Pore pressure and overburden pressure

Quick refresher

Hydrostatic pressure depends only on the density of the fluid and the vertical height of the column: $p = \rho g h$ . The shape of the wellbore and its length along the hole do not matter.
A gradient is the increase in pressure per metre of depth. Fresh water has 9.81 kPa/m (0.433 psi/ft). Normal pore pressure is usually 9.8–10.5 kPa/m (0.433–0.465 psi/ft). The overburden pressure averages about 22.6 kPa/m (1 psi/ft).
We can express any pressure at depth as an equivalent mud weight: the density of the fluid whose column would create that pressure. Then pore pressure, fracture pressure (the pressure at which the rock breaks) and mud density can be compared directly.
Overburden pressure = pore pressure + effective stress. The rock framework carries only the effective stress. So where the pore pressure is abnormally high, the rock is undercompacted (squeezed less than is normal for its depth).
We choose the mud density so that the pressure in the well is above the pore pressure by a margin (overbalance) but below the fracture pressure of the rocks.

Key relationships

p = \rho\, g\, h

hydrostatic pressure; $h$ is the true vertical depth, SI units

p\,[\text{MPa}] = 0.00981\,\rho\,[\text{g/cm}^3]\;h\,[\text{m}]

metric units

p\,[\text{psi}] = 0.052\,\rho\,[\text{ppg}]\;h\,[\text{ft}]

oilfield units

\sigma_v = p_p + \sigma'_v

the overburden pressure is shared between the fluid and the rock framework

\rho_b = \phi\,\rho_f + (1-\phi)\,\rho_{ma}

bulk density of a porous rock

Common pitfalls

Putting the measured depth (the length along the hole) into the hydrostatic formula instead of the true vertical depth.
Comparing gradients and equivalent mud weights measured from different datums (zero levels): the rotary table (on the rig floor), sea level or ground level.
Mixing up absolute and gauge pressure (gauge pressure is measured above the atmosphere). Downhole gauges often read absolute pressure, whereas the pressure of the mud column is calculated as a gauge pressure.
Treating 1 psi/ft (22.6 kPa/m) as an exact value or as a limit of the overburden gradient. It is a rough average.

After this lesson you can

calculate the hydrostatic pressure of a fluid column (the pressure caused by its weight) from the true vertical depth, the depth measured straight down, and express it in MPa, psi and kgf/cm²
explain where the constants 0.00981 MPa/m per g/cm³ and 0.052 psi/ft per ppg (pounds per US gallon) come from, and convert a pressure into a gradient (pressure per unit of depth) and an equivalent mud weight (the mud density that would produce that pressure)
distinguish normal pore pressure from abnormal pore pressure and say what the normal gradient depends on
calculate the overburden pressure from rock densities and find the effective stress
choose a mud density from the pore pressure and a given margin, and explain what sets its upper limit

Hydrostatic pressure

The pressure in a fluid at rest, at depth $h$ , comes from the weight of the column above that point. Take a vertical column of fluid with density $\rho$ and cross-sectional area $A$ . Its weight is $\rho g A h$ , and this weight is spread over the area $A$ , so the pressure at the base is $\rho g h$ . The area cancels. That is why the pressure depends on neither the width nor the shape of the vessel.

Two conclusions follow for a well. First, the drill pipe and the annulus (the space between the pipe and the wall of the hole) are connected at the bit, so they behave like a U-tube. With the same fluid in both, the pressure at a given depth is the same in each, however different their cross-sections are. Second, the formula contains only the true vertical depth (TVD), not the measured depth (MD). In a deviated well (a well that is not vertical) the mud column is longer, but it presses only with its vertical height.

Hydrostatic pressure of a fluid column

p = p_0 + \rho\, g\, h

where:

$p$ pressure at depth $h$ below the top of the column, Pa
$p_0$ pressure above the column: atmospheric pressure for an open well (zero in gauge units), the wellhead pressure for a closed well, Pa
$\rho$ fluid density, kg/m³
$g$ acceleration due to gravity, standard value 9.80665 m/s², m/s²
$h$ vertical height of the column — the true vertical depth below the top of the column, m

Units: SI

Holds when

The fluid is at rest. When the mud circulates, friction adds pressure losses to the hydrostatic pressure (see Circulating pressure losses).
The density is the same throughout the column. If it changes with depth (several fluids, gas, heating), we add up the pressure interval by interval: $p = p_0 + \sum_i \rho_i\, g\, \Delta h_i$ .
$h$ is specifically the vertical height, not the length along the hole.

Pressure rises in a straight line with depth, with a slope of $\rho g$ . The heavier the fluid, the faster the pressure rises. This is why the density of the drilling mud is the main tool for controlling pressure in the well.

In other unit systems (3)

Metric units

p\,[\text{MPa}] = 0.00981\,\rho\,[\text{g/cm}^3]\;h\,[\text{m}]

$\rho$ in g/cm³ (numerically equal to the specific gravity, which is the density relative to water), $h$ in m, $p$ in MPa (gauge pressure).

$0.00981 = 1000\ \text{kg/m}^3\cdot 9.80665\ \text{m/s}^2\cdot 10^{-6}$ : a column of fluid with a density of 1 g/cm³ and a height of 1 m produces a pressure of 9807 Pa. For a quick estimate in your head, use about 1 MPa per 100 m of water column (the error is about 2%).

Oilfield units

p\,[\text{psi}] = 0.052\,\rho\,[\text{ppg}]\;h\,[\text{ft}]

$\rho$ in pounds per US gallon (ppg, lb/gal), $h$ in feet, $p$ in psi.

$0.052 \approx 12/231 = 0.05195$ . A column 1 ft (12 inches) high with a cross-section of 1 square inch holds 12 cubic inches, and a US gallon is exactly 231 cubic inches. So the column holds $12/231$ of a gallon of fluid. The weight of that fluid in pounds, resting on 1 square inch, is the pressure in psi for each foot of height.

Technical units (kgf/cm²)

p\,[\text{kgf/cm}^2] = \frac{\rho\,[\text{g/cm}^3]\;h\,[\text{m}]}{10}

$\rho$ in g/cm³, $h$ in m; 1 kgf/cm² (the technical atmosphere) = 0.0981 MPa.

Dividing by 10 is exact. A column of water 10 m high with a cross-section of 1 cm² has a mass of 1 kg, and the kilogram-force is defined through the same standard acceleration, 9.80665 m/s². You will find these units in older design documents.

Worked exampleMud column pressure in a deviated well

Given (assumed data)

Drilling mud density $\rho$: 1.18 g/cm³
True vertical depth of the bottom of the hole (TVD): 2650 m
Measured depth to the bottom of the hole (MD): 3120 m

Solution

Bottomhole pressure in metric units: $p = 0.00981\cdot 1.18\cdot 2650 = 30.7$ MPa. Pressure gradient of the mud: $0.00981\cdot 1.18\cdot 1000 = 11.58$ kPa/m.
If we take the measured depth by mistake, we get $0.00981\cdot 1.18\cdot 3120 = 36.1$ MPa, which is 18% more than the true pressure.
In oilfield units: $\rho = 1.18\cdot 8.345 = 9.847$ ppg, $h = 2650/0.3048 = 8694$ ft, $p = 0.052\cdot 9.847\cdot 8694 = 4452$ psi. Direct conversion, with 1 MPa = 145.04 psi, gives $30.68\cdot 145.04 = 4450$ psi. The difference of a couple of psi comes from rounding of the constants.
In technical units: $p = 1.18\cdot 2650/10 = 312.7$ kgf/cm².

Answer: At the bottom of the hole the pressure is about 30.7 MPa (≈ 4452 psi, 312.7 kgf/cm²). A calculation with the measured depth would give a pressure that is 18% too high.

What it means. We need the measured depth to calculate the friction pressure losses during circulation and for volumes. For hydrostatic pressure only the true vertical depth counts. Along a horizontal section the pressure of the mud column hardly increases, even though the measured depth grows by hundreds of metres. The data in this example are assumed for illustration.

Pressure gradient and equivalent mud weight

Pressures at different depths are awkward to compare directly: 30 MPa at 2500 m and 30 MPa at 3000 m mean quite different things. So we often divide the pressure by the depth. This gives the pressure gradient, in kPa/m, MPa/m or psi/ft.

There are two kinds of gradient, and it is important not to confuse them. The average gradient from the surface, $p/h$ , is the pressure divided by the depth, measured from the chosen datum (the chosen zero level). The local gradient, $\dd p/\dd h$ , is the slope of the pressure curve at a given depth. It shows the density of the fluid in that particular interval. In drilling, the “pore pressure gradient” usually means the average gradient from the surface. We can compare it directly with the pressure gradient of the mud, whose column also starts at the surface.

The next step is to express the pressure as an equivalent mud weight (EMW). This is the density of a fluid whose column, from the datum down to the given depth, would produce that pressure. It is the same average gradient, only in density units. All mud planning is built on the equivalent mud weight: pore pressure, fracture pressure and mud density are plotted on one scale.

Equivalent mud weight

\rho_{eq} = \frac{p}{g\, h}

where:

$\rho_{eq}$ equivalent mud weight, kg/m³
$p$ gauge pressure at depth $h$ : pore pressure, fracture pressure, bottomhole pressure…, Pa
$g$ acceleration due to gravity, m/s²
$h$ true vertical depth from the chosen datum, m

Units: SI

Holds when

The pressure is a gauge pressure, and the datum is the same one from which the mud column is measured (usually the rotary table).
Equivalent mud weight is a notional (imaginary) quantity. It describes the pressure at one depth and does not mean that a fluid of that density actually lies above.

Suppose the equivalent mud weight of the pore pressure at some depth is 1.08 g/cm³. Then a mud with a density of 1.08 g/cm³ would just balance the formation at that depth: the mud pressure would equal the pore pressure.

In other unit systems (2)

Metric units

\rho_{eq}\,[\text{g/cm}^3] = \frac{p\,[\text{MPa}]}{0.00981\; h\,[\text{m}]}

$p$ in MPa, $h$ in m.

0.00981 is the same constant as in the formula for hydrostatic pressure.

Oilfield units

\rho_{eq}\,[\text{ppg}] = \frac{p\,[\text{psi}]}{0.052\; h\,[\text{ft}]}

$p$ in psi, $h$ in ft; the gradient in psi/ft equals $0.052\,\rho_{eq}$ .

0.052 comes from the oilfield formula for hydrostatic pressure.

The same quantities in different units

Item: Fresh water
g/cm³ (specific gravity): 1.00
kPa/m: 9.81
psi/ft: 0.433
ppg: 8.35

Item: Normal pore pressure, commonly assumed values
g/cm³ (specific gravity): 1.04–1.07
kPa/m: 10.2–10.5
psi/ft: 0.45–0.465
ppg: 8.7–8.9

Item: Very salty formation water
g/cm³ (specific gravity): up to ~1.15
kPa/m: up to ~11.3
psi/ft: up to ~0.50
ppg: up to ~9.6

Item: Overburden pressure, rough average
g/cm³ (specific gravity): ≈ 2.31
kPa/m: ≈ 22.6
psi/ft: ≈ 1.0
ppg: ≈ 19.2

Item	g/cm³ (specific gravity)	kPa/m	psi/ft	ppg
Fresh water	1.00	9.81	0.433	8.35
Normal pore pressure, commonly assumed values	1.04–1.07	10.2–10.5	0.45–0.465	8.7–8.9
Very salty formation water	up to ~1.15	up to ~11.3	up to ~0.50	up to ~9.6
Overburden pressure, rough average	≈ 2.31	≈ 22.6	≈ 1.0	≈ 19.2

1 g/cm³ = 1000 kg/m³ = 8.345 ppg = 62.43 lb/ft³; 1 psi/ft = 22.62 kPa/m; the gradient in kPa/m equals the density in g/cm³ multiplied by 9.81. Values are rounded.

Pore pressure: normal and abnormal

Sedimentary rocks (rocks built from layers of settled grains) are laid down in water. As the layer of sediment grows, the grains below are pressed together and pack more tightly, and some of the water is squeezed out of the pores and upwards. Suppose the pores and the channels between them are connected all the way to the surface (or to the sea floor), and the water has time to escape. Then the pressure in the pores at any depth equals the pressure of a plain column of formation water (the water in the pores of the rock) of the same height. The size and shape of the pores do not matter here. What matters is only that the fluid in the pores is connected.

This pressure is called normal, or hydrostatic, pore pressure. Its gradient depends on the density of the formation water, and that depends on its salinity (how much salt it holds). The gradient ranges from 9.8 kPa/m (0.433 psi/ft) for fresh water to 11–11.3 kPa/m (about 0.50 psi/ft) for strong brines (very salty water). In different areas the normal gradient usually lies within 9.8–10.5 kPa/m (0.433–0.465 psi/ft). Values of about 10.2 kPa/m (0.45 psi/ft) and 10.5 kPa/m (0.465 psi/ft) are often used. For a specific area we take the gradient from local formation-water data. The density of water also changes slightly with temperature, so the real gradient is not strictly constant, but for drilling purposes we usually treat it as constant.

Pressure above normal is called overpressure (abnormally high formation pressure), and pressure below normal is called subnormal pressure (abnormally low formation pressure). Overpressure arises, for example, when water has no time to escape from rapidly sinking shale sequences (thick layers of clay-rich rock), when fluids expand, or under tectonic compression (squeezing of the earth's crust). Subnormal pressure arises when sequences are pushed up and eroded, and in depleted reservoirs (reservoirs that have lost pressure because oil or gas was produced). The origin, prediction and recognition of abnormal pressures are covered in the lesson on abnormal formation pressures.

The terms pore pressure and formation pressure are often used as synonyms. Strictly speaking, formation pressure is the pressure in a permeable reservoir rock, a rock that holds oil, gas or water and lets it flow. We can measure it with a wireline formation tester (a tool lowered on a cable that measures the pressure in the rock) or during a well test (see pressure measurements with a formation tester). Shales give up almost no fluid from their pores. So we do not measure the pore pressure in shales directly. We estimate it indirectly, from well logs (measurements made along the hole) and drilling parameters.

The gradient is lower inside an accumulation

Inside an accumulation (a body of oil or gas trapped in the rock) the pores are filled with oil or gas, which are lighter than water. The local pressure gradient is roughly 6.5–8.5 kPa/m in an oil accumulation and about 1–3 kPa/m in a gas accumulation. The water sets the pressure at the base of the accumulation, and going up through the accumulation the pressure falls more slowly than it would in water. So at the top of a tall gas accumulation the pressure is noticeably higher than normal for that depth. For example, take an accumulation 150 m high, where the water and gas gradients differ by 8 kPa/m. The excess is about 1.2 MPa. A mud chosen from the normal gradient may be too light at the top of such an accumulation. More detail is in the lesson on pressure gradients and fluid contacts.

Worked exampleTwo pressure measurements: normal or abnormal?

Given (assumed data)

Measurement 1 (sandstone): gauge pressure 27.9 MPa at a true vertical depth of 2710 m below the rotary table
Measurement 2 (a sandstone deeper down): 33.6 MPa at a depth of 2980 m
Normal equivalent mud weight in the area: 1.05 g/cm³

Solution

Measurement 1: the average gradient is $27.9/2710 = 0.0103$ MPa/m, that is 10.3 kPa/m (0.455 psi/ft). The equivalent mud weight is $\rho_{eq} = \dfrac{27.9}{0.00981\cdot 2710} = 1.05$ g/cm³ (8.76 ppg).
Normal pressure at this depth: $0.00981\cdot 1.05\cdot 2710 = 27.9$ MPa. The measurement coincides with it, so the pressure is normal.
Measurement 2: $\rho_{eq} = \dfrac{33.6}{0.00981\cdot 2980} = 1.15$ g/cm³. Normal pressure at this depth is $0.00981\cdot 1.05\cdot 2980 = 30.7$ MPa, so the excess is 2.9 MPa. This is overpressure, with an abnormality coefficient of about 1.15.
Between the measurements the pressure rose by $33.6 - 27.9 = 5.7$ MPa over 270 m, so the local gradient is 21 kPa/m, twice that of a column of water.

Answer: The first formation has normal pressure (1.05 g/cm³). The second has abnormally high pressure (1.15 g/cm³, 2.9 MPa above normal).

What it means. A column of water cannot give an increase of 21 kPa/m. So the two formations are not connected: a seal (a layer that blocks the flow of fluid) lies between them and holds the elevated pressure. Note that the equivalent mud weight of the second measurement is only 1.15 g/cm³, because the average gradient from the surface “smears” the pressure jump over the whole depth. What matters for drilling is the equivalent mud weight of each formation. The data in this example are assumed for illustration.

Overburden pressure

Overburden pressure, also called geostatic pressure, is the vertical stress at depth that comes from the weight of everything above: the mineral grains, the fluids in the pores and, offshore, the column of sea water as well. It matters for drilling for two reasons. It decides the pressure at which the rock will fracture, and it sets an upper limit to the pore pressure.

The weight of a rock depends on its bulk density ( $\rho_b$ ). This is the average density of the mineral grains and the fluid that fills the pores, taken together.

Bulk density of a porous rock

\rho_b = \phi\,\rho_f + (1-\phi)\,\rho_{ma} = \rho_{ma} - (\rho_{ma} - \rho_f)\,\phi

where:

$\rho_b$ bulk density of the rock, g/cm³
$\phi$ porosity (the share of the rock volume taken up by pores), fraction
$\rho_f$ density of the pore fluid, g/cm³
$\rho_{ma}$ density of the mineral matrix (grains): about 2.65 g/cm³ for quartz, 2.71 for calcite, 2.87 for dolomite, g/cm³

Units: any density units, the same for all terms

Holds when

The pores are completely filled with fluid. If the pores contain several fluids, $\rho_f$ is their average density, weighted by the share of pore volume each one fills (saturation).
$\rho_{ma}$ is the average grain density, allowing for the mineral composition of the rock.

The higher the porosity, the more light fluid the rock contains and the lighter the rock is. With depth, rocks compact (the grains are pressed together) and porosity falls. So the bulk density rises, from 1.6–2.1 g/cm³ for loose, young sediments to 2.4–2.7 g/cm³ for dense rocks. The exception is rock salt: about 2.1–2.2 g/cm³ even without pores.

Density changes with depth, so we find the overburden pressure by adding up (integrating) the weight of all the layers from the top down. In practice we take density log data (Density log and photoelectric factor (Pe)), which are density measurements made along the hole, and sum the contributions of the intervals. Above the logged interval, we estimate the density from trends or from offset wells (nearby wells already drilled).

Overburden pressure as the weight of the overlying rock

\sigma_v = g\int_0^{h} \rho_b(z) \dd z \;\approx\; g \sum_i \rho_{b,i}\, \Delta h_i

where:

$\sigma_v$ overburden pressure (vertical stress) at depth $h$ , Pa
$\rho_b(z)$ bulk density of the rocks at depth $z$ , kg/m³
$\rho_{b,i},\ \Delta h_i$ average bulk density and vertical thickness of the $i$ -th interval, kg/m³, m
$g$ acceleration due to gravity, m/s²

Units: SI; in metric units $\sigma_v\,[\text{MPa}] = 0.00981\sum_i \rho_{b,i}\,[\text{g/cm}^3]\;\Delta h_i\,[\text{m}]$

Holds when

The vertical stress equals the weight of the overlying rock. The layers are roughly horizontal, and we can ignore the support from neighbouring rock masses.
On land the summation starts at the ground surface. Offshore, the first interval is the layer of sea water (about 1.03 g/cm³) from sea level to the sea floor.
Depths are true vertical depths. To compare with the mud pressure, we measure the depth from the rotary table for the overburden pressure, as we do for the pore pressure.

The average overburden gradient from the surface is often close to 22.6 kPa/m (1 psi/ft). This corresponds to an average bulk density of the overlying rock of about 2.3 g/cm³. Near the surface and in young, loose sediments the gradient is lower. In deep, compacted sections it is higher.

In other unit systems (1)

Oilfield units

\sigma_v\,[\text{psi}] = 0.433\sum_i \rho_{b,i}\,[\text{g/cm}^3]\;\Delta h_i\,[\text{ft}]

$\rho_b$ in g/cm³ (specific gravity), $\Delta h$ in ft.

0.433 psi/ft is the gradient of a water column of density 1 g/cm³ ( $0.052\cdot 8.345$ ).

Worked exampleOverburden pressure and effective stress from density log data

Given (assumed data)

Interval 0–450 m: average bulk density 2.02 g/cm³
Interval 450–1600 m: 2.26 g/cm³
Interval 1600–2980 m: average porosity 14%, grain density 2.65 g/cm³, formation water density 1.05 g/cm³
Pore pressure at 2980 m: 33.6 MPa (measurement 2 from the example above)

Solution

Bulk density of the lower interval: $\rho_b = 2.65 - (2.65 - 1.05)\cdot 0.14 = 2.426$ g/cm³.
Sum of the contributions of the intervals: $2.02\cdot 450 + 2.26\cdot 1150 + 2.426\cdot 1380 = 6856$ g/cm³·m.
Overburden pressure: $\sigma_v = 0.00981\cdot 6856 = 67.3$ MPa. The average gradient is 22.6 kPa/m (1.00 psi/ft), and the average bulk density of the overlying rock is 2.30 g/cm³.
Effective stress: $\sigma'_v = 67.3 - 33.6 = 33.7$ MPa. With normal pore pressure at this depth (30.7 MPa) it would be 36.6 MPa.

Answer: The overburden pressure at 2980 m is about 67.3 MPa (gradient 1.00 psi/ft). The mineral framework carries 33.7 MPa.

What it means. The average gradient came out at almost exactly 1 psi/ft, even though the upper intervals are lighter than the average density and the lower one is heavier. The rounded rule works as an average, not as a law. The abnormally high pore pressure reduced the effective stress by 8%. The framework carries less load, and the rock is less compacted than it would be at normal pressure. Overpressure is recognised by such signs: higher porosity, lower density and lower sonic velocity (the speed of sound in the rock). The data in this example are assumed for illustration.

Effective stress: what carries the weight of the rocks

The overburden pressure acts on the rock as a whole, but two “supports” share it: the mineral framework, where the grains touch one another, and the fluid in the pores. The fluid, compressed to the pore pressure $p_p$ , props the grains up from the inside. So the framework carries only the difference, which is the effective stress ( $\sigma'_v$ ). This stress, not the total overburden pressure, compacts the rock and decides its strength.

Effective vertical stress (Terzaghi's principle)

\sigma'_v = \sigma_v - \alpha\, p_p \;\approx\; \sigma_v - p_p

where:

$\sigma'_v$ effective vertical stress — the load on the mineral framework, MPa
$\sigma_v$ overburden pressure, MPa
$p_p$ pore pressure, MPa
$\alpha$ Biot coefficient, from 0 to 1; close to 1 for loose and moderately compacted rocks, —

Units: any pressure units, the same for all terms

Holds when

The pore space is connected, and the pore pressure is uniform within the volume of rock considered.
In pore pressure and drilling problems we normally assume $\alpha = 1$ . In calculations of how a reservoir deforms, we use a measured value.

At the same depth (the same overburden pressure), a rise in pore pressure reduces the effective stress. A fall in pore pressure, for example when a reservoir is produced, increases it.

Compaction. The porosity of sediments decreases with depth because the effective stress increases. If the pore pressure is abnormally high, the effective stress is lower than normal and the rock is undercompacted. It has higher porosity and lower density and sonic velocity. Methods of estimating pore pressure from well logs and from drilling data are based on this (Abnormal formation pressures: origin, prediction and recognition).
Upper limit of pore pressure. If the pore pressure approaches the overburden pressure, the effective stress falls almost to zero and the fluid begins to open fractures. This usually happens even earlier. A fracture opens against the smallest stress, which in most sedimentary basins (large areas where thick sediments have built up) is horizontal and smaller than the vertical stress. So in practice the fracture pressure of the rocks limits the pore pressure from above, and in such conditions the fracture pressure is lower than the overburden pressure.
Fracture pressure. The horizontal stresses in the rock are related to the effective vertical stress, so the fracture pressure also increases with depth and depends on the pore pressure (Fracture gradient and leak-off testing).
Reservoir depletion. When the reservoir pressure falls, the effective stress rises. The reservoir rock compacts, its porosity and permeability decrease slightly, and the surface may sink above large accumulations (Rock compressibility and the effect of stress on core properties).

Drilling mud versus pore pressure

The main way to stop formation fluid from flowing into the well is to keep the pressure of the mud column above the pore pressure of every permeable formation (one that lets fluid flow) that the well has drilled through. The amount by which the mud pressure exceeds the pore pressure at a given depth is called the overbalance on the formation. If the mud pressure is lower than the pore pressure, the condition is called underbalance. This is the basis of primary well control, which means holding the well under control with the mud column. See the lesson on well control for details.

The fracture pressure of the weakest rock in the open hole (the part of the hole that has no casing) sets the upper limit. This is usually the rock just below the shoe of the last casing string, the bottom end of the last steel pipe set in the well. If the mud pressure goes above the fracture pressure, the mud escapes into the fractures, the level in the well drops, and the well may kick. During circulation, friction in the annulus adds pressure losses to the hydrostatic pressure, so the bottom of the hole sees the equivalent circulating density (ECD), which is higher than the mud density. Sometimes the stability of the hole, not a formation, raises the lower limit. In stressed shales, we need a mud heavier than the pore pressure alone calls for, to prevent caving (pieces of rock falling from the wall into the hole). The allowed “window” of mud densities lies between these limits.

Mud density for a given overbalance

\rho_m = \frac{p_p + \Delta p_{ob}}{g\, h}

where:

$\rho_m$ drilling mud density, kg/m³
$p_p$ pore pressure of the formation with the highest equivalent mud weight in the open hole, Pa
$\Delta p_{ob}$ required overbalance — the margin above the pore pressure, Pa
$h$ true vertical depth of that formation below the rotary table, m

Units: SI

Holds when

The mud is at rest (static pressure). During circulation the bottomhole pressure is higher by the pressure losses in the annulus.
The mud column reaches the wellhead. If the level has dropped, the bottomhole pressure is lower.
After choosing a density, we check that the mud pressure, including the ECD, does not exceed the fracture pressure anywhere in the open hole.

The mud density must exceed the equivalent mud weight of the pore pressure in every permeable interval, and stay below the equivalent mud weight of the fracture pressure in every interval of the open hole.

In other unit systems (2)

Metric units

\rho_m\,[\text{g/cm}^3] = \frac{(p_p + \Delta p_{ob})\,[\text{MPa}]}{0.00981\; h\,[\text{m}]}

pressures in MPa, depth in m.

0.00981 comes from the formula for hydrostatic pressure.

Oilfield units

\rho_m\,[\text{ppg}] = \frac{(p_p + \Delta p_{ob})\,[\text{psi}]}{0.052\; h\,[\text{ft}]}

pressures in psi, depth in ft.

0.052 comes from the oilfield formula for hydrostatic pressure.

The well construction design (the project for building the well) and the safety regulations in force decide what margin to allow. In the regulatory documents used across the post-Soviet region, the overbalance is set as a percentage of the formation pressure, with an upper limit in megapascals, and the percentages differ between depth intervals. In English-language practice, about 1.4–2 MPa (200–300 psi) is often allowed against permeable formations. A margin is needed for two reasons. The pore pressure is not known exactly, and the bottomhole pressure briefly drops when the drill string is pulled out of the hole.

Why not simply use a mud with a large margin? Too much overbalance does harm in four ways. The risk of lost circulation (mud escaping into the rock) and of fracturing the rock rises. The pressure difference pins the pipe against the wall of the hole, which is called differential sticking. Filtrate (the liquid part of the mud that enters the rock) and solids go deeper into the reservoir and damage it. And the rate of penetration falls, because the excess pressure holds the drilled cuttings against the bottom of the hole. So we need enough density, not the maximum density. You can repeat the hydrostatic pressure and mud density calculations of this lesson with your own data in the engineering calculators on the site.

Worked exampleChoosing the mud density for an overpressured interval

Given (assumed data)

Measurement 1 (sandstone): 27.9 MPa at 2710 m
Measurement 2 (sandstone): 33.6 MPa at 2980 m — the bottom of the interval
Required overbalance: 1.5 MPa
Fracture pressure below the casing shoe at 1850 m: equivalent mud weight 1.62 g/cm³

Solution

We calculate the mud density for the formation with the highest equivalent mud weight, the lower one: $\rho_m = \dfrac{33.6 + 1.5}{0.00981\cdot 2980} = 1.201$ g/cm³. The mud density is adjusted in steps of 0.01 g/cm³, so we round up and take 1.21 g/cm³ (10.1 ppg).
Actual overbalance on the lower formation: $0.00981\cdot 1.21\cdot 2980 - 33.6 = 1.77$ MPa.
The same mud against the upper sandstone: $0.00981\cdot 1.21\cdot 2710 - 27.9 = 4.27$ MPa, which is 2.8 times the required margin.
Check against the upper limit: below the shoe, 1.21 g/cm³ is less than 1.62 g/cm³, a margin of 0.41 g/cm³ in equivalent mud weight. During circulation the ECD takes up part of this margin.

Answer: A mud with a density of 1.21 g/cm³ is needed. The overbalance is 1.77 MPa on the lower formation and 4.27 MPa on the upper one.

What it means. A single mud for the whole open hole gives different formations different overbalances. The upper sandstone, at normal pressure, receives about 4.27 MPa, which risks differential sticking and formation damage. Suppose the pressure difference were larger and the window below the shoe narrower. Then we would have to case off the upper interval (cover it with a casing string) before entering the overpressured zone. This is how pore pressure and fracture pressure determine the well design (Casing design). The data in this example are assumed for illustration.

CalculatedCalculated pressure–depth diagram for the data of this lesson's examples: normal pore pressure (1.05 g/cm³), the pressure of a 1.21 g/cm³ mud, and the overburden pressure from the rock densities. The orange dots are measurements 1 (2710 m) and 2 (2980 m): the first lies on the normal pressure line, the second to the right of it, which means overpressure. The horizontal distance between the mud line and the pore pressure line is the overbalance. The horizontal distance between the overburden pressure line and the pore pressure line is the effective stress.

How to build a pressure–depth diagram for a planned well

1
Datum
Choose the level from which you measure all true vertical depths. This is usually the rotary table of the rig planned for the well.
2
Normal pressure
Plot the line of normal hydrostatic pressure from the density of the formation waters of the area.
3
Pore pressure
Plot the pore pressure. In permeable formations, use measurements in offset wells. In shales, use estimates from well logs, seismic surveys and drilling parameters.
4
Overburden pressure
Calculate the overburden pressure from the density logs of offset wells.
5
Fracture pressure
Plot the fracture pressure of the rocks, using leak-off tests (pressure tests that show when the rock starts to take fluid) and calculation methods (Fracture gradient and leak-off testing).
6
Mud and casing strings
For each interval, choose a mud density between the pore pressure plus the margin and the fracture pressure. Where that is not possible, cover the interval with a casing string.

Terms and symbols

Pore pressure, $p_p$: The pressure of the fluid in the pores of a rock. In permeable formations it is also called formation pressure.
Normal (hydrostatic) pressure: Pore pressure equal to the pressure of a column of formation water from the surface. The gradient is 9.8 to 11.3 kPa/m, depending on the salinity of the water.
Overpressure and subnormal pressure: Abnormally high and abnormally low formation pressure, that is, above or below normal for the given depth.
Overburden (geostatic) pressure, $\sigma_v$: The vertical stress caused by the weight of the overlying rocks and fluids.
Effective stress, $\sigma'_v$: The part of the overburden pressure carried by the mineral framework: $\sigma_v - p_p$ .
Equivalent mud weight: The density of a fluid whose column from the datum would produce the given pressure at the given depth.
Overbalance and underbalance: Overbalance is the excess of the pressure in the well over the pore pressure. Underbalance is the shortfall of the pressure in the well relative to the pore pressure.
TVD and MD: True vertical depth and measured depth (the length along the hole). Hydrostatic pressure is calculated from the TVD.
ppg: Pounds per US gallon, the oilfield unit of mud density. 1 g/cm³ = 8.345 ppg.

Common mistakes and misconceptions

Wrong

“The bottomhole pressure of a deviated well is calculated from the measured depth.”

In fact

Hydrostatic pressure depends only on the true vertical depth. We need the measured depth for the friction pressure losses during circulation and for volumes.

Wrong

“The overburden gradient of 1 psi/ft is the maximum; it never gets any higher.”

In fact

It is a rounded average. In deep, dense sections the gradient is higher, and near the surface and offshore it is lower. For design, we calculate it from the density log.

Wrong

“Normal pressure is always 0.465 psi/ft.”

In fact

The normal gradient depends on the density of the formation water in the area. It runs from 0.433 psi/ft for fresh water to ~0.50 psi/ft for strong brines.

Wrong

“Equivalent mud weight can be calculated from any datum.”

In fact

It depends on the zero level we choose. To choose the mud density, we measure depths from the rotary table, where the mud column starts.

Wrong

“The heavier the mud, the safer it is.”

In fact

Too much overbalance brings the risk of lost circulation, differential sticking, formation damage and slow drilling. The density must lie in the window between the pore pressure plus the margin and the fracture pressure.

In depth: advanced detail

Compaction with depth: the porosity trend and the local overburden gradient

In a rock sequence that compacts normally, porosity often falls with depth along an exponential curve (it drops by the same factor over each equal step in depth): $\phi(z) = \phi_0\,\e^{-z/L}$ . Here $\phi_0$ is the porosity at the surface and $L$ is the characteristic compaction depth. If we put this into the bulk density formula and integrate, we get the overburden pressure in explicit form.

Overburden pressure for an exponential porosity trend

\sigma_v(z) = g\left[\rho_{ma}\, z - (\rho_{ma} - \rho_f)\,\phi_0\, L\left(1 - \e^{-z/L}\right)\right]

where:

$\phi_0$ porosity at the surface, fraction
$L$ characteristic compaction depth: at depth $L$ the porosity is reduced by a factor of $\e \approx 2.72$ , m
$\rho_{ma},\ \rho_f$ densities of the grains and of the pore fluid, kg/m³
$z$ true vertical depth from the surface, m

Units: SI; in metric units the factor $g$ is replaced by 0.00981 (densities in g/cm³, pressure in MPa)

Holds when

A normally compacted sequence of a single rock type, with no overpressured zones.
The exponential trend is an empirical approximation (it comes from observation, not from theory). We fit its parameters to the well logs of the area.

The local gradient $\dd\sigma_v/\dd z = g\,\rho_b(z)$ increases with depth, following the density of the rocks. The average gradient from the surface, $\sigma_v/z$ , is always smaller than the local gradient at the same depth.

Worked exampleAverage and local overburden gradient

Given (assumed data)

Porosity at the surface $\phi_0$: 0.40
Characteristic depth $L$: 2000 m
Density of the grains and of the water: 2.65 and 1.05 g/cm³

Solution

At a depth of 3000 m: $\sigma_v = 0.00981\left[2.65\cdot 3000 - 1.60\cdot 0.40\cdot 2000\,(1 - \e^{-1.5})\right] = 68.2$ MPa. The average gradient is 22.7 kPa/m.
The porosity at this depth is $0.40\,\e^{-1.5} = 0.089$ , the bulk density is $2.65 - 1.60\cdot 0.089 = 2.51$ g/cm³, and the local gradient is 24.6 kPa/m.
For comparison, at a depth of 500 m: $\sigma_v = 10.2$ MPa and the average gradient is 20.4 kPa/m.

Answer: The average gradient increases from 20.4 kPa/m at 500 m to 22.7 kPa/m at 3000 m, while the local gradient at 3000 m is already 24.6 kPa/m.

What it means. The “1 psi/ft” rule (22.6 kPa/m) matches the average gradient here only at around 3000 m. Above that depth the overburden pressure is lower than the rule gives, and below it the pressure is higher. So for shallow casing strings and for deep wells we calculate the overburden pressure from the rock sequence, not from the rounded rule. The data in this example are assumed for illustration.

Overburden pressure offshore

Offshore, a layer of water lies on top of the rocks. Its density is about 1.03 g/cm³, which is half that of the sediments. There is also an air gap between the rotary table and the sea. So the overburden pressure, with the depth measured from the rotary table, is noticeably lower offshore than on land at the same depth. The pore pressure hardly changes, because the water from sea level down is the normal hydrostatic column. The window between them narrows, and the deeper the sea, the more it narrows.

Worked exampleOverburden and pore pressure beneath deep water

Given (assumed data)

Height of the rotary table above sea level: 25 m
Water depth: 600 m, sea water 1.03 g/cm³
Sediments below the sea floor: average bulk density 2.0 g/cm³ in the first 1000 m
Pore water: 1.03 g/cm³, pressure normal

Solution

Depth of a point 1000 m below the sea floor, measured from the rotary table: $25 + 600 + 1000 = 1625$ m.
Overburden pressure: water $0.00981\cdot 1.03\cdot 600 = 6.06$ MPa plus sediments $0.00981\cdot 2.0\cdot 1000 = 19.62$ MPa, a total of 25.7 MPa. The equivalent mud weight relative to the rotary table is $25.7/(0.00981\cdot 1625) = 1.61$ g/cm³.
Pore pressure (a water column from sea level, 1600 m): $0.00981\cdot 1.03\cdot 1600 = 16.2$ MPa. Relative to the rotary table, this is 1.01 g/cm³.

Answer: At 1000 m below the sea floor the overburden pressure is equivalent to 1.61 g/cm³ (on land it would be about 2.2) and the pore pressure to 1.01 g/cm³.

What it means. The fracture pressure lies between the pore pressure and the overburden pressure. So beneath deep water the window of allowed mud densities in the upper intervals is narrow: the mud must be heavier than 1.01 g/cm³ but noticeably lighter than 1.61. This is one of the main difficulties of deepwater drilling (Offshore and subsea drilling). The data in this example are assumed for illustration.

A gas column: why it hardly “holds” any pressure

The density of a gas is proportional to pressure, so the pressure in a gas column does not rise in a straight line. For a gas at constant temperature and constant $Z$ , $\dd p/\dd z = \rho g = p\,M g/(Z R T)$ , and the pressure changes exponentially.

Pressure in a gas column (isothermal approximation)

p_{top} = p_{bot}\,\exp\!\left(-\frac{M\, g\, \Delta z}{Z\, R\, T}\right)

where:

$p_{top},\ p_{bot}$ absolute pressure at the top and at the bottom of the column, Pa
$M$ molar mass of the gas (about 0.0174 kg/mol for a natural gas with a specific gravity of 0.6 relative to air), kg/mol
$\Delta z$ vertical height of the column, m
$Z$ gas deviation factor (how far the real gas is from an ideal gas), averaged over the column, —
$R$ universal gas constant, 8.314 J/(mol·K), J/(mol·K)
$T$ average absolute temperature of the column, K

Units: SI; pressures are absolute

Holds when

The gas is at rest. The temperature and $Z$ are constant along the column (in practice we use average values).
The column is filled with gas only.

A column of gas weighs little, so the pressure at the top is only slightly lower than at the bottom. This leads to an important conclusion for well control: if the wellbore fills with gas, almost the full formation pressure reaches the wellhead.

Worked exampleHow much pressure a 1000 m column of gas “uses up”

Given (assumed data)

Pressure at the bottom of the column: 30 MPa (absolute)
Gas: $M = 0.0174$ kg/mol, $Z = 0.95$
Average temperature: 330 K (57 °C)
Height of the column: 1000 m

Solution

Exponent: $\dfrac{0.0174\cdot 9.81\cdot 1000}{0.95\cdot 8.314\cdot 330} = 0.0655$ .
Pressure at the top: $30\cdot \e^{-0.0655} = 28.1$ MPa. The gas column holds 1.9 MPa, which is 1.9 kPa/m on average.

Answer: The pressure at the top of the column is about 28.1 MPa. The gas “uses up” only 1.9 MPa per 1000 m, whereas a column of water uses up about 10 MPa.

What it means. The average gradient of the gas column here is about 1.9 kPa/m, five times smaller than that of water. The same calculation explains why the pressure at the top of a gas accumulation is higher than normal for that depth. The data in this example are assumed for illustration.

Connections to other topics

Reservoir pressure and temperature — reservoir pressure and temperature from the point of view of field development.
Abnormal formation pressures: origin, prediction and recognition — where overpressure and subnormal pressure come from and how they are recognised while drilling.
Fracture gradient and leak-off testing — the upper limit of mud density: fracture pressure and the leak-off test.
Well control: primary barrier and kick indicators — what happens when the mud pressure falls below the pore pressure.
Circulating pressure losses — why the bottomhole pressure during circulation is higher than the hydrostatic pressure.
Density log and photoelectric factor (Pe) — where the bulk densities of the rocks used to calculate the overburden pressure come from.

Check yourself

1
A vertical well and a deviated well have both reached the same formation at a true vertical depth of 2500 m. The measured depth of the deviated well is 3300 m. Both are filled with the same mud. In which well is the mud pressure opposite the formation higher?
1. A.In the deviated well: the mud column is longer
2. B.In the vertical well: the mud presses straight down
3. C.The same in both wells
4. D.It depends on the hole diameter
Show the answer
Correct answer: C. The same in both wells
Hydrostatic pressure depends only on the mud density and the vertical height of the column: $0.00981\cdot\rho\cdot 2500$ in both cases. The measured depth and the cross-section of the hole do not matter.
2
A formation tester recorded a gauge pressure of 31.5 MPa at a true vertical depth of 2650 m. What is the equivalent mud weight, and is this pressure normal if the normal equivalent mud weight in the area is 1.05 g/cm³?
Show the answer
$\rho_{eq} = 31.5/(0.00981\cdot 2650) \approx 1.21$ g/cm³. This is noticeably higher than the normal 1.05 g/cm³, so it is overpressure. The normal pressure at this depth would be about $0.00981\cdot 1.05\cdot 2650 \approx 27.3$ MPa, so the excess is about 4.2 MPa.
3
Where does the constant 0.052 in the formula $p = 0.052\,\rho\,h$ (psi, ppg, ft) come from?
1. A.It is an empirical coefficient fitted to measurements
2. B.It is the volume, in gallons, of a column of fluid 1 ft high with a cross-section of 1 square inch: 12/231
3. C.It is the acceleration due to gravity in feet per second squared, divided by 1000
4. D.It is the pressure gradient of fresh water in psi/ft
Show the answer
Correct answer: B. It is the volume, in gallons, of a column of fluid 1 ft high with a cross-section of 1 square inch: 12/231
A US gallon is 231 cubic inches, and a column 1 ft high with a cross-section of 1 square inch holds 12 cubic inches, which is 0.05195 gallon. The weight of this fluid in pounds, resting on 1 square inch, is the pressure in psi for each foot of height. The gradient of fresh water is a different number: $0.052\cdot 8.33 \approx 0.433$ psi/ft.
4
On an offshore platform the rotary table is 30 m above sea level. The pore pressure is 20.2 MPa at a depth of 2000 m below sea level. What is the equivalent mud weight relative to sea level and relative to the rotary table? Which of the two is compared with the mud density?
Show the answer
Relative to sea level: $20.2/(0.00981\cdot 2000) \approx 1.030$ g/cm³. Relative to the rotary table the depth is 2030 m: $20.2/(0.00981\cdot 2030) \approx 1.014$ g/cm³. We compare the mud density with the second one, because the mud column starts at the rotary table, not at sea level.
5
At the same depth in two wells the overburden pressure is the same, but the pore pressure in the second well is 8 MPa higher. What should you expect of the shales in the second well?
1. A.They are denser and stronger: the pressure is higher
2. B.There is no difference, because shales are impermeable
3. C.No difference: compaction depends only on depth
4. D.They are less compacted: the effective stress is 8 MPa lower
Show the answer
Correct answer: D. They are less compacted: the effective stress is 8 MPa lower
The rock framework carries only the effective stress, $\sigma_v - p_p$ . At a higher pore pressure this stress is lower, and the rock is less compacted: porosity is higher, and density and sonic velocity are lower. We recognise overpressure by such deviations from the normal compaction trend.
6
Why is the mud density not chosen “with a large margin”, when the main aim is to prevent an influx (formation fluid flowing into the well)?
Show the answer
Too much overbalance can push the mud pressure above the fracture pressure and cause lost circulation, after which the mud level drops and the well may kick. It pins the pipe against the wall (differential sticking). It damages the reservoir with filtrate and solids. And it lowers the rate of penetration. The mud density must lie in the window between the pore pressure plus the margin and the fracture pressure, allowing for the ECD during circulation.

Updated: 26 October 2026