Neutron log · MunaiHub

Quick refresher

A neutron loses the most energy in a collision with a hydrogen nucleus, because their masses are almost equal. To slow a neutron from 4 MeV to a thermal energy of about 0.025 eV takes about 19 collisions with hydrogen, about 160 with oxygen and about 380 with calcium. (Thermal energy is the energy that atoms have from heat.)
The more hydrogen there is, the closer to the source the neutrons slow down and are captured, and the fewer of them reach the far detector. The ratio of the near and far detector counts rises with the hydrogen content, and we convert it to porosity.
Hydrogen index (the amount of hydrogen per unit volume, compared with fresh water): fresh water 1, oil about 1, salt water slightly less than 1, gas 0.1–0.5 depending on pressure.
We usually express neutron porosity in limestone units. Clean limestone full of water shows the true porosity, sandstone shows less, and dolomite shows more.
Shale (with its bound water and hydroxyl, OH, groups) makes $\phi_N$ too high, and gas makes it too low. On compatible scales, gas makes the two curves cross over (the density porosity is larger than the neutron porosity). Shale makes them separate in the opposite direction.
Thermal neutrons are sensitive to absorbers (chlorine, boron, rare-earth elements), so a thermal-neutron tool gives a porosity that is too high. Epithermal neutrons (neutrons that are slowed but not yet at thermal energy) are almost insensitive.

Key relationships

n = \dfrac{\ln(E_0/E_{th})}{\xi}

average number of collisions needed to slow a neutron to thermal energy

\mathrm{HI}_h = \dfrac{9\,\rho_h\,x}{12 + x}

hydrogen index of the hydrocarbon $\mathrm{CH}_x$ of density $\rho_h$ (g/cm³)

\phi_N \approx \phi\,\mathrm{HI}_{fl}

neutron porosity of a clean formation in the correct matrix; $\mathrm{HI}_{fl}$ is the hydrogen index of the fluid in the flushed zone, and the excavation effect is ignored

\phi \approx \sqrt{(\phi_N^2 + \phi_D^2)/2}

quick estimate of the porosity of a gas-bearing formation

Common pitfalls

Reading neutron porosity in limestone units as the porosity of sandstone or dolomite.
Assuming that there is no gas if the curves do not cross over. Shale and deep flushing can hide the gas.
Taking a crossover in rock salt, or in sandstone on limestone scales, for gas.
Forgetting that a thermal-neutron tool gives a porosity that is too high in salt water and in shales that contain boron.

After this lesson you can

explain how neutrons are slowed down and captured in rock and why hydrogen plays the main role
calculate the hydrogen index of salt water, oil and gas and estimate how each fluid affects the reading
describe how the compensated tool is built, how thermal and epithermal neutrons differ, and what limestone units mean
explain how shale, gas and the matrix mineral affect neutron porosity, and recognise these cases on a combined log with the density log
estimate the porosity of a gas-bearing formation from the neutron and density porosity and understand the limits of such an estimate

How neutrons lose energy

The source of a neutron tool is usually a mixture of americium-241 and beryllium. The alpha particles of americium knock fast neutrons out of beryllium nuclei. Their energies go up to 10–11 MeV, and the average is about 4 MeV. A neutron is a particle with no electric charge. Its mass is almost equal to that of a proton, that is, of a hydrogen nucleus. The electrons of atoms do not slow it down, so it loses energy only in collisions with nuclei.

The ratio of the masses decides how much energy is lost in a single collision, as in billiards. A ball that hits an identical stationary ball head-on stops, because it passes all its speed to the other ball. The same ball bouncing off the cushion of the table loses almost no speed. For a neutron, a hydrogen nucleus is “an identical ball”. The nuclei of oxygen, silicon and calcium are 16–40 times heavier, and they behave more like the cushion.

Largest fraction of energy that a neutron loses in a single collision

\frac{\Delta E_{max}}{E} = \frac{4A}{(A + 1)^2}

where:

$\Delta E_{max}$ energy that the neutron loses in a head-on collision, eV
$E$ energy of the neutron before the collision, eV
$A$ mass number of the nucleus (its mass in units of the neutron mass), —

Units: energies in the same units; the fraction is dimensionless

Holds when

An elastic collision (the colliding bodies keep their total energy of motion) with a nucleus at rest. In a glancing collision the loss is smaller, on average half the largest.

Hydrogen ( $A = 1$ ) can take all of the neutron's energy, carbon up to 28%, oxygen up to 22%, silicon up to 13%, calcium up to 9.5%.

After each collision the energy falls by some factor, so it is convenient to count the losses in logarithms. The mean logarithmic energy loss per collision, $\xi$ , depends only on the mass of the nucleus. A neutron is slowed to thermal energy when it has about 0.025 eV at 20 °C (somewhat more in a hot borehole). The number of collisions needed for this equals the whole “logarithmic path” divided by $\xi$ .

Number of collisions needed to slow a neutron

n = \frac{\ln(E_0/E_{th})}{\xi}, \qquad \xi = 1 + \frac{(A-1)^2}{2A}\,\ln\frac{A-1}{A+1}

where:

$n$ mean number of elastic collisions, —
$E_0$ initial energy of the neutron, eV
$E_{th}$ thermal energy, about 0.025 eV at 20 °C, eV
$\xi$ mean logarithmic energy loss per collision, —
$A$ mass number of the nucleus, —

Units: energies in the same units; natural logarithms

Holds when

Elastic scattering that is isotropic (equal in all directions) in the centre-of-mass frame. This is a good approximation for light nuclei at the energies of logging sources.
For $A = 1$ the second term vanishes and $\xi = 1$ .
A medium made of nuclei of one kind. In a rock, the contributions of the nuclei add up according to their number and the probability of scattering on each.

For hydrogen $\xi = 1$ , for oxygen about 0.12 and for calcium about 0.05. Heavy nuclei need 8 to 20 times more collisions. Besides, at low energies a neutron is several times more likely to scatter on a hydrogen nucleus than on the nuclei of oxygen, silicon or calcium. So hydrogen mainly decides how neutrons slow down in rock.

Worked exampleHow many collisions it takes to slow a neutron

Given (assumed data)

Initial energy $E_0$: 4 MeV ( $4\cdot 10^6$ eV), the average for an americium–beryllium source
Thermal energy $E_{th}$: 0.025 eV
Nuclei: hydrogen ( $A = 1$ ), carbon (12), oxygen (16), silicon (28), calcium (40)

Solution

Logarithmic path: $E_0/E_{th} = 1.6\cdot 10^8$ , and the natural logarithm of this ratio is 18.9.
Mean loss per collision $\xi$ : hydrogen 1, carbon 0.158, oxygen 0.120, silicon 0.070, calcium 0.049.
Number of collisions $n = 18.9/\xi$ : hydrogen 19, carbon 120, oxygen 157, silicon 271, calcium 384.

Answer: Hydrogen needs about 19 collisions, oxygen about 157 and calcium about 384. Per collision, hydrogen slows neutrons down 8.3 times more effectively than oxygen and 20 times more effectively than calcium.

What it means. The matrix also slows neutrons down, because it contains a lot of oxygen, but it does so slowly. Every per cent of water in the pores adds the most effective moderator (a material that slows neutrons). This is why the detector count is so sensitive to hydrogen, and so to porosity. Differences between matrices remain: calcite and quartz slow neutrons slightly differently, which is why lithology corrections are needed (below). The calculation is simplified: it ignores the different chances of scattering on different nuclei.

CalculatedCalculated mean number of elastic collisions needed to slow a neutron from 4 MeV to 0.025 eV for the nuclei of the main elements of rocks and fluids (from the formulas for

n

and

\xi

above).

Once a neutron has become thermal, it does not slow down any further, because its energy has become equal to the energy of the thermal motion of atoms. It wanders about (diffuses) until some nucleus absorbs it. That nucleus immediately emits a capture gamma ray. Chlorine, boron and rare-earth elements such as gadolinium capture thermal neutrons especially readily. Hydrogen, iron and silicon capture them much more weakly. Neutrons of intermediate energies, above about 0.4 eV and not yet thermal, are called epithermal.

A rock rich in hydrogen works as a dense brake. Neutrons slow down and are absorbed within about ten centimetres of the source, and few of them reach the far detector. In a dense rock of low porosity, the cloud of slowing neutrons stretches out for several tens of centimetres, and the detector count rises. This is how we judge the hydrogen content of the rock from the count of detectors placed tens of centimetres from the source.

Hydrogen index

The hydrogen index $\mathrm{HI}$ is the number of hydrogen atoms per unit volume of a substance, compared with their number in fresh water at surface conditions. 1 cm³ of water contains 1/9 mole of hydrogen atoms. (A mole is a fixed number of particles. One mole of water weighs 18 g, and each water molecule holds two hydrogen atoms.) From this we get simple formulas for reservoir fluids.

Hydrogen index of a hydrocarbon

\mathrm{HI}_h = \frac{9\,\rho_h\,x}{12 + x}

where:

$\mathrm{HI}_h$ hydrogen index of the hydrocarbon, —
$\rho_h$ density of the hydrocarbon at reservoir conditions, g/cm³
$x$ number of hydrogen atoms per carbon atom in the average formula $\mathrm{CH}_x$ : 4 for methane, about 2 for oil, —

Units: $\rho_h$ in g/cm³; the index is dimensionless

Holds when

The atomic masses of carbon and hydrogen are rounded to 12 and 1.
The hydrocarbon is described by the average formula $\mathrm{CH}_x$ .

A mole of the group $\mathrm{CH}_x$ weighs $12 + x$ grams and contains $x$ moles of hydrogen. So 1 cm³ holds $\rho_h x/(12 + x)$ moles of hydrogen, against 1/9 for water. For oil ( $x \approx 2$ ), $\mathrm{HI}_h \approx 1.29\rho_h$ , which is about one at ordinary densities. For methane, $\mathrm{HI}_h = 2.25\rho_h$ : at a density of 0.1–0.25 g/cm³ this is 0.22–0.56.

Hydrogen index of salt water

\mathrm{HI}_w \approx \rho_w\,(1 - P)

where:

$\rho_w$ density of the salt water, g/cm³
$P$ mass fraction of dissolved salt, fraction

Units: $\rho_w$ in g/cm³; the index is dimensionless

Holds when

The salt contains no hydrogen; hydrogen belongs only to the water in the solution.

Dissolved salt displaces part of the water from a unit volume, so the hydrogen index of brines (salty waters) is slightly less than one.

Worked exampleHydrogen index of reservoir fluids

Given (assumed data)

Brine: NaCl 150 g per 1 kg of solution ( $P = 0.15$ ), density 1.11 g/cm³
Oil: $\rho_h = 0.80$ g/cm³ at reservoir conditions, average formula $\mathrm{CH_2}$
Gas of a deep reservoir: methane, $\rho_g = 0.18$ g/cm³
Gas of a shallow reservoir: methane, $\rho_g = 0.05$ g/cm³

Solution

Brine: $\mathrm{HI}_w = 1.11\cdot(1 - 0.15) = 0.94$ .
Oil: $\mathrm{HI}_h = 9\cdot 0.80\cdot 2/14 = 1.03$ .
Gas: $9\cdot 0.18\cdot 4/16 = 0.405$ in the deep reservoir and $9\cdot 0.05\cdot 4/16 = 0.113$ in the shallow one.

Answer: The hydrogen index is 0.94 for brine, 1.03 for oil and 0.405 for gas at depth, but only 0.113 in a shallow reservoir.

What it means. Oil and salt water hardly differ from fresh water. So in a clean formation full of liquid, the neutron porosity is close to the true porosity, and the neutron log cannot tell oil from water. That is a job for the resistivity methods. Gas has several times less hydrogen, and the lower the pressure, the less. A shallow gas reservoir shows up on the neutron log more strongly than a deep one. The data in this example are assumed for illustration.

Hydrogen index of substances (approximate)

Substance: Fresh water
HI: 1
Where the hydrogen comes from: the standard of the scale

Substance: Salt water
HI: 0.9–1
Where the hydrogen comes from: the higher the salinity, the lower the index

Substance: Oil
HI: 0.9–1.05
Where the hydrogen comes from: depends on density and composition

Substance: Gas
HI: 0.1–0.5
Where the hydrogen comes from: grows with pressure together with density

Substance: Coal
HI: about 0.4–0.7
Where the hydrogen comes from: organic matter

Substance: Gypsum
HI: about 0.5
Where the hydrogen comes from: water of crystallisation

Substance: Kaolinite
HI: about 0.36
Where the hydrogen comes from: hydroxyl groups

Substance: Quartz, calcite, dolomite, anhydrite, salt
HI: 0
Where the hydrogen comes from: no hydrogen

Substance	HI	Where the hydrogen comes from
Fresh water	1	the standard of the scale
Salt water	0.9–1	the higher the salinity, the lower the index
Oil	0.9–1.05	depends on density and composition
Gas	0.1–0.5	grows with pressure together with density
Coal	about 0.4–0.7	organic matter
Gypsum	about 0.5	water of crystallisation
Kaolinite	about 0.36	hydroxyl groups
Quartz, calcite, dolomite, anhydrite, salt	0	no hydrogen

Clay minerals contain hydrogen in the hydroxyl groups of their crystal lattice, and shales as a whole also hold bound water. So the neutron porosity of shales is usually 0.30–0.45, even though their effective porosity (the porosity that can hold fluid able to flow) is almost zero.

Tools: thermal and epithermal neutrons, one and two detectors

The first neutron tools came into use between the 1940s and the 1960s. They had a single detector and depended strongly on the borehole diameter, the mud, the mudcake and the salinity. In Russian-language practice, two methods have been and still are widely used. One is the neutron gamma log, which records capture gamma rays. The other is the neutron–neutron log, which records thermal or epithermal neutrons. The readings of the neutron gamma log depend noticeably on chlorine, because chlorine captures many neutrons and emits many capture gamma rays.

Later, pad tools with an epithermal neutron detector appeared. Pressing the pad against the wall reduces the influence of the mud, and absorbers hardly affect epithermal neutrons. Such tools work only in an open hole, but they also work in a hole that is not filled with liquid.

Today the main tool is the compensated neutron tool. It is a mandrel tool (a tool without a separate pad), and a spring presses it against the wall. It has two thermal-neutron detectors at different distances from the source. The borehole diameter, the mudcake and the mud change the count of both detectors in almost the same proportion, so the ratio of the counts hardly depends on them. The compensation is built on this. The ratio of the near count to the far count rises with the hydrogen content, and calibration converts it into neutron porosity (the NPHI curve). The mandrel tool is longer than the pad tool and sees deeper, 15–30 cm. Its vertical resolution is 30–60 cm. We can record it in open hole and behind casing, but the hole must be filled with liquid. The calibration does not apply to a hole filled with gas or air.

Epithermal neutrons are hardly captured at all. So some tools add a pair of epithermal detectors to the pair of thermal ones. They are placed closer to the source, because about ten times fewer epithermal neutrons arrive than thermal ones. Separate processing removes the influence of the borehole on them. The tool gives two porosities, and the difference between them carries information. In a clean formation with fresh water they are the same. If the thermal one is noticeably higher, there are absorbers in the zone of investigation: the boron and rare-earth elements of shales, or the chlorine of salt water. In such formations the epithermal porosity is closer to the density porosity, so gas in a shaly reservoir shows up better on it.

Varieties of neutron logging

Method: Neutron gamma log
What it records: capture gamma rays
Strengths: simple, works in casing
Limitations: strongly dependent on chlorine and on the borehole

Method: Thermal neutron–neutron log, compensated
What it records: thermal neutrons, two detectors
Strengths: the main porosity method, works in casing
Limitations: absorbers make the porosity too high; the hole must be filled with liquid

Method: Epithermal neutron–neutron log
What it records: epithermal neutrons
Strengths: hardly feels absorbers, works in a dry hole
Limitations: lower count, stronger influence of the borehole

Method: Neutron log while drilling (LWD)
What it records: thermal neutrons in a drill collar
Strengths: recorded before deep invasion
Limitations: the standoff between the tool and the wall changes with rotation

Method	What it records	Strengths	Limitations
Neutron gamma log	capture gamma rays	simple, works in casing	strongly dependent on chlorine and on the borehole
Thermal neutron–neutron log, compensated	thermal neutrons, two detectors	the main porosity method, works in casing	absorbers make the porosity too high; the hole must be filled with liquid
Epithermal neutron–neutron log	epithermal neutrons	hardly feels absorbers, works in a dry hole	lower count, stronger influence of the borehole
Neutron log while drilling (LWD)	thermal neutrons in a drill collar	recorded before deep invasion	the standoff between the tool and the wall changes with rotation

Limestone units and corrections

Calibration converts the ratio of the counts into porosity. The tool is calibrated in limestone saturated with fresh water, under standard conditions. These are: a hole of about 200 mm (7⅞ in); fresh water in the hole and in the formation; no mudcake or standoff; a temperature of about 24 °C; atmospheric pressure; and the tool pressed against the wall. That is why we usually express neutron porosity in limestone porosity units (limestone units for short). In clean limestone full of water, it equals the true porosity.

In other rocks the reading differs from the true porosity even if the pores are filled with water. The matrix also slows down and captures a few neutrons, and its contribution depends on the mineral. In limestone units, quartz sandstone shows a porosity several units lower than the true one, and dolomite shows a higher one. To convert to the porosity of sandstone or dolomite, we use charts for the particular tool (Porosity from the neutron log, Interpretation charts: how to read them). The log heading (the block of data at the top of the log) states the matrix in which the curve is recorded.

Conditions that differ from the standard ones also need corrections. We correct for these: the borehole diameter and the standoff between the tool and the wall (as a rule, the more mud around the tool, the higher the reading); mudcake; the salinity and density of the mud; temperature and pressure; and the salinity of the formation water. Taken one by one, most corrections are small. But in a wide, hot hole with a weighted mud, their sum reaches several porosity units (p.u.).

We usually show density and neutron porosity in one track (one column of the log display) on compatible scales. For limestone, the density scale runs from 1.95 to 2.95 g/cm³ and the neutron porosity scale from 0.45 to −0.15. Then in clean limestone full of water the curves coincide, and any separation immediately points to a difference from this standard. In siliciclastic sections we more often use the scales for sandstone: 1.90–2.90 g/cm³, with the same limits of neutron porosity converted to sandstone units.

Shale, gas and lithology on the neutron–density pair

Shale. The neutron tool sees all the hydrogen, not only the hydrogen of the mobile water in the pores. It also sees the bound water of shales and the hydroxyl groups of clay minerals. So in shales the neutron porosity is high, 0.30–0.45, although there are almost no effective pores. The density of shales is not especially low, so the density porosity in them is noticeably lower than the neutron porosity. This separation grows with shale volume, and we use it to estimate shale volume (Determining shale volume).

Gas. Part of the gas remains in the flushed zone of a gas-bearing formation (Borehole environment: invaded zone and mudcake). Its hydrogen index is small, so the neutron porosity is smaller than the true porosity. Its apparent density is small, so the density porosity is larger than the true porosity (Density log and photoelectric factor (Pe)). On compatible scales the curves change places: the density porosity is larger than the neutron porosity. This crossover is the main sign of gas on logs. Its size depends on the porosity, on the residual gas saturation of the flushed zone (the share of the pore space still filled with gas) and on the pressure. The deeper the invasion and the more complete the flushing, the weaker the effect.

The effects add up. Shale shifts the neutron porosity up and gas shifts it down. In a shaly gas-bearing formation they can cancel each other out, so there is no crossover although there is gas. So we look for gas using a combination of signs: crossover, high resistivity, gas shows while drilling, epithermal neutron porosity, nuclear magnetic resonance (NMR) logging and formation testing.

Neutron and density porosity on limestone scales

Rock or fluid: Limestone with water or oil
Relationship of the curves: coincide
How to check: —

Rock or fluid: Sandstone with water
Relationship of the curves: the neutron one is slightly lower: weak crossover
How to check: low GR, $P_e$ about 1.8

Rock or fluid: Dolomite with water
Relationship of the curves: the neutron one is higher than the density one
How to check: $P_e$ about 3

Rock or fluid: Shale
Relationship of the curves: the neutron one is much higher
How to check: high GR

Rock or fluid: Gas-bearing formation
Relationship of the curves: the neutron one is much lower: strong crossover
How to check: resistivity, gas shows

Rock or fluid: Rock salt
Relationship of the curves: neutron about 0, density about 0.4
How to check: very low GR, $P_e$ 4.7, washouts

Rock or fluid: Anhydrite
Relationship of the curves: neutron about 0, density negative
How to check: density 2.98, $P_e$ 5.1

Rock or fluid: Coal
Relationship of the curves: both very high
How to check: density 1.2–1.5, low GR

Rock or fluid	Relationship of the curves	How to check
Limestone with water or oil	coincide	—
Sandstone with water	the neutron one is slightly lower: weak crossover	low GR, $P_e$ about 1.8
Dolomite with water	the neutron one is higher than the density one	$P_e$ about 3
Shale	the neutron one is much higher	high GR
Gas-bearing formation	the neutron one is much lower: strong crossover	resistivity, gas shows
Rock salt	neutron about 0, density about 0.4	very low GR, $P_e$ 4.7, washouts
Anhydrite	neutron about 0, density negative	density 2.98, $P_e$ 5.1
Coal	both very high	density 1.2–1.5, low GR

A simple model shows how large the crossover in gas is. The neutron porosity of a clean formation, converted to the correct matrix, is proportional to the porosity and to the hydrogen index of the fluid in the flushed zone.

Neutron porosity of a clean formation (simplified model)

\phi_N \approx \phi\,\bigl[S_{xo}\,\mathrm{HI}_{mf} + (1 - S_{xo})\,\mathrm{HI}_h\bigr]

where:

$\phi_N$ neutron porosity in units of the correct matrix, fraction
$\phi$ true porosity, fraction
$S_{xo}$ fraction of the pores of the flushed zone occupied by filtrate, fraction
$\mathrm{HI}_{mf},\ \mathrm{HI}_h$ hydrogen index of the filtrate and of the residual hydrocarbons, —

Units: fraction

Holds when

A clean formation without shale. The curve is converted to the matrix of the formation and corrected for the borehole.
The excavation effect is ignored: in a gas-bearing formation a real tool shows an even lower porosity (advanced section).
Neutron absorbers have no influence (or the tool is epithermal).

For a formation with water or oil the bracket is close to one, and $\phi_N \approx \phi$ . Residual gas reduces the bracket, and with it the neutron porosity.

Quick estimate of the porosity of a gas-bearing formation

\phi \approx \sqrt{\frac{\phi_N^2 + \phi_D^2}{2}}

where:

$\phi_N,\ \phi_D$ neutron and density porosity calculated for the same (correct) matrix, fraction

Units: fraction

Holds when

An empirical rule (it comes from experience) for clean gas-bearing formations. Both curves are corrected for the borehole.
For formations with liquid on limestone scales, we use the arithmetic mean $(\phi_N + \phi_D)/2$ . It depends little on whether the formation is sandstone, limestone or dolomite.
Charts and calculations give an accurate estimate, with the excavation effect and the gas saturation of the flushed zone (Gas, shale, carbonate and pressure effects on porosity logs).

Gas shifts the two curves in opposite directions, and their root mean square (square each value, take the mean, then take the square root) brings the result back to the porosity of the formation. For a liquid both curves are close, and the formula gives almost the same as the arithmetic mean.

Worked exampleGas-bearing sandstone on the neutron and density logs

Given (assumed data)

Rock: clean quartz sandstone, $\rho_{ma} = 2.65$ g/cm³; both porosities are calculated for sandstone
True porosity: $\phi = 0.22$ , which is what we have to “find”
Flushed zone: 65% of the pores hold fresh filtrate (1.00 g/cm³, HI = 1), 35% residual gas
Gas: methane, $\rho_g = 0.18$ g/cm³; apparent density for the density tool 0.052 g/cm³ (Density log and photoelectric factor (Pe))

Solution

Hydrogen index of the gas: $\mathrm{HI}_g = 9\cdot 0.18\cdot 4/16 = 0.405$ .
Apparent density of the fluid: $0.65 + 0.35\cdot 0.052 = 0.668$ g/cm³. Bulk density: matrix $0.78\cdot 2.65$ plus fluid $0.22\cdot 0.668$ , in total $\rho_b = 2.214$ g/cm³.
Density porosity calculated for a liquid of density 1.00: $\phi_D = \dfrac{2.65 - 2.214}{1.65} = 0.264$ .
Hydrogen index of the fluid in the flushed zone: $0.65 + 0.35\cdot 0.405 = 0.792$ ; neutron porosity $\phi_N \approx 0.22\cdot 0.792 = 0.174$ .
Crossover: $\phi_D - \phi_N = 0.090$ , about 9 porosity units. Quick estimate: $\sqrt{(\phi_N^2 + \phi_D^2)/2} = 0.224$ ; the arithmetic mean is 0.219.

Answer: The tools show

\phi_D \approx 0.264

and

\phi_N \approx 0.174

, a crossover of about 9 porosity units. The quick estimate gives 0.224, against a true porosity of 0.22.

What it means. Each curve on its own is wrong by 4–5 porosity units, in opposite directions. Combined, they return the porosity to within about one percentage point. Because of the excavation effect, a real neutron tool shows a somewhat lower porosity in gas. So for a quantitative estimate we use charts and calculations that take this effect into account. The crossover appears only if the gas has stayed in the flushed zone. With deep flushing, or in a shaly formation, it may weaken or disappear. The data in this example are assumed for illustration.

CalculatedA calculated combined log of an assumed section, from the model and the data of the example: gas-bearing and water-bearing sandstone between shales, on sandstone scales. The curves are smoothed with a 0.6 m window, as the tool does. The orange fill is the crossover (gas), the grey fill is the separation in shales, and the dashed line is the gas–water contact (GWC).

Terms and symbols

Hydrogen index $\mathrm{HI}$: The amount of hydrogen per unit volume of a substance, compared with fresh water at surface conditions.
Neutron porosity $\phi_N$ (NPHI): The reading of the neutron log in porosity units. In limestone units it equals the porosity of clean limestone full of water.
Thermal and epithermal neutrons: Thermal neutrons (about 0.025 eV) have already slowed down and wander until they are captured. Epithermal neutrons (above about 0.4 eV) are still slowing down, and absorbers hardly affect them.
Compensated neutron tool: A tool with two detectors. We convert the ratio of their counts to porosity, and it depends little on the borehole.
Curve crossover: The density porosity is larger than the neutron porosity on compatible scales. It is the main sign of gas, but salt, and sandstone on limestone scales, also produce it.
Excavation effect: An extra lowering of the neutron porosity in gas. The gas-filled part of the pores holds almost no matter to slow neutrons down.

Common mistakes and misconceptions

Wrong

“Neutron porosity is the porosity of the formation.”

In fact

It is the hydrogen index expressed in limestone porosity units. It equals the porosity only in clean limestone with water or oil, or after conversion to the correct matrix. Shale, gas and absorbers distort it.

Wrong

“No crossover of the curves means no gas.”

In fact

Shale shifts the neutron porosity up and can cancel the gas effect. With deep flushing there is little gas in the zone of investigation. Check for gas with resistivity, gas shows, epithermal neutron logging and nuclear magnetic resonance logging.

Wrong

“A crossover is always gas.”

In fact

Rock salt, and sandstone on limestone scales, also give a crossover. Check the lithology first, from the GR, $P_e$ and the sonic log. Then compare with a water-bearing formation of the same rock.

Wrong

“The neutron log will tell oil from water.”

In fact

The hydrogen index of oil and water is almost the same, about 1. Resistivity tells oil from water. The neutron method separates only gas from liquid.

Wrong

“High neutron porosity in shales means there are many pores.”

In fact

The hydrogen of bound water and of the hydroxyl groups of clay minerals slows the neutrons down. Shales have almost no effective pores.

In depth: advanced detail

The excavation effect

The simple response model quietly assumes that gas only reduces the amount of hydrogen. It assumes that otherwise the rock slows neutrons down as if the missing hydrogen had been replaced by matrix material. In reality, the gas-filled part of the pores contains almost no matter, as if it had been dug out. So the rock slows neutrons down even less than the model predicts. Calculations of neutron transport (how neutrons move through the rock) that take this into account show that in gas-bearing formations the neutron porosity is lower than the simple estimate. The correction grows with porosity and with the gas saturation of the flushed zone, and it reaches several porosity units.

Without this correction, the joint interpretation of the neutron and density logs gives a porosity of a gas-bearing formation that is too low, and it puts too much gas in the flushed zone. Charts and approximate formulas give the corrections, and modern interpretation software solves the response equations of both tools together (Gas, shale, carbonate and pressure effects on porosity logs, Lithology and porosity from crossplots).

Neutron capture and pulsed methods

The macroscopic capture cross section $\Sigma$ describes how well a rock captures thermal neutrons. It is the probability of capture per unit path, and it is given in capture units (c.u.). Chlorine is a strong absorber, so salty formation water captures thermal neutrons much more strongly than oil and gas. This is the basis of the pulsed neutron methods (in Russian-language practice, pulsed neutron–neutron logging and pulsed neutron gamma logging). A generator in the tool emits short pulses of neutrons with an energy of 14 MeV. We measure $\Sigma$ of the formation from how fast the number of thermal neutrons or capture gamma rays dies away after each pulse.

Pulsed methods work through casing. When the formation water is salty, they let us follow water saturation during development. With fresh water, we use the spectrum of gamma rays from the inelastic scattering of fast neutrons (scattering in which the neutron loses part of its energy and the nucleus emits a gamma ray). The ratio of carbon to oxygen tells oil from water (the carbon–oxygen, or C/O, log). The generator emits only when it is powered. So such tools are safer to transport and store than tools with isotope sources.

Connections to other topics

Density log and photoelectric factor (Pe) — the paired method: together with the neutron log it identifies gas, shale and lithology.
Overview of logging tools — the place of the neutron log among other methods, depth of investigation and resolution.
Determining shale volume — shale volume from the separation of the neutron and density porosity.
Porosity — what porosity is and how it is estimated by different methods.
Borehole environment: invaded zone and mudcake — the flushed zone: which fluid the neutron tool sees.
Porosity from the neutron log — conversion of neutron porosity to different matrices and borehole corrections.
Gas, shale, carbonate and pressure effects on porosity logs — quantitative treatment of gas, shale and lithology in the porosity estimate.

Check yourself

1
Why is hydrogen the main moderator of neutrons in rock?
1. A.The mass of the hydrogen nucleus is almost equal to that of the neutron: in a single collision the neutron can give it all its energy
2. B.The hydrogen nucleus has the largest charge
3. C.Hydrogen captures neutrons more strongly than anything else
4. D.There is more hydrogen in rock by mass than other elements
Show the answer
Correct answer: A. The mass of the hydrogen nucleus is almost equal to that of the neutron: in a single collision the neutron can give it all its energy
The fraction of energy lost in a head-on collision is $4A/(A+1)^2$ . It is one for hydrogen and only 0.22 for oxygen. A neutron needs about 19 collisions with hydrogen, and hundreds with heavy nuclei, to become thermal. Hydrogen captures neutrons only weakly, and there is little of it in rock by mass.
2
How does residual gas in the flushed zone change the neutron and density porosity?
1. A.Neutron falls, density rises
2. B.Both rise
3. C.Neutron rises, density falls
4. D.Both fall
Show the answer
Correct answer: A. Neutron falls, density rises
The hydrogen index of gas is small, so the neutron porosity falls. The density of gas is small, so the bulk density decreases, and the density porosity, calculated for a liquid, rises. On compatible scales the curves change places. This is the crossover.
3
Clean sandstone, both porosities converted for sandstone: neutron 0.12, density 0.30; GR low, resistivity high. What kind of formation is this and what is its porosity?
Show the answer
A strong crossover in a clean formation with high resistivity is a sign of gas. Quick estimate: $\sqrt{(0.12^2 + 0.30^2)/2} \approx 0.23$ (the arithmetic mean would give 0.21). Because of the excavation effect, the neutron porosity is too low by more than the simple model says. So we make an accurate estimate with charts, or by solving the response equations together.
4
A shaly formation with salt formation water. Which neutron porosity will be higher: from thermal or from epithermal neutrons?
1. A.It depends only on porosity
2. B.From epithermal neutrons
3. C.They are the same
4. D.From thermal neutrons
Show the answer
Correct answer: D. From thermal neutrons
The chlorine of salt water and the boron of shales capture thermal neutrons, so the count at the thermal-neutron detectors is lower. The tool “sees” more hydrogen than there is. Epithermal neutrons are captured weakly, so the porosity from them is lower and closer to the density porosity.
5
On limestone scales, in an interval with a very low GR, the density porosity is about 0.40 and the neutron porosity about zero. Is this gas?
Show the answer
Most likely not. This is what rock salt looks like. Its apparent density of 2.03 g/cm³ gives about 0.40 on the limestone scale, and there is no hydrogen in salt. To check, look at these signs. $P_e$ is about 4.7 (for gas-bearing sandstone it is about 1.8). The density is steadily 2.03. The interval transit time is about 220 µs/m. There are often washouts where the salt has dissolved. No porous, permeable rock has this combination.
6
The hydrogen index of methane with a density of 0.15 g/cm³ is approximately…
1. A.1.0
2. B.0.68
3. C.0.34
4. D.0.15
Show the answer
Correct answer: C. 0.34
$\mathrm{HI} = 9\cdot 0.15\cdot 4/16 \approx 0.34$ . Methane has 2.25 times more hydrogen per unit mass than water, but the density of gas in the formation is several times lower.

Updated: 26 October 2026