Nuclear magnetic resonance (NMR) logging

Quick refresher

The tool sees only the hydrogen of the fluids in the pores. The hydrogen that is part of minerals loses its signal too quickly. So NMR porosity hardly depends on lithology, but gas and heavy oil make it too low.
Polarisation grows as $1 - e^{-t/T_1}$ . To see a fluid almost completely (95%), the polarisation time must be at least about $3T_1$ of the slowest component.
CPMG pulses give a train of echoes with the spacing $\mathrm{TE}$ . The decay of the train is a sum of exponentials (decaying curves) with different $T_2$ . We get the $T_2$ distribution by splitting the train into these exponentials.
In the wetting phase (the fluid that coats the pore walls), the relaxation rate is proportional to the ratio of pore surface to pore volume. Short times mean small pores, and long times mean large pores.
A boundary at about 3 ms separates clay-bound water. The cutoff time (typically 33 ms in sandstones and 90–100 ms in carbonates) separates capillary-bound water from free fluid. We refine these values on core (rock samples cut from the well).
We estimate permeability with the Coates model (from the ratio of free fluid to bound water) and the SDR model (from the logarithmic mean $T_2$ ). Both are empirical (they come from experience) and need calibration on core. The depth of investigation of NMR is only centimetres, so the tool sees the flushed zone (the zone next to the wall, flushed by mud filtrate).

Key relationships

f = \dfrac{\gamma}{2\pi}\,B_0

Larmor frequency; for protons $\gamma/2\pi = 42.58$ MHz/T

M = M_0\left(1 - e^{-\mathrm{TW}/T_1}\right)

polarisation during the time $\mathrm{TW}$

\dfrac{1}{T_2} = \dfrac{1}{T_{2B}} + \rho_2\dfrac{S}{V} + \dfrac{D\,(\gamma G\,\mathrm{TE})^2}{12}

bulk, surface and diffusion relaxation

\phi_\text{NMR} = \mathrm{CBW} + \mathrm{BVI} + \mathrm{FFI}

total NMR porosity; effective porosity $\phi_e = \mathrm{BVI} + \mathrm{FFI}$

k = \left[\left(\dfrac{\phi_e}{C}\right)^2\dfrac{\mathrm{FFI}}{\mathrm{BVI}}\right]^2

Coates model: $k$ in mD, $\phi_e$ in per cent

k = a\,\phi_e^4\,T_{2LM}^2

SDR model: $\phi_e$ as a fraction, $T_{2LM}$ in ms, $a$ found from core

Common pitfalls

Comparing NMR porosity with density porosity without allowing for gas, heavy oil and incomplete polarisation. An NMR shortfall often tells you about the fluid, not about a fault in the measurement.
Applying a cutoff of 33 ms to carbonates and to any sandstone without calibration on core.
Taking the free fluid seen by NMR for reservoir oil. The tool sees the flushed zone, which holds filtrate.
Taking permeability from the SDR model in an oil-bearing or gas-bearing interval without checking. Hydrocarbons shift $T_{2LM}$ .
Speeding up the recording with a short polarisation time, without checking $T_1$ of the slowest fluids.

After this lesson you can

explain how an NMR tool polarises protons, what the times $T_1$ and $T_2$ describe and how the CPMG pulse sequence produces a train of spin echoes
relate the time $T_2$ to pore size through surface relaxation and estimate the contributions of bulk and diffusion relaxation
split the $T_2$ distribution into clay-bound water, capillary-bound water and free fluid, and calculate total and effective porosity from NMR
estimate permeability with the Coates and SDR models and explain why their coefficients are calibrated on core
predict how gas, light and heavy oil and incomplete polarisation distort porosity and the $T_2$ distribution
choose the polarisation time, echo spacing and logging speed for the task and allow for the shallow depth of investigation of the method

Protons in a magnetic field: what the tool measures

The nucleus of the hydrogen atom, the proton, has its own angular momentum (spin) and a magnetic moment. In other words, it behaves like a tiny spinning magnet. Without an external field, the magnetic moments point in random directions, and their sum is zero. In a constant magnetic field $B_0$ , a small excess of protons lines up with the field. The fluid then acquires a magnetisation $M_0$ , which is proportional to the number of hydrogen nuclei per unit volume.

Protons that have been knocked out of equilibrium precess around the direction of the field (their axes sweep out a cone), like a tilted spinning top around the vertical. The frequency of this motion is the Larmor frequency, and it is proportional to the field. The tool is tuned to this frequency: it excites the protons on it and receives their signal on it.

Larmor frequency

f = \frac{\gamma}{2\pi}\,B_0

where:

$f$ Larmor frequency, the resonance frequency of protons, MHz
$\gamma$ gyromagnetic ratio of the proton (a constant of the proton), $2.675\cdot10^{8}$ rad/(s·T); $\gamma/2\pi = 42.58$ MHz/T, rad/(s·T)
$B_0$ strength (magnetic induction) of the constant magnetic field in the sensitive zone of the tool, T (1 T = $10^4$ G)

Units: $f$ in MHz when $B_0$ is in tesla

Holds when

Hydrogen nuclei: other nuclei have a different gyromagnetic ratio and do not respond at the frequency of the tool.
The design of the tool sets the field in the sensitive zone. In a non-uniform field, different distances from the tool correspond to different frequencies.

The fields of logging tools are of the order of hundredths of a tesla, and the working frequencies run from hundreds of kilohertz to about 2 MHz. At $B_0 = 0.05$ T, $f \approx 2.1$ MHz. In the Earth's magnetic field (about 50 µT) the same formula gives about 2 kHz. If the field of the tool gets weaker with distance from it, each frequency belongs to its own thin shell of rock. This is how multi-frequency tools measure several shells in turn.

There is hydrogen in the matrix of the rock too: in the hydroxyl groups (OH) of clay minerals and in the water of crystallisation of gypsum. But in a solid, protons lose their signal within microseconds, long before the first measurement. So the initial amplitude of the signal is proportional to the amount of hydrogen in the liquids and gases of the pores. We calibrate it on a container of water. For a rock full of water, the amplitude, as a fraction of the amplitude of pure water, equals the porosity. For other fluids, their hydrogen index enters the calculation. This is how NMR differs from the neutron log. The neutron log responds to all the hydrogen of the rock, including the hydrogen of clay minerals. NMR also differs from the density log, which needs the matrix density.

The measurement begins with polarisation. Magnetisation does not build up instantly. It builds up exponentially (fast at first, then more slowly), with a time constant $T_1$ , the longitudinal relaxation time (also called the spin–lattice relaxation time).

Polarisation of protons

M = M_0\left(1 - e^{-\mathrm{TW}/T_1}\right)

where:

$M$ magnetisation at the start of the measurement, relative units
$M_0$ equilibrium magnetisation, proportional to the amount of hydrogen, relative units
$\mathrm{TW}$ polarisation time (wait time): how long the rock stayed in the field before excitation, s
$T_1$ longitudinal relaxation time of the fluid in the rock, s

Units: $\mathrm{TW}$ and $T_1$ in the same units

Holds when

One fluid with a single value of $T_1$ . In a rock, $T_1$ has a spread of values, and we apply the formula to each part.
There is no magnetisation before polarisation starts: the rock has just entered the field of the tool, or the previous cycle has completely “erased” it.

After a time $\mathrm{TW} = T_1$ , 63% of the equilibrium magnetisation has built up. After $3T_1$ it is 95%, and after $5T_1$ more than 99%. Water in small pores polarises in tens of milliseconds, light oil in seconds and gas in many seconds. If the polarisation time is short, we see slow fluids only partly, and the porosity comes out too low.

CalculatedThe fraction polarised,

1 - e^{-t/T_1}

, for water in small pores (

T_1 = 0.05

s), residual light oil (0.8 s) and gas (4 s). The values are assumed, of a typical order of magnitude. The dashed line is the polarisation time of 2.4 s from the example on the recording mode: water is fully polarised, oil to 95%, and gas to less than half.

After polarisation, the antenna sends a short radio-frequency pulse at the Larmor frequency. It turns the magnetisation through 90°, into the plane perpendicular to the field. Rotating in this plane, the magnetisation induces a signal in the antenna. The signal quickly dies away. The field of the tool is non-uniform, so protons at different points precess at slightly different frequencies and drift out of phase. This loss of phase is reversible. A pulse that turns the spins through 180° swaps the laggards and the leaders. After the same time they come into phase again, and a spin echo appears.

The 180° pulses are repeated at equal intervals, the echo spacing $\mathrm{TE}$ , and a new echo appears after each pulse. Such a sequence is called a CPMG sequence, after the physicists Carr, Purcell, Meiboom and Gill. One sequence records hundreds or thousands of echoes. Their amplitudes get smaller because of irreversible losses, with a time constant $T_2$ . This is the transverse relaxation time (also called the spin–spin relaxation time).

SchematicThe NMR measurement cycle (a schematic drawing, not to scale): polarisation during the time

\mathrm{TW}

, a 90° pulse, 180° pulses at the interval

\mathrm{TE}

, and the spin echoes between them. The envelope of the echo amplitudes (the curve through their peaks) decays as

e^{-t/T_2}

Echo train as a sum of exponentials

A(t) = \sum_i a_i\,e^{-t/T_{2,i}},\qquad t = \mathrm{TE},\ 2\,\mathrm{TE},\ 3\,\mathrm{TE},\ \dots

where:

$A(t)$ echo amplitude at time $t$ after calibration on water, fraction of bulk volume
$a_i$ part of the porosity decaying with the time $T_{2,i}$ , fraction of bulk volume
$T_{2,i}$ transverse relaxation time of the $i$ -th group of pores, ms
$\mathrm{TE}$ echo spacing, ms

Units: amplitudes in porosity units (p.u.), times in ms

Holds when

The fluids are fully polarised and have a hydrogen index of 1. Otherwise the amplitudes are too low (see the section on hydrocarbons).
Components with $T_2$ comparable to $\mathrm{TE}$ or shorter die away before the first echoes, so they are poorly determined.
The amplitudes of the late echoes are small compared with noise. So for a reliable decomposition, we stack the signal over several cycles (add the cycles together).

The initial amplitude $A(0) = \sum a_i$ is the porosity filled with fluids. A rock contains pores of many sizes, so the decay is not a single exponential but a sum of exponentials. The job of the processing is to find the shares $a_i$ , that is, the $T_2$ distribution.

Why the signal decays: surface, bulk and diffusion

The signal is lost irreversibly in three ways. Their rates, which are the reciprocals of the times, add up.

Bulk relaxation is a property of the fluid itself: the protons interact with each other. In water it is slow, taking seconds. In oil it is faster the higher the viscosity (the thicker the oil).
Surface relaxation: a molecule of the wetting phase wanders about the pore, touches the wall again and again, and loses phase coherence and energy there. The larger the surface per unit pore volume, the faster the decay.
Diffusion relaxation: in a non-uniform field, a molecule moves between pulses to a point where the frequency is different. Then the 180° pulse can no longer bring the phases fully together. The effect grows with the diffusion coefficient, the field gradient and the square of the echo spacing.

Transverse relaxation rate

\frac{1}{T_2} = \frac{1}{T_{2B}} + \rho_2\,\frac{S}{V} + \frac{D\,(\gamma\,G\,\mathrm{TE})^2}{12}

where:

$T_2$ observed transverse relaxation time, s
$T_{2B}$ bulk relaxation time of the fluid, s
$\rho_2$ surface relaxivity of the rock (how strongly the pore wall speeds up relaxation), m/s (usually µm/s)
$S/V$ ratio of the pore surface to its volume, m⁻¹
$D$ self-diffusion coefficient of the fluid, m²/s
$\gamma$ gyromagnetic ratio of the proton, rad/(s·T)
$G$ gradient of the magnetic field in the sensitive zone (how fast the field changes with distance; 1 T/m = 100 G/cm), T/m
$\mathrm{TE}$ echo spacing, s

Units: SI; it is convenient to take $\rho_2$ in µm/s and $S/V$ in µm⁻¹, and their product comes out in s⁻¹

Holds when

Fast diffusion: a molecule travels around the pore many times during the relaxation time, $\rho_2 r/D \ll 1$ . The pore then relaxes as a whole.
The surface term refers to the wetting phase. A non-wetting phase (oil in a water-wet rock, or gas) does not touch the walls and relaxes only by bulk and diffusion relaxation.
The diffusion term is written for unrestricted diffusion in a uniform gradient. In small pores the walls limit the movement, and the term is smaller.
Neighbouring pores do not exchange molecules during the measurement. Otherwise their times average out.

For longitudinal relaxation the formula is the same, but without the diffusion term and with its own relaxivity: $1/T_1 = 1/T_{1B} + \rho_1 S/V$ . Here $\rho_1$ is usually smaller than $\rho_2$ . So for water in pores, $T_1$ is usually 1.5–2.5 times $T_2$ (diffusion adds part of the difference). For gas, which does not touch the walls, diffusion creates almost the whole difference between $T_1$ and $T_2$ .

In water that wets the pores, the surface term is usually the main one. Then $T_2 \approx (V/S)/\rho_2$ : the relaxation time is proportional to the pore size. For a spherical pore of radius $r$ , the ratio is $V/S = r/3$ . For a cylindrical pore it is $r/2$ . The $T_2$ distribution then becomes a pore size distribution, with the scale factor $\rho_2$ . This factor depends on the mineralogy of the surface: iron and manganese on the walls, chlorite and glauconite raise it. In sandstones, $\rho_2$ is usually from a few to several tens of micrometres per second. In carbonates it is more often lower, so in carbonates the same time $T_2$ corresponds to larger pores.

Worked exampleThe time

T_2

of water in pores of different size

Given (assumed data)

Pores: spherical, with radii of 0.5, 5 and 50 µm, filled with water
Surface relaxivity $\rho_2$: 8 µm/s
Bulk relaxation time of water $T_{2B}$: 2.5 s at formation temperature
Diffusion: $D = 5\cdot10^{-9}$ m²/s (water at 60–70 °C), $G = 0.2$ T/m (20 G/cm), $\mathrm{TE} = 0.6$ ms

Solution

Diffusion term: the product $2.675\cdot10^{8}\cdot 0.2\cdot 0.6\cdot10^{-3}$ is $\gamma G\,\mathrm{TE} = 32{,}100$ m⁻¹, and the rate is $D(\gamma G\,\mathrm{TE})^2/12 = 0.43$ s⁻¹. Bulk term: $1/2.5 = 0.40$ s⁻¹.
Surface term $3\rho_2/r$ : 48, 4.8 and 0.48 s⁻¹ for pores of 0.5, 5 and 50 µm.
Sums of the rates: 48.83, 5.63 and 1.31 s⁻¹. So $T_2$ is 20.5 ms, 178 ms and 0.76 s.
Share of the surface term in the sum: 98, 85 and 37%.
The pore that corresponds to $T_2 = 33$ ms if only the surface acts: $r = 3\rho_2 T_2 = 3\cdot 8\cdot 0.033 = 0.79$ µm.

Answer: Water in a pore of radius 0.5 µm decays with a time of 20.5 ms. In a pore of 5 µm the time is 178 ms, and in a pore of 50 µm it is 0.76 s. A time of 33 ms in this rock corresponds to pores with a radius of about 0.79 µm.

What it means. In small and medium pores, the time $T_2$ is almost proportional to the pore size. All NMR interpretation is built on this. In the largest pores, the surface term is no longer the main one, and a bigger pore lengthens $T_2$ less and less. The distribution “hits” a limit that the bulk relaxation of water and diffusion set. In practice we know $\rho_2$ poorly. So converting times to micrometres gives only an order-of-magnitude estimate, and we choose the boundaries between the types of water from core. The data in this example are assumed for illustration.

Pore body and pore throat

NMR measures the ratio of surface to volume, that is, the size of the pore bodies. But permeability and capillary pressure are set by the narrow constrictions, the throats (distribution from mercury injection). In sandstones the sizes of bodies and throats are usually related, and the $T_2$ distribution agrees well with mercury injection capillary pressure curves. In carbonates with vugs (large irregular holes), and in rocks with microporous grains, the relationship is worse. A large vug gives a long $T_2$ even if it connects with the other pores only through narrow channels. Iron-bearing minerals (for example siderite and iron-rich chlorites) raise the surface relaxation and create their own field gradients in the pores. This shortens $T_2$ further.

The $T_2$ distribution

A rock contains pores from fractions of a micrometre to hundreds of micrometres, and each group gives its own exponential. The processing solves an inverse problem (it works backwards from the measured signal to its causes). It fits the shares $a_i$ on a preset grid of times, usually from 0.3 ms to 3 s with a logarithmic step. The fit makes a sum of exponentials describe the recorded echo train. The result is the $T_2$ distribution: the porosity that falls in each interval of times. The area under the curve equals the porosity.

The inverse problem is unstable: because of noise, different sets of exponentials can describe the same echo train. So the solution is smoothed. Narrow peaks blur, and close peaks may merge. Separate peaks are meaningful if their times differ by a factor of several. The shape of the curve inside a peak is largely a result of the smoothing (more in the advanced section).

CalculatedThe

T_2

distribution of the assumed oil-bearing shaly sandstone of the lesson's examples, as the tool sees it in the flushed zone. The cutoffs of 3 and 33 ms divide the porosity into clay-bound water, capillary-bound water and free fluid. The free fluid consists of filtrate in the large pores and residual light oil (the dashed peak at about 700 ms).

Logarithmic mean time

T_2

T_{2LM} = \exp\!\left(\frac{\sum_i a_i \ln T_{2,i}}{\sum_i a_i}\right)

where:

$T_{2LM}$ logarithmic mean time $T_2$ , ms
$a_i$ porosity in the interval with the time $T_{2,i}$ , fraction of bulk volume
$T_{2,i}$ time of the interval of the distribution, ms

Units: ms

Holds when

The sum is taken over an agreed part of the distribution. In this lesson it is the effective porosity, $T_2 > 3$ ms. Use the same convention in the calculation and in the calibration on core.

This is the geometric mean of the times, weighted by porosity. It gives a typical pore size as a single number. On a logarithmic scale it lies at the “centre of gravity” of the distribution.

Total and effective porosity, bound and free water

We divide the $T_2$ distribution into three parts with cutoffs (boundary times). In a water-wet rock (a rock whose pore walls prefer water to oil), each part corresponds to its own state of water:

Clay-bound water (CBW): $T_2$ shorter than about 3 ms: Water in the electrical double layer at the surface of clay particles (a thin film held by electric charges), and in the tiniest pores between them. We see it only with a short echo spacing, 0.2–0.6 ms.
Capillary-bound water (bulk volume irreducible, BVI): from 3 ms to the cutoff: Water in small pores and in the corners of large ones. Capillary forces hold it in place at reservoir pressure drops. Together with clay-bound water it makes up the volume of irreducible water (water that cannot be pushed out). Its share of the pores is the irreducible water saturation.
Free fluid (free fluid index, FFI): $T_2$ longer than the cutoff: Fluid in the large pores that can move: water, oil, gas.

Total and effective porosity from NMR

\phi_\text{NMR} = \mathrm{CBW} + \mathrm{BVI} + \mathrm{FFI},\qquad \phi_e = \mathrm{BVI} + \mathrm{FFI}

where:

$\phi_\text{NMR}$ total NMR porosity, fraction of bulk volume
$\phi_e$ effective NMR porosity, without clay-bound water, fraction of bulk volume
$\mathrm{CBW},\ \mathrm{BVI},\ \mathrm{FFI}$ volumes of clay-bound water, capillary-bound water and free fluid, fraction of bulk volume

Units: fraction of bulk volume (or per cent, that is, porosity units)

Holds when

The hydrogen index of all fluids is about 1, and polarisation is complete. Otherwise parts of the distribution are too low.
The echo spacing is short enough to see clay-bound water. At a $\mathrm{TE}$ of about 1 ms the tool gives only effective porosity.
The cutoffs are chosen for this rock (typical values are in the table below).

The effective NMR porosity matches the effective porosity of the shaly-formation models (Water saturation in shaly sands): total porosity minus clay-bound water. The irreducible water saturation, as a fraction of total porosity, is $(\mathrm{CBW} + \mathrm{BVI})/\phi_\text{NMR}$ .

Typical

T_2

cutoffs

Boundary: Clay-bound water
Sandstones: about 3 ms
Carbonates: about 3 ms
How it is refined: from core and comparison with shale volume

Boundary: Capillary-bound water
Sandstones: about 33 ms
Carbonates: about 90–100 ms
How it is refined: by centrifuging core to irreducible water saturation

Boundary	Sandstones	Carbonates	How it is refined
Clay-bound water	about 3 ms	about 3 ms	from core and comparison with shale volume
Capillary-bound water	about 33 ms	about 90–100 ms	by centrifuging core to irreducible water saturation

These typical values are a guide, not norms. The cutoff depends on the surface relaxivity and on the pore structure, and it differs noticeably between formations. In carbonates with vugs and microporosity, a single boundary often works poorly. Then we use boundaries that depend on the shape of the distribution, or we calibrate separately for each rock type.

Why does a boundary in time separate water by mobility? In a water-wet rock, when oil or gas pushes the water out, the large pores empty first. Water stays in the small pores, where capillary pressure holds it (Capillary pressure and the transition zone). Bring a core sample to irreducible water saturation. In the laboratory this is done by centrifuging, usually at a pressure difference of about 0.69 MPa (100 psi) for the air–water pair. The water that remains then fills the pores with the shortest $T_2$ . The cutoff is the time below which exactly this share of the porosity lies. An example of calibration is in the advanced section.

Worked exampleBound water from NMR and water saturation from resistivity

Given (assumed data)

Formation: shaly sandstone, oil-bearing by the test of a neighbouring well; water-based mud
NMR (cutoffs 3 and 33 ms): $\mathrm{CBW} = 0.031$ ; $\mathrm{BVI} = 0.083$ ; $\mathrm{FFI} = 0.116$
Density log: total porosity $\phi_D = 0.232$
Resistivity: total water saturation of the uninvaded part of the formation $S_{wt} = 0.52$ from the shaly-formation model

Solution

Total NMR porosity: 0.031 + 0.083 + 0.116 = 0.230. It differs from the density porosity by 0.2 porosity units. So there is no gas or under-polarised fluid, and the hydrogen index is close to 1.
Effective porosity: $\phi_e = 0.083 + 0.116 = 0.199$ .
All the bound water: 0.031 + 0.083 = 0.114, that is, $0.114/0.230 = 0.496$ of the total porosity.
Volume of water in the uninvaded formation from resistivity: $0.230\cdot 0.52 = 0.1196$ .
Free water: $0.1196 - 0.114 = 0.0056$ , that is, 2.4% of the pore volume. Hydrocarbons occupy $0.230\cdot(1 - 0.52) = 0.1104$ , less than the volume of free fluid 0.116.

Answer: At a water saturation of 52%, almost all the water of the formation is bound. The free water is about 2.4% of the pore volume. We should expect an inflow of oil with a low water cut.

What it means. This is a typical low-resistivity pay. Small pores and shale hold a lot of water, so the resistivity is low. By water saturation alone the formation looks water-bearing, but NMR shows that the water cannot move. The comparison is valid because the volume of bound water is a property of the rock: filtrate hardly displaces it, so in the flushed zone it is the same as in the formation. The free fluid seen by NMR, on the contrary, belongs to the flushed zone. It is filtrate and residual oil, not reservoir fluids. We decide on a test with the errors in mind: an error of 0.03 in $S_{wt}$ changes the volume of free water by 0.007. The data in this example are assumed for illustration.

Permeability from NMR: the Coates and SDR models

Permeability (how easily fluid flows through the rock) grows with porosity and with the size of the pore channels. NMR gives both factors, though indirectly: porosity from the amplitude of the signal, and pore size from the times $T_2$ . Two widely used empirical models are built on this. Both continue the older correlations that relate permeability to porosity and residual water saturation (Timur correlation).

Coates model (free-fluid model)

k = \left[\left(\frac{\phi_e}{C}\right)^{2}\frac{\mathrm{FFI}}{\mathrm{BVI}}\right]^{2}

where:

$k$ permeability, mD
$\phi_e$ effective NMR porosity, % (porosity units)
$\mathrm{FFI},\ \mathrm{BVI}$ volumes of free fluid and capillary-bound water, any consistent units
$C$ calibration constant, %

Units: empirical: $k$ in mD, $\phi_e$ in per cent; with another way of writing porosity, $C$ changes

Holds when

This is an empirical relationship proposed for sandstones. We choose $C$ (one often starts from a value of 10) and the exponents from the core of the formation.
The cutoff separating BVI and FFI is calibrated for this rock.
BVI is water only, and FFI is all the mobile fluids, with corrections for the hydrogen index and incomplete polarisation.

The more free fluid there is compared with bound water, the larger the pores and channels, and the higher the permeability. Hydrocarbons occupy the large pores and fall into FFI. So with corrected porosity the model is not very sensitive to the type of fluid, and it is preferred in productive intervals.

SDR model (mean-

T_2

model)

k = a\,\phi_e^{4}\,T_{2LM}^{2}

where:

$k$ permeability, mD
$\phi_e$ effective NMR porosity, fraction
$T_{2LM}$ logarithmic mean $T_2$ (here over the part of the distribution above 3 ms), ms
$a$ calibration coefficient, mD/ms²

Units: empirical: $k$ in mD, $\phi_e$ as a fraction, $T_{2LM}$ in ms

Holds when

This is an empirical relationship. We choose $a$ and the exponents from the core of the formation. Without core, one often starts from $a$ of about 4 mD/ms² for sandstones. In carbonates the coefficients are different ( $a$ is much smaller).
The pores are filled with water, and the distribution reflects pore sizes, not fluid properties.
$T_{2LM}$ is calculated over the same part of the distribution as in the calibration.

The name comes from the research centre where the model was proposed. It does not need a cutoff, and it works well in water-bearing intervals. But oil with a long $T_2$ raises $T_{2LM}$ , which makes the permeability too high. Gas lowers the NMR porosity and usually makes the permeability too low.

Worked examplePermeability by two models

Given (assumed data)

Formation: the same shaly sandstone: $\phi_e = 0.199$ (19.9%), $\mathrm{BVI} = 0.083$ , $\mathrm{FFI} = 0.116$
Logarithmic mean $T_2$ (above 3 ms): 71 ms
Constants: $C = 10$ for the Coates model; $a = 4.5$ mD/ms² for the SDR model, chosen from the core of this formation
Core: permeability of samples from this interval is about 25 mD

Solution

Coates model: $(19.9/10)^2 = 3.960$ , $\mathrm{FFI}/\mathrm{BVI} = 0.116/0.083 = 1.398$ , the product is 5.536, and $k = 5.536^2 = 30.6$ mD.
SDR model: $\phi_e^4 = 0.199^4 = 0.001568$ , and $k = 4.5\cdot 0.001568\cdot 71^2 = 35.6$ mD.
The influence of oil. The distribution has a peak of residual light oil at about 700 ms. If filtrate filled the same large pores, $T_{2LM}$ would be 57 ms and $k = 4.5\cdot 0.001568\cdot 57^2 = 22.9$ mD. That is $(71/57)^2 = 1.55$ times less.

Answer: The Coates model gives about 30.6 mD and the SDR model 35.6 mD. Without the influence of oil, the SDR model would give 22.9 mD, which is closest to the core (about 25 mD).

What it means. The models disagree with core by a factor of 1.2–1.4. This is normal for estimates from logs. After calibration, permeability from NMR is reliable to within about a factor of two or three. Without calibration, the error reaches an order of magnitude (a factor of ten). In intervals with light oil, the SDR model consistently makes the permeability too high. There we rely on the Coates model, or we leave the oil peak out of the calculation of $T_{2LM}$ . The data in this example are assumed for illustration.

Oil and gas in the NMR signal

Hydrocarbons change the signal in three ways: through the hydrogen index, through the rate of polarisation and through their own relaxation times. The tool sees the flushed zone. So what it sees is residual hydrocarbons and filtrate, and with an oil-based mud also the filtrate of that mud (Oil-based muds).

How fluids look on NMR (approximate)

Fluid: Water in pores
Hydrogen index: about 1
$T_1$: tens of ms to seconds
$T_2$ and consequences: by pore size; the basis of interpretation

Fluid: Light oil
Hydrogen index: about 1
$T_1$: hundreds of ms to seconds
$T_2$ and consequences: long, bulk; shifts $T_{2LM}$ upward

Fluid: Heavy oil
Hydrogen index: about 1
$T_1$: short
$T_2$ and consequences: short: falls into bound water or is lost

Fluid: Gas
Hydrogen index: 0.2–0.5 at depth
$T_1$: seconds
$T_2$ and consequences: shortened by diffusion; porosity comes out too low

Fluid: Filtrate of an oil-based mud
Hydrogen index: about 1
$T_1$: hundreds of ms to seconds
$T_2$ and consequences: long, similar to light oil

Fluid	Hydrogen index	$T_1$	$T_2$ and consequences
Water in pores	about 1	tens of ms to seconds	by pore size; the basis of interpretation
Light oil	about 1	hundreds of ms to seconds	long, bulk; shifts $T_{2LM}$ upward
Heavy oil	about 1	short	short: falls into bound water or is lost
Gas	0.2–0.5 at depth	seconds	shortened by diffusion; porosity comes out too low
Filtrate of an oil-based mud	about 1	hundreds of ms to seconds	long, similar to light oil

This is for a water-wet rock. In an oil-wet rock, oil touches the walls and relaxes faster, and water relaxes more slowly (Wettability and interfacial phenomena).

NMR porosity with hydrocarbons in the flushed zone

\phi_\text{NMR} = \phi\left[S_{xo} + (1 - S_{xo})\,\mathrm{HI}_h\left(1 - e^{-\mathrm{TW}/T_{1h}}\right)\right]

where:

$\phi_\text{NMR}$ NMR porosity, fraction
$\phi$ true porosity, fraction
$S_{xo}$ water saturation of the flushed zone (the water is filtrate), fraction
$\mathrm{HI}_h$ hydrogen index of the hydrocarbon, —
$T_{1h}$ longitudinal relaxation time of the hydrocarbon, s
$\mathrm{TW}$ polarisation time, s

Units: porosity and saturation as fractions; $\mathrm{TW}$ and $T_{1h}$ in the same units

Holds when

Water-based filtrate with a hydrogen index of 1, fully polarised.
The hydrocarbon signal is not lost because $T_2$ is too short (for heavy oil this does not hold).
The hydrogen index of water is taken as 1; for salt water it is slightly lower.

The factor that multiplies $(1 - S_{xo})$ is the share of the hydrocarbon that the tool sees. For gas it is small, both because of the hydrogen index and because of incomplete polarisation. So in gas-bearing formations NMR gives a porosity that is too low. This is the mirror image of the density log, which gives one that is too high.

The two methods shift in opposite directions, and we use this in the same way as the separation of the density and neutron porosity (gas correction). Two curves that gas shifts differently give both the porosity and the volume of gas. Let the volume of gas in the rock be $V_g = \phi S_g$ (in the flushed zone, $S_g = 1 - S_{xo}$ ). Then $\phi_D = \phi + k_D V_g$ and $\phi_\text{NMR} = \phi - k_M V_g$ .

Porosity and gas volume from the density log and NMR

V_g = \frac{\phi_D - \phi_\text{NMR}}{k_D + k_M},\qquad \phi = \phi_D - k_D V_g,\qquad k_M = 1 - \mathrm{HI}_g\left(1 - e^{-\mathrm{TW}/T_{1g}}\right)

where:

$V_g$ volume of gas in the flushed zone per unit volume of rock, fraction
$\phi_D$ density porosity calculated with the density of the filtrate, fraction
$k_D$ sensitivity of density porosity to gas: $(\rho_{mf} - \rho_{a,g})/(\rho_{ma} - \rho_{mf})$ , —
$k_M$ sensitivity of NMR porosity to gas, —
$\mathrm{HI}_g,\ T_{1g}$ hydrogen index and longitudinal relaxation time of the gas, —, s

Units: fraction; $\mathrm{TW}$ and $T_{1g}$ in the same units

Holds when

A clean formation with a known matrix density; both tools see the same flushed zone.
Water is fully polarised, and the hydrogen index of the filtrate is 1.
The properties of the gas ( $\rho_g$ , $\mathrm{HI}_g$ , $T_{1g}$ ) are estimated for reservoir pressure and temperature.

The symbols $k_D$ and $\rho_{a,g}$ are the same as in the topic Gas, shale, carbonate and pressure effects on porosity logs. The solution is stable: gas shifts the two curves in opposite directions, so errors in the gas properties have little effect on the porosity.

Worked exampleGas-bearing sandstone: NMR together with the density log

Given (assumed data)

Formation: clean gas-bearing sandstone at a depth of about 2000 m, the same as in the example of a gas-bearing formation in the topic Resistivity–porosity crossplots; the neutron porosity there is 0.125
Readings: density porosity $\phi_D = 0.249$ (matrix 2.65 g/cm³, filtrate 1.00 g/cm³); NMR porosity $\phi_\text{NMR} = 0.138$
Gas: methane 0.13 g/cm³, as in the topic on gas corrections to porosity: hydrogen index 0.29, $k_D = 0.615$ ; $T_1 \approx 4$ s
Recording mode: polarisation time $\mathrm{TW} = 6$ s

Solution

Fraction of the gas polarised in 6 s: $1 - e^{-6/4} = 0.777$ .
Sensitivity of NMR porosity to gas: $k_M = 1 - 0.29\cdot 0.777 = 0.775$ ; the sum is $k_D + k_M = 1.390$ .
Volume of gas: $(0.249 - 0.138)/1.390 = 0.0799$ .
Porosity: $\phi = 0.249 - 0.615\cdot 0.0799 = 0.200$ ; gas occupies $0.0799/0.200 = 0.40$ of the pores of the flushed zone.
Stability check: with $T_1 = 3$ s instead of 4 s the porosity would change by only 0.001, and the volume of gas by 2%.

Answer: The porosity of the formation is about 0.200, and gas fills about 40% of the pores of the flushed zone. Taken separately, NMR would give a porosity that is 31.0% too low, and the density log would give one that is 24.5% too high.

What it means. Joint processing of the density and neutron logs gives the same porosity for this formation (gas correction). The sensitivity of NMR to gas is of the same order as that of the neutron method, but NMR is free from the influence of shale and mineral composition. The main assumptions are the hydrogen index and $T_1$ of the gas at reservoir conditions. As the check showed, errors in them hardly affect the result. The data in this example are assumed for illustration.

We find the type of fluid from differences that do not depend on pore size:

Two polarisation times (dual TW): fluids with a long $T_1$ , which are gas and light oil, are seen only partly at a short $\mathrm{TW}$ , while water is seen completely. The difference of the two distributions (the differential spectrum) leaves mainly the slowly polarising hydrocarbons.
Two echo spacings (dual TE): the diffusion term grows as $\mathrm{TE}^2$ and is proportional to the diffusion coefficient. Gas diffuses tens of times faster than water. At a larger $\mathrm{TE}$ its peak shifts noticeably towards short times, while viscous oil (thick oil) hardly shifts (the calculation is in the advanced section).
Two-dimensional maps (D– $T_2$ , $T_1$ – $T_2$ ): processing spreads the signal over two axes at once, and water, oil and gas occupy different areas of the map.
Oil viscosity: the more viscous the oil, the shorter its bulk relaxation time. We estimate viscosity from the position of the oil peak, using empirical relationships calibrated on samples.

Recording mode: polarisation time, echo spacing, speed

We choose the recording parameters in advance, to suit the task. They decide what the tool will be able to see.

Polarisation time $\mathrm{TW}$ must be at least about three times $T_1$ of the slowest fluid that we want to see almost completely (to 95%). For water in pores, a fraction of a second to a second is enough. Light oil needs seconds. Full polarisation of gas takes about ten seconds or more, so we usually pick out gas by comparing recordings made with different $\mathrm{TW}$ .
Echo spacing $\mathrm{TE}$ : the shorter it is, the better we see the short times. Clay-bound water, with $T_2$ of about a millisecond, is seen only at a $\mathrm{TE}$ of a fraction of a millisecond, 0.2–0.6 ms. We may choose a long $\mathrm{TE}$ on purpose, to get diffusion contrast.
Train length is the number of echoes multiplied by $\mathrm{TE}$ . It must be comparable to the longest times $T_2$ (hundreds of milliseconds to seconds). Otherwise the long components are poorly determined.
Stacking: the amplitude of one cycle is small compared with noise. To get an error of about one porosity unit, we repeat the cycles and average them. This is what makes NMR slow.

The rock must have time to polarise while it moves under the magnet towards the antenna. The tool must also not move far during the stacking time. This gives two limits on the logging speed.

Limits on the NMR logging speed

v \le \frac{L_m}{\mathrm{TW}},\qquad v \le \frac{\Delta z - L_a}{N\,(\mathrm{TW} + N_E\,\mathrm{TE})}

where:

$v$ logging speed, m/s (for m/h, multiply by 3600)
$L_m$ length of the magnet ahead of the antenna, the section of pre-polarisation, m
$\Delta z$ permissible averaging over depth (the desired vertical resolution), m
$L_a$ antenna length, m
$N$ number of cycles averaged into one point, —
$N_E$ number of echoes in a cycle, —
$\mathrm{TW},\ \mathrm{TE}$ polarisation time and echo spacing, s

Units: SI

Holds when

A tool with one working frequency. A multi-frequency tool, which measures several shells in turn, gathers data faster by about as many times as it has frequencies.
The cycle time is polarisation plus the echo train. Service pauses are ignored.
The distance the tool moves during stacking is simply added to the antenna length. This is a rough estimate of the vertical resolution.

The first condition is physical: otherwise the rock reaches the antenna under-polarised (not fully polarised). The second is a compromise between accuracy (the number of cycles $N$ ) and vertical resolution. It is usually the second one that limits the speed.

Worked examplePolarisation time and logging speed

Given (assumed data)

Slowest fluid: residual light oil of the formation of the examples (its volume 0.035, peak at about 700 ms): $T_1 \approx 0.8$ s
Tool (assumed): the magnet extends ahead of the antenna by $L_m = 0.9$ m, antenna length 0.4 m; one working frequency
Cycle: $\mathrm{TE} = 0.4$ ms, 1500 echoes; we average 8 cycles for the required accuracy
Required vertical resolution: about 1.0 m

Solution

Polarisation time for 95%: $\mathrm{TW} = 3T_1 = 2.4$ s.
Polarisation limit: $v \le 0.9/2.4 = 0.375$ m/s, or 1350 m/h.
Cycle: $2.4 + 0.4\cdot10^{-3}\cdot 1500 = 3.0$ s; eight cycles take 24 s per point.
Resolution limit: in 24 s the tool may move by no more than $1.0 - 0.4 = 0.6$ m, that is, $v \le 0.6/24 = 0.025$ m/s, or 90 m/h. A tool with four frequencies can go at about 360 m/h.
Suppose we cut the polarisation time to 1.0 s. The oil then polarises to $1 - e^{-1/0.8} = 0.713$ . Of its volume of 0.035, the tool sees 0.025, so the porosity is 1.0 porosity units too low. On the other hand, the cycle shortens to 1.6 s, the point takes 12.8 s, and the permissible speed rises to 169 m/h.

Answer: Polarisation does not limit the logging speed (it allows up to 1350 m/h). The stacking of the signal does: about 90 m/h for a one-frequency tool and about 360 m/h for a four-frequency tool.

What it means. That is why we record NMR several times more slowly than the ordinary suite of logs, and often only in the productive part of the section. A short polarisation time speeds up recording, but it makes the porosity too low in oil-bearing and especially gas-bearing intervals. The error is always in one direction. We choose the mode to suit the expected fluids. To look for gas, we order a recording with two polarisation times. The data in this example are assumed for illustration.

Depth of investigation. The sensitive zone of NMR lies a few centimetres from the borehole wall, about 3–10 cm (Overview of logging tools). This is in the flushed zone. Several things follow. The free fluid seen by NMR is filtrate and residual hydrocarbons, not reservoir fluids. Pad tools lose the signal in washouts and on a rough wall. For centralised tools, the sensitive shell lies at a set diameter, and in a strongly enlarged hole it may end up in the mud. Salty, highly conductive mud weakens the radio-frequency signal of centralised tools and raises the noise. Metallic particles and magnetic minerals distort the field.

Vertical resolution depends on the antenna length (from 15 cm for pad tools to several tens of centimetres for centralised ones) and on how far the tool moves during stacking. It is usually 0.3–1 m. NMR resolves thin beds worse than microresistivity tools.

Order of NMR interpretation

1
Quality control
Compare the total NMR porosity with the density porosity. In water-bearing intervals and with light oil, they should agree to within 1–2 porosity units. An NMR shortfall is a sign of gas, heavy oil, incomplete polarisation or a washout.
2
Bound water
Divide the distribution with cutoffs calibrated on core. Compare clay-bound water with shale volume (Determining shale volume).
3
Mobility of water
Compare the volume of water from resistivity with the volume of bound water from NMR. The excess is free water, and it means a risk of water production.
4
Permeability
Calculate it with the Coates and SDR models, using coefficients from core. In oil-bearing intervals, check how hydrocarbons affect $T_{2LM}$ .
5
Fluids
If gas or light oil is suspected, use recordings with different $\mathrm{TW}$ and $\mathrm{TE}$ and joint processing with the density log.

Terms and symbols

NMR: A logging method that records the signal of the hydrogen nuclei of fluids in the pores. A constant field first polarises the nuclei, and then radio-frequency pulses excite them.
$T_1$ , longitudinal relaxation time: The time constant of polarisation. It sets the polarisation time $\mathrm{TW}$ that we need.
$T_2$ , transverse relaxation time: The time constant of the decay of the echo train. In the wetting phase it is proportional to the pore size.
CPMG: A sequence of 90° and 180° pulses that gives a train of spin echoes at the echo spacing $\mathrm{TE}$ .
Surface relaxivity $\rho_2$: The coefficient of proportionality between the surface relaxation rate and the ratio of pore surface to volume, in µm/s.
$T_2$ cutoff: The boundary between capillary-bound water and free fluid. It is typically about 33 ms in sandstones and 90–100 ms in carbonates, and we refine it on core.
CBW, BVI, FFI: The volumes of clay-bound water, capillary-bound water and free fluid (the free fluid index).
$T_{2LM}$: The logarithmic mean $T_2$ , a typical pore size in the SDR model.

Common mistakes and misconceptions

Wrong

“NMR porosity is lower than density porosity, so the tool made a mistake.”

In fact

An NMR shortfall often carries information. Gas causes it (low hydrogen index and slow polarisation). So do heavy oil (times that are too short) and an insufficient polarisation time. Check the recording mode and the fluids before you reject the curve.

Wrong

“A cutoff of 33 ms suits any formation.”

In fact

It is a typical value for sandstones. In carbonates it is usually about 90–100 ms. In any particular rock the boundary depends on the surface relaxivity and the pore structure, so we calibrate it on core.

Wrong

“Free fluid on NMR is the oil that the formation will deliver.”

In fact

The tool sees the flushed zone, and the free fluid there is mostly filtrate and residual hydrocarbons. To judge whether formation water can move, compare the bound water from NMR with the water from resistivity.

Wrong

“Permeability from NMR has been measured.”

In fact

It is calculated with the empirical Coates or SDR models, whose coefficients are chosen from core. Without calibration the error can reach an order of magnitude (a factor of ten), and hydrocarbons and vugs shift the result.

Wrong

“A long polarisation time is a waste of time.”

In fact

If $\mathrm{TW}$ is less than about $3T_1$ of the slow fluids, we see too little of their signal, and the porosity in oil-bearing and gas-bearing intervals always comes out too low.

In depth: advanced detail

How the

T_2

distribution is obtained from the echo train

We must fit non-negative amplitudes $a_i$ on a grid of several tens of times $T_{2,i}$ , so that a sum of exponentials describes the thousands of measured echoes $A_k$ . Exponentials with close times are almost impossible to tell apart. With noise, the problem has many solutions that are almost equally good. A simple least-squares fit (the fit that makes the sum of the squared errors as small as possible) would give a “picket fence” of narrow peaks. So we add a penalty for a “non-smooth” solution to the misfit. This is called regularisation.

Inversion of the echo train with regularisation

\min_{a_i \ge 0}\ \sum_k\Bigl(A_k - \sum_i a_i\,e^{-t_k/T_{2,i}}\Bigr)^2 + \alpha\sum_i a_i^2

where:

$A_k$ amplitude of the $k$ -th echo, fraction of bulk volume
$t_k$ time of the $k$ -th echo, $k\cdot\mathrm{TE}$ , ms
$a_i$ porosity at the $i$ -th time of the grid $T_{2,i}$ , fraction of bulk volume
$\alpha$ regularisation parameter, —

Units: amplitudes in porosity units

Holds when

The noise is random and the same for all echoes.
The time grid is set in advance and covers the whole possible range, from fractions of a millisecond to several seconds.
A penalty in the form of a sum of squares is one common option. Penalties on the curvature of the distribution are also used.

A large $\alpha$ gives a smooth distribution with blurred peaks. A small $\alpha$ gives a solution that is unstable against noise. We choose $\alpha$ from the noise level. This matters for interpretation. The width of the peaks is largely a property of the processing. Peaks that are less than a factor of 2–3 apart in time are usually not resolved. The shares in wide intervals (bound water, free fluid) are more reliable than the shape of the curve.

The cutoff from core

We saturate a core sample with a model of formation water (a brine of similar composition) and record its $T_2$ distribution on a laboratory NMR instrument. Then we centrifuge the sample, or we push the water out with gas through a semipermeable membrane, until the sample reaches irreducible water saturation (Capillary pressure and the transition zone). The cutoff is the time at which the cumulative porosity of the saturated sample, counted from the short times, equals the volume of the remaining water $S_{wirr}\,\phi$ . A repeat NMR measurement after centrifuging shows which pores have emptied, and it serves as a check.

Worked exampleThe cutoff for the formation of the examples

Given (assumed data)

Sample: from the same shaly sandstone, saturated with water: NMR porosity 0.230
Irreducible water saturation: $S_{wirr} = 0.50$ after centrifuging at 0.69 MPa (air–water)
Cumulative porosity of the saturated sample: up to 20 ms: 0.1088; up to 33 ms: 0.1140; up to 40 ms: 0.1145; up to 50 ms: 0.1152; up to 60 ms: 0.1164; up to 70 ms: 0.1182

Solution

Volume of irreducible water: $0.50\cdot 0.230 = 0.1150$ .
It lies between the values at 40 and 50 ms. It exceeds the first by $0.1150 - 0.1145 = 0.0005$ , and the whole interval is $0.1152 - 0.1145 = 0.0007$ . So it lies at the fraction 0.714 of the interval. Interpolating in the logarithm of time gives $T_{2\,\text{cutoff}} = 40\cdot(50/40)^{0.714} = 47$ ms.
Sensitivity: at $S_{wirr} = 0.49$ , the volume 0.1127 lies between 20 and 33 ms, and the cutoff is about 29 ms. At 0.51 (0.1173) it lies between 60 and 70 ms, and the cutoff is about 65 ms.
Bound water at cutoffs of 33 and 47 ms: 0.1140 and 0.1150, a difference of only 0.0010.

Answer: The cutoff for this rock is about 47 ms. The typical value of 33 ms would give almost the same volume of bound water.

What it means. In a distribution with two distinct peaks and a trough between them, the cutoff is poorly determined. With $S_{wirr}$ only 0.01 lower or higher, it is 29 or 65 ms. But it also does not matter, because there is little porosity in the trough. In rocks with one wide peak (fine-grained and shaly sandstones, many carbonates) it is the other way round. The boundary is found with confidence, but every millisecond changes the volume of bound water noticeably, and calibration on core is a must. Laboratory measurements are made at room temperature, while in the formation the bulk relaxation and diffusion are different. So we carry the values over to logging with caution. The data in this example are assumed for illustration.

Gas and water at two echo spacings

Worked exampleHow the gas peak shifts when

\mathrm{TE}

is doubled

Given (assumed data)

Water: in pores of radius 5 µm from the first example: surface term 4.8 s⁻¹, bulk term 0.40 s⁻¹, diffusion term 0.43 s⁻¹ at $\mathrm{TE} = 0.6$ ms
Gas: methane at reservoir conditions: $D_g \approx 1\cdot10^{-7}$ m²/s, bulk relaxation time about 4 s; does not wet the walls
Recordings: $\mathrm{TE} = 0.6$ and 1.2 ms at the same gradient of 0.2 T/m

Solution

The diffusion term is proportional to $D\,\mathrm{TE}^2$ . For gas at 0.6 ms it is $10^{-7}/(5\cdot10^{-9}) = 20$ times larger than for water: $20\cdot 0.43 = 8.6$ s⁻¹. At 1.2 ms it is four times larger still: 34.4 s⁻¹.
Gas: $T_2 = 1/(0.25 + 8.6) = 113$ ms at 0.6 ms and $1/(0.25 + 34.4) = 29$ ms at 1.2 ms.
Water: the diffusion term at 1.2 ms is $4\cdot 0.43 = 1.72$ s⁻¹. The sum of the rates $4.8 + 0.40 + 0.43 = 5.63$ s⁻¹ gives $T_2 = 178$ ms, and the sum $4.8 + 0.40 + 1.72 = 6.92$ s⁻¹ gives $T_2 = 145$ ms.

Answer: When the echo spacing is doubled, the gas peak shifts from 113 to 29 ms (by a factor of 3.9), and the water peak only from 178 to 145 ms.

What it means. The difference in the shifts gives the gas away. When we compare the two recordings, we find the component that “runs away” towards short times. Viscous oil diffuses more slowly than water, so it shifts even less. The estimate is rough: in pores the walls limit the diffusion of gas, so the real shift is smaller than the calculated one. The data in this example are assumed for illustration.

Connections to other topics

Overview of logging tools — the place of NMR among other methods, its depth of investigation and resolution.
Neutron log — another method that responds to hydrogen, and why it sees the hydrogen of shales.
Gas, shale, carbonate and pressure effects on porosity logs — gas and hydrocarbons in the readings of the density and neutron logs.
Water saturation in shaly sands — total and effective porosity and clay-bound water.
Capillary pressure and the transition zone — why water stays in small pores and how pore sizes are related to pressure.
Rock permeability — what permeability is and how it is related to porosity.
Core analysis — the laboratory measurements on which NMR is calibrated.
Microresistivity tools and flushed-zone resistivity — saturation of the same flushed zone from resistivity.
Wettability and interfacial phenomena — wettability decides which fluid relaxes on the pore walls.

Check yourself

1
The field in the sensitive zone of an NMR tool is 0.047 T. At what frequency does it work?
1. A.About 0.2 MHz
2. B.About 2 MHz
3. C.About 2 kHz
4. D.About 42 MHz
Show the answer
Correct answer: B. About 2 MHz
$f = 42.58\cdot 0.047 \approx 2.0$ MHz. A frequency of about 2 kHz corresponds to the Earth's field (about 50 µT), and 42 MHz to a field of 1 T.
2
In clean sandstone the NMR porosity is 6 units lower than the density porosity, and the neutron porosity is also noticeably lower than the density porosity. What is the most likely cause?
1. A.Heavy oil
2. B.Gas in the flushed zone
3. C.Shale
4. D.An error in the NMR calibration
Show the answer
Correct answer: B. Gas in the flushed zone
Gas makes the NMR porosity too low (low hydrogen index, slow polarisation), and so it does the neutron porosity. It makes the density porosity too high. Shale would raise the neutron porosity above the density porosity. Heavy oil would lower the NMR porosity but would not pull the neutron and density curves apart.
3
The $T_2$ distribution of a water-bearing sandstone: 2.0% of the porosity is shorter than 3 ms, 6.0% is from 3 to 33 ms, 15.0% is longer than 33 ms. Find the total and effective porosity, the irreducible water saturation and the permeability by the Coates model at $C = 10$ .
Show the answer
Total porosity is 23.0% and effective porosity is 21.0%. The irreducible water saturation is $(2.0 + 6.0)/23.0 \approx 0.35$ of the total porosity. Permeability is $k = [(21/10)^2\cdot 15/6]^2$ , that is, $(4.41\cdot 2.5)^2 \approx 122$ mD. This is only an estimate, because we refine the constant $C$ on core.
4
In an oil-bearing interval the permeability by the SDR model came out twice as high as by the Coates model and by core. The most likely cause is:
1. A.Clay-bound water
2. B.High salinity of the formation water
3. C.Light oil with a long $T_2$ increased $T_{2LM}$
4. D.Too long a polarisation time
Show the answer
Correct answer: C. Light oil with a long $T_2$ increased $T_{2LM}$
Oil in a water-wet rock relaxes by bulk relaxation, with a long $T_2$ . It shifts the logarithmic mean time upward, and in the SDR model this time enters squared. The Coates model uses the volumes of bound water and free fluid, so it is less sensitive to the type of fluid in the large pores.
5
Why does comparing recordings with polarisation times of 1 and 8 s pick out gas, but give almost nothing for water? Take $T_1$ of water as 0.3 s and of gas as 4 s.
Show the answer
In 1 s water polarises to $1 - e^{-1/0.3} \approx 0.96$ , and in 8 s completely. The difference is about 4% of its signal. Gas polarises to $1 - e^{-1/4} \approx 0.22$ in 1 s and to $1 - e^{-2} \approx 0.86$ in 8 s. The difference is about 64% of its signal. So the difference of the two distributions consists almost entirely of gas.
6
Why is NMR porosity almost independent of the mineral composition of the matrix, while neutron porosity depends on it?
Show the answer
NMR receives a signal only from the hydrogen in the liquids and gases of the pores. Hydrogen in minerals (the hydroxyls of clays, water of crystallisation) loses its signal within microseconds. The neutron method responds to all the hydrogen of the rock, and to the way minerals scatter and absorb neutrons. So its readings depend on lithology and shale volume.

Updated: 19 October 2026

Quick refresher

After this lesson you can

Protons in a magnetic field: what the tool measures

Why the signal decays: surface, bulk and diffusion

The T2T_2T2​ distribution

Total and effective porosity, bound and free water

Permeability from NMR: the Coates and SDR models

Oil and gas in the NMR signal

Recording mode: polarisation time, echo spacing, speed

Terms and symbols

Common mistakes and misconceptions

In depth: advanced detail

Connections to other topics

Check yourself

The $T_2$ distribution