Question 12 of 12: Static SP — Ideal and Non-Ideal Shale Membrane

Question 12: Static SP — Ideal and Non-Ideal Shale Membrane (10 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. R_mf=0.6 Ω·m, R_w=0.12 Ω·m, T_f=200°F; non-ideal-membrane SP reading = −60 mV.

Find. (a) E_SSP for an ideal shale membrane. (b) R_sh, the adjacent shale's own resistivity, given the reduced (non-ideal) SP reading.

Approach. Since both R_mf and R_w exceed ≈0.1 Ω·m (the range in which the equivalent-resistivity chart converges toward the actual value), take (R_mf)_eq≈R_mf and (R_w)_eq≈R_w directly in the ESSP formula; then use the (R_mf/R_w) vs. E_SP chart (parameterized by R_sh/R_mf) to back out R_sh for the reduced, non-ideal reading.

1. Part (a) — ideal shale membrane. $$K = 61.3+0.133\,T_f = 61.3+0.133(200) = \boxed{87.9}$$ $$E_{SSP} = -K\log\left(\frac{R_{mf}}{R_w}\right)_{eq} = -87.9\log\left(\frac{0.6}{0.12}\right) = -87.9\log(5) = -87.9(0.699) = \boxed{-61.4\ \text{mV}}$$
2. Part (b) — non-ideal membrane, back out R_sh. With a non-ideal (partially conductive) shale membrane, the measured SP is reduced in magnitude below the ideal E_SSP ( −60 mV vs. the ideal −61.4 mV) by an amount that depends on how resistive the adjacent shale is relative to the mud filtrate. Compute the fixed ratio $$\frac{R_{mf}}{R_w} = \frac{0.6}{0.12} = \boxed{5.0}$$ and locate the point (E_SP=−60 mV, R_mf/R_w=5.0) on the family of R_sh/R_mf curves of the departure chart (source page 16). At E_SP=−60 mV the printed curves read R_mf/R_w = 11.4 (curve 1), 8.5 (curve 2), 6.0 (curve 4), 5.0 (curve 6) and 4.3 (curve 8), so the point falls essentially exactly on the $R_{sh}/R_{mf}=6$ curve, and $$R_{sh} = 6\times R_{mf} = 6(0.6) = \boxed{3.6\ \Omega\cdot\text{m}}$$