Question 7 of 10: Neutron-Density Porosity and Sxo in a Gas Sand

Question 7: Neutron-Density Porosity and Sxo in a Gas Sand (15 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. True porosity φ and flushed-zone filtrate saturation S_xo, both without and with the neutron excavation-effect correction.

Approach. Nothing here needs to be assumed. The attachment reproduces BOTH the φ_D–φ_N gas crossplot (source p.13, carrying true-porosity and S_xo contours and an annotation box reading ρ_mf=1, ρ_g=0, ρ_ma=2.65 g/cm³) and the formula sheet (source p.12), which prints the two relations this question is built on:

Part (i) is those two lines evaluated directly. Part (ii) re-solves the same pair with the Segesman–Liu excavation term added to the neutron response, which is exactly how the printed chart is constructed — so part (ii) can equally be read straight off the attachment.

1. Apparent density porosity from ρ_b (the chart entry point). $$\phi_{D}=\frac{\rho_{ma}-\rho_b}{\rho_{ma}-\rho_{mf}}=\frac{2.65-2.0}{2.65-1.00}=\frac{0.65}{1.65}=\boxed{39.4\%}$$ Against φ_N=5% that is a 34 p.u. separation — the classic neutron-density gas crossover — and the point (φ_N, φ_D) = (5, 39.4) is where the attachment chart is entered.
2. WITHOUT the excavation effect: the formula sheet in one line. With ρ_g=0 the three-component density mixing law $\rho_b=(1-\phi)\rho_{ma}+\phi S_{xo}\rho_{mf}+\phi(1-S_{xo})\rho_g$ collapses to $\rho_b=\rho_{ma}-\phi\rho_{ma}+\phi S_{xo}$, and because $\phi S_{xo}$ IS the neutron reading, no iteration is needed: $$\phi=\frac{\rho_{ma}-\rho_b+\phi_N}{\rho_{ma}}=\frac{2.65-2.00+0.05}{2.65}=\frac{0.70}{2.65}=\boxed{26.4\%}$$ $$S_{xo}=\frac{\phi_N}{\phi}=\frac{0.05}{0.2642}=\boxed{18.9\%}$$
3. WITH the excavation effect: what the correction actually is. Replacing liquid-filled pore volume with gas removes far more hydrogen than simple volumetric mixing allows for, so the neutron tool reads LOWER than $\phi S_{xo}$ by the Segesman–Liu excavation term: $$\phi_N=\phi S_{xo}-\Delta\phi_{ex},\qquad \Delta\phi_{ex}=\left(2\phi^2S_{xo}+0.04\phi\right)\left(1-S_{xo}\right)$$ The density leg picks up the companion electron-density effect: a gas of true density $\rho_g$ is seen by the density tool as an APPARENT density $\rho_{g,a}=1.335\rho_g-0.188$, i.e. $-0.19$ g/cm³ for the chart’s $\rho_g=0$, so $$\phi_D=\frac{\phi\left[\rho_{ma}-S_{xo}\rho_{mf}-(1-S_{xo})\rho_{g,a}\right]}{\rho_{ma}-\rho_{mf}}=\frac{\phi\left(2.84-1.19\,S_{xo}\right)}{1.65}$$
4. Solve the pair for φ and S_xo. Two equations, two unknowns, with $\phi_D=0.3939$ and $\phi_N=0.05$. Eliminating S_xo between them and solving by bisection: $$\phi=\boxed{26.6\%},\qquad S_{xo}=\boxed{33.3\%}$$
5. Cross-check against the printed chart. The same two equations reproduce the attachment’s own contours: along the φ_D=39.4 p.u. row the S_xo=10/20/30/40/50% curves compute to φ_N=0.50/2.20/4.24/6.71/9.71 p.u. against 0.49/2.18/4.29/6.80/9.81 p.u. read from the printed chart, and the φ=30% true-porosity contour to 12.0 p.u. against 11.8 p.u. read from the chart. Entering the chart at (5, 39.4) therefore lands between its 30% and 40% S_xo curves and just above its 25% porosity contour — a direct chart read of φ about 26–27% and S_xo about 33%, matching the algebra.