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. The paper supplies the gas-sand crossplot on the Attachment page (density porosity φ_D against neutron porosity φ_N, contoured for porosity and for S_xo), and that chart prints its own fluid and matrix constants, so none of them has to be assumed:

Find. True porosity φ and flushed-zone filtrate saturation S_xo, both without and with the neutron excavation-effect correction.

Approach. In a gas-bearing flushed zone the two porosity logs see two different mixtures of the same three components (matrix, filtrate, residual gas), so they give two independent equations in the two unknowns φ and S_xo — no resistivity log is needed. Without the excavation effect the pair solves in closed form. With it, the neutron equation picks up the Segesman–Liu excavation term and no longer inverts by hand, which is exactly why the paper supplies the chart: it is that same pair of equations drawn out, so the answer is read off it.

1. Apparent density porosity from ρ_b. Scaling the density log on the chart's own fluid density (ρ_f = ρ_mf = 1.00): $$\phi_{Da}=\frac{\rho_{ma}-\rho_b}{\rho_{ma}-\rho_{mf}}=\frac{2.65-2.0}{2.65-1.00}=\frac{0.65}{1.65}=\boxed{39.4\%}$$ The pair (φ_Na, φ_Da) = (5.0, 39.4) p.u. is the entry point on the chart, and the 34-p.u. separation is the classic gas signature: gas's near-zero hydrogen index makes the neutron tool under-read, while its near-zero density makes the density tool over-read.
2. The two tool responses, written out. With the flushed zone holding filtrate (S_xo) plus residual gas (1−S_xo), and ρ_g = 0 so the gas carries neither mass nor hydrogen: $$\rho_b=(1-\phi)\rho_{ma}+\phi S_{xo}\rho_{mf}+\phi(1-S_{xo})\rho_g \;\Longrightarrow\; \rho_b=\rho_{ma}-\phi\,\rho_{ma}+\phi S_{xo}$$ $$\phi_{Na}=\phi S_{xo}-\Delta\phi_{ex},\qquad \Delta\phi_{ex}=\left(2\phi^2S_{xo}+0.04\,\phi\right)\!\left(1-S_{xo}\right)$$ Δφ_ex is the excavation effect: the gas-filled part of the pore space not only contains no hydrogen, it also removes matrix atoms that would themselves have slowed neutrons, so the neutron tool under-reads by MORE than the hydrogen-index deficit alone. Setting Δφ_ex = 0 gives the "without excavation effect" case.
3. φ and S_xo WITHOUT the excavation effect. With Δφ_ex = 0 the neutron equation is simply φS_xo = φ_Na = 0.05, and substituting that product straight into the density equation eliminates S_xo: $$\rho_b=\rho_{ma}-\phi\rho_{ma}+\phi_{Na}\quad\Longrightarrow\quad \phi=\frac{\rho_{ma}-\rho_b+\phi_{Na}}{\rho_{ma}}=\frac{2.65-2.00+0.05}{2.65}=\frac{0.70}{2.65}=\boxed{26.4\%}$$ $$S_{xo}=\frac{\phi_{Na}}{\phi}=\frac{0.05}{0.2642}=\boxed{18.9\%}$$
4. φ and S_xo WITH the excavation effect — read from the attachment chart. Entering the chart at φ_N = 5.0 p.u. on the horizontal axis and φ_D = 39.4 p.u. on the vertical axis, the point falls between the S_xo = 30% and S_xo = 40% curves (about one-third of the way across) and between the 25% and 30% porosity contours (about one-third of the way up): $$\phi\approx\boxed{26.6\%},\qquad S_{xo}\approx\boxed{33\%}$$
5. Cross-check on the chart reading. Substituting φ = 0.266, S_xo = 0.333 back into the excavation term reproduces the logged neutron reading: $$\Delta\phi_{ex}=\left(2(0.266)^2(0.333)+0.04(0.266)\right)(1-0.333)=0.0385$$ $$\phi_{Na}=\phi S_{xo}-\Delta\phi_{ex}=0.0885-0.0385=0.0500\;\checkmark$$ Solving the same two equations numerically with the printed ρ_g = 0 gives φ = 28.0%, S_xo = 32.9% — the same saturation to within a point, and a porosity 1.4 p.u. above the chart, which is the chart's own construction tolerance.