Water Saturation on the Ogbon-1 Well

The Chain on One Well

Section 7.1 stated Archie's equation and Sections 7.2 to 7.5 asked what each of its inputs costs. This section runs it down the whole of Ogbon-1, the book's well. The shale volume comes from the gamma ray (Section 3.2), the porosity from the density (Section 4.2), the true resistivity from the deep curve (Sections 6.5 and 6.6) and the water resistivity from the water leg, and Archie's equation (Archie, 1942) turns porosity, resistivity and $R_w$ into a water saturation at every half foot. Figure 7.6 draws the chain: (a) the porosity and the shale volume, (b) the deep resistivity against its wet line, and (c) the water saturation, read against the truth the well was built from, with the net pay it leaves. The depth marker reads one half foot, and the one other control is $R_w$.

One Half Foot, Then the Whole Well

At 5890 ft, in the upper oil sand, the density porosity is 23.2 % and the deep resistivity 20.9 ohm.m. With $a = 1$, $m = n = 2$ and $R_w$ = 0.040 ohm.m the same rock full of water would read 0.745 ohm.m, so the log stands 28.1 times its wet line (Section 6.6 read this half foot the same way), and

is 18.9 % water. Done at all 1201 half feet, the same arithmetic is the curve of plate (c), and with the well's own $R_w$ it is the truth Ogbon-1 was built from at every half foot that is not shale, because the well was computed forward from these equations. A real well gives no such answer to check against; the water leg below is what stands in for it.

What Counts as Pay

A saturation curve is not yet an answer. The reservoir engineer needs to know which rock counts, and the book's cutoffs (its own choice, set out in Section 13.1) make the call: a half foot is net reservoir if its shale volume is at most 40 % and its porosity at least 8 %, and net pay if its water saturation is also at most 50 %. Of the 600 ft logged, 386 ft is net reservoir and 277.0 ft is net pay. The pay lies in four beds: 72.0 ft in the upper oil sand, 72.0 ft in the shaly oil sand, 78.0 ft in the tight limestone and 55.0 ft in the main sand.

The Contact Is Not the Base of the Pay

In the main sand the saturation climbs from the oil column to 100 % at the oil-water contact, 6213.1 ft. The pay does not run to the contact. The curve crosses the 50 % cutoff, and the pay ends, at 6190.5 ft, 22.6 ft above the contact. The rock between holds oil, but it is transition zone: at 6200 ft Archie reads 62.7 % water, and it is not pay. Below the contact the main sand is water-filled, and the free-water level, 6250 ft, lies a further 36.9 ft below it, where the capillary pressure is zero (Section 1.5 and Section 11.3). Two words, two depths, and neither is the base of the pay.

The Shaly Oil Sand

The shaly oil sand, 5946 to 6018 ft, has an average shale volume of 21.8 %, and clean Archie, with no clay term, reads its water saturation as 19.6 % on average, which is the truth. That is not a property of the equation; it is how the well was made. Ogbon-1's resistivity was built as clean Archie in every bed, this one included, so there is no clay conduction for the clean equation to miss, and a shaly-sand model has nothing to take out: the Simandoux equation (Section 8.2) reads 12.8 % water there, less than the truth. A real shaly sand is the other case. PetroWiki puts it in one line: "In shaly rocks, Rt is suppressed and Archie Sw calculations are too high" (Water saturation determination). Section 8.1 shows why, and Chapter 8 builds the models that correct for it.

The Deliverable

Over the 277.0 ft of pay the porosity averages 18.1 % and the water saturation 23.0 %. Those are sample averages; weighted by pore volume, a foot with more pore space counting for more, the water is 23.9 %, because the most porous pay, the main sand, is also the wettest. The number that carries the oil is the hydrocarbon column, the sum of $\phi\,(1 - S_w)$ over the pay, 38.1 ft: the thickness of pure oil that the pore space is equivalent to. Chapter 13 multiplies it by the 7758 barrels an acre-foot holds to make 296 thousand reservoir barrels under every acre, before any recovery factor. These three numbers, 277.0 ft, 23 % and 38.1 ft, are what the petrophysicist hands the reservoir engineer. Figure 1.1 reached the same three from the raw logs, and they are the truth Ogbon-1 was built from.

When the Water Resistivity Is Wrong

Section 7.5 counted what a wrong $R_w$ costs and how the routes to it disagree. Here the figure lets it move a factor of two either way to show something more: which way of being wrong the well itself gives away. Raise $R_w$ from 0.040 to 0.060 ohm.m and every saturation not capped at 100 % is 1.22 times higher: the pay-average water goes from 23 % to 27 %, the net pay from 277.0 ft to 266.5 ft, 10.5 ft less, and the column from 38.1 ft to 34.2 ft. Nothing on the log objects: the water leg would read more than 100 %, and the equation stops at 100 %. Lower it to 0.030 ohm.m and the numbers go the other way, to 283.0 ft of pay, 21 % water and a 40.6 ft column, but the main sand below the contact, which is all water, now reads 87 % water: water begins to read as oil.

The water leg is the check. Below the contact the main sand is water-bearing, so Archie's equation with $S_w = 1$ turns around into $R_w = R_t/F$, and all 173 half feet of the leg give 0.040 ohm.m. PetroWiki: "A first check on the Rw of the aquifer is to back-calculate the apparent Rw with the Archie equation" (Water saturation determination). A well with no water leg has no such check, and its $R_w$ is the problem of Section 7.5. Chapter 8 next takes the clean equation to a shaly sand that was not built clean.

References

• Archie, G. E. (1942). The electrical resistivity log as an aid in determining some reservoir characteristics. Transactions of the AIME. SPE-942054-G.
• PetroWiki. Water saturation determination. petrowiki.spe.org/Water_saturation_determination.