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

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.

Find. (i) The water saturation $S_{w,cut}$ at which the well would produce a 30 % water cut, and (ii) the corresponding minimum productive true resistivity $R_{mp}$ — the deep-resistivity value below which the formation is not worth completing.

Approach. Invert the fractional-flow equation to convert the 30 % water cut into a relative-permeability ratio, read the saturation off the supplied $k_w/k_o$ chart, then substitute that saturation into Archie's saturation equation to obtain the resistivity cutoff.

1. Write the fractional flow of water at reservoir conditions. Neglecting gravity and capillary terms, the ratio of volumetric rates from Darcy's law for two phases in the same rock is $q_o/q_w = (k_o/\mu_o)/(k_w/\mu_w)$, so $$f_w = \frac{q_w}{q_w + q_o} = \frac{1}{1 + \dfrac{k_o}{k_w}\cdot\dfrac{\mu_w}{\mu_o}}$$
2. Solve for the relative-permeability ratio at the target water cut. Rearranging and substituting $f_w = 0.30$ and $\mu_w/\mu_o = 0.25$, $$\frac{k_o}{k_w} = \frac{1}{\mu_w/\mu_o}\left(\frac{1}{f_w} - 1\right) = \frac{1}{0.25}\left(\frac{1}{0.30} - 1\right) = 4 \times 2.3333 = 9.333$$ so that $$\frac{k_w}{k_o} = \frac{1}{9.333} = 0.107$$
3. Read the saturation from the relative-permeability chart. Entering the attachment chart on its logarithmic $k_w/k_o$ axis at 0.107 and dropping to the saturation axis, $$\boxed{S_{w,cut} \approx 39\ \%\ \ (0.394)}$$ Above this saturation the water cut exceeds 30 %; below it the well produces acceptably dry oil.
4. Convert the saturation cutoff into a resistivity cutoff. Archie's saturation equation with $n = 2$ gives $$S_w^{\,n} = \frac{F R_w}{R_t} \quad\Longrightarrow\quad R_{mp} = \frac{F R_w}{S_{w,cut}^{\,n}}$$ Substituting $F = 18$, $R_w = 0.045\ \Omega \cdot \text{m}$ and $S_{w,cut} = 0.394$, $$R_{mp} = \frac{18 \times 0.045}{(0.394)^2} = \frac{0.81}{0.1552}$$ $$\boxed{R_{mp} = 5.2\ \Omega \cdot \text{m}}$$
5. Interpret the number. The wet-rock resistivity of this formation is $R_o = F R_w = 18 \times 0.045 = 0.81\ \Omega \cdot \text{m}$, so the cutoff sits at a resistivity index of $R_{mp}/R_o = 6.4$. Any interval whose deep resistivity reads below about 5.2 $\Omega \cdot \text{m}$ carries $S_w > 39\ \%$ and would come in above a 30 % water cut; intervals reading appreciably above it are the completion candidates.