Question 10 of 12: Humble's Correlation and Hydrocarbon Saturation vs. Depth

Question 10: Humble's Correlation and Hydrocarbon Saturation vs. Depth (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) Formation porosity via Humble's correlation. (b) The physical reason R_t rose to 7.5 Ω·m. (c) Hydrocarbon saturation for cases a and b.

Approach. Scale both R_o and R_w from surface temperature to the in-situ formation temperature with the Arps relation, form the formation resistivity factor F=R_o/R_w, invert Humble's correlation for φ, then use Archie's saturation equation on both the water-saturated and the later, higher-resistivity reading.

1. Formation temperature at 12,000 ft. $$T_f = T_s + \left(\frac{1.2\,{}^{\circ}\text{F}}{100\ \text{ft}}\right)(12{,}000\ \text{ft}) = 68 + 144 = \boxed{212\,{}^{\circ}\text{F}}$$
2. Scale R_w and R_o from 68°F to 212°F using $R_2=R_1(T_1+6.77)/(T_2+6.77)$ (formula sheet): $$R_w(212\,{}^{\circ}\text{F}) = 0.2\times\frac{68+6.77}{212+6.77}=0.2\times0.3418=\boxed{0.0684\ \Omega\cdot\text{m}}$$ $$R_o(212\,{}^{\circ}\text{F}) = 2.2\times\frac{68+6.77}{212+6.77}=2.2\times0.3418=\boxed{0.752\ \Omega\cdot\text{m}}$$ (this R_o=0.752 Ω·m matches the paper's own attachment hint, I_R=R_t/R_o=7.5/0.752, confirming the scaling.)
3. Formation resistivity factor and Humble's correlation. $$F=\frac{R_o}{R_w}=\frac{0.752}{0.0684}=\boxed{11.0}$$ Humble's correlation (formula sheet), $F=0.62/\phi^{2.15}$, inverted for φ: $$\phi = \left(\frac{0.62}{F}\right)^{1/2.15}=\left(\frac{0.62}{11.0}\right)^{1/2.15}=\boxed{26.2\%}$$

(b) Reason for the resistivity change

The original 2.2 Ω·m reading was measured on a sample 100% saturated with water (i.e. it is R_o, the water-resistivity-index baseline). The later reading of 7.5 Ω·m is more than nine times higher (I_R=R_t/R_o=7.5/0.752≈10 at formation temperature) with the SAME rock and water properties, i.e. the same F and R_w. Since resistivity in a clean formation rises only when non-conductive hydrocarbon displaces conductive brine (Question 1a), the only explanation consistent with unchanged rock/water properties is that the zone of interest is no longer 100% water-saturated — oil (or gas) has since moved into (or was always present in) the pore space, so the later reading of 7.5 Ω·m is the TRUE resistivity R_t of a hydrocarbon-bearing zone, not R_o. (The alternative explanation of a measurement/temperature error is ruled out by the fact that 7.5 Ω·m is exactly the paper's own attachment value for R_t, and the resulting S_w in part (c) is a physically reasonable oil saturation, not an artifact.)

(c) Hydrocarbon saturation, cases a and b

1. Case a (the original, water-saturated sample): here R_t=R_o by definition, so from Archie's simplified equation (formula sheet, n=2): $$S_w=\sqrt{\frac{R_o}{R_t}}=\sqrt{\frac{0.752}{0.752}}=\boxed{1.0\ (100\%)}\quad\Rightarrow\quad S_h = 0\%$$
2. Case b (R_t=7.5 Ω·m): $$S_w=\sqrt{\frac{R_o}{R_t}}=\sqrt{\frac{0.752}{7.5}}=\sqrt{0.1003}=\boxed{31.7\%}\quad\Rightarrow\quad S_h = 1-S_w=\boxed{68.3\%}$$