Archie's equation and saturation of clean reservoirs

Quick refresher

In a clean rock only the water in the pores conducts current. Mineral grains, oil and gas are insulators.
The formation resistivity factor $F = R_0/R_w$ shows how many times worse a water-saturated rock conducts current than the water itself. It depends on porosity and on how winding the paths of the current are: $F = a/\phi^m$ .
The resistivity index $I = R_t/R_0$ shows how many times hydrocarbons have raised the resistivity: $I = S_w^{-n}$ .
Together: $S_w = \left(a R_w / \phi^m R_t\right)^{1/n}$ . For sandstones, often $a \approx 1$ and $m \approx n \approx 2$ .
The uncertainty of $m$ and $n$ affects the result as much as an error in $R_w$ does. We fit the coefficients to the core of the particular bed.
Shale, fractures, vugs, oil-wet rock and thin beds break the assumptions of the equation. There we need corrections or other models.

Key relationships

F = \dfrac{R_0}{R_w} = \dfrac{a}{\phi^m}

formation resistivity factor: the definition and the link with porosity

I = \dfrac{R_t}{R_0} = S_w^{-n}

resistivity index

S_w = \left(\dfrac{a\,R_w}{\phi^m\,R_t}\right)^{1/n}

water saturation of the uninvaded formation

S_{xo} = \left(\dfrac{a\,R_{mf}}{\phi^m\,R_{xo}}\right)^{1/n}

water saturation of the flushed zone

Common pitfalls

Putting porosity in per cent instead of as a fraction.
Taking $R_w$ at laboratory temperature instead of formation temperature.
Mixing up $R_0$ , the resistivity of the water-saturated rock, and $R_w$ , the resistivity of the water itself.
Mixing $a$ from one relationship with $m$ from another: the coefficients are fitted as a pair.
Taking the reading of the deep sonde as $R_t$ without corrections.

After this lesson you can

explain why the resistivity of a clean rock is set by the water in its pores, and what the formation resistivity factor $F$ and the resistivity index $I$ describe
write Archie's equation in general form with the coefficients $a$ , $m$ and $n$ , and explain the meaning of each
calculate the water saturation of the uninvaded formation (the rock that the mud filtrate, the liquid that seeps in from the drilling mud, has not reached) and of the flushed zone (the zone next to the borehole wall, where the filtrate has pushed out most of the original fluids), and estimate the share of moveable oil
estimate how errors in $R_w$ , $R_t$ , porosity and the coefficients carry over into the error of water saturation
find $a$ and $m$ from laboratory data on core
recognise the conditions in which Archie's equation does not apply without corrections: shaly, fractured, vuggy (with rounded cavities) and oil-wet (hydrophobic) rocks, in which oil rather than water clings to the grains

Why resistivity tells us about saturation

Grains of quartz, feldspar and calcite practically do not conduct electric current, and neither do oil and gas. The formation water (the water that fills the pores underground) conducts. Salts are dissolved in it, and ions (charged particles) carry the current. So the resistivity of a clean reservoir rock (a rock that can hold oil or gas) depends on three things: how salty the water in the pores is, how much water there is, and how complicated the paths it forms are.

Electrical resistivity ( $R$ ) is measured in ohm-metres (Ω·m). Its inverse, $1/R$ , is the electrical conductivity, which is expressed in siemens per metre. Induction tools (logging tools that use coils instead of electrodes) often give it in millisiemens per metre: $\sigma\,[\text{mS/m}] = 1000/R\,[\text{Ω·m}]$ . Salty formation waters at formation temperature have a resistivity of roughly hundredths to tenths of an Ω·m. Water-bearing sandstones have from tenths of an Ω·m to a few Ω·m. Productive ones have from a few to hundreds of Ω·m.

Formation resistivity factor: water in a maze of pores

Take a sample of clean rock, completely saturated with water of resistivity $R_w$ , and measure its resistivity $R_0$ . The ratio of the two is the formation resistivity factor ( $F$ ), also called the formation factor. Experiments show that for a clean rock $F$ does not depend on how salty the water is that fills it. It describes the rock itself: how much of its cross section can carry current, and how the pore channels are shaped.

Formation resistivity factor

F = \frac{R_0}{R_w} = \frac{a}{\phi^m}

where:

$F$ formation resistivity factor, —
$R_0$ resistivity of the rock completely saturated with water, Ω·m
$R_w$ resistivity of this water at the same temperature, Ω·m
$\phi$ porosity, fraction
$a$ tortuosity factor (how winding the paths are), an empirical coefficient, —
$m$ cementation exponent, an empirical exponent, —

Units: dimensionless; $R_0$ and $R_w$ in the same units

Holds when

A clean rock: only the water in the pores conducts current, and the surface conductivity of the grains and of the shale is negligible.
Intergranular porosity, more or less uniform through the volume.
$a$ and $m$ are fitted for the given rock, and they are used in the range of porosity where they were found.
$R_0$ and $R_w$ refer to one temperature. For a bed, this is the formation temperature.

The first equality is a definition: we measure $F$ on core. The second is an empirical link with porosity. On logarithmic axes (axes where each step multiplies the value by the same factor), $\log_{10} F = \log_{10} a - m\log_{10}\phi$ is a straight line with a slope of $-m$ . The more complicated the path of the current, the larger $m$ is, and the higher $F$ is at the same porosity.

CalculatedFormation resistivity factor against porosity at

a = 1

(logarithmic axes). The numbers by the lines are the values of the cementation exponent

m

. The point is the rock of the example below:

\phi = 0.21

F \approx 22.7

m = 2

The physical meaning of the coefficients: at $\phi = 1$ (water alone, with no rock) $F$ must equal one, so ideally $a = 1$ . Values $a \ne 1$ appear when we fit the line to core in a limited range of porosity. Then $a$ and $m$ adjust to each other. So we use the pair $a$ – $m$ only together.

Approximate coefficients for different rocks

Rock: Loose (uncemented) sands
$a$: 1
$m$: 1.3–1.5
Comment: the paths of the current are simple, the channels wide

Rock: Cemented sandstones
$a$: 0.6–1
$m$: 1.8–2.2
Comment: often $F = 0.62/\phi^{2.15}$ (the Humble formula) or $F = 0.81/\phi^2$

Rock: Carbonates with intergranular and intercrystalline porosity
$a$: 1
$m$: about 2
Comment: $F = 1/\phi^2$

Rock: Tight carbonates, rocks with vugs and with pores where grains have dissolved
$a$: 1
$m$: 2.2–2.5, up to 3 and above
Comment: isolated pores add to the porosity but hardly conduct current

Rock: Fractured rocks
$a$: 1
$m$: 1.1–1.5
Comment: fractures are direct conducting paths

Rock	$a$	$m$	Comment
Loose (uncemented) sands	1	1.3–1.5	the paths of the current are simple, the channels wide
Cemented sandstones	0.6–1	1.8–2.2	often $F = 0.62/\phi^{2.15}$ (the Humble formula) or $F = 0.81/\phi^2$
Carbonates with intergranular and intercrystalline porosity	1	about 2	$F = 1/\phi^2$
Tight carbonates, rocks with vugs and with pores where grains have dissolved	1	2.2–2.5, up to 3 and above	isolated pores add to the porosity but hardly conduct current
Fractured rocks	1	1.1–1.5	fractures are direct conducting paths

Rough values, for guidance. For the estimate of hydrocarbons in place, we find $a$ , $m$ and $n$ on the core of the particular bed.

Resistivity index: what hydrocarbons change

When oil or gas takes up part of the pores, less water is left to conduct current, and the paths of the current become even more winding. In a water-wet rock (one where water is the liquid that clings to the grains) the water stays as films on the grains and in the small pores. The resistivity of the rock $R_t$ grows compared with $R_0$ , the resistivity of the same rock when fully saturated with water. Their ratio is the resistivity index ( $I$ ).

Resistivity index

I = \frac{R_t}{R_0} = S_w^{-n}

where:

$I$ resistivity index (the factor by which the resistivity increases), —
$R_t$ resistivity of the rock at water saturation $S_w$ , Ω·m
$R_0$ resistivity of the same rock at $S_w = 1$ , Ω·m
$S_w$ water saturation, fraction
$n$ saturation exponent, an empirical exponent, —

Units: dimensionless

Holds when

A clean rock, and only water conducts.
The water forms a connected system: films and filled small pores (a water-wet rock).
$R_t$ and $R_0$ refer to one rock, one water and one temperature.

For water-wet rocks $n$ is usually close to 2. In oil-wet (hydrophobic) rocks the water gathers into separate drops. The connectedness of the conducting network falls sharply, and $n$ can be 3–8 and more. See Wettability and interfacial phenomena.

CalculatedResistivity index against water saturation (logarithmic axes). The numbers by the lines are the values of the saturation exponent

n

. The point is the bed of the example below:

S_w \approx 0.31

I \approx 10.6

n = 2

. For the same

I = 25

, the water saturation is 0.20 at

n = 2

and 0.34 at

n = 3

Archie's equation and the calculation of water saturation

Combine the two definitions, $R_0 = F R_w$ and $R_t = I R_0$ . We get $R_t = F R_w S_w^{-n}$ , and the water saturation follows from it.

Archie's equation for water saturation

S_w = \left(\frac{F\,R_w}{R_t}\right)^{1/n} = \left(\frac{a\,R_w}{\phi^m\,R_t}\right)^{1/n}

where:

$S_w$ water saturation of the uninvaded part of the formation, fraction
$R_w$ resistivity of the formation water at formation temperature, Ω·m
$R_t$ resistivity of the uninvaded part of the formation (after corrections), Ω·m
$\phi$ porosity, fraction
$a,\ m,\ n$ the tortuosity factor, the cementation exponent and the saturation exponent, —

Units: dimensionless; all resistivities in the same units

Holds when

A clean (shale-free) bed with intergranular porosity. Only the formation water conducts current.
$R_t$ is the resistivity of the uninvaded part of the formation, with corrections for the borehole, the enclosing rocks and invasion (Borehole environment: invaded zone and mudcake).
$R_w$ is at formation temperature.
$a$ , $m$ and $n$ suit the rock (from core or from a checked analogy). The rock is water-wet.

Water saturation grows with $R_w$ , and it falls as $R_t$ and porosity grow. The dependence on $R_t$ is a power law: at $n = 2$ four times the resistivity gives only half the water saturation. The oil or gas saturation is $1 - S_w$ .

At $a = 1$ and $m = n = 2$ the equation takes the simple form $\phi\,S_w = \sqrt{R_w/R_t}$ . The product $\phi S_w$ is the volume fraction of water in the rock (bulk volume water). The resistivity–porosity crossplots (Resistivity–porosity crossplots) are based on this form.

The same expression is written for the flushed zone, where the water in the pores is mud filtrate. Instead of $R_t$ we put in the resistivity of the flushed zone $R_{xo}$ , from the microresistivity tools (tools with electrodes a few centimetres apart, on a pad pressed against the borehole wall). Instead of $R_w$ we put in the filtrate resistivity $R_{mf}$ . Comparing $S_{xo}$ and $S_w$ shows how many hydrocarbons the filtrate pushed out, that is, how many are moveable.

Archie's equation for the flushed zone

S_{xo} = \left(\frac{a\,R_{mf}}{\phi^m\,R_{xo}}\right)^{1/n}

where:

$S_{xo}$ water saturation of the flushed zone, fraction
$R_{mf}$ resistivity of the filtrate at formation temperature, Ω·m
$R_{xo}$ resistivity of the flushed zone (corrected for the mudcake), Ω·m

Units: dimensionless

Holds when

The same as for $S_w$ . The water in the flushed zone is practically only filtrate.
A water-based mud (with an oil-based mud the filtrate does not conduct current).

The difference $S_{xo} - S_w$ is the share of the pore volume occupied by moveable hydrocarbons. $1 - S_{xo}$ is the residual hydrocarbon saturation.

Order of calculation from logs

1
Choose a clean interval
Check by the gamma ray log (which reads high in shale) and other signs that the bed is shale-free. Otherwise we need models of shaly reservoirs (Water saturation in shaly sands).
2
Porosity
We take it from the density, neutron or sonic log, as a fraction (Porosity from the density log).
3
Formation resistivity factor
We find it from the coefficients $a$ and $m$ for the given rock. Without core, we use a checked analogy and estimate the uncertainty.
4
True resistivity
$R_t$ comes from a deep induction or laterolog sonde (tools that read the resistivity far from the borehole), with corrections for the borehole, the bed thickness and invasion.
5
Formation water resistivity
$R_w$ is taken at formation temperature. We check it in several ways: from the SP (the natural voltage in the borehole), from analyses of samples, from a water-bearing bed, from a catalogue of waters (Determining formation water resistivity).
6
Water saturation and a check
We calculate $S_w$ . In the water-bearing part of the same bed the calculation must give $S_w \approx 1$ . If it does not, we review $R_w$ , $a$ and $m$ .

Worked exampleWater saturation and moveable oil in a clean sandstone

Given (assumed data)

Porosity from the density log $\phi$: 0.21
Resistivity of the uninvaded formation $R_t$: 12.5 Ω·m (after corrections)
Resistivity of the formation water $R_w$: 0.052 Ω·m at formation temperature
Resistivity of the flushed zone $R_{xo}$: 6.5 Ω·m
Resistivity of the filtrate $R_{mf}$: 0.20 Ω·m at formation temperature
Coefficients: $a = 1$ , $m = 2$ , $n = 2$

Solution

The formation resistivity factor: $F = a/\phi^m = 1/0.21^2 = 22.68$ .
The resistivity of this rock when fully saturated with water: $R_0 = F R_w = 22.68\cdot 0.052 = 1.179$ Ω·m.
The resistivity index: $I = R_t/R_0 = 12.5/1.179 = 10.60$ .
The water saturation: $S_w = I^{-1/n} = 10.60^{-1/2} = 0.307$ . The oil saturation is $1 - S_w = 0.693$ .
The flushed zone: $S_{xo} = \left(\dfrac{F R_{mf}}{R_{xo}}\right)^{1/2} = \left(\dfrac{22.68\cdot 0.20}{6.5}\right)^{1/2} = 0.835$ .
The moveable oil is $S_{xo} - S_w = 0.53$ of the pore volume. The residual oil saturation is $1 - S_{xo} = 0.16$ .

Answer: The water saturation is about 31%, the oil saturation is about 69%, and about 0.53 of the pore volume is moveable.

What it means. The bed looks productive: water takes up less than a third of the pores, and the filtrate was able to move most of the oil. The volume fraction of water in the rock, $\phi S_w = 0.064$ , will be useful for comparison with neighbouring intervals. Before we include such a result in the estimate of hydrocarbons in place, we check $R_w$ against the water-bearing part of the bed and the coefficients against core. The next example shows how much the answer depends on them. The data in this example are assumed for illustration.

Worked exampleHow much the water saturation depends on the input data

Given (assumed data)

Base calculation: the bed of the previous example: $\phi = 0.21$ , $R_t = 12.5$ Ω·m, $R_w = 0.052$ Ω·m, $a = 1$ , $m = n = 2$
Variants: $m = 1.8$ and $2.2$ ; $n = 2.5$ ; $R_w$ 20% larger; porosity 0.02 smaller; the Humble formula ( $a = 0.62$ , $m = 2.15$ )

Solution

The base case: $S_w = \left(\dfrac{1\cdot 0.052}{0.21^2\cdot 12.5}\right)^{1/2} = 0.307$ .
At $m = 1.8$ , $S_w = 0.263$ . At $m = 2.2$ , $S_w = 0.359$ .
At $n = 2.5$ , $S_w = 0.389$ .
At $R_w = 0.0624$ Ω·m (20% larger), $S_w = 0.336$ .
At $\phi = 0.19$ , $S_w = 0.339$ .
The Humble formula: $F = 0.62/0.21^{2.15} = 17.8$ and $S_w = 0.272$ .

Answer: With reasonable variants of the input data, the water saturation changes from 0.26 to 0.39, against the base value of 0.307.

What it means. The uncertainty of the exponents $m$ and $n$ shifts the result as much as an error in $R_w$ does. And an error of only 0.02 in porosity acts almost like a 20% error in $R_w$ . So we find the coefficients on core, we check $R_w$ in several ways, and in the estimate of hydrocarbons in place we give the water saturation as a range, not as one number. The data in this example are assumed for illustration.

Where Archie's equation stops working

Archie's equation is empirical, and it rests on several assumptions. Only water conducts current. The pores are uniform and connected. The water forms a connected network. And $R_t$ was measured correctly. Each violation shifts the result in a known direction. It is useful to know this, so that we can tell whether an estimate is optimistic or pessimistic.

Typical violations of the assumptions and their consequences

Condition: Shaly sandstones
What happens: shale conducts current in parallel with the formation water (through clay-bound water and exchange cations, which are charged ions on the clay surface), and $R_t$ is lowered
How the $S_w$ calculation goes wrong: overestimates $S_w$ : a productive bed may be missed
What we do: models of shaly reservoirs (Water saturation in shaly sands)

Condition: Fresh formation waters
What happens: the surface conductivity of grains and shale is comparable to the conductivity of the water, and $F$ starts to depend on $R_w$
How the $S_w$ calculation goes wrong: overestimates $S_w$
What we do: calibration against core, models with surface conductivity

Condition: Vugs and isolated pores
What happens: part of the porosity hardly takes part in conduction, and the real $m$ is more than 2
How the $S_w$ calculation goes wrong: at $m = 2$ underestimates $S_w$
What we do: $m$ from core, allowing for the type of pores

Condition: Fractures
What happens: fractures are direct conducting paths, and the real $m$ is close to 1
How the $S_w$ calculation goes wrong: at $m = 2$ overestimates $S_w$
What we do: dual porosity models (which treat the rock matrix and the fractures separately), $m$ from core and tests

Condition: Oil-wet rock
What happens: the water is in separate drops, and the real $n$ is 3–8 and more
How the $S_w$ calculation goes wrong: at $n = 2$ underestimates $S_w$
What we do: $n$ from core with restored wettability

Condition: Thin beds and lamination
What happens: the deep sonde averages the bed with the enclosing rocks, and $R_t$ of a thin high-resistivity bed is too low
How the $S_w$ calculation goes wrong: overestimates $S_w$
What we do: corrections for the enclosing rocks, high-resolution tools

Condition: Conducting minerals (pyrite, magnetite, graphite)
What happens: the current also goes through the minerals, and $R_t$ is lowered
How the $S_w$ calculation goes wrong: overestimates $S_w$
What we do: allowing for the mineral composition

Condition: Core measured without load
What happens: in the bed the overburden pressure (the weight of the rocks above) squeezes the pores and raises the resistivity; $F$ on unloaded core is too low
How the $S_w$ calculation goes wrong: with too low an $m$ , underestimates $S_w$
What we do: measurements on core at the effective pressure of the formation

Condition	What happens	How the $S_w$ calculation goes wrong	What we do
Shaly sandstones	shale conducts current in parallel with the formation water (through clay-bound water and exchange cations, which are charged ions on the clay surface), and $R_t$ is lowered	overestimates $S_w$ : a productive bed may be missed	models of shaly reservoirs (Water saturation in shaly sands)
Fresh formation waters	the surface conductivity of grains and shale is comparable to the conductivity of the water, and $F$ starts to depend on $R_w$	overestimates $S_w$	calibration against core, models with surface conductivity
Vugs and isolated pores	part of the porosity hardly takes part in conduction, and the real $m$ is more than 2	at $m = 2$ underestimates $S_w$	$m$ from core, allowing for the type of pores
Fractures	fractures are direct conducting paths, and the real $m$ is close to 1	at $m = 2$ overestimates $S_w$	dual porosity models (which treat the rock matrix and the fractures separately), $m$ from core and tests
Oil-wet rock	the water is in separate drops, and the real $n$ is 3–8 and more	at $n = 2$ underestimates $S_w$	$n$ from core with restored wettability
Thin beds and lamination	the deep sonde averages the bed with the enclosing rocks, and $R_t$ of a thin high-resistivity bed is too low	overestimates $S_w$	corrections for the enclosing rocks, high-resolution tools
Conducting minerals (pyrite, magnetite, graphite)	the current also goes through the minerals, and $R_t$ is lowered	overestimates $S_w$	allowing for the mineral composition
Core measured without load	in the bed the overburden pressure (the weight of the rocks above) squeezes the pores and raises the resistivity; $F$ on unloaded core is too low	with too low an $m$ , underestimates $S_w$	measurements on core at the effective pressure of the formation

The directions of the errors are for the case where “standard” coefficients and an uncorrected $R_t$ are used in the calculation.

Terms and symbols

Formation resistivity factor $F$: The ratio of the resistivity of a water-saturated rock to the resistivity of the water, $F = R_0/R_w = a/\phi^m$ .
Resistivity index $I$: The ratio of the resistivity of a rock at a given water saturation to its resistivity when fully saturated with water, $I = R_t/R_0 = S_w^{-n}$ .
Cementation exponent $m$: The slope of the $F(\phi)$ relationship on logarithmic axes. It reflects how winding and connected the pores are.
Saturation exponent $n$: The slope of the $I(S_w)$ relationship on logarithmic axes. It is about 2 in water-wet rocks.
Tortuosity factor $a$: An empirical multiplier in $F = a/\phi^m$ . It is fitted together with $m$ .
$R_0$: The resistivity of the same rock when fully saturated with formation water.
Bulk volume water $\phi S_w$: The share of the rock volume occupied by water.

Common mistakes and misconceptions

Wrong

“Porosity is 21%, so we put in 21.”

In fact

In Archie's equation, porosity and saturation are fractions: 0.21. Putting in per cent gives a meaningless result.

Wrong

“ $R_w$ was taken from a sample analysis at 20 °C.”

In fact

The resistivity of water falls strongly as the temperature rises. We bring it to formation temperature (Formation water and formation temperature). Otherwise $S_w$ will be too high.

Wrong

“Let us take $a = 0.62$ from the Humble formula and $m = 2$ : it is simpler.”

In fact

The coefficients $a$ and $m$ are fitted as a pair. If we mix them, we get a relationship that describes no rock. We use either $0.62/\phi^{2.15}$ , or $0.81/\phi^2$ , or a pair from core.

Wrong

“ $m = n = 2$ for all rocks.”

In fact

This is a reasonable first approximation for water-wet sandstones. In fractured, vuggy, oil-wet and shaly rocks the coefficients are different, and the error in saturation can be tens of per cent.

Wrong

“A high $S_w$ from Archie's equation means the bed is water-bearing.”

In fact

A low resistivity can come from shale, thin lamination, conducting minerals or fresh water. We draw the conclusion after checking the assumptions and comparing with core and testing.

In depth: advanced detail

How

a

m

and

n

are found on core

To find $a$ and $m$ , we saturate several samples of different porosity with brine (salty water) of known resistivity and measure their resistivity $R_0$ . We plot the points $(\phi,\ F = R_0/R_w)$ on logarithmic axes and draw a straight line. Its slope is $-m$ , and its value at $\phi = 1$ is $a$ . We find $n$ on the same samples. We gradually push the brine out with oil or gas (for example, in a capillary pressure apparatus) and measure $R_t$ at several values of $S_w$ . The slope of the line of $\log_{10} I$ against $\log_{10} S_w$ is $-n$ .

Worked exampleThe coefficients

a

and

m

from five samples

Given (assumed data)

Brine resistivity $R_w$: 0.10 Ω·m
Porosity of the samples $\phi$: 0.12; 0.16; 0.20; 0.24; 0.28
Resistivity of the saturated samples $R_0$: 5.8; 3.1; 2.15; 1.42; 1.09 Ω·m

Solution

The formation resistivity factor $F = R_0/R_w$ : 58; 31; 21.5; 14.2; 10.9.
The least-squares line (the straight line with the smallest sum of squared deviations from the points) in the coordinates $\log_{10}\phi$ and $\log_{10} F$ : the slope is $-1.968$ , so $m = 1.97$ . The intercept is $\log_{10} a = -0.057$ , so $a = 0.88$ .
If we force $a = 1$ (a line through the point $\phi = 1$ , $F = 1$ ), the best exponent is $m = 1.89$ .
A comparison at $\phi = 0.20$ : the pair $a = 0.88$ , $m = 1.97$ gives $F = 20.8$ , and the pair $a = 1$ , $m = 1.89$ gives $F = 21.0$ .

Answer: From core,

a \approx 0.88

and

m \approx 1.97

. With

a

fixed at 1,

m \approx 1.89

What it means. Both pairs describe the samples almost equally well in the measured range of porosity, although the coefficients are different. This is the reason for the rule “ $a$ and $m$ only as a pair”. Outside the range that was studied, the pairs diverge more. So it is risky to extrapolate them to a markedly lower or higher porosity. The data in this example are assumed for illustration.

How errors of the input data carry over into water saturation

Take the logarithm of Archie's equation: $\ln S_w = \tfrac{1}{n}\left(\ln a + \ln R_w - m\ln\phi - \ln R_t\right)$ . Differentiating with respect to each quantity gives a linear estimate of the relative error.

Relative error of water saturation by Archie's equation

\frac{\delta S_w}{S_w} \approx \frac{1}{n}\left(\frac{\delta a}{a} + \frac{\delta R_w}{R_w} - \frac{\delta R_t}{R_t} - m\,\frac{\delta\phi}{\phi}\right) - \frac{\ln\phi}{n}\,\delta m - \frac{\ln S_w}{n}\,\delta n

where:

$\delta S_w,\ \delta a,\ \delta R_w,\ \delta R_t,\ \delta\phi$ small errors (deviations) of the corresponding quantities, in the units of the quantity
$\delta m,\ \delta n$ errors of the cementation exponent and the saturation exponent, —
$\ln$ natural logarithm, —

Units: dimensionless

Holds when

The errors are small compared with the quantities themselves (a linear approximation).
The errors are independent. When we count them together, we add them with their signs, or, for random errors, in quadrature (as the square root of the sum of squares).

The errors of $R_w$ and $R_t$ enter with the weight $1/n$ , and the error of porosity enters with the weight $m/n$ . So at $m = n$ the relative error of porosity passes into $S_w$ one to one. Since $\ln\phi$ and $\ln S_w$ are negative, an increase of $m$ or $n$ increases the calculated $S_w$ , and the more so, the lower the porosity and the water saturation.

Worked exampleLinear estimate against exact recalculation

Given (assumed data)

Base calculation: $\phi = 0.21$ , $S_w = 0.307$ , $m = n = 2$
Deviations: $\delta m = +0.2$ ; separately, $\delta n = +0.5$

Solution

For $m$ : $\dfrac{\delta S_w}{S_w} \approx -\dfrac{\ln 0.21}{2}\cdot 0.2 = 0.156$ , so $S_w \approx 0.355$ . The exact recalculation gives 0.359.
For $n$ : $\dfrac{\delta S_w}{S_w} \approx -\dfrac{\ln 0.307}{2}\cdot 0.5 = 0.295$ , so $S_w \approx 0.398$ . The exact recalculation gives 0.389.

Answer: The linear estimate is wrong by thousandths. For a quick estimate of the uncertainty it is enough.

What it means. We can use the formula to see which input parameter affects the result most in a given bed, and to focus the laboratory work on it. The data in this example are assumed for illustration.

Connections to other topics

Borehole environment: invaded zone and mudcake — why the reading of the deep sonde is not $R_t$ , and what the flushed zone is.
Determining formation water resistivity — how to find the resistivity of the formation water, the main input of the equation.
Formation water and formation temperature — how to bring the resistivities to formation temperature.
Resistivity–porosity crossplots — the graphical solution of Archie's equation, and the estimate of $R_w$ and $m$ from logs.
Resistivity ratio methods and movable oil — saturation without porosity, from the ratio of the resistivities of the zones.
Water saturation in shaly sands — what to do when shale also conducts current.
Wettability and interfacial phenomena — why the saturation exponent depends on wettability.
Volumetric estimation of hydrocarbons in place — how water saturation enters the estimate of hydrocarbons in place.

Check yourself

1
A rock with a porosity of 0.25 is completely saturated with water of resistivity 0.05 Ω·m; $a = 1$ , $m = 2$ . What is the resistivity of the rock $R_0$ ?
1. A.3.2 Ω·m
2. B.0.0125 Ω·m
3. C.0.2 Ω·m
4. D.0.8 Ω·m
Show the answer
Correct answer: D. 0.8 Ω·m
$F = 1/0.25^2 = 16$ , and $R_0 = F R_w = 16\cdot 0.05 = 0.8$ Ω·m. The rock conducts current 16 times worse than the water in its pores.
2
The calculation uses $m = 2$ , but for a rock with vugs the real value is $m = 2.4$ . How is the calculated water saturation wrong?
1. A.It depends only on $R_w$
2. B.Too high
3. C.Too low: the estimate is too optimistic
4. D.It does not change: $m$ affects only $F$
Show the answer
Correct answer: C. Too low: the estimate is too optimistic
At $\phi < 1$ a larger $m$ gives a larger $F$ , and $S_w \propto F^{1/n}$ . A too low $m$ gives a too low $F$ and a too low water saturation: the bed looks more productive than it is.
3
In the water-bearing part of a bed the deep sonde reads 2.0 Ω·m. In the upper, oil-bearing part of the same bed, with the same porosity, it reads 20 Ω·m. Estimate the water saturation of the upper part at $n = 2$ .
Show the answer
The water-bearing part gives $R_0 = 2.0$ Ω·m, so $I = 20/2 = 10$ and $S_w = 10^{-1/2} \approx 0.32$ . We did not need to know the porosity or $R_w$ : they are the same in both parts and cancel. The condition is a uniform clean bed and correct values of $R_t$ .
4
Why does Archie's equation usually overestimate the water saturation in a shaly sandstone?
Show the answer
Shale conducts current in addition to the formation water, through bound water and the exchange cations on the surface of clay particles. The bed resistivity $R_t$ turns out lower than it would be in a clean rock with the same saturation. Archie's equation assigns all the conductivity to the formation water, so it “sees” more water than there is.
5
Which coefficient of Archie's equation depends most on the wettability of the rock?
1. A. $m$
2. B. $a$
3. C. $R_w$
4. D. $n$
Show the answer
Correct answer: D. $n$
The saturation exponent $n$ describes how the connectedness of the conducting water changes as hydrocarbons push the water out. In an oil-wet rock the water gathers into isolated drops, and $n$ rises to 3–8 and higher. $a$ and $m$ belong to a fully water-saturated rock.
6
For a bed there are two relationships: $F = 0.62/\phi^{2.15}$ and $F = 1/\phi^{1.9}$ . Can we take $a = 0.62$ and $m = 1.9$ for the calculation?
Show the answer
No. $a$ and $m$ were fitted together, and each pair describes the data only as a whole. The pair $0.62$ and $1.9$ at a porosity of 0.20 gives $F \approx 13$ , instead of about 20 for the original relationships. So the water saturation would be underestimated by about 20%.

Updated: 19 October 2026