Robert C. Ransom
What are Archie’s Basic Relationships
What is Meant by the Plot of Rt versus Swtϕt
Parallel Resistivity Equations Used in Resistivity Interpretation
What is the Formation Resistivity Factor
Table of Retrievable Contents:
APPENDIX
The model appearing in Figure 1 incorporates Figure 2. The origin always is Rwe . But, Rwe is the same as Rw in clean rocks and different from Rw in heterogeneous rocks.
(A) Figure 2. In this figure are five variables relative to ϕe and ϕt . They are: Rw , R0 , Rwe , R0 corrected , and a . Input variables to the figure are Rw and ϕe and ϕt . 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. Variable R0 is from Eq.(2a) with Rw as input, Rwe is from Eq(1a), R0 corrected from Eq.(2a) with Rwe as input, all values are at 100% water saturation in this figure. Coefficient a is shown as 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). Coefficient a may be calculated for comparison purposes only.
(B) Figure 1. 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 is 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 ( AE )
-m2 ( log10 ( AE ) ) = log10 ( EG ) = log10 Ft
Ft = 1.0 / ( AE ) m2
In equation form, this is
-m2 = log10 Ft / ( 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 ( 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 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 above.
Therefore, ( Swtϕt ) m2 = (Swt )n ( ϕt )m1
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 “unreasonable”. When the corrected n value is used, the slope representing n will intercept the level of Rt at some point G, 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.
(C) Figure 5. This figure is a plot of Rwa vs Clayiness (% clay). It could have been a plot of Rt vs Clayiness where R0 and R0 corrected could have been estimated in exactly the same manner as Rw and Rwz . For estimating Swt, the ratios R0 corrected / Rt and Rwz / Rwa are equal. The plot of Rwa vs Clayiness yields more information relative to secondary constituents.
Assuming that the resistivity values are reliable, the most critical variable in this plot is Clayiness, which should be determined from the best clay-shale-responsive indicators available.
When calculating clayiness for this plot, clayiness should be determined from the best clay-shale-responsive indicators available, but not 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, if water saturation is 100%, porosity is 20% and m is uniform throughout, then, when the porosity is reduced to 0.06, 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 predicted value of Rt always becomes less than R0 at equivalent water volumes. 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, 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 by becoming more tube-like or less tortuous by whatever means. This, the presence of a continuous phase of oil or gas in water-wet rock cannot do. In this model, 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 actual value of R0 for the same quantity of water in the same sample. But, can there be an exception to this rule? It depends on whether or not the oil can occupy pores in a form where the interface with water is smoother and less irregular than the equivalent volume of host rock exhibiting slope m. See APPENDIX (E) for further discussion.
When n < m the slope representing the n is so low that for whatever value of Rt exists, the slope representingan aberrant n will intercept that level of Rt far to the right on Figures 1 and 6 at an artificially low water saturation value. 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.
(E) A question often posed is, "Why cannot the value of exponent n be lower than the value of exponent m?" The main reason it is most unlikely was explained in APPENDIX (D) immediately above. That explanation is virtually universal. Can there be an exception? The answer to that question lies with the saturation form that the oil phase takes. Can oil reside in a saturation form that presents less electrical interference than an equivalent volume of rock exhibiting slope m? Under ideal conditions there is a possibility for that condition to exist. A theory will be proposed below.
Our discussion applies only to water-wet rocks where the surface of all grains or particles will be covered with a film of water. In this case, water or oil can exist in three different forms of saturation. These forms, from Levorsen (1967), can be summarized as:
1. Funicular saturation: Related to the nonwetting phase when the nonwetting phase exists as a continuous web throughout the interstices. In this case the nonwetting phase is oil, and if it is mobile it can be produced. Water might or might not exist at irreducible saturation.
2. Pendular saturation: Related to the wetting phase when the wetting phase is continuous throughout the interstices. It is possible that either the wetting phase (water) or the nonwetting phase (oil), or both can be produced. Oil saturation might or might not be at irreducible levels.
3. Insular saturation: Related to the nonwetting phase when it exists as discontinuous or isolated spherical micro-globules throughout the interstices. The wetting phase can be produced, but the insular globules of oil would exist at irreducible saturation and probably could not be produced unless the globules could be distorted to pass through pore throats and be reunited to form a continuous phase. How this can be done without damage to the formation is problematic.
In the proposed theory where we explore the conditions that might make it possible for n to have a value lower than the m of the host rock, we consider the condition where oil is at irreducible saturation in a water-wet rock. The oil must reside as insular globules. The interfacial tension between oil and water must cause the insular oil to be drawn nearly into a sphere with a surface that is significantly smoother and rounder than the irregular interface between water and host rock. This would reduce electrical interference and improve the electrical efficiency of the water path for the volume of water present. Can that happen? Yes, it can happen under the conditions that are theorized here. But, there is another condition: The value of m for the rock must be relatively high for the presence of a small saturation of oil globules to have a significant influence.
Recall that the exponent m is related to the efficiency for a confined electrolyte to conduct electricity, and in unfractured rock samples the highest efficiency has been found where solid, inert, insulating grains take on a spherical shape and m has a value of about 1.3. Insulating spherical globules of oil should present the same interference to electrical current flow as inert insulating spherical grains of rock and, therefore, should demonstrate the same value for m of about 1.3.
Can the value of the n exponent be lower than the host m where oil exists as insular, smooth, spherical micro-globules at irreducible oil saturation? If the host rock has an m value significantly higher than 1.3, and if there are enough undistorted spherical globules of oil at the existing irreducible oil saturation, then the m value related to the total electrical interference within the sample might be decreased. Therefore, because water saturation is < 1.0, the value of the exponent n as seen by the water path might be lower than the host rock value of m. Where the walls confining the pores are smooth and regular and the value of m for the host rock is correspondingly lower, it is unlikely that the presence of spherical micro-globules of oil will make a recognizable difference. Can the value of the n exponent be lower than m in a producible oil zone where the oil phase is continuous? This is unlikely because a continuous phase of oil is subject to all of the factors that cause it to increase its electrical interference and to cause it to decrease the electrical efficiency of the water paths throughout the interstices. Here, n will increase.
The scenario might play out as follows: First, mobile oil in a continuous nonwetting phase migrates into the water-filled pores and causes the value of n to increase over m. Later, as the oil departs, the value of n decreases. After the oil has completed its emigration as a continuous phase, a trail of insular globules is left in the pores as a remnant of the oil that now resides somewhere else, and the value of n has been decreased further. Here, if the value of the host m is high relative to 1.3, and the residual oil saturation is significant, the value of n might become lower than the m. If there is any oil staining of the rock, it could add another dimension to electrical interference. Grain surface staining can alter the value of n in this scenario depending on its influence on the electrical efficiency of the water paths.
The consequence of an exponent n that has been reduced in this manner, although the n would be valid, is: For whatever value of Rt exists, the calculated water saturation will be decreased and the calculated residual oil saturation will be increased as can be seen in Figure 6 . Unfortunately, this value of n does not represent the degree of interference where the oil phase is continuous. A corollary from this is that the n determined from a sample where oil is at irreducible saturation, and oil exists as insular globules, probably should not be used where oil is mobile.
A CLARIFYING CONCEPT OF ARCHIE'S RESISTIVITY RELATIONSHIPS AND PARAMETERS.
A MODEL AND DISCUSSION
Within the text, clicking on the bold terms FIGURE (*) and APPENDIX (*) will take you to the specific page of interest. Use your browser's back button to return to your previous page.
