranular porosity and resistivity in Archie`s
equation in order to get a better bulk volume of intergranular
water.Interpretation Of the Effect of Fractures on Resistivity
Factor.Fig. 2 show us that the major effect of fracture porosity is
on the measured resistivity. To illustrate this let`s look at the
effect of fractures in a rock with an intergranular porosity of
0.05. if there was only 0.01 pore volume of fracture porosity
present the total porosity in then 0.06 and the resistivity would
read 80 ohms. This gives us a calculated water saturation of
183.3%. One would have to use a entrapment of 1.56 in order to
compute Swt=100.0% .Notice that the addition of fractures at all
porosities is to reduce the measured resistivity and thus the
apparent entrapment exponent. The effect is more pronounced at
lower intergranular porosities. In order to compute the apparent
entrapment exponent n a fractured reservoir it would be necessary
to measure the fracture cvolume directly. This is very difficult
with radioactive type measurements because of their statistical
nature in the low porosities normally encountered in naturally
fractured reservoirs. It is possible to estimate indirectly the
fracture intensity or volume from combinations of various well
logs. Knowledge of the approximate fracture volume allows one to
use Eq. 12 to compute a entrapment exponent. Another possible way
would be to use the in-situ measured resistivities and total
porosities in Eq. 12 to calculate the fracture volume. This would
only be appropriate if both the intergranular and fracture pore
volumes were water bearing. More work needs to be done in order to
accurately measure fracture volumes in situ to better predict their
influence on resistivities.This effect of natural fractures in the
lab is generally not seen on resistivity factor plots for the
reason that a core plug that contains a fracture is generally not
selected for the measurements. However, micro fractures induced by
stress relief in samples can severely affect a lab measurement of
resistivity factor. It is therefore wise to perform the resistivity
factor measurements under simulated overburden pressures.Fig. 2
also demonstrates that fractures present in reservoirs of
significant porosity (greater then 0.20) will not significantly
influence the measured resistivity.
INTERPRETATION OF DATA CONCERNINGRESISTIVITY INDEX AND
SATURATION EXPONENT.Interpretation of the Effect of Vugs on
Resistivity Index
Condition When Wardlaw, using etched glass micromodels has show
a rock that is water wet will contain nearly all the irreducible
water in the smaller capillaries (intergranular porosity), while
the larger capillaries (vugs) will contain nearly 100% oil. When a
water-wet vuggy rock is desaturated, the oil will then
preferentially invade the larger capillaries (vugs) first giving us
the condition of Condition when .In fig. 3 the rib for is curved,
which indicates that the resistivity is responding to the placement
of the oil in the vugs. What is significant is the fact the
resulting total saturations are lower than one would expect for the
give resistivity index. The net result is that the saturation
exponent is lower than 2.2. Fig. 1 showed us that cementation
exponent m is great than 2.0 in vuggy rocks. It is therefore not
valid in this case to set m=n in vuggy rocks that are water wet
when working in the total porosity system.In order to illustrate
the magnitude of the error involved and the usefulness of the
resistivity porosity concept, lets investigate the hypothetical
rock pore volumes discussed in the section interpretation of the
effect of Vugs on Formation Factor whit the vugs filled with water,
the cementation exponent to be uses whit total porosity was found
to be 3.37. This will be the m we will utilize here since it
defines the resistivity response at full saturation. Now, consider
the case where the vugs desaturate fully before the intergranular
pores desaturate to a value of 50.0% saturation. Fig. 3 show us
that the saturation exponent for this case () is 1.23. Taking fig.
3 gives as 9.47 so putting these numbers into an Archie
equation
One gets an of 0.165, which is the actual total water
saturation. Notice that m is not equal to n in this case.For
comparison, lets assume that n was taken to equal m (3.37). The
resulting Archie expression is:
Then equals 51.3%! One can see that setting n=m in the total
porosity system cause erroneous saturation to be computed.Lets
determine the effects of setting n=m=2.0 as if only intergranular
porosity were present.
Would then be 0.14. The resulting BV W ( ) of 0.042 is close to
the actual intergranular pore water volume of 0.05. Why is the
saturation erroneous but the BVW approximately correct? Rearrange
Eq. 16 in the following manner:
The term is our resistivity porosity mentioned early. This is
also the approximate intergranular bulk volume water since the
resistivity predominantly responds to this pore volume.Notice that
the computed is a total water saturation and the porosity is a
total porosity. Why is equal to , the intergranular bulk volume
water? Because is too low by an amount that is approximately equal
to the value that is too high by, in other words .
Condition when In an oil-wet rock, the vugs will desature last
from a capillary pressure stand point. This causes the total water
saturation to be higher than that expected for the given
resistivity response and a high n results. We should now recompute
our hypothetical rock saturation using the oil-wet n values.
Cementation exponent is again 3.37 since the rock is initially
fully saturated. Fig 3 show us that the saturation exponent for
this case (=0.10) is 7.53. Taking is 58 ohms as before. Fig 3 gives
as 3.95 so ohms. As a check on the math, lets put these numbers in
an Archie calculation.
Which gives . This checks as our total water saturation.Lets
assume that n was taken equal to m (3.37).
now equals 0.665, which is quite far from the actual saturation
of 0.833. Again, setting n=m has shown to be an invalid
assumption.Lets now assume that n=m=2.0. The resulting
calculation
Gives a saturation of 0.22. Notice again that the total
saturation is erroneous but the BVW () is 0.066, which is close to
the actual BVW of 0.5. The same reasoning applies to this situation
as we found for the case when was 0.0. The resistivity measurement
is shown again to be responding mainly to the intergranular pore
water volume.
SummaryThe previous calculations demonstrate the difficulty in
applying lab-derived values of n and m to use with total porosity.
It was shown that letting m=n=2.0 will fairly accurately predict
the intergranular pore water volume, such as sonic porosity, to
obtain an intergranular saturation would be the most beneficial for
determining the production potential of a reservoir since fluids in
the vug must first pass through the intergranular porosity to reach
the wellbore. In most cases the relative permeability of the
intergranular pores control the type of fluid production.An actual
measurement of the in situ fluid content in the vugs could only be
determined with sigma, carbon-oxygen, or dielectric constant
measurements. The measurements are volumetric in nature and not as
sensitive as resistivity to the spatial distribution of the fluid.
The in-situ fluid saturations of the vugs are of course important
from a reserves standpoint.Interpretation of the effect of
fractures on resistivity indexThe effect of fracture at low
Porosities
Condition when Fig. 4a illustrates the influence of a fracture
volume of 0.01 when the intergranular porosity is 0.05. The m for
this case was given early as 1.56 from fig. 2. Lets look at the
situation where the intergranular saturation () has desaturated to
a value of 0.20 while the fracture saturation () remains equal to
1.0. The total water saturation would then be 0.33. when and , the
n value from fig. 4a is 0.19. From fig. 2, resistivity index is
1.24 from fig. 4a so is ohms. As a check, the Archie calculation
is
From this we calculate , which is the actual .Lets now consider
the case where n is taken equal to m.
is now 0.87. Again setting m equal to n creates errors.The
resistivity porosity concept (m=n=2.0) does not work when the
fracture contain water. The Archie calculation.
Gives a saturation of 166% and BVW of 0.12. This is to be
compared with the actual total porosity of 0.06 and the
intergranular porosity of 0.5. It is interesting to note that the
resistivity porosity was 0.11 when the rock was fully saturated.
Adding 0.04 pore volumes of hydrocarbons into the intergranular
porosity has only dropped the BVW by 0.01 pore volume and is
insensitive to the intergranular pore water.In the lab, the
measurement of resistivity index cannot be performed under
simulated overburden pressure. Microfractures induced by stress
relief could very likely remain water bearing as the core is
desaturated. This condition would cause abnormally low values of
saturation exponent to be computed from the resistivity index
data.
Condition when When the intergranular porosity is low some
naturally fractured reservoirs contain only water in the
intergranular porosity, with the hydrocarbons residing only in the
fracture porosity. If this situation occurred near enough to the
wellbore so that a logging measurement saw only hydrocarbons in the
fracture, then the resistivity would be 404 ohms from eq. 11 for
the above case of and the total saturation is then (0.05/0.06) 0.83
and the n value from fig. 4a. As a check, the Archie calculation
is
Which gives the correct saturation of 0.83.Lets consider the
case where n is taken equal to m.
Which give an optimistic of 0.36.When the resistivity concept
(m=n=2.0) is applied the resulting equation would
Which gives the correct of 0.83. Should be expected since the
resistivity sees the hydrocarbonbearing fracture as matrix the
resulting BVW (0.83*0.06) of 0.05 is the intergranular BVW. With
the total porosity at 0.06 it would be hard distinguish the bulk
volume of hydrocarbons (0.01). In normal practices, significant
flushing of the hydrocarbons occurs away from the wellbore.If one
could obtain a resistivity measurement that responded deeper than
any mud filtrate invasion into the fracture system and had a
fracture porosity measurement the resistivity could be used to
quantitatively estimate the fluid content of the fracture.
The effect of fracture at high porosities
Fig. 2 showed us that at porosities greater the about 0.20, the
formation factor is only slightly affected by fracture volumes. The
cementation exponent therefore remains approximately 2.0 fig. 4b
illustrates that n can drop below 1.0 at low intergranular
saturation a high fracture saturations in the lab. Under in-situ
conditions, it is quite probable that the fractures present in this
type of reservoir contain hydrocarbons in the virgin zone and mud
filtrate in the near wellbore zone. If m=n=2.0 was assumed for
calculation of the virgin zone saturation using a deep reading
resistivity, an accurate total water saturation would result. A
shallow reading device may be influenced by the filtrate in the
fracture a lead to abnormally high flushed zone water saturations.
Summary We have seen that the measured resistivity responds
entirely to the fracture pore water volume. In the lab,
microfracture induced by stress relief can occur. The desaturation
analysis cannot be done at simulated overburden pressures to
eliminate this problem. A microfracture, whether induce or real,
will probably not desaturate as easily as the intergranular
porosity and its saturation will therefore remain close to 100.0%.
Thus the effect on the measured resistivity index will be to reduce
it to a valued lower than the intergranular resistivity index. The
n value is subsequently low and seen to vary with the fracture
saturation as well as with the intergranular saturation. Notice
that the apparent n values can reach values lower than 1.0, the
saturation exponent for an open volume.It is therefore not possible
to evaluate fracture and intergranular saturations individually
given only the total resistivity and total porosity. Eq 11 show us
that the fracture pore volume must be known as well as the
intergranular saturation in order to determine the fracture
saturations.Conclusion1. When secondary porosity is present within
intergranular porosity, the resistivity is relatively insensitive
to the type of fluid contained in the secondary porosity. The
resistivity responds primarily to the intergranular pore water
volume. The concept of resistivity porosity can then be used to
compute the bulk volume water of the intergranular porosity. This,
in conjunction with a measurement of intergranular porosity, allows
one to compute accurate intergranular saturations to predict the
type of fluid production expected. Lab- derived cementation
exponents will always be greater than 2.0 when working with total
porosity (secondary plus intergranular) in order to accommodate the
resistivity response.2. The presence of secondary porosity
overexaggerates the effects of rock wettability. The lab-derived
saturation exponent will be a function of the saturation of the
secondary porosity. The saturation exponent will not be equal to
the cementation exponent when working in the total porosity
system.3. When fracture porosity is present within intergranular
porosity, the measured resistivity is almost totally dominated by
the volume of fracture porosity present at the lower intergranular
porosities. Fracture porosity contained within the higher
intergranular porosities has only a slight effect on the measured
resistivity. Cementation exponents will always be less than 2.0 in
order to accommodate the resistivity response. The saturation
exponent is a function of the saturation of the fracture porosity.
This is true regardless of the volume of intergranular porosity
present. The saturation exponent will not be equal to the
cementation exponent when fractures are present.AcknowledgementsI
would like to thank Don Fergus for sharing his library of reference
papers on the subjects of cementation and saturation exponents, as
well as his interest and comments on these subjects. I would also
like to thank Neil Pashak for his programming efforts in computing
the equations used in the text for all the combinations of variable
present.BiographyJohn C. Rasmus joined Schlumberger after receiving
a B.S. degree in mechanical engineering from Iowa State University
in 1975. He worked as a field engineer in Utah, North Dakota, and
Wyoming until 1979. At that time he was assigned the position of
recruiting engineer for the Western Unit until 1981. From 1981
until 1984 he was applications development engineer, first in the
rocky Mountain Division, the in the Kansas Division. From 1984
until present he has been assigned as a product development manager
for Schlumberger of Canada.