Resistivity on the Ogbon-1 Well

The Chapter on One Well

Section 1.6 laid the log display of Ogbon-1 beside a wet line and moved the wet line's $R_w$. This section takes the wet line apart the way Chapter 6 built it. Archie's second law (Section 6.1) says that a rock with a fraction $S_w$ of its pores full of water has the resistivity $R_t = R_o/S_w^{\,n}$, where $R_o$ is what the same rock would read full of water. The wet line (Section 6.2) is the formation factor times the water, $R_o = F\,R_w$, and the factor comes from the porosity and the cementation exponent, $F = a/\phi^{m}$. Figure 6.6 draws it on the well: (a) the deep resistivity with its wet line, (b) the formation factor, and (c) the water saturation the display reads, against the truth the well was built from. Two controls set the wet line, the exponent $m$ and the $R_w$ put into it.

What the Log Is

Ogbon-1's resistivity is the true resistivity of a clean Archie rock. The well was computed forward from its truth: in its sands and its limestone the log is Archie's equation with $R_w$ = 0.040 ohm.m, $a = 1$ and $m = n = 2$, and in its shales it is 2.20 ohm.m, give or take 10 %. It carries no invasion. The tools of Sections 6.3 to 6.5 exist because a real deep curve is dragged toward the flushed zone, and Section 6.5 corrects it with a tornado chart, whose answer is only as good as the step profile it assumes; here there is nothing to correct, so the log can be set against a wet line as it stands. On a real well it cannot, and every number below would sit on top of that correction.

The Ratio Is the Saturation

At 5890 ft, in the upper oil sand, the density porosity is 23.2 %, so the formation factor is 18.6 and the wet line, for an $R_w$ of 0.040 ohm.m, is 0.745 ohm.m. The log reads 20.9 ohm.m, 28.1 times the line. Archie's second law read backwards says that ratio is the saturation to the power minus $n$, $R_t/R_o = S_w^{-n}$, so

is 18.9 % water: 1/28.1 is 3.6 %, and the square root of that is 18.9 %. At 6200 ft, in the transition zone, the log reads 1.52 ohm.m, only 2.55 times its line, which is 62.7 % water. A water saturation 3.32 times higher gives a ratio 11.0 times lower, not 3.32 times, and that square is why pay shows so plainly on a resistivity log. Because $S_w \le 50\%$ is the same as $R_t \ge 4\,R_o$, a log must stand 4.00 times its wet line to pass the book's saturation cutoff (the cutoffs are the book's own choice, set out in Section 13.1).

The Wet Line Is the Factor Times the Water

Two numbers make the line. The first is the factor $F = a/\phi^{m}$, which only the porosity and the exponent set. At 6080 ft, in the tight limestone, the density porosity is 8.5 % and the factor is 139, 7.5 times the upper sand's, so the line stands at 5.57 ohm.m against the sand's 0.745 ohm.m. A high reading is not a high reading against the line, as Section 1.6 showed, and plate (b) shows why: the factor is large exactly where the porosity is small. The second number is $R_w$, the resistivity of the formation water.

The exponent is the less certain of the two. Archie proposed $a = 1.0$ and $m = 1.8$ to $2.0$ for his data, the Humble formula has $m = 2.15$ for sandstones, with a coefficient printed from 0.61 to 0.65 (Section 6.2), and "m is not a constant, but varies with rock type" (PetroWiki, Water saturation determination). Because the porosity is raised to the power minus $m$, a change in the exponent moves the factor of a tight rock more than that of a porous one: from $m$ = 2.0 to 2.2 the factor grows 1.34 times at 23.2 % porosity and 1.64 times at 8.5 %. The figure sweeps $m$ from 1.6 to 2.4, wider than the published values, so that a wrong exponent shows.

The Water Leg Is the Test

One place on the well has a known answer. Below the oil-water contact, 6213.1 ft, the main sand is all water: 173 half feet, 6213.5 to 6300 ft, with a porosity of 25.0 to 27.0 %. There the log is its own wet line, so it must read exactly 1 times it. With the well's own exponent and $R_w$ it does, and that is how Section 1.6 found the $R_w$ of 0.040 ohm.m. PetroWiki gives the same check for an aquifer: back-calculate the apparent $R_w$ with Archie's equation, in which $n$ drops out because the water saturation is 100 %.

Hold $R_w$ at 0.040 ohm.m and change the exponent, and the water leg gives the error away. At $m$ = 2.2 the leg reads 0.76 times its wet line, below it, which a wet sand cannot do; the line is drawn too high, and the net pay falls from 277.0 to 270.5 ft. At $m$ = 1.8 the leg reads 1.31 times its line and the display calls it 87 % water in a sand that is all water; water begins to read as oil, and the net pay grows to 282.5 ft. The leg fails the test in both directions, and it fails without any appeal to the truth the well was built from.

Refitting Hides the Exponent

On a real well $R_w$ is rarely known so well, and a common remedy is to refit it in the water leg for the exponent in use. The refit takes the mean of $R_t/F$ over the leg, which is 0.0306 ohm.m at $m$ = 2.2 and 0.0233 ohm.m at $m$ = 2.4, and by construction the leg then reads 1.00 times its line. The net pay is back at 277.0 ft for both. Yet the exponent is still wrong, and the saturation shows it. The leg fixes only the product $F\,R_w$ at its own porosity, 25.0 to 27.0 %; in rock of another porosity the wet line is off by about $(\phi_{leg}/\phi)^{\Delta m}$. At $m$ = 2.4 the tight limestone at 6080 ft reads 27 % water against the 22 % it was built with, the largest error in the well is 6.4 saturation units at 6114.5 ft, where the limestone's porosity is lowest, 8.0 %, and across the sweep its saturation at 6080 ft wanders from 17.5 % at $m$ = 1.6 to 27.4 % at 2.4. A wrong exponent that the water leg cannot see costs saturation, not pay, and the saturation is what the volumes of Chapter 13 are made from.

Ogbon-1 was built with $m = 2$ in every rock, so the right exponent returns the truth exactly, and it was built as clean Archie even in its shaly oil sand, so the shale volume plays no part in these wet lines; Chapter 8 asks what changes when it does. A real well offers no such answer: its exponent is the subject of Section 7.4 and its $R_w$ the problem of Section 7.5. Chapter 7 now makes the rearrangement official.

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.