Water-Saturation Cutoff and Minimum Productive Resistivity: 24-Pet-B1 Natural Gas Engineering, May 2014, Q9

Question 9: Water-Saturation Cutoff and Minimum Productive Resistivity (6 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
Water cut, f_w	40%
μ_o / μ_w	4
Formation-water resistivity, R_w	0.06 Ω·m
Formation resistivity factor, F	18
Saturation exponent, n	2
K_w/K_o vs. S_w curve	Attachment (p.18)

Find. The water-saturation cutoff S_cw at 40% water cut, and the minimum productive resistivity R_mp.

Approach. Invert the fractional-flow equation for the target k_rw/k_ro, read the corresponding S_w off the digitized K_w/K_o chart, then substitute that S_w as S_cw into the R_mp formula from the attachment.

Invert the fractional-flow equation for k_ro/k_rw. $f_w=\frac{1}{1+\dfrac{k_{ro}\mu_w}{k_{rw}\mu_o}} \;\Rightarrow\; \frac{k_{ro}}{k_{rw}}=\left(\frac{1}{f_w}-1\right)\frac{\mu_o}{\mu_w}=\left(\frac{1}{0.40}-1\right)(4)=\boxed{6.0}$ so the target relative-permeability RATIO plotted on the chart is $\frac{k_{rw}}{k_{ro}}=\frac{1}{6.0}=0.167$
Read S_w off the K_w/K_o-vs.-S_w chart at ratio = 0.167. Digitizing the printed curve against its own grid gives the anchor points (33.0%, 0.01), (47.5%, 1.0), (55%, 10), (63%, 100) — it is very nearly a straight line on the semilog grid. Interpolating log-linearly between the 33.0%/47.5% anchors for a ratio of 0.167 (log₁₀=−0.778): $S_w = 33.0 + \frac{-0.778-(-2)}{0-(-2)}\,(47.5-33.0) = \boxed{41.9\%\approx 42\%}$

Water-oil relative permeability ratio vs. water saturation, read from the Attachment chart, with the 40%-water-cut operating point marked.

Minimum productive resistivity. Taking S_cw=0.419 as the saturation cutoff and substituting into the attachment's R_mp formula: $R_{mp}=\frac{FR_w}{(S_{cw})^n}=\frac{(18)(0.06)}{(0.419)^2}=\frac{1.08}{0.1752}=\boxed{6.16\ \Omega\cdot\text{m}}$

Quantity	Result
k_ro/k_rw at 40% water cut	6.0
Water-saturation cutoff, S_cw	≈ 41.9% (0.419)
Minimum productive resistivity, R_mp	6.16 Ω·m

Check: S_w=41.9% is read from the printed semilog relative-permeability chart by reading the curve against its own decade lines and 2%-S_w grid; a ±1 percentage-point reading tolerance moves R_mp by about ±0.3 Ω·m through the squared denominator.

Topic: Water cut, fractional flow, and the resistivity productivity cutoff.

Key relations: $f_w=1/(1+k_{ro}\mu_w/(k_{rw}\mu_o))$ , $R_{mp}=FR_w/S_{cw}^n$ .

Why this works: A well produces at an uneconomic water cut once the mobility ratio k_rwμ_o/(k_roμ_w) reaches the value implied by that cut; converting the corresponding S_w into a resistivity via Archie gives a single number (R_mp) a log analyst can compare directly against R_t to flag productive vs. sub-economic pay without re-running the fractional-flow calculation on every foot of log.

Common pitfall: Plugging S_w=40% directly as the water CUT rather than first inverting f_w through the mobility ratio — the two are numerically different quantities that happen to be easy to conflate.

Source: Bassiouni, Ch. 9 (productivity cutoffs); Craft & Hawkins, Applied Petroleum Reservoir Engineering, Ch. 6 (relative permeability).

Question 9 of 12: Water-Saturation Cutoff and Minimum Productive Resistivity

Question 9: Water-Saturation Cutoff and Minimum Productive Resistivity (6 marks)