The Pickett Plot

Archie as a Straight Line

Section 7.1 solved Archie's law for the water saturation. G. R. Pickett saw that the same law is also a picture. Take the logarithm of $R_t = a R_w/(\phi^{m} S_w^{n})$ and solve for the porosity:

Plot the porosity up against the true resistivity to the right, both axes logarithmic, and a fixed water saturation is a straight line of slope $-1/m$. Only the last term changes from one saturation to the next, and on any one line it is a constant, so changing $S_w$ slides the line without turning it: the lines of constant water saturation are parallel. The SLB Energy Glossary defines the Pickett plot as "A double logarithmic plot of a resistivity measurement on the x-axis versus a porosity measurement on the y-axis". Pickett described the method in 1966 and the pattern recognition the plot allows in 1973.

The Wet Line Gives m and Rw

The line for $S_w = 100\%$ is the wet line, and it is the rock's own. Its slope is $-1/m$, and where it meets 100 % porosity, at the top of the plot, $R_t = a R_w$: with $a = 1$ the resistivity at the top of the line is the water resistivity itself (AAPG Wiki, Pickett plot construction). So the wet line gives the cementation exponent from its slope and the water resistivity from its height, neither assumed in advance, provided it is drawn through rock that is full of water.

Ogbon-1 has such rock. Below the oil-water contact at 6213.1 ft the main sand is all water: 173 half feet, 6213.5 to 6300 ft, at porosities of 25.0 to 27.0 %. A least-squares straight line of log porosity on log resistivity through them has $m$ = 2.00 and meets 100 % porosity at 0.0400 ohm.m, which is $R_w$ because $a$ is 1. Those are the values Ogbon-1 was built with, and the cover of this book draws the same kind of fit and finds the same pair; Section 6.6 found the same $R_w$ by another route. That is a luxury of a synthetic well, whose water leg is exact: a real leg scatters. This one also covers only a short stretch of the line, 25.0 to 27.0 % porosity, and the line is carried from there to 100 % porosity, a factor of 16 in resistivity, to read $R_w$, so an error of slope is magnified (Section 6.6 follows a wrong exponent through).

Reading a Saturation Off the Lines

Every other line is the wet line moved to the right by a fixed factor. A half foot with water saturation $S_w$ has the resistivity $R_t = R_o/S_w^{\,n}$, so it stands $S_w^{-n}$ times to the right of the wet line at its porosity: with $n = 2$ the 50 % line stands 4.00 times, the 30 % line 11.1 times, the 20 % line 25.0 times and the 10 % line 100 times. At 5890 ft, in the upper oil sand, the density porosity is 23.2 % and the log reads 20.9 ohm.m, 28.1 times the wet line there, so the point falls between the 20 % and 10 % lines, and the water saturation is 18.9 %. A point on the wet line is all water, as every half foot of the water leg is: at 6260 ft the log is 1.00 times the wet line, and the plot reads 100 %.

Read every clean half foot that way and the lines give the whole well. With the book's cutoffs (the cutoffs are the book's own choice, set out in Section 13.1) the 50 % line is the saturation cutoff, and the lines read 277.0 ft of net pay with a pay-average water saturation of 23.0 %, the truth Ogbon-1 was built from.

The spacing of the lines is set by $n$, and the plot does not show it. The SLB Energy Glossary says that the lines for the other saturations are usually drawn assuming an exponent $n$ of 2, and PetroWiki that Archie "recommended that n = 2.0, and this value is still widely used when no experimental data are available". Raise $n$ to 2.5 and the 50 % line stands 5.66 times the wet line; the point at 5890 ft now reads 26 % water, the pay average rises to 30.0 % and the net pay falls to 268.0 ft. Ogbon-1's rock was built with 2, so this is the wrong exponent brought to a rock of 2, and it reads too much water; Section 7.4 asks the reverse question, a rock whose exponent is 2.5 read with 2, and is where the exponent is measured.

The Line Needs Wet Points

The water leg works because it lies on one line. Rock that holds oil does not. The 599 clean half feet above the contact all hold oil, at built saturations from 18.7 % to 99.6 %, so they lie across many parallel lines, and a straight line fitted through them takes its slope from how the porosity and the saturation vary together, not from the rock. It has $m$ = 3.68 and meets 100 % porosity at 0.0354 ohm.m, and read as a wet line it puts every half foot at 58.6 % water or more: the net pay is 0.0 ft, against the 277.0 ft. Fit all 772 clean half feet of the well and $m$ = 5.87, the line meets 100 % porosity at 0.000571 ohm.m and the net pay is 91.0 ft. Turning the regression round, resistivity on porosity, does not rescue it: the oil-bearing rock then gives $m$ = 2.65 and 0.215 ohm.m. The wet line has to come from water-bearing rock, and a well with no water leg has to take its $R_w$ from elsewhere (Section 7.5).

One Family of Lines, or Scatter

On Ogbon-1 every clean half foot falls on the line of its own saturation, sand and limestone alike. At 6080 ft the tight limestone has a porosity of 8.5 % and reads 116 ohm.m, 20.9 times its wet line, between the 30 % and 20 % lines: 22 % water. It sits on the same family as the upper sand because Ogbon-1 was built with one $m$ in both. A real carbonate does not oblige: PetroWiki says that "carbonate pore and fracture networks vary greatly, and m values from 1.0 to 3.0 may be required", so sand and limestone plotted together do not fall on parallel lines of one slope. That is the pattern recognition of Pickett's second paper: a cloud on parallel lines is one rock behaving as Archie's law says, and a cloud that is not is a warning that the model, or the mixture of rocks, is wrong.

The plot leaves the shales off, the 429 half feet of the well that are not clean (shale volume over 40 %). A shale conducts through its clay as well as through its brine, so the same saturation reads a lower resistivity than clean Archie gives, and a point plotted from it falls left of the clean-rock line and reads too wet; Chapter 8 is where that is modelled. Ogbon-1's shaly oil sand, at 5980 ft with a shale volume of 18.7 %, is on the plot and on the lines with the clean sands, because the well was built as clean Archie (Section 6.6): a real shaly sand would not be so obliging.

Section 7.3 takes the saturation the lines give and turns it into bulk-volume water.

References

• Pickett, G. R. (1966). A review of current techniques for determination of water saturation from logs. Journal of Petroleum Technology, November 1966, 1425-1435. SPE 1446.
• Pickett, G. R. (1973). Pattern recognition as a means of formation evaluation. The Log Analyst, 14(4), 3-11.
• SLB Energy Glossary. Pickett plot. glossary.slb.com.
• AAPG Wiki. Pickett plot construction. wiki.aapg.org.
• PetroWiki. Water saturation determination. petrowiki.spe.org.