Water saturation in shaly sands

Quick refresher

The layers of clay minerals carry a negative charge, which exchangeable cations (positive ions that can be swapped for others) balance. In water the cations can move, and they carry current in parallel with the formation water. The measure of this ability is the cation exchange capacity. For smectite it is about ten times larger than for kaolinite.
The extra conductivity from shale (its extra ability to carry current) falls with water saturation more slowly than the conductivity of the water itself. So shale distorts the calculation most where there is little water and it is fresh. Archie's equation gives too high a water saturation first of all in pay zones.
Laminated shale reduces the share of reservoir rock in the bed but does not spoil the sandy layers. Dispersed shale fills the pores and sharply lowers permeability. Structural shale replaces part of the grains. On a plot of total porosity against shale volume, these forms lie on different lines.
Total porosity includes the clay-bound water, and effective porosity does not: $\phi_t = \phi_e + V_{sh}\phi_{t,sh}$ . The volume of hydrocarbons is the same in both systems, so we must not multiply the porosity of one system by the saturation of the other.
The models add up an Archie term and a shale term. The Simandoux and Indonesia models describe the shale through shale volume and the resistivity of the neighbouring shales. The Waxman–Smits model uses the cation exchange capacity. The dual-water model uses the volume and conductivity of the bound water. Without shale, all the models reduce to Archie's equation.
On the same data the models disagree: in the lesson's example, from 0.30 to 0.45 in water saturation. The water-bearing part of the bed must give $S_w \approx 1$ . This check is needed, but it is not enough. The final choice of model rests on core and well tests.

Key relationships

\phi_t = \phi_e + V_{sh}\,\phi_{t,sh}, \qquad S_{we} = \dfrac{S_{wt} - S_{wb}}{1 - S_{wb}}

total and effective porosity and saturation

\dfrac{1}{R_t} = \dfrac{\phi_e^m S_w^n}{a R_w} + \dfrac{V_{sh}\,S_w}{R_{sh}}

the Simandoux equation in its common form

\dfrac{1}{\sqrt{R_t}} = \left(\dfrac{V_{sh}^{\,1 - V_{sh}/2}}{\sqrt{R_{sh}}} + \dfrac{\phi_e^{m/2}}{\sqrt{a R_w}}\right) S_w^{n/2}

the Indonesia equation

C_t = \dfrac{S_{wt}^{\,n^*}}{F^*}\left(C_w + \dfrac{B\,Q_v}{S_{wt}}\right)

the Waxman–Smits model; $C = 1/R$ , $F^* = a/\phi_t^{m^*}$

C_t = \dfrac{\phi_t^m S_{wt}^n}{a}\left[C_w + \dfrac{S_{wb}}{S_{wt}}\left(C_{wb} - C_w\right)\right]

the dual-water model

Common pitfalls

Taking the reading of any shale bed of the section as the shale resistivity $R_{sh}$ , instead of the neighbouring shales that resemble the shale material in the bed.
Putting total porosity into models written for effective porosity (Simandoux, Indonesia), and the other way round.
Trusting the name of a model in the software: different formulas appear under the same name.
Cutting $S_w > 1$ off at one in water-bearing shaly beds instead of revising the model parameters.
Treating a model that gave $S_w = 1$ in the water-bearing part as checked for the oil-bearing part as well.

After this lesson you can

explain why shale conducts current, and why Archie's equation therefore gives too high a water saturation, most of all in pay zones and with fresh formation water
tell laminated, dispersed and structural shale apart by their effect on porosity, permeability and resistivity, and recognise them on a plot of total porosity against shale volume
convert between total porosity (all the pore space, including the space held by clay-bound water) and effective porosity (the pore space that oil and gas can use), and between the matching saturations, without counting the clay-bound water twice
write down the laminated-sand model, the Simandoux equation and its modified form, the Indonesia equation, and the Waxman–Smits and dual-water models, and say what data each one needs and where it applies
calculate the water saturation of one bed with several models, check them in the water-bearing part of the bed, and explain why this check is not enough

Why shale conducts current

Archie's equation (Archie's equation and saturation of clean reservoirs) rests on a simple picture. The grains of the rock are insulators (they do not carry current). Only the ions (electrically charged particles) of the formation water in the pores carry the current. For clean sandstones and carbonates this is a good approximation. Shale material breaks this picture, and the reason lies in the structure of the clay minerals.

Clay minerals are layered aluminosilicates (minerals built from aluminium, silicon and oxygen). In their crystal lattice (the regular pattern of atoms in the crystal), some of the cations are replaced by cations of a smaller charge. For example, aluminium $\mathrm{Al^{3+}}$ takes the place of silicon $\mathrm{Si^{4+}}$ , and magnesium or divalent iron (iron with a charge of 2+) takes the place of aluminium. As a result, the layer as a whole carries a negative charge. Cations of sodium, calcium and magnesium balance this charge, and they are held on the surface of the particles. In water, these exchangeable cations form a cloud around the particle, the electrical double layer. They can move along the surface under an electric field (more on the charge of clays).

Two things follow. First, the rock gets a second path for the current: along the surface of the clay particles, in parallel with the formation water. Second, some of the water is held at the surface of the clay so firmly that oil and gas do not push it out. Logs see this clay-bound water as porosity, but hydrocarbons cannot use that space.

The cation exchange capacity (CEC) shows how many cations a clay can hold. It is measured in milliequivalents per 100 g of dry rock (meq/100 g; a milliequivalent is a thousandth of a mole of charge). For smectite (montmorillonite) it is 80–150 meq/100 g, for illite and chlorite 10–40, and for kaolinite 3–15. For the conductivity calculation, what matters is not the capacity per gram of rock but the concentration of cations in the pore water. This quantity is called $Q_v$ .

Cation exchange capacity per unit of pore volume

Q_v = \frac{\mathrm{CEC}\,(1 - \phi_t)\,\rho_{gr}}{100\,\phi_t}

where:

$Q_v$ amount of exchangeable cations per unit of pore volume, meq/cm³
$\mathrm{CEC}$ cation exchange capacity of the dry rock, meq/100 g
$\phi_t$ total porosity, fraction
$\rho_{gr}$ density of the solid phase (grains and clays), g/cm³

Units: $Q_v$ is in meq per cm³ of pores. The 100 converts the CEC from 100 g to 1 g of rock.

Holds when

The CEC is measured on a sample of the same rock. Chemical methods can do this (for example, the uptake of methylene blue, a dye). So can electrical measurements at several salinities (salt contents of the water), described in the advanced section.
All the exchangeable cations are in the pore water, that is, they belong to the whole pore volume $\phi_t$ .

$Q_v$ is the concentration of “extra” cations in the pore water. The same clay in a less porous rock gives a larger $Q_v$ : the same number of cations, but less water. In shaly sandstones $Q_v$ is usually a few tenths of a meq/cm³, and in shales it is a few meq/cm³.

How shale distorts an Archie calculation

Almost all shaly-sand models are built in the same way: a “shale” term is added to the “water” term of Archie's equation. With a saturation exponent $n = 2$ (the exponent of water saturation in Archie's equation), such a model reduces to a quadratic equation (an equation with a squared unknown) for water saturation.

General form of shaly-sand models

\frac{1}{R_t} = \alpha\,S_w^2 + \gamma\,S_w \quad\Rightarrow\quad S_w = \frac{-\gamma + \sqrt{\gamma^2 + 4\alpha/R_t}}{2\alpha}

where:

$R_t$ resistivity of the uninvaded part of the bed (the part that the mud filtrate has not reached), Ω·m
$S_w$ water saturation, fraction
$\alpha$ the “water” term: the conductivity of the rock at $S_w = 1$ without shale; in Archie's equation $\alpha = \phi^m/(a R_w)$ , S/m
$\gamma$ coefficient of the shale term; it depends on the amount of shale and its conductivity, S/m

Units: conductivities in siemens per metre, S/m $= 1/(\text{Ω·m})$

Holds when

The saturation exponent is $n = 2$ . For another $n$ , we solve the equation numerically (by repeated trial calculations).
$\alpha$ and $\gamma$ do not depend on water saturation.
At $\gamma = 0$ we get Archie's equation.

The water term falls as the square of water saturation, and the shale term only as the first power. The less water there is in the pores, the larger the share of the conductivity that comes from the shale, and the bigger the error of a calculation that ignores it. The particular models differ in their expressions for $\alpha$ and $\gamma$ , and in the porosity to which $S_w$ refers.

Worked exampleThe extra conductivity from shale in a pay zone

Given (assumed data)

Bed: a sandstone with effective porosity $\phi_e = 0.18$ , shale volume $V_{sh} = 0.25$ and water saturation $S_w = 0.30$
Formation water and shale: $R_w = 0.08$ Ω·m, neighbouring shales $R_{sh} = 2.5$ Ω·m, both at the formation temperature of 80 °C
Model: the simplest one: $\alpha = \phi_e^2/R_w$ , $\gamma = V_{sh}/R_{sh}$ (this is how the Simandoux equation, covered below, is built); $a = 1$ , $m = n = 2$ (the Archie coefficients: $a$ is a constant, $m$ is the cementation exponent and $n$ is the saturation exponent)

Solution

Coefficients: $\alpha = 0.18^2/0.08 = 0.405$ S/m, $\gamma = 0.25/2.5 = 0.100$ S/m.
Conductivity of the bed: $1/R_t = 0.405\cdot 0.30^2 + 0.100\cdot 0.30 = 0.03645 + 0.030 = 0.06645$ S/m, that is, $R_t = 15.0$ Ω·m. Shale gives 45% of the conductivity.
A clean sandstone with the same porosity and saturation: $R_t = 1/0.03645 = 27.4$ Ω·m.
Archie's calculation for the shaly bed: $S_w = \sqrt{0.08/(0.18^2\cdot 15.0)} = 0.41$ .
The water-bearing part of the same bed ( $S_w = 1$ ): $1/R_0 = 0.405 + 0.100 = 0.505$ S/m, $R_0 = 1.98$ Ω·m. Here shale gives 20% of the conductivity, and Archie's equation would show $S_w = 1.12$ .

Answer: A shaly bed with a water saturation of 30% shows 15.0 Ω·m instead of the 27.4 Ω·m of a clean one, and Archie's equation gives 0.41 for it.

What it means. In the water-bearing part, shale adds a fifth of the conductivity. In the pay zone it adds almost a half. Its contribution matters more the less water there is in the pores. So the error of Archie's equation is largest in oil-bearing beds. In the water-bearing part it is also visible, as an impossible water saturation above one. This is a convenient sign that the model ignores the shale. The data in this example are assumed for illustration.

CalculatedResistivity of a clean and a shaly sandstone with an effective porosity of 0.18 against water saturation (

R_w = 0.08

Ω·m, the shale term from the example, logarithmic scales). The horizontal dashed line shows how Archie's calculation “reads” the 15 Ω·m of the shaly bed: it finds 0.41 instead of 0.30.

The second factor is the salinity (salt content) of the formation water. The fresher it is, the lower the conductivity of the water, and the more noticeable the same shale is against it. In the table, we calculate the same bed and the same shale ( $\gamma = 0.10$ S/m, $S_w = 0.30$ ) for three different values of $R_w$ .

How far Archie's equation puts the water saturation too high at different water salinities (true

S_w = 0.30

)

$R_w$ , Ω·m: 0.03
Water: strong brine (very salty water)
$R_t$ of the bed, Ω·m: 7.9
Share of shale in the conductivity: 24%
$S_w$ by Archie: 0.34

$R_w$ , Ω·m: 0.08
Water: salty
$R_t$ of the bed, Ω·m: 15.0
Share of shale in the conductivity: 45%
$S_w$ by Archie: 0.41

$R_w$ , Ω·m: 0.25
Water: slightly salty
$R_t$ of the bed, Ω·m: 24.0
Share of shale in the conductivity: 72%
$S_w$ by Archie: 0.57

$R_w$ , Ω·m	Water	$R_t$ of the bed, Ω·m	Share of shale in the conductivity	$S_w$ by Archie
0.03	strong brine (very salty water)	7.9	24%	0.34
0.08	salty	15.0	45%	0.41
0.25	slightly salty	24.0	72%	0.57

Calculated with the simplest model of the example ( $\phi_e = 0.18$ , $V_{sh}/R_{sh} = 0.10$ S/m, $a = 1$ , $m = n = 2$ ). The salinity labels are approximate, and the resistivities are at formation temperature. The data are assumed for illustration.

Laminated, dispersed and structural shale

The same shale volume $V_{sh}$ (Determining shale volume) affects a bed in different ways, depending on how the shale material is distributed in it. There are three forms of shale distribution. In real rocks they often occur together.

Three forms of shale distribution

Form: Laminated
Where the shale is: layers between the layers of clean sandstone
Porosity and permeability: the sandy layers do not change; their share in the bed falls
How it conducts current: the shale layers are parallel conductors along the bedding (the layering of the rock)

Form: Dispersed
Where the shale is: in the pores: films on the grains, filling of small pores
Porosity and permeability: effective porosity, and especially permeability, fall sharply
How it conducts current: together with the pore water, inside the pore space

Form: Structural
Where the shale is: grains and fragments of shale rocks in place of some of the frame grains
Porosity and permeability: effective porosity hardly changes
How it conducts current: like shale particles surrounded by pores

Form	Where the shale is	Porosity and permeability	How it conducts current
Laminated	layers between the layers of clean sandstone	the sandy layers do not change; their share in the bed falls	the shale layers are parallel conductors along the bedding (the layering of the rock)
Dispersed	in the pores: films on the grains, filling of small pores	effective porosity, and especially permeability, fall sharply	together with the pore water, inside the pore space
Structural	grains and fragments of shale rocks in place of some of the frame grains	effective porosity hardly changes	like shale particles surrounded by pores

Dispersed clay in sandstones is often authigenic: it grew in the pores after the sand was deposited. Examples are the “booklets” of kaolinite, the fibres of illite and the films of chlorite. Its properties may differ noticeably from those of the neighbouring shale beds.

A plot of total porosity against shale volume helps to recognise the dominant form. Let the clean sandstone of the interval have porosity $\phi_{sd}$ , and let the shales have a total porosity $\phi_{t,sh}$ , filled entirely with clay-bound water. Laminated shale replaces sandstone with shale, so the point moves along a straight line towards the shale point. Dispersed shale takes the place of water in the pores, so total porosity falls faster. Structural shale takes the place of grains and brings its own porosity, so total porosity rises.

Total porosity for three forms of shale distribution (the Thomas–Stieber diagram)

\begin{aligned} &\text{laminated:} && \phi_t = \phi_{sd} - V_{sh}\,(\phi_{sd} - \phi_{t,sh}) \\ &\text{dispersed:} && \phi_t = \phi_{sd} - V_{sh}\,(1 - \phi_{t,sh}), \quad V_{sh} \le \phi_{sd} \\ &\text{structural:} && \phi_t = \phi_{sd} + V_{sh}\,\phi_{t,sh}, \quad V_{sh} \le 1 - \phi_{sd} \end{aligned}

where:

$\phi_t$ total porosity of the bed, including the clay-bound water, fraction
$\phi_{sd}$ porosity of the clean sandstone of the interval, fraction
$\phi_{t,sh}$ total porosity of the shales (all of it is bound water), fraction
$V_{sh}$ shale volume: the share of the volume taken up by shale material together with its water, fraction

Units: fractions

Holds when

The clean sandstone has the same porosity throughout the interval, and the shale material in the bed is the same as in the neighbouring shales.
Dispersed shale fills the pores between the grains without pushing the grains apart. At $V_{sh} = \phi_{sd}$ , the pores are completely filled.
Structural shale replaces grains without touching the pores between the grains.
Shale volume and total porosity are found independently, for example from the gamma ray log and the density log (Gas, shale, carbonate and pressure effects on porosity logs).

The laminated and dispersed lines, together with the line from the limit of dispersed shale ( $V_{sh} = \phi_{sd}$ ) to the shale point, form a triangle. The structural line runs above the laminated one. A bed point on the laminated line means laminated shale. A point below it means dispersed shale, and a point above it means structural shale. Positions in between are mixed forms.

CalculatedTotal porosity against shale volume for a clean sandstone with a porosity of 0.26 and shales with a total porosity of 0.133: the lines of laminated, dispersed and structural shale, and the points of beds A and B from the example.

Worked exampleThe same shale volume, different reservoirs

Given (assumed data)

Interval: clean sandstones $\phi_{sd} = 0.26$ ; neighbouring shales $\phi_{t,sh} = 0.133$ (density 2.48 g/cm³; solid part 2.70 g/cm³, pore water 1.05 g/cm³)
Bed A: $V_{sh} = 0.15$ from the gamma ray log, total porosity $\phi_t = 0.241$
Bed B: $V_{sh} = 0.15$ , $\phi_t = 0.130$

Solution

The lines at $V_{sh} = 0.15$ : laminated $0.26 - 0.15\cdot(0.26 - 0.133) = 0.241$ ; dispersed $0.26 - 0.15\cdot(1 - 0.133) = 0.130$ ; structural $0.26 + 0.15\cdot 0.133 = 0.280$ .
Bed A lies on the laminated line. Its sandy layers kept their porosity of 0.26 and take up $1 - 0.15 = 0.85$ of the volume. The effective porosity of the bed is $0.85\cdot 0.26 = 0.221$ .
Bed B lies on the dispersed line. Shale has filled the pores, the effective porosity is $0.26 - 0.15 = 0.11$ , and 58% of the space between the grains is filled with shale.

Answer: For the same shale volume of 0.15, the effective porosity of bed A is 0.221 and that of bed B is 0.11.

What it means. The shale volume from the gamma ray log is the same, but the reservoirs are different. In bed A the sandy layers are good, and we calculate its saturation with the laminated-sand model. In bed B the pores are more than half filled with shale, the permeability may be orders of magnitude lower, and the bed may turn out to be non-reservoir. The diagnosis depends on the input data. An error in shale volume shifts the point horizontally, and an error in porosity shifts it vertically. So we draw conclusions from the cloud of points of the whole interval and check them with core. The data in this example are assumed for illustration.

Total and effective porosity and saturation

Clay-bound water takes up part of the pores, but oil and gas cannot use that space. So in shaly sands we work with two systems of quantities. Total porosity $\phi_t$ includes this water, and effective porosity $\phi_e$ does not. The saturations in each system refer to its own pore volume: total water saturation $S_{wt}$ includes the bound water, and effective water saturation $S_{we}$ includes only the free water (the water that is not bound to the clay).

Total and effective porosity of a shaly sand

\phi_t = \phi_e + V_{sh}\,\phi_{t,sh}, \qquad S_{wb} = \frac{V_{sh}\,\phi_{t,sh}}{\phi_t}

where:

$\phi_t,\ \phi_e$ total and effective porosity, fraction
$V_{sh}$ shale volume, fraction
$\phi_{t,sh}$ total porosity of the neighbouring shales, for example from their density, fraction
$S_{wb}$ share of the total pore volume taken up by clay-bound water, fraction

Units: fractions

Holds when

All the porosity of the shales is filled with bound water: the shales have no effective porosity.
The shale material in the bed is the same as in the neighbouring shales.
$V_{sh}$ is the volume of shale material together with its water, as in the topic Determining shale volume.

In a shaly sandstone, the density log gives the total porosity most closely, if the density of the dry shale is close to the density of the matrix. The density and neutron logs corrected for shale volume give the effective porosity (more). The nuclear magnetic resonance log separates bound and free water by their relaxation times (how fast the signal from the water fades) (Nuclear magnetic resonance (NMR) logging).

Total and effective water saturation

S_{we} = \frac{S_{wt} - S_{wb}}{1 - S_{wb}}, \qquad \phi_t\,(1 - S_{wt}) = \phi_e\,(1 - S_{we})

where:

$S_{wt}$ total water saturation: bound and free water in the total pore volume, fraction
$S_{we}$ effective water saturation: free water in the effective pore volume, fraction
$S_{wb}$ share of clay-bound water, fraction

Units: fractions

Holds when

Hydrocarbons do not push out the clay-bound water, so $S_{wb} \le S_{wt} \le 1$ .

The second equality is the main one: the volume of hydrocarbons in a unit volume of rock is the same in both systems. The models based on total porosity (Waxman–Smits, dual water) give $S_{wt}$ , and those based on effective porosity (Simandoux, Indonesia) give $S_{we}$ . We can compare their results only when both are expressed in the same system.

Worked exampleWhy mixing the systems is dangerous

Given (assumed data)

Bed: $\phi_e = 0.20$ , $V_{sh} = 0.30$ , shales $\phi_{t,sh} = 0.133$
A model based on total porosity gave: $S_{wt} = 0.45$

Solution

Total porosity: $\phi_t = 0.20 + 0.30\cdot 0.133 = 0.2399$ . Bound water: $S_{wb} = 0.0399/0.2399 = 0.166$ .
Effective water saturation: $S_{we} = (0.45 - 0.166)/(1 - 0.166) = 0.341$ .
The volume of hydrocarbons: $\phi_t(1 - S_{wt}) = 0.2399\cdot 0.55 = 0.132$ and $\phi_e(1 - S_{we}) = 0.20\cdot 0.659 = 0.132$ . The two are equal.
Mixing the systems: $\phi_e(1 - S_{wt}) = 0.20\cdot 0.55 = 0.110$ , which is 17% less. $\phi_t(1 - S_{we}) = 0.158$ is 20% more.

Answer: The correct volume of hydrocarbons is 0.132 of the rock volume. Mixing the systems gives a value that is 17% too low or 20% too high.

What it means. An error of this kind is not visible on any curve: every number is correct by itself. We avoid it with a rule: we take porosity and saturation from the same system, and we pass a matching pair to the hydrocarbons-in-place calculation. The data in this example are assumed for illustration.

Shaly-sand models

Many models have been proposed. They differ in three ways. The first is the porosity to which the quantities refer (effective or total). The second is how the shale is described: by shale volume and the resistivity of the neighbouring shales, by cation exchange capacity, or by bound water. The third is how the shale term depends on water saturation. We start with a model in which the form of shale distribution is given explicitly.

Laminated-sand model (parallel conductors)

\frac{1}{R_t} = \frac{(1 - V_{lam})\,\phi_{sd}^m\,S_w^n}{a\,R_w} + \frac{V_{lam}}{R_{sh}}

where:

$R_t$ resistivity of the bed along the bedding, Ω·m
$V_{lam}$ share of the volume taken up by shale laminae (thin layers), fraction
$\phi_{sd}$ porosity of the sandy layers, fraction
$S_w$ water saturation of the sandy layers, fraction
$R_{sh}$ resistivity of the shale layers along the bedding, taken from the neighbouring thick shales, Ω·m

Units: conductivities in S/m, resistivities in Ω·m

Holds when

The layers are parallel to the bedding and thinner than the vertical resolution of the sonde (the measuring part of the logging tool), and the current flows along them. In a vertical borehole, an induction sonde (a resistivity tool that works with magnetic fields) measures this resistivity $R_h$ (more).
The sandy layers are clean and obey Archie's equation, and the shale layers are the same as the neighbouring shales.
In a deviated borehole, or with dipping beds, the sonde sees a mixture of resistivities along and across the layers, and the model must be extended.

The shale layers are a constant parallel conductor that does not depend on the saturation of the sandstone. The more oil there is in the sandy layers, the higher their resistivity, and the more current goes into the shale. In terms of the porosity of the whole bed, $\phi = (1 - V_{lam})\phi_{sd}$ , the same equation is written as $1/R_t = \phi^m S_w^n/[(1 - V_{lam})^{m-1} a R_w] + V_{lam}/R_{sh}$ . The volume of hydrocarbons in a unit volume of the bed is $(1 - V_{lam})\,\phi_{sd}\,(1 - S_w)$ .

Worked exampleLaminated bed A: saturation of the sandy layers

Given (assumed data)

Bed A from the previous example: $V_{lam} = 0.15$ , sandy layers $\phi_{sd} = 0.26$ , effective porosity of the bed 0.221
Resistivity along the layers (deep induction sonde): $R_t = 9.5$ Ω·m
Shale and water: $R_{sh} = 2.5$ Ω·m, $R_w = 0.08$ Ω·m; $a = 1$ , $m = n = 2$

Solution

The conductivity of the bed is $1/9.5 = 0.1053$ S/m. Of this, the shale layers give $0.15/2.5 = 0.060$ S/m.
That leaves $0.1053 - 0.060 = 0.0453$ S/m for the sandy layers. Their own resistivity is $R_{sd} = 0.85/0.0453 = 18.8$ Ω·m.
Water saturation of the sandy layers by Archie: $S_w = \sqrt{0.08/(0.26^2\cdot 18.8)} = 0.251$ .
Archie for the bed as a whole: $S_w = \sqrt{0.08/(0.221^2\cdot 9.5)} = 0.415$ .
The volume of oil in a unit volume of the bed: $0.221\cdot(1 - 0.251) = 0.166$ , against $0.221\cdot(1 - 0.415) = 0.129$ by Archie.
Across the layers, the resistivities add up: $R_v = 0.85\cdot 18.8 + 0.15\cdot 2.5 = 16.4$ Ω·m. The anisotropy is $R_v/R_t = 1.7$ .

Answer: The sandy layers are 25% saturated with water, while Archie's calculation for the bed as a whole gives 42%, and so a volume of oil that is 22% too low.

What it means. The current bypasses the high-resistivity oil-bearing layers through the shale layers, and the deep sonde “sees” the bed as worse than it is. A sign of a laminated bed is anisotropy (a different resistivity along and across the layers). A tool with coils of different orientations would show 16.4 Ω·m across the layers. The model works if we can find $V_{lam}$ and $\phi_{sd}$ reliably: from the diagram of the previous section, from borehole imagers (tools that make a picture of the borehole wall) or from core. The data in this example are assumed for illustration.

For dispersed and mixed shale, the form of distribution is not specified explicitly. The most common models of this kind in practice are the Simandoux equation and the Indonesia equation. They describe the shale with the same quantities as the laminated model: shale volume $V_{sh}$ and the resistivity of the neighbouring shales $R_{sh}$ . They work with effective porosity.

Simandoux equation

\frac{1}{R_t} = \frac{\phi_e^m\,S_w^n}{a\,R_w} + \frac{V_{sh}\,S_w}{R_{sh}}

where:

$\phi_e$ effective porosity, fraction
$S_w$ effective water saturation, fraction
$V_{sh}$ shale volume, fraction
$R_{sh}$ resistivity of the neighbouring shales, Ω·m
$R_w,\ a,\ m,\ n$ formation water resistivity and the Archie coefficients for the clean sandstone of this bed, Ω·m; —

Units: resistivities in Ω·m, conductivities in S/m

Holds when

It is an empirical equation: the shale term is proportional to shale volume and water saturation.
For $n = 2$ , it is a quadratic equation of the general form with $\alpha = \phi_e^m/(a R_w)$ and $\gamma = V_{sh}/R_{sh}$ .
We take $R_{sh}$ from the neighbouring shales. In practice people often adjust it so that water-bearing shaly beds give $S_w \approx 1$ . Then it is a parameter of the model, not the resistivity of the shales.

Without shale, we get Archie's equation. In pure shale ( $\phi_e = 0$ , $V_{sh} = 1$ , $S_w = 1$ ), $R_t = R_{sh}$ . We use the equation for moderate shale volumes and salty formation waters.

Modified Simandoux equation

\frac{1}{R_t} = \frac{\phi_e^m\,S_w^n}{a\,R_w\,(1 - V_{sh})} + \frac{V_{sh}\,S_w}{R_{sh}}

where:

$\phi_e,\ S_w,\ V_{sh},\ R_{sh}$ as in the Simandoux equation, —; Ω·m
$1 - V_{sh}$ share of the rock outside the shale material, fraction

Units: resistivities in Ω·m, conductivities in S/m

Holds when

The same as for the Simandoux equation. The factor $1/(1 - V_{sh})$ comes from the laminated-sand model at $m = 2$ : the effective porosity is concentrated in the share $1 - V_{sh}$ of the rock volume.
In the version with $m = n = 2$ it is also called the total shale equation: it does not assume a particular form of shale distribution.

For the same data, the water term is larger than in the Simandoux equation, so the water saturation comes out lower. The difference grows with shale volume.

A name does not guarantee a formula

P. Simandoux found his relationship in 1963 from laboratory measurements on artificial mixtures of sand and montmorillonite clay. In the original form, as it is usually quoted, the shale term did not depend on $S_w$ . The factor $S_w$ was added later by other authors. The literature and software contain variants. The shale term may have the factor $S_w$ or not. The water term may have the factor $1/(1 - V_{sh})$ or not. The saturation exponent may be $n$ or 2 (a square). The names “Simandoux” and “modified Simandoux” are attached to them in different ways. On the data of the example below, the variant without $S_w$ in the shale term gives a water saturation almost twice as low. Before comparing results, we check the formulas themselves.

Indonesia equation

\frac{1}{\sqrt{R_t}} = \left(\frac{V_{sh}^{\,1 - V_{sh}/2}}{\sqrt{R_{sh}}} + \frac{\phi_e^{m/2}}{\sqrt{a\,R_w}}\right) S_w^{n/2}

where:

$V_{sh},\ R_{sh}$ shale volume and the resistivity of the neighbouring shales, fraction; Ω·m
$\phi_e,\ S_w$ effective porosity and water saturation, fraction
$R_w,\ a,\ m,\ n$ as in Archie's equation, Ω·m; —

Units: resistivities in Ω·m

Holds when

It is an empirical equation fitted to sections with fresh and slightly salty formation waters and high shale volume.
The exponent $1 - V_{sh}/2$ changes from 1 at low shale volume to 0.5 in pure shale.
It can be solved directly for $S_w$ : $S_w = \left[R_t^{-1/2}/(\ldots)\right]^{2/n}$ .

What is added is not the conductivities but their square roots. After squaring, a cross term appears: the interaction of water and shale. Both terms are multiplied by $S_w^{n/2}$ . So in a given bed the ratio $R_t/R_0$ follows the same law $S_w^{-n}$ as in Archie's equation. Shale lowers the resistivity by the same number of times at any saturation.

The models of the second group, Waxman–Smits and dual water, describe the shale not by the resistivity of neighbouring beds but by the physics of the conductivity of the exchangeable cations. They work with total porosity.

Waxman–Smits model

C_t = \frac{S_{wt}^{\,n^*}}{F^*}\left(C_w + \frac{B\,Q_v}{S_{wt}}\right), \qquad F^* = \frac{a}{\phi_t^{\,m^*}}

where:

$C_t,\ C_w$ conductivity of the bed and of the formation water, $C = 1/R$ , S/m
$S_{wt}$ total water saturation, fraction
$Q_v$ cation exchange capacity per unit of pore volume, meq/cm³
$B$ equivalent conductivity of the exchangeable cations, (S/m)/(meq/cm³)
$F^*,\ m^*,\ n^*$ formation resistivity factor (the ratio of the resistivity of the fully water-saturated rock to that of the water), cementation exponent and saturation exponent, found with allowance for shale and based on total porosity, —

Units: conductivities in S/m

Holds when

The exchangeable cations carry current in parallel with the pore water. When hydrocarbons push the water out, the cations stay in the water that remains, and their concentration grows as $Q_v/S_{wt}$ .
$m^*$ and $n^*$ are measured on core with a correction for shale (advanced section). They are larger than the “apparent” Archie exponents of the same core.
$B$ grows with temperature and falls in fresh water. At 25 °C in salty water it is about 4 (S/m)/(meq/cm³), and at 80 °C about three times larger. For the formation, we take it from published correlations for the formation temperature, not from the value at laboratory temperature.
$Q_v$ is taken from the CEC of core, or from log estimates calibrated to core.

At $Q_v = 0$ we get Archie's equation for total porosity. The relative weight of the shale term, $B Q_v/(S_{wt} C_w)$ , grows as water saturation and salinity fall. The model rests on laboratory measurements on core, and it serves as the standard against which log-based models are compared.

Dual-water model

C_t = \frac{\phi_t^m\,S_{wt}^n}{a}\left[C_w + \frac{S_{wb}}{S_{wt}}\left(C_{wb} - C_w\right)\right]

where:

$C_{wb}$ conductivity of the clay-bound water, S/m
$S_{wb}$ share of the total pore volume taken up by bound water, fraction
$S_{wt},\ \phi_t$ total water saturation and total porosity, fraction
$C_w,\ a,\ m,\ n$ conductivity of the formation water and the Archie coefficients, S/m; —

Units: conductivities in S/m

Holds when

There are two kinds of water in the pores: clay-bound water (volume $S_{wb}$ , conductivity $C_{wb}$ ) and free formation water. Hydrocarbons displace only the free water, so $S_{wt} \ge S_{wb}$ .
From logs, we take $S_{wb}$ and $C_{wb}$ from the neighbouring shales: $S_{wb} = V_{sh}\phi_{t,sh}/\phi_t$ , $R_{wb} = 1/C_{wb} = R_{sh}\,\phi_{t,sh}^m/a$ .
In very fresh water the layer of bound water becomes thicker, so we increase its share. In the examples of this lesson the water is salty enough to ignore this.

The expression in square brackets is the average conductivity of a mixture of the two waters. If $C_{wb} > C_w$ , which is usual with fresh and moderately salty water, shale raises the conductivity. If $C_{wb} < C_w$ (a strong brine), the bound water conducts worse than the free water. Then a water-bearing shaly bed can have a higher resistivity than a clean bed of the same total porosity. This is what happened to the shaly bed in the example with the apparent water resistivity (more).

The dual-water model was proposed in 1977 by C. Clavier, G. R. Coates and J. L. Dumanoir as a development of the Waxman–Smits model. If the $Q_v$ of the bed is estimated from the neighbouring shales, so that $B Q_v = S_{wb}(C_{wb} - C_w)$ , then the Waxman–Smits equation with $m^* = m$ and $n^* = n$ is the same as the dual-water model. They differ when $Q_v$ and $B$ are taken from core and correlations, and in fresh waters.

Shaly-sand models

Model: Laminated sand
How the shale is described: layers with resistivity $R_{sh}$
What is needed: $V_{lam}$ , $\phi_{sd}$ , $R_t$ along the layers
When it suits: thin alternating layers of sandstone and shale

Model: Simandoux and modified Simandoux
How the shale is described: $V_{sh}/R_{sh}$ , the term grows with $S_w$
What is needed: $\phi_e$ , $V_{sh}$ , $R_{sh}$
When it suits: moderate shale volume, salty waters

Model: Indonesia
How the shale is described: square root of the shale conductivity
What is needed: $\phi_e$ , $V_{sh}$ , $R_{sh}$
When it suits: fresh waters, high shale volume

Model: Waxman–Smits
How the shale is described: $B Q_v$ : the exchangeable cations
What is needed: $\phi_t$ , $Q_v$ and $m^*$ , $n^*$ from core, $B$ from temperature
When it suits: core is available; the standard for calibration

Model: Dual water
How the shale is described: volume $S_{wb}$ and conductivity $C_{wb}$ of the bound water
What is needed: $\phi_t$ , $V_{sh}$ and the properties of the neighbouring shales
When it suits: logs without core, total porosity

Model	How the shale is described	What is needed	When it suits
Laminated sand	layers with resistivity $R_{sh}$	$V_{lam}$ , $\phi_{sd}$ , $R_t$ along the layers	thin alternating layers of sandstone and shale
Simandoux and modified Simandoux	$V_{sh}/R_{sh}$ , the term grows with $S_w$	$\phi_e$ , $V_{sh}$ , $R_{sh}$	moderate shale volume, salty waters
Indonesia	square root of the shale conductivity	$\phi_e$ , $V_{sh}$ , $R_{sh}$	fresh waters, high shale volume
Waxman–Smits	$B Q_v$ : the exchangeable cations	$\phi_t$ , $Q_v$ and $m^$ , $n^$ from core, $B$ from temperature	core is available; the standard for calibration
Dual water	volume $S_{wb}$ and conductivity $C_{wb}$ of the bound water	$\phi_t$ , $V_{sh}$ and the properties of the neighbouring shales	logs without core, total porosity

All the models reduce to Archie's equation when there is no shale. At low shale volume the differences between them are usually small. They grow with shale volume and with fresh water.

One bed, five answers

Let us compare the models on the same data. The bed is a shaly sandstone with dispersed shale. Its oil-bearing part and its water-bearing part below the oil–water contact (the level below which the pores hold only water) have the same porosity and shale volume.

Worked exampleThe oil-bearing part of the bed by five models

Given (assumed data)

Bed: $\phi_e = 0.18$ , $V_{sh} = 0.25$ (from several indicators), $R_t = 8.0$ Ω·m
Neighbouring shales: $R_{sh} = 2.5$ Ω·m, total porosity $\phi_{t,sh} = 0.133$
Formation water: $R_w = 0.08$ Ω·m at 80 °C, that is, $C_w = 12.5$ S/m
Coefficients: $a = 1$ , $m = n = 2$

Solution

Archie with the effective porosity: $S_w = \sqrt{0.08/(0.18^2\cdot 8.0)} = 0.556$ .
Simandoux: $\alpha = 0.18^2/0.08 = 0.405$ S/m, $\gamma = 0.25/2.5 = 0.100$ S/m, $1/R_t = 0.125$ S/m. By the general formula, $S_w = \dfrac{-0.100 + \sqrt{0.100^2 + 4\cdot 0.405\cdot 0.125}}{2\cdot 0.405} = 0.446$ .
Modified Simandoux equation: $\alpha = 0.405/0.75 = 0.540$ S/m with the same $\gamma$ . $S_w = 0.397$ .
Indonesia: the shale term is $0.25^{0.875}/\sqrt{2.5} = 0.188$ , and the water term is $0.18/\sqrt{0.08} = 0.636$ . $S_w = 0.3536/(0.188 + 0.636) = 0.429$ (for $n = 2$ the exponent $2/n = 1$ , and $0.3536 = 1/\sqrt{8.0}$ ).
Dual water: $\phi_t = 0.18 + 0.25\cdot 0.133 = 0.213$ , $S_{wb} = 0.03325/0.213 = 0.156$ . $R_{wb} = 2.5\cdot 0.133^2 = 0.0442$ Ω·m, $C_{wb} = 22.6$ S/m.
This is the same quadratic form for $S_{wt}$ : $\alpha = \phi_t^2 C_w = 0.567$ S/m, $\gamma = \phi_t^2 S_{wb}(C_{wb} - C_w) = 0.0715$ S/m. $S_{wt} = 0.411$ , and $S_{we} = (0.411 - 0.156)/(1 - 0.156) = 0.302$ .
The volume of oil $\phi_e(1 - S_{we})$ in a unit volume of rock: Archie 0.080, Simandoux 0.100, modified 0.109, Indonesia 0.103, dual water 0.126.
A variant of the Simandoux formula without $S_w$ in the shale term: $S_w = \sqrt{0.08\cdot(0.125 - 0.100)/0.18^2} = 0.248$ .

Answer: The effective water saturation of the same bed is from 0.302 to 0.446 by the shaly-sand models, and 0.556 by Archie. The largest volume of oil is 1.6 times the smallest.

What it means. The order of the results follows from how the models are built. Archie's equation does not see the shale and gives the most water. In the Simandoux formulas, the shale term grows with $S_w$ only to the first power, and the modified version also strengthens the water term. In the Indonesia equation, shale lowers the resistivity by the same number of times at any saturation. The dual-water model works in the total system: the bound water stays in the pores at any oil saturation and explains part of the conductivity, so less free water is needed. Which answer is closer to the truth cannot be decided from one pay zone. The data in this example are assumed for illustration.

CalculatedEffective water saturation of the bed from the example (

\phi_e = 0.18

V_{sh} = 0.25

) against resistivity, for the five models. The vertical lines are the readings in the water-bearing (1.56 Ω·m) and the oil-bearing (8.0 Ω·m) parts of the bed.

Worked exampleCheck in the water-bearing part of the bed

Given (assumed data)

The water-bearing part of the same bed below the contact: $\phi_e = 0.18$ , $V_{sh} = 0.25$ , $R_t = 1.56$ Ω·m
Other data: as in the previous example

Solution

Archie: $S_w = \sqrt{0.08/(0.18^2\cdot 1.56)} = 1.26$ .
Simandoux ( $1/R_t = 0.641$ S/m): $S_w = 1.14$ . Modified: $S_w = 1.00$ .
Indonesia: $S_w = 0.8006/(0.188 + 0.636) = 0.97$ .
Dual water: $S_{wt} = 1.002$ , $S_{we} = 1.00$ .
Let us adjust the Simandoux equation: which $R_{sh}$ gives $S_w = 1$ ? From $0.405 + 0.25/R_{sh} = 0.641$ we get $0.25/R_{sh} = 0.236$ S/m and $R_{sh} = 1.06$ Ω·m.
With this parameter, the oil-bearing part gives $\gamma = 0.236$ , $1/R_t = 0.125$ S/m, $S_w = 0.34$ .

Answer: The modified Simandoux equation and the dual-water model passed the check in the water-bearing part (about 1.00 and 1.00). The Indonesia equation is close to them (0.97). The Simandoux equation with the resistivity of the neighbouring shales gives 1.14, and Archie's equation gives 1.26. The adjusted Simandoux equation gives 0.34 in the oil-bearing part.

What it means. The check ruled out Archie's equation, and showed that with $R_{sh} = 2.5$ Ω·m the Simandoux equation gives too much water. But two models that pass the check equally well differ in the oil-bearing part by 0.10 in water saturation (0.40 against 0.30). The water-bearing part fixes only the point $S_w = 1$ , and the models differ in how the shale term changes with saturation. We calculated the resistivities of this bed with the dual-water model at $S_{we} = 0.30$ , which is why that model “guesses” right. In a real bed the answer is not known in advance. Several data decide the choice: the water saturation of core cut with oil-based mud (which adds no water to the core), CEC and measurements at several salinities, capillary curves (laboratory curves that show how much water the rock holds at each pressure), and well tests. The fitted $R_{sh} = 1.06$ Ω·m is no longer the resistivity of the shale, but a parameter of the model. The data in this example are assumed for illustration.

The dual-water model has one more useful check. The bound water is not displaced, so the resistivity of the bed cannot exceed the value at $S_{wt} = S_{wb}$ : $R_{t,\max} = a/(\phi_t^m S_{wb}^n C_{wb})$ . For the bed of the example this is about 40 Ω·m. If the log shows more, then the shale volume or the bound water in the calculation is too high, or the resistivity has been measured with an error.

How to choose a model and check the result

1
Shale properties
From the thick shales next to the bed, take the resistivity $R_{sh}$ , the density and total porosity, and the neutron reading. Choose shales that resemble the shale material in the bed. Uranium-rich and carbonate packages are not suitable.
2
Shale volume
Use several indicators based on different physics, and check them against core (Determining shale volume). An error in $V_{sh}$ passes into both porosity and saturation.
3
Porosity
Find total and effective porosity consistently, with corrections for shale and gas (Gas, shale, carbonate and pressure effects on porosity logs). Find the form of shale distribution from the plot of total porosity against shale volume.
4
Water resistivity
Use water samples, the spontaneous potential (SP, the natural voltage in the borehole) and clean water-bearing beds (Determining formation water resistivity). Do not use shaly water-bearing beds to find $R_w$ : their apparent water resistivity $R_{wa}$ (the water resistivity that Archie's equation gives for the bed) absorbs the effect of the shale.
5
Model
For laminated shale, use the laminated-sand model. For dispersed and mixed shale, use the Simandoux, Indonesia or dual-water models. If core with CEC data is available, use the Waxman–Smits model as the standard.
6
Water-bearing shaly beds
The model with the chosen parameters must give $S_w \approx 1$ in them. If it does not, revise $R_{sh}$ , shale volume, $R_w$ and $m$ , and do not cut the result off at one.
7
Calibration with core and well tests
Use the water saturation of core cut with oil-based mud, capillary curves, CEC and measurements at several salinities, and the results of formation testing. These are what tell the models apart in the pay zone.
8
Uncertainty
Recalculate the volume of hydrocarbons for reasonable choices of $V_{sh}$ , $R_{sh}$ , $n$ and the model. Pass a consistent “porosity and saturation” pair from one system to the hydrocarbons-in-place calculation (Volumetric estimation of hydrocarbons in place).

Terms and symbols

Shaly sand: A reservoir in which shale material noticeably affects porosity, permeability and resistivity.
Clay-bound water: Water in the double layer at the surface of the clay particles. It is part of the total porosity, and hydrocarbons do not push it out.
Effective porosity $\phi_e$: In shaly-sand models, the total porosity minus the clay-bound water (PHIE).
$Q_v$: The cation exchange capacity per unit of pore volume, meq/cm³.
$B$: The equivalent conductivity of the exchangeable cations in the Waxman–Smits model. It depends on temperature and salinity.
$S_{wb}$ , $C_{wb}$: The share of the pore volume taken up by bound water, and its conductivity, in the dual-water model.
Laminated, dispersed and structural shale: Shale in layers, in the pores, and in place of the grains of the frame.
Low-resistivity pay zone: A bed that gives hydrocarbons at a resistivity close to that of a water-bearing bed.

Common mistakes and misconceptions

Wrong

“The bed has almost the same resistivity as a water-bearing one, so it is water-bearing.”

In fact

Shaly pay zones and thinly laminated pay zones look like this too. Before writing a bed off, we calculate the saturation with shaly-sand models and look at the resistivity anisotropy, the core and the results of well testing.

Wrong

“The Simandoux equation in one program is the same as in another.”

In fact

Formulas with $S_w$ in the shale term and without it, and with the factor $1/(1 - V_{sh})$ and without it, are found under this name. On the same data the results differ almost twofold. We compare the formulas, not the names.

Wrong

“Let us take the total porosity, and the saturation from the Simandoux equation.”

In fact

The Simandoux equation works with effective porosity and gives effective saturation. A pair from different systems gives a volume of hydrocarbons that is too low or too high by tens of per cent.

Wrong

“In a water-bearing shaly bed we got $S_w = 1.15$ , so we will cut it off at one.”

In fact

This is a signal that the model parameters do not fit the bed: $R_{sh}$ or $R_w$ is too high, or the shale volume is too low. We correct the parameters, and then the result in the pay zones changes too.

Wrong

“The model passed the check in the water-bearing part, so it is correct.”

In fact

The check fixes only the point $S_w = 1$ . Models that pass it equally well can differ by 0.1 in water saturation in the pay zone. Core and well tests tell them apart.

In depth: advanced detail

Waxman–Smits parameters from core

We measure the parameters of the model on samples saturated with brines (salt solutions) of several concentrations. At $S_{wt} = 1$ , the model gives a straight line when we plot the sample conductivity against the brine conductivity.

Conductivity of a water-saturated shaly sample against the brine conductivity

C_0 = \frac{1}{F^*}\,C_w + \frac{B\,Q_v}{F^*}

where:

$C_0$ conductivity of a sample fully saturated with brine, S/m
$C_w$ conductivity of the brine, S/m
$F^*$ formation resistivity factor with allowance for shale, —
$B\,Q_v$ extra conductivity from the exchangeable cations, S/m

Units: conductivities in S/m at the measurement temperature

Holds when

The sample is fully saturated with brine, $S_{wt} = 1$ .
$B$ is constant. This holds for brines that are salty enough. In fresh brines $B$ decreases, and the points fall below the line, so we find the slope from the salty brines.

The slope of the line is $1/F^*$ , and the intercept on the $C_0$ axis is $B Q_v/F^*$ . From $F^*$ and the total porosity we find $m^*$ . From the intercept and $Q_v$ (from the CEC) we find $B$ . The exponent $n^*$ is found in the same way, but at several saturations.

Worked exampleA sample at three salinities

Given (assumed data)

Sample: $\phi_t = 0.21$ , solid-phase density 2.65 g/cm³, CEC 4.0 meq/100 g
Brine $C_w$ , S/m (at 25 °C): 5; 10; 20
Sample $C_0$ , S/m: 0.3240; 0.5623; 1.039

Solution

$Q_v = 4.0\cdot(1 - 0.21)\cdot 2.65/(100\cdot 0.21) = 0.399$ meq/cm³.
A line through the two points with salty brines: the slope is $(1.039 - 0.5623)/(20 - 10) = 0.04767$ , that is, $F^* = 20.98$ . The intercept is $0.5623 - 10\cdot 0.04767 = 0.0856$ S/m.
The extra conductivity of the cations is $B Q_v = 0.0856\cdot 20.98 = 1.796$ S/m. $B = 1.796/0.399 = 4.50$ (S/m)/(meq/cm³).
Cementation exponent with allowance for shale: $m^* = \log_{10}\!20.98/(-\log_{10}\!0.21) = 1.95$ .
Check with the third point: $(5 + 1.796)/20.98 = 0.3239$ S/m, against the measured 0.3240.
If the formation factor had been found from one brine of 5 S/m, as for a clean rock: $F = 5/0.3240 = 15.43$ and $m = \log_{10}\!15.43/(-\log_{10}\!0.21) = 1.75$ .

Answer:

F^* \approx 20.98

and

m^* \approx 1.95

. The extra conductivity of the cations is 1.796 S/m, and

B \approx 4.50

. The “apparent” exponent from one brine is 1.75.

What it means. A value of $B$ of about 4.5 is of the same order as the published correlations give for 25 °C and salty brines. The main practical conclusion is in the last step. If we measure a shaly core by the usual method for clean rocks, the conductivity of the cations “hides” in the formation factor. The exponent $m$ then comes out too low, and even lower with a fresh brine. Such an $m$ must not be put into Archie's equation or into the shaly-sand models. The data in this example are assumed for illustration.

The dispersed-shale model

An early model of dispersed shale, proposed by L. de Witte in the 1950s, treats the clay in the pores as one more “electrolyte” (a conducting solution) mixed with the formation water. The volume between the grains, the intergranular porosity $\phi_{im}$ , is shared between the hydrocarbons, the water and the dispersed shale.

Dispersed-shale model

\frac{1}{R_t} = \frac{\phi_{im}^2\,S_{im}}{a}\left(\frac{q}{R_{shd}} + \frac{S_{im} - q}{R_w}\right), \qquad S_w = \frac{S_{im} - q}{1 - q}

where:

$\phi_{im}$ intergranular porosity: the volume between the grains, filled by fluids and dispersed shale, fraction
$S_{im}$ share of the intergranular volume taken up by the mixture of water and dispersed shale, fraction
$q$ share of the intergranular volume taken up by dispersed shale, fraction
$R_{shd}$ resistivity of the dispersed shale; usually taken equal to the resistivity of the neighbouring shales, Ω·m
$S_w$ water saturation of the effective pore volume, fraction

Units: resistivities in Ω·m

Holds when

The shale is spread evenly in the pores and conducts current together with the water, like a mixture of electrolytes.
The cementation and saturation exponents equal 2.
We estimate $\phi_{im}$ from the sonic log, which records the speed of sound in the rock ( $\phi_{im} \approx \phi_S$ ). For the sonic log, the dispersed shale “looks like” pore water. We estimate $q$ from the gap between density and sonic porosity. If the density of the shale is close to the density of the matrix, the density porosity is close to the effective porosity $\phi_{im}(1 - q)$ , and so $q \approx (\phi_S - \phi_D)/\phi_S$ .

The equation is quadratic in $S_{im}$ . When the resistivity of the dispersed shale is much larger than $R_w$ , its solution hardly depends on $R_{shd}$ (see below). The model is rarely used now: in practice the Simandoux, Indonesia and dual-water models have replaced it.

Dispersed-shale model for

R_w \ll R_{shd}

S_w \approx \frac{\sqrt{\dfrac{a R_w}{\phi_{im}^2 R_t} + \dfrac{q^2}{4}} - \dfrac{q}{2}}{1 - q}

where:

$S_w$ water saturation of the effective pore volume, fraction
$\phi_{im},\ q$ intergranular porosity and the share of it taken up by dispersed shale, fraction
$R_w,\ R_t$ resistivity of the formation water and of the bed, Ω·m

Units: resistivities in Ω·m

Holds when

The resistivity of the dispersed shale is several times larger than $R_w$ , and the bed is not too shaly.
The other assumptions are the same as for the full model.

At $q = 0$ the formula becomes Archie's equation with $m = n = 2$ . Compared with Archie's equation applied to the effective porosity $\phi_{im}(1 - q)$ , it gives a smaller water saturation, because the shale in the pores conducts current together with the water.

Connections to other topics

Archie's equation and saturation of clean reservoirs — the equation for clean rocks to which all the models reduce, and its assumptions.
Determining shale volume — how to find shale volume, the main input of shaly-sand models.
Gas, shale, carbonate and pressure effects on porosity logs — porosity corrections for shale, and effective porosity from logs.
Determining formation water resistivity — formation water resistivity from clean beds, and why shaly beds are not suitable for it.
Resistivity logs: electrical and induction tools — resistivity along and across the layers, and low-resistivity laminated beds.
Resistivity–porosity crossplots — how shaly beds look on resistivity–porosity plots.
Lithology and porosity from crossplots — the shale point and shaly beds on the neutron–density crossplot.
Porosity — total, open and effective porosity, and the confusion in the names.
Water-based muds and clay chemistry — the structure and charge of clay minerals, the electrical double layer.
Volumetric estimation of hydrocarbons in place — how porosity and saturation of one system enter the calculation of hydrocarbons in place.
Nuclear magnetic resonance (NMR) logging — bound and free water from the nuclear magnetic resonance log.

Check yourself

1
In which bed (with the same shale volume) will Archie's calculation put the water saturation too high by the largest amount?
1. A.A water-bearing bed with slightly salty water
2. B.An oil-bearing bed with slightly salty formation water
3. C.An oil-bearing bed with a strong brine
4. D.A water-bearing bed with a strong brine
Show the answer
Correct answer: B. An oil-bearing bed with slightly salty formation water
The extra conductivity from shale falls with saturation more slowly than the conductivity of the water, and it is more noticeable against fresh water. So its share of the conductivity, and with it the error of Archie's equation, is largest at low water saturation and low salinity.
2
The total porosity of a bed is 0.25, and the clay-bound water takes up a share $S_{wb} = 0.2$ of it. The dual-water model gave $S_{wt} = 0.52$ . Find the effective porosity and the effective water saturation, and check that the volume of hydrocarbons is the same in both systems.
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$\phi_e = 0.25\cdot(1 - 0.2) = 0.20$ ; $S_{we} = (0.52 - 0.2)/(1 - 0.2) = 0.40$ . The volume of hydrocarbons is $0.25\cdot(1 - 0.52) = 0.12$ and $0.20\cdot(1 - 0.40) = 0.12$ . They are the same.
3
On the plot of total porosity against shale volume, the point of a bed lies clearly below the straight line that joins the clean-sandstone point to the shale point. What does this tell us?
1. A.The shale is dispersed: it fills the pores, and effective porosity and permeability are reduced
2. B.The shale is structural: it has replaced part of the grains
3. C.The shale is laminated, and the sandy layers are unchanged
4. D.The bed contains gas
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Correct answer: A. The shale is dispersed: it fills the pores, and effective porosity and permeability are reduced
Dispersed shale takes the place of water in the pores and brings only its own small porosity, so total porosity falls faster than with laminated shale. Structural shale would put the point above the line. Gas affects the density and neutron readings in different ways and would shift the porosity estimate itself. We check for that before drawing the plot.
4
Which model needs the cation exchange capacity or $Q_v$ measured on core?
1. A.The Indonesia equation
2. B.The laminated-sand model
3. C.The Simandoux equation
4. D.The Waxman–Smits model
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Correct answer: D. The Waxman–Smits model
In the Waxman–Smits model, the shale is described by the product $B Q_v$ : $Q_v$ comes from the CEC of core, and $B$ from the temperature and salinity. The other models use shale volume and the resistivity of the neighbouring shales. When the dual-water model is run on logs, it takes the bound-water parameters from the neighbouring shales, although it can be calibrated with the same core $Q_v$ values.
5
In a water-bearing shaly bed, the Simandoux equation with the resistivity of the neighbouring shales gave $S_w = 1.15$ . What does this mean, and what should we do?
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A water saturation above one is impossible: the model parameters do not fit the bed. The model gives the shale too little conductivity ( $R_{sh}$ is too high), or the shale volume, $R_w$ or $m$ is wrong. We check the parameters and, if necessary, adjust $R_{sh}$ so that the water-bearing part gives one, and then recalculate the pay zones. We must not cut the result off at 1: the error would remain in the pay zone.
6
Why can two models that both gave $S_w = 1$ in the water-bearing part of a bed differ by 0.1 in water saturation in its oil-bearing part?
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The check fixes only one point: full water saturation. The models differ in how the shale term changes with saturation. In the Indonesia equation it falls in the same way as the water term. In the Simandoux formulas it falls more slowly. In the dual-water model the bound water is not displaced at all. So at low water saturation they turn the same resistivity into different saturations. Core that keeps its original saturation, CEC measurements and well tests help to tell them apart.

Updated: 28 October 2026