Gas, shale, carbonate and pressure effects on porosity logs

1. Oil: $x \approx 4 - 2.5\cdot 0.75 = 2.125$. The hydrogen index is $\dfrac{9\cdot 0.75\cdot 2.125}{12 + 2.125} = 1.015$, and the apparent density from the formula above is 0.734 g/cm³.
2. The sensitivities for oil are $k_D = \dfrac{1.00 - 0.734}{2.65 - 1.00} = 0.161$ and $k_N = 1 - 1.015 = -0.015$. The density porosity rises by 1.6 units, and the neutron porosity hardly changes (0.15 units).
3. Gas: $\mathrm{HI}_h = 2.25\cdot 0.13 = 0.2925$ and $\rho_{a,h} = 1.334\cdot 0.13 - 0.188$ = −0.015 g/cm³. Then $k_D = \dfrac{1.00 - (-0.015)}{1.65} = 0.615$ and $k_N = 0.7075$.
4. The shifts due to gas: the density porosity goes up by $+6.2$ units. The neutron porosity goes down by $-7.1$ because of the lack of hydrogen, and by another $-2.8$ because of the excavation effect (at $S_{wH} = 0.717$).

What it means. Oil hardly differs from filtrate in either hydrogen or apparent density, so an oil-bearing formation on the density–neutron pair looks almost like a water-bearing one. Resistivity tells them apart (Archie's equation and saturation of clean reservoirs). Gas that takes up 10% of the rock volume separates the curves by 16 units, and 18% of this gap comes from the excavation effect. The coefficients depend on the gas density: deeper, at higher pressure, both are smaller (see the section on pressure below). The data in this example are assumed for illustration.

1. Density porosity calculated for filtrate: $\phi_D = \dfrac{2.65 - 2.145}{1.65} = 0.3061$. The difference from the neutron porosity is 0.1701, a clear crossover.
2. First approximation, without excavation. The sum of the sensitivities is $k_D + k_N = 1.3225$. The weights of the curves are $k_N/1.3225 = 0.535$ for density and $k_D/1.3225 = 0.465$ for neutron. So $\phi = 0.535\cdot 0.3061 + 0.465\cdot 0.136$ = 0.2270. The gas volume is $0.1701/1.3225 = 0.1286$, and $S_{hr} = 0.1286/0.2270 = 0.567$.
3. Excavation correction: $S_{wH} = 1 - 0.7075\cdot 0.567 = 0.599$ and $\Delta\phi_{ex} = 0.028$. The corrected neutron porosity is $\phi_N' = 0.136 + 0.028 = 0.164$.
4. The second approximation gives $\phi = 0.240$ and $S_{hr} = 0.45$. The third changes the porosity by less than 0.1 unit: $\phi = 0.240$, $S_{hr} = 0.45$, that is, $S_{xo} = 0.55$.
5. For comparison, two quick estimates: the root mean square of the two curves, $\sqrt{(\phi_D^2 + \phi_N^2)/2} = 0.237$, and the arithmetic mean, 0.221.

What it means. From the density curve alone, calculated as if the pores held water, the porosity would be 30.6%, and from the neutron curve alone 13.6%. These are errors of almost 7 and 10 units in opposite directions. Without the excavation correction, the porosity would come out 1.3 units too low, and the gas saturation of the flushed zone too high, at 57%. The root-mean-square rule (quick estimate) gave 0.237, close to the solution. The root mean square is larger than the arithmetic mean, so it partly makes up for the lowering of the neutron curve by excavation. But it is an empirical rule. It knows nothing of the gas coefficients, and we cannot find the gas saturation with it. The gas saturation of the formation from resistivity is 75%, while 45% is only the gas that remained at the borehole wall after the filtrate invaded. So the estimate from the density–neutron pair is a lower limit. The data in this example are assumed for illustration.

1. Taken as they are: $\phi_D = \dfrac{2.65 - 2.286}{1.65} = 0.221$ and $\phi_N = 0.230$. The neutron porosity is even higher than the density porosity, and there is no crossover. The shale volume from the disagreement of the curves would come out as 0.03, against 0.30 from GR. This contradiction is the first sign that something else is bringing the curves together.
2. Corrections for shale: $\phi_{D,sh} = \dfrac{2.65 - 2.48}{1.65} = 0.103$, then $\phi_{D,c} = 0.221 - 0.30\cdot 0.103$ = 0.190 and $\phi_{N,c} = 0.230 - 0.30\cdot 0.38$ = 0.116. Now the density porosity is higher than the neutron porosity by 0.074. This is gas.
3. The solution for gas with the excavation correction, as in the previous example, gives the effective porosity $\phi_e = 0.160$ and $S_{hr} = 0.30$.
4. Check with a third method. In a compacted sandstone gas hardly affects the interval transit time (the time a sound wave needs to cross a unit length of rock). So $\phi_S = (287 - 182)/438 = 0.240$, the “porosity of shale” is $(300 - 182)/438 = 0.269$, and the effective porosity is $0.240 - 0.30\cdot 0.269 = 0.159$.

What it means. Without the shale corrections the formation would have looked like a slightly shaly water-bearing one: shale “hid” the gas. If only the density curve had been corrected, and the porosity calculated as if the pores held water, the result would be 19%, which is 3 units too high. From the corrected neutron curve alone it would be 12%. The gas solution agrees with the sonic porosity, which gas hardly affects in a compacted rock. This is a strong argument for the model of the formation. Resistivity and testing finally confirm the gas. The data in this example are assumed for illustration.

1. The density porosity on the limestone scale is $\dfrac{2.71 - 2.526}{1.71} = 0.1076$. The neutron porosity is 0.163, higher by 0.0554. GR is low, so this is not shale. The matrix is heavier than calcite and slows neutrons more, so it is dolomite.
2. The total porosity, as the mean of the two readings, is $(0.1076 + 0.163)/2 = 0.1353$.
3. The sonic porosity with limestone parameters is $(198 - 156)/464 = 0.0905$.
4. The secondary porosity index is $0.1353 - 0.0905 = 0.0448$, about 33% of all the voids.

What it means. The formation in the example was set up as follows: 40% dolomite in the matrix, a total porosity of 14%, of which 4 units are vugs. The mean of the two curves was off by 0.5 units, although each curve on its own was off by 3 and 2 units in opposite directions. The secondary porosity index is too high by 0.5. Dolomite is faster than calcite (a sound wave travels faster through it), and with limestone parameters the intergranular porosity comes out too low. If we knew the dolomite fraction, from cuttings, core or a crossplot, we would take the matrix transit time 151 µs/m and get an intergranular porosity of 0.101, almost exact. The data in this example are assumed for illustration.

1. The shale porosity is $\phi_{sh} = (2.70 - \rho_{sh})/1.65$: $0.400/1.65 = 0.242$ at 1500 m, $0.276/1.65 = 0.167$ at 2400 m and $0.356/1.65 = 0.216$ at 3000 m. Below 2400 m the porosity did not decrease. It increased.
2. We get the normal trend from the two upper points (the ratio of porosities equals the ratio of the differences $2.70 - \rho_{sh}$): $c_z = \dfrac{\ln(0.400/0.276)}{0.9}$ = 0.4123 km⁻¹. At 3000 m the trend gives a density of $2.70 - 0.276\cdot \e^{-0.4123\cdot 0.6}$ = 2.484 g/cm³, which is 0.140 g/cm³ more than the measured value. The porosity from the trend is 0.131.
3. The equivalent depth is the depth where the trend gives the same porosity as the measured one at 3000 m: $z_e = 2.4 - \dfrac{\ln(0.356/0.276)}{0.4123}$ = 1.783 km.
4. Pore pressure by the equivalent depth method. The overburden pressure at 3000 m is $0.00981\cdot 2.25\cdot 3000 = 66.22$ MPa. The effective stress of normal shale at the equivalent depth is $0.00981\cdot 1.20\cdot 1783 = 20.99$ MPa. So $p_p = 66.22 - 20.99 = 45.2$ MPa, an equivalent density of 1.54 g/cm³ against the normal 1.05.

What it means. The tight layer above the interval is a seal, and the pressure began to grow under it. On the log it looks like a bulge before the trend turns around. The estimate is valid only if undercompaction created the pressure, the shales are the same, and the trend is extended reliably. That is why we check it against the d-exponent (a number calculated from the drilling rate, which changes with pore pressure), gas shows and direct pressure measurements (Abnormal formation pressures: origin, prediction and recognition). For the porosity of sands in this interval the conclusion is the same as above: the sonic log without a large correction will overestimate it, and the density log will not. The data in this example are assumed for illustration.