Modified Simandoux

Summary

The modified Simandoux equation divides the water term of the Simandoux equation by the clay-free fraction of the rock, one minus the clay volume. It is a quadratic in Water saturation for n = 2, solved in closed form. Use it as an alternative to Simandoux when the clean-sand fraction should carry the formation-water conduction, and expect it to give lower water saturation than the original.

Inputs and outputs

Equations

The water term is divided by the clay-free fraction of the rock, with \(n = 2\):

Define \(c = \phie^{\,m} / \left[\aTort\,\Rw\,(1 - \Vcl)\right]\) and \(b = \Vcl / \Rcl\). The positive root of the quadratic is:

The result is limited to the interval 0 to 1, and \(\Sw = 1\) is returned when \(\Vcl \ge 1\), where the equation has no solution. With \(\Vcl = 0\) the equation reduces to Archie with \(n = 2\).

Single-value calculator

Behavior

The modified equation always gives a lower water saturation than the original, and the difference grows with clay volume. At \(R_t\) = 10 ohm·m, \(R_w\) = 0.05 ohm·m, 18% porosity and a clay resistivity of 2.5 ohm·m, the modified form gives 0.393, 0.305, 0.239 and 0.185 at clay volumes of 0, 0.2, 0.4 and 0.6, against 0.393, 0.336, 0.288 and 0.249 from Simandoux. At 0.6 clay the modified result is a quarter lower than the original. The two curves meet at zero clay, and both meet Archie there. Because the extra division makes the water term larger, the modified form is the most optimistic of the three closed-form shaly-sand equations (Simandoux, modified Simandoux and Indonesian).

Parameter guidance

The inputs are the same as for Simandoux. The clay resistivity Clay resistivity and clay volume Clay volume carry more weight here, because the clay term and the \(1/(1 - V_{cl})\) scaling both respond to them. Check that clay volume is capped below 1 in the data feeding the equation: the equation has no solution in 100% clay, and the common handling is to return a water saturation of 1. See Rw Determination and Cementation and Saturation Exponents for the shared inputs, and the Clay Volume step for \(V_{cl}\).

Worked example

The same shaly sand as on the Simandoux page, with the results of both equations side by side and the clean-sand check:

import math

def sim(rt, rw, phie, vcl, rcl, a=1.0, m=2.0, scale=False):
    c = phie ** m / (a * rw * ((1.0 - vcl) if scale else 1.0))
    b = vcl / rcl
    return min(1.0, max(0.0, (math.sqrt(b * b + 4.0 * c / rt) - b) / (2.0 * c)))

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, rcl = 10.0, 0.05, 0.18, 2.5
print(f"{'Vcl':>5} {'Archie':>8} {'Simandoux':>10} {'Modified':>9}")
for vcl in (0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6):
    print(f"{vcl:5.2f} {archie(rt, rw, phie):8.3f} {sim(rt, rw, phie, vcl, rcl):10.3f} {sim(rt, rw, phie, vcl, rcl, scale=True):9.3f}")

# clean-sand limit
worst = max(abs(sim(r, rw, p, 0.0, rcl, scale=True) - archie(r, rw, p))
            for r in (2.0, 5.0, 20.0, 100.0) for p in (0.08, 0.15, 0.25))
print(f"clean-sand check: largest |modified - Archie| = {worst:.2e}")

Output

  Vcl   Archie  Simandoux  Modified
 0.00    0.393      0.393     0.393
 0.10    0.393      0.363     0.346
 0.20    0.393      0.336     0.305
 0.30    0.393      0.311     0.270
 0.40    0.393      0.288     0.239
 0.50    0.393      0.268     0.211
 0.60    0.393      0.249     0.185
clean-sand check: largest |modified - Archie| = 2.78e-17

Assumptions and limitations

• The saturation exponent is 2, as for Simandoux.
• The formation water conducts only through the clay-free fraction of the rock, and the clay conducts in parallel. This is an empirical adjustment, not a derivation.
• A single, constant clay resistivity and an accurate clay volume are known. Both feed the equation twice.
• Effective porosity is used. At very high clay volume the equation has no solution and is cut off at 1.
• The equation is the most optimistic of the three closed-form shaly-sand equations, so an overestimated clay volume turns directly into extra hydrocarbon.

QC checks

• At Vcl = 0 the result equals the Archie value with n = 2.
• The result is never above the Simandoux result at the same inputs, and never above Archie.
• Check the water leg: Sw near 1 with the same picks.
• Compare with Indonesian. If the modified result is far lower in the clay-rich intervals, treat the hydrocarbon in those intervals as unproven.
• Check that no output equals exactly 1 because of the Vcl cutoff in non-reservoir shale. Mask these as non-reservoir.

Going Deeper

The factor \(1/(1 - V_{cl})\) is often described as a way of making the water term refer to the clay-free rock, so that the porosity and water are weighted by the fraction of the rock they occupy. The weighting is commonly attributed to Bardon and Pied. Different texts and tools use different versions of both the Simandoux and the modified equations, so the same name does not guarantee the same numbers: write down the equation, and check the clay volume the version expects. The computed difference between the two is largest at high clay volume, which is where neither has been shown to be reliable.

References

1. Bardon, C. and Pied, B., 1969. Formation water saturation in shaly sands. Transactions of the SPWLA 10th Annual Logging Symposium, Paper Z.

Python reference implementation

Related methods

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