Porosity from the neutron log

Quick refresher

Neutron porosity is the amount of hydrogen, written as the porosity of a reference rock. It equals the true porosity only in a clean formation of the same matrix, with pores full of water or oil, under standard borehole conditions.
Order of processing: environmental corrections (borehole, mudcake, mud, temperature, pressure, salinity) → conversion to the formation matrix → allowance for shale and gas.
Taken one by one, the environmental corrections reach 4–5 porosity units. But their signs differ. The temperature correction is positive, while the corrections for the salinity of the water in the zone of investigation and for the borehole diameter are negative. Their sum is often small, but not always.
For the same reading in limestone units, the porosity of sandstone is 2–5 units higher (2–3 near zero, 4–5 at medium porosity) and that of dolomite is 2–8 units lower. The conversion does not follow a straight line, and it depends on the tool.
Shale raises the reading: in shale the tool reads 0.30–0.45. Effective porosity is $\phi_N - V_{sh}\phi_{N,sh}$ .
Gas lowers the reading twice over. It contains little hydrogen, and the pores it fills hold almost no material that slows the neutrons (the excavation effect).

Key relationships

\phi_N \approx \phi\,\mathrm{HI}_{fl} + V_{sh}\,\phi_{N,sh}

the reading in units of the formation matrix as the sum of the contributions of fluid and shale

\phi_e \approx \phi_N - V_{sh}\,\phi_{N,sh}

effective porosity of a shaly formation with liquid in the pores

S_{wH} = S_{xo}\,\mathrm{HI}_{mf} + (1 - S_{xo})\,\mathrm{HI}_g

hydrogen index of the flushed-zone fluid with residual gas

\Delta\phi_{ex} \approx K\,(2\phi^2 S_{wH} + 0.04\,\phi)(1 - S_{wH})

excavation-effect correction; $K = 1$ for sandstone

Common pitfalls

Reading the scale of the neutron curve as porosity without checking which matrix it is recorded in.
Applying corrections that have already been applied during recording or processing, or applying them selectively.
Adding “four units for sandstone” to the reading at any porosity and for any tool.
Assuming that agreement between neutron and density porosity proves there is no gas: shale and gas cancel each other out.

After this lesson you can

explain what a reading in limestone units means and what it is made up of
list the standard calibration conditions, and state the sign and the size (order of magnitude) of the environmental corrections
convert a reading from limestone units to a sandstone or dolomite matrix and estimate the error when the wrong matrix is assumed
correct neutron porosity for shale and compare the result with density porosity
calculate the effect of residual gas, including the excavation effect, and recover the porosity when the saturation of the flushed zone is known

What is recorded on the log

A neutron tool counts the neutrons that the rock has slowed down. The more hydrogen the rock contains, the fewer neutrons reach the far detector (Neutron log). Calibration converts the ratio of the counts of the two detectors into neutron porosity. The meaning of this number is as follows. It is the porosity of a water-saturated limestone that would, under standard conditions, give the same count ratio. These are limestone porosity units. On English-language logs the curve is labelled NPHI, and its scale is marked in fractions or in porosity units (p.u., one percentage point of porosity).

The log heading says in units of which matrix the curve is recorded (limestone, sandstone or dolomite) and which corrections have already been applied. Any work with neutron porosity starts there. The same curve, recorded in sandstone units and in limestone units, differs by several porosity units.

After the corrections and the conversion to the formation matrix, it is convenient to think of the reading as a sum of contributions. Each component adds a share proportional to the volume it occupies.

Neutron porosity as the sum of the contributions of fluid and shale

\phi_N \approx \phi\,\mathrm{HI}_{fl} + V_{sh}\,\phi_{N,sh}

where:

$\phi_N$ reading after environmental corrections, converted to the formation matrix, fraction
$\phi$ porosity outside the shale material (effective), fraction
$\mathrm{HI}_{fl}$ hydrogen index of the fluid in the zone of investigation (its hydrogen content relative to water): about 1 for water and oil, less for gas, —
$V_{sh}$ shale volume, fraction
$\phi_{N,sh}$ tool reading in pure shale of the same section, fraction

Units: fraction

Holds when

The contributions add up in proportion to the volumes. This is an approximation that suits fluid and shale.
The matrix itself contributes nothing when the reading is in units of its own matrix. If the matrix is different, first convert the reading using the tool's conversion curve.
Gas lowers the reading further because of the excavation effect (see the section on gas).
There are no hydrogen-bearing minerals, such as gypsum or coal, in the formation.

For a clean formation with water or oil, only $\phi_N = \phi$ remains. This is the very case for which the scale is marked in porosity units. Each departure from it, shale or gas, adds or subtracts its own term.

What changes if…

…the curve is recorded in limestone units and the formation is sandstone, the reading is lower than the porosity. If the formation is dolomite, the reading is higher.
…the borehole is wider than nominal, or the tool has moved away from the wall, the reading rises, because there is more mud, and so more hydrogen, around the tool.
…the well is hot, the reading of a thermal neutron tool (one that counts slowed-down neutrons) is too low, and the more so the higher the porosity.
…the water in the zone of investigation is salty, the reading of the same tool is too high, because chlorine absorbs thermal neutrons.
…there is shale in the formation, the reading rises by the shale volume multiplied by the reading in shale.
…gas has remained in the flushed zone, the reading falls by more than the lack of hydrogen alone would suggest.

Standard conditions and environmental corrections

The calibration of a compensated thermal neutron tool (two detectors, counting slowed-down neutrons) applies to particular conditions: a borehole about 200 mm (7⅞ in) in diameter, fresh water in the borehole and in the pores, no mudcake and no standoff (gap) between the tool and the borehole wall, a temperature of 24 °C, atmospheric pressure, and the tool pressed against the wall. In a real well everything is different, and each difference shifts the reading. We remove these shifts with environmental corrections, taken from charts or from the tool manufacturer's software (Interpretation charts: how to read them).

Corrections to the reading of a compensated thermal neutron tool

Departure from the standard: Borehole wider than 200 mm
What happens: more mud around the tool: the reading is too high
Correction: minus
Order of magnitude: 1–3 units per 100 mm

Departure from the standard: Standoff (gap) from the wall
What happens: mud between the tool and the formation: the reading is too high
Correction: minus
Order of magnitude: a few units already at 1–2 cm

Departure from the standard: Mudcake
What happens: a layer with a different hydrogen content in front of the tool
Correction: usually minus
Order of magnitude: fractions of a unit

Departure from the standard: Salty mud in the borehole
What happens: chlorine absorbs thermal neutrons near the tool: the reading is too low
Correction: plus
Order of magnitude: up to 1–2 units

Departure from the standard: Weighted mud (heavy mud with added solids)
What happens: the solid phase displaces water from the borehole: the reading is too low
Correction: plus
Order of magnitude: up to 1–2 units

Departure from the standard: High temperature
What happens: the reading is too low, more so at high porosity
Correction: plus
Order of magnitude: 4–5 units at 110 °C and a reading of about 20

Departure from the standard: High pressure
What happens: the liquid is denser: the reading is too high
Correction: minus
Order of magnitude: about half a unit at 40 MPa

Departure from the standard: Salty water in the zone of investigation
What happens: chlorine absorbs thermal neutrons in the formation: the reading is too high
Correction: minus
Order of magnitude: 2–4 units at 100–200 g/kg

Departure from the standard	What happens	Correction	Order of magnitude
Borehole wider than 200 mm	more mud around the tool: the reading is too high	minus	1–3 units per 100 mm
Standoff (gap) from the wall	mud between the tool and the formation: the reading is too high	minus	a few units already at 1–2 cm
Mudcake	a layer with a different hydrogen content in front of the tool	usually minus	fractions of a unit
Salty mud in the borehole	chlorine absorbs thermal neutrons near the tool: the reading is too low	plus	up to 1–2 units
Weighted mud (heavy mud with added solids)	the solid phase displaces water from the borehole: the reading is too low	plus	up to 1–2 units
High temperature	the reading is too low, more so at high porosity	plus	4–5 units at 110 °C and a reading of about 20
High pressure	the liquid is denser: the reading is too high	minus	about half a unit at 40 MPa
Salty water in the zone of investigation	chlorine absorbs thermal neutrons in the formation: the reading is too high	minus	2–4 units at 100–200 g/kg

The signs and orders of magnitude come from the chart of one common tool. For another tool, or with other processing, they differ. The unit is one porosity unit (0.01). The water in the zone of investigation is mainly filtrate with some formation water mixed in: the tool sees 15–30 cm of rock behind the wall.

The two largest corrections, for temperature and for the salinity of the water in the zone of investigation, have different signs. In a hot formation with salty water they almost cancel each other, and the sum of all the corrections is small. But when one of the conditions changes (fresh water in a hot formation, or brine in a cold one), the total correction reaches several porosity units.

Some of the corrections are applied during recording. The tool applies the borehole-diameter correction itself if it is recorded together with the caliper of the density tool (the caliper measures the hole diameter). Modern processing adds the others. The heading and the notes of the log say which corrections the curve already contains. Applying a correction a second time is as bad as not applying it at all.

Worked exampleNeutron porosity of a hot, deep formation

Given (assumed data)

Uncorrected reading: 0.175 in limestone units
Borehole: diameter 251 mm, mudcake 6 mm
Mud: density 1.45 g/cm³ (weighted with barite), 80 g of NaCl per kilogram
Formation: 108 °C, 42 MPa; water in the zone of investigation about 100 g/kg
Corrections from the tool chart for the borehole and the mudcake, porosity units: −1.2 and −0.4
For mud salinity and mud density: +0.6 and +0.5
For temperature and pressure: +4.6 and −0.5
For the salinity of the water in the zone of investigation: −3.1

Solution

The negative corrections add up to $-1.2 - 0.4 - 0.5 - 3.1 = -5.2$ , and the positive ones to $0.6 + 0.5 + 4.6 = 5.7$ .
The total correction is 0.5 porosity units. The corrected reading is $0.175 + 0.005 = 0.180$ in limestone units.
If the water in the zone of investigation were fresh, there would be no last correction: the sum would be 3.6 units and the reading 0.211. If the formation were cold, with the same brine, there would be no temperature correction: the sum would be −4.1 units and the reading 0.134.

Answer: The corrected reading is 0.180 in limestone units: all the corrections together came to only 0.5 porosity units.

What it means. The two largest corrections, for temperature and for the salinity of the water in the zone of investigation, almost cancelled each other. This is a common case, but not a rule. With fresh water in the same hot formation the total correction would have been 3.6 units, and with the same brine in a cold formation −4.1. That is why we do not apply corrections selectively: apply either all the missing ones, or none if the recording has already been corrected. The correction values in this example are assumed for illustration. In practice they are read from a chart or calculated by software for the particular tool.

From limestone units to the formation matrix

The rock matrix also slows down and absorbs neutrons, though much more weakly than water. Quartz does this less than calcite. So in sandstone, at the same porosity, more neutrons reach the detector, and a tool calibrated on limestone shows a lower porosity. Dolomite does the opposite: it acts on neutrons more strongly than calcite, and the reading in it is higher than the true porosity. In tight, non-porous quartzite the tool reads about −0.02, and in tight dolomite from +0.01 to +0.03.

Porosity of the formation for the same reading in limestone units

Reading: 0
Sandstone: 2–3%
Limestone: 0
Dolomite: —

Reading: 10%
Sandstone: 14–15%
Limestone: 10%
Dolomite: 4–8%

Reading: 20%
Sandstone: 24–25%
Limestone: 20%
Dolomite: 12–17%

Reading: 30%
Sandstone: 34–35%
Limestone: 30%
Dolomite: 22–26%

Reading	Sandstone	Limestone	Dolomite
0	2–3%	0	—
10%	14–15%	10%	4–8%
20%	24–25%	20%	12–17%
30%	34–35%	30%	22–26%

These are guide values for compensated thermal neutron tools. The spread for dolomite reflects different tools and processing methods. A zero reading does not occur in dolomite: in a tight rock the tool shows 1 to 3%. For epithermal neutron tools the differences are smaller: about +3 units for sandstone and −2 to −3 for dolomite. For calculations, use the conversion curve of your own tool.

The table shows two properties of the conversion. First, it does not follow a straight line. The correction for sandstone grows from two or three units near zero to four or five at medium porosity, so the rule “add four units” is only a rough guess. Second, it depends on the tool. Tools of different types and generations respond differently to the slowing down and absorption of neutrons by the matrix (why).

Worked exampleClean water-bearing sandstone: neutron porosity and a check against density

Given (assumed data)

Reading after environmental corrections: 0.180 in limestone units (previous example)
Tool conversion curve for sandstone: a reading of 0.10 corresponds to a porosity of 0.145; a reading of 0.20 to 0.245 (assumed values)
Density log in the same formation: $\rho_b = 2.28$ g/cm³; matrix 2.65, filtrate 1.03 g/cm³
Formation: clean sandstone, pores filled with filtrate and formation water

Solution

Convert to sandstone by linear interpolation between the points of the conversion curve (we read along a straight line between them, which has unit slope): $0.145 + (0.180 - 0.10) = 0.225$ .
The density porosity is $\phi_D = \dfrac{2.65 - 2.28}{2.65 - 1.03} = 0.228$ . The discrepancy is 0.3 porosity units.
If the recorded reading of 0.175 were taken as the porosity, the error would be 5.0 units: 0.5 from the environmental conditions that were not corrected for, and 4.5 from the matrix.
If the formation were mistaken for dolomite, the same reading would give 10 to 15%, one and a half to two times lower.

Answer: The sandstone porosity is 22.5% from the neutron log and 22.8% from the density log.

What it means. In a clean water-saturated formation of known lithology, the two methods must agree to within one or two porosity units. This is the main check of both the matrix and the corrections. If the neutron porosity is clearly higher than the density porosity, look for shale or for a matrix that is “heavier” for neutrons. If it is lower, look for gas. The data in this example are assumed for illustration.

Shale

A neutron tool sees all the hydrogen: the water in the pores, the water bound to the surface of the clay particles, and the hydroxyl groups (OH) in the crystal lattice of the clay minerals. In pure shale it reads 0.30–0.45, although the effective porosity is almost zero. Kaolinite and chlorite give more, illite less. In a shaly reservoir a share of this “shale hydrogen” is added to the reading, and we subtract it.

Effective porosity of a shaly formation from the neutron log

\phi_e \approx \phi_N - V_{sh}\,\phi_{N,sh}

where:

$\phi_e$ effective porosity: the pore volume outside the shale material, as a fraction of the rock volume, fraction
$\phi_N$ reading in units of the formation matrix after environmental corrections, fraction
$V_{sh}$ shale volume, fraction
$\phi_{N,sh}$ reading in the adjacent pure shales, in the same units, fraction

Units: fraction

Holds when

The pores are filled with water or oil (hydrogen index about 1); there is no gas.
The shale material in the formation is the same as in the reference shale unit.
The shale volume has been found independently, for example from the gamma ray log (GR) or SP (Determining shale volume).

The correction here is several times larger than for the density log: the “porosity of shale” is 0.30–0.45 for the neutron tool and about 0.1 for the density tool. So in a shaly formation the neutron porosity is much higher than the density porosity, and the gap between them is what we use to estimate the shale volume itself.

In other unit systems (1)

Laminated formation: reading in the sand laminae

\phi_{N,sd} = \frac{\phi_N - V_{sh}\,\phi_{N,sh}}{1 - V_{sh}}

$\phi_{N,sd}$ is the neutron porosity of the sand laminae themselves, as a fraction of their volume; $V_{sh}$ is the fraction of shale laminae.

There is no constant. The contribution of the shale laminae is taken out of the reading, and the remainder is related to the volume of the sand laminae. Then $\phi_e = (1 - V_{sh})\,\phi_{N,sd}$ .

Worked exampleShaly sandstone: neutron and density estimates

Given (assumed data)

Formation: shaly water-bearing sandstone, $V_{sh} = 0.25$ from the gamma ray and SP logs
Neutron porosity in sandstone units: 0.257; in the adjacent shales 0.38
Density porosity (sandstone matrix): 0.188; in the adjacent shales 0.103

Solution

From the neutron log: $\phi_e = 0.257 - 0.25\cdot 0.38 = 0.162$ .
From the density log: $\phi_e = 0.188 - 0.25\cdot 0.103 = 0.162$ .
Before the corrections the curves were separated by $0.257 - 0.188 = 0.069$ , and in the shales by 0.277. The ratio of the two gaps is the shale volume from the two methods: 0.25.

Answer: Both estimates give an effective porosity of about 16%, although the original readings were 25.7% and 18.8%.

What it means. The neutron correction for shale is 9.5 porosity units, and the density correction only 2.6. The two corrected porosity estimates agree, and so does the shale volume: from the gamma ray and SP logs it is the same as from the gap between the two curves. So the model of the formation is consistent. If there were gas in the pores, the corrected neutron porosity would come out lower than the density porosity. The density part of the example is worked through in the topic Porosity from the density log. The data in this example are assumed for illustration.

Gas: hydrogen index and the excavation effect

In the flushed zone of a gas-bearing formation, residual gas occupies part of the pores. The hydrogen index of gas is low (0.1–0.5, depending on pressure), so the reading falls. A simple estimate replaces the hydrogen index of the fluid with the average over filtrate and gas. But neutron transport calculations (computer models of how neutrons move through rock) have shown that the actual reading is lower still. The simple estimate quietly assumes that the pores occupied by gas behave like a matrix that merely lacks hydrogen. In fact there is almost no material in them at all. It is as if the matrix had been dug out, hence the name excavation effect. So the rock slows neutrons less.

Neutron porosity of a clean gas-bearing formation

\phi_N \approx \phi\,S_{wH} - \Delta\phi_{ex}, \qquad S_{wH} = S_{xo}\,\mathrm{HI}_{mf} + (1 - S_{xo})\,\mathrm{HI}_g

where:

$\phi_N$ reading in units of the formation matrix, fraction
$\phi$ porosity, fraction
$S_{wH}$ hydrogen index of the flushed-zone fluid: the same $\mathrm{HI}_{fl}$ , written for a mixture of filtrate and gas, —
$S_{xo}$ fraction of the pore space of the flushed zone occupied by filtrate, fraction
$\mathrm{HI}_{mf},\ \mathrm{HI}_g$ hydrogen indices of the filtrate and of the gas, —
$\Delta\phi_{ex}$ correction for the excavation effect, fraction

Units: fraction

Holds when

The formation is clean; the reading has been corrected for environmental conditions and converted to the formation matrix.
The zone of investigation contains only filtrate and residual gas.
The hydrogen index of methane is $\mathrm{HI}_g = 2.25\,\rho_g$ , with the gas density in g/cm³ (derivation).

The first term accounts for the lack of hydrogen, and the second for the additional lowering. When there is no gas in the zone of investigation, $S_{wH} = \mathrm{HI}_{mf}$ is close to one, the correction becomes zero, and the reading returns to the porosity.

Excavation-effect correction (approximate formula)

\Delta\phi_{ex} \approx K\,\bigl(2\,\phi^2 S_{wH} + 0.04\,\phi\bigr)\,(1 - S_{wH})

where:

$\Delta\phi_{ex}$ the amount by which the reading is below the simple estimate $\phi S_{wH}$ , fraction
$\phi$ porosity, fraction
$S_{wH}$ hydrogen index of the flushed-zone fluid, —
$K$ lithology coefficient: 1 for sandstone, about 1.046 for limestone and 1.173 for dolomite, —

Units: fraction

Holds when

It approximates the results of neutron transport calculations. The numbers 2 and 0.04 are fitted to those results.
Enter porosity and $S_{wH}$ as fractions.
The formula describes the additional lowering. The lack of hydrogen itself is accounted for by the factor $S_{wH}$ .

The correction grows roughly as the square of the porosity. It is greatest when the hydrogen index of the fluid is close to one half, and at $S_{wH} = 1$ it is zero. The second term in the brackets is small and is often ignored. At a porosity of 30% the correction reaches five units.

Worked exampleGas-bearing sandstone: what the tool will show and how to get back to porosity

Given (assumed data)

Formation: clean quartz sandstone from the example in the topic Neutron log: $\phi = 0.22$
Flushed zone: 65% of the pores is fresh filtrate (HI = 1), 35% is methane of density 0.18 g/cm³
Density porosity (calculation “for water”): 0.264

Solution

The hydrogen index of the gas is $\mathrm{HI}_g = 2.25\cdot 0.18 = 0.405$ , and that of the fluid is $S_{wH} = 0.65 + 0.35\cdot 0.405 = 0.792$ .
The simple estimate is $\phi S_{wH} = 0.22\cdot 0.792 = 0.174$ .
Excavation ( $K = 1$ ): $2\phi^2 S_{wH} = 0.0766$ and $0.04\phi = 0.0088$ . Their sum, multiplied by $1 - 0.792$ , gives $\Delta\phi_{ex} = 0.018$ .
The reading is $\phi_N = 0.174 - 0.018 = 0.156$ , which is 6.4 porosity units below the true porosity.
Back-calculation. When $S_{xo}$ is known, the expression for $\phi_N$ is a quadratic equation in the porosity. Its solution for $\phi_N = 0.156$ and $S_{wH} = 0.792$ gives $\phi = 0.22$ .
A quick empirical estimate from the two methods (the root mean square of the two porosities): $\sqrt{(0.156^2 + 0.264^2)/2} = 0.217$ .

Answer: The tool will show about 0.156 instead of 0.22: 4.6 units are lost because of the low hydrogen index of the gas and another 1.8 because of the excavation effect.

What it means. Excavation added almost 39% to the lowering. If we ignore it, the reading 0.156 will suggest that there is more gas in the flushed zone, and less porosity, than there really is. To calculate back from the neutron log alone, we need the saturation of the flushed zone, which microresistivity tools provide. Without it, we find porosity and saturation together from the neutron and density logs, on which gas shows with opposite signs (Gas, shale, carbonate and pressure effects on porosity logs). The quick estimate by the empirical root-mean-square rule gave 0.217 here. The data in this example are assumed for illustration.

CalculatedNeutron porosity of the gas-bearing sandstone of the example (true porosity 0.22), plotted against the fraction of the flushed-zone pore space occupied by filtrate. The dashed line allows only for the hydrogen index of the gas, and the solid line adds the excavation effect. Calculated from the formulas above; the point is the example.

What the reading is made of

In a real formation everything happens at once: the matrix is not the one on the scale, the rock contains shale, and gas has remained in the pores. The effects have different signs and can almost cancel each other. It is useful to follow the path once, from the properties of the formation to the reading, and to see how many porosity units each effect costs.

Worked exampleShaly gas-bearing sandstone on a log in limestone units

Given (assumed data)

Effective porosity: 0.20
Shale volume and reading in shale: $V_{sh} = 0.15$ ; $\phi_{N,sh} = 0.36$ in sandstone units
Flushed zone: 60% of the pores is filtrate (HI = 1), 40% is methane of density 0.16 g/cm³
Tool conversion curve: a sandstone porosity of 0.145 corresponds to a reading of 0.10 in limestone units, a porosity of 0.245 to 0.20 (assumed values)

Solution

Hydrogen index: the gas has $2.25\cdot 0.16 = 0.360$ , and the fluid has $S_{wH} = 0.60 + 0.40\cdot 0.360 = 0.744$ .
Gas, lack of hydrogen: $0.20\cdot(1 - 0.744) = 0.051$ . The reading is lower by this amount.
Gas, excavation: $\Delta\phi_{ex} = 0.017$ . The reading is lower still.
Shale: $0.15\cdot 0.36 = 0.054$ . The reading is higher.
In sandstone units: 0.20 minus 0.051 and 0.017, plus 0.054, gives 0.186 in all.
In limestone units, using the conversion curve of the tool, the reading is 0.141, which is another 0.045 lower.

Answer: The tool will record about 0.141 in limestone units at an effective porosity of 0.20.

What it means. Gas took away 6.8 porosity units, shale gave back 5.4, and the limestone scale took away another 4.5. Note the intermediate result. In sandstone units the reading 0.186 happens to be close to the true porosity, because gas and shale almost cancelled each other. One curve cannot tell you this. The gamma ray log gives the shale, core and cuttings give the matrix, and the density log and resistivity give the gas. The data in this example are assumed for illustration.

CalculatedStep-by-step calculation of the reading for the formation of the example, from an effective porosity of 0.20 to the recorded 0.141. The orange bars are effects that lower the reading (gas, excavation, limestone scale). The grey bar is the effect that raises it (shale).

Calculation and checking procedure

1
Log heading
Find the type of tool, the matrix in whose units the curve is recorded, and the corrections already applied.
2
Quality of the recording
In washouts and where the tool moves away from the wall, the reading is too high. Neutron counting is random, so at a high logging speed the curve is noisy. Mark such intervals using the caliper log and the repeat section (a second run over the same interval).
3
Environmental conditions
Apply the missing corrections from charts or with software for the particular tool: borehole, mudcake, mud, temperature, pressure, salinity.
4
Matrix
Convert the reading to the formation matrix using the tool's conversion curve. Take the lithology from core, cuttings and the photoelectric factor. For a mixed composition, use the density log as well (Lithology and porosity from crossplots).
5
Shale and gas
In shaly formations, subtract the contribution of the shale. The signs of gas are a reading below the density porosity, high resistivity, and gas shows while drilling. Find the porosity of a gas-bearing formation from two methods.
6
Check
In a clean water-bearing formation, neutron and density porosity should agree, and both should agree with core (Porosity from the density log).

Terms and symbols

Neutron porosity $\phi_N$ (NPHI): The reading of the neutron log in porosity units of a reference rock.
Limestone units: A scale on which the reading equals the porosity of clean water-saturated limestone under standard conditions.
Environmental corrections: Corrections for the ways in which the borehole, mud, temperature, pressure and salinity differ from the calibration conditions of the tool.
Conversion to the formation matrix: The step from limestone units to the porosity of sandstone or dolomite, using the tool's conversion curve. It does not follow a straight line.
Reading in shale $\phi_{N,sh}$: The neutron porosity of the pure shales of the section, usually 0.30–0.45. It comes from bound water and hydroxyl groups.
Fluid hydrogen index $S_{wH}$: The average hydrogen index of the filtrate and residual hydrocarbons of the flushed zone.
Excavation effect: The additional lowering of neutron porosity in a gas-bearing formation, over and above that caused by the low hydrogen index of gas.

Common mistakes and misconceptions

Wrong

“NPHI = 0.18 on the log means a porosity of 18%.”

In fact

This is true only in clean limestone with liquid in the pores, if the curve is recorded in limestone units and corrected for environmental conditions. In sandstone the porosity is about 22–23%, and in dolomite 10–15%.

Wrong

“The environmental corrections are small and can be ignored.”

In fact

In total they are often small, because the largest ones, for temperature and for the salinity of the water, have different signs. But each reaches 4–5 porosity units, and with fresh water in a hot formation, or brine in a cold one, nothing cancels them.

Wrong

“For sandstone it is enough to add four units.”

In fact

The conversion does not follow a straight line, and it depends on the tool. Near zero porosity the correction is 2–3 units, and at medium porosity 4–5. For epithermal neutron tools it is about three. For dolomite the sign is the opposite.

Wrong

“The correction for shale is to subtract $V_{sh}$ .”

In fact

We subtract $V_{sh}\phi_{N,sh}$ : the shale volume multiplied by the reading in the pure shales of the section. Take the reading in shale from the same log, not from a reference book.

Wrong

“Neutron and density porosity agree, so there is no gas.”

In fact

In a shaly gas-bearing formation, shale raises the neutron curve and gas lowers it, so the neutron curve can match the density porosity by chance. Check for gas using resistivity, gas shows and curves corrected for shale.

In depth: advanced detail

Why the matrix correction is non-linear and depends on the tool

The ratio of the counts of the two detectors depends on how far from the source the neutrons get while they slow down. For thermal neutrons it also depends on how far they wander before they are captured. Both distances depend on all the nuclei in the rock, not only on hydrogen. In a tight rock there is no hydrogen, so the matrix decides everything. Quartz, calcite and dolomite differ in density, in the number of light nuclei (oxygen and carbon) per unit volume, and in their ability to capture thermal neutrons. As porosity grows, hydrogen takes over more and more of the slowing down, but the matrix still occupies most of the volume and keeps affecting the reading.

In addition, the scale of the tool is marked for limestone, and the link between the count ratio and porosity does not follow a straight line. The same shift in the count ratio means a different number of porosity units in different parts of the scale. So the matrix correction, written in porosity units, changes with porosity, and it changes differently for different tools. For some tools it is almost constant, and for others it grows or passes through a maximum. Epithermal neutron tools do not respond to capture, so the difference between matrices is smaller for them. Thermal neutron tools respond to both slowing down and capture. For them the difference is larger, and it depends more strongly on the processing method and on the salinity of the water, especially for dolomite. So the practical rule is to convert to the matrix only with the conversion curve of the same tool and the same processing that recorded the curve.

Behind casing and while drilling

A compensated neutron tool also works in a cased well, because neutrons pass through steel. But now a casing (the steel pipe that lines the well) and a cement sheath lie between the tool and the formation. Steel scatters and partly absorbs neutrons, and set cement contains much bound water. So corrections for the thickness of the casing wall and of the cement are added to the usual ones, and the cement thickness is poorly known. The absolute porosity behind casing is less accurate than in open hole. On the other hand, repeat runs in the same well are easy to compare. The casing and cement do not change, so a change in the reading with time reflects a change of fluid, for example the appearance of gas as the gas–oil contact moves down.

In logging-while-drilling tools the source and detectors are in a drill collar (a heavy section of the drill string), which sits in the borehole without being pressed against the wall. The standoff correction becomes the main one, and the standoff changes as the drill string rotates. On the other hand, the recording is made soon after the bit penetrates the formation, when the filtrate has not yet pushed the gas away. So the gas effect on it is usually stronger than on a wireline log (a log run on a cable).

Connections to other topics

Neutron log — the physics of the method, the hydrogen index, the tools and the signs of gas and shale.
Porosity from the density log — the second porosity estimate: the general check in a clean water-bearing formation.
Density log and photoelectric factor (Pe) — density and the photoelectric factor for choosing the matrix.
Determining shale volume — shale volume for the correction and its estimate from the separation of two curves.
Borehole environment: invaded zone and mudcake — the borehole, mudcake and flushed zone: what surrounds the tool.
Gas, shale, carbonate and pressure effects on porosity logs — joint treatment of gas, shale and lithology from several methods.
Lithology and porosity from crossplots — lithology and porosity from the neutron–density pair.
Interpretation charts: how to read them — how to read the charts of corrections and of matrix conversion.

Check yourself

1
In clean water-saturated dolomite, a curve recorded in limestone units reads 0.20. What is the true porosity?
1. A.It cannot even be estimated
2. B.About 12–17%, or more exactly, as given by the tool's conversion curve
3. C.Exactly 20%
4. D.About 24–25%
Show the answer
Correct answer: B. About 12–17%, or more exactly, as given by the tool's conversion curve
Dolomite acts on neutrons more strongly than calcite, and a tool calibrated on limestone overestimates its porosity. For compensated thermal neutron tools, a reading of 0.20 corresponds to 12–17%, depending on the tool and the processing. The option of 24–25% would be right for sandstone.
2
The reading without corrections is 0.150 in limestone units. Chart corrections: for the borehole −1.0, for temperature +2.8, for pressure −0.6 porosity units. The formation is clean water-bearing sandstone; by the tool's conversion curve a reading of 0.10 corresponds to 0.145 and a reading of 0.20 to 0.245. What is the porosity?
Show the answer
The corrections add up to $-1.0 + 2.8 - 0.6 = +1.2$ units, so the corrected reading is 0.162. Converted to sandstone: $0.145 + (0.162 - 0.10) = 0.207$ . The porosity is about 21%, which is 5.7 units more than the recorded number.
3
Shaly water-bearing sandstone: neutron porosity in sandstone units 0.30, shale volume 0.40, reading in the adjacent shales 0.40. Find the effective porosity. What is the porosity of the sand laminae if the shale is laminated?
Show the answer
$\phi_e = 0.30 - 0.40\cdot 0.40 = 0.14$ . The sand laminae occupy 60% of the volume, and their porosity is $0.14/0.60 \approx 0.23$ . Both quantities describe the same pore volume, but they refer to different volumes of rock.
4
The caliper reads 280 mm with a bit diameter of 216 mm. How does this affect the reading of the neutron tool if the borehole correction has not been applied?
1. A.The reading is too low
2. B.The reading is too high: there is more mud, and so more hydrogen, around the tool
3. C.The direction of the effect depends on the lithology
4. D.It does not: the tool is compensated
Show the answer
Correct answer: B. The reading is too high: there is more mud, and so more hydrogen, around the tool
Compensation with two detectors weakens the effect of the borehole but does not remove it. In an enlarged borehole, and even more when the tool moves away from the wall, mud enters the zone of investigation and the reading rises. The diameter correction is negative.
5
Clean gas-bearing sandstone: $\phi = 0.25$ , the filtrate occupies half of the pores of the flushed zone (HI = 1), the hydrogen index of the gas is 0.30. What will the tool show in sandstone units?
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$S_{wH} = 0.5 + 0.5\cdot 0.30 = 0.65$ . The simple estimate is $0.25\cdot 0.65 = 0.1625$ . Excavation: $2\cdot 0.25^2\cdot 0.65 \approx 0.081$ and $0.04\cdot 0.25 = 0.01$ . Their sum, multiplied by $1 - 0.65$ , gives about 0.032. The reading is about 0.13, which is almost half the true porosity, and roughly a quarter of the lowering is due to the excavation effect.
6
In a shaly sandstone, neutron porosity (in sandstone units) and density porosity coincide. Does it follow that there is no gas in the formation?
1. A.No: shale raises the neutron curve, gas lowers it, and they can cancel each other
2. B.Yes, if the borehole is of nominal diameter
3. C.Yes: gas always produces a crossover of the curves
4. D.No: the neutron log does not respond to gas at all
Show the answer
Correct answer: A. No: shale raises the neutron curve, gas lowers it, and they can cancel each other
In a water-bearing shaly formation, the neutron porosity should be clearly higher than the density porosity, by $V_{sh}$ multiplied by the difference between the readings in shale. If the curves coincide, something has lowered the neutron reading and raised the density reading. That something is a sign of gas. Check it using resistivity and curves corrected for shale.

Updated: 19 October 2026