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24-Pet-B1 Natural Gas Engineering · December 2016

Question 9 of 11: ESSP with an ideal and a nonideal shale membrane

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2016. 98-Pet-B1, Well Logging and Formation Evaluation (every question is log-interpretation content, not gas-engineering material). 3-hour closed-book exam, 11 questions, all marked, calculators and attached graphs/formula sheet permitted.

Reference texts: Bassiouni, Theory, Measurement, and Interpretation of Well Logs (SPE Textbook Series Vol. 4); Asquith & Krygowski, Basic Well Log Analysis, 2nd ed.; Ellis & Singer, Well Logging for Earth Scientists, 2nd ed.; Schlumberger, Log Interpretation Charts.

Check: Q3, Q7, Q8, Q9(b), Q10 and Q11 are built on the paper's printed logs and attached charts. Values printed as annotations on the logs (Q11's SSP, PSP and GR labels) are used exactly as printed. Values read off a curve or a chart (Q9(b) and the Q8 chart check, Q10's track readings) are read from the printed figure and flagged inline with their precision. All arithmetic that follows is exact.

Question 9: ESSP with an ideal and a nonideal 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.

[Figure not reproduced: Attached nonideal-membrane SP chart with the Q9(b) reading marked. See the official exam paper or the cited reference text.]

Fig. Q9 — The paper's attached nonideal shale-membrane chart (page 23): $R_{mf}/R_w$ against $E_{SP}$, with curves for $(R_{sh}/R_{mf})$ at formation temperature. The marked point is part (b): the $R_{sh}/R_{mf}=4$ curve at $R_{mf}/R_w=5$.

Given. $T_f=200$ °F, so $K=87.9$ (from Q4). $R_{mf}=0.5\ \Omega\cdot\text{m}$ and $R_w=0.1\ \Omega\cdot\text{m}$, both already at formation temperature. Part (b) adds $R_{sh}=2\ \Omega\cdot\text{m}$ at 200 °F.

(a) Perfect (ideal) shale membrane. An ideal membrane passes only Na⁺ ions (it is fully cation-selective), so the standard ESSP formula applies directly with no correction: $$ESSP=-K\log_{10}\!\left(\frac{R_{mf}}{R_w}\right)=-87.9\log_{10}\!\left(\frac{0.5}{0.1}\right)=-87.9\log_{10}(5)$$ $$\boxed{ESSP_{ideal}\approx -61.4\ \text{mV}}$$

(b) Nonideal shale membrane. A real shale is only partly cation-selective. Its departure from ideal behaviour is read from the attached two-parameter chart:

  1. Curve parameter. $R_{sh}/R_{mf}=2/0.5=4$.
  2. Chart entry. $R_{mf}/R_w=0.5/0.1=5$ on the log-scaled vertical axis.
  3. Read $E_{SP}$. Follow the horizontal line $R_{mf}/R_w=5$ across to the curve labelled 4, then drop down to the $E_{SP}$ axis. Curve 4 is fairly flat here: reading the chart places it at −22 mV on $R_{mf}/R_w=4.8$ and at −35 mV on 5.2. The reading is therefore Check: chart reading, about ±5 mV $$\boxed{ESSP_{nonideal}\approx -28\ \text{mV}}$$
CaseESSP
(a) Ideal shale membrane≈ −61.4 mV
(b) Nonideal membrane, $R_{sh}=2\ \Omega\cdot$m≈ −28 mV (chart read, about ±5 mV)
Check: part (b) is read off the attached chart, not calculated from a closed form. Because curve 4 is shallow at $R_{mf}/R_w=5$, a small misreading of the vertical position moves $E_{SP}$ by several millivolts. The conclusion does not depend on that precision: a shale only 4 times more resistive than the mud filtrate leaks enough current to cut the SP to under half of its ideal magnitude (−28 vs. −61 mV).