Archie's equation assumes that the only thing in the rock that conducts electricity is the brine in the pores. In a clean sand that is close enough to true. In a shaly sand it is not: clay minerals carry exchangeable cations on their surfaces, and those cations conduct. Archie reads that extra conductivity as extra water, so in a shaly sand it overstates water saturation, and the interval it overstates most is the low-resistivity pay you most needed to see.
Every shaly-sand model fixes this by adding a clay conductivity term. They differ in two ways: how they describe the clay's contribution, and which porosity they are written on. Those two choices matter more than the details of any one equation, and the worked example below shows why.
Two families
Effective-porosity models treat the shale as a separate conductor in parallel with the clean sand. They take the shale volume Vsh and the resistivity of a nearby shale, Rsh, and use effective porosity, the pore space outside the clay. Simandoux (1963), its modified forms, and the Indonesia equation of Poupon and Leveaux (1971) belong here. Their inputs come straight from the logs, which is their practical attraction.
Total-porosity models treat the clay's conductivity as a property of the water close to the clay surfaces. Waxman and Smits (1968) express it through Qv, the clay's cation exchange capacity expressed per unit of pore volume, and a counterion conductance B. The dual water model of Clavier, Coates and Dumanoir (1984) splits the pore water into clay-bound water and free water, each with its own conductivity. Both use total porosity, and both give a total water saturation that includes the bound water.
One rock, five answers
Take a sand with these properties. They are illustrative, chosen to be typical rather than taken from any well:
Rt = 10 ohm·m deep resistivity
Rw = 0.05 ohm·m formation water at reservoir temperature (Cw = 20 S/m)
φe = 0.20 effective porosity
Vsh = 0.25 shale volume
Rsh = 2 ohm·m resistivity of the adjacent shale
φsh = 0.16 total porosity of that shale
a = 1, m = 2, n = 2
The shale's porosity is clay-bound, so total porosity is φt = φe + Vsh·φsh = 0.20 + 0.25 × 0.16 = 0.24, and the bound-water fraction of the total pore space is Swb = 0.04 / 0.24 = 0.167.
Archie
Sw = ( a·Rw / (φe^m · Rt) )^(1/n)
= ( 0.05 / (0.04 × 10) )^0.5 = 0.125^0.5 = 0.354
Simandoux
In the form most software implements:
1/Rt = φe^m · Sw^n / (a·Rw) + Vsh · Sw / Rsh
0.1 = (0.04 / 0.05) · Sw² + (0.25 / 2) · Sw
0.1 = 0.8 · Sw² + 0.125 · Sw
A quadratic in Sw, with the positive root:
Sw = ( −0.125 + √(0.125² + 4 × 0.8 × 0.1) ) / (2 × 0.8)
= ( −0.125 + √0.335625 ) / 1.6
= ( −0.125 + 0.5793 ) / 1.6 = 0.284
Indonesia
1/√Rt = [ Vsh^(1 − Vsh/2) / √Rsh + φe^(m/2) / √(a·Rw) ] · Sw^(n/2)
Vsh^(1 − Vsh/2) = 0.25^0.875 = 0.2973; 0.2973 / √2 = 0.2102
φe^(m/2) = 0.20; 0.20 / √0.05 = 0.8944
bracket = 0.2102 + 0.8944 = 1.1046
1/√Rt = 1/√10 = 0.3162
Sw = 0.3162 / 1.1046 = 0.286 (n = 2, so Sw^(n/2) = Sw)
Waxman–Smits
Waxman–Smits needs Qv and B, which really should come from core. For the example take Qv = 0.3 meq/cm³ and B = 4.0 (S/m) per meq/cm³, so B·Qv = 1.2 S/m. Its exponents m* and n* are set to 2 here so the comparison is like for like; more on that below.
Ct = φt^m* · Swt^n* · ( Cw + B·Qv / Swt ) / a
0.1 = 0.0576 · ( 20 · Swt² + 1.2 · Swt )
20 · Swt² + 1.2 · Swt − 1.7361 = 0
Swt = ( −1.2 + √(1.44 + 4 × 20 × 1.7361) ) / 40
= ( −1.2 + √140.33 ) / 40
= ( −1.2 + 11.846 ) / 40 = 0.266
Dual water
Dual water needs the bound-water conductivity Cwb. A practical way to get it is from the shale itself: if the shale's pore space is all bound water, then 1/Rsh = φsh^m · Cwb, so
Cwb = 1 / (Rsh · φsh^m) = 1 / (2 × 0.0256) = 19.53 S/m
Ct = φt^m · Swt^n · [ Cw + (Swb / Swt) · (Cwb − Cw) ] / a
0.1 = 0.0576 · [ 20 · Swt² + 0.167 × (19.53 − 20) · Swt ]
0.1 = 0.0576 · ( 20 · Swt² − 0.078 · Swt )
Swt = ( 0.078 + √(0.078² + 4 × 20 × 1.7361) ) / 40
= ( 0.078 + 11.785 ) / 40 = 0.297
Compare hydrocarbon volume, not Sw
The five saturations, 0.354, 0.284, 0.286, 0.266 and 0.297, cannot be compared directly, because the first three are fractions of effective porosity and the last two are fractions of total porosity. The quantity that matters for volumetrics is the hydrocarbon pore volume per unit rock volume, φ·(1 − Sw), and that is basis-independent:
Archie 0.20 × (1 − 0.354) = 0.129
Simandoux 0.20 × (1 − 0.284) = 0.143
Indonesia 0.20 × (1 − 0.286) = 0.143
Dual water 0.24 × (1 − 0.297) = 0.169
Waxman–Smits 0.24 × (1 − 0.266) = 0.176
Three things stand out.
First, the two effective-porosity shaly-sand models agree closely with each other and both add about 11 percent to the hydrocarbon volume Archie gives. That is the clay correction doing its job.
Second, the total-porosity models sit noticeably higher, and most of that gap is not the clay term. Archie run on total porosity, with no clay term at all, gives √(0.05 / (0.0576 × 10)) = 0.295 and a hydrocarbon volume of 0.24 × 0.705 = 0.169, the same as dual water. In this saline water the shale's bound water is almost as conductive as the free water, so dual water's correction is tiny; the jump comes from applying m = 2 to a larger porosity.
Third, that tells you where the real uncertainty sits. Cementation and saturation exponents belong to the porosity they were measured on. Waxman–Smits m* and n* are measured on total porosity with the clay conduction removed, and on the same rock they normally come out higher than Archie's m and n. Repeat the Waxman–Smits calculation with m* = 2.2 and the total saturation rises to 0.311 and the hydrocarbon volume falls to 0.165. Borrowing exponents across bases is one of the quieter ways to bias a shaly-sand interpretation.
From lowest to highest, the five answers span 36 percent in hydrocarbon volume, with every input held fixed. That spread is model uncertainty, and it belongs in the volumetric range, not in a footnote.
A wet-sand check
The most useful test of a set of parameters needs no core: put Sw = 1 into each model and ask what resistivity it predicts for a water-bearing sand of the same porosity and shale content. For this rock:
Archie Ro = 1 / 0.8 = 1.25 ohm·m
Simandoux Ro = 1 / (0.8 + 0.125) = 1.08 ohm·m
Dual water Ro = 1 / (0.0576 × 19.92) = 0.87 ohm·m
Indonesia Ro = 1 / 1.1046² = 0.82 ohm·m
Waxman–Smits Ro = 1 / (0.0576 × 21.2) = 0.82 ohm·m
If a water-bearing interval of the same sand, at similar porosity and shale volume, reads about 1.1 ohm·m, the Simandoux parameters are consistent with this rock and the total-porosity set is not; if it reads about 0.85, the total-porosity models and Indonesia fit and Simandoux does not. The check does not prove that a model is right in the hydrocarbon leg, but it exposes a parameter set that cannot reproduce the one interval where the answer is known, and it costs nothing.
Choosing, in practice
There is no universally best model. A defensible choice usually follows from a few questions.
- Is there core with measured cation exchange capacity? Then Waxman–Smits can be calibrated rather than assumed, with m* and n* from the same plugs, and it is hard to beat.
- No core, but good shale picks? Dual water and the effective-porosity models can be parameterised from the logs. Dual water's Cwb and Swb come from the shale's resistivity and porosity, as above.
- How is the clay distributed? The models above assume clay dispersed through the sand. Laminated sand–shale sequences behave differently: the resistivity is dominated by the sand laminae, and a laminated approach, with sand properties recovered by something like Thomas–Stieber analysis, is the better starting point.
- How fresh is the water? The fresher the formation water, the larger the clay's share of the total conductivity and the more the models diverge. Indonesia was developed for fresh-water, shaly reservoirs, which is where it is most often used.
- What independent saturation evidence exists? Saturation from preserved core, a capillary-pressure saturation-height model tied to a pressure-derived free water level, or a production test will discriminate between models far better than any argument about equations.
Whichever model is chosen, record the alternatives and the spread they give. A reviewer who sees one Sw curve cannot tell whether the model was tested; one who sees the comparison and the wet-sand check can.
The Saturation Model Comparison module of the Intermediate Petrophysics track covers this ground in more depth, and the advanced saturation and wettability module picks up where the models stop working: low-resistivity pay, conductive minerals and oil-wet rock.
In the platform
The Sw workspace in Formation Evaluation offers Archie alongside Simandoux and its modified forms, Indonesia, dual water and Waxman–Smits, with the equation and inputs of the selected model shown in the case, and a saturation-height model from capillary pressure as the independent check. Zone averages reach Volumetrics through an explicit handoff that you review before it is applied.
References
- Archie, G. E. (1942). The electrical resistivity log as an aid in determining some reservoir characteristics. Transactions of the AIME, 146, 54–62.
- Simandoux, P. (1963). Dielectric measurements on porous media: application to the measurement of water saturations, study of the behaviour of argillaceous formations. Revue de l'Institut Français du Pétrole, 18, supplementary issue.
- Waxman, M. H., and Smits, L. J. M. (1968). Electrical conductivities in oil-bearing shaly sands. SPE Journal, 8(2), 107–122.
- Poupon, A., and Leveaux, J. (1971). Evaluation of water saturation in shaly formations. SPWLA 12th Annual Logging Symposium.
- Clavier, C., Coates, G., and Dumanoir, J. (1984). Theoretical and experimental bases for the dual-water model for interpretation of shaly sands. SPE Journal, 24(2), 153–168.