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.