Resistivity ratio methods and movable oil

Quick refresher

The rock is the same in the flushed and the uninvaded zones, and the formation resistivity factor $F$ (how many times a water-filled rock resists current more than the water alone) is shared. So in a clean bed $R_{xo}/R_t = (R_{mf}/R_w)\,(S_w/S_{xo})^n$ . The porosity, $a$ and $m$ are not needed, but the exponent $n$ stays.
The resistivity ratio directly measures the moveable hydrocarbon index $S_w/S_{xo}$ . $S_w$ itself comes only from a second relation that closes the system, the closure relation. There are two options. One is the empirical relation $S_{xo} = S_w^{1/5}$ . At $n = 2$ it gives the water saturation as the normalised ratio to the power 5/8 (the normalised ratio is the resistivity ratio divided by the water contrast). The other is the residual oil saturation of the flushed zone from core. (The residual oil saturation is the share of the pores in which oil is left after the filtrate has passed. Core means rock samples taken from the well.)
An $S_w/S_{xo}$ near 1 means that the filtrate moved nothing: the bed is water-bearing, or the oil is immobile, or the rock is tight. By an empirical rule, 0.7 and below in sandstones (0.6 and below in carbonates) points to moveable hydrocarbons.
Quick-look method (a fast check made at the rig): the curve $-K\lg(R_{xo}/R_t)$ , matched to the SP in a water-bearing bed, departs from the SP opposite beds with moveable hydrocarbons. The separation $-nK\lg(S_w/S_{xo})$ is about 25 mV at $S_w/S_{xo} = 0.7$ and $K \approx 80$ mV.
The method needs an invasion “window”: the shallow sonde sees the flushed zone, the deep sonde sees the uninvaded formation, and the profile is a step. Shallow invasion, too deep invasion, incomplete flushing and shale with a fresh mud (a mud with little salt) give an optimistic error: the oil looks more moveable than it is. An annulus gives a pessimistic one: it hides moveable oil.
The check is the porosity balance: the saturation from the resistivity ratio must give the same porosity as the porosity methods, that is, it must agree with the saturation from Archie's equation.

Key relationships

\dfrac{R_{xo}}{R_t} = \dfrac{R_{mf}}{R_w}\left(\dfrac{S_w}{S_{xo}}\right)^{n}

ratio of the resistivities of the zones of a clean bed

Q = \dfrac{R_{xo}/R_t}{R_{mf}/R_w},\qquad S_w = Q^{5/8}

the normalised ratio, and the water saturation at $S_{xo} = S_w^{1/5}$ , $n = 2$

S_w = (1 - S_{hr})\,Q^{1/n}

water saturation with a known residual hydrocarbon saturation of the flushed zone

\Delta E = -nK\lg\dfrac{S_w}{S_{xo}}

separation of the curve $-K\lg(R_{xo}/R_t)$ and the SP, in mV

Common pitfalls

Taking $S_w/S_{xo} \approx 1$ as a sign of water: immobile oil and tight rock look the same.
Dividing $R_{mf}$ at the measurement temperature by $R_w$ at formation temperature: the water contrast is calculated at one temperature.
Taking the ratio of the readings of the shallow and deep sondes, without corrections for invasion, and calling it $R_{xo}/R_t$ .
Using the empirical relation $S_{xo} = S_w^{1/5}$ as a law: it describes an “average” oil-bearing sandstone.
Using the method in shaly beds, with an oil-based mud, and in fractured rocks that have no flushed zone.

After this lesson you can

derive the ratio $R_{xo}/R_t$ from Archie's equation for the two zones, and explain why the porosity and the coefficients $a$ and $m$ cancel, but the saturation exponent $n$ does not
calculate the water saturation from the resistivity ratio, using an empirical relation between the saturations of the zones or a known residual oil saturation, and estimate how the choice of relation changes the answer
estimate the moveability of hydrocarbons from the index $S_w/S_{xo}$ , and tell a water-bearing bed from a bed with immobile oil
compare the curve calculated from $R_{xo}/R_t$ with the SP curve, and convert their separation into $S_w/S_{xo}$
list the conditions in which the ratio methods go wrong (invasion, incomplete flushing, which means that some formation water is left in the flushed zone, shale, an annulus, which is a ring of lowered resistivity at the edge of the invaded zone, and the type of mud), with the direction of the error, and check the result with the porosity balance (a check against the porosity from the porosity logs)

Two zones of one bed: where the ratio comes from

Recall the picture around the borehole (resistivity profile). In the flushed zone, the mud filtrate has replaced almost all the formation water and the moveable part of the hydrocarbons. The pores hold filtrate of resistivity $R_{mf}$ and residual hydrocarbons. Beyond the invasion front (the outer edge of the zone that the filtrate has entered) lies the uninvaded formation, with formation water of resistivity $R_w$ and the original hydrocarbons. The rock in both zones is the same: the same grains, the same pores and the same formation resistivity factor $F$ (Archie's equation and saturation of clean reservoirs).

Write Archie's equation for each zone: $R_t = F R_w S_w^{-n}$ and $R_{xo} = F R_{mf} S_{xo}^{-n}$ . If we divide the second by the first, the formation resistivity factor cancels, and with it the porosity and the coefficients $a$ and $m$ :

Ratio of the resistivities of the flushed zone and the uninvaded formation

\frac{R_{xo}}{R_t} = \frac{R_{mf}}{R_w}\left(\frac{S_w}{S_{xo}}\right)^{n}

where:

$R_{xo}$ resistivity of the flushed zone, Ω·m
$R_t$ resistivity of the uninvaded part of the formation, Ω·m
$R_{mf}$ resistivity of the mud filtrate, Ω·m
$R_w$ resistivity of the formation water, Ω·m
$S_w$ water saturation of the uninvaded formation, fraction
$S_{xo}$ water saturation of the flushed zone (the water in it is filtrate), fraction
$n$ saturation exponent, —

Units: any identical units of resistivity (usually Ω·m); $R_{mf}$ and $R_w$ at the same temperature

Holds when

A clean bed with intergranular porosity: only the water in the pores conducts current.
Archie's equation with the same formation resistivity factor and the same exponent $n$ in both zones.
The water of the flushed zone is filtrate, and that of the uninvaded formation is formation water. The mud is water-based.
$R_{xo}$ and $R_t$ are the resistivities of the zones after corrections for the borehole, the mudcake, the enclosing rocks and invasion, not the readings of the sondes.

At $n = 2$ this is the formula from the topic on the resistivity profile. Everything that describes the pore space cancels, but the saturation exponent does not. It describes how the saturation changes the paths of the current, and the saturations of the zones are different. In a water-bearing bed $S_w = S_{xo} = 1$ , and the resistivity ratio equals the water contrast.

It is convenient to divide the resistivity ratio by the water contrast. This gives a quantity that we call the normalised ratio $Q$ in this lesson. It no longer depends on the properties of the waters, only on the saturations.

Normalised resistivity ratio and the moveable hydrocarbon index

Q = \frac{R_{xo}/R_t}{R_{mf}/R_w} = \left(\frac{S_w}{S_{xo}}\right)^{n}

where:

$Q$ normalised resistivity ratio, —
$S_w/S_{xo}$ index of the moveable hydrocarbons (the moveable hydrocarbon index), —
$R_{xo},\ R_t,\ R_{mf},\ R_w,\ n$ as in the resistivity ratio above, Ω·m; —

Units: dimensionless

Holds when

The same as for the ratio of the resistivities of the zones.

$Q = 1$ means that the filtrate did not move the hydrocarbons, or that there are none. $Q < 1$ means that it moved them. The $n$ -th root of $Q$ is the ratio $S_w/S_{xo}$ , the index of moveable hydrocarbons. The more hydrocarbons the filtrate pushed out, the smaller $Q$ is.

The water contrast $R_{mf}/R_w$ does not depend on temperature, if both resistivities refer to the same one. By the Arps formula (the rule for converting the resistivity of salty water from one temperature to another; resistivity and temperature), both are converted by the same factor, and it cancels. An error appears when $R_{mf}$ from the log heading (the block of data at the top of the log), taken at the measurement temperature, is divided by $R_w$ brought to formation temperature.

We take the contrast itself from two sources. The first is measured values: $R_{mf}$ from the log heading, and $R_w$ from water samples, the SP and water-bearing beds (Determining formation water resistivity). The second is the SP curve. In a clean water-bearing bed its static amplitude (the full size of its deflection) equals $-K\lg(R_{mfe}/R_{we})$ (static amplitude). The ratio of the equivalent resistivities (the values to which the SP responds) is close to the ratio of the true ones, although not equal to it. The equivalent resistivities of the filtrate, and especially of brines, differ from the true ones (more). How this is handled is shown in the quick-look method below.

Worked exampleWater contrast at one temperature, and a check against a water-bearing bed

Given (assumed data)

Filtrate resistivity (log heading): 0.90 Ω·m at 20 °C
Formation water sample (the composition has stabilised): 0.130 Ω·m at 20 °C
Formation temperature: 74 °C
Bed 1, water-bearing, after corrections: $R_{xo}$ = 8.08 Ω·m, $R_t$ = 1.17 Ω·m

Solution

The resistivities at formation temperature by the Arps formula: $R_{mf} = 0.90\cdot\dfrac{20 + 21.5}{74 + 21.5} = 0.391$ Ω·m, $R_w = 0.130\cdot\dfrac{41.5}{95.5} = 0.0565$ Ω·m.
The water contrast at 74 °C: $0.391/0.0565 = 6.92$ . At 20 °C: $0.90/0.130 = 6.92$ . It is the same: the Arps factor cancelled.
Bed 1: $R_{xo}/R_t = 8.08/1.17 = 6.91$ and $Q = 6.91/6.92 = 0.999$ . This is a water-bearing bed, as it should be.
The error: $R_{mf}$ at 20 °C was divided by $R_w$ at 74 °C. The “contrast” is $0.90/0.0565 = 15.9$ , which is 2.3 times too large. For the same bed, $Q = 6.91/15.9 = 0.435$ and $S_w/S_{xo} = \sqrt{0.435} = 0.66$ .

Answer: The water contrast is 6.92 at any common temperature. In the water-bearing bed 1,

Q

= 0.999. Mixing the temperatures would raise the contrast 2.3 times, and would give “moveable oil” in a water-bearing bed (

S_w/S_{xo}

= 0.66).

What it means. For the ratio method, temperature matters in one way only: both resistivities must refer to the same temperature. If both were measured in the laboratory at one temperature, we can divide them at once. A water-bearing bed with a known water is the best check. If $Q$ in it differs markedly from one, the error is in the water contrast or in $R_{xo}$ and $R_t$ , and until we remove it, it makes no sense to calculate the saturations of other beds. The data in this example are assumed for illustration.

The second relation: how much oil is left in the flushed zone

One equation links two unknown saturations. The resistivity ratio gives only their ratio $S_w/S_{xo}$ . To find $S_w$ , we need one more relation, one that tells how many hydrocarbons the filtrate left at the wall. It is called the closure relation, because it closes the system: two unknowns need two equations. There are two ways to get it.

An empirical relation. For oil-bearing beds with moderate invasion and an ordinary residual oil saturation, a relation has been published: $S_{xo} = S_w^{1/5}$ . This is an empirical average, not a law. In a particular bed, noticeably more or less oil may be left at the wall (see the example below). We put it into the normalised ratio, and we get the water saturation without porosity:

Water saturation from the resistivity ratio, with the empirical relation between the saturations of the zones

S_w = Q^{5/8} = \left(\frac{R_{xo}/R_t}{R_{mf}/R_w}\right)^{5/8}

where:

$S_w$ water saturation of the uninvaded formation, fraction
$Q$ normalised resistivity ratio, —
$R_{xo}/R_t$ ratio of the resistivities of the zones after corrections, —
$R_{mf}/R_w$ water contrast at one temperature, —

Units: dimensionless

Holds when

The empirical relation $S_{xo} = S_w^{1/5}$ : an oil-bearing bed, moderate invasion, an ordinary residual oil saturation. It has not been checked for gas, heavy oil, carbonates with vugs and oil-wet rocks.
The saturation exponent is $n = 2$ . At another $n$ the power is $5/(4n)$ (advanced section).
All the assumptions of the ratio of the resistivities of the zones.

The substitution gives $(S_w^{4/5})^2 = Q$ , and so the power 5/8. The relation sets the residual hydrocarbon saturation of the flushed zone, $1 - S_w^{1/5}$ : about 0.28 at $S_w = 0.2$ , 0.13 at $S_w = 0.5$ and 0.04 at $S_w = 0.8$ . By this relation, the less water there is in the bed, the more oil is left at the wall.

A known residual saturation. Suppose we know the residual hydrocarbon saturation of the flushed zone, $S_{hr}$ . We may have it from core, where experiments push oil out of the rock with filtrate or water (Relative permeability). Or we may have it from neighbouring wells, where $S_{xo}$ was found from microresistivity tools and porosity (more). Then $S_{xo} = 1 - S_{hr}$ , and the relation is exact in form:

Water saturation from the resistivity ratio, with a known residual saturation of the flushed zone

S_w = (1 - S_{hr})\,Q^{1/n}

where:

$S_{hr}$ residual hydrocarbon saturation of the flushed zone, $1 - S_{xo}$ , fraction
$Q$ normalised resistivity ratio, —
$n$ saturation exponent, —

Units: dimensionless

Holds when

$S_{hr}$ belongs to this bed and to the same process: displacement by filtrate during drilling. The value from waterflooding experiments can differ.
All the assumptions of the ratio of the resistivities of the zones.

An error in $S_{hr}$ passes into $S_w$ through the factor $1 - S_{hr}$ . If in fact 40% of oil was left at the wall and we took 30%, $S_w$ is overestimated by the factor $0.7/0.6$ , that is, by 17%.

CalculatedThe saturations of the zones and the measured ratio. The dashed ray shows all the pairs

S_w

S_{xo}

with the ratio

S_w/S_{xo} = 0.5

, as in bed 2 of the example below. The ray crosses the empirical curve

S_{xo} = S_w^{1/5}

S_w \approx 0.42

, and the line of residual oil from core,

S_{xo} = 0.64

, at

S_w = 0.32

. The diagonal

S_{xo} = S_w

means that the filtrate moved nothing.

Worked exampleWater saturation of an oil-bearing bed by two closure relations

Given (assumed data)

Bed 2 after corrections for invasion: $R_{xo}$ = 23.9 Ω·m, $R_t$ = 13.8 Ω·m
Water contrast: $R_{mf}/R_w$ = 6.92 (the example above)
Saturation exponent: $n = 2$
Residual oil saturation of the flushed zone from core (assumed): 0.36

Solution

$R_{xo}/R_t = 23.9/13.8 = 1.732$ . The normalised ratio is $Q = 1.732/6.92 = 0.2503$ . The moveable hydrocarbon index is $S_w/S_{xo} = \sqrt{0.2503} = 0.500$ , which is well below 0.7: the oil is moveable.
The empirical relation: $S_w = 0.2503^{5/8} = 0.421$ . Then $S_{xo} = 0.421^{1/5} = 0.841$ . That is, the relation assumes a residual oil saturation of the flushed zone of 0.159.
Residual oil from core: $S_{xo} = 1 - 0.36 = 0.64$ and $S_w = 0.64\cdot 0.500 = 0.320$ .
The empirical relation gave a water saturation 32% larger: $0.421/0.320 = 1.32$ .

Answer: The oil of bed 2 is moveable (

S_w/S_{xo}

= 0.500). The water saturation is 0.320 with the residual oil from core, and 0.421 with the empirical relation.

What it means. The moveable hydrocarbon index is a direct result of the log, while the water saturation depends on the relation we adopt. The empirical relation assumes that 0.159 of oil was left at the wall, while core shows 0.36. The filtrate moved less oil than the average relation says it should, and the relation put the difference down to water. If the residual oil had been less than average, the error would have been the other way, and optimistic. So for the parameters of the estimate of hydrocarbons in place we take $S_{hr}$ from core, and we keep the empirical relation for quick estimates. The data in this example are assumed for illustration.

Moveable hydrocarbons: what the ratio shows

The index $S_w/S_{xo}$ answers a question that neither the porosity nor the water saturation itself can settle: has the filtrate moved the hydrocarbons? The flushed zone is a kind of small waterflood (pushing oil out of the rock with water) at the borehole wall (more). If it pushed out part of the oil, the oil is moveable under the pressure differences that drilling creates.

How to read the moveable hydrocarbon index

$S_w/S_{xo}$: about 1
What happened in the flushed zone: the filtrate moved nothing
Possible beds: water-bearing; viscous or immobile oil; tight rock with no invasion

$S_w/S_{xo}$: from 0.7 to 1
What happened in the flushed zone: a little was moved
Possible beds: transition zone (the interval just above the water zone, where the water saturation falls gradually with height), barely moveable oil, or an error: check the invasion and the flushing

$S_w/S_{xo}$: 0.7 and below (0.6 in carbonates)
What happened in the flushed zone: the filtrate pushed out a noticeable part of the hydrocarbons
Possible beds: moveable oil or gas

$S_w/S_{xo}$	What happened in the flushed zone	Possible beds
about 1	the filtrate moved nothing	water-bearing; viscous or immobile oil; tight rock with no invasion
from 0.7 to 1	a little was moved	transition zone (the interval just above the water zone, where the water saturation falls gradually with height), barely moveable oil, or an error: check the invasion and the flushing
0.7 and below (0.6 in carbonates)	the filtrate pushed out a noticeable part of the hydrocarbons	moveable oil or gas

The thresholds are empirical rules of interpreters, not physical boundaries. We apply them to the ratio found from corrected $R_{xo}$ and $R_t$ , and we refine them for our area from tests.

The main trap is the first row. The resistivity ratio “sees” only the hydrocarbons that were moved. If the oil is viscous (thick) and the filtrate did not move it, $S_{xo} = S_w$ and $Q = 1$ . With the empirical relation, the method then shows $S_w = 1$ : a “water-bearing” bed, although oil fills two thirds of the pores. The level of the resistivity gives such a bed away. In the lesson's assumed section, bed 3 with immobile oil has $R_t$ = 10.5 Ω·m. That is 9.0 times more than the water-bearing bed 1 of similar porosity, and its $R_{xo}/R_t$ = 6.90 is almost equal to the water contrast (6.92). If the bed has a water-bearing part of the same porosity, the ratio of $R_t$ to its resistivity is the resistivity index, and we find $S_w$ without porosity (resistivity index).

Gas is usually moveable, but the filtrate does not push it out completely: residual gas stays in the flushed zone. The relation $S_{xo} = S_w^{1/5}$ is for oil with an ordinary residual saturation. For gas, heavy oil and carbonates with vugs, it is better to take the residual saturation of the flushed zone from core or from neighbouring wells.

Quick-look methods: comparison with the SP and apparent water resistivities

At the rig we need an answer fast: where the moveable hydrocarbons are, and which beds to test. The resistivity ratio suits this. It needs no porosity, and the water contrast can be taken from the same log, from the SP (quick-look interpretation).

Opposite a clean water-bearing bed, the SP deflection is close to $-K\lg(R_{mf}/R_w)$ , and the resistivity ratio equals the water contrast. So if we convert $R_{xo}/R_t$ to millivolts with the same coefficient $K$ , in water-bearing beds we get a curve that coincides with the SP (comparison with the SP). Opposite a bed with moveable hydrocarbons, $R_{xo}/R_t$ is smaller, and the calculated curve departs from the SP towards the shale baseline (the SP reading opposite clean shales) and beyond.

The curve from the resistivity ratio and its separation from the SP

E_R = -K\lg\frac{R_{xo}}{R_t},\qquad \Delta E = E_R - E_{SP} = -nK\lg\frac{S_w}{S_{xo}}

where:

$E_R$ the curve calculated from the ratio of the resistivities of the zones, mV
$E_{SP}$ deflection of the SP from the shale baseline opposite the same bed, with its sign, mV
$K$ coefficient of the static SP amplitude, $65 + 0.24\,T$ with $T$ in °C (more), mV per decade
$\Delta E$ separation of the curves after matching in a water-bearing bed, mV
$\lg$ base-10 logarithm, —

Units: mV; $T$ in °C

Holds when

Clean beds with the same formation water. The SP is not distorted by the bed thickness and by streaming potentials (voltages caused by the flow of filtrate through the rock), or it is distorted in the same way in the beds that we compare.
We shift the curve $E_R$ so that in a bed that is certainly water-bearing it coincides with the SP. The shift absorbs the difference between the equivalent and the true water resistivities, and constant errors of the ratio.
The mud is fresh ( $R_{mf} > R_w$ ), and $R_{xo}/R_t$ was found with a correction for invasion.
The second formula follows from the ratio of the resistivities of the zones, if the SP of the bed equals the SP of the water-bearing bed.

The separation grows with moveability (table below). So the rule for reading is this: curves that coincide in a permeable bed mean water or immobile oil. A separation of tens of millivolts means moveable hydrocarbons.

Separation of the curves at

K

= 82.76 mV (74 °C) and

n = 2

$S_w/S_{xo}$: 0.9
$\Delta E$ , mV: 7.6
$Q$: 0.81

$S_w/S_{xo}$: 0.8
$\Delta E$ , mV: 16.0
$Q$: 0.64

$S_w/S_{xo}$: 0.7
$\Delta E$ , mV: 25.6
$Q$: 0.49

$S_w/S_{xo}$: 0.5
$\Delta E$ , mV: 49.8
$Q$: 0.25

$S_w/S_{xo}$: 0.3
$\Delta E$ , mV: 86.5
$Q$: 0.09

$S_w/S_{xo}$	$\Delta E$ , mV	$Q$
0.9	7.6	0.81
0.8	16.0	0.64
0.7	25.6	0.49
0.5	49.8	0.25
0.3	86.5	0.09

A separation of 25 mV corresponds to the moveability threshold of 0.7. An error of a few millivolts, which is usual for reading the SP, changes the index by a few hundredths.

CalculatedAn assumed section 2140–2186 m (fresh mud, 74 °C). On the left are the SP and the curve

E_R = -K\lg(R_{xo}/R_t)

, matched in the water-bearing bed 1. The shading is the separation. On the right are

R_t

and

R_{xo}

. Bed 2 has moveable oil. Bed 3 has oil that the filtrate did not move: the curves coincide, but

R_t

is high. Bed 4 is shaly and water-bearing: both curves are reduced, and the separation is small. The tight streak (a thin layer of dense rock) in bed 1 gives a false separation.

Worked exampleMoveability from the separation of the curve from the SP

Given (assumed data)

Formation temperature: 74 °C
Bed 1, water-bearing, thick: $R_{xo}/R_t$ = 6.91; SP −68 mV
Bed 2: $R_{xo}/R_t$ = 1.732; SP −65 mV (a little less in the oil-bearing bed: assumed)
Saturation exponent: $n = 2$

Solution

The coefficient at 74 °C: $K = 65 + 0.24\cdot 74 = 82.76$ mV.
Bed 1: $E_R = -82.76\lg 6.91 = -69.5$ mV, and the SP is −68 mV. The matching shift $-68 - (-69.5) = 1.5$ mV is added to the whole curve $E_R$ .
Bed 2: $E_R = -82.76\lg 1.732 = -19.7$ mV, and after the shift −18.2 mV. The separation from the SP is $\Delta E = -18.2 - (-65) = 46.8$ mV.
The moveable hydrocarbon index: $S_w/S_{xo} = 10^{-\Delta E/(2K)} = 10^{-46.8/(2\cdot 82.76)} = 0.522$ .
If the SP of bed 2 were the same as in the water-bearing bed (−68 mV), the separation would be 49.8 mV and $S_w/S_{xo}$ = 0.500, as from the measured water contrast in the previous example.

Answer: The comparison with the SP gives

S_w/S_{xo} \approx

0.522: the oil is moveable, in agreement with the calculation from the measured water contrast (0.500).

What it means. Matching in the water-bearing bed removed everything that is the same for the beds of the section: the difference between the equivalent and the true water resistivities, the approximation $R_{mfe} \approx 0.85\,R_{mf}$ (more), and constant errors of the ratio. The small underestimate of the SP in the oil-bearing bed hardly changed the conclusion. But in a thin bed the SP is underestimated more, and without a correction for thickness the separation, and with it the moveability, is underestimated. The data in this example are assumed for illustration.

The second quick-look technique uses apparent water resistivities (more). From the deep sonde and the porosity we calculate $R_{wa} = R_t/F$ , the water resistivity that the bed would have if it were water-bearing. From the microresistivity tool, in the same way, we calculate the apparent filtrate resistivity $R_{mfa} = R_{xo}/F$ . In a water-bearing bed $R_{wa} = R_w$ and $R_{mfa} = R_{mf}$ . The ratio $R_{wa}/R_w = S_w^{-n}$ tells whether there are hydrocarbons. The ratio $R_{mfa}/R_{mf} = S_{xo}^{-n}$ tells how many of them are left at the wall. The quotient of these two ratios does not depend on porosity, and it equals $1/Q$ .

A pair of apparent water resistivities

$R_{wa}/R_w$: about 1
$R_{mfa}/R_{mf}$: about 1
Conclusion: a water-bearing bed

$R_{wa}/R_w$: much more than 1
$R_{mfa}/R_{mf}$: about 1 or a little more
Conclusion: hydrocarbons, and the filtrate pushed out most of them: they are moveable

$R_{wa}/R_w$: much more than 1
$R_{mfa}/R_{mf}$: about the same
Conclusion: the hydrocarbons were not moved: immobile oil

$R_{wa}/R_w$: about 1
$R_{mfa}/R_{mf}$: much more than 1
Conclusion: impossible in a clean bed: an error in $R_{xo}$ or $R_{mf}$

$R_{wa}/R_w$	$R_{mfa}/R_{mf}$	Conclusion
about 1	about 1	a water-bearing bed
much more than 1	about 1 or a little more	hydrocarbons, and the filtrate pushed out most of them: they are moveable
much more than 1	about the same	the hydrocarbons were not moved: immobile oil
about 1	much more than 1	impossible in a clean bed: an error in $R_{xo}$ or $R_{mf}$

“Much more” means three or higher: at $n = 2$ this is a saturation of about 0.58 and below. Errors in the porosity and in the cementation exponent shift both ratios in the same way, and they cancel in their quotient.

When the method goes wrong: invasion, flushing, shale, mud

All the formulas of the lesson assume that $R_{xo}$ and $R_t$ are the resistivities of the zones, not the readings of the sondes, and that the flushed zone really is flushed. In practice each assumption is violated. It is useful to know in which direction the answer shifts.

The invasion window. The shallow sonde must see only the flushed zone, and the deep sonde only the uninvaded formation. This happens only when the invasion front lies beyond the depth of investigation of the shallow sonde (how far into the rock it reads), and clearly closer than the depth of investigation of the deep one. With shallow invasion the shallow sonde reaches into the uninvaded formation. With deep invasion the deep sonde reaches into the flushed zone. In both cases the ratio of the readings is closer to one than the true $R_{xo}/R_t$ . With a fresh mud, the true ratio in a water-bearing bed is much larger than one. So its “squeezing” towards one looks like moveable oil: the error is optimistic.

CalculatedThe apparent moveable hydrocarbon index from a shallow electrode sonde and a deep induction sonde, against the invasion diameter: the water-bearing bed 1 (true value 1) and the oil-bearing bed 2 (0.50). The dashes show the ratio of the readings without a correction. The solid lines show the linear formula for two sondes, fitted to the water-bearing bed at

d_i

= 0.75 m. The radial responses of the sondes (how much each one reads at each distance from the hole) are assumed. The dotted line is the moveability threshold of 0.7.

So we convert the ratio of the readings into $R_{xo}/R_t$ . We do it from three curves with an invasion correction chart (Interpretation charts: how to read them, more), or with a linear formula for two sondes fitted for a typical invasion (advanced section). The formula is right only in its “window” of diameters. Outside it, it too gives a false moveability.

Incomplete flushing. If part of the formation water stayed in the flushed zone, the water there is a mixture, and its resistivity is lower than $R_{mf}$ . For solutions of one salt, the conductivities of a mixture add up approximately by volume:

Resistivity of the water of the flushed zone with incomplete flushing

\frac{1}{R_z} = \frac{z}{R_w} + \frac{1 - z}{R_{mf}}

where:

$R_z$ resistivity of the mixture of filtrate and formation water in the flushed zone, Ω·m
$z$ share of formation water in the water of the flushed zone, fraction
$R_w,\ R_{mf}$ resistivities of the formation water and of the filtrate at one temperature, Ω·m

Units: Ω·m; $z$ is a fraction

Holds when

The conductivity of the mixture equals the volume-weighted conductivity of the waters. This is an approximation for solutions of one salt. Real mixtures conduct a little better, and the effect is slightly stronger.
The share $z$ is the same through the whole flushed zone. Its values in the example are assumed.

In a water-bearing bed with such a flushed zone, the normalised ratio equals $R_z/R_{mf}$ , and it is less than one although there are no hydrocarbons.

Normalised ratio of a water-bearing bed with incomplete flushing

Q = \frac{R_z}{R_{mf}} = \frac{1}{z\,R_{mf}/R_w + 1 - z}

where:

$Q$ normalised resistivity ratio, —
$z$ share of formation water in the water of the flushed zone, —
$R_{mf}/R_w$ water contrast, —

Units: dimensionless

Holds when

A water-bearing bed ( $S_w = S_{xo} = 1$ ), and the waters mix by the formula above.

For the same share $z$ , the larger the water contrast, the larger the error. Fresh filtrate is especially sensitive to remnants of salty water.

Worked exampleIncomplete flushing of a water-bearing bed

Given (assumed data)

Bed: water-bearing bed 1: $R_w$ = 0.0565 Ω·m, $R_{mf}$ = 0.391 Ω·m, contrast 6.92
Share of formation water in the flushed zone (assumed): $z = 0.05$
For comparison: the same bed with a fresher mud: contrast 20

Solution

The water of the flushed zone: $1/R_z = 0.05/0.0565 + 0.95/0.391$ , so $R_z = 0.3017$ Ω·m, which is 23% below $R_{mf}$ .
The normalised ratio: $Q = 1/(0.05\cdot 6.92 + 0.95) = 1/1.296 = 0.772$ . $S_w/S_{xo} = 0.879$ . With the empirical relation, $S_w = 0.772^{5/8} = 0.851$ .
At a contrast of 20: $Q = 1/(0.05\cdot 20 + 0.95) = 1/1.950 = 0.513$ . $S_w/S_{xo} = 0.716$ . $S_w = 0.659$ .

Answer: Five per cent of formation water in the flushed zone makes a water-bearing bed look like a bed with 15% hydrocarbons at a contrast of 6.92. At a contrast of 20 it looks like a bed with 34% hydrocarbons, and the moveable hydrocarbon index of 0.716 is already at the threshold of 0.7.

What it means. From the resistivity ratio alone we cannot tell incomplete flushing from moveable oil, and nobody knows the share $z$ in advance. The error grows with the water contrast. So with a very fresh mud and salty formation water, we check the signs of moveability in the water-bearing beds of the same section. If $Q$ in them is markedly less than one, the flushing is incomplete, and we shift the threshold for the section. The data in this example are assumed for illustration.

Shale. Shale conducts current in both zones, but in the flushed zone the water is fresh, and the contribution of the shale there is relatively larger. With a fresh mud $R_{xo}$ falls more than $R_t$ , and the ratio becomes smaller. This is one more optimistic error (Water saturation in shaly sands). In bed 4 of the assumed section (shale volume 0.25, that is, a quarter of the rock is shale; the Simandoux model of conduction in shaly sand), the ratio with the measured water contrast gives $S_w/S_{xo}$ = 0.70. That is at the moveability threshold, although the bed is water-bearing. The SP deflection in a shaly bed is also smaller (shale volume from the SP; in the lesson's section we assume it is 0.75 of the water-bearing one). So when we compare with the SP, the error is mostly cancelled: the separation is 8.9 mV, that is, 0.88. The statement that the comparison with the SP is not very sensitive to shale is true only approximately.

The annulus. In highly permeable oil-bearing beds, soon after the bed is drilled, the displaced formation water collects at the outer edge of the invaded zone (annulus). An induction sonde (a coil tool that adds up the conductivities of the zones) sees this conducting ring, and $R_t$ is underestimated. The ratio $R_{xo}/R_t$ is overestimated, and the moveable oil is underestimated: the error is pessimistic.

Mud and rock. In an oil-based mud the filtrate does not conduct current. $R_{xo}$ is high in every bed, there is no SP, and the ratio methods do not apply. In a salty mud the water contrast is close to one, and the SP hardly deflects, so there is nothing to compare with. The ratio of the corrected $R_{xo}$ and $R_t$ still tells about the moveability. In fractured rocks the filtrate goes into the fractures, and the matrix blocks (the rock between the fractures) are hardly flushed at all (more). There is no flushed zone in the sense of the lesson's formulas. In thin beds the shallow and deep sondes average different intervals, and their ratio loses its meaning.

Where the answer shifts

Condition: Shallow invasion: the shallow sonde sees the formation
$R_{xo}/R_t$ from the data: closer to 1 than the true value
Consequence: in a water-bearing bed, a false moveability

Condition: Deep invasion: the deep sonde sees the flushed zone
$R_{xo}/R_t$ from the data: closer to 1 than the true value
Consequence: the same

Condition: Incomplete flushing
$R_{xo}/R_t$ from the data: too low
Consequence: a false moveability, stronger at a large water contrast

Condition: Shale
$R_{xo}/R_t$ from the data: too low
Consequence: a false moveability; in a comparison with the SP it is mostly cancelled

Condition: Annulus, with $R_t$ from an induction sonde
$R_{xo}/R_t$ from the data: too high
Consequence: moveability is underestimated

Condition: A tight streak opposite a deflected SP
$R_{xo}/R_t$ from the data: about 1
Consequence: a false moveability in a comparison with the SP

Condition: Oil-based mud
$R_{xo}/R_t$ from the data: meaningless
Consequence: the method does not apply

Condition	$R_{xo}/R_t$ from the data	Consequence
Shallow invasion: the shallow sonde sees the formation	closer to 1 than the true value	in a water-bearing bed, a false moveability
Deep invasion: the deep sonde sees the flushed zone	closer to 1 than the true value	the same
Incomplete flushing	too low	a false moveability, stronger at a large water contrast
Shale	too low	a false moveability; in a comparison with the SP it is mostly cancelled
Annulus, with $R_t$ from an induction sonde	too high	moveability is underestimated
A tight streak opposite a deflected SP	about 1	a false moveability in a comparison with the SP
Oil-based mud	meaningless	the method does not apply

The directions are for a fresh water-based mud, when $R_{mf}/R_w$ is much larger than one. Several causes can act at once and partly compensate each other.

A check: the porosity balance and comparison with Archie's equation

The ratio method does not need porosity, but if it is known, we use it for a check. This is the porosity balance. From the saturation $S_{wR}$ found from the resistivity ratio and Archie's equation for the uninvaded formation, we calculate the formation resistivity factor, and from it the porosity. We compare it with the porosity from the porosity methods or from core.

Porosity from the saturation found by the resistivity ratio

F_c = \frac{S_{wR}^{\,n}\,R_t}{R_w},\qquad \phi_c = \left(\frac{a}{F_c}\right)^{1/m}

where:

$F_c$ formation resistivity factor calculated from $S_{wR}$ , —
$S_{wR}$ water saturation from the resistivity ratio, fraction
$R_t,\ R_w$ resistivities of the uninvaded formation and of the formation water, Ω·m
$\phi_c$ calculated porosity, fraction
$a,\ m,\ n$ coefficients of Archie's equation, —

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

Holds when

A clean bed, and the coefficients $a$ , $m$ and $n$ are known.
The porosity $\phi_t$ that we compare with is independent: from the density or neutron log, or from core.

If $\phi_c \approx \phi_t$ , the resistivity ratio and Archie's equation agree, and $S_{wR}$ equals the saturation by Archie's equation, $S_{wA}$ . If $\phi_c < \phi_t$ , then $S_{wR}$ is overestimated. If $\phi_c > \phi_t$ , it is underestimated.

We can also compare the saturations directly: $S_{wA}$ from Archie's equation with the porosity, and $S_{wR}$ from the resistivity ratio. If they agree, that confirms that the invasion profile is a step, the corrections are right, and the closure relation suits the bed. A disagreement points to a broken assumption, but does not say which one.

What a disagreement of the saturations means

Observation: $S_{wA} \approx S_{wR}$
Possible causes: the assumptions hold
What we do: we accept the result

Observation: $S_{wA} > S_{wR}$
Possible causes: very shallow invasion or a transitional profile; incomplete flushing; less oil was left in the flushed zone than the relation assumes
What we do: we trust $S_{wA}$ , and check the flushing and the relation

Observation: $S_{wA} < S_{wR}$
Possible causes: an annulus with an induction $R_t$ ; more oil was left in the flushed zone than the relation assumes
What we do: we look for an annulus with repeat or multi-sonde logs, and for the residual oil from core

Observation	Possible causes	What we do
$S_{wA} \approx S_{wR}$	the assumptions hold	we accept the result
$S_{wA} > S_{wR}$	very shallow invasion or a transitional profile; incomplete flushing; less oil was left in the flushed zone than the relation assumes	we trust $S_{wA}$ , and check the flushing and the relation
$S_{wA} < S_{wR}$	an annulus with an induction $R_t$ ; more oil was left in the flushed zone than the relation assumes	we look for an annulus with repeat or multi-sonde logs, and for the residual oil from core

For the case of an annulus, an empirical estimate has been published that puts the answer below the saturation by Archie's equation:

An empirical estimate of water saturation with an annulus

S_w \approx S_{wA}\left(\frac{S_{wA}}{S_{wR}}\right)^{1/4}

where:

$S_{wA}$ water saturation by Archie's equation, fraction
$S_{wR}$ water saturation from the resistivity ratio, fraction

Units: fraction

Holds when

An empirical correction without a rigorous derivation. We use it only when the disagreement is really caused by an annulus.
$S_{wA}$ and $S_{wR}$ were found from the same corrected $R_t$ and $R_{xo}$ .

The correction rests on the idea that an annulus lowers the induction $R_t$ . Then both estimates are too high, and $S_{wR}$ more so. If the cause of the disagreement is something else, for example more residual oil than the empirical relation assumes, the correction takes the answer the wrong way (example below).

Worked examplePorosity balance in an oil-bearing bed

Given (assumed data)

Bed 2: $R_t$ = 13.8 Ω·m; $S_{wR}$ = 0.421 with the empirical relation and 0.320 with the residual oil from core (the example above)
Density porosity (independent): $\phi_t = 0.20$
Parameters: $R_w$ = 0.0565 Ω·m, $a = 1$ , $m = n = 2$

Solution

The saturation by Archie's equation: $S_{wA} = \sqrt{0.0565/(0.20^2\cdot 13.8)} = 0.320$ .
The balance for $S_{wR}$ = 0.421: $F_c = 0.421^2\cdot 13.8/0.0565 = 43.3$ , and $\phi_c = 1/\sqrt{43.3} = 0.152$ . This is less than the density porosity of 0.20: $S_{wR}$ is overestimated.
The correction for an annulus would give $S_w = 0.320\cdot(0.320/0.421)^{1/4} = 0.299$ .
With the residual oil from core, $S_{wR}$ = 0.320: $F_c = 0.320^2\cdot 13.8/0.0565 = 25.0$ and $\phi_c = 0.200$ . The balance closes.

Answer: The porosity balance revealed that

S_{wR}

was overestimated (0.152 against 0.20). The cause is the closure relation: with the residual oil from core, both estimates coincide, at about 0.320.

What it means. A disagreement $S_{wA} < S_{wR}$ points to a broken assumption, but does not say which one. We confirm an annulus with repeat or multi-sonde logs (there is none in this bed: assumed), and we check the closure relation against core. A correction for an annulus, applied by habit, would have taken the answer to 0.299. That is below the value at which all the data agree. The data in this example are assumed for illustration.

Terms and symbols

Resistivity ratio method: An estimate of $S_w$ and of the moveability of hydrocarbons from the ratio $R_{xo}/R_t$ and the water contrast $R_{mf}/R_w$ , without porosity.
Water contrast $R_{mf}/R_w$: The ratio of the resistivities of the filtrate and the formation water at one temperature. It equals $R_{xo}/R_t$ of a water-bearing bed.
Normalised ratio $Q$: The resistivity ratio divided by the water contrast: $(R_{xo}/R_t)/(R_{mf}/R_w) = (S_w/S_{xo})^n$ . A value of 1 means that the hydrocarbons were not moved.
Moveable hydrocarbon index $S_w/S_{xo}$: About 1 means that the filtrate did not move the hydrocarbons. By an empirical rule, 0.7 and below (0.6 in carbonates) means that they are moveable.
Closure relation: The second relation between the saturations of the zones, needed to find $S_w$ : the empirical $S_{xo} = S_w^{1/5}$ , or $S_{xo} = 1 - S_{hr}$ from core.
The curve $E_R$: The curve $-K\lg(R_{xo}/R_t)$ , matched to the SP in a water-bearing bed. Its separation from the SP is a sign of moveable hydrocarbons.
Incomplete flushing: A remnant of formation water in the flushed zone. It lowers $R_{xo}$ and imitates moveable oil.
Porosity balance: A check of the method: the porosity calculated from $S_{wR}$ and Archie's equation is compared with an independent porosity.

Common mistakes and misconceptions

Wrong

“The ratio $R_{xo}/R_t$ is the ratio of the readings of the shallow and deep sondes.”

In fact

The readings mix the zones. Without a correction for invasion, their ratio is closer to one than the true one, and with a fresh mud a water-bearing bed looks like a bed with moveable oil.

Wrong

“ $S_w/S_{xo} = 1$ means water.”

In fact

Immobile oil and tight rock look the same. We look at the level of $R_t$ compared with a water-bearing bed of the same porosity, and at the signs of permeability.

Wrong

“The empirical relation $S_{xo} = S_w^{1/5}$ is right for any bed.”

In fact

It is an average relation for oil-bearing beds with moderate invasion. If more oil stays at the wall, $S_w$ is overestimated, and if less, it is underestimated. The residual saturation from core is more reliable.

Wrong

“We take the water contrast from the log heading and the water sample as they are.”

In fact

In the heading $R_{mf}$ is given at the measurement temperature, and $R_w$ at its own. We calculate the contrast at one temperature. Mixing the temperatures makes the answer wrong by a factor of several.

Wrong

“If $S_{wA} < S_{wR}$ , it is an annulus, and we need a correction.”

In fact

This is only one of the causes. The same picture appears when the empirical relation underestimates the residual oil. First we establish the cause, with a repeat or multi-sonde log or with core, and only then do we correct.

In depth: advanced detail

The saturation exponent and the power 5/8

The power 5/8 is right only at $n = 2$ . In general, we put $S_{xo} = S_w^{1/5}$ into the normalised ratio: $S_w/S_{xo} = S_w^{4/5}$ , and $Q = S_w^{4n/5}$ . From this:

Water saturation with the empirical relation at any saturation exponent

S_w = Q^{\frac{5}{4n}}

where:

$Q$ normalised resistivity ratio, —
$n$ saturation exponent, —

Units: dimensionless

Holds when

The empirical relation $S_{xo} = S_w^{1/5}$ and all the assumptions of the resistivity ratio. The relation was published for ordinary sandstones, and at an unusual $n$ its applicability is also in doubt.

The larger $n$ is, the smaller the power, and the closer $S_w$ is to one at the same $Q$ . In oil-wet rocks $n$ is often 3 and more (Wettability and interfacial phenomena), and a calculation with $n = 2$ underestimates $S_w$ .

Worked exampleThe same bed at a saturation exponent of 2.5

Given (assumed data)

Normalised ratio of bed 2: $Q$ = 0.2503
Saturation exponent: 2 (chosen) and 2.5 (from core, assumed)
Residual oil of the flushed zone from core: 0.36

Solution

The empirical relation: at $n = 2$ , $S_w = 0.2503^{0.625} = 0.421$ . At $n = 2.5$ the power is $5/(4\cdot 2.5) = 0.5$ , and $S_w = 0.2503^{0.5} = 0.500$ .
The residual oil from core: at $n = 2$ , $S_w = 0.64\cdot 0.2503^{1/2} = 0.320$ . At $n = 2.5$ , $S_w = 0.64\cdot 0.2503^{0.4} = 0.368$ .

Answer: With

n = 2.5

the water saturation is larger: 0.500 instead of 0.421 by the empirical relation, and 0.368 instead of 0.320 by core.

What it means. The saturation exponent does not cancel in the resistivity ratio, and it affects the answer with any closure relation: by 19% and 15% in this example. So for the ratio method we take the same $n$ as for Archie's equation, from core. The data in this example are assumed for illustration.

R_{xo}/R_t

from two curves: where the linear formulas come from

Before programs began to process three curves, the ratio $R_{xo}/R_t$ for a quick comparison with the SP was calculated from two curves, a shallow one and a deep one, by a simple linear formula. It is easy to derive from the model of step invasion (more). Suppose the deep sonde sees almost only the uninvaded formation ( $R_d \approx R_t$ ), and the share of the flushed zone in the reading of the shallow sonde, $J_s$ , is constant. Then $R_s = J_s R_{xo} + (1 - J_s) R_t$ , and after dividing by $R_d$ :

Ratio of the resistivities of the zones from the readings of two sondes

\frac{R_{xo}}{R_t} \approx A\,\frac{R_s}{R_d} - (A - 1),\qquad A = \frac{1}{J_s}

where:

$R_s,\ R_d$ readings of the shallow and deep sondes after corrections for the borehole and the enclosing rocks, Ω·m
$J_s$ share of the flushed zone in the reading of the shallow sonde (the pseudo-geometric factor), —
$A$ coefficient of the formula, —

Units: dimensionless

Holds when

A step profile. The deep sonde hardly sees the flushed zone, and $J_s$ is the same for all the beds.
In practice $A$ is chosen so that the formula agrees with the invasion correction chart in the chosen range of diameters. This also absorbs the contribution of the flushed zone to the deep sonde.

At $R_s = R_d$ the formula gives 1: with no contrast of the zones there is no ratio either. For particular pairs of sondes, formulas of this form were published with $A$ from 1.4 to 1.9. By the formula $A = 1/J_s$ this is a share of the flushed zone in the reading of the shallow sonde of 0.5–0.7. But the fitted $A$ is larger than $1/J_s$ when the deep sonde also sees part of the flushed zone (example below). Outside the range for which $A$ was fitted, the formula goes wrong (the figure in the section on errors).

Worked exampleThe coefficient of the formula for two sondes, and a check on an oil-bearing bed

Given (assumed data)

Sondes: a shallow electrode sonde (the zones add up as resistivities) and a deep induction sonde (as conductivities)
Shares of the flushed zone in the readings at $d_i$ = 0.75 m (assumed): $J_s$ = 0.737, $G_d$ = 0.240
Beds: water-bearing bed 1 ( $R_{xo}/R_t$ = 6.92) for the fitting, and oil-bearing bed 2 ( $R_{xo}/R_t$ = 1.732) for the check

Solution

Bed 1, the readings in fractions of $R_t$ : $R_s = 0.737\cdot 6.92 + 0.263 = 5.363$ ; $R_d = 1/(0.240/6.92 + 0.760) = 1.258$ ; the ratio of the readings is $5.363/1.258 = 4.263$ .
The coefficient at which the formula is exact for bed 1: $A = (6.92 - 1)/(4.263 - 1) = 1.814$ .
Bed 2: $R_s = 0.737\cdot 1.732 + 0.263 = 1.539$ ; $R_d = 1/(0.240/1.732 + 0.760) = 1.113$ ; the ratio of the readings is 1.383.
The estimate by the formula: $1.814\cdot 1.383 - 0.814 = 1.695$ instead of 1.732 (−2.1%). The moveable hydrocarbon index is 0.495 instead of 0.500.

Answer:

A \approx

1.814. At the same invasion, the formula estimates

R_{xo}/R_t

of the oil-bearing bed with an error of −2.1%, so the moveability is hardly distorted.

What it means. Without recalculation, the ratio of the readings 1.383 would give a moveable hydrocarbon index of 0.45, and the water-bearing bed would show 0.78, that is, “moveable oil”. The formula corrects this, but only for an invasion close to the one for which $A$ was fitted: at 0.3 m the water-bearing bed again looks oil-bearing (the figure in the section on errors). The shares $J_s$ and $G_d$ are assumed. The data in this example are assumed for illustration.

Connections to other topics

Borehole environment: invaded zone and mudcake — the zones around the hole, the resistivity profile and the annulus.
Microresistivity tools and flushed-zone resistivity — $R_{xo}$ and $S_{xo}$ with porosity, and the volumes of moveable and residual oil.
Archie's equation and saturation of clean reservoirs — Archie's equation and the resistivity index, on which the ratio rests.
Determining formation water resistivity — the resistivity of the formation water, $R_{wa}$ and the calculation from the SP.
Spontaneous potential (SP) log — the static SP amplitude and the coefficient $K$ .
Resistivity logs: electrical and induction tools — how to get $R_t$ and $R_{xo}$ from three curves.
Interpretation charts: how to read them — charts of invasion and saturation.
Water saturation in shaly sands — why shale distorts the resistivity ratio.
Relative permeability — the residual oil saturation from experiments on core.
Log interpretation workflow — where the check of moveability fits in the order of interpretation.

Check yourself

1
A clean water-bearing sandstone is drilled with a fresh mud; the water contrast $R_{mf}/R_w = 9$ . What should $R_{xo}/R_t$ be after all the corrections?
1. A.About 3
2. B.About 9
3. C.About 1
4. D.It depends on porosity
Show the answer
Correct answer: B. About 9
In a water-bearing bed $S_w = S_{xo} = 1$ , and the ratio of the resistivities of the zones equals the water contrast: the formation resistivity factor of the zones is shared, and it cancels. If after the corrections the ratio is markedly less than 9, the cause is invasion, incomplete flushing, shale or an error in the water contrast.
2
$R_{xo}/R_t = 1.2$ , $R_{mf}/R_w = 12$ , $n = 2$ . Find $S_w/S_{xo}$ and $S_w$ with the empirical relation. Is the oil moveable?
Show the answer
$Q = 1.2/12 = 0.10$ , and $S_w/S_{xo} = \sqrt{0.10} \approx 0.32$ , which is much less than 0.7: the oil is moveable. $S_w = 0.10^{5/8} \approx 0.24$ , and the relation assumes $S_{xo} = 0.24^{1/5} \approx 0.75$ , that is, a residual oil of about 0.25. If core gives another residual oil, we recalculate $S_w$ : $S_w = (1 - S_{hr})\cdot 0.32$ .
3
The resistivity of a bed is 8 times that of a water-bearing bed of the same porosity, and the ratio $R_{xo}/R_t$ equals the water contrast. What is this most likely?
1. A.A gas-bearing bed with moveable gas
2. B.Oil that the filtrate did not move: viscous or immobile oil
3. C.A water-bearing bed
4. D.An error of the microresistivity tool
Show the answer
Correct answer: B. Oil that the filtrate did not move: viscous or immobile oil
The resistivity index $R_t/R_0 = 8$ means $S_w \approx 8^{-1/2} \approx 0.35$ at $n = 2$ : there are many hydrocarbons. But $Q = 1$ : the filtrate did not move them, and $S_{xo} = S_w$ . This is how viscous oil and immobile hydrocarbons look. Moveable gas would give a $Q$ markedly less than one.
4
The curve $-K\lg(R_{xo}/R_t)$ is matched to the SP in a water-bearing bed. Opposite a sandstone above it, it departs from the SP towards the shale baseline by 30 mV. The formation temperature is 60 °C. Estimate $S_w/S_{xo}$ . What should we check before we recommend a test?
Show the answer
$K = 65 + 0.24\cdot 60 = 79.4$ mV, and $S_w/S_{xo} = 10^{-30/(2\cdot 79.4)} \approx 0.65$ : moveable hydrocarbons. Before this, we check the permeability (mudcake, microresistivity tools: a tight streak gives the same separation), the bed thickness (the SP of a thin bed is underestimated), the shale volume, and that $R_{xo}/R_t$ was recalculated with a correction for invasion.
5
Which cause gives a pessimistic error, one that hides moveable oil?
1. A.An annulus, when $R_t$ is taken from an induction sonde
2. B.Incomplete flushing
3. C.Shallow invasion with a fresh mud
4. D.Shale with a fresh mud
Show the answer
Correct answer: A. An annulus, when $R_t$ is taken from an induction sonde
The conducting ring of the annulus lowers the induction $R_t$ , the ratio $R_{xo}/R_t$ is overestimated, and $S_w/S_{xo}$ comes out closer to one. The other causes make $R_{xo}/R_t$ smaller and imitate moveable oil where there is none.
6
From the resistivity ratio, $S_{wR} = 0.45$ . From Archie's equation with the density porosity, $S_{wA} = 0.34$ . How should we interpret this, and what should we do?
Show the answer
The saturations do not agree: $S_{wR}$ is larger. The possible causes are an annulus with an induction $R_t$ , or more residual oil in the flushed zone than the empirical relation assumes. The porosity calculated from $S_{wR}$ will be less than the density porosity. We must not correct for an annulus at once. First we check the invasion profile (a repeat or multi-sonde log) and the residual oil from core. Until then, $S_{wA}$ is more reliable.

Updated: 17 October 2026