Horizontal and Vertical Resistivity Sw

Summary

In a laminated sequence the water saturation that matters is that of the sand layers: Sand water saturation, from the sand porosity and the Sand resistivity, with the shale laminae taken out of the Horizontal resistivity. The sand fraction, Net-to-gross ratio, then carries the result to the whole interval. Where a triaxial or anisotropic resistivity tool is available, the vertical resistivity gives the laminated volume and the sand resistivity directly.

Inputs and outputs

Equations

Sand resistivity and Sw. The sand resistivity comes from the parallel model on the sand properties page, and Archie's equation (with \(a = 1\)) is applied to the sand layers with the sand porosity:

limited to 0 to 1. The calculator uses the clean-sand porosity for the sand porosity. In a log analysis it is the sand-only porosity computed from the logs.

Conventional Sw, for comparison. The same Archie equation with the bulk porosity of the laminated rock, from the laminated shale line, and the unseparated horizontal resistivity:

Net-to-gross. If the laminations are too thin to be resolved, the sand fraction is the net-to-gross factor:

and the hydrocarbon pore volume per unit area of the interval of thickness \(h\) is \(h\,\CpNTG\,\phiSa\,(1 - \SwSand)\).

Using both horizontal and vertical resistivity. If a triaxial tool gives \(\Rhz\) and \(\Rvt\), and the shale resistivities \(\RshH\) and \(\RshV\) are known, the parallel and series models together give two equations for the two unknowns \(\Vlam\) and \(\Rsand\). Eliminating \(\Rsand\) gives a quadratic in \(\Vlam\):

The root between 0 and 1 that is the smaller of the two is the laminated volume. \(A\) is zero when the shale is isotropic, and the equation is then linear.

Dip correction. In a deviated well or dipping bed, an induction-type tool measures an apparent resistivity \(\Rapp\) between the horizontal and vertical resistivity. A simple correction to the horizontal resistivity with the relative dip \(\thetaDip\) (0 for a well normal to the bedding) is:

where \(\RvRh\) is the anisotropy ratio of the rock. At \(\thetaDip = 0\) the horizontal resistivity is the measured value, and at \(\thetaDip = 90^\circ\) it is \(\Rapp\sqrt{\Rhz/\Rvt}\), which corresponds to an apparent resistivity of \(\sqrt{\Rhz\Rvt}\).

Single-value calculator

Behavior

The plot holds the horizontal resistivity at 6 ohm·m and increases the laminated shale volume. The conventional Sw rises, from 0.365 with no laminated shale to 0.445 at a volume of 0.3 and 0.500 at 0.45, because the bulk porosity falls from 0.25 to 0.205 and 0.183 and the resistivity does not change. The sand Sw falls, from 0.365 to 0.276 at 0.3 and 0.156 at 0.45, because the shale laminae take current and the sand is more resistive than the log says (sand resistivity 6.0, 10.5 and 33 ohm·m). The two agree with no laminated shale. At a laminated volume of 0.3 the conventional answer is 0.445 and the sand answer is 0.276, so the conventional answer would put the interval close to a water saturation cutoff while the sand is clearly hydrocarbon-bearing.

Parameter guidance

Sand porosity and laminated volume come from the previous pages. Rw, m and n are the same as for the ordinary Archie calculation and are covered in Water Saturation and Cementation and Saturation Exponents. Use the water resistivity of the sand zone. Shale resistivity. Read it in thick shale next to the sand. Tool type. Where a triaxial or anisotropic resistivity tool exists, use the horizontal and vertical measurement and solve for the laminated volume as a check on the volume from porosity logs. Otherwise use the horizontal-only equations with the volume from triangulation. Relative dip. The well inclination minus the bed dip, along the same azimuth. A dip correction is only needed for relative dips above about 30 degrees, and it needs an anisotropy ratio. The step page, Laminated Sand/Shale Analysis, gives the default approach.

Worked example

First the horizontal-only case, with the numbers of the calculator. Then a triaxial measurement is generated from a known laminated volume (0.30) and sand resistivity (20 ohm·m) and the quadratic recovers both. Last, the dip correction for an apparent resistivity of 5 ohm·m and an anisotropy ratio of 2:

import math
# 1. horizontal resistivity only
rh, vl, rsh_h, ps, ph, rw, m, n = 6.0, 0.30, 3.0, 0.25, 0.10, 0.05, 2.0, 2.0
rs = (1 - vl) * rh * rsh_h / (rsh_h - vl * rh)
sw_sand = (rw / (ps ** m * rs)) ** (1 / n)
phit = (1 - vl) * ps + vl * ph
sw_conv = (rw / (phit ** m * rh)) ** (1 / n)
print(f"Rsand = {rs:.2f} ohm.m, Sw sand = {sw_sand:.3f}, bulk porosity = {phit:.3f}, Sw conventional = {sw_conv:.3f}, NTG = {1 - vl:.2f}")
print(f"HCPV per ft of interval = {(1 - vl) * ps * (1 - sw_sand):.4f} ft (sand) against {phit * (1 - sw_conv):.4f} ft (conventional)")
# 2. triaxial: forward model, then solve the quadratic
rsh_v = 6.0
v_true, rs_true = 0.30, 20.0
Rh = 1 / (v_true / rsh_h + (1 - v_true) / rs_true)
Rv = v_true * rsh_v + (1 - v_true) * rs_true
A = Rh * (rsh_v - rsh_h)
B = 2 * Rh * rsh_h - Rv * Rh - rsh_v * rsh_h
C = rsh_h * (Rv - Rh)
disc = math.sqrt(B * B - 4 * A * C)
roots = sorted(((-B - disc) / (2 * A), (-B + disc) / (2 * A)))
v_sol = [r for r in roots if 0 <= r < 1][0]
rs_sol = (1 - v_sol) * Rh * rsh_h / (rsh_h - v_sol * Rh)
print(f"triaxial: Rh = {Rh:.3f}, Rv = {Rv:.3f}; roots {roots[0]:.4f} and {roots[1]:.4f}; Vlam = {v_sol:.4f}, Rsand = {rs_sol:.2f}")
# 3. dip correction
ra, aniso = 5.0, 2.0
for th in (0, 30, 60, 90):
    t = math.radians(th)
    print(f"relative dip {th:2d}: Rh = {ra * math.sqrt(math.cos(t) ** 2 + math.sin(t) ** 2 / aniso):.3f} ohm.m")

Output

Rsand = 10.50 ohm.m, Sw sand = 0.276, bulk porosity = 0.205, Sw conventional = 0.445, NTG = 0.70
HCPV per ft of interval = 0.1267 ft (sand) against 0.1137 ft (conventional)
triaxial: Rh = 7.407, Rv = 15.800; roots 0.3000 and 3.7767; Vlam = 0.3000, Rsand = 20.00
relative dip  0: Rh = 5.000 ohm.m
relative dip 30: Rh = 4.677 ohm.m
relative dip 60: Rh = 3.953 ohm.m
relative dip 90: Rh = 3.536 ohm.m

Assumptions and limitations

• Archie's equation holds in the sand layers with constant exponents, and the water resistivity is the same in all sand layers.
• The shale laminae contribute only through their resistivity and volume, and the shale resistivity is constant. Sand layers are free of clay, or the sand porosity and resistivity already account for any clay in them.
• The laminations are thinner than the log resolution, so the sand fraction is the net-to-gross factor for the interval.
• The horizontal resistivity is the measured value corrected for dip, borehole and invasion. A deep-reading tool that is invaded or in a washout reads the wrong volume.
• The triaxial equations assume the same laminated volume for the horizontal and vertical measurement. Vertical resolution and sensitivity differ in practice, and the vertical measurement is the less well determined.
• The dip correction is a simplified low-frequency induction approximation. It does not apply to all tool types, and needs an anisotropy ratio that is itself uncertain.

QC checks

• With zero laminated volume, the sand Sw equals the conventional Sw, and the net-to-gross factor is 1.
• The sand resistivity is greater than the horizontal resistivity where the horizontal resistivity is above the shale resistivity, and does not exist (NaN) where the horizontal resistivity is above the shale resistivity divided by the laminated volume.
• In a hydrocarbon sand, the sand Sw is below the conventional Sw. If it is higher, check the shale resistivity and the sand porosity.
• The laminated volume from the triaxial quadratic is close to the volume from the porosity logs. A large difference points to wrong shale resistivities or a wrong shale type.
• The sand Sw in clean water-bearing intervals is close to 1, and the net-to-gross factor is close to 1 in thick clean sand.
• Core, image logs or a thin-bed resolution analysis agree with the net-to-gross factor.

Going Deeper

The idea that a laminated sand needs separate sand and shale properties goes back to the laminated shaly-sand model of Poupon and others in the 1950s. The conventional shaly-sand equations (Simandoux and others) treat the shale as dispersed in the sand, which reduces the effect of the shale on resistivity to a correction of the same size as for dispersed clay and misses the strong resistivity contrast of laminated beds. In the anisotropic approach, the horizontal resistivity is the parallel mean, dominated by the conductive shale. Hydrocarbon sand layers are masked by it, and the vertical resistivity is the series mean, dominated by the resistive layers. A tool that measures both gives the sand resistivity without the need for a laminated volume from porosity logs. Tool resolution, borehole effects and the dip of the beds are the limits in practice, and are the reason that triaxial results are compared with the porosity-based volume and not used alone. The net-to-gross factor of a sub-resolution interval is the sand volume. It is not the same as the net reservoir of the Cutoffs Analysis, which is a resolved count of the beds that pass a cutoff.

References

1. Poupon, A., Loy, M.E. and Tixier, M.P., 1954. A contribution to electric log interpretation in shaly sands. Journal of Petroleum Technology, 6(6), 27–34.
2. Klein, J.D., Martin, P.R. and Allen, D.F., 1997. The petrophysics of electrically anisotropic reservoirs. The Log Analyst, 38(3), 25–36.
3. Moran, J.H. and Gianzero, S., 1979. Effects of formation anisotropy on resistivity-logging measurements. Geophysics, 44(7), 1266–1286.
4. Archie, G.E., 1942. The electrical resistivity log as an aid in determining some reservoir characteristics. Transactions of the AIME, 146(1), 54–62.

Python reference implementation

Related methods

• Thomas-Stieber
• Triangulation
• Shale Volume Options
• Sand-Only Porosity and Resistivity