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

Question 7: Neutron-Density Porosity and S_xo in a Gas Sand (14 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, without and with the excavation effect.

Approach. In a gas sand each tool responds to both unknowns. The neutron sees only the hydrogen of the filtrate, so φ_N depends on φ and S_xo. The density tool sees the average pore-fluid density, which is also set by S_xo. Write one response equation per tool and solve the pair. Without the excavation effect the pair is linear, and the paper's formula sheet (page 13) prints its solution. With the excavation effect the neutron equation gains the excavation term, and the attached φ_D-versus-φ_N chart (page 14) is the graphical solution of that 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. WITHOUT the excavation effect. With a gas hydrogen index of 0 and ρ_g = 0, the tool responses are $$\phi_N=\phi\,S_{xo}\qquad \rho_b=(1-\phi)\rho_{ma}+\phi\,S_{xo}\,\rho_{mf}$$ Substituting φS_xo = φ_N into the density equation (with ρ_mf = 1) gives the formula printed on the attachment: $$\phi=\frac{\rho_{ma}-\rho_b+\phi_N}{\rho_{ma}}=\frac{2.65-2.0+0.05}{2.65}=\frac{0.70}{2.65}\qquad \boxed{\phi_{noEE}=0.264\ (26.4\%)}$$ $$S_{xo}=\frac{\phi_N}{\phi}=\frac{0.05}{0.264}\qquad \boxed{S_{xo,noEE}=0.189\ (18.9\%)}$$
3. The excavation effect. Residual gas occupies pore volume that, in a liquid-filled rock of the same bulk density, would hold hydrogen-bearing liquid and matrix. The neutron therefore reads lower still than φS_xo, by $$\Delta\phi_{Nex}=K\left(2\phi^2S_{xo}+0.04\,\phi\right)\left(1-S_{xo}\right),\qquad K=1\ \text{(sandstone)}$$ so φ_N = φS_xo − Δφ_Nex. The chart is also drawn with the density tool's own response to zero-density gas, an apparent (log) density of about −0.19 g/cm³ (from ρ_log = 1.0704ρ_e − 0.1883), so $$\phi_D=\frac{\phi\left[\rho_{ma}-S_{xo}\rho_{mf}+0.19(1-S_{xo})\right]}{\rho_{ma}-\rho_{mf}}$$ These two relations reproduce the chart's printed porosity curves and S_xo lines: every porosity curve meets the S_xo = 100% diagonal at φ_N = φ_D = φ, and the 30% porosity curve crosses the S_xo = 50% line near φ_N = 10%.
4. WITH the excavation effect. Enter the chart at φ_N = 5%, φ_D = 39.4% (figure below). The point lies between the 25% and 30% porosity curves, next to the S_xo = 30% line. Solving the two chart relations together gives $$\boxed{\phi_{EE}=0.266\ (26.6\%)}\qquad \boxed{S_{xo,EE}=0.333\ (33.3\%)}$$
5. What the excavation effect changes. Keeping the chart's density basis but switching the excavation term off gives φ = 25.0% and S_xo = 20.0%. The excavation term alone therefore moves porosity by only about 1.6 porosity units, but moves S_xo by about 13 saturation points. Leaving it out attributes the whole neutron deficit to gas and makes 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 about 14 points, because the whole neutron deficit is then blamed on gas in the pores. A quick-look root-mean-square average, √[(φ_D² + φ_N²)/2] = 28.1%, lands near the right porosity but gives no S_xo at all, which is why the paper supplies the response equations and the chart.