Question 10 of 12: Humble Porosity and Hydrocarbon Saturation from a Core Sample

Question 10: Humble Porosity and Hydrocarbon Saturation from a Core Sample (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) the porosity from the Humble correlation; (b) the physical reason a rock resistivity of 2.2 $\Omega \cdot \text{m}$ becomes 7.5 $\Omega \cdot \text{m}$; (c) the hydrocarbon saturation in each case.

Approach. Obtain the formation resistivity factor from the pair of laboratory measurements, invert the Humble correlation for porosity, and then use the resistivity index to convert the later resistivity into a saturation with Archie's saturation equation.

(a) Porosity from the Humble correlation

1. Extract the formation resistivity factor. By definition $F$ is the ratio of the resistivity of the rock when it is 100 % saturated with a water of resistivity $R_w$ to the resistivity of that water: $$F = \frac{R_o}{R_w} = \frac{2.2}{0.2} = 11.0$$ $F$ is a property of the pore geometry alone. It is dimensionless and, crucially, independent of temperature, because $R_o$ and $R_w$ shift by the same Arps factor.
2. Confirm that the depth does not change $F$. The formation temperature at 12 000 ft is $$T_f = 68 + \left(\frac{1.2\ ^{\circ}\text{F}}{100\ \text{ft}}\right)(12\,000\ \text{ft}) = 68 + 144 = 212\ ^{\circ}\text{F}$$ At that temperature the Arps relation gives $$R_{w@212} = 0.2 \times \frac{68 + 6.77}{212 + 6.77} = 0.0684\ \Omega \cdot \text{m}, \qquad R_{o@212} = F R_{w@212} = 0.752\ \Omega \cdot \text{m}$$ and their ratio is again 11.0. The 12 000 ft datum therefore does not alter the porosity calculation, though it is essential in part (c) if resistivities at depth are wanted.
3. Invert the Humble correlation. For clean, consolidated sandstone the Humble (Winsauer) form supplied on the formula sheet is $$F = \frac{0.62}{\phi^{2.15}} \quad\Longrightarrow\quad \phi = \left(\frac{0.62}{F}\right)^{1/2.15}$$ Substituting $F = 11.0$, $$\phi = \left(\frac{0.62}{11.0}\right)^{0.4651} = (0.05636)^{0.4651} = 0.2625$$ $$\boxed{\phi = 26.2\ \%}$$ A check by substitution returns $0.62/(0.2625)^{2.15} = 11.0$, as required. The value is consistent with a clean, well-sorted, consolidated sandstone.

(b) Why the rock resistivity rose to 7.5 $\Omega \cdot \text{m}$

Part (a) fixes the rock: $F = 11.0$, and at the 212 °F of the zone of interest the same rock fully saturated with the same water must read $R_o = F\,R_{w@212} = 0.752\ \Omega \cdot \text{m}$ — exactly the value the paper's own formula sheet prints as the denominator of $I_R = R_t/R_o = 7.5/0.752$. The later 7.5 $\Omega \cdot \text{m}$ is measured in the zone of interest, so it is compared with that in-situ $R_o$, not with the 2.2 $\Omega \cdot \text{m}$ laboratory value at 68 °F. Pore geometry cannot change on its own and the water resistivity is stated to be unchanged, so a reading about ten times $R_o$ can only mean that the pore space is no longer fully water-saturated: a non-conducting fluid now occupies part of it. Conduction in a clean sandstone is entirely electrolytic, through the connate water; replace some of that water with an insulator and both the cross-sectional area available for current and the tortuosity of the remaining paths get worse, so the resistivity rises steeply.

Two readings of that statement are available, and both are worth giving:

• In the reservoir — the zone of interest has become hydrocarbon-bearing: oil or gas has migrated into the rock, which was water-filled when first drilled, and displaced part of the formation water. Oil and gas are effectively infinite resistors, so $R_t > R_o$ is the fundamental hydrocarbon indicator, and the ratio $R_t/R_o$ is precisely the resistivity index of Archie's second equation. The magnitude here, a resistivity index of about 10, corresponds to roughly two-thirds of the pore volume being occupied by hydrocarbon.
• In the laboratory — if the elevated value came from a repeat measurement on the core after it had been standing, the same physics applies to a different insulator: the sample has partially dried out or been desaturated, with air (also non-conducting) replacing evaporated pore water. A secondary contributor is the loss of the thin conductive water film and any salt precipitation as the brine concentrated. This is why core resistivity measurements must be made on freshly preserved, fully re-saturated plugs.

What the change is not: it is not a temperature effect (both $R_o = 0.752$ and $R_t = 7.5\ \Omega \cdot \text{m}$ refer to the 212 °F zone, and heating a brine-filled rock would in any case lower its resistivity), and it is not a change in $R_w$ or in porosity, both of which are given as unchanged.

(c) Hydrocarbon saturation in each case

1. Case (a) — fully water-saturated rock. The sample of part (a) is stated to be 100 % saturated with water, which is what makes its resistivity $R_o$ by definition: $$S_w = 1.00 \quad\Longrightarrow\quad \boxed{S_h = 1 - S_w = 0\ \%}$$ This is the reference state, not a result to be computed — and recognising that is part of the question.
2. Case (b) — form the resistivity index. Both the later reading and the reference wet-rock resistivity refer to the zone of interest at 212 °F, which is the form the formula sheet prints: $$I_R = \frac{R_t}{R_o} = \frac{7.5}{0.752} = 9.97$$
3. Apply Archie's saturation equation. With the saturation exponent $n = 2$ used throughout this paper, $$S_w = I_R^{-1/n} = \left(\frac{R_o}{R_t}\right)^{1/2} = \left(\frac{0.752}{7.5}\right)^{1/2} = (0.1003)^{1/2} = 0.3166$$ so that $$\boxed{S_w = 31.7\ \%, \qquad S_h = 1 - S_w = 68.3\ \%}$$
4. Sanity-check the result. A hydrocarbon saturation of 68 % in a 26 % porosity sand is a good hydrocarbon-bearing zone; whether it would flow acceptably depends on a relative-permeability cutoff of the kind computed in Question 9. The step that decides the answer is using a consistent reference: dividing the in-situ 7.5 $\Omega \cdot \text{m}$ by the 2.2 $\Omega \cdot \text{m}$ laboratory value at 68 °F would mix two temperatures, give $I = 3.4$ and $S_h = 46\ \%$, and understate the hydrocarbon saturation by more than 20 saturation units — the formula sheet's own $I_R = 7.5/0.752$ confirms the in-situ reference is the intended one.