Archie's Equation

Part 7, Chapter 7: Archie's Equation and Water Saturation

Learning objectives

Write Archie's saturation equation and name each term's source
Read it as the two-step Ro then Sw
Compute the water saturation from Rt, porosity, and Rw
Find the bulk-volume hydrocarbon phi*(1-Sw)

The Whole Chain in One Formula

Here is where the course pays off. Every log of the last six chapters feeds one equation, Archie's law for water saturation:

S_w = \left(\frac{a\,R_w}{\phi^{\,m}\,R_t}\right)^{1/n}.

The porosity $\phi$ came from the density, neutron, or sonic; the water resistivity $R_w$ from the SP; the true resistivity $R_t$ from the resistivity tools; and $a$ , $m$ , $n$ from the formation factor. Archie just combines them into the one number the well was drilled to find.

Two Steps

It reads most clearly as two steps, the resistivity ratio of the last chapter in full. First build the wet-rock resistivity from the formation factor and the water, $R_o = a,R_w/\phi^m$ ; then compare it with the measured $R_t$ :

S_w = \left(\frac{R_o}{R_t}\right)^{1/n}.

When the rock is fully wet, $R_t = R_o$ and $S_w = 1$ ; the further $R_t$ climbs above $R_o$ , the lower the water saturation.

What Is Left Is Pay

The figure builds a unit of rock: grain, then the pore split into water ( $\phi S_w$ ) and hydrocarbon ( $\phi(1-S_w)$ ). That last slice, the bulk-volume hydrocarbon, is the prize. In the Ogbon-1 pay it is about 19 percent of the pore, so most of the porosity holds oil. Higher $R_t$ , lower $R_w$ , and lower porosity all push $S_w$ down and the pay up. The rest of the chapter sharpens this: the Pickett plot, the bulk-volume check, the exponents, and the Rw the answer hangs on.

References

Archie, G. E. (1942). The electrical resistivity log as an aid in determining some reservoir characteristics. Transactions of the AIME, 146(1).
Asquith, G. and Krygowski, D. (2004). Basic Well Log Analysis, 2nd ed. AAPG Methods in Exploration 16.