24-Pet-B1 Natural Gas Engineering · December 2016

Question 8 of 11: Neutron-density porosity and S_xo in a gas sand

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 8: Neutron-density porosity and S_xo 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 $\phi$ and flushed-zone filtrate saturation $S_{xo}$, first without and then with the excavation effect.

Approach. In a gas sand, each tool responds to both unknowns. The neutron sees only the filtrate's hydrogen, so $\phi_N$ depends on $\phi$ and $S_{xo}$. The density sees the average pore-fluid density, which is also set by $S_{xo}$. Write one response equation per tool and solve the pair simultaneously. Without the excavation effect the pair is linear. With it, the neutron equation gains the excavation term, and the paper's attached $\phi_D$-vs-$\phi_N$ chart (drawn for $\rho_{mf}=1$, $\rho_g=0$, $\rho_{ma}=2.65$) is the graphical solution of the same pair.

1. Apparent density porosity (fresh-water scale). $$\phi_{D}=\frac{\rho_{ma}-\rho_b}{\rho_{ma}-\rho_{mf}}=\frac{2.65-2.0}{2.65-1.0}=\frac{0.65}{1.65}=0.394\ (39.4\%)$$
2. Tool response equations WITHOUT excavation effect. Take $HI_{gas}=0$ and $\rho_g=0$: $$\phi_N=\phi\,S_{xo}\qquad \phi_D=\frac{\phi\,(\rho_{ma}-S_{xo}\rho_{mf})}{\rho_{ma}-\rho_{mf}}=\frac{\phi\,(2.65-S_{xo})}{1.65}$$ Multiply the density equation by 1.65 and add the neutron equation. The $S_{xo}$ terms cancel: $$1.65\,\phi_D+\phi_N=2.65\,\phi\;\Rightarrow\;\phi=\frac{0.65+0.05}{2.65}$$ $$\boxed{\phi_{noEE}\approx 0.264\ (26.4\%)}$$ $$S_{xo}=\frac{\phi_N}{\phi}=\frac{0.05}{0.264}\qquad \boxed{S_{xo,noEE}\approx 0.189\ (18.9\%)}$$
3. Excavation effect. Residual gas displaces formation that would otherwise contain hydrogen-bearing solid and liquid. The neutron therefore reads lower still than $\phi S_{xo}$, by $$\Delta\phi_{Nex}=K\big(2\phi^2S_{xo}+0.04\,\phi\big)\big(1-S_{xo}\big),\qquad K=1\ \text{(sandstone)}$$ so $\phi_N=\phi S_{xo}-\Delta\phi_{Nex}$. The attached chart also plots the density tool's own response to zero-density gas. That response is an apparent density of $-0.19\ \text{g/cm}^3$ (from $\rho_{log}=1.0704\rho_e-0.1883$), so $\phi_D=\phi\,[2.65-S_{xo}+0.19(1-S_{xo})]/1.65$. Overlaying these two relations on the printed chart on page 23 reproduces both the printed porosity curves and the printed $S_{xo}$ lines, so they are the chart in equation form.
4. Solve WITH excavation effect. Enter the chart at $\phi_N=5\%$, $\phi_D=39.4\%$. The point falls between the 25% and 30% porosity curves, next to the $S_{xo}=30\%$ line (see the figure below). Solving the two chart relations simultaneously gives $$\boxed{\phi_{EE}\approx 0.266\ (26.6\%)}\qquad \boxed{S_{xo,EE}\approx 0.333\ (33.3\%)}$$
5. Isolate the excavation term. Keep the chart's density basis but switch the excavation term off. That gives $\phi=25.0\%$ and $S_{xo}=20.0\%$. The excavation effect alone therefore moves porosity by only about 1.6 porosity units, but moves $S_{xo}$ by about 13 saturation points. Leaving it out would make the flushed zone look far more gas-saturated than it is.

Porosity is robust: both treatments give about 26–27%, far below the 39.4% the density log shows on its own. The flushed-zone saturation is not robust. Ignoring the excavation effect understates $S_{xo}$ by roughly 14 points, because the whole neutron deficit is then attributed to gas in the pores.