24-Pet-B1 Natural Gas Engineering · December 2016

Question 4 of 11: SSP for a clean water-bearing sand

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 4: SSP for a clean water-bearing sand (4 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.

Quantity	Value
Formation temperature, $T_f$	200 °F
$R_{mf}$ at 68 °F	0.31 Ω·m
$R_w$ at 68 °F	0.054 Ω·m

Find. The static spontaneous potential, SSP, at formation temperature.

Approach. Apply the standard SSP relation $SSP=-K\log_{10}(R_{mf}/R_w)$ at formation temperature, with $K=61.3+0.133\,T_f({}^{\circ}\mathrm{F})$ (this exam's own attached formula sheet).

Temperature coefficient K. $$K = 61.3+0.133(200) = 87.9$$
Resistivity ratio at formation temperature. $R_{mf}$ and $R_w$ are both measured at the same reference temperature (68 °F) and converted to the same target temperature (200 °F) by the identical Arps factor $(T_1+6.77)/(T_2+6.77)$ — that factor cancels in the ratio, so $$\frac{R_{mf}}{R_w}\bigg|_{200^\circ F}=\frac{R_{mf}}{R_w}\bigg|_{68^\circ F}=\frac{0.31}{0.054}=5.741$$
SSP. $$SSP=-K\log_{10}\!\left(\frac{R_{mf}}{R_w}\right)=-87.9\log_{10}(5.741)=-87.9(0.7590)$$ $$\boxed{SSP \approx -66.7\ \text{mV}}$$

Quantity	Result
K	87.9
$R_{mf}/R_w$	5.74
SSP	≈ −66.7 mV

Topic: Static spontaneous potential (SSP) of a clean NaCl sand.

Key relations: $SSP=-K\log_{10}(R_{mf}/R_w)$, $K=61.3+0.133\,T({}^{\circ}\mathrm{F})$; Arps: $R_2=R_1\dfrac{T_1+6.77}{T_2+6.77}$.

Why this works: SSP is an electrochemical potential generated across the shale membrane and at the mud-filtrate/formation-water liquid junction, driven purely by the salinity (activity) contrast between $R_{mf}$ and $R_w$; because both values here share the same 68→200 °F conversion factor, that factor drops out of the ratio and only $K$ (which is itself temperature-dependent) needs the formation temperature.

Common pitfall: Converting $R_{mf}$ and $R_w$ to formation temperature individually and then taking the ratio — correct, but unnecessary extra arithmetic; recognizing the cancellation saves a step and a rounding-error source.

Source: Bassiouni Ch. 5 (SP Log); Schlumberger Log Interpretation Charts (SP chart Sw-1).

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