Abnormal formation pressures: origin, prediction and recognition

Quick refresher

Normal pressure is the pressure of a column of formation water from the surface. It is usually 9.8–10.5 kPa/m. In equivalent mud weight, which is the density of a mud column that would give the same pressure, this is 1.00–1.07 g/cm³. Above it is overpressure, below it is subnormal pressure. Overpressure has an upper limit: the pressure at which fractures open in the seal.
Overpressure is “locked in” under a barrier of very low permeability: a thick shale sequence, salt, cemented rocks or a sealing fault. Truly impermeable seals hardly exist, so overpressure is a balance between a source of pressure and its leakage.
The most common mechanism in young, rapidly sinking shale sequences is undercompaction. The water has no time to escape, the fluid takes over the load, and the shales stay porous. Its “trace” is the departure of the porosity, density, sonic velocity (the speed of sound in the rock), resistivity (how strongly the rock resists electric current) and drillability (how easily the rock drills) of shales from the normal trend.
Unloading mechanisms (the generation of hydrocarbons and the release of clay-bound water) raise the pressure after the compaction and hardly change the porosity. Methods based on compaction trends underestimate such pressure.
Signs of a transition to overpressure while drilling: a rise in the corrected rate of penetration (a fall of the d-exponent); a rise in background gas (the steady level of gas in the mud) and in gas after connections (when a pipe is added to the string); a steeper rise in the temperature of the mud at the outlet; a lower density of shale cuttings; large splinters of cavings (pieces of rock that fall from the wall of the hole); and overpull (extra pull needed to move the string up).
The shales of the transition zone (the zone where the pressure rises from normal to high) are almost impermeable and give no kick themselves. The influx starts when the bit opens a permeable formation under them. The signs in the transition zone are the last chance to make the mud heavier or to run casing.

Key relationships

d = \dfrac{\log_{10}\left(R/(60N)\right)}{\log_{10}\left(12W/(10^{6}B)\right)}

d-exponent in oilfield units: $R$ in ft/h, $N$ in rpm, $W$ in lbf, $B$ in inches

d_c = d\,\dfrac{\rho_n}{\rho_m}

correction for the mud density

p_p = \sigma_v - (\sigma_v - p_n)\left(\dfrac{d_{c,o}}{d_{c,n}}\right)^{1.2}

Eaton method: the pressures at one depth (equivalent mud weights may be used)

p_p(z) = \sigma_v(z) - \left[\sigma_v(z_e) - p_n(z_e)\right]

equivalent depth method

Common pitfalls

Thinking that overpressure can always be “seen” in advance. Under salt and cemented seals the transition is sharp, and compaction trends underestimate the pressure created by unloading or by transfer from another place.
Building the d-exponent trend from sandstones and limestones, or not rebuilding it after a change of the type of bit.
Waiting for an influx from the shales of the transition zone. The permeable formation below them is the danger.
Forgetting the delay: the gas and the cuttings need time to rise from the bottom, so they reach the surface when the bit has already gone further.

After this lesson you can

explain why overpressure needs a source of pressure and a seal, and list the main mechanisms of overpressure and of subnormal pressure
explain why undercompaction (a rock that holds more water and pores than is normal for its depth) leaves a “trace” in the properties of shales, and why methods based on compaction trends underestimate the pressure created by unloading mechanisms (processes that raise the pressure after compaction, so that the rock grains carry less load)
name the signs of a transition to overpressure in the drilling parameters, the mud and the cuttings, and explain the physics of each
calculate the d-exponent (a number that corrects the drilling rate for weight, rotary speed and bit size) with a correction for the mud density, and use it to estimate the pore pressure by the Eaton method
describe how pressure is forecast before drilling and confirmed after it, and where each way has its limits

Normal and abnormal pressure

Normal, or hydrostatic, pore pressure is pore pressure that equals the pressure of a column of formation water from the surface to the given depth. Its gradient (how fast the pressure grows with depth) depends on the salinity of the water of the area (how salty it is). It is usually 9.8–10.5 kPa/m (0.433–0.465 psi/ft), which is an equivalent mud weight of 1.00–1.07 g/cm³ (the details are in the lesson on pore and overburden pressure). Pressure above normal is called overpressure (abnormally high formation pressure, also called geopressure). Pressure below normal is called subnormal pressure (abnormally low formation pressure, also called underpressure).

It is convenient to express the size of the anomaly by the abnormality coefficient $k_a$ , the ratio of the pore pressure to the pressure of a column of fresh water at the same depth. Numerically it equals the equivalent mud weight in g/cm³. Overpressure has an upper limit. When the pore pressure comes close to the pressure at which fractures open in the seal, the fluid starts to escape upwards, and the pressure does not grow any further. This pressure is usually lower than the overburden pressure (the pressure from the weight of all the rock above), so even the strongest overpressure lies between the normal pressure and the overburden pressure. In some basins the equivalent mud weight of the pore pressure is above 2 g/cm³.

Any overpressure is pressure “locked in” under a barrier. If the pores were connected with the surface, the excess pressure would disperse. A seal holds it: a thick sequence of shale, rock salt, tightly cemented carbonates or a sealing fault. Truly impermeable rocks hardly exist. A seal only slows down the equalising of pressure to geological time scales. So overpressure is a dynamic balance between a source of pressure and its leakage. If the source dries up, the pressure disperses over millions of years. An area that is surrounded by barriers and has a pressure of its own is called a pressure compartment.

Undercompaction: the water has no time to escape

Clay sediments are very loose when they pile up. The porosity of soft clay ooze at the sea floor is 60–80%. As the sediment sinks, the weight of the new layers squeezes it, the particles come closer, and water is pressed out of the pores. The water has to go upwards, or into neighbouring sand layers that carry it away. Shales have very low permeability, and the water is squeezed out of them slowly.

While the sediments sink slowly, the water has time to escape. The effective stress (the part of the load that the rock skeleton carries) grows with depth, the rock compacts along the normal trend, and the pore pressure stays hydrostatic. If the sinking is fast, and the shale sequence is thick and has almost no permeable layers, the water has no time to escape. Below some depth, the top of the overpressure zone, the pore water takes almost all the added load. The effective stress stops growing, the porosity “freezes”, and the pore pressure then rises almost in parallel with the overburden pressure. This mechanism is called undercompaction (also called disequilibrium compaction). For an analogy with a spring in a cylinder from which the water cannot escape, see the lesson on pore and overburden pressure.

Normal compaction: porosity as a function of effective stress

\phi = \phi_0\,\e^{-c\,\sigma'_v},\qquad \sigma'_v = \sigma_v - p_p

where:

$\phi$ porosity of the shale, fraction
$\phi_0$ porosity at zero effective stress (at the surface), fraction
$c$ compaction coefficient of the given shale, MPa⁻¹
$\sigma'_v$ effective vertical stress: the overburden pressure $\sigma_v$ minus the pore pressure $p_p$ , MPa

Units: $\phi$ in fraction, $\sigma'_v$ in MPa, $c$ in MPa⁻¹

Holds when

an empirical relation for one lithology: the shales of the given basin. The parameters are fitted to logs (measurements made along the hole) and to cores (rock samples cut from the well);
it describes only compaction as the effective stress grows (the “loading curve”). When the stress falls, the porosity hardly recovers.

Porosity is a “memory” of the highest effective stress that the rock has felt. If the effective stress stops growing, the porosity stops falling. So from the porosity, and from the density, sonic velocity, resistivity and drillability of shales that go with it, we can estimate the effective stress. From it we find the pore pressure, $p_p = \sigma_v - \sigma'_v$ .

This is where the “trace” of undercompaction comes from. Below the top of the overpressure zone the properties of shales depart from the normal trend. The porosity is higher, the density and the sonic velocity are lower, the electrical resistivity is lower (there is more conducting water), and the bit goes deeper faster. All trend-based methods of forecasting and recognising overpressure rest on this: those that use seismic, log and drilling data.

CalculatedAn illustrative loading curve of a shale,

\phi = 0.45\,\e^{-0.035\,\sigma'_v}

, and an unloading branch. A is a normally compacted shale. B is an undercompacted one: the effective stress is lower and the porosity is higher, but the point stays on the loading curve. C is unloading: the pore pressure rose after the compaction, the effective stress fell from 36 to 24 MPa, and the porosity hardly changed. By the loading curve, the porosity of point C “means” about 35 MPa, and the pressure would be underestimated by more than 10 MPa.

Other sources of overpressure and the causes of subnormal pressure

It is convenient to divide the mechanisms of overpressure into three groups. Loading mechanisms: the load on the rock grows faster than the water escapes. They are undercompaction and tectonic compression (squeezing of the earth's crust). Volume mechanisms (unloading mechanisms): an extra volume of fluid appears in a closed pore space. Redistribution: the pressure is passed on from another part of the rock sequence, along faults, leaking wells or along a dipping (tilted) bed.

Mechanisms of overpressure and their trace in the properties of shales

Mechanism: Undercompaction
What happens: rapidly sinking shales have no time to give up their water; the fluid carries part of the weight of the rocks
Trace in shales: porosity above the trend; density, velocity and resistivity below it

Mechanism: Tectonic compression
What happens: side squeezing loads the rock faster than the water escapes; an uplifted block keeps its former pressure
Trace in shales: as with undercompaction, often together with faults and fractures

Mechanism: Hydrocarbon generation
What happens: organic matter turns into oil and gas, and the volume of fluid grows, especially strongly when gas forms
Trace in shales: almost none: this is unloading

Mechanism: Mineral transformation
What happens: smectite changes into illite (both are clay minerals) and gypsum into anhydrite, and clay-bound water is released
Trace in shales: weak: this is unloading, and in addition the permeability falls

Mechanism: Heating of water
What happens: water in a closed volume expands as it sinks deeper
Trace in shales: none; the mechanism is usually secondary

Mechanism: Redistribution
What happens: pressure is passed up along faults, leaking wells and a dipping bed, and into the top of a tall accumulation
Trace in shales: none: a sand is “charged” with pressure without any signs in the shales

Mechanism: Salt
What happens: an impermeable seal; domes disturb the bedding of the layers
Trace in shales: a sharp transition under the salt

Mechanism	What happens	Trace in shales
Undercompaction	rapidly sinking shales have no time to give up their water; the fluid carries part of the weight of the rocks	porosity above the trend; density, velocity and resistivity below it
Tectonic compression	side squeezing loads the rock faster than the water escapes; an uplifted block keeps its former pressure	as with undercompaction, often together with faults and fractures
Hydrocarbon generation	organic matter turns into oil and gas, and the volume of fluid grows, especially strongly when gas forms	almost none: this is unloading
Mineral transformation	smectite changes into illite (both are clay minerals) and gypsum into anhydrite, and clay-bound water is released	weak: this is unloading, and in addition the permeability falls
Heating of water	water in a closed volume expands as it sinks deeper	none; the mechanism is usually secondary
Redistribution	pressure is passed up along faults, leaking wells and a dipping bed, and into the top of a tall accumulation	none: a sand is “charged” with pressure without any signs in the shales
Salt	an impermeable seal; domes disturb the bedding of the layers	a sharp transition under the salt

In real basins the mechanisms often act together. Which of them is the main one is decided from the geology of the area and by comparing different methods of estimating the pressure.

Subnormal pressure is less frequent than overpressure, but for drilling it matters no less. Its causes:

depletion of a reservoir during production: the most frequent and the strongest source. The pressure in the reservoir may fall several times over, while the shales and seals around it keep their former pressure;
uplift and erosion of rock sequences: when the load is removed, the pores widen a little, and the fluid cools and contracts, so the pressure in an isolated bed falls. If these effects are weaker, an uplifted bed, on the contrary, keeps the pressure that it gained at depth and ends up in the overpressure zone (see the table above);
a low potentiometric surface: the level to which the water from the bed would rise lies below the wellhead. This happens in arid areas with a deep level of groundwater, or when the bed comes to the surface below the wellhead and discharges there.

The opposite case, where a bed comes to the surface above the wellhead, in the mountains, gives overpressure of the artesian type. The potentiometric surface lies above the wellhead, and water from such a bed may flow out of the well by itself.

The transition zone: a warning before a dangerous bed

Between an interval with normal pressure and a permeable bed with overpressure lies the transition zone: the seal itself, in which the pore pressure rises with depth. In a thick shale sequence the pressure rises gradually, over tens and hundreds of metres. As the bit approaches the bed with overpressure, the properties of the shales change. This gives us time to notice the danger. If the seal is salt or a dense cemented rock, the transition is sharp, and we cannot see it in advance from the trends.

The shales of the transition zone are almost impermeable. Even if the pressure in them is higher than the pressure of the mud, they give no noticeable influx. A kick begins when the bit opens a permeable bed with overpressure under the seal. So the transition zone is the last warning. While the bit is in it, we can still make the mud heavier, or stop and run casing, to isolate the interval with normal pressure from the beds with overpressure.

As the bit goes deeper into the transition zone, the overbalance on the rock (how far the mud pressure is above the pore pressure) gets smaller, and then the mud pressure becomes lower than the pore pressure of the shales. This strengthens the signs of the transition themselves: the bit goes deeper faster, gas from the shales enters the mud, and fragments break off the walls. The smaller the overbalance, the earlier and the clearer the signs. But the margin in case a permeable bed is opened gets smaller too. This compromise is settled by the well design and the safety rules, and not by the drilling crew while the work is under way.

Why abnormal pressures are dangerous in drilling

The mud pressure must be higher than the pore pressure of the permeable beds and than the pressure in the shales (otherwise the walls cave in). It must also be lower than the fracture pressure (the pressure at which the rock cracks) of the weakest rock in the open hole (the part of the well not yet lined with casing). Abnormal pressures narrow this window or close it, and the main problems come from that.

Problems linked with abnormal pressures

Situation: A bed with overpressure that was not predicted is opened
What happens: formation fluid flows in: a kick
How the risk is reduced: a pressure forecast, watching the signs of a transition, making the mud heavier in time

Situation: A weak bed is open above the overpressure zone
What happens: the mud that holds the overpressure fractures the weak bed: lost circulation, a fall of the mud level, a kick
How the risk is reduced: cover the weak interval with casing before entering the overpressure zone

Situation: The casing was run too shallow
What happens: after the well is shut in on a kick, the pressure under the casing shoe (the bottom end of the casing) exceeds the fracture pressure
How the risk is reduced: choose the depth of the shoe by the fracture pressure

Situation: Heavy mud against beds with normal or low pressure
What happens: differential sticking (the pipe is pressed to the wall by the pressure difference and gets stuck), lost circulation, damage to the reservoir, slow drilling
How the risk is reduced: do not make the density higher than needed, cover such intervals with casing

Situation: The mud is lighter than the pore pressure of the shales
What happens: caving of the walls, overpull, tight spots when running in, the bottom filling with cuttings
How the risk is reduced: keep the density no lower than the pressure in the shales

Situation: A depleted bed (subnormal pressure)
What happens: lost circulation and sticking; the fracture pressure is also lowered
How the risk is reduced: light muds, plugging additives, covering with casing

Situation	What happens	How the risk is reduced
A bed with overpressure that was not predicted is opened	formation fluid flows in: a kick	a pressure forecast, watching the signs of a transition, making the mud heavier in time
A weak bed is open above the overpressure zone	the mud that holds the overpressure fractures the weak bed: lost circulation, a fall of the mud level, a kick	cover the weak interval with casing before entering the overpressure zone
The casing was run too shallow	after the well is shut in on a kick, the pressure under the casing shoe (the bottom end of the casing) exceeds the fracture pressure	choose the depth of the shoe by the fracture pressure
Heavy mud against beds with normal or low pressure	differential sticking (the pipe is pressed to the wall by the pressure difference and gets stuck), lost circulation, damage to the reservoir, slow drilling	do not make the density higher than needed, cover such intervals with casing
The mud is lighter than the pore pressure of the shales	caving of the walls, overpull, tight spots when running in, the bottom filling with cuttings	keep the density no lower than the pressure in the shales
A depleted bed (subnormal pressure)	lost circulation and sticking; the fracture pressure is also lowered	light muds, plugging additives, covering with casing

Primary well control, the first protection against a kick, is the pressure of the mud column. It is covered in detail in the lesson on well control. How the depths of casing strings are chosen from the pore pressure and the fracture pressure is explained in the lesson on the fracture gradient.

Forecast before drilling and checking afterwards

We estimate the pressure in a planned well in advance from three groups of data. The geology of the area: where the seals and the salt lie, which sequences sank fast, where the source rocks (the rocks that generate oil and gas) entered the generation zone, and which reservoirs are already produced and depleted. Offset wells: pressure measurements in beds, mud densities, kicks, lost circulation, stuck pipe, gas after connections, leak-off tests (tests that find the pressure at which the rock starts to take mud), logs of shales. Seismic data: interval velocities of waves (the speed of sound in each layer), found by processing data over the whole area.

The sonic velocity in undercompacted shales is lower than the normal trend. So from seismic velocities we can see the overpressure zone even before drilling, and estimate the pressure by the same formulas as from logs (sonic and density). The depth resolution of seismic data is hundreds of metres, and the pressure is sensitive to an error in the velocity. In the Eaton method the effective stress is proportional to the cube of the ratio of velocities, and a velocity error of 5% gives an error of the effective stress of about 15%. So a seismic forecast must always be calibrated against wells.

The cycle of work with pressure

1
Forecast
From the geology, offset wells and seismic data, we draw the curves of the pore pressure and the fracture pressure, with a range of uncertainty. The choice of mud and the well design are based on them.
2
Control while drilling
We compare the trends of the d-exponent, gas, cuttings and measurements while drilling (Measurement while drilling) with the forecast. A difference is a reason to review the mud density and the casing depths.
3
Confirmation
After drilling we refine the pressure in shales from logs (sonic, density, neutron, resistivity). In permeable beds we measure it directly with a wireline formation tester (a tool lowered on a cable that measures the pressure in the rock; Pressure measurement with the wireline formation tester), with a drill stem test (DST) tool run on pipe, or in a well test.
4
Updating the model
The measurements calibrate the trends and the parameters of the methods for the next wells of the area.

Recognition while drilling: drilling parameters, mud, cuttings

While drilling, we look for a transition to overpressure in three sources of data: the drilling parameters, which are recorded at once, the mud and the cuttings. The last two reach the surface only after the time they need to rise from the bottom. In a deep well that is tens of minutes, or even more than an hour, and in that time the bit may drill several metres more. The data are collected by the mud logging unit (the unit that records the drilling parameters and examines the mud and cuttings at the surface). Measurements while drilling add logs of the shales just behind the bit.

Signs of a transition to overpressure

Sign: Corrected rate of penetration (d-exponent)
How it changes: the rate rises, $d_c$ falls below the trend
Why: the rock is more porous and weaker, the overbalance is smaller, and the drilled particles come off the bottom more easily

Sign: Background gas
How it changes: rises
Why: gas escapes from the pores of the drilled shale, and at a small overbalance, from the walls as well

Sign: Gas after a connection and after trips (pulling the string out and running it back in)
How it changes: peaks of gas that come from the bottom after each stop of circulation
Why: when circulation stops there are no friction losses in the annulus (the space between the pipe and the wall of the hole), and when the string is pulled the pressure at the bottom drops; for a while it falls below the pore pressure of the shales

Sign: Temperature of the mud at the outlet
How it changes: rises more steeply with depth
Why: in porous water-filled shales the thermal conductivity (how easily heat flows through the rock) is lower, and the temperature of the rocks rises faster with depth

Sign: Density of shale cuttings
How it changes: lower than the trend
Why: undercompacted shale is more porous

Sign: Size and shape of cuttings
How it changes: large concave splinters
Why: when the mud pressure is lower than the pore pressure, the shale breaks off the wall

Sign: Torque, overpull, tight spots
How it changes: rise
Why: caving of the walls and narrowing of the hole

Sign	How it changes	Why
Corrected rate of penetration (d-exponent)	the rate rises, $d_c$ falls below the trend	the rock is more porous and weaker, the overbalance is smaller, and the drilled particles come off the bottom more easily
Background gas	rises	gas escapes from the pores of the drilled shale, and at a small overbalance, from the walls as well
Gas after a connection and after trips (pulling the string out and running it back in)	peaks of gas that come from the bottom after each stop of circulation	when circulation stops there are no friction losses in the annulus (the space between the pipe and the wall of the hole), and when the string is pulled the pressure at the bottom drops; for a while it falls below the pore pressure of the shales
Temperature of the mud at the outlet	rises more steeply with depth	in porous water-filled shales the thermal conductivity (how easily heat flows through the rock) is lower, and the temperature of the rocks rises faster with depth
Density of shale cuttings	lower than the trend	undercompacted shale is more porous
Size and shape of cuttings	large concave splinters	when the mud pressure is lower than the pore pressure, the shale breaks off the wall
Torque, overpull, tight spots	rise	caving of the walls and narrowing of the hole

Each sign may have other causes: a change of lithology, wear of the bit, a change of the pump rate and of the mud properties. Only a consistent shift of several signs in uniform shales is reliable.

A rise in the volume of mud in the pits (the tanks at the rig that hold the mud) and an overflow (the well flows by itself) are no longer signs of a transition, but signs of a kick: fluid is entering the well. They are covered in the lesson on well control. The same lesson explains why a sudden rise in the rate of penetration (a drilling break) is a reason to be on guard and to check the well.

The d-exponent and the Eaton method for estimating the pressure

In normally compacted shales the rate of penetration falls with depth, because the rock is denser and stronger. In undercompacted shales it rises, because the rock is more porous, and the smaller overbalance presses the drilled particles to the bottom less. But the rate also depends on the weight on bit, the rotary speed and the bit diameter, so we cannot compare the rate itself. We normalise it, which means that we correct it for these factors. The basis is an empirical equation of the drilling rate (one based on observation), $R = a\,N^{e}\,(W/B)^{d}$ . At a given weight per unit of diameter $W/B$ and a given rotary speed $N$ , the rate $R$ is set by the drillability of the rock. If we take the coefficients $a$ and $e$ as one (a single lithology, which means one type of rock, and a rate proportional to the rotary speed), the equation gives the exponent $d$ , the d-exponent.

d-exponent (the normalised rate of penetration)

d = \frac{\log_{10}\left(\dfrac{R}{60\,N}\right)}{\log_{10}\left(\dfrac{12\,W}{10^{6}\,B}\right)}

where:

$R$ rate of penetration, ft/h
$N$ rotary speed of the bit, rpm
$W$ weight on bit, pound-force (lbf)
$B$ bit diameter, inches

Units: oilfield units: $R$ in ft/h, $N$ in rpm, $W$ in lbf, $B$ in inches

Holds when

an empirical formula for one lithology, uniform shales. The other influences (the type of bit and its wear, the hydraulics of flushing, the properties of the mud) are taken as unchanged. If they change sharply, we build the trend again;
we hold the weight and the rotary speed as constant as possible, and we take points at small steps of depth;
the logarithms in the numerator and the denominator are taken to one base (usually base 10). Both arguments are less than one, both logarithms are negative, and $d$ is positive;
the constants inside the logarithms are not dimensionless: the formula is used strictly in the units for which the constants are written.

$R/(60N)$ is the advance in one revolution of the bit, and $12W/(10^6 B)$ is the weight per inch of bit diameter, with a scale factor of $12/10^6$ . The more slowly the rock drills at the same weight and rotary speed, the greater $d$ is. In normally compacted shales $d$ grows with depth. If it falls below the trend, the rock drills more easily than it “should” at this depth. Here the weight and the bit diameter are $W$ and $B$ . In the lesson Bit performance and cost per metre the same quantities are $G$ and $D$ .

In other unit systems (2)

Metric units (weight in kN)

d = \frac{\log_{10}\left(\dfrac{R}{18.29\,N}\right)}{\log_{10}\left(\dfrac{0.0685\,W}{B}\right)}

$R$ in m/h, $N$ in rpm, $W$ in kN, $B$ in mm.

$18.29 = 60\cdot 0.3048$ (m/h into ft/h); $0.0685 = 12\cdot 224.8\cdot 25.4\cdot 10^{-6}$ (1 kN = 224.8 lbf, 1 inch = 25.4 mm). This form gives the same value of $d$ as the original.

Metric units (weight in tonnes-force)

d = \frac{\log_{10}\left(\dfrac{R}{18.29\,N}\right)}{\log_{10}\left(\dfrac{0.672\,W}{B}\right)}

$R$ in m/h, $N$ in rpm, $W$ in tf (1 tf = 9.807 kN), $B$ in mm.

$0.672 = 12\cdot 2204.6\cdot 25.4\cdot 10^{-6}$ (1 tf = 2204.6 lbf).

The formula does not take the mud density into account, but the density affects the rate: the heavier the mud, the more the pressure at the bottom presses the drilled particles down, and the lower the rate. When the mud is made heavier, $d$ rises and may hide the transition. So we use the corrected d-exponent $d_c$ :

Corrected d-exponent

d_c = d\,\frac{\rho_n}{\rho_m}

where:

$d_c$ d-exponent corrected for the mud density, —
$\rho_n$ equivalent mud weight of the normal pore pressure of the area, g/cm³
$\rho_m$ mud density (sometimes the equivalent circulating density is taken, which adds the effect of friction while the mud flows), g/cm³

Units: densities in any identical units

Holds when

a simple empirical correction, not a law of physics. It is enough to remove from the plot the jumps caused by making the mud heavier.

With a mud heavier than the normal pressure needs, $d_c < d$ . We build the trend from $d_c$ and look for departures from it.

We plot the values of $d_c$ in uniform shales against depth. In the interval with normal pressure the points lie on the normal trend: a smooth curve that rises with depth (often a straight line on a semilog plot, a plot with a logarithmic scale on one axis). We continue it downwards. The depth from which the points shift systematically to the left of the trend is taken as the top of the transition zone.

CalculatedAn illustrative plot of the corrected d-exponent in shales. Down to a depth of about 2900 m the points follow the normal trend (the dashed line). Below that they go to the left: the rock drills more easily than a normally compacted one. The orange point is the example below (3150 m).

A departure from the trend can be turned into pressure. The Eaton method starts from the same idea as the loading curve: the more a property of the shale departs from normal, the smaller the actual effective stress is compared with the normal one.

Pore pressure from the departure from the normal trend (the Eaton method)

p_p = \sigma_v - (\sigma_v - p_n)\left(\frac{d_{c,o}}{d_{c,n}}\right)^{1.2}

where:

$p_p$ pore pressure at the depth of the estimate, MPa
$\sigma_v$ overburden pressure at this depth, MPa
$p_n$ normal hydrostatic pressure at this depth, MPa
$d_{c,o}$ observed corrected d-exponent, —
$d_{c,n}$ value of the normal trend at the same depth, —

Units: any identical units of pressure; all the pressures refer to one depth, so the formula holds both for gradients and for equivalent mud weights

Holds when

the high pressure was created by undercompaction. With unloading and transfer mechanisms the method underestimates the pressure;
the normal trend is built from uniform shales with normal pressure and is reliably continued down to the depth of the estimate;
the exponent 1.2 is empirical. For other data we use other exponents: 1.2 for the ratio of the resistivities of shales $R_o/R_n$ , and 3.0 for the ratio of the interval transit times $\Delta t_n/\Delta t_o$ from the sonic log (the time that a sound wave takes to cross a unit length of rock). In each area the exponent and the trend are checked against pressure measurements.

The bracket $\sigma_v - p_n$ is the effective stress that the rock would have at this depth at normal pressure. The factor $(d_{c,o}/d_{c,n})^{1.2}$ shows what share of it really acts. At $d_{c,o} = d_{c,n}$ the pressure is normal. The further the points go from the trend, the higher the pore pressure is.

Worked exampleThe d-exponent and the pore pressure in the shales of the transition zone

Given (assumed data)

True vertical depth (uniform shale): 3150 m
Rate of penetration $R$: 10.2 m/h
Rotary speed $N$ and weight on bit $W$: 110 rpm and 110 kN
Bit diameter $B$: 215.9 mm
Mud density $\rho_m$ and normal equivalent mud weight $\rho_n$: 1.15 and 1.05 g/cm³
Normal trend $d_{c,n}$ at this depth: 1.60
Overburden pressure at this depth: an equivalent mud weight of 2.30 g/cm³

Solution

Numerator: $\log_{10}\dfrac{R}{18.29\,N} = \log_{10}\dfrac{10.2}{18.29\cdot 110} = \log_{10} 0.00507 = -2.295$ .
Denominator: $\log_{10}\dfrac{0.0685\,W}{B} = \log_{10}\dfrac{0.0685\cdot 110}{215.9} = \log_{10} 0.0349 = -1.457$ . So $d = \dfrac{-2.295}{-1.457} = 1.575$ .
Correction for the mud density: $d_c = 1.575\cdot\dfrac{1.05}{1.15} = 1.438$ . The ratio to the trend is $1.438/1.60 = 0.899$ : the shale drills more easily than a normally compacted one at this depth.
The Eaton method in equivalent mud weights: $\rho_p = 2.30 - (2.30 - 1.05)\cdot 0.899^{1.2} = 2.30 - 1.25\cdot 0.880 = 1.20$ g/cm³. The pore pressure is $p_p = 0.00981\cdot 1.20\cdot 3150 = 37.1$ MPa.
The mud pressure at this depth is $0.00981\cdot 1.15\cdot 3150 = 35.5$ MPa, which is 1.6 MPa less than the pore pressure. For a permeable bed with such a pressure to get an overbalance of 1.5 MPa, we need a mud of $\dfrac{37.1 + 1.5}{0.00981\cdot 3150} = 1.249$ g/cm³.

Answer: The corrected d-exponent is 1.438 against a trend of 1.60. The pore pressure in the shale is about 37.1 MPa (1.20 g/cm³), which is 1.6 MPa above the mud pressure.

What it means. The shale at 3150 m is already in the transition zone, and the mud presses on it less than the fluid in its pores. So we should expect gas after connections and cavings, but no influx from the shale itself, because it is almost impermeable. The danger is a permeable bed below: if its pressure is the same, a mud of 1.15 g/cm³ will not hold it. We can make the mud heavier, up to about 1.25 g/cm³, only after checking the fracture pressure under the casing shoe (the bottom end of the last casing; Fracture gradient and leak-off testing). If the window (the gap between the pore pressure and the fracture pressure) is not wide enough, we run casing. We decide not from one point, but from the trend of tens of points and from agreement between the gas, the cuttings and the logs. The data in this example are assumed for illustration.

When the d-exponent misleads

A change of lithology: in sandstones and limestones $d_c$ behaves differently from shales. Such points are not included in the trend.
A change of bit: PDC bits (bits that shear the rock with polycrystalline diamond cutters) and roller-cone bits have a different dependence of the rate on the weight. After a change we build the trend again. Wear of the bit overstates $d_c$ and may mask the transition.
Changes in the hydraulics (the flow of mud through the bit), sharp changes in the weight and the rotary speed, and the start of an inclined hole (kicking off) distort the rate of penetration.
A change of the type of mud shifts the trend. In soft rocks, where the rate of penetration is limited by the cleaning of the bottom and not by the strength of the rock, $d_c$ hardly feels the pressure. So we always check the results of the method against the gas, the cuttings and the logs.

Terms and symbols

Overpressure and subnormal pressure: Abnormally high and abnormally low formation pressure: above or below the normal hydrostatic pressure for the given depth.
Abnormality coefficient $k_a$: The ratio of the pore pressure to the pressure of a column of fresh water at the same depth. Numerically it equals the equivalent mud weight in g/cm³.
Seal: A rock of low permeability (shale, salt, dense carbonate) that slows down the equalising of pressure and holds the overpressure.
Undercompaction: A mechanism of overpressure: the water has no time to escape from rapidly sinking shales, the fluid carries part of the weight of the rocks, and the porosity stays high.
Unloading: A rise of the pore pressure after the rock has been compacted (hydrocarbon generation, release of clay-bound water). The effective stress falls, and the porosity hardly changes.
Transition zone: The interval of the seal in which the pore pressure rises from normal to abnormally high.
d-exponent, $d_c$: An index of drillability worked out from the rate of penetration, corrected for the weight, the rotary speed, the bit diameter and (for $d_c$ ) the mud density.
Normal trend: The dependence of a property of shales (porosity, velocity, resistivity, $d_c$ ) on depth in an interval with normal pressure, continued downwards.
Connection gas: A peak of gas in the mud after the pumps are stopped. It is a sign that, without friction losses, the pressure at the bottom is close to the pressure in the shales, or lower.

Common mistakes and misconceptions

Wrong

“Overpressure can always be noticed in advance from the d-exponent.”

In fact

Only when the transition is gradual, in thick shales. Under salt and dense carbonates the transition is sharp, and pressure created by unloading or by transfer along a fault does not show in the trends. Such zones are forecast from the geology and offset wells.

Wrong

“The d-exponent has decreased, so overpressure has begun.”

In fact

The cause may be a sandstone, a change of bit, a change of hydraulics or of drilling parameters. We confirm the transition by a consistent shift of several signs in uniform shales: gas, the density and shape of cuttings, logs.

Wrong

“The pressure in the shales is higher than the mud pressure, so an influx will come from them now.”

In fact

Shales are almost impermeable. They give gas to the mud and cavings, but not a kick. The danger is a permeable bed with the same pressure, which may lie below.

Wrong

“Subnormal pressure is not dangerous for drilling: the bed is weaker than the mud and will give nothing.”

In fact

Depleted beds are a frequent cause of lost circulation and differential sticking, and their fracture pressure is lowered too. A loss of mud into such a bed lowers the level in the well and may cause a kick from another bed.

Wrong

“The pore pressure equals the density of the mud with which a neighbouring well was drilled.”

In fact

The mud density is only an upper estimate of the pressure, if there were no kicks, and often an overstated one. For a forecast we need direct pressure measurements and the signs recorded while drilling.

In depth: advanced detail

The equivalent depth method

If a shale at a depth $z$ has the same porosity (density, sonic velocity, resistivity) as a normally compacted shale at a smaller depth $z_e$ , then by the loading curve they have the same effective stress. At the depth $z_e$ the pressure is normal, and the effective stress there is known: $\sigma_v(z_e) - p_n(z_e)$ . The rest of the overburden pressure at the depth $z$ is carried by the fluid.

Pore pressure by the equivalent depth method

p_p(z) = \sigma_v(z) - \left[\sigma_v(z_e) - p_n(z_e)\right]

where:

$z$ depth of the estimate (true vertical), m
$z_e$ equivalent depth: at it a normally compacted shale has the same value of the property as the shale at the depth $z$ , m
$\sigma_v,\ p_n$ overburden pressure and normal hydrostatic pressure at the corresponding depth, MPa

Units: any identical units of pressure; true vertical depths

Holds when

the pressure was created by undercompaction: the shale “got stuck” on the loading curve. With unloading the method, like the Eaton method, underestimates the pressure;
we compare the same lithology, and the normal trend is reliably continued.

The same properties mean the same effective stress: $\sigma'(z) = \sigma'_n(z_e)$ . The Eaton method is an empirical way to do the same without searching for the equivalent depth explicitly.

Worked exampleThe same shale by the equivalent depth method

Given (assumed data)

Depth of the estimate $z$: 3150 m; the overburden pressure is an equivalent mud weight of 2.30 g/cm³
Equivalent depth $z_e$ from the density and the sonic velocity of the shale: 2820 m; the overburden pressure there is 2.28 g/cm³
Normal equivalent mud weight of the pore pressure: 1.05 g/cm³

Solution

Overburden pressure at 3150 m: $\sigma_v = 0.00981\cdot 2.30\cdot 3150 = 71.07$ MPa.
Effective stress of the normal shale at 2820 m: $0.00981\cdot (2.28 - 1.05)\cdot 2820 = 34.03$ MPa.
Pore pressure: $p_p = 71.07 - 34.03 = 37.0$ MPa, an equivalent mud weight of 1.20 g/cm³.

Answer: About 37.0 MPa (1.20 g/cm³), which is practically the same as from the d-exponent in the example of the main section.

What it means. Two ways based on independent data gave almost the same result. This checks the data, but not the mechanism, because both start from undercompaction. If the pressure had been created by unloading, both would have shown about the same value, and both would have been wrong on the low side. So in areas with gas generation and with transfer along faults, we check the results against direct pressure measurements. The data in this example are assumed for illustration.

Density of shale cuttings on a mud balance

A simple field way to follow compaction is to measure the bulk density of shale cuttings (the density of the rock together with its pores) on the same lever balance (the mud balance) that we use to measure the density of the mud. We wash the cuttings, sieve out the pieces of cavings, and pour the cuttings into the cup of the balance until it shows the density of water. The mass of the cuttings then equals the mass of the water that would fill the cup, $\rho_s V_s = \rho_w V$ . Then we top up the cup with water and balance it again: $\rho V = \rho_s V_s + \rho_w (V - V_s)$ . We eliminate the volume of cuttings $V_s$ and get:

Bulk density of cuttings from the reading of the balance

\rho_s = \frac{\rho_w^2}{2\rho_w - \rho}

where:

$\rho_s$ bulk density of the shale cuttings, g/cm³
$\rho_w$ density of the water used to balance and to top up the cup, g/cm³
$\rho$ reading of the balance after topping up with water, g/cm³
$V,\ V_s$ volume of the cup and volume of the cuttings particles, cm³

Units: any identical units of density

Holds when

the pores of the cuttings are filled with water, and the cuttings do not swell or dissolve during the measurement;
there are no pieces of cavings or particles of mud in the sample.

The method is not very sensitive: $\dd\rho/\dd\rho_s = (\rho_w/\rho_s)^2 \approx 0.19$ at $\rho_s \approx 2.3$ g/cm³. A change of the density of the shale by 0.1 g/cm³ changes the reading of the balance by only 0.02 g/cm³. So we take several samples at each depth and look at the trend, not at single values. Today we more often find the density of shales from logs recorded while drilling.

Connections to other topics

Pore pressure and overburden pressure — normal pressure, overburden pressure, effective stress and equivalent mud weight.
Well control: primary barrier and kick indicators — what happens if a bed with overpressure is opened with a mud that is not heavy enough.
Fracture gradient and leak-off testing — the upper limit of the mud density and the choice of casing depths.
Sonic log — the sonic velocity in shales as an indicator of undercompaction.
Seismic interpretation — seismic velocities and the structure of seals in a pressure forecast.
Source rocks, maturation and migration — hydrocarbon generation is one of the unloading mechanisms.
Casing design — how pressures decide the well design.

Check yourself

1
In what situation do the properties of shales (porosity, sonic velocity) hardly show overpressure, although the pressure is high?
1. A.In all cases the properties of shales reflect the pressure exactly
2. B.With undercompaction of rapidly sinking shales
3. C.When the shales are normally compacted and the pressure is hydrostatic
4. D.When the pressure rose after compaction, for example during the generation of gas
Show the answer
Correct answer: D. When the pressure rose after compaction, for example during the generation of gas
In unloading the effective stress falls after the compaction, and the porosity hardly recovers: the rock stays on the flat unloading branch. Methods that convert the properties of shales into effective stress by the loading curve then underestimate the pressure.
2
Why does a kick usually not happen when we drill in the shales of the transition zone, where the pore pressure is already higher than the mud pressure?
Show the answer
The permeability of shales is very small, and fluid enters the hole from them in negligible amounts: we see gas in the mud and cavings from the walls, but not an influx. A kick begins when we open a permeable bed with the same pressure. So the signs of the transition zone are a signal to make the mud heavier or to run casing before such a bed is opened.
3
The mud was made heavier from 1.10 to 1.20 g/cm³, and the d-exponent rose from 1.50 to 1.62, although the rock did not change. How did the corrected $d_c$ change at $\rho_n = 1.05$ g/cm³?
1. A.We need the bit diameter to answer
2. B.It halved
3. C.It rose in the same way, by 8%
4. D.It hardly changed
Show the answer
Correct answer: D. It hardly changed
$d_c = 1.50\cdot 1.05/1.10 \approx 1.43$ before the mud was made heavier and $1.62\cdot 1.05/1.20 \approx 1.42$ after. The correction removes the rise of $d$ caused by the heavier mud, and the trend stays smooth.
4
At a depth of 2900 m the overburden pressure is equivalent to 2.25 g/cm³, the normal pressure to 1.05 g/cm³, and the ratio of the observed $d_c$ to the trend is 0.85. Estimate the equivalent mud weight of the pore pressure by the Eaton method.
Show the answer
$0.85^{1.2} \approx 0.823$ ; $\rho_p = 2.25 - (2.25 - 1.05)\cdot 0.823 \approx 2.25 - 0.99 = 1.26$ g/cm³. This is an estimate on the condition that the pressure was created by undercompaction, and that the trend and the exponent have been checked for the area.
5
Which of the signs does NOT, by itself, show an approach to the overpressure zone, but means that a kick is already happening?
1. A.A fall in the density of shale cuttings
2. B.A gain in the volume of mud in the receiving pits
3. C.A rise in background gas
4. D.A steeper rise of the temperature of the mud at the outlet
Show the answer
Correct answer: B. A gain in the volume of mud in the receiving pits
A gain of volume in the pits means that formation fluid is entering the well. This is a direct sign of a kick. The other signs are indirect: they point to a change in the properties of shales and a smaller overbalance, that is, to an approach to the overpressure zone.
6
Shale cuttings balanced the mud balance at a reading of 1.00 g/cm³ (like a cup of water). After topping up with water, the balance showed 1.57 g/cm³. What is the density of the cuttings?
Show the answer
$\rho_s = \rho_w^2/(2\rho_w - \rho) = 1/(2 - 1.57) \approx 2.33$ g/cm³. The sensitivity of the method is small: a reading of 1.55 would give 2.22 g/cm³, so at each depth we take several samples and follow the trend.

Updated: 19 October 2026