Question 12 of 12: Ideal and Nonideal Shale-Membrane SP

Question 12: Ideal and Nonideal Shale-Membrane SP (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.

Find. (a) the static SP an ideal (perfectly cation-selective) shale membrane would produce; (b) the resistivity of the adjacent shale that would reduce the SP to the measured −45 mV through a nonideal membrane.

Approach. For (a), convert both resistivities to their NaCl equivalents and apply the electrochemical SP equation. For (b), the ideal equation no longer holds — a leaky membrane passes some anions and generates less potential — so enter the two-parameter departure chart supplied on attachment page 16 with the measured SP and the resistivity ratio, and read the shale-to-filtrate resistivity ratio off the curve family.

(a) Static SP for an ideal shale membrane

1. Evaluate the SP coefficient at formation temperature. From the formula sheet, $$K = 61.3 + 0.133\,T_f = 61.3 + 0.133(180) = 85.24\ \text{mV}$$
2. Convert both resistivities to NaCl equivalents. The formation water is stated to be fresh, and referred back to 75 °F both fluids exceed 0.1 $\Omega \cdot \text{m}$ by a wide margin ($R_{mf@75} = 1.28$, $R_{w@75} = 0.37\ \Omega \cdot \text{m}$), so the equivalent-resistivity conversion of chart SP-2 applies to both. In that regime the chart is well represented by the standard factor $$(R_{mf})_{eq} = 0.85\,R_{mf} = 0.85(0.56) = 0.476\ \Omega \cdot \text{m}, \qquad (R_w)_{eq} = 0.85\,R_w = 0.85(0.16) = 0.136\ \Omega \cdot \text{m}$$ Because the same factor applies to both fluids, it cancels in the ratio: $$\frac{(R_{mf})_{eq}}{(R_w)_{eq}} = \frac{R_{mf}}{R_w} = \frac{0.56}{0.16} = 3.50$$
3. Apply the electrochemical SP equation. $$E_{SSP} = -K \log_{10}\!\left[\frac{(R_{mf})_{eq}}{(R_w)_{eq}}\right] = -85.24 \times \log_{10}(3.50) = -85.24 \times 0.5441$$ $$\boxed{E_{SSP} = -46.4\ \text{mV}}$$

This is the deflection a thick, clean, permeable bed bounded by a perfectly cation-selective shale would produce — the theoretical maximum for this fluid pair at 180 °F. The measured −45 mV of part (b) is 97 % of it, which is the first sign that the membrane is close to, but not quite, ideal.

(b) Resistivity of the adjacent shale for a nonideal membrane

A real shale is a leaky membrane: it is cation-selective only in proportion to its clay-bound charge density, and the more conductive it is relative to the borehole fluid, the more anions leak through and the smaller the potential it generates. The relevant parameter is the ratio of shale resistivity to mud-filtrate resistivity at formation temperature, $(R_{sh}/R_{mf})_{F.T.}$, and it cannot be recovered from the ideal equation — which is precisely why the exam supplies the departure chart on page 16.

[Figure not reproduced: Schematic of the nonideal-shale-membrane departure chart supplied on attachment page 16, redrawn from the digitised curve family. Entering at $E_{SP} = -45$ mV and $R_{mf}/R_w = 3.5$ falls between the curves for $R_{sh}/R_{mf}$ of 8 and 10. See the official exam paper.]

1. Form the chart entry values. The chart's ordinate is the actual resistivity ratio and its abscissa is the measured SP: $$\frac{R_{mf}}{R_w} = \frac{0.56}{0.16} = 3.50, \qquad E_{SP} = -45\ \text{mV}$$
2. Enter the chart and read the curve parameter. Moving vertically from $E_{SP} = -45$ mV to an ordinate of 3.50, the point falls between the $(R_{sh}/R_{mf}) = 8$ curve (which passes at 3.63 there) and the $(R_{sh}/R_{mf}) = 10$ curve (at 3.08). Logarithmic interpolation between them gives $$\left(\frac{R_{sh}}{R_{mf}}\right)_{F.T.} \approx 8.4$$
3. Convert to the shale resistivity. Multiplying by the mud-filtrate resistivity at formation temperature, $$R_{sh} = 8.4 \times R_{mf} = 8.4 \times 0.56$$ $$\boxed{R_{sh} \approx 4.7\ \Omega \cdot \text{m}}$$
4. Check that the answer is physically sensible. A shale roughly eight times more resistive than the mud filtrate is a firm, low-porosity shale, and the SP it supports, −45 mV, is a little smaller in magnitude than the ideal −46.4 mV — the right direction, since a leaky membrane can only reduce $|E_{SP}|$, never increase it. Had the reading been much smaller, say −25 mV, the same chart would have returned a far less resistive (more conductive, more leaky) shale.