PO.PP.Sw.Simandoux
Description
Water saturation in a shaly sand from the Simandoux total-shale equation in the form Bassiouni publishes. The shale conducts, so Archie read on the same logs overstates Sw. This solution is the quadratic for saturation exponent 2, with tortuosity 0.8 and cementation exponent 2 already substituted, which is why the leading factor is 0.4 Rw / φe² and the term under the square root carries a 5. Sw = (0.4 Rw / φe²) * (−Vsh/Rsh + sqrt((Vsh/Rsh)² + 5 φe² / (Rw Rt))). Set Vsh to 0 and it is Archie at a = 0.8, m = 2, n = 2.
Formula
=LAMBDA(Phi_e, Rt, Rw, Vsh, Rsh, (0.4*Rw/Phi_e^2)*(-Vsh/Rsh+SQRT((Vsh/Rsh)^2+5*Phi_e^2/(Rw*Rt))))Parameters
| Parameter | Unit | Description |
|---|---|---|
Phi_e |
fraction | Effective porosity |
Rt |
ohm·m | True formation resistivity |
Rw |
ohm·m | Formation-water resistivity |
Vsh |
fraction | Bulk volume fraction of shale |
Rsh |
ohm·m | Resistivity of an adjacent shale |
Examples
| Invocation | Expected |
|---|---|
=PO.PP.Sw.Simandoux(0.2, 20, 0.05, 0, 2) |
0.2236068 |
=PO.PP.Sw.Simandoux(0.2, 10, 0.05, 0.2, 2) |
0.27015621 |