Indonesian

Summary

The Indonesian equation of Poupon and Leveaux combines a clay term and a formation-water term under a square root, and has a closed-form solution for any saturation exponent. It was developed for fresh-water shaly sands with high clay volume and, in fresh water, gives less optimistic results than the Simandoux forms. Use it as the default shaly-sand model when core calibration of the Waxman-Smits or dual-water inputs is not available.

Inputs and outputs

Equations

The Indonesian equation is written in terms of the square root of conductivity:

Solving for the water saturation:

The result is limited to the interval 0 to 1. With \(\Vcl = 0\) the clay term vanishes and the equation reduces to Archie: \(\Sw = \left[\aTort\,\Rw / (\phie^{\,m}\,\Rt)\right]^{1/n}\).

Single-value calculator

Behavior

Water saturation falls as clay volume rises, but more slowly than in the Simandoux forms. At \(R_t\) = 10 ohm·m, \(R_w\) = 0.05 ohm·m, 18% porosity and a clay resistivity of 2.5 ohm·m, the Indonesian equation gives 0.393, 0.332, 0.285 and 0.254 at clay volumes of 0, 0.2, 0.4 and 0.6. In this saline case it is almost the same as Simandoux (0.336, 0.288, 0.249), within 0.005. The gap to Simandoux opens at high resistivity and fresh water: with \(R_w\) = 0.5 ohm·m, \(R_t\) = 30 ohm·m and a clay resistivity of 3 ohm·m, 30% clay gives Archie 0.717, Indonesian 0.395, Simandoux 0.282 and modified Simandoux 0.267. The exponent works as it does in Archie: at 30% clay and the base case, \(n\) of 1.8, 2.0 and 2.2 give 0.269, 0.306 and 0.341.

Parameter guidance

Clay resistivity Clay resistivity and clay volume Clay volume are as for Simandoux. Because the clay volume appears as a power \(V_{cl}^{1-V_{cl}/2}\), the correction is larger than a straight Vcl at low clay volume and tapers off at high clay volume. n is an input and need not be 2: use the value from core, or see Cementation and Saturation Exponents. Water resistivity: Rw Determination. Clay volume: Clay Volume. Porosity: effective porosity from Porosity. The step page (Water Saturation) explains why this is the recommended default.

Worked example

A shaly sand with \(R_t\) = 10 ohm·m, \(R_w\) = 0.05 ohm·m, 18% effective porosity, 30% clay, a clay resistivity of 2.5 ohm·m and \(n = 2\), with the clean-sand check:

import math

def indonesian(rt, rw, phie, vcl, rcl, a=1.0, m=2.0, n=2.0):
    term = vcl ** (1.0 - vcl / 2.0) / math.sqrt(rcl) + phie ** (m / 2.0) / math.sqrt(a * rw)
    return min(1.0, max(0.0, ((1.0 / math.sqrt(rt)) / term) ** (2.0 / n)))

def archie(rt, rw, phie, a=1.0, m=2.0, n=2.0):
    return min(1.0, (a * rw / (phie ** m * rt)) ** (1.0 / n))

rt, rw, phie, vcl, rcl = 10.0, 0.05, 0.18, 0.30, 2.5
clay_term = vcl ** (1.0 - vcl / 2.0) / math.sqrt(rcl)
water_term = phie / math.sqrt(rw)
print(f"clay term  = {vcl:g}^{1 - vcl / 2:.2f} / sqrt({rcl:g}) = {clay_term:.4f}")
print(f"water term = {phie:g}^(m/2) / sqrt(a Rw) = {water_term:.4f}")
print(f"Archie     Sw = {archie(rt, rw, phie):.3f}")
print(f"Indonesian Sw = {indonesian(rt, rw, phie, vcl, rcl):.3f}")

# clean-sand limit, for several exponents
print()
worst = 0.0
for n in (1.8, 2.0, 2.4):
    for rt_ in (2.0, 5.0, 20.0, 100.0):
        for phi_ in (0.08, 0.15, 0.25):
            d = abs(indonesian(rt_, rw, phi_, 0.0, rcl, n=n) - archie(rt_, rw, phi_, n=n))
            worst = max(worst, d)
print(f"clean-sand check: largest |Indonesian - Archie| over 36 cases = {worst:.2e}")
assert worst < 1e-12

Output

clay term  = 0.3^0.85 / sqrt(2.5) = 0.2273
water term = 0.18^(m/2) / sqrt(a Rw) = 0.8050
Archie     Sw = 0.393
Indonesian Sw = 0.306

clean-sand check: largest |Indonesian - Archie| over 36 cases = 1.11e-16

Assumptions and limitations

• Clay and formation water conduct in parallel, and the clay conduction is described by an empirical power of the clay volume and the clay resistivity.
• The exponents m and n are the same for the clean and shaly parts of the rock. The same n is used for both terms through the single saturation power.
• Effective porosity is used, and the clay volume is a clay volume (not a shale volume that includes silt).
• A single clay resistivity applies. Where the shale resistivity varies between zones, use one per zone.
• The relation was derived from fresh-water shaly sands, and in very saline water the clay term is small and the difference between methods is too.

QC checks

• At Vcl = 0 the result equals Archie for any n. The worked example runs this check.
• The result is at or below the Archie value and at or above the modified Simandoux value at the same inputs. If it is outside that range, check the inputs.
• Water leg Sw is close to 1 with the same picks.
• Sensitivity to the clay resistivity is moderate. Re-run with the clay resistivity changed by 30% and confirm the pay thickness is stable.
• Compare with a dual-water or Waxman-Smits result at core calibration points, if any.

Going Deeper

The equation was published by Poupon and Leveaux in 1971 for the shaly sands of Indonesia, which have fresh formation water and high clay content, and in which the conventional models overestimated hydrocarbon saturation. The exponent \(1 - V_{cl}/2\) on the clay volume was fitted so that the equation reproduced the observed saturations, and the square-root form follows from writing the conductivity as a sum of the two terms raised to a half power. The result is that the clay correction saturates at high clay volume instead of growing without bound. The approach is empirical, but it has been used widely because it requires only the clay volume and the clay resistivity and handles an exponent other than 2 without iteration.

References

1. Poupon, A. and Leveaux, J., 1971. Evaluation of water saturation in shaly formations. The Log Analyst, 12(4), 3–8.

Python reference implementation

Related methods

• Archie
• Simandoux
• Modified Simandoux
• Dual Water
• Waxman-Smits
• Total vs Effective Sw