Determining formation water resistivity

Quick refresher

$R_w$ is the resistivity of the water in the pores at formation temperature. A log does not measure it. In Archie's equation with $n = 2$ , water saturation is proportional to $\sqrt{R_w}$ . So an error in $R_w$ shifts all the beds of an interval in the same way.
There are four sources: a formation water sample, regional data (water catalogues, neighbouring wells), the SP, and a bed that is known to be water-bearing. Reliability comes from independent estimates that agree, not from any single method.
From the SP: take the deflection in a thick, clean (shale-free), water-bearing bed and correct it for the thickness of the bed. Turn it into the ratio $R_{mfe}/R_{we}$ of the equivalent resistivities of the filtrate and the water, then into $R_{we}$ , and then into $R_w$ . Take all the resistivities at formation temperature.
For brines (very salty waters), the equivalent resistivity $R_{we}$ is several times smaller than the true $R_w$ . The equivalent resistivity is the one that matches the chemical activity of the water, which is what the SP responds to. We convert one into the other with a chart or a formula that approximates it. The resistivity of a NaCl (common salt) solution never falls below about 0.04 Ω·m at 24 °C, which gives a handy check of the result.
In a clean water-bearing bed, $R_{wa} = R_t/F$ (the true resistivity of the bed divided by the formation resistivity factor) equals $R_w$ . In a bed with hydrocarbons it is larger: $R_{wa} = R_w/S_w^n$ . We take as $R_w$ the level where the lowest $R_{wa}$ values of clean reservoir beds (porous, permeable beds) line up. An $R_{wa}$ greater than about $3R_w$ is a sign of hydrocarbons.
$R_{wa}$ goes together with the values of $a$ and $m$ that we use. It is a calibration of Archie's equation on a water-bearing bed (an adjustment of the equation to a bed where we know the answer), not only a property of the water.

Key relationships

R_{we} = R_{mfe}\cdot 10^{\,\mathrm{SSP}/K},\quad K = 65 + 0.24\,T

equivalent water resistivity from the static SP amplitude (put in SSP with its sign; $T$ in °C)

R_{w,24} \approx \dfrac{77\,R_{we,24} + 5}{146 - 337\,R_{we,24}}

converting the equivalent resistivity of a brine to the true one at 24 °C ( $R_{we,24} \le 0.12$ Ω·m)

R_{wa} = \dfrac{R_t}{F} = \dfrac{R_t\,\phi^m}{a}

apparent water resistivity; in a clean water-bearing bed it equals $R_w$

S_w = \left(\dfrac{R_w}{R_{wa}}\right)^{1/n}

water saturation from the apparent water resistivity

Common pitfalls

Taking the equivalent resistivity of a brine, $R_{we}$ , as $R_w$ . The result is several times too low.
Taking the lowest $R_{wa}$ of the whole section (the rock sequence the well crosses) as $R_w$ , even when it comes from shales, tight streaks (dense layers with almost no pores) or intervals with unreliable porosity.
Using a water sample that filtrate has diluted, condensed water from a gas well, or water from another horizon (another layer of the section).
Using one value of $R_w$ for an interval hundreds of metres thick without converting it to the temperature of each bed.

After this lesson you can

name the sources of data on $R_w$ — water samples, regional data, the SP and water-bearing beds — and say when each one is reliable
convert the mud and filtrate resistivities from the log heading to formation temperature, and estimate $R_{mf}$ when it has not been measured
follow the full procedure for calculating $R_w$ from the SP, starting with the shale baseline and the bed-thickness correction and ending with the step from the equivalent resistivity to the true one
calculate the apparent water resistivity $R_{wa}$ for a series of beds, choose $R_w$ from these values and pick out the beds that contain hydrocarbons
estimate how an error in $R_w$ becomes an error in water saturation, and combine the results of different methods into one value

One number for the whole interval

Archie's equation links the resistivity of a clean bed (a bed with no shale in it) to its porosity and water saturation: $S_w^n = a R_w/(\phi^m R_t)$ (Archie's equation and saturation of clean reservoirs). Porosity $\phi$ is the share of the rock that is pore space. Logging tools give us the porosity $\phi$ and the bed resistivity $R_t$ (the resistivity of the part of the bed that the mud has not reached). Core (rock samples brought up from the well) gives the coefficients $a$ , $m$ and $n$ , numbers that describe how the pores of the rock affect resistivity. What is left is $R_w$ , the formation water resistivity at formation temperature (Russian-language literature writes it $\rho_\text{v}$ , with the subscript transliterated from Russian). No tool measures it. A tool “sees” the water in the pores of the uninvaded part of the bed only together with the rock.

$R_w$ sets the scale of the whole calculation. The lower the value of $R_w$ that we use, the lower the water saturation we get for the same bed resistivity. The equation then treats the water as a better conductor, so it needs less room in the pores to explain the measured resistivity. At $n = 2$ the water saturation is proportional to $\sqrt{R_w}$ . Another point matters too. We use one value of $R_w$ for all the beds of an interval. So its error is not random but systematic (it always acts in the same direction). It shifts all the saturations the same way, and it passes in full into the estimate of hydrocarbons in place.

How an error in the water resistivity shifts the water saturation (saturation exponent 2)

Chosen $R_w$: half the true value
$S_w$ multiplier: 0.71
Instead of the true 35%: 25%

Chosen $R_w$: 20% lower
$S_w$ multiplier: 0.89
Instead of the true 35%: 31%

Chosen $R_w$: 25% higher
$S_w$ multiplier: 1.12
Instead of the true 35%: 39%

Chosen $R_w$: twice the true value
$S_w$ multiplier: 1.41
Instead of the true 35%: 49%

Chosen $R_w$: three times the true value
$S_w$ multiplier: 1.73
Instead of the true 35%: 61%

Chosen $R_w$	$S_w$ multiplier	Instead of the true 35%
half the true value	0.71	25%
20% lower	0.89	31%
25% higher	1.12	39%
twice the true value	1.41	49%
three times the true value	1.73	61%

The multiplier is the square root of the ratio of the chosen $R_w$ to the true $R_w$ . An error of two or three times is not rare. It happens when the resistivity of a sample is not converted to formation temperature, or when the equivalent resistivity of a brine is taken as the true one.

Where the water resistivity comes from

There are four sources of data on $R_w$ . The direct one is a water sample taken from the formation. Two indirect ones come from logging. The spontaneous potential depends on how different the salinity of the filtrate is from the salinity of the formation water. And the resistivity of a water-bearing bed whose porosity is known tells us the resistivity of the water itself. The fourth source is the data that the region has already collected: catalogues of formation waters and neighbouring wells.

Sources of data on formation water resistivity

Source: Formation water sample
What is needed: a resistivity measurement at a known temperature, or a chemical analysis
Reliable when: the water comes from the bed under study, after the filtrate has been flushed out
Main pitfalls: filtrate, condensed water, injected water, a mixture of waters

Source: Water catalogue, neighbouring wells
What is needed: data from the same horizon and block, converted to the right temperature
Reliable when: the aquifer (the water-bearing layer) is the same over the whole area
Main pitfalls: salinity that changes across the area and with depth; measurements recorded without a temperature

Source: SP
What is needed: the static amplitude, $R_{mf}$ and the formation temperature
Reliable when: a thick, clean, water-bearing bed; fresh mud; water of the NaCl type (mostly sodium chloride)
Main pitfalls: thin, shaly and oil-bearing beds; a wrong $R_{mf}$ ; brines

Source: Water-bearing bed: $R_{wa}$
What is needed: $R_t$ , porosity, and the coefficients $a$ and $m$
Reliable when: a clean bed that is 100% saturated with water
Main pitfalls: residual hydrocarbons (oil or gas left behind), shale, errors in porosity and in $m$

Source	What is needed	Reliable when	Main pitfalls
Formation water sample	a resistivity measurement at a known temperature, or a chemical analysis	the water comes from the bed under study, after the filtrate has been flushed out	filtrate, condensed water, injected water, a mixture of waters
Water catalogue, neighbouring wells	data from the same horizon and block, converted to the right temperature	the aquifer (the water-bearing layer) is the same over the whole area	salinity that changes across the area and with depth; measurements recorded without a temperature
SP	the static amplitude, $R_{mf}$ and the formation temperature	a thick, clean, water-bearing bed; fresh mud; water of the NaCl type (mostly sodium chloride)	thin, shaly and oil-bearing beds; a wrong $R_{mf}$ ; brines
Water-bearing bed: $R_{wa}$	$R_t$ , porosity, and the coefficients $a$ and $m$	a clean bed that is 100% saturated with water	residual hydrocarbons (oil or gas left behind), shale, errors in porosity and in $m$

A fifth method is a crossplot, a graph of resistivity against porosity for a series of beds. It is in essence the graphical form of the water-bearing bed method (Resistivity–porosity crossplots).

The rule is simple: never trust one source on its own. If two or three independent estimates agree within 10–20%, we can accept the value as well supported. If they differ, the difference itself shows where to look for the error: in the sample, in the SP amplitude, or in the coefficients of Archie's equation. We show how to bring the estimates together in the last section.

Input data: the log heading and the temperature

The indirect methods need the resistivities of the mud $R_m$ , of its filtrate $R_{mf}$ and of the mudcake $R_{mc}$ . Mud is the drilling fluid. The filtrate is its liquid part, which soaks into the rock. The mudcake is the layer of solid mud particles that builds up on the wall of the hole. The crew measures these resistivities at the rig before logging. They take a sample of mud from the flowline or the suction pit (the channel and the tank that the mud passes through). They make the filtrate and the cake in a filter press, a device that squeezes the liquid out of a mud sample. They write the values in the log heading, the block of information at the top of the log, together with the temperature of the measurement (Borehole environment: invaded zone and mudcake). These are resistivities at surface temperature. In the hotter formation, the same liquids conduct current two to three times better.

So the first step of any calculation is to bring $R_{mf}$ (and $R_m$ , for the corrections) to the temperature of the bed under study. We do this with the Arps formula: the resistivity of a mud is inversely proportional to the sum $T + 21.5$ , where $T$ is the temperature in °C. In other words, when $T + 21.5$ doubles, the resistivity halves. We find the formation temperature itself from the geothermal gradient (how fast the temperature rises with depth) and the bottom-hole temperature (measured at the bottom of the well), corrected for the cooling that circulating mud causes. The topic Formation water and formation temperature covers both steps.

Sometimes $R_{mf}$ has not been measured, or we doubt the entry in the heading. Then we estimate it from the mud resistivity.

Estimating the filtrate and mudcake resistivities from the mud resistivity

R_{mf} \approx 0.75\,R_m, \qquad R_{mc} \approx 1.5\,R_m

where:

$R_m$ mud resistivity, Ω·m
$R_{mf}$ resistivity of the mud filtrate, Ω·m
$R_{mc}$ resistivity of the mudcake, Ω·m

Units: Ω·m; all three quantities at the same temperature

Holds when

A statistical relationship for water-based muds in which NaCl is the main salt. It is unreliable for gypsum muds, for muds treated heavily with chemicals, and for heavily weighted muds (muds made dense with added solids).
The scatter is large: tens of per cent. The heavier the mud, the smaller the ratio $R_{mf}/R_m$ usually is. Correlations that take the mud density into account exist.
If a sample has been measured, use the measurement.

The solid part of the mud hardly conducts current. The filtrate has none of it, so it conducts better than the mud. The cake is full of it, so it conducts worse. So usually $R_{mf} < R_m < R_{mc}$ . If the heading shows the opposite order, check the entry again.

Water resistivity from the SP: the full procedure

The static SP amplitude (SSP) is the full deflection that the SP curve would show if no current flowed. For a clean water-bearing bed it depends only on two things: the temperature, and the ratio of the equivalent resistivities of the filtrate and the formation water. The formula is $\mathrm{SSP} = -K\lg(R_{mfe}/R_{we})$ , where $K = 65 + 0.24\,T$ in mV, $T$ is in °C, and $\lg$ is the base-10 logarithm (Spontaneous potential (SP) log). So if we know the filtrate and the temperature, we can find the water from the amplitude. In practice the chain is longer, and every link in it is a source of error.

Calculation procedure

1
Reference bed
Choose a reference water-bearing zone: a thick, clean bed that holds only water, as close as possible to the interval under study. It must contain the same water. Check that it is clean on the gamma ray (GR) log, which records the natural radioactivity of the rock and reads high in shale. Check that it holds water: its resistivity is low and there are no signs of hydrocarbons.
2
Formation temperature
Find it at the depth of the reference bed. Use the temperature gradient between the mean annual temperature at the surface and the corrected bottom-hole temperature.
3
Filtrate
Bring $R_{mf}$ from the heading to formation temperature, and also to 24 °C. If it is greater than 0.1 Ω·m at 24 °C, take $R_{mfe} = 0.85\,R_{mf}$ . This is an empirical rule (it comes from measurements, not from theory) for ordinary fresh muds. If $R_{mf}$ is smaller, the mud is salty, and you find $R_{mfe}$ from the same relationship that links $R_{we}$ and $R_w$ (below).
4
Shale baseline
The shale baseline is the reference level of the SP curve. Draw it through the readings opposite the thick shales above and below the bed. With a fresh mud, it is the far right-hand position of the curve. The baseline can drift slowly with depth. If it shifts noticeably, use the stretch next to the bed.
5
Boundaries and thickness
You need the bed thickness for the correction. Take it from the SP curve itself: the boundaries are where the curve is steepest (the inflection points). Shaly streaks inside a sandstone make the choice harder. Then refine the boundaries from the GR and the resistivity curves.
6
Amplitude
Read the deflection in millivolts, with its sign, from the shale baseline to the largest deflection in the bed.
7
Correction for thickness and resistivity
Multiply the amplitude by the correction factor from the chart. The factor depends on three things: the bed thickness; how many times the resistivity of the invaded zone (the part of the bed around the hole that the filtrate has entered) exceeds the mud resistivity; and the diameter of invasion. The result is the static SP amplitude, SSP.
8
Equivalent water resistivity
Calculate $K$ , then the ratio $R_{mfe}/R_{we} = 10^{-\mathrm{SSP}/K}$ , and divide $R_{mfe}$ by it to get $R_{we}$ .
9
True resistivity
Go from $R_{we}$ to $R_w$ with a chart or a formula that approximates it. For waters of moderate salinity the two are almost equal. For brines they differ several times over.
10
Check
Compare the result with the limit for a saturated brine (a brine that cannot dissolve any more salt), with water samples, and with the $R_{wa}$ of water-bearing beds.

We need the seventh step because the SP current splits between the mud, the bed and the shales. Opposite a bed of limited thickness, the curve records only part of the static amplitude (more). The correction charts are empirical, which means they come from measurements. They use three inputs: the bed thickness; the ratio $R_i/R_m$ of the resistivity of the invaded zone to the mud resistivity; and the diameter of the invaded zone. We take the resistivity of the invaded zone from a shallow sonde, a tool that looks only a short way into the rock.

What share of the static amplitude is recorded on the SP curve

Bed thickness: 6 m or more
Low-resistivity invaded zone: almost 100%
High-resistivity invaded zone: about 90%

Bed thickness: 3 m
Low-resistivity invaded zone: 94–97%
High-resistivity invaded zone: 70–80%

Bed thickness: 1.5 m
Low-resistivity invaded zone: 83–92%
High-resistivity invaded zone: 40–55%

Bed thickness: 0.9 m
Low-resistivity invaded zone: 60–80%
High-resistivity invaded zone: 20–30%

Bed thickness	Low-resistivity invaded zone	High-resistivity invaded zone
6 m or more	almost 100%	about 90%
3 m	94–97%	70–80%
1.5 m	83–92%	40–55%
0.9 m	60–80%	20–30%

These are rough values from one of the empirical charts, for a borehole about 200 mm across. “Low-resistivity” means $R_i/R_m$ of about 5. “High-resistivity” means 100–200, with an invasion diameter of 0.75–1 m. The correction factor is the reciprocal of this share (one divided by the share). For the calculation, use a chart or a program (Interpretation charts: how to read them).

The thinner the bed and the higher the resistivity of the invaded zone, the larger the factor, and the less we can trust it. With a factor of 1.5–2, an error of half a metre in the thickness changes the amplitude by tens of per cent. That is why we choose a thick reference bed.

The last link is the step from the equivalent resistivity to the true one. The EMF (electromotive force, the voltage source) of the SP comes from the chemical activity of the ions in the water, not from how well the solution conducts electricity. The equivalent resistivity is the resistivity that goes with this chemical activity. In dilute solutions (solutions with little salt) activity and conductivity go together, and $R_{we} \approx R_w$ . In brines the picture changes. As the concentration rises, the chemical activity grows faster than the conductivity. The ions hinder one another's movement more and more, but they “act” chemically with more and more force. So the equivalent resistivity of a brine is noticeably lower than the true one. The conversion charts come from measurements on solutions. For calculations, a formula that approximates them is more convenient.

Converting the equivalent resistivity of a brine to the true one (a formula fitted to the chart)

R_{w,24} \approx \frac{77\,R_{we,24} + 5}{146 - 337\,R_{we,24}}

where:

$R_{w,24}$ true resistivity of NaCl-type water at 24 °C, Ω·m
$R_{we,24}$ equivalent water resistivity converted to 24 °C with the Arps formula, Ω·m

Units: empirical: both resistivities in Ω·m at 24 °C (75 °F)

Holds when

The water is close to a NaCl solution. The formula applies when $R_{we,24}$ is not greater than 0.12 Ω·m, that is, when the salinity is above about 50 g/kg.
The procedure is: bring $R_{we}$ from formation temperature to 24 °C, convert it, and bring $R_w$ back to formation temperature. The real chart for different temperatures is more complicated. At formation temperatures up to 150–200 °C the formula differs from it by up to about 10%, and at higher temperatures by more.
The numbers 77, 5, 146 and 337 are fitting coefficients. They were chosen to make the curve fit the chart, and they have no physical meaning.
When $R_{we,24}$ is greater than 0.12 Ω·m, take $R_w \approx R_{we}$ for NaCl-type waters, to within about 10%. For fresh waters of complex composition the relationship is different (see the advanced section).

As $R_{we,24}$ gets smaller, the result does not go to zero. It tends to $5/146 \approx 0.034$ Ω·m, the resistivity of an almost saturated solution (one that is nearly full of salt). The true resistivity of a brine is “pressed” against this limit, however large the SP amplitude. At the edge of the range where the formula applies, at 0.12 Ω·m, the correction is about 12%.

In other unit systems (1)

Filtrate of a salty mud: resistivity at 24 °C below 0.1 Ω·m

R_{mfe,24} \approx \frac{146\,R_{mf,24} - 5}{337\,R_{mf,24} + 77}

$R_{mf,24}$ and $R_{mfe,24}$ are the filtrate resistivity and its equivalent resistivity at 24 °C, Ω·m.

The same relationship, solved the other way round to give the equivalent resistivity. At $R_{mf,24} = 0.1$ Ω·m it gives 0.087, almost the same as the rule $0.85\,R_{mf}$ .

CalculatedHow many times the true resistivity of a NaCl solution is larger than the equivalent one, plotted against salinity. We calculated it from the formula above and from the relation between resistivity and concentration at 24 °C. The point is the water of the example below. Beyond about 200 g/kg both approximations become rough, so the end of the curve is only approximate. For a saturated solution (about 260 g/kg) the ratio from the chart is between four and five.

Worked exampleFormation brine resistivity from the SP: a full calculation

Given (assumed data)

Reference bed: clean water-bearing sandstone 4 m thick (from the SP)
SP deflection from the shale baseline: −95 mV
Correction factor for thickness and resistivity (from the chart): 1.10
Formation temperature $T$: 95 °C
Filtrate resistivity (measured on a sample): 0.46 Ω·m at 20 °C
Mud: fresh, water-based, with mostly NaCl as the salt

Solution

Static amplitude: $\mathrm{SSP} = -95\cdot 1.10 = -104.5$ mV.
Filtrate at formation temperature: $R_{mf} = 0.46\cdot\dfrac{20 + 21.5}{95 + 21.5} = 0.164$ Ω·m. At 24 °C it would be 0.42 Ω·m. That is more than 0.1, so $R_{mfe} = 0.85\,R_{mf} = 0.139$ Ω·m.
Coefficient: $K = 65 + 0.24\cdot 95 = 87.8$ mV. Resistivity ratio: $R_{mfe}/R_{we} = 10^{104.5/87.8} = 15.5$ .
Equivalent water resistivity: $R_{we} = 0.139/15.5 = 0.0090$ Ω·m at 95 °C. At 24 °C this is 0.0230 Ω·m, far below 0.12, so the water is a brine.
True resistivity at 24 °C: $R_{w,24} = \dfrac{77\cdot 0.0230 + 5}{146 - 337\cdot 0.0230}$ , that is 0.0490 Ω·m. At 95 °C, $R_w = 0.0191$ Ω·m, which is 2.1 times the equivalent resistivity.
Checks. The salinity from $R_{w,24}$ is about 171 g/kg: a strong brine, but not a saturated one. A saturated NaCl solution has about 0.04 Ω·m at 24 °C, that is, about 0.016 Ω·m at 95 °C. The $R_w$ found is above this limit, while $R_{we}$ is below it, so $R_{we}$ cannot be the resistivity of real water.

Answer: The formation water resistivity from the SP is about 0.0191 Ω·m at 95 °C.

What it means. If we had stopped at the equivalent resistivity, we would have underestimated $R_w$ by a factor of 2.1, and the water saturation from Archie's equation by a factor of 1.46. The thickness correction changed the equivalent resistivity by 28%, but the true resistivity by only 9%. For brines the true resistivity depends only weakly on the amplitude, because it is pressed against the limit of a saturated solution. The same limit is a quick check of any value of $R_w$ . The chart approximation has its own error, up to 10%, so we check the result against a water sample and against the water-bearing beds (next section). The data in this example are assumed for illustration.

Water resistivity from water-bearing beds

The second indirect method uses Archie's equation itself. Take a clean bed that is fully saturated with water. The resistivity of the rock is $R_0 = F R_w$ , where $F = a/\phi^m$ is the formation resistivity factor (how many times a water-filled rock resists current more than the water alone does). So if we measure the resistivity and porosity of such a bed, we can calculate the water resistivity: $R_w = R_t/F$ . The difficulty is that we do not know in advance which beds are 100% water-bearing. So we do it differently. We calculate the same quantity for all the reservoirs in turn, as if they were all water-bearing. We call it the apparent water resistivity $R_{wa}$ .

Apparent water resistivity

R_{wa} = \frac{R_t}{F} = \frac{R_t\,\phi^m}{a}

where:

$R_{wa}$ apparent water resistivity, Ω·m
$R_t$ resistivity of the uninvaded part of the bed: the reading of the deep sonde (a tool that looks far into the rock) after corrections, Ω·m
$F$ formation resistivity factor, $F = a/\phi^m$ , —
$\phi$ porosity, fraction
$a,\ m$ tortuosity factor (a measure of how winding the paths through the pores are) and cementation exponent, —

Units: Ω·m; the result refers to the bed temperature

Holds when

A clean bed with intergranular porosity (pores between the grains). Shale and conductive minerals distort the result.
$R_t$ has been corrected for the borehole, the enclosing rocks and invasion (Resistivity logs: electrical and induction tools). With shallow invasion, take the reading of the deep sonde.
Use the same porosity and the same $a$ and $m$ that will later go into the saturation calculation.
The rock type (lithology) and the water are the same in all the beds that we compare.

Archie's equation gives $R_t = F R_w S_w^{-n}$ , so $R_{wa} = R_w/S_w^n$ . In a water-bearing bed, $R_{wa} = R_w$ . The less water in the pores, the larger $R_{wa}$ is. We take as $R_w$ the level where the lowest values in several clean reservoirs line up.

Water saturation from the apparent water resistivity

S_w = \left(\frac{R_w}{R_{wa}}\right)^{1/n}

where:

$S_w$ water saturation, fraction
$R_w$ chosen formation water resistivity at bed temperature, Ω·m
$R_{wa}$ apparent water resistivity in the bed under study, Ω·m
$n$ saturation exponent, —

Units: fraction; the resistivities in the same units

Holds when

The same as for $R_{wa}$ . We find the exponent $n$ from core; without core data we usually assume 2.
$R_w$ and $R_{wa}$ refer to the same temperature. If the depths differ a lot, convert $R_w$ to the temperature of the bed.

This is Archie's equation written in terms of $R_{wa}$ . At $n = 2$ , an $R_{wa}$ three times larger than $R_w$ means $S_w \approx 0.58$ . Four times larger gives 0.5, and ten times larger gives about 0.32. This gives a practical rule: an $R_{wa}$ greater than about $3R_w$ is a sign of possible hydrocarbons.

Working procedure

1
Reservoirs
Pick out the permeable, clean beds (beds that let fluid flow through) from the GR, the SP and the caliper log, which records the diameter of the hole. Leave shales and tight streaks out of the comparison.
2
Resistivity
For each bed, take $R_t$ from the deep sonde, with the necessary corrections.
3
Porosity and formation resistivity factor
Take the porosity from the density, neutron or sonic log (Porosity from the density log). Take $F$ from the relationship that we have chosen for this rock.
4
Apparent water resistivity
Calculate $R_{wa}$ for each bed and plot the values against depth.
5
Comparison
Find the level where the lowest values of several clean beds line up. This is $R_w$ . Beds where $R_{wa}$ is three or more times larger are possible accumulations of oil or gas.

Worked exampleApparent water resistivity in seven beds

Given (assumed data)

Section: sandstones and shales at 2770–2820 m and 95 °C, the same interval as in the SP example
Coefficients: $a = 1$ , $m = n = 2$
Bed A, 2772–2776 m: $\phi = 0.22$ ; $R_t = 8.6$ Ω·m
Bed B, 2780–2783 m: $\phi = 0.19$ ; $R_t = 4.9$ Ω·m
Bed C, 2786–2788 m: $\phi = 0.21$ ; $R_t = 1.15$ Ω·m
Bed D, 2790–2795 m: $\phi = 0.23$ ; $R_t = 0.37$ Ω·m
Bed E, 2799–2803 m: $\phi = 0.20$ ; $R_t = 0.49$ Ω·m
Bed F, 2808–2812 m: $\phi = 0.17$ ; $R_t = 0.70$ Ω·m
Bed G, 2816–2819 m: $\phi = 0.18$ ; $R_t = 0.95$ Ω·m; shaly on the GR log

Solution

Formation resistivity factor $F = 1/\phi^2$ : A — 20.7; B — 27.7; C — 22.7; D — 18.9; E — 25.0; F — 34.6; G — 30.9.
Apparent water resistivity $R_{wa} = R_t/F$ , Ω·m: A — 0.416; B — 0.177; C — 0.0507; D — 0.0196; E — 0.0196; F — 0.0202; G — 0.0308.
The three clean beds D, E and F have different porosities but give almost the same value: from 0.0196 to 0.0202 Ω·m, a spread of 3%. This is how water-bearing beds behave. We take their mean: $R_w = 0.0198$ Ω·m.
The ratio $R_{wa}/R_w$ : A — 21; B — 8.9; C — 2.6; G — 1.6. In beds A and B it is much greater than three, so these beds contain hydrocarbons. Bed C is close to the boundary value of three.
Water saturation $S_w = \sqrt{R_w/R_{wa}}$ : A — 0.22; B — 0.33; C — 0.62.

Answer:

R_w \approx 0.0198

Ω·m at formation temperature. Beds A and B are productive (

S_w

about 22 and 33%). Bed C is a transition zone, the interval just above the oil–water contact where the water saturation falls gradually with height (

S_w

about 62%).

What it means. The value is close to the one from the SP (0.0191 Ω·m): two independent methods agree to within 3.5%. The shaly bed G has an $R_{wa}$ that is 1.6 times higher than the water level. But this is not a sign of hydrocarbons. In brines, a shaly rock conducts worse than a clean water-bearing sandstone, and Archie's equation for clean rocks does not apply to it. We leave such beds out of the comparison. The data in this example are assumed for illustration.

CalculatedApparent water resistivity of each bed in the example. The solid vertical line is the chosen

R_w

, and the dashed line is

3R_w

. The water-bearing beds D, E and F lie on the

R_w

line. The productive beds A and B are far to the right. The transition zone C is at the boundary. We leave the shaly bed G out of the comparison.

The method has a big advantage: it needs nothing but the ordinary set of logs. In interpretation programs, $R_{wa}$ is often calculated as a continuous curve over the whole section. But the method has strict conditions.

Clean beds. With fresh formation water, shales conduct better than a water-bearing sandstone, so they lower $R_{wa}$ . With brines it is the other way round: a shaly rock conducts worse, so it raises $R_{wa}$ . In both cases we leave the shaly beds out of the comparison (Water saturation in shaly sands).
Reliable $R_t$ and porosity. If fresh filtrate has invaded deeply, the deep sonde reads too high, and so $R_{wa}$ comes out too high. A washout (a place where the hole is wider than the bit) makes the density log read too much porosity, and this also raises $R_{wa}$ . In the opposite direction, an $R_t$ that is too low (in a thin bed between conductive shales) and a porosity that is too low make $R_{wa}$ too low.
One water and one rock. Further down the section, the water may have a different salinity, or a bed may be a different rock with different $a$ and $m$ . Then you cannot compare the $R_{wa}$ levels.
Truly water-bearing beds. Residual oil (oil that stays behind in the pores) raises $R_{wa}$ . It is found below the oil–water contact (the level where oil gives way to water) and in a zone flushed by water during development. So we look at the lowest values and check that there are several of them.

Water samples and regional data

A formation water sample is the only direct source. The laboratory measures its resistivity at a known temperature, and we convert the result to formation temperature. If we have only a chemical analysis, we first find from it the equivalent NaCl concentration (the concentration of a NaCl solution that conducts the same as this water), and from that the resistivity (Formation water and formation temperature). A measurement is more reliable than a calculation from an analysis, because it does not depend on how closely the water resembles a NaCl solution.

The main question about a sample is whether it is the right water. A sample is representative (it truly stands for the formation water) if it was taken from the formation under study and its composition stopped changing during sampling.

Filtrate. The first liquid that the bed gives up in a well test is mud filtrate from the invaded zone. So we take several samples as the inflow continues, and we wait until the salinity and the resistivity settle at a constant level.
Condensed water. In gas and gas-condensate wells, the gas carries out almost fresh water that has condensed from water vapour. This water has nothing to do with the water in the pores of the formation.
Injected water. In fields that are being produced, the water that comes out of the well contains a share of injected water, usually less salty. Such a sample is no good for beds that injection has not reached. In beds that injection has flushed, the proportion of the two waters is itself unknown.
A mixture of waters. A well that produces from several beds gives a mixture. A sample from the wellhead (the top of the well) belongs to none of the beds.
Water from the wellbore. If the inflow is weak, the sample picks up mud and service water (water used in well operations) that were left in the casing (the steel pipe that lines the well).

When there is no sample from the well, we turn to regional data: catalogues of formation waters, reports on neighbouring wells, and salinity maps for each horizon. Use them with care. Salinity changes across the area and with depth, sometimes abruptly, at faults and between blocks that are hydrodynamically isolated (no fluid flows between them). In salt basins (basins that contain thick salt layers), the waters near the salt are brines close to saturation, and farther from the salt they are noticeably fresher. So take the value for the same horizon and block, and always convert it to the temperature of your own bed. Old records often give the resistivity at laboratory temperature, and sometimes with no temperature at all.

How to bring the estimates together, and what an error costs

Archie's equation shows that the relative error of the water saturation (the error as a fraction of the value) equals the relative error of $R_w$ divided by the exponent $n$ (derivation). At $n = 2$ , a 10% error in $R_w$ changes $S_w$ by 5% of its value. At $S_w = 0.30$ that is 1.5 percentage points (for example, from 30% to 31.5%). That is not much, as long as the error is a few per cent and not several times over. The aim of bringing the estimates together is to remove exactly the gross errors (the big ones).

Bringing the results together

1
One temperature
Compare only values that have been converted to the temperature of the same bed.
2
Judging each value
For a sample, check that it is representative. For the SP, check the thickness and cleanness of the reference bed and the reliability of $R_{mf}$ . For $R_{wa}$ , check that the beds are clean, and check the porosity and the coefficients $a$ and $m$ .
3
Comparison
A difference of up to 10–20% is usually within the accuracy of the methods. If it is larger, look for the cause. Do not just average.
4
Sense check
Does the salinity fit the region and the depth? Is $R_w$ at least as high as the limit for a saturated brine? Does the calculation give $S_w \approx 1$ in beds that are known to be water-bearing?
5
Value and range
Put the value into the calculation together with its uncertainty (the range in which it may lie). In a long interval, convert $R_w$ to the temperature of each bed.

What a disagreement between the estimates says

What is seen: $R_w$ from the SP is larger than $R_w$ from the sample and from $R_{wa}$
Likely cause: the amplitude is too small: the reference bed is thin, shaly or oil-bearing; or $R_{mf}$ was not converted to formation temperature
What to check: the thickness and cleanness of the bed, the correction, the temperature at which $R_{mf}$ was measured

What is seen: $R_w$ from the SP is several times smaller than the others
Likely cause: the equivalent resistivity of a brine was taken as the true one; or a streaming potential made the amplitude too large
What to check: the step from $R_{we}$ to $R_w$ , and the overbalance (how much the mud pressure exceeds the formation pressure)

What is seen: The sample gives a larger $R_w$ than the logs
Likely cause: fresh filtrate or condensed water has diluted the sample
What to check: how the salinity changed during sampling

What is seen: $R_{wa}$ of water-bearing beds is smaller than $R_w$ from the sample
Likely cause: $m$ is too high, the porosity or $R_t$ is too low, or there is shale with fresh water
What to check: the core, the lithology, the resistivity corrections

What is seen: $R_{wa}$ of “water-bearing” beds is larger than $R_w$ from the sample
Likely cause: residual hydrocarbons, $m$ too low, porosity too high, or deep invasion of fresh filtrate
What to check: signs of oil in the core and in the cuttings (rock chips brought up by the mud), the caliper log, the invaded zone

What is seen	Likely cause	What to check
$R_w$ from the SP is larger than $R_w$ from the sample and from $R_{wa}$	the amplitude is too small: the reference bed is thin, shaly or oil-bearing; or $R_{mf}$ was not converted to formation temperature	the thickness and cleanness of the bed, the correction, the temperature at which $R_{mf}$ was measured
$R_w$ from the SP is several times smaller than the others	the equivalent resistivity of a brine was taken as the true one; or a streaming potential made the amplitude too large	the step from $R_{we}$ to $R_w$ , and the overbalance (how much the mud pressure exceeds the formation pressure)
The sample gives a larger $R_w$ than the logs	fresh filtrate or condensed water has diluted the sample	how the salinity changed during sampling
$R_{wa}$ of water-bearing beds is smaller than $R_w$ from the sample	$m$ is too high, the porosity or $R_t$ is too low, or there is shale with fresh water	the core, the lithology, the resistivity corrections
$R_{wa}$ of “water-bearing” beds is larger than $R_w$ from the sample	residual hydrocarbons, $m$ too low, porosity too high, or deep invasion of fresh filtrate	signs of oil in the core and in the cuttings (rock chips brought up by the mud), the caliper log, the invaded zone

Worked exampleThree estimates of the water resistivity and the cost of an error

Given (assumed data)

From the SP (example above): 0.0191 Ω·m at 95 °C
From water-bearing beds D, E and F: 0.0198 Ω·m
Water sample from bed D during a test: 0.052 Ω·m at 20 °C; the resistivity of the samples has stabilised
Apparent water resistivity in beds A and C: 0.416 and 0.0507 Ω·m
Saturation exponent: $n = 2$

Solution

The sample at formation temperature: $0.052\cdot\dfrac{20 + 21.5}{95 + 21.5} = 0.0185$ Ω·m.
The three estimates — 0.0185, 0.0191 and 0.0198 Ω·m — differ by 7% from the smallest to the largest. We take the mean: $R_w = 0.0192$ Ω·m, with an uncertainty of about ±3%.
Productive bed A: $S_w = \sqrt{0.0192/0.416} = 0.215$ . At the two ends of the range, $S_w$ is 0.211 and 0.218. That is ±1.7% of the value, half the uncertainty of $R_w$ .
Suppose we had taken the equivalent resistivity, 0.0090 Ω·m, as $R_w$ . Then bed A would give 0.15, and the transition zone C would give 0.42 instead of 0.61.
Suppose we had believed the first sample, which the filtrate had diluted (0.11 Ω·m at 20 °C, that is 0.0392 Ω·m in the bed). Then bed A would give 0.31 and bed C 0.88.

Answer:

R_w = 0.0192

Ω·m with an uncertainty of about ±3%. The water saturation of bed A is about 21%, and because of

R_w

it is uncertain by less than half a percentage point.

What it means. Three independent methods agree, so the contribution of $R_w$ to the error in saturation is small. It is smaller than the error that the exponents $m$ and $n$ usually bring. The two gross errors, in contrast, change the conclusion about the bed. With the equivalent resistivity instead of the true one, the transition zone C looks productive (42% water instead of 61%). With the diluted sample, bed C is “watered out” to 88%, and the good bed A looks mediocre. The data in this example are assumed for illustration.

In a long interval the temperature changes too, and so does $R_w$ . With a gradient of 3 °C per 100 m, five hundred metres of section means 15 °C of temperature change. At a formation temperature of about 100 °C, the resistivity of the same water falls by 11–13% over this distance. So we do not put one number into the calculation. We put in the salinity of the water (or its resistivity at 24 °C) and the temperature of each bed.

Terms and symbols

Formation water resistivity $R_w$: The resistivity of the water in the pores at formation temperature. Russian-language literature writes it $\rho_\text{v}$ (the subscript is transliterated from Russian).
Equivalent resistivity $R_{we}$: The resistivity that is inversely proportional to the chemical activity of the water. For brines it is several times smaller than the true resistivity.
Apparent water resistivity $R_{wa}$: $R_t/F$ . It is the water resistivity that would explain the resistivity of the bed if the bed were 100% water-bearing.
Reference water-bearing zone: A thick, clean bed that holds the same water as the interval under study. We find the static SP amplitude and $R_{wa}$ from it.
$R_m$ , $R_{mf}$ , $R_{mc}$: The resistivities of the mud, its filtrate and the mudcake. They are measured on a sample and written in the log heading with the temperature of the measurement.
SP correction for bed thickness and resistivity: A factor that converts the recorded SP deflection into the static amplitude. It is close to 1 in thick, low-resistivity beds.
Formation water catalogue: A summary of the analyses and resistivities of waters in a region, listed by horizon and by area.

Common mistakes and misconceptions

Wrong

“The SP gave $R_{we}$ — that is $R_w$ .”

In fact

That holds only for waters of moderate salinity. For brines the equivalent resistivity is several times smaller than the true one. Without the conversion, $R_w$ and the water saturation come out too low, and the hydrocarbons in place come out too high.

Wrong

“ $R_w$ is the lowest $R_{wa}$ in the section.”

In fact

Take the lowest values only among clean reservoirs with reliable $R_t$ and porosity, and there must be several of them. A single minimum more often points to an error in porosity or resistivity.

Wrong

“A water sample from the well is the final truth.”

In fact

A sample may turn out to be filtrate, condensed water, injected water or a mixture of waters. Check it against the SP, $R_{wa}$ and regional data, and see whether the composition settled during sampling.

Wrong

“ $R_w$ was taken from the neighbouring well — no need to check.”

In fact

Make sure that it is the same horizon and block, and convert the value to the temperature of your own bed. Salinity changes across the area and with depth, sometimes abruptly.

Wrong

“The $R_{wa}$ of the water-bearing beds did not match the sample — so the sample is bad.”

In fact

Perhaps the coefficients $a$ and $m$ or the porosity do not suit the rock. $R_{wa}$ absorbs their errors. The difference is a reason to check the core, not to throw out one of the results.

In depth: advanced detail

Why apparent water resistivity is a calibration, not only a property of the water

Suppose the true formation resistivity factor of the rock is $1/\phi^{m_0}$ , but we have put a different exponent $m$ into the calculation. In a water-bearing bed of porosity $\phi_1$ we then get $R_{wa} = R_w\,\phi_1^{\,m - m_0}$ , which is not the true water resistivity. But suppose we put this $R_{wa}$ , together with the same $m$ , into Archie's equation for a productive bed of porosity $\phi_2$ . Then the error almost cancels out.

Water saturation with a wrong cementation exponent, calibrated on a water-bearing bed

S_{w,\text{calc}} = S_w\left(\frac{\phi_1}{\phi_2}\right)^{(m - m_0)/n}

where:

$S_{w,\text{calc}}$ calculated water saturation of the bed under study, fraction
$S_w$ its true water saturation, fraction
$\phi_1,\ \phi_2$ porosity of the reference water-bearing bed and of the bed under study, fraction
$m,\ m_0$ the cementation exponent used in the calculation and the true one, —
$n$ saturation exponent, —

Units: dimensionless

Holds when

Both beds are clean, with the same water and the same relationship between the formation resistivity factor and porosity.
The coefficient $a$ is the same in both calculations, and we have taken the exponent $n$ correctly.
We take as $R_w$ the $R_{wa}$ of the reference bed, calculated with the same $m$ .

When $\phi_2 = \phi_1$ there is no error at all, whatever $m$ we use: the wrong exponent distorted the reference bed and the bed under study in the same way. The more the porosity of the bed under study differs from that of the reference bed, the worse this calibration works.

Worked exampleA wrong cementation exponent: with and without calibration

Given (assumed data)

True properties: $m_0 = 1.8$ ; $R_w = 0.019$ Ω·m (from a reliable sample)
Used in the calculation: $a = 1$ , $m = 2$ , $n = 2$
Reference water-bearing bed: $\phi_1 = 0.23$
Productive beds: $\phi = 0.22$ with a true $S_w = 0.25$ and $\phi = 0.14$ with $S_w = 0.35$

Solution

Apparent water resistivity in the reference bed: $R_{wa} = 0.019\cdot 0.23^{\,0.2} = 0.0142$ Ω·m. This is 25% lower than the value from the sample.
With this $R_{wa}$ and $m = 2$ , the bed with $\phi = 0.22$ gives $S_w = 0.251$ and the bed with $\phi = 0.14$ gives $S_w = 0.368$ .
With the true $R_w$ from the sample but the same wrong $m = 2$ , the two beds give 0.291 and 0.426.

Answer: The calibrated calculation gives 0.251 and 0.368, against the true 0.25 and 0.35. The calculation with the correct

R_w

and the wrong

m

gives 0.291 and 0.426.

What it means. Here, the difference between the $R_{wa}$ of the water-bearing beds and the reliable sample is not an error in one of the methods. It is a signal that the exponent $m$ does not suit the rock. The right solution is to refine $m$ from core. Until there is core, the consistent pair “ $R_{wa}$ and the same $m$ ” makes a smaller error than the pair “correct $R_w$ and wrong $m$ ”. But this holds only in beds whose porosity is close to that of the reference bed: at $\phi = 0.14$ the error is already noticeable. The data in this example are assumed for illustration.

The same technique on a graph: resistivity against porosity

If we take logarithms, Archie's equation becomes the equation of a straight line.

Archie's equation in logarithmic form

\lg R_t = \lg(a R_w) - m\lg\phi - n\lg S_w

where:

$R_t$ resistivity of the uninvaded part of the bed, Ω·m
$\phi$ porosity, fraction
$S_w$ water saturation, fraction
$a,\ m,\ n$ coefficients of Archie's equation, —
$R_w$ formation water resistivity, Ω·m

Units: resistivities in Ω·m; $\lg$ is the base-10 logarithm

Holds when

Clean beds of one rock type (lithology) with the same water.
$R_t$ and porosity are corrected in the same way as for the calculation of $R_{wa}$ .

For water-bearing beds the last term is zero. Their points on a graph of $\lg R_t$ against $\lg\phi$ lie on a straight line with a slope of $-m$ . At $\phi = 1$ the line gives the value $aR_w$ . Beds with hydrocarbons lie above this line. The graph gives both $m$ and $aR_w$ at once, if the water-bearing beds cover a wide range of porosity (Resistivity–porosity crossplots).

Worked exampleA straight line through three water-bearing beds

Given (assumed data)

Water-bearing beds D, E and F from the example above: $\phi$ : 0.23; 0.20; 0.17 and $R_t$ : 0.37; 0.49; 0.70 Ω·m
Productive bed A: $\phi = 0.22$ ; $R_t = 8.6$ Ω·m; $n = 2$

Solution

A least-squares line (the straight line that fits the points best) on the graph of $\lg R_t$ against $\lg\phi$ has a slope of −2.11, that is $m = 2.11$ . The value at $\phi = 1$ is $aR_w = 0.0165$ Ω·m.
The calculation with $m = 2$ gave $R_w = 0.0198$ Ω·m, which is 20% more.
Water saturation of bed A: 0.217 with the pair “ $m = 2.11$ , $aR_w = 0.0165$ ”, and 0.218 with the pair “ $m = 2$ , $R_w = 0.0198$ ”.

Answer: Both pairs of coefficients give practically the same water saturation for bed A, about 0.217.

What it means. The spread of $R_{wa}$ between the three beds is only 3%. But when we extend the line from a porosity of 0.17–0.23 to a porosity of 1, this small spread becomes a noticeable difference in $aR_w$ . This is the same calibration idea. The pair “ $m$ and $aR_w$ ” is reliable inside the range of porosity it was built on. From a narrow range we cannot find $m$ or $R_w$ accurately, each on its own. The data in this example are assumed for illustration.

Fresh waters and waters of complex composition

The SP formulas were derived for NaCl solutions. In fresh formation waters, calcium, magnesium, bicarbonate and sulfate make up a noticeable share. Divalent cations (ions with two positive charges) create the membrane potential, the voltage across shale, differently from sodium. So, at the same electrical conductivity, such water acts as more “active” in the SP. Its equivalent resistivity is smaller than the true one, and the fresher the water, the stronger the effect. The charts allow for this with an averaged curve for “average” fresh formation waters.

Converting the equivalent resistivity to the true one for fresh formation waters (a formula fitted to the chart)

R_{w,24} \approx 10^{\,0.69\,R_{we,24} - 0.24} - 0.58

where:

$R_{w,24}$ true water resistivity at 24 °C, Ω·m
$R_{we,24}$ equivalent water resistivity brought to 24 °C, Ω·m

Units: empirical: both resistivities in Ω·m at 24 °C (75 °F)

Holds when

$R_{we,24}$ is greater than 0.12 Ω·m, and the water is an “average” fresh formation water with a noticeable share of divalent ions.
The formula averages over waters of different composition, so for one particular water the error may be large.
The numbers 0.69, 0.24 and 0.58 are fitting coefficients. The difference from the chart is about 10%.

At about 0.15 Ω·m the formula gives $R_w \approx R_{we}$ . At 0.3 Ω·m it gives about 15% more, and at 1 Ω·m more than twice as much.

For a pure NaCl solution of the same conductivity there would be no correction. The correction reflects the composition of the water, and we cannot learn that composition from logs. There is more. With fresh formation water, the salinity contrast with the filtrate is small or of opposite sign, so the SP amplitude is small. And in shaly beds the conductivity of the shale is comparable to that of the water itself. So for fresh waters we treat the SP estimate as a rough guide and rely on samples.

Connections to other topics

Spontaneous potential (SP) log — where the SP comes from, what reduces its amplitude, and the formula for the static amplitude.
Formation water and formation temperature — salinity, converting resistivities to formation temperature, and water resistivity from the analysis of a sample.
Archie's equation and saturation of clean reservoirs — how $R_w$ enters the water saturation calculation, and how errors carry over into it.
Resistivity logs: electrical and induction tools — how to get $R_t$ , without which we can find neither $R_{wa}$ nor saturation.
Borehole environment: invaded zone and mudcake — filtrate, cake and the invaded zone; what is written in the log heading.
Porosity from the density log — porosity for the formation resistivity factor, and its error.
Resistivity–porosity crossplots — finding $R_w$ and $m$ from a series of beds on a graph.
Water saturation in shaly sands — why $R_{wa}$ is not equal to $R_w$ in shaly beds.

Check yourself

1
The SP gives $R_{we} = 0.010$ Ω·m at a formation temperature of 90 °C. What can we say about $R_w$ ?
1. A.The water is fresh, and the SP cannot be used for it
2. B. $R_w = 0.010$ Ω·m
3. C. $R_w$ is less than 0.010 Ω·m
4. D. $R_w$ is about twice as large: this is a brine, so we must convert the equivalent resistivity to the true one
Show the answer
Correct answer: D. $R_w$ is about twice as large: this is a brine, so we must convert the equivalent resistivity to the true one
At 24 °C this equals $0.010\cdot 111.5/45.5 \approx 0.0245$ Ω·m. That is far below 0.12, so the water is a brine. The formula fitted to the chart gives $R_{w,24} \approx 0.050$ Ω·m, and at 90 °C about 0.020 Ω·m. Check: a saturated NaCl solution at 90 °C has about 0.016 Ω·m, so 0.010 Ω·m cannot be the resistivity of real water.
2
A clean sandstone has $\phi = 0.25$ , $R_t = 0.80$ Ω·m, $a = 1$ and $m = 2$ . The $R_w$ chosen for the section is 0.045 Ω·m. Find $R_{wa}$ and estimate the water saturation at $n = 2$ .
Show the answer
$F = 1/0.25^2 = 16$ and $R_{wa} = 0.80/16 = 0.050$ Ω·m, only 11% more than $R_w$ . $S_w = \sqrt{0.045/0.050} \approx 0.95$ , so the bed is water-bearing. A difference like this is within the accuracy of the method. Hydrocarbons are suspected when $R_{wa}$ is about three times $R_w$ .
3
At the wellhead of a gas well, a water sample had a salinity of 2 g/L, although in neighbouring wells in this horizon the salinity is about 120 g/L. What kind of water is it?
1. A.Condensed water that dropped out of the gas
2. B.A saturated brine
3. C.Formation water: the water in the horizon is fresher here
4. D.A laboratory error — the sample should simply be thrown away
Show the answer
Correct answer: A. Condensed water that dropped out of the gas
The gas in the formation is saturated with water vapour. As the pressure and temperature fall in the wellbore, the vapour condenses into almost fresh water. This water has nothing to do with the water in the pores of the formation, so we cannot use its resistivity as $R_w$ . It is also wrong to throw the sample away without explanation: finding out what exactly was sampled is part of the job.
4
Opposite a clean water-bearing bed 2 m thick, the SP deflection is −50 mV and the correction factor from the chart is 1.3. The formation temperature is $T = 70$ °C, and $R_{mfe} = 0.20$ Ω·m at formation temperature. Find $R_{we}$ . By what factor does the result change if we leave out the correction?
Show the answer
$\mathrm{SSP} = -50\cdot 1.3 = -65$ mV; $K = 65 + 0.24\cdot 70 = 81.8$ mV; $10^{65/81.8} \approx 6.2$ ; $R_{we} = 0.20/6.2 \approx 0.032$ Ω·m. Without the correction: $10^{50/81.8} \approx 4.1$ and $R_{we} \approx 0.049$ Ω·m, which is one and a half times more. At 24 °C the value 0.032 Ω·m becomes about 0.065 Ω·m. That is a brine, so the true $R_w$ is about 25% higher than $R_{we}$ , about 0.040 Ω·m.
5
Three clean water-bearing beds have porosities of 0.16, 0.21 and 0.26. At $m = 2$ they give $R_{wa}$ values of 0.029, 0.030 and 0.032 Ω·m. A reliable water sample gives 0.045 Ω·m at formation temperature. What does this mean, and what should we do?
Show the answer
$R_{wa}$ is systematically lower than the sample, and it grows with porosity. This happens when the exponent $m$ that we used is greater than the true one: $R_{wa} = R_w\,\phi^{\,m - m_0}$ . Here the data fit a true $m$ of about 1.75. The right step is to refine $a$ and $m$ from core. Until then, we calculate saturation in one of two ways. One is to use $R_w$ from the sample with a corrected $m$ . The other is to use the consistent pair “ $R_{wa}$ and the same $m$ ”. Remember that the second way suits only beds whose porosity is close to that of the reference bed.
6
Why do we take the level of the lowest $R_{wa}$ in clean reservoirs as $R_w$ , and not simply the lowest value in the section?
Show the answer
The formula $R_{wa} = R_t/F$ holds for a clean rock with intergranular porosity. In shales, in shaly beds with fresh water, in thin beds among conductive rocks, and where the porosity is too low, we get values below $R_w$ that have nothing to do with water. The sign of a real water level is that $R_{wa}$ agrees in several clean beds of different porosity.

Updated: 28 October 2026