Archie's Equation

Two Laws, One Equation

Chapter 6 left two results. Section 6.1 gave the resistivity index, $I = R_t/R_o = S_w^{-n}$, which says how far oil lifts a rock's resistivity above its wet value, and Section 6.2 gave that wet value, $R_o = F\,R_w$ with $F = a/\phi^{m}$. Solve the first for the saturation and put the second in, and the two relations of Archie (1942) become one equation that turns a resistivity log into a water saturation:

Three measurements and three constants go in. The porosity $\phi$ comes from the density, neutron or sonic log (Chapters 4 and 5), the water's resistivity $R_w$ from the SP or a sample of the water (Sections 3.5 and 2.2), and the true resistivity $R_t$ from the resistivity tools (Chapter 6). The constants $a$, $m$ and $n$ are those of the formation factor and the resistivity index (Sections 6.2 and 6.1): Archie proposed $a = 1$ and $m$ of 1.8 to 2.0 for his data set, and $n = 2$ is still widely used when no experimental data are available (PetroWiki, Water saturation determination). Ogbon-1 was built with $a = 1$, $m = 2$ and $n = 2$, and Figure 7.1 puts it through the equation.

Two Steps

The equation reads most clearly as two steps. The first uses the rock alone: the porosity and the water give the wet-rock resistivity, what the rock would read full of brine. At 5890 ft, in the upper oil sand, the density porosity is 23.2 %, so the formation factor is 18.6 and, with $R_w$ of 0.040 ohm.m, $R_o$ is 0.745 ohm.m. The second uses the log: it reads 20.9 ohm.m there, 28.1 times $R_o$, and with $n$ of 2 the saturation is the index to the power $-1/2$, 18.9 %. Fully wet rock is the other end: in the water leg at 6260 ft the log reads 0.621 ohm.m, exactly its wet-rock resistivity, so the index is 1.00 and the saturation is 100 %. Ogbon-1 was built forward with this same equation in every sand and limestone bed, so with the inputs it was built with the equation returns the built saturation exactly. The figure shows what it does with wrong ones.

What Is Left Is Oil

As Section 1.3 set out, the water takes $\phi\,S_w$ of the rock's volume and the hydrocarbon takes the rest of the pore space, $\phi\,(1-S_w)$. At 5890 ft the sand's 18.9 % water leaves 81.1 % of the pore space as oil, which is 18.8 % of the rock, against the 4.4 % that is water. Plate (c) draws that quantity along the well; Section 7.3 takes up its counterpart, the bulk-volume water.

Each Input Has Its Own Power

Taking logarithms of the equation shows what a mistake in an input costs:

The saturation changes by an input's power times the input's fractional error: the porosity at $-m/n$, $R_w$ at $1/n$ and $R_t$ at $-1/n$ (the derivatives of the logarithm above). With Ogbon-1's $m = n = 2$ the powers are -1, 0.5 and -0.5. A porosity put in 2.3 porosity units too high, 10 % of the 23.2 % at 5890 ft, reads 17.2 % water instead of 18.9 %, 0.910 times as much, while an $R_w$ 10 % too high reads 19.8 %, 1.05 times as much: the porosity counts about twice as much as either resistivity. And the water's resistivity and the log enter only as the ratio $R_w/R_t$: put in an $R_w$ of 0.080 ohm.m, twice the well's, and read the log twice too high, and the saturation is the built 18.9 % again.

A porosity error costs most where the porosity is least. Three porosity units are 13 % of the upper sand's 23.2 % and 35 % of the tight limestone's 8.5 % at 6080 ft, and they move the sand from 18.9 % water to 16.7 % and the limestone from 21.9 % to 16.1 %. An error that applies to every bed, such as a matrix density that is wrong everywhere, also moves the answer the well is drilled for: with the porosity 3.0 porosity units high throughout, the pay reads 19.5 % water on average and 282.0 ft of net pay, against the 23.0 % and 277.0 ft built.

Where the Equation Stops

Archie's equation is for clean rock in which the brine is the only conductor. PetroWiki says that in the clean-sand model the formation water is the only electrically conductive medium, and that in shaly rocks Rt is suppressed and the Archie Sw calculation comes out too high. The figure therefore reads no saturation in the shales, where the well built its resistivity at 2.20 ohm.m whatever the porosity; Chapter 8 is the shaly sands. Ogbon-1 was built clean even in its shaly oil sand, whose shale volume is 22 % and leaves no mark on its resistivity, so on this well clean Archie is exact, which no real well allows.

The equation has limits in its numbers too. When the log does not stand above its wet-rock resistivity the index is 1 or less and the equation gives 100 % water or more; the kernel and the figure stop at 100 %, and a reading like that says the inputs are wrong, a porosity put in too low or an $R_w$ too high, not that the rock is wetter than water. And Archie (1942) gave the equation for water saturations "down to about 0.15 or 0.20" (Section 6.1); below that it is an extrapolation, and the figure says so under 15 %.

The inputs are never as good as the well's. The sections that follow take them in turn: Section 7.2 is the Pickett plot, 7.3 the bulk-volume water, 7.4 the saturation exponents, 7.5 the Rw problem, and 7.6 puts the equation on the whole of Ogbon-1.

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.