Robert C. Ransom

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What are Archie’s Basic Relationships

The Graphical Model

What is Meant by the Plot of Rt versus Swtϕt

Summary of Equations

Parallel Resistivity Equations Used in Resistivity Interpretation

What is the Formation Resistivity Factor

The m Exponents

How is Exponent n Related to Exponent m

The a Coefficient

The Saturation Evaluation

Challenging Well-Log Examples

Observations and Conclusions from Figure 10 about Exponent n

Are There Limitations to Archie's Relationships Developed in this Model?




Symbols Defined



All Figures

About the Author

Table of Retrievable Contents:


PREFACE. The model appearing in Figure 1 incorporates Figure 2. The origin always is Rwe , the apparent or equivalent resistivity of formation water. But, Rwe is the same as Rw in clean rocks and different from Rw in heterogeneous rocks, and in the presence of hydrocarbon saturation. The influence of hydrocarbon saturation on Rwe can be observed in Equation (1b) found under PARALLEL RESISTIVITY EQUATIONS USED . . .

The Y-axis is Rt , the resistivity of rock containing water. The X- axis is Swtϕt , the bulk volume of water, consisting of two parts, ϕt and Swt . It must be recognized that ϕt and Swt have no units, but are fractions and in themselves are not volumes. As fractions, neither ϕt nor Swt has resistivity because neither is dimensional. Only bulk volume water can have volume and, therefore, can have resistivity. This could be confusing, because the fraction that is bulk volume water in literature must be multiplied by a dimensional conversion factor that has a value of 1.0 and appropriate three-dimensional units. Now being dimensional, the bulk volume water can have resistivity.

In the discussion of the Y-axis, the basic resistance-resistivity equation must be introduced. That equation is:               resistancewater              =        (L/A)              ×     resistivitywater

Dimensionally,                ohms           =           meters/meters2    ×    ohm meters2/meter

                                         (water)                                 (F)                              (water)

The expression (L/A) is a formation resistivity factor in its simplest form. It converts resistivity to resistance, and resistance to resistivity. So, how does resistance of water become resistivity of rock?

In Figure 1, R0 and Rt are resistances of water per unit length, and at the same time resistivities of the unit volume of rock that contains that water. In inert, electrically non-conductive rock, water is contained in the pore spaces of the rock. The water volume occupies pore paths and creates electrically conductive paths that exhibit various levels of electrically conductive efficiency. These paths can have various degrees of resistance as the dimensions of (L/A) are varied in the resistance-resistivity equation above. However, the resistance of the water over the length of rock in a unit volume of insulating rock constitutes the resistivity exhibited by that rock. Numerically, the resistance of water is the same as the resistivity of rock.

In Figure 1, the resistance of water over a unit volume of rock is the resistivity of the rock. The Y-axis in Figure 1 is resistivity of rock except at 100% porosity where it is the resistivity of water.

(A) Figure 2. This figure shows the function of coefficient a. In the section above related to coefficient a an equation was written showing all the variables used in the calculation of a. One of the variables was water saturation in the effective pore space. Why is that necessary?

The presence of hydrocarbon decreases the volume of water in the effective pore space. This changes the proportionality of water volume with resistivity Rw relative to the volume of pseudo water with resistivity Rwb that is associated with clay shales. In a parallel conductivity relationship this moves Rwe closer to the value of Rwb as can be seen in Figure 2.

Input value Rw is from the best known source or from Figure 5 after clayiness has been determined by the best clay indicators available. Porosity values, ϕe and ϕt , are determined by traditional methods, or from Ransom (1977, 1995), after employing the appropriate matrix values for the host rock.

Coefficient a is shown to be a multiplier of Rw but is never calculated independently for use in an interpretation program when it is an integral part of Rwe , as seen in Eqs. (1a) and (1b). In dual-water dual-porosity methods, coefficient a may be calculated only as a matter of interest, but must never appear in a saturation equation with Rwe .

(B) Figure 1 illustrates the model that Archie’s relationships obey. It is intended for informative and illustrative use, only. It is not drawn to scale. The inclinations of the straight lines representing slopes are determined by trigonometric tangents. The value of a specific trigonometric tangent is the value of the specific m or n. The steeper is the slope, the greater is the interference and resistance to electrical-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.

(1) On the X - axis, it is seen that Swtϕt decreases to the right. A study of the logarithmic scales will show, for example, that ϕt = 0.2 and Swt = 0.3 ; and, as a result, their product Swtϕt = 0.06.

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

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

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 will be negative. Therefore, on the log-log plot such as Figure 1,

(tangent of acute angle) ( log10 ( side adjacent ) ) = log10 ( side opposite )

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

Trigonometry was created for solving problems. Express these trigonometric functions in equation form and the Formation Resistivity Factor and an improved Archie’s water saturation relationship will emerge.

In Figure 1, the tangent of the acute angle β of the right triangle AEG is represented by

tan β = m2 = log10 ( EG ) / ( log10 1 - log10 ( AE ) )

-m2 ( log10 ( AE ) ) = log10 ( EG ) = log10 Ft

Ft = 1.0 / ( AE ) m2

In equation form, this is

m2 = log10 Ft / ( log10 1 - log10 ( Swtϕt  ) )

-m2 ( log10 ( Swtϕt ) ) = log10 Ft

Ft = 1.0 / ( Swtϕt )m2                                                            . . . (3b)

In Figure 1, the tangent of the acute angle ϒ of the right triangle CDG is represented by

tan ϒ = n = log10 ( DG ) / ( log10 1 - log10 ( CD ) )

-n ( log10 ( CD ) ) = log10 ( DG ) = ( log10 Rt - log10 R0  corrected )

( CD )n = R0  corrected / Rt

In equation form, this is

n = ( log10 Rt - log10 R0 corrected ) / ( log10 1 - log10 Swt )

-n ( log10 Swt ) = ( log10 Rt - log10 R0  corrected )

( Swt )n = R0  corrected / Rt                                                      . . . (4b)

(3) ( Swtϕt )m2 has the same function as and is equal to ( Swt )n ( ϕt )m1 .

In triangle AEG,         -m2 ( log10 ( Swtϕt ) ) = log10 Rt - log10 Rwe

( Swtϕt )m2 = Rwe / Rt

In triangle CDG,       -n ( log10 Swt ) = log10 Rt - log10 R0  corrected

In triangle ABC,       -m1 ( log10 ϕt ) = log10 R0  corrected - log10 Rwe

adding the two equations that involve n and m1 , yields

-n ( log10 Swt ) + ( -m1 ( log10 ϕt ) ) = log10 Rt - log10 Rwe

Swtnϕt m1 = Rwe / Rt

But,                             ( Swtϕt )m2 = Rwe / Rt , from triangle AEG above.

Therefore,               ( Swtϕt ) m2 = ( Swt )n ( ϕt )m1                                              . . . same as (3c)

This equivalence also was proved graphically in Eq. (8) on page 7 in Ransom (1974).

(4) In addition, in partially oil-wet or oil-wet rocks, the assumption that n = m can lead to calculated water saturations that are too low. In Figure 1, it can be seen that when the default value of n equals m the line representing n will intercept the Rt level far to the right at H. The resulting calculated water saturation will be too low and might even be “unreasonable”. When the corrected n value is used, the slope representing n will intercept the level of Rt at point G somewhere near point H or to the left of H, which will yield a “reasonable” value for Swt depending on the value of n. Straight lines representing values of n can rotate along the arc δ as either n or Swt varies. The slopes representing n should intercept the corrected value of Rt at calculated water saturations within the saturation range between irreducible water and irreducible hydrocarbon.

(5) In oil-wet or partially oil-wet rocks the presence of oil will increase the interference to the flow of electrical current, causing the value of the saturation exponent n to increase. In Figure 1, it can be seen that as the value of the saturation exponent n increases, the slope of n increases. In Figure 1, under the condition that there will be no increase in Rt , as the slope of n increases, the calculated water saturation will be seen to increase. In oil-wet rock, a film of viscous adhesive oil coats the walls of the pores and pore throats. The value of n is related to the degree of electrical interference caused by the adhesive oil, and that degree of interference can be caused by adhesive oil films whether or not the oil saturation and oil distribution is sufficient for oil production to take place. Figure 1 shows that the saturation exponent n is the resistivity gradient that Rt employs between R0  corrected and Rt relative to changes in the saturation distribution of oil, not to the value of oil saturation. It is related to the value of oil saturation only through Rt. The saturation exponent n is not to be confused with the saturation value of oil. Figure 1 suggests that with no increase in Rt and with varying degrees of effectiveness of the coating of pore walls and pore throats with adhesive, viscous oil, this condition can allow a high water saturation to exist with the result that no oil will be produced, or oil will be produced with a high water cut. If oil is to be produced with a low water cut or with zero water cut then the oil must reside in a continuous phase that allows the oil to move and in such quantity and distribution that it causes discontinuity in the water phase. If the presence of oil resides in a continuous phase in oil-wet rock, this condition will present further increased interference to the flow of electrical survey current with the result that the value of Rt will increase to uncommonly high resistivity levels.

It might be inferred from Figure 1 that the ideal condition for oil production to take place in oil-wet rocks would be for the value of n to be as low as possible and the value of Rt to be high. Under this combination the slope of n will intercept the level of Rt farther to the right on the X-axis. Calculated water saturation will be lower and oil might be produced with a reduced water cut or no water cut. Where are these conditions likely to be found? Examples might be in rock containing relatively large, well sorted grains or in rock with well interconnected dissolution porosity. These conditions will be found in good quality reservoir rock where the value of porosity exponent m is low.

(C) Figure 5. This figure is a plot of Rwa vs Clayiness (% clay). It could have been a plot of Rt vs Clayiness. The plot of Rwa vs Clayiness yields more information relative to secondary constituents. The plot also has uses in focusing attention on data and depths where unexpected events or mineralization occurs.

When calculating clayiness for this plot, clayiness should be determined from the best clay-shale-responsive indicators available, but never from an 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 is not intended to be a working graphical procedure, it has been exaggerated to illustrate detail, and it is intended only to be informative and explanatory. In Figure 6 it is shown that the same rock can contain the same volume fraction of water, Swtϕt , whether or not oil or gas is present. In Figure 6, the example uses the same porosity and saturation values as seen depicted in Figure 1. In this example, m is uniform throughout the host rock. If water saturation is 100% when the porosity is reduced to 0.06, then Swtϕt = 1.0 x 0.06 = 0.06. This is the value of the total water volume as a fraction at J. The same volume of water can be observed in the same rock when the porosity is not reduced, but oil or gas displaces 70% of the water volume: Swtϕt = 0.30 x 0.20 = 0.06. This is the total water volume as a fraction at K. This common value of Swtϕt is shown at E at the base of Figure 6 on the logarithm scale for Swtϕt .

In Figure 6, where the example rock shows a water volume fraction of 0.06, it can be seen that when n < m, R0 at resistivity level J will be greater than Rt at resistivity level K, for all values of porosity and all values of saturation. In real life this cannot happen. But, when the value 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.

When n is said to have a lower value than the host value of m, the associated values of Rt always becomes less than R0 at equivalent water volumes in the same rock. And that associated value is profoundly incorrect. For Rt at K to have a lower value than R0 at J, the migration of oil or gas into the interstitial water volume must make the electrical paths through the remaining water volume more efficient. This enhancement of the electrical conductivity of the water paths through the interstices by the presence of oil, must be done with no change in salinity. Hypothetically, for oil or gas to migrate into the undisturbed effective pore space and cause the value of n to be < m, the presence of oil or gas must cause the electrically conductive paths of water to become more efficient. This, the presence of a continuous phase of oil or gas in water-wet rock cannot do. In the model in this paper, which is faithful to laws of physics, when all other things remain unchanged, it is shown that it is impossible for the actual value of Rt to have a lower value than the extrapolated value of R0 for the same quantity of water in the same sample. But, can there be an exception to this rule? Probably not. To do so would be in conflict with physics. Any occupation of pore space by oil decreases the electrically conductive volume of water and, therefore, increases the resisitivity gradient.

When n < m the slope representing the n becomes so low that for whatever value of Rt exists, the slope representing an aberrant n will intercept that level of Rt far to the right at an artificially low water saturation value on the right-facing Figure 1 and Figure 6. The consequence is that the calculated water saturation will be too low and the hydrocarbon saturation will be over estimated indiscriminately in producible and non-producible hydrocarbon-bearing beds alike. This will be 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 the migration of oil or gas into rock will cause more electrical interference than water alone. That denial contradicts basic physics and violates the fundamental principles of resistivity analysis.

How the presence of oil can make water more electrically conductive will not be explained in this paper. For explanantions why values of n lower than values of m have been measured, the reader will have to look elsewhere.



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