Lithology from the photoelectric factor and spectral gamma ray

The photoelectric factor $P_e$ is the cross section for photoelectric absorption of gamma rays per electron, in barns per electron (how it is measured). (A cross section shows how likely a gamma ray is to be absorbed; the barn is its unit.) The tool senses how many photons are absorbed per unit path. This number is proportional to the number of electrons per unit volume times the cross section of one electron, which is the volumetric cross section $U = P_e\rho_e$ (barn/cm³). The $P_e$ recorded on the log is $U$ divided by the electron density index $\rho_e$ of the rock, which we find from the density measurement of the same tool.

This gives the mixing rule. Each component of the rock (mineral, water, oil, gas) with a volume fraction $V_i$ adds $V_iU_i$ to the absorption and $V_i\rho_{e,i}$ to the number of electrons. So it is $U$ that adds up by volume. $P_e$ is the ratio of two volume sums, that is, an average weighted by the share of electrons of each component.

On this crossplot, $P_e$ is plotted horizontally and the bulk density $\rho_b$ vertically, with heavy rocks at the bottom. The line of each mineral shows the points of a formation made of that mineral, with fresh filtrate in the pores, at porosities from zero to 40%. From zero to 40% pores, the density along the line changes by 0.65–0.75 g/cm³, but $P_e$ changes by only about a fifth of its value. So the lines are steep, almost vertical, and spread out horizontally: quartz at about 1.8, dolomite at about 3.1, calcite at about 5.1. Rock salt (4.65; 2.03) and anhydrite (5.05; 2.98) lie apart. Gypsum (3.99; 2.35) falls between the dolomite and calcite lines, and it looks like a porous mixture of two carbonates.

If two minerals are known, we read the point of the formation as on the neutron–density crossplot: find the isoporosity segment that passes through the point, and take the mineral fractions from the position on it. With a calculator we can solve this problem exactly: in the coordinates “density–$U$” all components mix by volume, and the equations are linear.

It is useful to compare the porosity from density and $P_e$ with the neutron–density porosity. In a clean formation with liquid, and with the minerals chosen correctly, they agree. If they do not, try another pair of minerals. But suppose the neutron–density porosity is noticeably lower than the porosity from density and $P_e$, and the apparent matrix density from neutron and density is lower than the matrix density found from $P_e$. Then there is gas in the flushed zone: it lowers the density and hardly changes $P_e$.

If there are three minerals, two equations from density and $P_e$ are not enough. There are four unknowns: the volumes of three minerals, and the porosity. The neutron log gives the missing equation. It is most convenient to use the porosity read from the neutron–density crossplot, which hardly depends on which minerals make up the matrix (why). Once we know the porosity, we can “subtract” the pores from the readings and get the apparent matrix parameters: what the density and the volumetric cross section of the rock would be without pores.

These formulas have a simple geometric meaning. In the coordinates “density–$U$” all components mix by volume. So all rocks with one matrix and different porosities lie on the segment from the matrix point to the fluid point, and the porosity is the fraction of the way from the matrix to the fluid. The apparent parameters are the continuation of the line from the fluid point through the rock point to zero porosity. If the apparent porosity is correct, we get the matrix point. If it is too high, the point goes further, and if it is too low, the point falls short.

Where does $\phi_{ta}$ come from? Plot the point of the formation on the neutron–density crossplot with sandstone, limestone and dolomite lines, and read the porosity along the isoporosity lines of the pair between whose lines the point lies. For points above the sandstone line (gas-bearing rocks and salt), the usual rule is to take the porosity of sandstone with the same neutron reading (the vertical projection onto its line). All this assumes a formation with liquid in the pores. In a gas-bearing formation the apparent porosity is only approximate.

On the $U_{maa}$–$\rho_{maa}$ crossplot, the points of quartz (4.80; 2.65), calcite (13.76; 2.71) and dolomite (8.97; 2.87) form a triangle. Both coordinates mix by volume of the matrix, so a mixture of three minerals lies inside the triangle. We find its fractions from three linear equations: the fractions add up to one, the mean density equals $\rho_{maa}$, and the mean cross section equals $U_{maa}$. The triangle is wide in $U$ (from 4.8 to 13.8) and narrow in density (from 2.65 to 2.87). So $U$ mainly separates quartz from calcite, and density separates dolomite from both.

Where the other minerals and effects lie on this crossplot (values from the lesson model):

• Potassium feldspar (7.26; 2.53) lies above the quartz–calcite side: lighter than quartz with a larger $U$. Arkosic sandstones shift towards it from quartz.
• Anhydrite (14.95; 2.98) lies below and to the right of calcite, and rock salt (9.64; 2.03) far at the top. A carbonate with anhydrite leaves the triangle beyond the calcite–dolomite side.
• Heavy minerals go far to the right and down: siderite (55; 3.84), pyrite (82; 4.99), barite (1070; 4.10). Even a few per cent of pyrite shift the point noticeably to the right.
• Shale lies below dolomite. Its neutron porosity includes bound water, so the apparent porosity comes out large, and subtracting the pores “overcorrects” both coordinates. The shale of the lesson ($\rho_b = 2.48$, $\phi_N = 0.38$, $P_e = 3.0$) gives the point (10.0; 3.01).
• Vugs do not shift the point: both density and neutron see all the pores. The vuggy limestone from the topic on crossplots gives (13.75; 2.71), the point of calcite.
• Gas raises the point almost vertically; see the example below.

How strongly does the answer depend on the apparent porosity? An error $\delta\phi$ shifts the point along the ray from the fluid point. $\rho_{maa}$ changes by $\dfrac{\rho_{maa} - \rho_f}{1 - \phi_{ta}}\,\delta\phi$, that is, by about 0.02 g/cm³ for an error of 0.01. $U_{maa}$ changes by 0.05–0.16 barn/cm³ (from quartz to calcite). For $U$ this is little, but for density it is a lot, because quartz and calcite differ in density by only 0.06 g/cm³. So the fractions of quartz and calcite rest on $U$, while the fractions of dolomite and anhydrite, which differ mainly in density, are sensitive to the porosity. When a carbonate contains anhydrite, the neutron–density porosity read from the limestone and dolomite lines comes out too low (in the lesson model by about half a unit for every 10% of anhydrite in the matrix), and so does the anhydrite fraction.

$P_e$ is measured from low-energy photons, which do not penetrate far into the rock. So it sees an even thinner layer behind the borehole wall than density does, and it responds more strongly than other curves to mudcake, mud and borehole roughness. Before calculating from it, check the recording conditions.

The spectral gamma ray log gives the contents of potassium, in mass per cent, and of thorium and uranium, in g/t, that is, ppm (parts per million) (how they are separated). Potassium is found in potassium feldspars, micas, illite, glauconite and potassium salts. Thorium is almost insoluble, and only mineral particles carry it: clay minerals, heavy minerals (monazite, zircon) and volcanic ash hold it. Uranium is soluble and mobile, and it has a separate section below. Pure quartz, calcite, dolomite, anhydrite and rock salt are almost non-radioactive.

The tool measures contents per unit mass of rock, so they mix by mass. Suppose one mineral creates the radioactivity and the rest of the rock is “empty”. Then the potassium and thorium of the rock are proportional to the mass fraction of that mineral, and their ratio equals the ratio of the mineral. On the potassium–thorium crossplot such rocks lie on a straight line through the origin. The slope says which mineral is radioactive, and the distance from the origin says how much of it there is.

The Th/K ratios of different minerals overlap, so we combine the ratio with $P_e$. Glauconite and muscovite have similar ratios, but their $P_e$ is about 6 and 2.4. Chlorite and kaolinite both have large ratios, but their $P_e$ is about 6 and 1.8. Illite and smectite differ in both ratio and $P_e$ (3.5 and 2). In a formation, $P_e$ is diluted by quartz and carbonate. So we recognise clay minerals from it in thick shale units and by trends, not from a single sandstone point.

Each element in spectrometry tells its own story, and this helps to read the log in a section. Thorium tells of detrital shale admixture (clay carried in as particles): in sandstones and carbonates it grows with shale volume, if there are no heavy minerals or volcanic ash. Potassium tells of potassium-bearing minerals. In sandstones these are often feldspars and micas (as in the arkose above), and in shales they are illite and micas, so there potassium reflects the type of clay. Uranium in carbonates usually points not to shale but to organic matter, phosphates and stylolites. Chemogenic rocks (rocks that formed by chemical precipitation: salt, anhydrite, clean limestone) are almost non-radioactive unless they are potassium salts. Weathering zones betray themselves by a sharp change of thorium and potassium while their ratio stays almost the same.

Uranium behaves differently from potassium and thorium. In oxidising waters (waters that contain oxygen) it is soluble and is carried far. In a reducing environment (one without oxygen) it turns into an insoluble form and is deposited, above all on organic matter. So marine shales that built up where oxygen was scarce are rich in both organic carbon and uranium, and their thorium/uranium ratio is low (Th/U). High uranium in a shale unit is a hint that it is a source rock (Source rocks, maturation and migration).

• Carbonates: uranium is a sign of organic matter, phosphates and stylolites, not of shale. On the total GR such intervals look shaly, so calculate the shale volume from thorium or from GR without uranium.
• Groundwater pathways: uranium is deposited on faults, in fractures and in the formations through which waters have flowed. A uranium anomaly in a tight rock may point to open fractures.
• Producing wells: in watered-out intervals, salts containing radium from the formation water settle on pipes and in the near-wellbore zone. Spectrometry sees them by the lines of the uranium series, and the GR behind casing grows with time. This is not lithology but production history. Such deposits are radioactive, and radiation safety regulations govern how they are handled.