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    DUAL-WATER ARCHIE, and the IMPORTANCE of WATERGEOMETRY. 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 Archies relationships.The geometry of water occupying void space in rock is shown to be themost important factor in controlling efficiency of the flow of

    electrical-survey current through interstitial-water paths in rock,thus giving value to Archies 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 saturation

    distributions makes a better understanding of the analytical processimperative.

    WHAT ARE ARCHIES BASIC RELATIONSHIPS

    Most water saturation equations used in resistivity well-loginterpretation are based in some way on Archies relationships.Resolve those equations to clean rocks, and Archies basic water-saturation equation emerges. Archies 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. Archies basic relationships are:

    w 0 t measured S =R /Rn

    F = 1.0/()m

    0 w R =FR

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

    t t F =a/() m

    where the acoefficient 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

    Archies classicrelationships usually are considered to be clean-sand relationships. In shaly sands where clay shales produceadditional electrical conductivity, Archies relationships are saidnot to apply. Relative to Archies original concept and its popularuse, this pronouncement is correct. In the concept born of thismodel, an overall version of Archies relationships emerges that

    shows that Archies 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 inArchies relationships, and the parameters derived and defined applyto any other methodology that makes use of Archies parameters.

    Figure 1is 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 thowwater (S ) is related to a unit volume of rock with resistivity

    t(R ). The model in Figure 1 is illustrative in nature and is not drawnto 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 (and

    t tF= 1.0) when is 1.0 or 100%. Where it commonly has been believedthere are only two slopes in Archies 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 that

    tpertains to the single parameter . This is the familiar porosityexponentmknown in industry. Sometimes for the sake of clarity the

    1 2term m will be used instead of m. The second exponent is m which

    t wtpertains to two parameters, and S, and is the exponent for the

    wt t wtproduct S . The third is the exponent nthat 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 mand n, drawn from 100% water at R to the intersectionof the line representing slope nextrapolated to intercept

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

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    wt tand the fraction of water in the total rock volume, S , are

    10 t 10 wtdepicted on the X-axis by log and log S. The fraction of totalrock volume that is water provides the electrical conductivity to the

    trock resulting in resistivity R . As oil or gas displaces water, the

    wtwater saturation, S, decreases to the right as the saturation of oil

    wtor 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 nwill 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= mare employed, the line representing the slopeof exponent nwill be extended far to the right to intersect the

    tlevel of the measured or derived value of R at an unreasonable and

    wtunacceptably 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 nis equal to m,

    tthe usualnintercepts 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 nisemployed. 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 RVERSUS S

    t wt tIt must be recognized that and S have no units, but are

    t wt tfractions and in themselves are not volumes. Only when or S aremultipliers 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 at

    we100% porosity, but R is one of the products of Figure 2. Figure 2is

    wa detailed view of an interior part of Figure 1 showing how R in the

    wepresence of dual water becomes R to become the origin of the

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

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    wt e tdetermined from Eq.(1b) when S is 1.0. Within Figure 2, and

    wexist simultaneously, and when they have different values, R becomes

    we 0 0 corrected R ,and R becomes R .

    tIn Figure 1, R is seen plotted on a log-log plot versus the volume

    wt t tS , of which is a part. In resistivity well-log interpretation,themor nslope in every case represents only the slope between two

    points: 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 when S= 1.0 and slope m= m , or it can

    wt t wt 2be S when S1.0 and slope m= m . Slope nrepresents the casewhere 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 mornevaluationwhen the two end points are known. The resulting slope, be it mor 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 nrotates

    throughout the range of arc shown in the diagram as either or bothwtnand Svary. The steeper the slope for either mor n, the more

    inefficient will be the water path in the rock for conductingelectrical-survey current, the greater will be the values of mandn,

    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 mor nare trigonometric tangentialratios of the side opposite (on Y-axis) over the side adjacent (on X-axis) of any right triangle in the figure. Exponents mand n

    trepresent rates of change in resistivity R relative to water volume

    t wt tor S , and reflect the efficiency or inefficiency of that watervolume to conduct electrical-survey current. In the followingexamples, it is shown how each mand nis 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 ))

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

    t wt t F =1.0/(S ) . . . (3b)m2

    tThe formation factor, F, as it is derived here from Figure 1, applies

    wtto all values of S1.0 and is represented on the resistivity axisas the difference between the logarithms of resistivity of the

    weequivalent 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 nis the tangentthat

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    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

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

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

    See theAPPENDIX (B)for a more detailed explanation of Figures 1 and

    t wt2 and the derivations of mand n, and Fand S.

    In addition, in an examination of the line slopes in Figure 1, it can

    wt tbe seen that (S ) has the same function as and is equivalent tom2

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

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

    The line representing either slope, mor 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 mand nfor several rock samples canbe curvilinear and can represent a numberof different slopes as the porosities, saturations, and proportions ofminerals vary. The locus of points for either mor ndepends 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 shaleis 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

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

    w nevoids remains 100% and Sis always 1.0 in .

    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 parallel

    conductivity 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 =(/)1/R +(/)1/R (1a)

    t e newhere = +.

    For 100% water-bearing rock the conductivity relationship in Eq.(1a)

    w ne weyields 1/R (in clean rock) where is 0.0, and yields 1/R (in shaly

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

    It should be noted that in this equation there are no limitations on

    wthe resistivity values of formation waterR 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 be

    wb 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 which

    we 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 =(/)1/R =1/R

    0and the corrected R in either event is

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

    where, from Figure 1,

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

    wt 2 1The value of Sis 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 the

    displacement volume of oil, and Eq.(1a) becomes

    we we e wt t w ne wt t wb 1/R = ((S )/(S ))1/R + (/(S ))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 ) (3b)m2

    2and mis 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 w the two waters have resistivities of R and pseudo R , and R formerly

    wefrom 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 be

    werelegated to terms in the R equation.

    wt wt tEquation (3b) applies to all cases where S1.0, and the term (S )

    tin the denominator of the Frelationship must be compatible with the

    wesame term in the denominator in the calculatedR 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 Sand have the same exponentm. In this case, the graphical model shows this

    2 2mto be m because m is the combination exponent equivalent to the

    1individual exponents m and n. It is illustrated in the trigonometry of

    wt tFigure 1 that the same result would be realized if Sand were to

    1use their individual exponents nand m , respectively, because

    tcalculated water saturation is related to the resistivity level ofR

    wt t wt tand this remains unchanged. Again, the proof that (S ) = S ism2 n m1

    shown inAPPENDIX (B)(3).

    It can be seen that Eq.(1b) is a representation of Eq.(1a) when

    thydrocarbons are present, and can be used in the calculation of R .

    wt t 0 0 corrected When Sis 1.0, the calculated R becomes a corrected R , or R .These relationships will be revisited later.

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    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 , 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 2that as R is corrected to R ,

    0 0 corrected eR is corrected to R , both by the same factora. Just as and

    tare two intrinsic properties of the rock and exist simultaneously,

    w weR and R exist simultaneously and relate to the same exponent m. The

    mexponent describes an intrinsic property of the rock and producesthe parallelism seen in the Figure 2. As a result, the sloperepresented by mis 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 acoefficient 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

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    where ris resistance, Lis length, Ais the cross-sectional area of astraight electrically-conductive path of length L, and Ris thefamiliar resistivity. The term L/Ain the equation is similar to aformation factor and describes a fractional volume having length Landcross-sectional area Athat is 100 percent occupied by a single,homogeneous, electrically-conductive medium, either an electrolyte orsolid. This equation represents 100% efficiency in the conversion from

    resistivity to resistance, and the converse.

    Figure 3is 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 Lequal 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 apparentformation 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/= 1.0/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 mor m is 1.0 at the highest

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    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 mmight 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, used

    wt tin calculations and derivations, will involve the term S. Fis notintrinsic.

    To carry this demonstration one step further, in the same rock wheretF= 1.0/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 now becomes S and Fnow becomes 1.0/(S ) for

    wt tall values of water saturation and porosity. The fraction S now hasbecome the fractional cross-section of area in Figure 3for allelectrically-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 is the fractional cross-section area for allelectrically-conductive water paths at their highest efficiency, as in

    wt ta bundle of straight tubes, (S ) 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 nin Figure 1. As

    wt t wt ta result, 1.0/(S ) becomes 1.0/(S ) 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 ) (3b)m2

    as it was derived from the trigonometrics of Figure 1.

    The Formation Resistivity Factor is aptly named. Resistance rof the

    tinterstitial-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

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    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 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 pathway

    wt tresistance, the greatest efficiency at any given value of S occurs

    1 2when exponent mor m equals 1.0 and m approaches 1.0. And, the

    wtgreatest effectiveness in one unit of total volume occurs when both S

    t t wand are 100% or 1.0. When that happens, F= 1.0 and r= R= R . At

    tany given saturation, overall efficiency of water paths decreases as

    1 2decreases and/or as m , m or nincreases.

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

    THEmEXPONENTS

    The porosity exponent mis 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 mis 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

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    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 in

    solid matter, not related to pore geometry, that can influence rockresistivity. The electrical conductivity of such minerals must beaccommodated through the acoefficient in proportionalityrelationships in Eq.(1a) and (1b), or an equivalent.

    In practice, the mexponent usually has a default value of about 2.0.

    1In the laboratory, the minimum value for m(or m ) in homogeneousgranular 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 mwill 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 was

    further demonstrated that mhas an absolute minimum value of 1.0 at100% efficiency.

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

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

    10 0 corrected 10 we 10 tand -m = (log R - log R )/(log ) (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 larger

    warange of data. Turn to Figure 5. This figure is a plot of R vs

    t tClayiness. The assumption is made that , R , andClayiness are

    wa treliable. The value R is determined by dividing R by the formationfactor in Eq.(3a). In interactive computer methods, the plot isentered by assuming a reasonable value for the mexponent. 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 w Both 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 of

    wthe logging tool environment at the depth of interest. If the R fromthe plot differs from the known value, iterate by program subroutine

    w wbetween the known R and the derived R by incrementally changing minEq.(3a) until the two values agree. The value of mdetermined 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 mof Eq.(5a) does not agree with min Eq.(5b),other methods for singular values of mare based on Figure 2. Onemethod is based on the solution ofmin similar triangles, and another

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

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    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 or

    0 0 tcorrected values of R (or C) versus , as in Figure 2, values of msometimes can be derived for each specific set of log data or rock

    we 0 corrected sample. Individual values of R and R can continually changewith changing clay shale content. As a result, a reliable trendspecific to comparable rock samples might not be observed for munless

    ne tthe ratio of /, or other appropriate discriminator, is held nearlyconstant to ensure a nearly constant relationship between theproportions of mineral constituents as porosity changes. And,laboratory measurements formmust show repeatability in both mandthe 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 nRELATED TO EXPONENTm

    The saturation exponent nis 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 nis 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 nin water-wetrocks has been the same as for m,

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

    1other things remaining the same. Because nis 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 inAPPENDIX (D), based on Figure 6,why the actual

    1value of ncannot be lower than the actual value of m in the samesample, in situ or in the laboratory. Exponent ncan 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 nas 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 fraction

    tof oil causes the electrical interference to change, then R willchange, and this will result in a corresponding change in the value of

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    n.

    The maximum value for nin 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 noccurs where the presenceof hydrocarbon produces the least interference. This minimum should

    occur at the water saturation where the hydrocarbon saturation isirreducible. These statements suggest that there might be a causative

    rorelationship with nin a crossplot of the relative permeabilities (k ,

    rw ro o rw w wtk ) or of the mobilities (k / , k / )for oil and water vs S.

    The exaggerated influence due to the presence of oil will increase

    2 wt tboth the combination exponent m for the product S and the usual

    wtexponent nfor S. For any given porosity and any given oil

    2saturation, the exponents m and nwill increase with those propertiesof 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 of

    grain-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 niswith 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 aCOEFFICIENT

    Historically, the acoefficient 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 awith a constant value in the formationfactor. The formation factor is intrinsic to the rock at 100% watersaturation. The acoefficient is not an intrinsic property, but isdependent on variables not related to the electrically inert rock orto the physical geometries of its pores. The ais related neither tovolume nor shape imparted to the network of electrically-conductivewater paths. The acoefficient contributes nothing to the efficiencyof the conversion of water resistivity to resistance, or vice versa.In the model, the shift in resistivity corresponding to aalways isfound in the resistivity axis, and the shift varies from data to datawith depth. Coefficient aemerges as an inherent and inseparablefactor of the complex resistivity relationship, as in Eqs.(1a) and

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

    weIt can be seen in the graphics of Figure 2that R is the product of a

    w w we weand R , therefore aR =R . R varies from depth to depth, and so doescoefficient 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 acoefficient cannot be used as an

    independent parameter. If the ais used in single-water, single-porosity methods, it must be calculated properly. And that isdifficult to do when the correct value of avaries 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 ais associated with R , as in the equation developments of (1a)

    wand (1b), and as a multiplier or reduction factor for R , the acoefficient more readily can be recognized as being an indispensablefactor that accounts for those heterogeneities that produce additionalelectrical conductivity. Coefficient ais 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 acan 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 , the value of the compound coefficient

    wawould be reduced. These influences, all, can change R to an apparent

    we 0or pseudo value R with a corresponding change in R to

    0 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/())R = (1.0/())aR = (1.0/())R m m m

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

    1. The acoefficient is a required proportionality factor, related

    wto 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 for

    weconductive influences. Duplication occurs when both aand R are usedin the same water-saturation equation, or when aappears as anindependent variable in dual-water, dual-porosity methods. Anysaturation relationship involving resistivity can employ either

    wecoefficient aor R , but not both.

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    3. To prevent the use of a constant artificial and unwarranted

    correction factor.

    4. Although the formation factor is meaningless at 100% porosity,

    tthe value of Falways must be 1.0 when both porosity and water

    tsaturation in the Fequation 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 afor the usual shalysand:

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

    This relationship is shown for comparison purposes or informativepurposes only. It is not to be calculated and used independently in

    dual-water, dual-porosity methods because it already has beenweincorporated 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 of

    wt tboth the volume of water S and the inefficiency with whichelectrical 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 mand nof the expression (S) () . In then m

    10 wtmodel it is shown that log Sis the length of the projection along

    10 t 10 tthe X-axis between log and the intercept of slope nwith 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, when

    we 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 ) (3b)m2

    tAfter combining Eq.(2b)and (3b) when either the measured or actual R

    tis substituted for the calculated R , then

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

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    However, it was illustrated in Figure 1in triangle ACG, and in

    wt t wt tAPPENDIX (B)(3), that (S ) is equivalent to (S) () , therefore,m2 n m1

    Eq.(4a) resolves to

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

    t measuredThe R in Eq.(4b) must be corrected for environmental conditions

    and tool-measurement characteristics before water saturation iscalculated.

    It can be seen in Eqs.(1b) and (4a) that the acoefficient is variable

    we wewith depth and mineralization and is included as part of R . When Rhas 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 model

    are 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 nfor 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 nin 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 nis the most difficult of all the parameters toevaluate. If a valid value of oil saturation is known, or can bederived, the value of exponent ncan be estimated by substitution inEq.(4b) or (4c). When the actual values of mand nare known, or can be

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    derived, either or both can be important mappable parameters, and amathematical relationship between mand nnot 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 Salso

    weappears in the proportionality terms within the R equation, Eq.(1b), an

    wtalgebraic solution for Sin Eq.(4b) is not viable and is notconsidered. Probably the simplest and best method for all anticipatedinteger and non-integer values of nis 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 Sis varied, the individualcurves for the left and right sides of Eq.(4b) will converge and cross

    wtat the Svalue that will satisfy the equation.

    Figure 4shows a crossplot of example data to demonstrate theequivalence of the graphical solution to the iterations performed by a

    computer-program subroutine. The following input values are forillustration purposes only.

    wm = 2.17 R = 0.30

    wb n = 2.92 R = 0.08

    t t = 0.22 R = 20.00

    ne = 0.09

    weFor Eq.(4b), the equivalent water resistivity R has been evaluated by

    we eEq.(1b) after substitution for S has been made from the volumetricmaterial balance equation for water:

    wt t we e ne S =S +(1.0)

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

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

    wtproduce the same Sof 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 goes to zero the absolute minimum value for S in this

    neexample is 0.09, the value of . The mathematical minimum water

    wtsaturation Sthat can exist in this hypothetical reservoir is

    ne t wb /= 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.

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    wt weFor the conversion of Sto 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)= (1.0 - S)

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

    we t e wt S= 1.0- (/)(1.0 - S)

    weand Snow 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 Sin turn, both R and R must

    w wbbe known. In the event neither R nor R is known, these values mostoften can be estimated by interactive computer graphics from a crossplot

    wa waof R (or C) versus Clayiness(% clay) as shown in Figure 5, fromRansom(1995), where clayiness is estimated by appropriate clay-shale

    wa t tindicators. R is determined by dividing the corrected value of R by F

    t wt t t t wtwhere F =1.0/(S ) and Fhas resolved to 1.0/( ) because Salwaysm2 m1

    is 1.0 (for 100% water saturation) for this determination. The value of

    tR 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 be

    wz100% water-filled often describe a trend or curve identified as an Rtrend 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 clay

    wzand its distribution varies. Once the R trend has been established it

    wcan be extrapolated to 0% clay for R , and extrapolated to 100% clay for

    wb w wb wzR at in situ conditions. These values of R , R , and R have beenestimated 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 in

    w 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 the

    wz 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

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    w wb wz w for 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 mto vary the value of R from the plot inFigure 5until there is agreement. This is another means for the

    we wevaluation of R in terms of calculated clayiness. The resulting R and

    wb weR can be used in Eq.(1a) in the calculation of R in terms ofporosities, if desired.

    It is interesting to note, however, that from Figure 5alone, once an

    wzacceptable 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, evenR and R , all in terms of an estimated clayiness value. Thevalue of exponent nlies between its minimum value ofm, corrected or asestablished above, and its maximum value determined from irreduciblewater saturation. Furthermore, unlike Eq.(4b), once exponent nhas been

    wtestablished, Scan 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 inAPPENDIX (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 Indexfor 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 Archies relationships, whether new or old. Above, it was mentionedthat Archies 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

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    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

    Archies 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 throughwhich the electrical-survey current must flow. The basic electricalresistance equation is the key to understanding the FormationResistivity Factor.

    Archies 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 Archies 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.

    Archies 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 users 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.

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    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 cleanrocks, 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 saturation

    equation. Related to the electrical interference caused by thepresence of hydrocarbon in electrolyte-filled paths. A form of mafter hydrocarbon has displaced water.

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

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

    ne Is the noneffective porosity in clay shale and is directlyrelated to the fraction of clay shale present. A porosity volumethat 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 conductive

    wt t wtwater volume fraction always is S , where S1.0. In thispaper, 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 by

    t 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 resistivity

    when secondary conductive influences are present. In this paper,

    wthe product of coefficient aand R as it is incorporated in the

    weequation 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

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    we waequivalent to R and a special condition of R .

    0R The true resistivity of the uninvaded rock when water saturation is

    we1.0. 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 .

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

    wt t wtS The total conductive water volume fraction where S1.0. 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 RockResistivity 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

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    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 FallMeeting of the Petroleum Branch, AIME, held at Houston, TX.,October, 1-3.

    APPENDIX

    The model appearing in Figure 1incorporates Figure 2. The origin always

    we we w wis R . But, R is the same as R in clean rocks and different from R inheterogeneous rocks.

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

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

    e tavailable. Porosity values, and ,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 as

    we 0 corrected weinput, R is from Eq(1a), R from Eq.(2a) with R as input, allvalues are at 100% water saturation in this figure. Coefficient ais

    wshown 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 amay be calculated forcomparison purposes only.

    (B)Figure 1. Figure 1 illustrates the model that Archies 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 mor 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 mor n.

    wt t(1)On the X-axis, it is seen that S decreases to the right. A study

    t wtof the logarithmic scales will show, for example, that = 0.2 and S=

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

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    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 Archies water saturation relationship willemerge.

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

    2 10 10 tan = -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 ))

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

    t wt t F =1.0/(S ) . . .(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 /Rn

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

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

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    wt t we t (S ) =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 ) =log R - log R

    1adding the two equations that involve nand m , yields

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

    wt t we t S = R /Rn m1

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

    wt t wt tTherefore, (S ) = S m2 n m1

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

    tline representing nwill intercept the R level far to the right at H.The resulting calculated water saturation will be too low and

    unreasonable. When the corrected nvalue is used, the slopetrepresenting nwill intercept the level of R at some point G, which

    wtwill yield a reasonable value for Sdepending on the validity of n.Straight lines representing values of ncan rotate along the arc as

    wteither nor Svaries.

    wa(C)Figure 5. This figure is a plot of R vs Clayiness. It could have

    t 0 0 corrected been a plot of R vs Clayinesswhere R and R could have been

    w wz 0 corrected testimated in the same manner as R and R . The ratios R /R and

    wz wa waR /R would still be the same. The plot of R vs Clayinessyields moreinformation.

    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 6is 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 6it is shown that the same rock

    wt tcan contain the same volume of water, S , 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 mis uniform throughout, then,

    wt twhen the porosity is reduced to 0.06, S = 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 = 0.30 x 0.20 = 0.06.

    wt tThis is the total water volume at K. This common value of S is shown

    wt tat E at the base of Figure 6 on the logarithm scale for S .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

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    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 nis found to be or made to be < min the same sample of rock,most users will not recognize the impossibility of this condition.

    tWhen nis said to be < m, the predicted value of R always becomes less

    0 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 nto 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< mthe slope representing

    tthe aberrant nis so low that for whatever value of R exists, the slope

    trepresenting nwill 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 nthat 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.

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    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 an

    Honorary 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 of

    Elizabeth, Colorado.

    http://www.interpretivegroundwaterglossary.com./http://www.interpretivegroundwaterglossary.com./http://www.interpretivegroundwaterglossary.com./
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    Figure 1. A graphical representation of the model. In this diagram the

    tmodel illustrates oil-bearing sandstone. R varies with oil saturationand distribution. Exponent nvaries with the electrical interference

    wtcaused by the presence of oil regardless of saturation value. Sisdetermined by the intersection of the slope representing nwith

    tresistivity level R . See text for full explanation. Based on Ransom(1974,1995).

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    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).

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    wtFigure 4.A crossplot illustrating the iteration of Sfor the value of

    wtSthat satisfies both sides of Equation (4b). See the text fordiscussion.

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    waFigure 5.The crossplot of R versus Clayinesswhere the identified

    w wbtrend is extrapolated in both directions to evaluate R and R at 0% and100% clayiness, respectively. See text for discussion. From Ransom(1995), courtesy of John Wiley & Sons, Inc.

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    tFigure 6. This figure, similar to Figure 1, shows that when n< m, R at

    0K has a lower value than R at J for equivalent water volumes at allvalues of porosity and saturation. Any condition for nto be < miscontrary to the laws of physics and the fundamentals of resistivityanalysis. SeeAPPENDIX (D)for a detailed explanation.