The Saturation Exponents

Chapter 7, Archie's Equation and Water Saturation

Learning objectives

  • Read the saturation exponent nn as the slope of the resistivity index I=Rt/RoI = R_t/R_o against the water saturation on logarithmic axes, and say why a log cannot supply it
  • Say what moves nn, wettability above all, and what a rock of another nn does to a reading made with Archie's 2
  • Say which way a wrong nn and a wrong mm each push the water saturation, and what the two do together
  • Say where the saturation is most sensitive to each exponent: nn in the oil and not at all in water, mm in tight rock
  • Follow an exponent error through to the water saturation, the net pay and the hydrocarbon column of a well

The Slope of the Index

Section 6.1 wrote Archie's second law, Rt=Ro/Sw nR_t = R_o/S_w^{\,n}: a rock whose pores hold a fraction SwS_w of water reads Swβˆ’nS_w^{-n} times what it would read full of brine. That ratio is the resistivity index, I=Rt/Ro=Swβˆ’nI = R_t/R_o = S_w^{-n}, and on logarithmic axes it is a straight line through the point (100 % water, 1) whose slope is βˆ’n-n: the saturation exponent. Sections 6.2 and 6.6 dealt with the other exponent, mm, which a log can partly check against a water leg. Nothing in a log checks nn. PetroWiki says so: "Although cementation exponents can be determined from log analysis, saturation exponents cannot and, therefore, require external information from core data." The laboratory measures it on core plugs held at a series of brine saturations, and "On the basis of experimental data, Archie recommended that n = 2.0, and this value is still widely used when no experimental data are available" (PetroWiki, Water saturation determination).

At 5890Β ft, in the upper oil sand, oil-bearing, the porosity is 23.2Β %. The poreshold oil here, at a built water saturation of 18.9Β %. A rock with a saturationexponent of 2.00 would log 20.9Β ohm.m against its wet-rock resistivity of0.745Β ohm.m, a resistivity index of 28.1, which is Ogbon-1's own log. With your n of2.00 and m of 2.00 Archie reads 18.9Β % water, the saturation the well was builtwith. Over the well this pair reads the built saturation: 23.0Β % on average in thenet pay, 277.0Β ft of net pay and a column of 38.1Β ft.(a) Index against saturation102030501001101001000Water saturation (%)both n 2.0018.9 %(b) Saturation read02550751005800590060006100620063006400Water saturation (%)Depth (ft)(c) Rise per 0.2051015205800590060006100620063006400Rise in saturation (points)Key: in (a) the dashed line is the rock's own resistivity index against the water saturation and thesolid line the one you read with, so far the same; the grey dots are the oil-bearing half feet ofOgbon-1 as a core would measure them, one every 5 ft; the ring is the rock here and the dot what youcompute from the log. In (b) the line is the water saturation built in clean rock, the dots thesaturation you read and the dashed line the book's 50 % cutoff. In (c) the rust line is how far thereading rises for a saturation exponent 0.2 higher and the blue line for a cementation exponent 0.2higher, in points of saturation. The shaded beds are the four reservoir zones, and the dotted line isthe depth read.Log would read 20.9 ohm.m (a rock of n 2.00; the well's own 20.9Β ohm.m). Water saturation read 18.9 %(built 18.9Β %). Pay water saturation 23.0 % (net pay average; built 23.0Β %). Net pay 277.0 ft (built277.0Β ft). Hydrocarbon column 38.1 ft (built 38.1Β ft).

Ogbon-1's Own Line

Ogbon-1 is synthetic and was built with n=2n = 2 in every clean bed, so it offers what a real well cannot: the resistivity index of its oil-bearing rock, one value at each of 599 half feet, against the water saturation it was built with. Plate (a) sets them on the plot. A straight line fitted to them on logarithmic axes has a slope of βˆ’2.00-2.00 and a height of 1.00 at 100 % water, and the line the figure draws for the rock is that line. At 5890 ft, in the upper oil sand, the porosity is 23.2 % and the water saturation the well was built with is 18.9 %. The wet-rock resistivity is 0.745 ohm.m and the log reads 20.9 ohm.m, an index of 28.1, which is one over the square of 0.189.

What Moves n

The slope is not a property of the grains; it follows how the water is arranged in the pores as it drains, and that depends on which fluid wets the rock. Abdassah and others (1996), modelling twenty thin sections of limestone and sandstone with a fractal pore model, found that "The exponent consistently increases as the wetting condition is shifted from strongly water-wet toward oil-wet", and gave ranges of 1.6 to 2.0 for strongly water-wet, 1.9 to 2.8 for mixed-wet and 3.4 to 5.0 for strongly oil-wet systems. Saadat and others (2024) measured carbonate plugs by the porous plate method and found exponents from 0.75 to 3.8; they report that "pore connectivity and wettability are the main factors affecting the saturation exponent" and that "oil-wetting conditions lead to an increase in the saturation exponent". They also recall Anderson's (1986) review of earlier measurements, which showed that "the saturation exponent for oil-wet systems increases dramatically when the brine saturation is below a certain value", so the index is then not a straight line on the plot; PetroWiki says: "A straight line-fit is usually used, but curved line-fits can be considered where necessary." The figure keeps the straight line.

Plate (a) lets you ask what Ogbon-1 would have logged had its rock another wettability. The figure keeps the saturations the well was built with, the capillary column of Chapter 11, and changes only the exponent of the rock; that is the figure's own simplification, since in a real rock the wettability also changes where the water sits. A mixed-wet rock with an exponent of 2.50 would log 48.2 ohm.m in the upper oil sand, against the well's 20.9 ohm.m, an index of 64.7 against 28.1. This is the reverse of Section 7.2's question: there the rock kept Archie's 2 and the lines were drawn with 2.5, which reads too much water; here the rock is given 2.5 and is read with 2, which reads too little.

Which Way Each Error Pushes

The analyst brings an exponent nun_u and a cementation exponent mum_u to a log that a rock with its own nn and mm made, and Archie reads

Sw,read=(ϕ mβˆ’mu Sw n)1/nu.S_{w,\text{read}} = \left(\phi^{\,m-m_u}\,S_w^{\,n}\right)^{1/n_u}.

With the right mm this is Sw n/nuS_w^{\,n/n_u}. The saturation is under 1, so a larger nun_u makes a larger reading and a smaller one a smaller reading. On the right rock an nun_u of 2.20 reads 22.0 % water at 5890 ft against the 18.9 % built, 3.1 points too high, and an nun_u of 1.80 reads 15.7 %, 3.2 points too low. Where the rock is not water-wet the usual error is the second: the mixed-wet rock of 2.50 read with Archie's 2 gives 12.4 % at 5890 ft, 6.4 points too low, and over the net pay an average of 16.1 % against the 23.0 % built. The error flatters the well, and it does so by more the more oil-wet the rock: at 3.40, the bottom of the oil-wet range, the same reading gives 5.9 % at 5890 ft and 8.6 % over the net pay. An mum_u off the built value pushes the same way: above it reads more water, below it less, which Section 6.6 followed through the well.

The net pay hides most of it. An nun_u of 3.00 on the right rock raises the pay-average saturation from 23.0 % to 37.3 %, but the net pay falls only from 277.0 ft to 256.0 ft, 21.0 ft or 7.6 %, because most of the oil sand reads well under the book's 50 % cutoff. The hydrocarbon column, the sum of the porosity times one minus the saturation over the net pay, falls from 38.1 ft to 28.6 ft, a quarter of it. A wrong nn costs volume far more than it costs pay, and an nun_u of 2.20 costs 3.0 ft of net pay where an mum_u of 2.20 costs 6.5 ft.

Where the Saturation Is Most Sensitive

Differentiating Archie's reading gives how far the reading moves, in proportion, for one unit of each exponent:

βˆ‚ln⁑Swβˆ‚n=βˆ’ln⁑Swn,βˆ‚ln⁑Swβˆ‚m=βˆ’ln⁑ϕn.\dfrac{\partial \ln S_w}{\partial n} = -\dfrac{\ln S_w}{n}, \qquad \dfrac{\partial \ln S_w}{\partial m} = -\dfrac{\ln \phi}{n}.

The first grows as the saturation falls, the second as the porosity falls. At 5890 ft, where the saturation is 18.9 % and the porosity 23.2 %, they are 0.83 and 0.73 per unit, so the two exponents move the reading about alike: a rise of 0.20 adds 3.1 saturation points through nn and 3.0 through mm. In the tight limestone at 6080 ft, where the saturation is 21.9 % and the porosity 8.5 %, they are 0.76 and 1.23 per unit, and plate (c) shows the difference: the same rise adds 3.2 points through nn and 6.1 through mm. In absolute points the rise for a given change in nn is βˆ’Swln⁑Sw/n-S_w \ln S_w/n times that change, which peaks at Sw=1/eS_w = 1/e, 36.8 %, and the main sand's pay, at 34.5 %, sits near it. In water, where the saturation is 100 %, nn drops out: the index is 1 for every exponent, and a mixed-wet rock read with Archie's 2 reads 100 % at 6260 ft, as the right rock does. A wrong mm spoils that, as Section 6.6 showed: an mum_u of 1.80 reads the water leg 87.2 % water, and then nn matters a little, 1.1 points for a rise of 0.20, against 12.8 points for mm. In the transition zone, at 6200 ft with a saturation of 62.7 %, the two are 0.23 and 0.68 per unit, and the rise is 2.7 points through nn against 9.1 through mm.

Two Errors at Once

Both exponents appear in the reading, and an error in one can hide an error in the other. An nun_u of 2.20, too high, with an mum_u of 1.77, too low, reads 18.8 % at 5890 ft against 18.9 % built: for that half foot ϕ mβˆ’mu\phi^{\,m-m_u} has cancelled Sw nβˆ’nuS_w^{\,n-n_u} exactly. It is only one half foot. In the tight limestone at 6080 ft the same pair reads 19.4 % against 21.9 % built, 2.5 points too low, and the pay-average reading is 21.9 % against 23.0 %. The cancellation holds where the porosity and the saturation are what they were at 5890 ft, and nowhere else; a bed at one porosity and one saturation fixes one number, as Section 6.2 found for the pair of aa and mm. A real well offers no truth to check against. Ogbon-1 was built with n=m=2n = m = 2, so the right exponents return its saturations exactly, but the exponents of a real rock come from core, and Section 7.5 turns to the water resistivity, the other input a log analyst has to bring.

References

  • Archie, G. E. (1942). The electrical resistivity log as an aid in determining some reservoir characteristics. Transactions of the AIME. SPE-942054-G.
  • PetroWiki. Water saturation determination. petrowiki.spe.org/Water_saturation_determination.
  • Abdassah, D., Permadi, P., Sumantri, Y. and Sumantri, R. (1996). Fractal modeling of thin-sections to determine saturation exponent for various wetting conditions. Society of Core Analysts, SCA 9622.
  • Saadat, K., Rahimpour-Bonab, H., Tavakoli, V. et al. (2024). Experimental investigation and prediction of saturation exponent in carbonate rocks: the significance of rock-fluid properties. Journal of Petroleum Exploration and Production Technology, 14, 149-163. doi:10.1007/s13202-023-01714-2.

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