DUAL-WATER ARCHIE, and the IMPORTANCE of WATER …archieparameters.com/Resources/Archie's...

DUAL-WATER ARCHIE, and the IMPORTANCE of WATER GEOMETRY. A MODEL and a DISCUSSION.

Robert C. Ransom

ABSTRACT

A dual-water Archie model for the interpretation of traditionalresistivity well logs is developed. From that development, aphilosophy for Archie's parameters emerges that is quite differentfrom some literature in our industry. In this paper, electrical law,mathematical proofs and derivations are employed in order to promotea better understanding and a better use of Archie’s relationships. The geometry of water occupying void space in rock is shown to be themost important factor in controlling efficiency of the flow ofelectrical-survey current through interstitial-water paths in rock,thus giving value to Archie’s parameters.

The classic Archie saturation equation (1942) emerges from the modelpresented herein, and in doing so is shown to be a dual-water, dual-porosity relationship, and is extended to address shaly sands andadditional levels of heterogeneity and practical reservoircomplexity. Additionally, the development illustrates howheterogeneities such as clay shale and semi-conductive mineralsinfluence resistivity relationships. The model further illustratesboth the resistivity behavior in the presence of hydrocarbon and theproblems of interpretation in partially oil-wet and oil-wet porousenvironments. The reality that many oil- and gas- formations arecomplex in terms of mineralogy, lithology and saturationdistributions makes a better understanding of the analytical processimperative.

WHAT ARE ARCHIE’S BASIC RELATIONSHIPS

Most water saturation equations used in resistivity well-loginterpretation are based in some way on Archie’s relationships.Resolve those equations to clean rocks, and Archie’s basic water-saturation equation emerges. Archie’s basic relationships (1942) mostoften are modified by location-specific empirical coefficients andexponents that are not compatible with basic electricity or rockproperties. Whether or not any of these modifications providesolutions to local problems is of no interest in these discussions.The purpose of this paper is to dispel misunderstandings, preventmisuse, and promote better interpretations through betterunderstanding. Archie’s basic relationships are:

w 0 t measured S = R /R n

F = 1.0/(Ø)m

0 w R = FR

of 34

where, at the present time, the formation factor equation in many loganalysis applications has been modified to

t t F = a/(Ø ) m

where the a coefficient most often is given a value lower than 1.0.

Each parameter in these equations will be developed graphically andmathematically, and through this development a conception of whateach represents will emerge.

THE GRAPHICAL MODEL

Archie’s classic relationships usually are considered to be clean-sand relationships. In shaly sands where clay shales produceadditional electrical conductivity, Archie’s relationships are saidnot to apply. Relative to Archie’s original concept and its popularuse, this pronouncement is correct. In the concept born of thismodel, an overall version of Archie’s relationships emerges thatshows that Archie’s concept is a dual-porosity, dual-water conceptthat also applies to shaly sands and other heterogeneous rocks thatexhibit uniformity within the depth of investigation of the loggingtools. This model is used to derive and define each of the terms inArchie’s relationships, and the parameters derived and defined applyto any other methodology that makes use of Archie’s parameters.

Figure 1 is an illustration of the basic resistivity model used inthis concept. Figure 1 is not intended to be a working graphicalprocedure, it is informative and explanatory. This figure illustrates

wt thow water (S Ø ) is related to a unit volume of rock with resistivityt(R ). The model in Figure 1 is illustrative in nature and is not drawn

to scale. The line drawing in the X-axis has been expanded so thatdetail can be observed and discussed. The figure is designedprimarily to illustrate the electrical behavior of a volume offormation water as its environment changes with variations ofinsulating rock and insulating fluid.

weIn Figure 1, the origin of the diagram is represented by R (andt tF = 1.0) when Ø is 1.0 or 100%. Where it commonly has been believed

there are only two slopes in Archie’s concept, the model demonstratesthat there actually are three slopes, each representing exponents in

1 2 1Archie's concept. They are m , m , and n. The first is m thattpertains to the single parameter Ø . This is the familiar porosity

exponent m known in industry. Sometimes for the sake of clarity the1 2term m will be used instead of m. The second exponent is m which

t wtpertains to two parameters, Ø and S , and is the exponent for thewt t wtproduct S Ø . The third is the exponent n that pertains only to S ,

and is the saturation exponent commonly known in industry.

From the diagram, the total fractional volume of water in the rock isrepresented by the projection on the X-axis under the two slopes,

werepresenting m and n, drawn from 100% water at R to the intersectionof the line representing slope n extrapolated to intercept

t wtresistivity level R . The fraction of water in total pore volume, S ,

of 34

wt tand the fraction of water in the total rock volume, S Ø , are10 t 10 wtdepicted on the X-axis by log Ø and log S . The fraction of total

rock volume that is water provides the electrical conductivity to thetrock resulting in resistivity R . As oil or gas displaces water, the

wtwater saturation, S , decreases to the right as the saturation of oilwtor gas, (1.0 - S ), increases.

It is illustrated in the model that there can be conditions relatedto the presence of hydrocarbon, oil in particular, that cause in siturock resistivities to increase to extraordinarily high values. Inoil-wet rocks the values of n will increase greatly (Keller, 1953;Sweeney and Jennings, 1960). The presence of oil in oil-wet rocksproduces a very exaggerated interference to the flow of electrical-

tsurvey current. Resistivity R then will be increased correspondinglywith the wettability to oil and resulting interference. The figurealso shows that, under these conditions, when the commonly useddefault values of n = m are employed, the line representing the slopeof exponent n will be extended far to the right to intersect the

tlevel of the measured or derived value of R at an unreasonable andwtunacceptably low value of S .

This illustrates the resistivity interpretation problem in oil-wetrocks or other rocks where resistivity is exaggerated by theproperties of oil. Here, when the default value of n is equal to m,

tthe usual n intercepts R far to the right in the model in Figure 1,suggesting that water saturation is very low. In such cases, thederived saturations from the basic Archie equation, or from any morecomprehensive equation, cannot be correct and cannot be used in anyfield or reservoir description when the common default value for n isemployed. This illustrates a problematic but common occurrence whenusing the usual resistivity interpretation methods by both privateand commercial organizations. The cause of this problem and othersmust be recognized before they can be addressed.

An exploration of the graphic model will promote a betterunderstanding of resistivity behavior in rocks. This exploration willbe accompanied by parallel development of the terms and parameters inthe basic Archie relationships. What each term or parameterrepresents is the key to understanding where to begin to solveinterpretation problems.

t wt tWHAT IS MEANT BY THE PLOT OF R VERSUS S Ø

t wt tIt must be recognized that Ø and S Ø have no units, but aret wt tfractions and in themselves are not volumes. Only when Ø or S Ø are

multipliers of cross-sectional area (m )or total volume (m ) do these2 3

fractions become dimensional. In this paper there will beapplications where they will be fractional multipliers of both areaand volume. In Figure 1, they are fractions of total volume.

weIn Figure 1, again, the origin of the diagram is represented by R atwe100% porosity, but R is one of the products of Figure 2. Figure 2 is

wa detailed view of an interior part of Figure 1 showing how R in thewe presence of dual water becomes R to become the origin of the

wediagram. The value R , as an equivalent water resistivity, is

of 34

wt e tdetermined from Eq.(1b) when S is 1.0. Within Figure 2, Ø and Øwexist simultaneously, and when they have different values, R becomes

we 0 0 correctedR , and R becomes R .

tIn Figure 1, R is seen plotted on a log-log plot versus the volumewt t tS Ø , of which Ø is a part. In resistivity well-log interpretation,

the m or n slope in every case represents only the slope between twopoints: the value of the resistivity of the equivalent water, and thevalue of the resistivity of the total rock volume filled with thesame water in whatever fraction and physical distribution. This

t wt 1fractional volume can be Ø when S = 1.0 and slope m = m , or it canwt t wt 2be S Ø when S # 1.0 and slope m = m . Slope n represents the case

where the two resistivity end points for the slope are theresistivity of a rock completely filled with the equivalent water,and the resistivity of the same rock and the same water afterhydrocarbon has displaced some of the water. There is noextrapolation and no interpolation involved in the m or n evaluationwhen the two end points are known. The resulting slope, be it m or n,is a measure of the difficulty and interference electrical-surveycurrent experiences as it is forced to flow through the water in therock. In Figure 1, the line representing the slope n rotatesthroughout the range of arc ä shown in the diagram as either or both

wtn and S vary. The steeper the slope for either m or n, the moreinefficient will be the water path in the rock for conductingelectrical-survey current, the greater will be the values of m and n,

tand the greater will be the value of R . Each value applies only tothe individual sample of interest, whether in situ or in thelaboratory.

The slopes representing values of m or n are trigonometric tangentialratios of the side opposite (on Y-axis) over the side adjacent (on X-axis) of any right triangle in the figure. Exponents m and n

trepresent rates of change in resistivity R relative to water volume t wt tØ or S Ø , and reflect the efficiency or inefficiency of that water

volume to conduct electrical-survey current. In the followingexamples, it is shown how each m and n is derived as trigonometrictangents and calculated from their respective right triangles, andwhat can be derived from each.

2In the triangle AEG where m is the tangent

2 10 t 10 wt t -m = log F /(log (S Ø ))

2 10 wt t 10 t -m (log (S Ø )) = log F

t wt t F = 1.0/(S Ø ) . . . (3b) m2

tThe formation factor, F , as it is derived here from Figure 1, applieswtto all values of S # 1.0 and is represented on the resistivity axis

as the difference between the logarithms of resistivity of theweequivalent water in the rock (R ) and the resistivity of the rock

tcontaining that water (R ).

Also, it can be seen from the triangle CDG where n is the tangentthat

of 34

10 t 10 0 corrected 10 wt -n = (log R - log R )/log S

10 wt 10 t 10 0 corrected -n(log S ) = (log R - log R )

wt 0 corrected t (S ) = R /R . . . (4b)n

Archie’s saturation relationship is a straightforward derivation fromthe model, but is improved for use in dual-water, dual-porosity

w we 0 0 correctedmethodology by the correction of R to R and R to R . Bothcorrections are from Equations (1a) and (2a), respectively, butobserved in graphic form in Figure 2.

See the APPENDIX (B) for a more detailed explanation of Figures 1 andt wt2 and the derivations of m and n, and F and S .

In addition, in an examination of the line slopes in Figure 1, it canwt tbe seen that (S Ø ) has the same function as and is equivalent tom2

wt t(S ) (Ø ) . It is important to remember this in the developments thatn m1

follow. For proof of this equality, see the APPENDIX (B)(3).

The line representing either slope, m or n, does not represent a locusof points as either porosity or saturation changes, and there shouldbe no reference to boundary or unlikely regions or conditions becausethese have nothing to do with the conductive efficiency of thespecific rock sample of interest, whether in situ or in thelaboratory. If the locus of points were shown, it might or might notrepresent a constant slope. The locus of data points for both m and nfor several rock samples can be curvilinear and can represent a numberof different slopes as the porosities, saturations, and proportions ofminerals vary. The locus of points for either m or n depends entirelyon the geometry of the pores and pore paths and the saturationdistributions imposed on the conductive water volume by thenonconductive environment, whether that environment is electricallyinert rock or hydrocarbon.

PARALLEL RESISTIVITY EQUATIONS USED IN RESISTIVITY INTERPRETATIONS

The basic premise is that the host rock is an insulator. Equationswill be demonstrated for the case of shaly sand with a uniformdistribution of constituents and pore space. Shale is a term used torefer to fine grained, fissile, sedimentary rock. The term shale is adescriptive property of rock, not a mineral, therefore, shale may bereferred to as "clay shale" when the predominant mineral constituentis clay. The shale in this presentation is not a source for oil orkerogen, but is shale or mudstone that has been compacted byoverburden to the degree that any form of porosity has becomenoneffective, and migrating oil cannot or has not penetrated the voidspace.

Connate water is water entrapped within the pores or spaces betweenthe grains or particles of rock minerals, muds, and clays at the timeof their deposition. The water is derived from sea water, meteoricwater, or ground surface water. Other investigators have shown that asclay shales and mudstones are compacted, a fresh water component of

of 34

the original connate water is expelled and an ion-concentratedcomponent remains in the voids. Therefore, water saturation in the

w nevoids remains 100% and S is always 1.0 in Ø .

The relationship in Eq.(1a) below, presented in Ransom (1977), wasdeveloped to show that conductivity in water-filled voids in the hostrock and the conductivity in clay shale can be represented by parallelconductivity relationships commonly used in basic physics. Therelationship applies to heterogeneous rocks with a uniformdistribution of minerals and porosity. In resistivity and conductivityrelationships the dimensional units for a reservoir bed (m/m ) must be2

made to be unity so that bed dimensions will not be a factor.

For shaly sand it was shown that

we e t w ne t wb 1/R = (Ø /Ø )1/R + (Ø /Ø )1/R (1a)

t e newhere Ø = Ø + Ø .

For 100% water-bearing rock the conductivity relationship in Eq.(1a)w ne weyields 1/R (in clean rock) where Ø is 0.0, and yields 1/R (in shaly

nesands) when conductive clay is present and Ø is greater than 0.0.

It should be noted that in this equation there are no limitations onwthe resistivity values of formation water R and bound water in shale,

wb wbR . However, nature does set limits. The bound water resistivity R inthe mudstone and clay shale is related to the original connate water.Waters entrapped at the time of deposition do vary, but usually do notvary greatly from shale to shale. Sea water has an average salinity ofabout 35,000 ppm, other surface waters probably are significantlylower. However, as indicated above, the ion concentration in clay-bound water can increase as fresh water is expelled from the clayswith increasing depth and compaction.

wUnlike connate water, interstitial-water resistivity R in thereservoir can vary over a wide range. Interstitial water might haveundergone many changes through the dissolution and/or precipitation ofminerals throughout geologic history. It can vary from supersaturatedsalt solutions to very fresh potable waters. Waters in the Salinadolomite (Silurian) in Michigan have specific gravities of 1.458 and areported salinity of 642,798 ppm. In other formations, the waters vary

wto as little as a few hundred parts per million. Most often R will bewb we w wb wgreater than R ; and R , a mixture of R and R , will be lower than R .

w wbBut, it is quite common for R to have a lower value than R , in whichwe wcase R will have some value greater than R .

In water-filled dirty sands, Eq.(1a) applies. In clean sands, Eq. (1a)becomes

we e t w w 1/R = (Ø /Ø )1/R = 1/R

0and the corrected R in either event is

0 corrected t we R = F R (2a)

where, from Figure 1,

of 34

t wt t t F = 1.0/(S Ø ) = 1.0/(Ø ) (3a)m2 m1

wt 2 1The value of S is 1.0, therefore, the exponent m becomes m describedabove and in Figure 1.

However, in the presence of oil, water saturations are lower than 1.0and the conductive water-filled volumes are reduced by thedisplacement volume of oil, and Eq.(1a) becomes

we we e wt t w ne wt t wb 1/R = ((S Ø )/(S Ø ))1/R + (Ø /(S Ø ))1/R (1b)

O corrected t weand either R = F R (2a)

t calculated t weor R = F R (2b)

depending on the water saturation values in Eq.(1b),

t wt twhere this time F = 1.0/(S Ø ) (3b)m2

2and m is a form of m .

Equation (1b) converts the basic Archie method from a single-porosity,single-water method to the dual-porosity, dual-water methodology where

w wb wthe two waters have resistivities of R and pseudo R , and R formerlywefrom the single-porosity, single-water method becomes R , as seen in

wb wbEq.(1b). The R has been referred to as a pseudo R because some ofthe conductivity attributed to the clay shale might not be in the formof water, but the usual resistivity-measuring logging tools do notknow the difference.

In dual-porosity, dual-water methodology, the numerator in Eq.(3b)always must be 1.0 because the compensation for all electrically

wconductive influences, along with formation water R , should bewerelegated to terms in the R equation.

wt wt tEquation (3b) applies to all cases where S # 1.0, and the term (S Ø )tin the denominator of the F relationship must be compatible with the

wesame term in the denominator in the calculated R relationship ineither Eq.(1a) or (1b) in whatever application either is used. Thefractional volume of conductive water represented in the formationfactor must always be the same fractional volume of water that

we wt texhibits the equivalent resistivity R . Also, note that both S and Øhave the same exponent m. In this case, the graphical model shows this

2 2m to be m because m is the combination exponent equivalent to the1individual exponents m and n. It is illustrated in the trigonometry of

wt tFigure 1 that the same result would be realized if S and Ø were to1use their individual exponents n and m , respectively, because

tcalculated water saturation is related to the resistivity level of Rwt t wt tand this remains unchanged. Again, the proof that (S Ø ) = S Ø ism2 n m1

shown in APPENDIX (B)(3).

It can be seen that Eq.(1b) is a representation of Eq.(1a) whenthydrocarbons are present, and can be used in the calculation of R .

wt t 0 0 correctedWhen S is 1.0, the calculated R becomes a corrected R , or R .These relationships will be revisited later.

of 34

In Figure 2, adapted from Ransom (1974), it is illustrated that theresistivity value of water-filled rock at any specific depth can bethe result of a shift by the influence of a variable a, and the causeof that influence must be explained.

In these equations the accountability or compensation for theinfluence of clay shale in shaly sands resides in the proportionality

w wbterms involving the conductivity counterparts of R and R observed inEq.(1b). In this relationship the proportionality terms are functionsof both porosity and saturation as well as clayiness. It further canbe seen that for every value of clayiness or “shaliness” and resulting

nevolume of bound water within Ø , the prevailing values of effectiveporosity and water saturation can vary the conductive water volume ineffective pore space and, therefore, can vary the relative proportions

w wb weof R and R for the mixture R . And, this can and does vary depth bydepth.

w weIt is further illustrated in Figure 2 that as R is corrected to R ,0 0 corrected eR is corrected to R , both by the same factor a. Just as Ø andtØ are two intrinsic properties of the rock and exist simultaneously,w weR and R exist simultaneously and relate to the same exponent m. The

m exponent describes an intrinsic property of the rock and producesthe parallelism seen in the Figure 2. As a result, the sloperepresented by m is independent of the conductivity of the waters in

w wethe pores. In this figure, R is related to R by the factor a, orcoefficient a, that varies depth by depth. As a consequence, when

wethese relationships for R are implemented, or their derivatives orequivalent relationships are used in any resistivity-basedinterpretation, it is important to recognize that the accountabilityor compensation by the a coefficient must be removed from the modifiedFormation Resistivity Factor. And, the numerator of the FormationResistivity Factor must always be equal to 1.0 to prevent duplicationin the accountability for the secondary conductivity provided by clayshale or other conductive constituents. This perspective will bediscussed in detail below.

WHAT IS THE FORMATION RESISTIVITY FACTOR

The Formation Resistivity Factor, F, is an intrinsic property of aporous insulating medium, related to the degree of efficiency orinefficiency for the electrolyte-filled paths to conduct electricalcurrent through the medium. The formation factor pertains to and isintrinsic to the insulating medium only. It is independent of theelectrical conductivity of the electrolyte in its pores. Anyrecognizable, valid extraneous electrical conductivity that sometimesis seen to influence the value of the formation factor must berelegated to appropriate conductivity equations, such as Eq.(1a) andEq.(1b), where conductivities can be accommodated by discrete terms.

An equation that demonstrates the purpose of the formation factor isthe very basic resistance equation that converts resistivity toresistance:

r ohms = ((L/A)m/m )(R ohms m /m) (3c)2 2

of 34

where r is resistance, L is length, A is the cross-sectional area of astraight electrically-conductive path of length L, and R is thefamiliar resistivity. The term L/A in the equation is similar to aformation factor and describes a fractional volume having length L andcross-sectional area A that is 100 percent occupied by a single,homogeneous, electrically-conductive medium, either an electrolyte orsolid. This equation represents 100% efficiency in the conversion fromresistivity to resistance, and the converse.

Figure 3 is similar to a figure used earlier by Ransom (1984, 1995).This figure represents a unit volume of insulating solid matter whereeach side of the cube has a length L equal to 1.0 meter. Remove 0.20 ofthe insulating cube from the center and fill the void with water of

wresistivity R . Electrical current now can pass from top to bottom ofthe cube through the water without deviation or interference. This formof electrical path through the insulating matter has the highestefficiency.

A measurable resistance implies a measurable resistivity, andresistivity implies conductivity. And, within this insulating cube, theonly electrical conductivity occurs in the fraction that is water.Therefore, the measured resistance across the example cube is due tothe resistance of the water. Eq.(3c) now can be written

cube water w r = (r ohms) = ((L/0.2L )m/m )(R ohms m /m)2 2 2

If this cube were a unit volume of rock where all void volume wasrepresented by 20% porosity, then, the relationship would be

rock water w w r = r = (L/0.2L )(R ) = (1.0/0.2L)(R )2

The dimensional analysis of this relationship is

(ohms) = (m/m )(ohms m /m) = (ohms)2 2

and the apparent formation factor is (1.0/0.2L) with units ofreciprocal meters (m ). After dividing both sides of the equation by-1

the volume represented by L/A, as observed in Eq.(3c), resistancebecomes resistivity and the relationship becomes

rock rock t w (A/L)r = R = R = (A/L)(1.0/0.2L)(R )

Here, the dimensional analysis is

(m /m)(ohms) = (m /m)(1/m)(ohms m /m)2 2 2

(ohms m /m) = (m /m )(ohms m /m) = (ohms m /m) 2 2 2 2 2

and the formation factor now is dimensionless and has become

t t F = 1.0/(0.2) = 1.0/Ø = 1.0/Ø m

in its simplest form, where exponent m = 1.0.

In addition to showing the relationship between the formation factorand the void space in the rock, this exercise demonstrates that the

1absolute minimum value for exponent m or m is 1.0 at the highest

of 34

electrical-path efficiency possible of 100%. Theoretically, thisdegree of efficiency can be duplicated by the presence of an openfracture or other similar water-filled void aligned favorably with theelectrical-survey current flow. Although the value of m might neverreach 1.0 in practice, the presence of fractures and similar voids canand do reduce the value of m.

There is a fine distinction between the formation factor that is anintrinsic property of rock related to the shape of its voids, and theformation factor that is related to the shape of the water thatoccupies the voids. The intrinsic formation factor is directly relatedto the solid insulating framework of rock and how it shapes theelectrically-conductive water volume when water saturation is 100%,and nothing else. This by definition is intrinsic. But, the size andshape of the conductive water volume also is influenced by thepresence of insulating fluids, such as oil and gas, that can displacewater and occupy part of that pore volume. Therefore, the distinction

tis made that the actual formation factor, that will be called F , usedwt tin calculations and derivations, will involve the term S . F is not

intrinsic.

To carry this demonstration one step further, in the same rock wheretF = 1.0/Ø at 100% efficiency, if part of the rock is electrically

inert, heterogeneous, porous, insulating rock framework, with auniform distribution of constituents and porosity, and the remainderis formation water partly displaced by hydrocarbon, then the former

t wt t wt tvolume of water Ø now becomes S Ø and F now becomes 1.0/(S Ø ) forwt tall values of water saturation and porosity. The fraction S Ø now has

become the fractional cross-section of area in Figure 3 for all electrically-conductive water paths where the electrical efficiency is100%.

Not all water-filled electrically-effective pore paths are 100%efficient because the network of interstitial water within the rockframework can take on many different shapes and configurations imposedby the many and varied properties of the pore walls and rock

wt tframework. Where S Ø is the fractional cross-section area for allelectrically-conductive water paths at their highest efficiency, as in

wt ta bundle of straight tubes, (S Ø ) is an equivalent cross-sectionalm2

area resulting from all factors that impede the flow of electricalcurrent or increase the resistance-to-flow through the rock. Thisinterference to electrical-current flow is reflected in the steepness

2 1of the slopes of the tangential exponents m , m , and n in Figure 1. Aswt t wt ta result, 1.0/(S Ø ) becomes 1.0/(S Ø ) to accommodate the variedm2

geometries of pores and paths for all conditions of pore pathinefficiency and interference, and the Formation Resistivity Factoronce again becomes

t wt t F = 1.0/(S Ø ) (3b) m2

as it was derived from the trigonometrics of Figure 1.

The Formation Resistivity Factor is aptly named. Resistance r of thetinterstitial-water network, and therefore resistivity R of the rock,

is a function of the size and geometric dimensions of configurationand shape imparted to the water volume within the network of

of 34

interconnected pores. In one unit of total volume, size or cross-sectional area through which electrical current must flow is relatedto both porosity and water saturation as a fraction of that one unitarea. Tortuous length, configuration, saturation distributions, and

wt tshape of the electrical pathway filled with water volume S Ø determinethe efficiency or inefficiency of in-place water to conduct electrical

2current and, therefore, provide a value for exponent m .

In this conversion of formation water resistivity to pathwaywt tresistance, the greatest efficiency at any given value of S Ø occurs

1 2when exponent m or m equals 1.0 and m approaches 1.0. And, thewtgreatest effectiveness in one unit of total volume occurs when both S

t t wand Ø are 100% or 1.0. When that happens, F = 1.0 and r = R = R . Attany given saturation, overall efficiency of water paths decreases as Ø

1 2decreases and/or as m , m or n increases.

All factors that influence the parameters referred to in thisdiscussion determine the efficiency for electrical-survey current toflow through the rock. These factors, through Archie’s parameters,ultimately determine the value given to the Formation ResistivityFactor. Next, we look at those factors.

THE m EXPONENTS

The porosity exponent m is an intrinsic property of the rock relatedto the geometry of the electrically-conductive water network imposedby the pore walls or surfaces of solid insulating materials. This hasbeen verified by investigators of porous media in extremely meticulousand controlled laboratory investigations, particularly by Atkins andSmith (1961). All minerals have characteristic crystalline or particleshapes, whether the minerals are electrically conductive or not, andcontribute to the geometry of the electrically conductive waterresiding in the pores through shape, physical dimensions of the poresand pore throats, tortuosity, configuration, continuity, poreisolation, orientation, irregularity, surface roughness, angularity,sphericity, and anisotropy. In addition, allogenic minerals, andauthigenic mineral growths such as quartz or calcite or dolomite orclays, all, contribute to the water geometry within the pores. And,the electrical pathways through the insulating rock are configuredfurther by secondary porosity in all its shapes and forms, such as:dissolution porosity, replacement porosity, fissures, fractures,micro-cracks, and vugs in their various orientations within the rock.Exponent m is related to all these features acting separately or inconcert through the various inefficiencies (or efficiencies as thecase might be) imparted to the electrically-conductive paths by thediversified void geometries, and this results in variations inelectrolyte-filled path resistance and consequent rock resistivity.The greater is the electrical inefficiency of the shape of theresulting water volume, the greater will be the value of exponent m,

0 tand the greater will be R and/or R . This, too, can be seen in Figure1.

Heterogeneous solid matter, if it is electrically inert, constitutespart of the framework of the rock only and has no other effect onresistivity other than by occupying a position in the rock framework

of 34

and possibly displacing interstitial water and influencing the poreshape and pore geometry.

Heterogeneous electrically-conductive minerals such as pyrite andsiderite present a different scenario. Their presence not only canaffect the shape of the conductive water paths in the same manner aselectrically inert minerals, but can provide electrical conductance insolid matter, not related to pore geometry, that can influence rockresistivity. The electrical conductivity of such minerals must beaccommodated through the a coefficient in proportionalityrelationships in Eq.(1a) and (1b), or an equivalent.

In practice, the m exponent usually has a default value of about 2.0.1In the laboratory, the minimum value for m (or m ) in homogeneous

granular media has been determined to be about 1.3 for sphericalgrains regardless of uniformity of grain size or packing (Atkins andSmith, 1961; Fricke, 1931; Pirson,1947; Wyllie and Gregory, 1952). Itwas shown in the development of the formation factor, above, and inRansom (1984, 1995),that m will decrease in the presence of an openfracture, dissolution porosity, or fissure where the continuous voidspace is aligned favorably with the survey-current flow; and it wasfurther demonstrated that m has an absolute minimum value of 1.0 at100% efficiency.

1In the concept in this paper, the porosity exponent m can be estimatedfrom rock containing 100% water by

10 0 10 w 10 e -m = (log R - log R )/(log Ø ) (5a)

10 0 corrected 10 we 10 tand -m = (log R - log R )/(log Ø ) (5b)

These relationships were derived from Figure 2 and Figure 1 and arequalified for use in dual-porosity, dual-water methods ofinterpretation.

1Additionally, a value for m can be determined that applies to a largerwarange of data. Turn to Figure 5. This figure is a plot of R vs

t tClayiness. The assumption is made that Ø , R , and Clayiness arewa treliable. The value R is determined by dividing R by the formation

factor in Eq.(3a). In interactive computer methods, the plot isentered by assuming a reasonable value for the m exponent. Select aninterval that is believed to contain some wet zones, if possible. Most

wzoften an R trend can be recognized in this plot, as seen in Figure 5.w wb wBoth an R and R often emerge. If an independent R value is known

wfrom a reliable source, convert that R to the in situ temperature ofwthe logging tool environment at the depth of interest. If the R from

the plot differs from the known value, iterate by program subroutinew wbetween the known R and the derived R by incrementally changing m in

Eq.(3a) until the two values agree. The value of m determined in thismanner is independent of temperature and is compatible with the logdata over the depth range covered by the estimated clayiness values.

In the event that the m of Eq.(5a) does not agree with m in Eq.(5b),other methods for singular values of m are based on Figure 2. Onemethod is based on the solution of m in similar triangles, and another

w weis based on the dual salinities of R and R in a method similar to the

of 34

laboratory method proposed by Worthington (2004). Both methods requireconsiderable iteration in the evaluation of unknown variables in theirrespective equations.

It should be noted that on a log-log plot for either measured or0 0 tcorrected values of R (or C ) versus Ø , as in Figure 2, values of m

sometimes can be derived for each specific set of log data or rockwe 0 correctedsample. Individual values of R and R can continually change

with changing clay shale content. As a result, a reliable trendspecific to comparable rock samples might not be observed for m unless

ne tthe ratio of Ø /Ø , or other appropriate discriminator, is held nearlyconstant to ensure a nearly constant relationship between theproportions of mineral constituents as porosity changes. And,laboratory measurements for m must show repeatability in both m andthe resistivity of the influent and effluent after significant timelapses to prove that the measurement process, or sample degradation,does not influence the value of the measurement being made.

HOW IS EXPONENT n RELATED TO EXPONENT m

The saturation exponent n is related to both pore geometry and theinterference to electrical current flow within the complex water-filled paths remaining in the pores after displacement by hydrocarbonhas taken place. In Figure 1 and Eq.(3b) it is shown that n is ageometrical element similar to m. In Figure 1, it can be seen thatslope n is what slope m becomes after hydrocarbon has migrated intothe pores and has displaced a fraction of the water.

Historically, and in the absence of better information, the usualdefault value for n in water-wet rocks has been the same as for m,

1i.e. exponent m . Also see discussion in APPENDIX (B)(4). However, thepresence of oil at any saturation displaces some water volume andproduces some electrical interference and, therefore, must increasethe minimum value of n to some value higher than the value of m, all

1other things remaining the same. Because n is what m has become afterhydrocarbon has occupied a fraction of the pore space, exponent n

1cannot have a lower value than the actual value of m . A detailedexplanation appears in APPENDIX (D), based on Figure 6, why the actual

1value of n cannot be lower than the actual value of m in the samesample, in situ or in the laboratory. Exponent n can and will increase

1over m in the presence of oil or gas as the presence of hydrocarbondecreases the volume of electrically conductive water and changes thedimensions of electrical paths, or otherwise impedes the flow of theelectrical-survey current. Additionally, there can be multiple values

2for both m and n as oil saturation changes and/or wettability to oilchanges. For oil-wet and partially-oil-wet rocks this effect can bequite significant. When oil is present, in partially oil-wet and oil-wet rocks, for any given water saturation the saturation exponent ncan vary from as low as 3.0, or lower, to as high as 9.0 or moredepending on the degree of wettability to oil, physical distributionsof oil and water, oil properties, and rock-framework surfaceproperties and characteristics. In Figure 1, it can be seen that at

wtany constant value of S , if the redistribution of a constant fractiontof oil causes the electrical interference to change, then R will

change, and this will result in a corresponding change in the value of

of 34

n.

The maximum value for n in any specific rock is that value where thepresence of oil or gas produces the greatest interference. Thetheoretical maximum should occur where water saturation is at itsirreducible value. The minimum value for n occurs where the presenceof hydrocarbon produces the least interference. This minimum shouldoccur at the water saturation where the hydrocarbon saturation isirreducible. These statements suggest that there might be a causative

rorelationship with n in a crossplot of the relative permeabilities (k ,rw ro o rw w wtk ) or of the mobilities (k /ì , k /ì )for oil and water vs S .

The exaggerated influence due to the presence of oil will increase2 wt tboth the combination exponent m for the product S Ø and the usual

wtexponent n for S . For any given porosity and any given oil2saturation, the exponents m and n will increase with those properties

of the rock and pore walls that when covered with adhesive oil filmsincrease the interference to the flow of electrical current throughthe conductive paths. These factors increase in severity with theincrease in wettability to oil, finer grained sandstone (increasedsurface area), increased efficiency of packing, increased number ofgrain-to-grain contacts, finer pores and pore throats, properties ofoil, interfacial tension between oilfield brine and crude oil,isolation of pores, and the physical saturation distributions of boththe wetting- and nonwetting- phases whether oil or water. All theseinfluences act in concert at their respective levels of severity tocause or alter interference to electrical flow. These featuresrelative to the presence of oil, and sometimes gas, must berecognized. Gas usually does not have the same exaggerated effect on nunless the reservoir has been filled with oil at some former time ingeologic history and an adhesive film of remnant oil precedes theoccupation by gas. The resistivity of a gas-bearing zone can increase,however, due to the decrease in irreducible water saturation. This,too, can be demonstrated in Figure 1. The primary exaggeration in n iswith partially oil-wet and oil-wet rocks that are filled with oil orhave been filled with oil at a former time whether as a reservoir oras a migration path.

THE a COEFFICIENT

Historically, the a coefficient always has appeared in the numeratorof the modified Formation Resistivity Factor, and has been perpetuatedin industry with no defined purpose. In this model there is no supportfor the appearance of an a with a constant value in the formationfactor. The formation factor is intrinsic to the rock at 100% watersaturation. The a coefficient is not an intrinsic property, but isdependent on variables not related to the electrically inert rock orto the physical geometries of its pores. The a is related neither tovolume nor shape imparted to the network of electrically-conductivewater paths. The a coefficient contributes nothing to the efficiencyof the conversion of water resistivity to resistance, or vice versa.In the model, the shift in resistivity corresponding to a always isfound in the resistivity axis, and the shift varies from data to datawith depth. Coefficient a emerges as an inherent and inseparablefactor of the complex resistivity relationship, as in Eqs.(1a) and

of 34

(1b), and is related to the proportions of all secondary electrically-conductive constituents and influences.

weIt can be seen in the graphics of Figure 2 that R is the product of aw w we weand R , therefore aR = R . R varies from depth to depth, and so does

coefficient a. It can be seen in this relationship that a varying acoefficient technically can be used in single-water, single-porosity

wemethods where R is not calculated. But, in dual-water, dual-porositywemethods where R is calculated, the a coefficient cannot be used as an

independent parameter. If the a is used in single-water, single-porosity methods, it must be calculated properly. And that isdifficult to do when the correct value of a varies with mineralogy anddepth, and only one water and one porosity is available to work with.That is one of the reasons why dual-water, dual-porosity methods wereconceived.

wWhen a is associated with R , as in the equation developments of (1a)wand (1b), and as a multiplier or reduction factor for R , the a

coefficient more readily can be recognized as being an indispensablefactor that accounts for those heterogeneities that produce additionalelectrical conductivity. Coefficient a is the composite factor that

w w wbrelates R to the resistivity value of the combination of R and R ,and all other natural electrically-conductive influences, in such

weproportions to result in R .

Secondary conductive influences that produce coefficient a can be:clay shale, surface conductance, or solid semi-conductors such aspyrite and siderite as separate influences or in concert. These inturn, and perhaps others, can produce a change in rock resistivity.

The occurrence of one additional electrically-conductive influencewould add one more term to Eqs.(1a) and (1b), or their equivalent, and

wt tif there were no change in S Ø , the value of the compound coefficientwa would be reduced. These influences, all, can change R to an apparent

we 0or pseudo value R with a corresponding change in R to0 correctedR at their respective porosities.

Technically, in dual-water, dual-porosity methods, the coefficient ashould be transposed from the modified Formation Resistivity Factorterm to the water resistivity relationship. For example:

0 corrected t we t w t w t we R = F R = (a/(Ø ))R = (1.0/(Ø ))aR = (1.0/(Ø ))R m m m

The question might be asked, "What is the difference?" There are anumber of reasons, four of which are:

1. The a coefficient is a required proportionality factor, relatedwto electrically-conductive influences, that converts resistivity R to

we weR , and is calculated along with R at each depth increment.

2. To prevent duplication by the user in the correction forweconductive influences. Duplication occurs when both a and R are used

in the same water-saturation equation, or when a appears as anindependent variable in dual-water, dual-porosity methods. Anysaturation relationship involving resistivity can employ either

wecoefficient a or R , but not both.

of 34

3. To prevent the use of a constant artificial and unwarrantedcorrection factor.

4. Although the formation factor is meaningless at 100% porosity,tthe value of F always must be 1.0 when both porosity and water

tsaturation in the F equation are 1.0.

As explained above, and in Figure 2,

we w a = R /R (1c)

After substituting Eq.(1c) in Eq.(1b), a relationship is derivedshowing the proportionality terms in coefficient a for the usual shalysand:

we e wt t ne wt t w wb 1/a = (S Ø )/(S Ø ) + (Ø /(S Ø ))(R /R )

This relationship is shown for comparison purposes or informativepurposes only. It is not to be calculated and used independently indual-water, dual-porosity methods because it already has been

weincorporated in the calculation of R as can be seen in Eq.(1b).

THE SATURATION EVALUATION

tThe model in Figure 1 is a diagram showing that R is a function ofwt tboth the volume of water S Ø and the inefficiency with which

electrical current passes through that water. The inefficiency of theelectrical current flow is related to the distribution of the waterand the interference to that flow within the water network as it is

wt treflected in the exponents m and n of the expression (S ) (Ø ) . In then m

10 wtmodel it is shown that log S is the length of the projection along10 t 10 tthe X-axis between log Ø and the intercept of slope n with log R .

This length also is represented by length CD of triangle CDG in Figure1.

Revisiting Eq.(1b), (2b), and (3b), the reader already might havededuced that water saturations can be estimated from these equations.Keeping faithful to the self-evident truth that the volume of waterreferred to in the denominator of the formation factor must be the

wesame volume of water that provides electrical conductivity in the Requation, then Eq.(1b) can be used only with Eq.(3b). Therefore, whenwe wtS (or S ) is less than 100%, the product resulting from Eq.(1b) and

(3b) is

t calculated t we R = F R (2b)

t wt twhere F = 1.0(S Ø ) (3b)m2

tAfter combining Eq.(2b)and (3b) when either the measured or actual Rtis substituted for the calculated R , then

t measured wt t w wt t we R = (1.0/((S Ø ) ))aR = (1.0/((S Ø ) ))R (4a)m2 m2

of 34

However, it was illustrated in Figure 1 in triangle ACG, and inwt t wt tAPPENDIX (B)(3), that (S Ø ) is equivalent to (S ) (Ø ) , therefore,m2 n m1

Eq.(4a) resolves to

wt t we t measured 0 corrected t measured S = ((1.0/(Ø ))R )/R = R /R (4b)n m1

t measuredThe R in Eq.(4b) must be corrected for environmental conditionsand tool-measurement characteristics before water saturation iscalculated.

It can be seen in Eqs.(1b) and (4a) that the a coefficient is variablewe wewith depth and mineralization and is included as part of R . When R

has been calculated, and is used, the appearance of a constant acoefficient in the formation factor, usually as a fraction less than1.0, would be gratuitous and would artificially increase the calculatedhydrocarbon saturation in productive and nonproductive zones alike;and, in this model, would be both logically and mathematicallyincorrect.

Figure 1, together with Figure 2, is a concept model that hassignificant informative and educational value. The graphics of the modelare meant primarily to illustrate, to develop, or to explain what iscalculated blindly by algebraics in computer-program subroutines.

In an interactive computer program, irreducible water saturation orother core-derived information can be input for the purposes ofexamining the plausibility, validity, and integrity of certainparameters. On a well log above the transition zone in oil-bearingreservoir rock, for example, tthe intersection of R with a laboratoryvalue of irreducible water saturation fixes the upper limiting value ofsaturation exponent n for that specific set of data. However, whenirreducible water saturation is known, this upper limit of exponent nshould be calculated from the algebraics of Eq.(4b), or Eq.(4c) as willbe shown below. The same can be said for the lower limit of n in thesame rock which could be estimated by inserting water saturation whenoil saturation is irreducible. But, in either exercise, no value of n

1 tcan be lower than m . The actual value of R is required for each ofthese procedures, whether derived from the well-log or rock sample.

The water saturation equation, Eq.(4b), has been developed from thetrigonometric model in Figure 1 and again corroborated by the algebraicdevelopment of Eq.(4b), all, for certain heterogeneous, but uniform,environments. And, each development herein shows that it authenticatesArchie's basic relationships presented in 1942, and further refinesthese relationships in the developments and discussions.

It has been said that Archie's relationships are empirical developments.Whether or not this is true, it has been shown here that Archie'sclassic relationships and parameters have a mathematical basis, and haveforthright and substantive relevance to rock properties that is quitedifferent from many accepted theories and usages in industry literature.

Saturation exponent n is the most difficult of all the parameters toevaluate. If a valid value of oil saturation is known, or can bederived, the value of exponent n can be estimated by substitution inEq.(4b) or (4c). When the actual values of m and n are known, or can be

of 34

derived, either or both can be important mappable parameters, and amathematical relationship between m and n not only can be an importantmappable parameter, but can be a possible indicator to the degree ofwettability to oil or distribution of oil under in situ conditions. Thisinformation not only can be important in resistivity log analysis butcan be important in the design of recovery operations.

The calculation of Eq.(1b) is required in the solution of the waterwtsaturation relationship in Eq.(4b). Because water saturation S also

weappears in the proportionality terms within the R equation, Eq.(1b), anwtalgebraic solution for S in Eq.(4b) is not viable and is not

considered. Probably the simplest and best method for all anticipatedinteger and non-integer values of n is an iterative solution performedby a computer-program subroutine. Graphically the iteration process canbe demonstrated by a system of coordinates where both sides of Eq.(4b)

wt wtare plotted versus input values of S . As S is varied, the individualcurves for the left and right sides of Eq.(4b) will converge and cross

wtat the S value that will satisfy the equation.

Figure 4 shows a crossplot of example data to demonstrate theequivalence of the graphical solution to the iterations performed by acomputer-program subroutine. The following input values are forillustration purposes only.

w m = 2.17 R = 0.30 wb n = 2.92 R = 0.08

t t Ø = 0.22 R = 20.00 neØ = 0.09

weFor Eq.(4b), the equivalent water resistivity R has been evaluated bywe eEq.(1b) after substitution for S Ø has been made from the volumetric

material balance equation for water:

wt t we e ne S Ø = S Ø + (1.0)Ø

0 corrected t stFor the calculation of R , F was evaluated from Eq.(3a) where Sis 1.0. In Figure 4 it can be observed that when values from each side

wtof Eq.(4b) are plotted versus S the two curves have a common value at awater saturation of about 0.485. The iteration by subroutine will

wtproduce the same S of about 0.485 or 48.5% for the same basic inputdata.

It is worthy of note that in the volumetric material balance for water, we e wt twhen S Ø goes to zero the absolute minimum value for S Ø in this

neexample is 0.09, the value of Ø . The mathematical minimum waterwtsaturation S that can exist in this hypothetical reservoir is

ne t wbØ /Ø = 0.4091 or 40.91% where R becomes 0.08. The minimum saturationof 40.91% is related only to the pseudo bound water in clay shale andtells us nothing about irreducible water saturation in the effectiveporosity. If this were an actual case in a water-wet sand, at 48.5%water saturation, water-free oil might be produced because the onlywater in the effective pore space might be irreducible. Grain size andsurface area would be a consideration. Any water saturation below 40.91%cannot exist and is imaginary.

of 34

wt weFor the conversion of S to S , either the material balance for water(shown above) or the material balance for hydrocarbon (Ransom, 1995),can be used. In terms of hydrocarbon fractions, the material balance forthe amount of hydrocarbon in one unit volume of rock is

wt t we e (1.0 - S )Ø = (1.0 - S )Ø

weFor the calculation of S the balance can be re-arranged to read:

we t e wt S = 1.0 - (Ø /Ø )(1.0 - S )

weand S now can be estimated.

When the material balance equation for hydrocarbon is multiplied by thetrue vertical thickness of the hydrocarbon-bearing layer, either side ofthis equation produces the volume of hydrocarbon per unit area at insitu conditions of temperature and pressure.

we wt w wbFinally, for this evaluation of R , and S in turn, both R and R mustw wbbe known. In the event neither R nor R is known, these values most

often can be estimated by interactive computer graphics from a crossplotwa waof R (or C ) versus Clayiness (% clay) as shown in Figure 5, from

Ransom(1995), where clayiness is estimated by appropriate clay-shalewa t tindicators. R is determined by dividing the corrected value of R by F

t wt t t t wtwhere F = 1.0/(S Ø ) and F has resolved to 1.0/(Ø ) because S alwaysm2 m1

is 1.0 (for 100% water saturation) for this determination. The value oftR may be taken from wireline tools or measured while drilling. As

tpointed out earlier, the measured R should be corrected forenvironmental conditions and measuring-tool characteristics. In a figure

wasuch as Figure 5, R values from zones known to be or believed to bewz100% water-filled often describe a trend or curve identified as an R

trend where hydrocarbon saturation is zero. This trend can takevirtually any curvature, steep, convex, concave, or flat, depending on

w wbthe resistivity relationship between R and R as the fraction of claywzand its distribution varies. Once the R trend has been established it

wcan be extrapolated to 0% clay for R , and extrapolated to 100% clay forwb w wb wzR at in situ conditions. These values of R , R , and R have been

estimated from preliminary formation factor and clay indicatorinformation and are subject to examination by the analyst.

By using the method such as found in Figure 5, all values derived inw wb t 0this manner are compatible. That is, the derived R , R , R , hence R ,

are compatible because they have been derived by the same method frommeasurements by the same resistivity measuring device at the sameenvironmental conditions of temperature, pressure, and invasion profileat the same moment in time. As a result, the ratios of certain fractions

0 t wz wasuch as R /R and R /R at any specific clayiness value are fixed andwill not change regardless of the formation factor used. Independentmeasurements or determinations of any or all of these variables might ormight not be compatible. This is the risk the analyst must consider whenusing data from different or mixed sources.

In an interactive computer program, an equation can be derived for thewz wzR trend in the plot for the evaluation of R in terms of clayiness. A

wecurve for R from Eq.(1a) can be superimposed on the same plot forcomparison and analysis. If the analyst is not satisfied with the values

of 34

w wb wz wfor R , R , and R , or if R from the plot does not agree with anaccepted value, the analyst can employ the iteration subroutine as

wdescribed above for exponent m to vary the value of R from the plot inFigure 5 until there is agreement. This is another means for the

we wevaluation of R in terms of calculated clayiness. The resulting R andwb weR can be used in Eq.(1a) in the calculation of R in terms of

porosities, if desired.

It is interesting to note, however, that from Figure 5 alone, once anwzacceptable R trend has been established

wt 0 corrected t t wz t wa wz wa S = R /R =(F R )/(F R ) = R /R (4c)n

wtIt can be seen here that S can be estimated independently of porosity,n

w wbm, even R and R , all in terms of an estimated clayiness value. Thevalue of exponent n lies between its minimum value of m, corrected or asestablished above, and its maximum value determined from irreduciblewater saturation. Furthermore, unlike Eq.(4b), once exponent n has been

wtestablished, S can be calculated directly from Eq.(4c)and there is noneed for the iteration process described above or as seen in Figure 4. Acautionary note appears in APPENDIX (C).

wz wbAlthough, in Figure 5, the end point of the R trend for R is said tobe defined at 100% clayiness, 100% clayiness might not exist for theformation. The actual percentage of clay might be 70%, for example,instead of 100%. This will not affect the water saturations calculatedfrom Figure 5 by Eq.(4c). It is important to use the best clay-indicatormethods available, and to use them correctly for providing informationrelative to the presence of clays for other calculations; but whenFigure 5 is used, it is not important to know the actual amount of claypresent for the estimation of water saturations. The X-axis could berenamed Clay Index for this work. Only the relationship between the

0 t wz wacorrected R and the corrected R , and thus R and R , is important forsaturation calculations, and this relationship in the vertical axis willnot change if the scale on the X-axis is changed. In Figure 5, theclayiness fraction or calculated clay percentage is used only as adepth-related index for the saturation estimations to be calculated andrecorded. However, it is important that the clayiness measurementmethods and resulting clayiness estimations be consistent andrepeatable.

ARE THERE LIMITATIONS

Are there rocks that complicate resistivity interpretations? Of course.As in every discipline and in every analytical method, there can beproblems that complicate the interpretation process. That is not uniqueto Archie’s relationships, whether new or old. Above, it was mentionedthat Archie’s relationships, as they have been developed in this paper,apply to heterogeneous rocks with uniform distributions of porosity,saturations, and electrically-conductive constituents. That is arequirement for all matter subject to electrical investigations. Problemrocks are those that exhibit nonuniform distributions of causativefactors that influence resistivity. Such factors or conditions can be,but are not limited to: anisotropy, beds considerably thinner than thevertical resolution of the logging tools, bedding planes at high angles

of 34

relative to the borehole axis, human error in running the wrongresistivity-measuring tool for the existing resistivity profile in themud-filtrate invaded zone, etc. Are these Archie problems? No. These areproblems that affect all related analytical resistivity methods. Theseare problems that either nature or the shortcomings of man have dealt tothe analyst. These conditions are part of the interpretation process andrequire the attention of qualified well-log analysts.

CONCLUSIONS

Archie’s fundamental saturation relationship has been derived from thetrigonometrics of the model and corroborating algebraics, and has beenshown to be a dual-water methodology.

The concept described herein is based on the very fundamental conversionof water resistivity - to water resistance - to rock resistivity, and onthe efficiency of the network of pores and pathways in the rock through which the electrical-survey current must flow. The basic electricalresistance equation is the key to understanding the FormationResistivity Factor.

Archie’s relationships, as the parameters have been defined in thismodel, have been shown to apply in dual-water, dual-porosity conditionsand in shaly sands and other rocks of limited heterogeneity. Themethodology developed herein can serve as the spine to which numerouscorrectional and mineral specific subroutines can be attached to addanalytical refinements. The model and the foregoing developments in thispaper have shown that Archie’s relationships have a straightforward,physical and mathematical basis, and, in this concept, have been shownto apply to heterogeneous rocks with uniform distributions of porosity,saturations, and electrically-conductive constituents.

Archie’s relationships are fundamental and the foregoing developmentsare deemed necessary for the relationships to be understood, used inliterature, and employed in saturation calculations. An understanding ofwhat influences these parameters and what they represent will providethe user with a place to start and a recognition of what is requiredfrom other disciplines for the solution of interpretation problems, andwill complement the user’s inventory of tools for describing rocks asthey reside in nature.

EPILOGUE

A good well-logging interpretation program is like a double-edged sword.It must cut both ways. The program must find hydrocarbons where theyexist and must not imply their existence where they do not exist. Thepurpose is to be accurate with consistent reliability. Together withinteractive computer graphics, the author has used the foregoing logicand fundamentals as the skeletal framework to guide resistivityinterpretations for many years in all parts of the world.

of 34

ACKNOWLEDGMENT

I wish to acknowledge Ben W. Ebenhack, Associate Professor of PetroleumEngineering at Marietta College, founder and Chairman of the Board ofthe non-profit AHEAD Energy Corp. Mr. Ebenhack reviewed this paper andoffered suggestions and encouragement for its publication.

SYMBOLS DEFINED

a Historically, the coefficient of the Formation Resistivity Factor.wIn this paper, the coefficient of R . Usually equal to 1.0 in clean

rocks, and different from 1.0 in the presence of any conductiveinfluence other than formation water. Often lower than 1.0. When

weincorporated in comprehensive R relationships, the numerator inthe Formation Resistivity Factor must always be 1.0.

m The porosity exponent in the Formation Resistivity Factor. Relatedto the shape and dimensions of electrolyte-filled paths.

n The saturation exponent found in Archie's water saturationequation. Related to the electrical interference caused by thepresence of hydrocarbon in electrolyte-filled paths. A form of mafter hydrocarbon has displaced water.

tØ Total porosity fraction consisting of the sum of effective andnoneffective porosity fractions.

eØ That part of the total porosity fraction that is hydrodynamicallyeffective and can be occupied by hydrocarbon.

neØ Is the noneffective porosity in clay shale and is directly related to the fraction of clay shale present. A porosity volume that has not been penetrated by hydrocarbon. Water saturation inthe noneffective pore space is 1.0 and resistivity is represented

wbby R .

tF The Formation Resistivity Factor of the rock, where the conductivewt t wtwater volume fraction always is S Ø , where S # 1.0. In this

paper, the numerator always is equal to 1.0.

wR Formation water resistivity. Interstitial-water resistivity.

wa tR The apparent resistivity of formation water found by dividing R byt wt 2F , where S = 1.0 and m = m.

wbR The apparent resistivity of water bound in clay shales.

weR A calculated value for the equivalent or pseudo water resistivitywhen secondary conductive influences are present. In this paper,

wthe product of coefficient a and R as it is incorporated in theweequation for the calculation of R .

wz waR Is the R value determined from a zone known to be or believed tobe completely filled with water or nearly saturated with water.

wzHydrocarbon saturation is zero or near zero. R is a graphical

of 34

we waequivalent to R and a special condition of R .

0R The true resistivity of the uninvaded rock when water saturation iswe1.0. The water or pseudo water can have the resistivity value R .

Sometimes calculated, sometimes read directly from the deepresistivity curve, sometimes derived from the deep resistivity

0 tcurve. R is a special condition of R .

tR The true resistivity of the uninvaded rock, whether or not itcontains hydrocarbon. Sometimes read directly from the deepresistivity curve, sometimes derived from the deep resistivity

0 tcurve, sometimes calculated. R is a special condition of R .

we eS Water saturation as a fraction of the effective porosity Ø .

wt tS Water saturation as a fraction of the total porosity Ø .

wt t wtS Ø The total conductive water volume fraction where S # 1.0. Afraction of one unit of formation volume.

REFERENCES

Archie, G. E., 1942, The Electrical Resistivity Log as an Aid inDetermining Some Reservoir Characteristics, Trans., AIME, vol.146.

Atkins, E. R., and Smith, G. H., 1961, The Significance of ParticleShape in Formation Resistivity Factor - Porosity Relationships,Journal of Petroleum Technology, AIME, vol. 13.

Fricke, H., 1931, The Electric Conductivity and Capacity of DisperseSystems, Physics, August.

Keller, G. V., 1953, Effect of Wettability on the Electrical Resistivityof Sand, Oil and Gas Journal, Jan. 5; 51: 62-65.

Pirson, S. J., 1947, Factors Which Affect True Formation Resistivity,Oil and Gas Journal, November 1.

Ransom, Robert C., 1974, The Bulk Volume Water Concept of ResistivityWell Log Interpretation. A Theory Based on a New Reservoir Rock Resistivity Model, The Log Analyst, SPWLA (Jan.-Feb.)

Ransom, Robert C., 1977, Methods Based on Density and Neutron Well-Logging Responses to Distinguish Characteristics of ShalySandstone Reservoir Rock, The Log Analyst, SPWLA (May-June)

Ransom, Robert C., 1984, A Contribution to a Better Understanding of theModified Archie Formation-Resistivity-Factor Relationship, The LogAnalyst, SPWLA (March-April)

Ransom, Robert C., 1995, Practical Formation Evaluation, John Wiley &Sons, Inc., New York.

Sweeney, S. A. and Jennings, H. Y., 1960, The Electrical Resistivity of

of 34

Preferentially Water-Wet and Preferentially Oil-Wet CarbonateRocks, Producers Monthly, 24, 29-32.

Worthington, P. F., 2004, Characterization of the Intrinsic PorosityExponent through Dual-salinity Measurements of ElectricalConductivity, Petrophysics, SPWLA, (November-December, pp. 499-516)

Wyllie, M. R. J. and Gregory, A. R.; 1952, Formation Factors ofUnconsolidated Porous Media; Influence of Particle Shape andEffect of Cementation; Paper Number 223-G presented at Fall Meeting of the Petroleum Branch, AIME, held at Houston, TX., October, 1-3.

APPENDIX

The model appearing in Figure 1 incorporates Figure 2. The origin alwayswe we w wis R . But, R is the same as R in clean rocks and different from R in

heterogeneous rocks.

e t(A) Figure 2. In this figure are five variables relative to Ø and Ø .w 0 we 0 correctedThey are: R , R , R , R , and a. Input variables to the figure are

w e t wR and Ø and Ø . Input value R is from the best known source or fromFigure 5 after clayiness has been determined by the best clay indicators

e tavailable. Porosity values, Ø and Ø , are determined by traditionalmethods, or from Ransom (1977, 1995), after employing the appropriate

0 wmatrix values for the host rock. Variable R is from Eq.(2a) with R aswe 0 corrected weinput, R is from Eq(1a), R from Eq.(2a) with R as input, all

values are at 100% water saturation in this figure. Coefficient a iswshown as a multiplier of R but is never calculated independently for

weuse in an interpretation program when it is an integral part of R , asseen in Eqs.(1a) and (1b). Coefficient a may be calculated forcomparison purposes only.

(B) Figure 1. Figure 1 illustrates the model that Archie’s relationshipsobey. It is intended for informative and illustrative use, only. It isnot drawn to scale. The inclinations of the straight lines representingslopes are determined by trigonometric tangents. The value of a specifictrigonometric tangent is the value of the specific m or n. The steeperis the slope, the greater is the interference and resistance toelectrical-current flow through the pores and pore paths in the rock,the greater is the inefficiency for the transmission of electrical-survey current, and the greater will be the value of m or n.

wt t(1) On the X-axis, it is seen that S Ø decreases to the right. A studyt wtof the logarithmic scales will show, for example, that Ø = 0.2 and S =

wt t0.3; and, as a result, their product S Ø = 0.06.

(2) There are three right triangles of interest in Figure 1. They aretriangles ABC, CDG, and AEG.

In any triangle shown in the figure, the slope or tangent of an angle isdescribed as the side opposite divided by the side adjacent. Thetrigonometric law pertaining to tangents, that right triangles follow,can be described as:

of 34

tangent of acute angle = (side opposite)/(side adjacent)

Because the X-axis is descending in value to the right of the origin,the sign of the tangent is negative. Therefore, on the log-log plot suchas Figure 1,

10 10 (-)(tangent of acute angle)(log (side adjacent)) = log (side opposite)

and, 1/(side adjacent) = (side opposite)tangent of acute angle

Trigonometry was created for solving problems. Express thesetrigonometric functions in equation form and the Formation ResistivityFactor and an improved Archie’s water saturation relationship willemerge.

In Figure 1, the tangent of the acute angle â of the right triangle AEGis represented by

2 10 10 tan â = -m = log (EG)/log (AE)

2 10 10 10 t -m (log (AE)) = log (EG) = log F

t F = 1.0/(AE) m2

In equation form, this is

2 10 t 10 wt t -m = log F /(log (S Ø ))

2 10 wt t 10 t -m (log (S Ø )) = log F

t wt t F = 1.0/(S Ø ) . . .(3b) m2

In Figure 1, the tangent of the acute angle of the right triangle CDGis represented by

10 10 tan = -n = log (DG)/log (CD)

10 10 10 t 10 0 corrected -n(log (CD)) = log (DG) = (log R - log R )

0 corrected t (CD) = R /R n

In equation form, this is

10 t 10 0 corrected 10 wt -n = (log R - log R )/log S

10 wt 10 t 10 0 corrected -n(log S ) = (log R - log R )

wt 0 corrected t (S ) = R /R . . .(4b)n

wt t wt t(3) (S Ø ) has the same function as and is equal to (S ) (Ø ) .m2 n m1

2 10 wt t 10 t 10 weIn triangle AEG, -m (log (S Ø )) = log R - log R

of 34

wt t we t (S Ø ) = R /Rm2

10 wt 10 t 10 0 correctedIn triangle CDG, -n(log S ) = log R - log R

1 10 t 10 0 corrected 10 weIn triangle ABC, -m (log Ø ) = log R - log R

1adding the two equations that involve n and m , yields

10 wt 1 10 t 10 t 10 we -n(log S )+(-m (log Ø )) = log R - log R

wt t we t S Ø = R /Rn m1

wt t we tBut, (S Ø ) = R /R , from above.m2

wt t wt tTherefore, (S Ø ) = S Ø m

2 n m1

(4) In addition, in partially oil-wet or oil-wet rocks, the assumptionthat n = m can lead to calculated water saturations that are too low. InFigure 1, it can be seen that when the default value of n equals m the

tline representing n will intercept the R level far to the right at H.The resulting calculated water saturation will be too low and“unreasonable”. When the corrected n value is used, the slope

trepresenting n will intercept the level of R at some point G, whichwtwill yield a “reasonable” value for S depending on the validity of n.

Straight lines representing values of n can rotate along the arc ä aswteither n or S varies.

wa(C) Figure 5. This figure is a plot of R vs Clayiness. It could havet 0 0 correctedbeen a plot of R vs Clayiness where R and R could have been

w wz 0 corrected testimated in the same manner as R and R . The ratios R /R andwz wa waR /R would still be the same. The plot of R vs Clayiness yields more

information.

When calculating clayiness for this plot, clayiness should be determinedfrom the best clay-shale-responsive indicators available, but not froman average of several. Consider the influences on each of the clay-responsive measuring methods before each is entered into the clay-estimation process.

(D) Figure 6 is nearly the same as Figure 1. Like Figure 1, Figure 6 isnot intended to be a working graphical procedure, it has beenexaggerated to illustrate detail, and it is intended only to beinformative and explanatory. In Figure 6 it is shown that the same rock

wt tcan contain the same volume of water, S Ø , whether or not oil or gas ispresent. In Figure 6, the example uses the same porosity and saturationvalues as seen depicted in Figure 1. In this example, if watersaturation is 100%, porosity is 20% and m is uniform throughout, then,

wt twhen the porosity is reduced to 0.06, S Ø = 1.0 x 0.06 = 0.06. This isthe value of the total water volume at J. The same volume of water canbe observed in the same rock when the porosity is not reduced, but oil

wt tor gas displaces 70% of the water volume: S Ø = 0.30 x 0.20 = 0.06.wt tThis is the total water volume at K. This common value of S Ø is shown

wt tat E at the base of Figure 6 on the logarithm scale for S Ø . In Figure 6, where the example rock shows a water volume fraction of

00.06, it can be seen that when n < m, R at resistivity level J will be

of 34

tgreater than R at resistivity level K, for all values of porosity andall values of saturation. In real life this cannot happen. But, when thevalue of n is found to be or made to be < m in the same sample of rock,most users will not recognize the impossibility of this condition.

tWhen n is said to be < m, the predicted value of R always becomes less0 tthan R at equivalent water volumes. It cannot be otherwise. For R at K

0to have a lower value than R at J, the migration of oil or gas into theinterstitial water volume must make the electrical paths through theremaining water more efficient. This enhancement of the electricalconductivity of the water paths through the void space, must be donewith no change in salinity. Hypothetically, for oil or gas to migrateinto the undisturbed effective pore space and cause the value of n to be< m, the presence of oil or gas must cause the electrically conductivepaths of water to become more tube-like or to become less tortuous bywhatever means. This, the mere presence of oil or gas cannot do. In thismodel, which is faithful to laws of physics, when all other thingsremain unchanged, it is shown that it is impossible for the actual value

t 0of R to have a lower value than the actual value of R for the samequantity of water in the same sample. Therefore, the actual value of ncannot have a value lower than the actual value of m.

Not only is this condition impossible, when n < m the slope representingtthe aberrant n is so low that for whatever value of R exists, the slope

trepresenting n will intersect that level of R far to the right onFigures 1 and 6 at an artificially low water saturation value. Theconsequence is that the calculated water saturation will be too low andthe hydrocarbon saturation will be over estimated indiscriminately inproducible and non-producible hydrocarbon-bearing beds alike. This willbe an insidious by-product that most users will not anticipate.

To accept a value for n that is lower than m, is to deny that themigration of oil or gas into rock will cause more electricalinterference than water alone. That denial contradicts basic physics andviolates the fundamental principles of resistivity analysis.

of 34

ABOUT THE AUTHOR

Robert C. Ransom is a retired consultant in formation evaluation throughthe use of well-log analysis. Formerly he was Senior Research Associatein Formation Evaluation for UNOCAL Corporation's research center atBrea, California. He graduated in 1951 with a B. E. Degree in ChemicalEngineering from the School of Engineering at Yale University. He is anHonorary Member of the Society of Petrophysicists and Well Log Analysts(SPWLA). He joined Schlumberger Well Surveying Corp. in 1951 and was onleave of absence to serve with the Chemical Corps of the U.S. Armyduring the Korean War. He left Schlumberger to research and developwell-log analysis methods at UNOCAL's research center and to work withthe field department on problems of practical importance. Bob compiledand edited both the 1st and 2nd editions of the SPWLA Glossary. He hasauthored and published a number of technical papers, a textbook, and anInterpretive Groundwater Glossary atwww.interpretivegroundwaterglossary.com. He taught formation evaluationat the graduate level in the Petroleum Engineering Department of theUniversity of Southern California for nearly a decade prior to his movefrom California to Colorado. He and his wife of over 50 years haveretired to their home in the countryside outside of the town ofElizabeth, Colorado.

http://www.interpretivegroundwaterglossary.com.

of 34

Figure 1. A graphical representation of the model. In this diagram thetmodel illustrates oil-bearing sandstone. R varies with oil saturation

and distribution. Exponent n varies with the electrical interferencewtcaused by the presence of oil regardless of saturation value. S is

determined by the intersection of the slope representing n withtresistivity level R . See text for full explanation. Based on Ransom

(1974,1995).

of 34

weFigure 2. A detailed portion of the graphical model showing how Rw wbresults from a mixture of waters R and R in a shaly sand. The effect

wbof the more conductive pseudo water represented by R produces thewtypical a multiplier of R . Exponent m is an intrinsic property of the

rock. Based on Ransom (1974).

of 34

Figure 3. An insulating cube with a 20% void filled with water.Electrical-survey current is flowing through the cube from top tobottom. This cube is a visual aid in the development of the formationfactor. See text for discussion. From Ransom (1984, 1995).

of 34

wtFigure 4. A crossplot illustrating the iteration of S for the value ofwtS that satisfies both sides of Equation (4b). See the text for

discussion.

of 34

waFigure 5. The crossplot of R versus Clayiness where the identifiedw wbtrend is extrapolated in both directions to evaluate R and R at 0% and

100% clayiness, respectively. See text for discussion. From Ransom(1995), courtesy of John Wiley & Sons, Inc.

of 34

t Figure 6. This figure, similar to Figure 1, shows that when n < m, R at0K has a lower value than R at J for equivalent water volumes at all

values of porosity and saturation. Any condition for n to be < m iscontrary to the laws of physics and the fundamentals of resistivityanalysis. See APPENDIX (D) for a detailed explanation.

DUAL-WATER ARCHIE, and the IMPORTANCE of WATER …archieparameters.com/Resources/Archie's...

Documents

Transcript of DUAL-WATER ARCHIE, and the IMPORTANCE of WATER …archieparameters.com/Resources/Archie's...