Borehole environment: invaded zone and mudcake

Quick refresher

During drilling, the pressure of the mud column is usually higher than the formation pressure (this is overbalance). The filtrate goes into permeable rocks, and the solid particles of the mud settle on the wall and form a mudcake.
The cake is many times less permeable than the reservoir (by orders of magnitude), so once it has formed, filtration slows sharply. The cake, not the formation, mainly sets the volume of filtrate.
Around the borehole lie the flushed zone ( $R_{xo}$ : filtrate and residual hydrocarbons, the oil and gas that the filtrate could not push out), the transition zone ( $R_i$ ) and the uninvaded formation ( $R_t$ : formation water and original hydrocarbons).
Microresistivity tools and nuclear methods (those that use radioactivity) see mainly the flushed zone. Only deep resistivity sondes “reach” the uninvaded formation, and even they need corrections.
The contrast $R_{mf}/R_w$ and the saturation set the ratio of $R_{xo}$ to $R_t$ . With fresh mud in a water-bearing bed, $R_{xo} \gg R_t$ (much greater). In an oil-bearing bed this ratio is much smaller.
Interpretation starts with a check of the borehole: compare the caliper log with the bit size, find the mud type, and work out the expected invasion profile.

Key relationships

d_i = \sqrt{d_h^2 + \dfrac{4\,V_f}{\pi\,h\,\phi\,S_{xo}}}

diameter of the invaded zone from the volume of filtrate (step model: a sharp boundary between the invaded and uninvaded zones)

\dfrac{R_{xo}}{R_t} = \dfrac{R_{mf}}{R_w}\left(\dfrac{S_w}{S_{xo}}\right)^2

a clean formation (no shale), Archie's law with $n = 2$ ; in a water-bearing bed $S_w = S_{xo} = 1$

h_{mc} = \dfrac{d_{bit} - d_{cal}}{2}

mudcake thickness from the caliper log

Common pitfalls

Taking the reading of a deep sonde for $R_t$ without estimating the invasion.
Treating the separation of the resistivity curves as a sign of oil: with fresh mud it is greatest in water-bearing beds.
Confusing the transition zone of invasion (around the borehole) with the transition zone above the oil–water contact (in the vertical).
Forgetting that the density and neutron methods see the fluid of the flushed zone, not the formation fluid.

After this lesson you can

explain how overbalance (mud pressure above formation pressure) creates the mudcake and the invaded zone, and why the cake limits further filtration
name the zones around the borehole (flushed, transition and uninvaded) and the resistivities $R_m$ , $R_{mc}$ , $R_{mf}$ , $R_{xo}$ , $R_i$ , $R_t$ , $R_w$ that describe them
predict how the readings of shallow and deep resistivity sondes compare in a water-bearing bed and in an oil-bearing bed, with fresh mud and with salt mud
estimate the diameter of the invaded zone from the volume of filtrate and explain why invasion is deeper in low-porosity rocks
make the first checks before interpretation: the caliper log (a record of the hole diameter) against the bit size, the mud type, and the expected invasion profile

What drilling does to the formation

We drill a well with circulation. The drilling mud carries the cuttings (rock chips) out of the hole, and it cools and lubricates the bit. The column of mud also presses on the walls of the hole. To keep formation fluids from entering the well, we usually keep the pressure of the column above the pore pressure (the pressure of the fluid in the rock pores). This excess is called overbalance. Pressures are covered in more detail in the topic Pore pressure and overburden pressure.

Overbalance pushes the liquid part of the mud, the filtrate, into permeable rocks. The solid particles (clay, weighting material that makes the mud heavy, and fine cuttings) are held back on the wall, and they form the mudcake (also called filter cake). The cake is a filter that the mud builds for itself. Its permeability is of the order of $10^{-2}$ – $10^{-4}$ mD (millidarcy, a unit of permeability). That is thousands to millions of times lower than the permeability of a typical reservoir. Once the cake has formed, it, not the formation, sets the rate of filtration.

Muds are either water-based (the filtrate is water with dissolved salts and additives) or oil-based (OBM; the filtrate is a hydrocarbon liquid). We divide water-based muds into fresh and salt muds. The salinity (salt content) sets the resistivity of the filtrate $R_{mf}$ , and that in turn sets how invasion shows up on the resistivity curves. We measure the resistivities of the mud, filtrate and cake on samples, and we write them in the log heading together with the temperature of measurement.

Zones around the borehole

The filtrate pushes the formation fluids away from the wall, deeper into the formation. The displacement is incomplete and uneven, so ring-shaped zones form around the borehole. We describe each zone by its electrical resistivity and its saturation. These are the main language in which the log curves speak.

SchematicA sector of the section around the well, in an oil-bearing bed drilled with water-based mud. The drops show hydrocarbons. Only residual hydrocarbons remain at the wall. Beyond the invasion front (the outer edge of the invaded zone) the original saturation is preserved. The diagram is not to scale.

Borehole: The borehole is filled with mud of resistivity $R_m$ . The bit sets the nominal diameter $d_h$ , and the caliper log shows the actual diameter.
Mudcake: A layer on the wall opposite permeable beds. Its thickness is $h_{mc}$ (usually millimetres, sometimes more than a centimetre) and its resistivity is $R_{mc}$ .
Flushed zone: It adjoins the wall. The filtrate has displaced almost all the formation water and the moveable part of the hydrocarbons. The pores hold filtrate ( $R_{mf}$ ) and residual hydrocarbons. The zone has resistivity $R_{xo}$ and water saturation $S_{xo}$ (the water here is mainly filtrate). In a water-bearing bed $S_{xo} = 1$ . In a productive bed $S_{xo} = 1 - S_{hr}$ , where $S_{hr}$ is the residual hydrocarbon saturation.
Transition zone: The share of filtrate decreases with distance from the borehole, and the resistivity $R_i$ changes from $R_{xo}$ to $R_t$ . The flushed and transition zones together are called the invaded zone. The idealised step model has no transition zone: the resistivity $R_{xo}$ is maintained up to the invasion diameter $d_i$ , and beyond it the uninvaded formation begins.
Uninvaded zone: The filtrate has not reached this zone. The pores hold formation water of resistivity $R_w$ and the original hydrocarbons. The resistivity $R_t$ (true resistivity) and the water saturation $S_w$ are exactly what we need to estimate the hydrocarbons in place.

Resistivities in and around the borehole

What: Drilling mud
Symbol: $R_m$
Russian-language notation (subscripts transliterated): $\rho_\text{s}$
How it is found: measurement on a sample (log heading) or with a downhole mud resistivity meter

What: Mud filtrate
Symbol: $R_{mf}$
Russian-language notation (subscripts transliterated): $\rho_\text{f}$
How it is found: measurement on a filtrate sample

What: Mudcake
Symbol: $R_{mc}$
Russian-language notation (subscripts transliterated): $\rho_\text{gk}$
How it is found: measurement on a cake sample; needed for the corrections of microresistivity tools

What: Flushed zone
Symbol: $R_{xo}$
Russian-language notation (subscripts transliterated): $\rho_\text{pp}$
How it is found: microresistivity tools with a correction for the cake

What: Invaded zone
Symbol: $R_i$
Russian-language notation (subscripts transliterated): $\rho_\text{zp}$
How it is found: shallow and medium sondes

What: Uninvaded formation
Symbol: $R_t$
Russian-language notation (subscripts transliterated): $\rho_\text{p}$
How it is found: deep sondes with corrections for the borehole, the enclosing rocks and invasion

What: Formation water
Symbol: $R_w$
Russian-language notation (subscripts transliterated): $\rho_\text{v}$
How it is found: water samples, water-bearing beds, the spontaneous potential (SP) method; see Determining formation water resistivity

What	Symbol	Russian-language notation (subscripts transliterated)	How it is found
Drilling mud	$R_m$	$\rho_\text{s}$	measurement on a sample (log heading) or with a downhole mud resistivity meter
Mud filtrate	$R_{mf}$	$\rho_\text{f}$	measurement on a filtrate sample
Mudcake	$R_{mc}$	$\rho_\text{gk}$	measurement on a cake sample; needed for the corrections of microresistivity tools
Flushed zone	$R_{xo}$	$\rho_\text{pp}$	microresistivity tools with a correction for the cake
Invaded zone	$R_i$	$\rho_\text{zp}$	shallow and medium sondes
Uninvaded formation	$R_t$	$\rho_\text{p}$	deep sondes with corrections for the borehole, the enclosing rocks and invasion
Formation water	$R_w$	$\rho_\text{v}$	water samples, water-bearing beds, the spontaneous potential (SP) method; see Determining formation water resistivity

The resistivity of muds depends strongly on temperature. We can compare values only after we bring them to one temperature, usually the formation temperature (Formation water and formation temperature).

The picture becomes more complicated in two common cases.

Fractured rocks. The filtrate easily goes into the fractures, but it hardly enters the tight matrix blocks (the rock between the fractures). Only a small part of the formation fluids is displaced, even right at the wall. There is no real flushed zone, so techniques that assume one give a distorted result.
Oil-based muds. The filtrate does not conduct current, so the resistivity of the flushed zone is high in both water-bearing and productive beds. Methods that need a conductive mud (SP, laterolog, most microresistivity tools) do not work. Because the fluids differ in density, the filtrate and the formation fluids can separate vertically (gravity segregation: the lighter fluid rises and the heavier one sinks).

Depth of invasion

How far the filtrate goes depends on the properties of the mud (its fluid loss and how quickly it builds a cake), on the porosity and permeability of the formation, on the overbalance, and on the time since the bit penetrated the formation. Intuition says that invasion is deeper in a more permeable formation, but often the opposite is true.

Once the cake has formed, it mainly decides the volume of filtrate. In the same time, about the same amount of filtrate goes into neighbouring permeable beds. This volume has to fit into the pores. In a high-porosity rock a thin ring around the borehole is enough. A low-porosity rock needs a wider ring. So the lower the porosity, the deeper the invasion. In a productive bed the filtrate can fill only the part of the pores that the moveable fluids have left ( $S_{xo} < 1$ ). For the same volume of filtrate, the invaded zone is then even wider.

Diameter of the invaded zone from the volume of filtrate

d_i = \sqrt{d_h^2 + \frac{4\,V_f}{\pi\,h\,\phi\,S_{xo}}}

where:

$d_i$ diameter of the invaded zone (step model), m
$d_h$ borehole diameter, m
$V_f$ volume of filtrate that has gone into the formation, m³
$h$ thickness of the bed into which this volume has gone, m
$\phi$ porosity, fraction
$S_{xo}$ fraction of the pore volume occupied by filtrate in the flushed zone: 1 in a water-bearing bed, $1 - S_{hr}$ in a productive one, fraction

Units: any consistent system (here SI)

Holds when

Piston-like displacement (the filtrate pushes the fluids ahead of it like a piston): beyond the filtrate front the formation is unchanged, and we neglect the transition zone.
Invasion is uniform around the circumference and over the thickness of the bed, without gravity segregation.
The filtrate displaces the formation water and the moveable hydrocarbons rather than mixing with them.
We take the volume of filtrate as given. In reality the mud, the cake, the overbalance and time determine it.

This is a volume balance: the filtrate fills the pores of the ring between $d_h$ and $d_i$ , so $V_f = \tfrac{\pi}{4}(d_i^2 - d_h^2)\,h\,\phi\,S_{xo}$ . For the same $V_f$ , the diameter of invasion grows when the porosity or the filtrate saturation gets smaller.

Worked exampleThe same volume of filtrate in beds of different porosity

Given (assumed data)

Borehole diameter $d_h$: 215.9 mm (8½ in bit)
Volume of filtrate per 1 m of thickness $V_f/h$: 25 L/m (0.025 m³/m)
Porosity of the beds: $\phi_1 = 0.25$ and $\phi_2 = 0.08$
Saturation: both beds are water-bearing, $S_{xo} = 1$

Solution

First bed: $d_i = \sqrt{0.2159^2 + \dfrac{4\cdot 0.025}{\pi\cdot 0.25}} = \sqrt{0.0466 + 0.1273} = 0.417$ m.
Second bed: $d_i = \sqrt{0.0466 + \dfrac{4\cdot 0.025}{\pi\cdot 0.08}} = \sqrt{0.0466 + 0.3979} = 0.667$ m.
Depth of invasion beyond the wall, $(d_i - d_h)/2$ : 10 cm and 23 cm. The ratio $d_i/d_h$ : 1.9 and 3.1.

Answer: The filtrate goes about 10 cm beyond the wall in the bed with 25% porosity and 23 cm in the bed with 8% porosity.

What it means. With a porosity about three times lower, the same filtrate has gone 2.2 times deeper. For logging this means that in tight beds, shallow and medium sondes see almost only the invaded zone, so it is harder to find $R_t$ . If the second bed had been oil-bearing with $S_{xo} = 0.75$ , the front would have gone still farther. The data in this example are assumed for illustration.

CalculatedDiameter of the invaded zone against porosity for the data of the example (25 L of filtrate per metre of thickness, 215.9 mm borehole). The lower, teal curve is a water-bearing bed (

S_{xo} = 1

), and the points on it are the two beds of the example. The upper, grey curve is an oil-bearing bed (

S_{xo} = 0.75

). The dashed line is the borehole diameter.

The resistivity profile: what to expect from shallow and deep sondes

The water in the pores sets the resistivity of a clean (non-shaly) rock. Grains of quartz and calcite hardly conduct current, and neither do oil and gas. A rock whose pores are entirely full of water has a resistivity $F$ times that of the water itself. This formation resistivity factor $F$ depends only on the porosity and the structure of the pore space. If hydrocarbons fill part of the pores, the resistivity rises. Under typical conditions it rises roughly in inverse proportion to the square of the water saturation. This is Archie's law, covered in detail in the topic Archie's equation and saturation of clean reservoirs.

In the flushed zone the water in the pores is filtrate ( $R_{mf}$ ). In the uninvaded formation it is formation water ( $R_w$ ). The factor $F$ is the same for both, because the rock is the same. So in the ratio of the resistivities of the two zones, $F$ cancels, and we do not need to know the porosity.

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)^{2}

where:

$R_{xo},\ R_t$ resistivity of the flushed zone and of the uninvaded formation, Ω·m
$R_{mf},\ R_w$ resistivity of the filtrate and of the formation water at formation temperature, 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

Units: any, the same for all the resistivities (usually Ω·m)

Holds when

A clean formation with intergranular porosity (pores between the grains): only the water in the pores conducts current.
Archie's law with a saturation exponent $n = 2$ and the same formation resistivity factor in both zones.
A step profile. We obtained $R_{xo}$ and $R_t$ after corrections for the borehole, the cake and invasion.
$R_{mf}$ and $R_w$ have been brought to formation temperature.

In a water-bearing bed ( $S_w = S_{xo} = 1$ ) the ratio $R_{xo}/R_t$ equals the water contrast $R_{mf}/R_w$ . Moveable hydrocarbons reduce the ratio by a factor of $(S_w/S_{xo})^2$ . The more hydrocarbons the filtrate has displaced, the further the ratio moves from the “water” value. Quick methods of finding productive beds are built on this (Resistivity ratio methods and movable oil).

Typical resistivity profiles in clean beds

Mud and bed: Fresh mud, water-bearing bed
Water contrast: $R_{mf} \gg R_w$
Expected relation: $R_{xo} \gg R_t$ : shallow sondes read much higher than deep ones
What it is called in Russian-language practice: resistivity-increasing invasion

Mud and bed: Fresh mud, productive bed
Water contrast: $R_{mf} \gg R_w$
Expected relation: the ratio $R_{xo}/R_t$ is several times smaller than in a water-bearing bed: the curves come closer or change places
What it is called in Russian-language practice: resistivity-decreasing invasion, or its weak form

Mud and bed: Salt mud, water-bearing bed
Water contrast: $R_{mf} \approx R_w$ or less
Expected relation: $R_{xo} \approx R_t$ or $R_{xo} < R_t$ : the separation is small
What it is called in Russian-language practice: invasion hardly shows in the resistivity

Mud and bed: Salt mud, productive bed
Water contrast: $R_{mf} \lesssim R_w$
Expected relation: $R_{xo} < R_t$ : shallow sondes read lower than deep ones
What it is called in Russian-language practice: resistivity-decreasing invasion

Mud and bed: Oil-based mud
Water contrast: the filtrate does not conduct current
Expected relation: $R_{xo}$ is high in any bed; the contrast between the zones says little about the saturation
What it is called in Russian-language practice: —

Mud and bed	Water contrast	Expected relation	What it is called in Russian-language practice
Fresh mud, water-bearing bed	$R_{mf} \gg R_w$	$R_{xo} \gg R_t$ : shallow sondes read much higher than deep ones	resistivity-increasing invasion
Fresh mud, productive bed	$R_{mf} \gg R_w$	the ratio $R_{xo}/R_t$ is several times smaller than in a water-bearing bed: the curves come closer or change places	resistivity-decreasing invasion, or its weak form
Salt mud, water-bearing bed	$R_{mf} \approx R_w$ or less	$R_{xo} \approx R_t$ or $R_{xo} < R_t$ : the separation is small	invasion hardly shows in the resistivity
Salt mud, productive bed	$R_{mf} \lesssim R_w$	$R_{xo} < R_t$ : shallow sondes read lower than deep ones	resistivity-decreasing invasion
Oil-based mud	the filtrate does not conduct current	$R_{xo}$ is high in any bed; the contrast between the zones says little about the saturation	—

This is a qualitative picture: real relations depend on the saturations, the depth of invasion and the shale volume.

Worked exampleWhat shallow and deep sondes show in the water-bearing and oil-bearing parts of a bed

Given (assumed data)

Porosity $\phi$: 0.22; the formation resistivity factor is $F = 1/\phi^2$
Formation water resistivity $R_w$: 0.06 Ω·m (at formation temperature)
Filtrate resistivity $R_{mf}$: 0.60 Ω·m (fresh mud, at formation temperature)
Water-bearing part: $S_w = S_{xo} = 1$
Oil-bearing part: $S_w = 0.20$ ; residual oil saturation of the flushed zone 0.20, that is $S_{xo} = 0.80$

Solution

Formation resistivity factor: $F = 1/0.22^2 = 20.7$ .
Water-bearing part: $R_t = F R_w = 1.24$ Ω·m, $R_{xo} = F R_{mf} = 12.4$ Ω·m, ratio $R_{xo}/R_t = 10$ .
Oil-bearing part: $R_t = F R_w/S_w^2 = 31.0$ Ω·m, $R_{xo} = F R_{mf}/S_{xo}^2 = 19.4$ Ω·m, ratio $R_{xo}/R_t = 0.625$ .
Check by the ratio formula: $\dfrac{0.60}{0.06}\cdot\left(\dfrac{0.20}{0.80}\right)^2 = 0.625$ , the same, and the factor $F$ was not needed.

Answer: In the water-bearing part the shallow sonde reads a resistivity about 10 times higher than the deep one. In the oil-bearing part it reads a lower one (ratio 0.625).

What it means. A sharp drop of the ratio $R_{xo}/R_t$ , compared with the water-bearing part of the same bed (here from 10 to 0.625), is a sign of moveable oil. The separation of the curves by itself shows only invasion, that is, permeability: with fresh mud it is greatest in a water-bearing bed. Real sondes read values between $R_{xo}$ and $R_t$ , depending on the depth of invasion, so we find $R_{xo}$ and $R_t$ with corrections. The data in this example are assumed for illustration.

Which zone each sonde sees

Every tool has its own depth of investigation. Its reading is made up of the signals of different zones, each with a weight that depends on the design of the sonde. Roughly, a tool sees the rock out to a certain distance from the wall. We must compare this distance with the depth of invasion.

Depth of investigation of the methods and the influence of borehole conditions

Method: Microresistivity tools (pad tools, pressed against the wall)
Depth of investigation (approximate): the first few centimetres from the wall
What it mainly sees: the flushed zone, $R_{xo}$
How the borehole and invasion affect it: contact with the wall is needed; a thick cake and washouts (enlarged parts of the hole) distort the readings

Method: Gamma-gamma density log
Depth of investigation (approximate): about 10 cm
What it mainly sees: the flushed zone
How the borehole and invasion affect it: a pad tool: sensitive to the cake and to irregularities of the wall; barite in the mud distorts the density and especially the photoelectric factor

Method: Neutron log
Depth of investigation (approximate): 10–30 cm
What it mainly sees: the flushed zone
How the borehole and invasion affect it: corrections for the borehole diameter, the cake and the salinity of the mud

Method: Sonic log
Depth of investigation (approximate): from centimetres to tens of centimetres
What it mainly sees: the near-borehole part of the formation
How the borehole and invasion affect it: washouts; rocks loosened by drilling

Method: Shallow and medium resistivity sondes
Depth of investigation (approximate): tens of centimetres
What it mainly sees: the invaded zone
How the borehole and invasion affect it: the reading lies between $R_{xo}$ and $R_t$

Method: Deep induction and laterolog sondes
Depth of investigation (approximate): of the order of 1–2 m from the borehole axis
What it mainly sees: mainly the uninvaded formation
How the borehole and invasion affect it: with deep invasion a correction is needed; $R_t$ is obtained by calculation

Method: Gamma ray log (GR)
Depth of investigation (approximate): 15–30 cm
What it mainly sees: the natural radioactivity of the rock; independent of the fluids
How the borehole and invasion affect it: potassium in the mud (KCl) and a thick layer of mud in washouts change the readings

Method: Spontaneous potential (SP)
Depth of investigation (approximate): —
What it mainly sees: the potential at the contacts of filtrate, formation water and shales
How the borehole and invasion affect it: the amplitude depends on $R_{mf}/R_w$ ; at $R_{mf} = R_w$ there is no deflection, and in oil-based muds the SP is not recorded

Method	Depth of investigation (approximate)	What it mainly sees	How the borehole and invasion affect it
Microresistivity tools (pad tools, pressed against the wall)	the first few centimetres from the wall	the flushed zone, $R_{xo}$	contact with the wall is needed; a thick cake and washouts (enlarged parts of the hole) distort the readings
Gamma-gamma density log	about 10 cm	the flushed zone	a pad tool: sensitive to the cake and to irregularities of the wall; barite in the mud distorts the density and especially the photoelectric factor
Neutron log	10–30 cm	the flushed zone	corrections for the borehole diameter, the cake and the salinity of the mud
Sonic log	from centimetres to tens of centimetres	the near-borehole part of the formation	washouts; rocks loosened by drilling
Shallow and medium resistivity sondes	tens of centimetres	the invaded zone	the reading lies between $R_{xo}$ and $R_t$
Deep induction and laterolog sondes	of the order of 1–2 m from the borehole axis	mainly the uninvaded formation	with deep invasion a correction is needed; $R_t$ is obtained by calculation
Gamma ray log (GR)	15–30 cm	the natural radioactivity of the rock; independent of the fluids	potassium in the mud (KCl) and a thick layer of mud in washouts change the readings
Spontaneous potential (SP)	—	the potential at the contacts of filtrate, formation water and shales	the amplitude depends on $R_{mf}/R_w$ ; at $R_{mf} = R_w$ there is no deflection, and in oil-based muds the SP is not recorded

These are rough values for orientation. The depth of investigation depends on the design of the tool, the properties of the rocks and the resistivity contrast.

Two consequences follow. First, we do not measure $R_t$ directly; we calculate it. Suppose we have recorded three resistivity curves of different depths of investigation. Then we use invasion correction charts (called “tornado” charts because of their shape) to find $R_t$ , $R_{xo}$ and $d_i$ at once, assuming a step profile. In Russian-language practice the same task is solved by lateral logging sounding (BKZ, from its Russian name) and high-frequency induction isoparametric logging sounding (VIKIZ). A set of sondes of different lengths gives the radial profile of resistivity (how the resistivity changes with distance from the borehole). More detail is in the topics Resistivity logs: electrical and induction tools and Interpretation charts: how to read them.

Second, nuclear methods and sonic logging see the fluid of the flushed zone, not the formation fluid. The density and hydrogen index (the hydrogen content) of the fluid in their calculations are the properties of filtrate with residual hydrocarbons. Gas still shows up on the density and neutron logs, because part of it stays in the flushed zone as residual gas. With good flushing this effect gets weaker.

The first steps of interpretation: checking the borehole conditions

What to check before calculating

1
Caliper log against the bit
If the diameters agree, the hole is in good condition. In weak (low-strength) and shaly rocks the diameter can increase. This is a washout. Pad tools then lose contact with the wall, the density log underestimates density, and the neutron log overestimates porosity. If the diameter decreases opposite a bed, the cause is mudcake, which is a sign of permeability. For a uniform cake the thickness is $h_{mc} = (d_{bit} - d_{cal})/2$ .
2
Mud and its resistivities
Find in the log heading the mud type (water-based or oil-based, fresh or salt), the values of $R_m$ , $R_{mf}$ and $R_{mc}$ , and the temperature at which they were measured. Bring the resistivities to formation temperature. Note the additives: barite interferes with the density log, and potassium chloride raises the GR readings.
3
Expected profile
Use the ratio $R_{mf}/R_w$ to predict how the shallow and deep curves should compare in a bed that we know is water-bearing. This is the reference for comparing the other beds.
4
Separation of the curves
Separation of curves of different depths of investigation points to invasion, that is, to permeability. A departure of $R_{xo}/R_t$ from the water-bearing reference is a sign of moveable hydrocarbons. Cake on the caliper log, and microlog separation (the two microlog curves read differently), confirm permeability.
5
Choice of curves
Decide which curves are suitable for $R_t$ and $R_{xo}$ , and which intervals to exclude from the quantitative interpretation (washouts, thin beds, an unclear invasion profile). The further order of work is in the topic Log interpretation workflow.

Worked exampleMudcake thickness from the caliper log

Given (assumed data)

Bit diameter $d_{bit}$: 215.9 mm (8½ in)
Caliper diameter opposite the sandstone $d_{cal}$: 206.5 mm
Caliper diameter opposite the shales: 262 mm

Solution

Opposite the sandstone the hole is narrowed: $h_{mc} = (215.9 - 206.5)/2 = 4.7$ mm, or 0.19 in.
Opposite the shales the hole is enlarged by 46 mm in diameter, that is by 21%.

Answer: Opposite the sandstone there is a cake about 4.7 mm thick; opposite the shales, a washout.

What it means. The cake means that the sandstone is permeable and has an invaded zone. We correct the readings of microresistivity tools there for the cake. In washouts pad tools may not touch the wall, so opposite the shales we check the density log and microresistivity tools with special care, or exclude them. The data in this example are assumed for illustration.

Terms and symbols

Overbalance: The excess of the pressure in the well over the formation pressure. It is the driving force of filtrate invasion.
Mud filtrate, $R_{mf}$: The liquid part of the mud that has passed through the cake into the formation, and its resistivity.
Mudcake, $h_{mc}$ , $R_{mc}$: A layer of the solid particles of the mud on the wall opposite permeable beds, with its thickness and resistivity.
Flushed zone, $R_{xo}$ , $S_{xo}$: The part of the formation next to the borehole, where filtrate has replaced almost all the formation fluids except the residual hydrocarbons.
Invaded zone, $d_i$ , $R_i$: The flushed and transition zones together, with their diameter and resistivity.
Uninvaded zone, $R_t$ , $S_w$: The formation beyond the invasion front, with its original fluids; its true resistivity and its water saturation.
Annulus: A ring of raised water saturation and lowered resistivity at the boundary of the invaded zone in productive beds.
Depth of investigation: The distance from the borehole wall from which a tool receives most of its signal.

Common mistakes and misconceptions

Wrong

“A deep resistivity sonde reads $R_t$ .”

In fact

It reads an apparent resistivity (the value the tool measures, which mixes several zones) that includes the invaded zone. We find $R_t$ with corrections for the borehole, the enclosing rocks and invasion.

Wrong

“The resistivity curves have separated, so the bed contains oil.”

In fact

Separation means invasion, that is, permeability. With fresh mud it is greatest in water-bearing beds. Oil is indicated by a departure of $R_{xo}/R_t$ from the value in a water-bearing bed.

Wrong

“The more permeable the bed, the deeper the invasion.”

In fact

The cake mainly sets the volume of filtrate, and in a permeable bed the cake forms quickly. The depth depends on where this volume fits: in low-porosity rocks invasion is usually deeper.

Wrong

“The density and neutron logs see the formation oil and water.”

In fact

They see the flushed zone: filtrate and residual hydrocarbons. We allow for this by setting the fluid properties in the porosity calculation.

Wrong

“The caliper log is only needed by drillers.”

In fact

It is the first check of the quality of all the other curves: washouts spoil the readings of pad tools, and the cake points to permeable beds.

In depth: advanced detail

How the mudcake grows: static and dynamic filtration

At rest, without circulation, each new volume of filtrate leaves a new portion of solid particles on the wall, and the cake thickens. The thickness of the cake is proportional to the volume of filtrate that has passed: $h_{mc} = c\,V/A$ . By Darcy's law (the law of flow through porous media), the flow through the cake is inversely proportional to its thickness: $\dfrac{\dd V}{\dd t} = \dfrac{k_{mc}\,A\,\Delta p}{\mu_f\,h_{mc}}$ . We put the first relation into the second and get $V\,\dd V = \dfrac{k_{mc} A^2 \Delta p}{\mu_f\,c}\,\dd t$ . After integration:

Static filtration through a growing cake

V = A\sqrt{\frac{2\,k_{mc}\,\Delta p\,t}{\mu_f\,c}}

where:

$V$ volume of filtrate that has passed through the cake in time $t$ , m³
$A$ filtration area (of the wall), m²
$k_{mc}$ permeability of the cake, m²
$\Delta p$ pressure drop across the cake (approximately equal to the overbalance), Pa
$t$ filtration time, s
$\mu_f$ viscosity of the filtrate, Pa·s
$c$ volume of cake formed per unit volume of filtrate, —

Units: SI (any consistent system)

Holds when

Filtration at rest, without circulation.
The cake is incompressible, and its permeability and composition are constant.
The resistance of the formation is small compared with the resistance of the cake.
The cake is thin compared with the borehole radius (a planar problem).
We neglect the instant initial filtration (spurt loss). In practice an initial volume $V_{sp}$ is added to the formula.

The volume of filtrate grows as the square root of time. In the first minutes a noticeable share of the hourly total already goes into the formation. So the standard 30-minute fluid loss is sometimes estimated as twice the volume collected in 7.5 minutes: $\sqrt{30/7.5} = 2$ . Real cakes are compressible, and the volume grows with the pressure drop more weakly than $\sqrt{\Delta p}$ .

During circulation the flow of mud in the annular space (the gap between the drill string and the borehole wall) washes away the outer layer of the cake. The cake stops growing in thickness, and filtration continues at an almost constant rate: the volume grows roughly in proportion to time. So most of the filtrate goes into the formation during drilling and circulation. In periods of rest (tripping, logging) the cake builds up and filtration slows.

How zones combine in one reading: laterolog and induction log

The reading of a resistivity sonde is a mixture of the contributions of the zones. The way they mix depends on how the current flows. In a laterolog, the current from the electrodes goes from the borehole into the formation along the radius, and it crosses the cake, the flushed zone and the uninvaded formation in turn. The resistivities of sections connected in series (one after another) add up.

Laterolog reading with step invasion

R_{LL} \approx J\,R_{xo} + (1 - J)\,R_t

where:

$R_{LL}$ apparent resistivity from the laterolog sonde, Ω·m
$J$ pseudo-geometric factor: the share of the invaded zone in the reading; it grows with $d_i$ and differs between shallow and deep sondes, —
$R_{xo},\ R_t$ resistivity of the invaded zone and of the uninvaded formation, Ω·m

Units: any resistivity units, the same for all

Holds when

A step invasion profile.
Corrections for the borehole and the enclosing rocks have already been applied.
A conductive mud (the laterolog does not work in oil-based muds).

A weighted mean of resistivities is “pulled” towards the larger value: a high-resistivity invaded zone strongly overstates the reading.

In an induction log, the alternating magnetic field of the coils induces ring currents in the rock around the axis of the borehole. The zones act in parallel, so their conductivities add up.

Induction sonde reading with step invasion

\frac{1}{R_{IL}} \approx \frac{G}{R_{xo}} + \frac{1 - G}{R_t}

where:

$R_{IL}$ apparent resistivity from the induction sonde, Ω·m
$G$ geometric factor of the invaded zone: its share in the signal, —
$R_{xo},\ R_t$ resistivity of the invaded zone and of the uninvaded formation, Ω·m

Units: any resistivity units, the same for all

Holds when

A step invasion profile.
Corrections for the borehole, the enclosing rocks and the skin effect have already been applied.

A weighted mean of conductivities is “pulled” towards the lower resistivity: a low-resistivity invaded zone strongly understates the reading.

Worked exampleOne bed through the eyes of a laterolog and an induction sonde

Given (assumed data)

Case A: fresh mud, water-bearing bed: $R_{xo} = 12.4$ Ω·m, $R_t = 1.24$ Ω·m (as in the example above)
Case B: salt mud, productive bed: $R_{xo} = 2.0$ Ω·m, $R_t = 20$ Ω·m
Share of the invaded zone in the reading: for clarity, the same for both sondes: $J = G = 0.25$

Solution

Case A, laterolog sonde: $R_{LL} = 0.25\cdot 12.4 + 0.75\cdot 1.24 = 4.03$ Ω·m; induction sonde: $1/R_{IL} = 0.25/12.4 + 0.75/1.24$ , whence $R_{IL} = 1.60$ Ω·m.
Case B, laterolog sonde: $R_{LL} = 0.25\cdot 2.0 + 0.75\cdot 20 = 15.5$ Ω·m; induction sonde: $1/R_{IL} = 0.25/2.0 + 0.75/20$ , whence $R_{IL} = 6.15$ Ω·m.
Departure from $R_t$ . In case A the laterolog overstates it by a factor of 3.25, and the induction sonde overstates it by 29%. In case B the laterolog understates it by 22.5%, and the induction sonde understates it by a factor of 3.25.

Answer: When the invaded zone has a high resistivity (

R_{xo} > R_t

), the induction sonde is closer to the truth. When it has a low resistivity (

R_{xo} < R_t

), the laterolog is.

What it means. This gives a practical rule. With fresh mud, and with oil-based mud, we usually find $R_t$ from the induction log. With salt mud and high formation resistivities, we usually find it from the laterolog. Using the same share of the invaded zone for the two tools is a simplification: real sondes differ in their radial response, and we correct from charts for the specific tool. The data in this example are assumed for illustration.

Connections to other topics

Open-hole logging: why, how and when — why and when open-hole logging is carried out and what the suite consists of.
Resistivity logs: electrical and induction tools — how resistivity sondes of different depths of investigation are built and how $R_t$ is found from them.
Microresistivity tools and flushed-zone resistivity — how the resistivity of the flushed zone and the cake thickness are measured.
Archie's equation and saturation of clean reservoirs — how to calculate water saturation from $R_t$ and $R_{xo}$ .
Resistivity ratio methods and movable oil — saturation and moveable oil from the ratio of the resistivities of the zones.
Drilling fluids: functions and properties — what a drilling fluid consists of and what its fluid loss depends on.
Formation damage and damage skin — how filtrate and the solid particles of the mud reduce the permeability of the near-wellbore zone.
Supercharging and permeability from formation-tester pretests — why the pressure measured by a formation tester in a tight bed can be overstated.

Check yourself

1
A water-bearing sandstone has been drilled with fresh mud, and at formation temperature $R_{mf}/R_w = 8$ . What is the most likely relation between $R_{xo}$ and $R_t$ ?
1. A. $R_{xo}$ is about 8 times smaller than $R_t$
2. B.The relation is determined by the porosity alone
3. C. $R_{xo}$ is about 8 times greater than $R_t$
4. D. $R_{xo} \approx R_t$
Show the answer
Correct answer: C. $R_{xo}$ is about 8 times greater than $R_t$
In a clean water-bearing bed $R_{xo} = F R_{mf}$ and $R_t = F R_w$ with the same formation resistivity factor $F$ , so $R_{xo}/R_t = R_{mf}/R_w = 8$ (with complete flushing and after corrections). Porosity enters both resistivities in the same way and cancels.
2
Which of the methods mainly “sees” the flushed zone rather than the uninvaded formation?
1. A.None: all methods see the uninvaded formation
2. B.Deep induction log
3. C.Gamma-gamma density log
4. D.Deep laterolog
Show the answer
Correct answer: C. Gamma-gamma density log
The depth of investigation of the density method is about 10 cm, and almost all its signal comes from the flushed zone. So the fluid density in the porosity calculation is the density of filtrate with residual hydrocarbons, not of the formation fluid.
3
Two neighbouring water-bearing beds were drilled at the same time with the same mud; their porosities are 24% and 9%. In which bed is invasion most likely deeper, and why?
Show the answer
In the bed with 9% porosity. As long as the cake controls filtration, about the same volume of filtrate goes into both beds. In the low-porosity bed the filtrate needs more rock to fit into. By the volume balance, $d_i^2 - d_h^2$ is inversely proportional to $\phi\,S_{xo}$ .
4
Opposite a sandstone the caliper log shows a diameter smaller than the bit size, and opposite shales a noticeably larger one. What does this say about the beds and about the quality of the curves?
Show the answer
The narrowing opposite the sandstone is mudcake. The sandstone is permeable and has an invaded zone, so the readings of microresistivity tools have to be corrected for the cake. The enlargement opposite the shales is a washout. Pad tools (the density log, microresistivity tools) may lose contact with the wall there, so we check their readings or exclude them from the quantitative interpretation.
5
In a tight, low-permeability bed a wireline formation tester showed a pressure noticeably higher than expected. What is the most likely cause?
1. A.The influence of the mudcake on the resistivity
2. B.A depth-matching error
3. C.Supercharging: the filtrate has no time to dissipate, and the pressure at the wall is higher than the formation pressure
4. D.Abnormally high pressure throughout the bed
Show the answer
Correct answer: C. Supercharging: the filtrate has no time to dissipate, and the pressure at the wall is higher than the formation pressure
In a tight rock the cake isolates the formation poorly, compared with the rock's own resistance to filtration, and the incoming filtrate raises the pressure at the wall. Before concluding that there is overpressure, we check for signs of supercharging, for example whether the points agree with the pressure gradient in neighbouring permeable beds.
6
A well was drilled with oil-based mud. Why does the suite have no SP curve and no laterolog, and why was the resistivity measured by the induction method?
Show the answer
An oil-based mud does not conduct current. The SP potential arises and is recorded only in a conductive mud, and a laterolog has to pass current from the borehole into the formation. The induction method induces currents in the rock with a magnetic field and does not require a conductive mud.

Updated: 19 October 2026