Porosity · MunaiHub

Quick refresher

Porosity $\phi = V_p/V_b$ is the fraction of the rock's volume that is empty space. Reports give it as a percentage; in formulas we use a fraction (20% → 0.20).
Total porosity counts all the voids. Open porosity (called effective porosity in English-language literature) counts only the voids that are connected to one another. Open porosity is what sets the hydrocarbons in place and the flow: fluid cannot be recovered from isolated pores.
Primary porosity forms when the sediment is deposited — it is the space between the grains. Secondary porosity forms later, through dissolution of minerals, fractures and dolomitization (limestone turning into dolomite).
The porosity of an ideal packing of equal spheres does not depend on the size of the spheres: it is 47.6% for cubic packing and 26.0% for the tightest packing. Real porosity is reduced by poor sorting (grains of mixed sizes), compaction and cementation.
On core (rock samples cut from the well), porosity is measured directly, from two of the three volumes (bulk, grain, pore). From logs (measurements made down the well) it is found indirectly, through density, hydrogen content or the speed of sound, and is calibrated against core.
Average the porosity of a reservoir with the reservoir thickness as the weight, not as a plain average over the number of samples.

Key relationships

\phi = \dfrac{V_p}{V_b} = \dfrac{V_b - V_{gr}}{V_b}

definition; volumes in any units, as long as all are the same

\phi_D = \dfrac{\rho_{ma} - \rho_b}{\rho_{ma} - \rho_f}

porosity from the density log

\bar\phi = \dfrac{\sum_i \phi_i h_i}{\sum_i h_i}

average porosity over the net thickness

N = \dfrac{A\,h\,\phi\,(1 - S_w)}{B_o}

oil in place is directly proportional to porosity

Common pitfalls

Putting porosity in per cent into a formula written for fractions. The result is wrong by a factor of 100.
Assuming that a gas porosimeter (an instrument that measures grain volume with gas), or saturating a sample with liquid, gives total porosity. Both methods see only the connected pores.
Confusing the English-language effective porosity (open porosity) with the “effective porosity” of Russian-language petrophysics, which has already had the irreducible (unremovable) water taken out.
Calculating porosity from the density log with a wrong matrix density (the density of the solid grains). The error is several percentage points.
Averaging porosity over the number of core samples instead of over thickness, and including non-reservoir rock in the average.

After this lesson you can

explain how total, open and isolated porosity differ, and which of them you need to calculate hydrocarbons in place
tell primary porosity from secondary porosity, and name the geological factors that increase it and decrease it
calculate porosity from laboratory measurements of sample volumes, and know which kind of porosity each method gives
estimate porosity from the density log, and correct for the effect of the rock type (lithology)
average porosity correctly over the well section, and estimate how an error in porosity turns into an error in hydrocarbons in place

The rock: solid framework and voids

A sedimentary rock (a rock formed from deposited sediment, such as sand or clay) is made of a solid framework and of the voids between its parts. The framework is the mineral grains, the cement that binds them and the clay particles. The bulk volume of a sample, $V_b$ (its whole volume, solid and voids together), is the grain volume $V_{gr}$ (the volume of the solid part) plus the pore volume $V_p$ : $V_b = V_{gr} + V_p$ . If you know any two of these volumes, you can find the third — and with it the porosity.

Porosity

\phi = \frac{V_p}{V_b} = \frac{V_b - V_{gr}}{V_b} = \frac{V_p}{V_{gr} + V_p}

where:

$\phi$ porosity, fraction (in reports, %)
$V_p$ volume of the voids that the given kind of porosity counts, cm³, m³
$V_b$ volume of the sample or rock element including the pores, cm³, m³
$V_{gr}$ volume of the solid phase: grains, cement and clay minerals, cm³, m³

Units: any units of volume, the same for all three volumes; porosity has no units

Holds when

All three volumes refer to the same sample at the same conditions (pressure, temperature, load).
What counts as the pore volume depends on the kind of porosity: all the voids (total) or only the connected ones (open); see the next section.

Porosity is how much fluid each unit of rock can hold: 1 m³ of rock with a porosity of 0.20 holds 0.20 m³ (200 L) of fluid. The saturation shows what share of this volume is filled with oil or gas.

Total, open and isolated porosity

Not all the voids in a rock are equally useful. A rock is compacted (squeezed by the weight above) and cemented (glued by minerals that grow in the pores). As this happens, some pores end up completely surrounded by cement or by crystals that grew on the grains (overgrowths). They lose their connection to the rest of the pore network. The fluid in such pores can neither flow into the well nor be pushed out by injected water.

Total porosity $\phi_t$: The volume of all the voids, both connected and isolated, divided by the volume of the rock.
Open porosity $\phi_e$: The volume of the connected pores alone, divided by the volume of the rock (English-language usage calls it effective or connected porosity). This is the porosity that decides how much fluid can be recovered, and it is what standard laboratory measurements give.
Isolated (closed) porosity: Total porosity minus open porosity. In clean sandstones (with little clay) it is usually too small to matter. It can be significant in volcanic rocks, in some carbonates (limestones and dolomites) and in strongly cemented rocks.
Dead-end pores: Pores that connect to the network on one side only. They count as open porosity and hold fluid that can be recovered. But almost no fluid flows through them, so when the oil is displaced they give it up more slowly and less completely.

SchematicPore space in a section of sandstone: the connected pores form a network that runs right through the rock, a dead-end pore is open on one side only, an isolated pore is completely surrounded by cement.

One name, different quantities: what is called “effective porosity”

Where it occurs: English-language core analysis (effective porosity)
What it means: the volume of the connected pores; in Russian-language practice this is open porosity $K_{\text{p.o}}$
What is excluded: isolated pores

Where it occurs: English-language log interpretation (effective porosity, PHIE)
What it means: total porosity minus the volume of water bound to clay minerals
What is excluded: clay-bound water

Where it occurs: Russian-language petrophysics and calculation of hydrocarbons in place ( $K_{\text{p.ef}}$ )
What it means: the part of the open porosity that oil or gas can occupy: $K_{\text{p.ef}} = K_{\text{p.o}}\,(1 - K_{\text{vo}})$ , where $K_{\text{vo}}$ is the residual water saturation
What is excluded: isolated pores and all the residual (irreducible) water

Where it occurs	What it means	What is excluded
English-language core analysis (effective porosity)	the volume of the connected pores; in Russian-language practice this is open porosity $K_{\text{p.o}}$	isolated pores
English-language log interpretation (effective porosity, PHIE)	total porosity minus the volume of water bound to clay minerals	clay-bound water
Russian-language petrophysics and calculation of hydrocarbons in place ( $K_{\text{p.ef}}$ )	the part of the open porosity that oil or gas can occupy: $K_{\text{p.ef}} = K_{\text{p.o}}\,(1 - K_{\text{vo}})$ , where $K_{\text{vo}}$ is the residual water saturation	isolated pores and all the residual (irreducible) water

Before you use someone else's “effective porosity”, check which definition they used. In shaly (clay-rich) reservoirs the difference between the definitions can reach tens of per cent of the pore volume. The symbols $K_{\text{p.o}}$ , $K_{\text{p.ef}}$ and $K_{\text{vo}}$ are the Russian-language notation, with the subscripts spelled in Latin letters.

In clean, well-sorted sandstones, total and open porosity are almost the same. The difference matters in carbonates with isolated vugs and in shaly reservoirs. In shaly reservoirs a noticeable part of the pores holds water that clings tightly to the surface of the clay particles.

Primary and secondary porosity: where pores come from

Primary porosity forms when the sediment is deposited. It is the space between sand grains (intergranular porosity) or between particles of carbonate mud (lime mud). After burial it usually only gets smaller. The weight of the rocks above pushes the grains closer together and deforms them. At the contacts between grains the mineral dissolves under pressure and is deposited again in the pores as cement.

Secondary porosity appears later, when the rock has already formed:

Dissolution. Groundwater dissolves and washes out unstable grains (feldspars, shell fragments) or the carbonate framework itself. This is how mouldic pores (pores that keep the shape of a dissolved grain or shell) and vugs appear.
Fracturing. Tectonic stresses (forces in the earth's crust) split dense rock. The fractures take up little volume, but they raise permeability (the ability of the rock to let fluid flow) sharply.
Dolomitization. Calcite (the mineral of limestone) is replaced by dolomite. If the replacement is ideal, mole for mole, the volume of the solid part shrinks by about 13%, and the space that is freed becomes pores.

Secondary porosity is especially typical of carbonates (limestones and dolomites). There it is often larger than primary porosity. It is spread less predictably, and it has a stronger effect on flow: connected vugs and fractures carry fluid far better than intergranular pores (pores between grains) of the same volume.

Several factors set how large the porosity is:

Sorting (how similar the grains are in size) is the main factor. If the grains differ in size, the small ones fill the spaces between the large ones, and porosity falls. For a mixture with a very wide spread of sizes it can fall towards zero.
Grain shape and packing. Angular and flattened grains pack more tightly than rounded ones when the rock is compacted, so they leave fewer voids.
Grain size does not affect porosity in an ideal packing (see the example below), but it strongly affects permeability.
Compaction with depth and age. Porosity usually decreases, and it decreases faster in clays than in sands.
Cement and clay minerals. Quartz, calcite and clays that are deposited in the pores reduce porosity. Cement that coats the pore walls evenly lowers both porosity and permeability. Cement or fibrous clays that block the pore throats (the narrow passages between pores) can leave porosity almost unchanged, yet cut permeability by several times.

Worked examplePorosity of an ideal packing of equal spheres

Given (assumed data)

Grains: equal spheres of radius $r$
Packing 1: cubic: the spheres stand in columns, each touching six neighbours
Packing 2: tightest (rhombohedral): each sphere sits in the hollow between spheres of the layer below

Solution

Cubic packing. Join the centres of eight neighbouring spheres. This gives a cube with edge $2r$ and volume $V_b = (2r)^3 = 8r^3$ . One eighth of each of the eight spheres falls inside the cube, which is exactly one sphere: $V_{gr} = \tfrac43\pi r^3$ .
Porosity: $\phi = 1 - \dfrac{V_{gr}}{V_b} = 1 - \dfrac{4\pi r^3/3}{8r^3} = 1 - \dfrac{\pi}{6} = 0.476$ , that is 47.6%.
Tightest packing. The spheres fill a fraction $\pi/(3\sqrt2)$ of the volume. This is a well-known result from the geometry of packing spheres as tightly as possible. The porosity is $\phi = 1 - \dfrac{\pi}{3\sqrt2} = 0.260$ , that is 26.0%.
In both cases the radius $r$ cancels out: the porosity of an ideal packing does not depend on grain size.

Answer: Cubic packing gives 47.6% and the tightest packing gives 26.0%, whatever the size of the spheres.

What it means. Real sediments are packed at random. A freshly deposited, well-sorted sand has a porosity of about 40–45%; a poorly sorted sand has noticeably less. Compaction and cementation then reduce it to the 10–30% typical of sandstone reservoirs. Grain size sets the width of the pore channels, and so the permeability: a fine-grained rock with the same porosity has a much lower permeability. The example is purely geometrical.

Approximate values of porosity

Rock: Loose modern sands
Typical porosity: 30–45%
Comment: not compacted or cemented

Rock: Sandstone reservoirs
Typical porosity: 10–30%
Comment: most often 15–25%; lower for tight, cemented ones

Rock: Limestones and dolomites
Typical porosity: 3–25%
Comment: a large share of secondary porosity; values vary very widely

Rock: Chalk
Typical porosity: 15–45%
Comment: high porosity with low permeability: the pores are very small

Rock: Clays and claystones
Typical porosity: from a few per cent at depth to 40% and more in young sediments
Comment: the pores are very small, the water in them is tightly bound, and fluid hardly flows

Rock: Tight fractured rocks
Typical porosity: a few per cent or less; fractures usually account for fractions of a per cent
Comment: the storage capacity is small, but the permeability along fractures can be high

Rock	Typical porosity	Comment
Loose modern sands	30–45%	not compacted or cemented
Sandstone reservoirs	10–30%	most often 15–25%; lower for tight, cemented ones
Limestones and dolomites	3–25%	a large share of secondary porosity; values vary very widely
Chalk	15–45%	high porosity with low permeability: the pores are very small
Clays and claystones	from a few per cent at depth to 40% and more in young sediments	the pores are very small, the water in them is tightly bound, and fluid hardly flows
Tight fractured rocks	a few per cent or less; fractures usually account for fractions of a per cent	the storage capacity is small, but the permeability along fractures can be high

These are rough figures for guidance, not a classification. The actual values depend on the basin, the depth and how the rocks have changed over time.

The geological causes are covered in more detail in the topics Diagenesis: compaction, cementation and dissolution, Reservoir quality: geological controls on porosity and permeability and Carbonate sediments, environments and carbonate porosity.

Measuring porosity on core

In the laboratory, porosity is measured on cylindrical samples called plugs. A plug is usually 25–38 mm in diameter and is drilled out of the core that was brought up from the well. First, the sample is washed with solvents to remove oil and salts, and it is dried. Then two of the three volumes (bulk, grain, pore) are measured, and the porosity is calculated.

How each volume is found

1
Bulk volume
For a cylinder with a regular shape, use its dimensions: $V_b = \pi d^2 L/4$ . For a sample of irregular shape, measure the volume of liquid it displaces, or the buoyant force on it (Archimedes' principle). The liquid must not soak into the pores. So either the sample is saturated with that liquid beforehand, or a non-wetting liquid (one that does not soak into the rock) is used. Mercury used to be widely used for this, but it is toxic, so such methods are being phased out.
2
Grain volume
Use a gas porosimeter, which works by Boyle's law (see below). The gas is usually helium: its atoms are small enough to enter the tiniest pores, it is chemically inert, and it hardly sticks (adsorbs) to the minerals. The gas cannot reach isolated pores, so they are counted as part of the “grain volume”.
3
Pore volume
Saturate the sample with a liquid. The dry sample is put under vacuum to remove the air and is then filled with a liquid of known density, such as brine (salt water) or kerosene. The gain in mass divided by the density equals the volume of the connected pores. Another way is to inject helium straight into the pores of a sample held in a core holder (a cell that squeezes the sample). This gives the porosity at once, under a load close to the load in the reservoir.
4
Total porosity
Crush the sample to the size of single grains. This opens the isolated pores. Then measure the volume of the solid part of the powder. Comparing the result with the open porosity of the same sample shows the share of isolated pores.

SchematicA two-chamber gas porosimeter: gas from the reference chamber (volume

V_r

, pressure

p_1

) expands into the chamber that holds the sample. The grain volume is found from the final, equalised pressure

p_2

Grain volume by Boyle's law (two-chamber gas porosimeter)

V_{gr} = V_c - V_r\,\frac{p_1 - p_2}{p_2 - p_0}

where:

$V_{gr}$ volume of the sample that the gas cannot reach: grains together with isolated pores, cm³
$V_c$ volume of the chamber with the sample (including the channels up to the valve), cm³
$V_r$ volume of the reference chamber (including the channels up to the valve), cm³
$p_1$ initial gas pressure in the reference chamber, kPa, absolute
$p_0$ initial pressure in the chamber with the sample, usually atmospheric, kPa, absolute
$p_2$ final (equalised) pressure after the valve is opened, kPa, absolute

Units: all volumes in the same units; absolute pressures, all in the same units

Holds when

The process is isothermal (the temperature does not change): $p_2$ is read after the gas temperature has settled.
The gas is ideal (it obeys the ideal gas law) and does not adsorb (stick to the minerals). For helium at pressures of about 1 MPa this holds with high accuracy.
The volumes of the chambers and connecting channels are calibrated with steel cylinders of known volume.
If the chamber with the sample starts at atmospheric pressure, you can use gauge pressures (read from a gauge, relative to the atmosphere) instead. Then $p_0 = 0$ , and the denominator equals the gauge reading $p_2$ .

Gas from the reference chamber expands into the free space of the chamber with the sample. The larger the grain volume, the less room there is for the gas, and the higher the final pressure $p_2$ . From $V_{gr}$ and $V_b$ we get the open porosity: $\phi = (V_b - V_{gr})/V_b$ .

Worked exampleSample porosity from a gas porosimeter and from saturation with brine

Given (assumed data)

Sample dimensions: diameter 3.80 cm, length 4.50 cm
Mass of the dry sample: 108.3 g
Chamber volumes: reference $V_r = 40.0$ cm³, with the sample $V_c = 70.0$ cm³
Gauge pressures: $p_1 = 700$ kPa, after the valve is opened $p_2 = 405$ kPa; the chamber with the sample starts at atmospheric pressure
Then: mass of the sample saturated with brine: 118.9 g; brine density 1.05 g/cm³

Solution

Bulk volume: $V_b = \pi d^2 L/4 = \pi\cdot 3.80^2\cdot 4.50/4 = 51.04$ cm³.
Free volume of the chamber with the sample (gauge pressures, $p_0 = 0$ ): $V_c - V_{gr} = 40.0\cdot\dfrac{700 - 405}{405} = 29.14$ cm³, so $V_{gr} = 70.0 - 29.14 = 40.86$ cm³.
Open porosity from gas: $\phi = \dfrac{51.04 - 40.86}{51.04} = 0.199$ , that is 19.9%.
Check with the grain density: $\rho_{gr} = 108.3/40.86 = 2.65$ g/cm³, which matches quartz (2.65 g/cm³).
From saturation with brine: $V_p = \dfrac{118.9 - 108.3}{1.05} = 10.10$ cm³, $\phi = 10.10/51.04 = 19.8$ %.

Answer: The open porosity of the sample is 19.9% from gas and 19.8% from saturation with brine.

What it means. Both methods measure the connected pores, so the results are close. Saturation with a liquid usually gives slightly less, because the smallest pores are not filled completely and some liquid is lost from the surface during weighing. The grain density is a handy quality check. A value clearly lower than the density of the mineral may mean isolated pores (the gas cannot see them, so they are added to the grain volume) or a measurement error. The data in this example are assumed for illustration.

Porosity from logs

Core is taken only in some wells and intervals. Logs are recorded in almost every well, over the whole section. But no logging tool measures porosity directly. It measures a physical property of the rock together with the fluid in its pores, and porosity is then calculated with a rock model. So the result depends on the assumed lithology (rock type) and fluid, and it must be calibrated against core.

Three main methods of estimating porosity from logs

Method: Density (gamma-gamma) log
What it measures: how strongly the electrons of the rock scatter gamma rays. This gives its electron density, which is almost proportional to the bulk density $\rho_b$
What it is sensitive to: the matrix minerals (the minerals of the grains); gas in the pores makes the porosity too high

Method: Neutron log
What it measures: how much neutrons are slowed down by hydrogen nuclei. This gives the hydrogen content of the rock
What it is sensitive to: water in clays makes the porosity too high, gas makes it too low

Method: Sonic log
What it measures: the time a compressional (sound) wave takes to travel a unit length of rock, $\Delta t$
What it is sensitive to: mainly reflects intergranular porosity and underestimates vugs and fractures; in unconsolidated (loose) rocks a correction is needed

Method	What it measures	What it is sensitive to
Density (gamma-gamma) log	how strongly the electrons of the rock scatter gamma rays. This gives its electron density, which is almost proportional to the bulk density $\rho_b$	the matrix minerals (the minerals of the grains); gas in the pores makes the porosity too high
Neutron log	how much neutrons are slowed down by hydrogen nuclei. This gives the hydrogen content of the rock	water in clays makes the porosity too high, gas makes it too low
Sonic log	the time a compressional (sound) wave takes to travel a unit length of rock, $\Delta t$	mainly reflects intergranular porosity and underestimates vugs and fractures; in unconsolidated (loose) rocks a correction is needed

Details are in the topics on each method and on the effects of gas, clay and lithology: Gas, shale, carbonate and pressure effects on porosity logs.

Porosity is most often calculated from the density log. The bulk density of a rock is the matrix density (the density of the solid grains) plus the fluid density, each weighted by its share of the volume: $\rho_b = (1 - \phi)\,\rho_{ma} + \phi\,\rho_f$ . From this we get:

Porosity from the density log

\phi_D = \frac{\rho_{ma} - \rho_b}{\rho_{ma} - \rho_f}

where:

$\phi_D$ porosity from the density log. It is total porosity, because isolated pores also lower the density, fraction
$\rho_b$ bulk density of the rock measured by the tool, g/cm³
$\rho_{ma}$ grain (matrix) density of the framework: quartz 2.65, calcite 2.71, dolomite 2.87, g/cm³
$\rho_f$ density of the fluid in the zone the tool reads, usually mud filtrate (the liquid from the drilling mud that has seeped into the rock): about 1.0 for fresh mud, up to 1.1–1.2 for salty mud, g/cm³

Units: g/cm³ or kg/m³ — any units of density, the same for all three densities

Holds when

The solid framework is one known mineral, or a mixture with a known average density, and has no clay.
The pores in the zone the tool reads (about 10 cm from the borehole wall) are filled with a fluid of known density — usually mud filtrate.
Corrections for mudcake (a layer of drilling-mud solids on the borehole wall) and for the roughness of the borehole wall have been applied.

For the same minerals, the lower the measured density, the more pores there are. When $\rho_b = \rho_{ma}$ the porosity is zero; when $\rho_b = \rho_f$ it is one.

Worked exampleOne density log reading, three different porosities

Given (assumed data)

Bulk density from the log $\rho_b$: 2.32 g/cm³
Filtrate density $\rho_f$: 1.00 g/cm³
Possible framework: quartz (2.65), calcite (2.71) or dolomite (2.87 g/cm³)

Solution

Quartz sandstone: $\phi_D = \dfrac{2.65 - 2.32}{2.65 - 1.00} = 0.200$ .
Limestone: $\phi_D = \dfrac{2.71 - 2.32}{2.71 - 1.00} = 0.228$ .
Dolomite: $\phi_D = \dfrac{2.87 - 2.32}{2.87 - 1.00} = 0.294$ .

Answer: Depending on the assumed lithology, the porosity comes out as 20.0, 22.8 or 29.4%.

What it means. The porosity varies by 9.4 percentage points. The pore volume, and with it the hydrocarbons in place, is 1.47 times larger for a dolomite framework than for a quartz one. So the lithology must be found before the porosity is calculated. Three ways to find it are the photoelectric factor (a log reading that depends on the minerals), the joint interpretation of the density and neutron logs (Lithology and porosity from crossplots), and core. The data in this example are assumed for illustration.

CalculatedPorosity from the density log against the measured density, for

\rho_f = 1.0

g/cm³ and three framework minerals: the solid line is quartz, the dashed line is calcite and the orange line is dolomite. The dots mark the 2.32 g/cm³ reading from the example.

For the sonic log in cemented rocks we use the Wyllie time-average equation. It is the same mixing rule, but applied to the travel time of the wave: $\Delta t = (1 - \phi)\,\Delta t_{ma} + \phi\,\Delta t_f$ , where $\Delta t_{ma}$ is the travel time in the matrix and $\Delta t_f$ in the fluid. For quartz, $\Delta t_{ma} \approx 182$ µs/m (55.5 µs/ft); for calcite it is about 156 µs/m (47.5 µs/ft), and for water about 620 µs/m (189 µs/ft). The neutron log is calibrated directly in porosity units (p.u.), usually for limestone filled with water.

Log estimates are calibrated against core. First the core depths are matched to the log depths. Then the porosity of the samples is compared with the calculated porosity, and the matrix and fluid densities are adjusted. Modern nuclear magnetic resonance (NMR) logging (Nuclear magnetic resonance (NMR) logging) gives a porosity that hardly depends on the minerals. It also splits the porosity into bound (held in the rock) and moveable (free to flow) parts.

Average porosity and hydrocarbons in place

To calculate the hydrocarbons in place and to build models, we need the average porosity of the reservoir. We calculate it over the net thickness only. The net thickness is made up of the layers that count as reservoir rock because they pass cut-off values (limits) of porosity, permeability, shale volume and saturation. The ratio of net thickness to gross (total) thickness is called the net-to-gross ratio (NTG; also the reservoir fraction or sand ratio).

Average porosity over thickness

\bar\phi = \frac{\sum_i \phi_i\,h_i}{\sum_i h_i}

where:

$\phi_i$ porosity of the $i$ -th layer (interval), fraction
$h_i$ net thickness of the $i$ -th layer, m

Units: all thicknesses in the same units; porosity as a fraction

Holds when

Only reservoir layers enter the sum.
Each $\phi_i$ represents its whole interval. For logs this holds automatically; for core it holds only if the samples are spread evenly.

The sum $\sum \phi_i h_i$ is the pore volume under each unit of area of the reservoir. The average porosity is the single constant value that would give the same pore volume for the same thickness. For a whole accumulation (one deposit of oil or gas), the values from the wells are averaged with the rock volume as the weight, using maps.

Volumetric calculation of oil in place

N = \frac{A\,h\,\phi\,(1 - S_w)}{B_o}

where:

$N$ stock-tank oil initially in place (STOIIP): the oil in place, converted to standard conditions (surface pressure and temperature), m³
$A$ oil-bearing area, m²
$h$ average net oil-saturated thickness, m
$\phi$ average open porosity, fraction
$S_w$ average water saturation (see Fluid saturation), fraction
$B_o$ oil formation volume factor — the number of reservoir cubic metres needed for one cubic metre of oil at standard conditions (see Black oil model: solution gas–oil ratio and formation volume factors), m³/m³

Units: SI: area in m², thickness in m; the result is m³ of oil at standard conditions

Holds when

The parameters are averages over the whole accumulation, and they match each other: porosity and saturation come from the same definition.
The accumulation contains oil and irreducible water only, with no free gas.

The oil in place is directly proportional to porosity. So the relative error in porosity becomes, in full, the relative error in the oil in place. At $\phi = 0.15$ , an error of one percentage point is 6.7% of the oil in place.

In other unit systems (2)

Oilfield units

N\,[\text{STB}] = \frac{7758\,A\,[\text{acre}]\;h\,[\text{ft}]\;\phi\,(1 - S_w)}{B_o\,[\text{rb/STB}]}

$A$ in acres, $h$ in feet, $B_o$ in reservoir barrels per barrel at standard conditions; $N$ in barrels of oil at standard conditions (STB).

$7758$ is the number of barrels in one acre-foot (a volume 1 acre in area and 1 ft deep): $43{,}560\ \text{ft}^3 / 5.6146\ \text{ft}^3$ .

Mass form (Russian-language practice of calculating hydrocarbons in place)

Q = F\,h\,K_{\text{p}}\,K_{\text{n}}\,\theta\,\rho

$Q$ in t; $F$ is the area, m²; $h$ in m. $K_{\text{p}}$ is the open porosity and $K_{\text{n}} = 1 - S_w$ is the oil saturation, both as fractions (these are the Russian-language symbols, with the subscripts spelled in Latin letters). $\theta = 1/B_o$ is the shrinkage factor (it converts reservoir volume to stock-tank volume). $\rho$ is the density of the stock-tank oil, t/m³.

There is no constant. It is the same formula, multiplied by the density of the oil to give tonnes.

Worked exampleAverage over samples and average over thickness

Given (assumed data)

Upper layer: thickness 2.0 m, porosity 25%, 8 core samples
Lower layer: thickness 8.0 m, porosity 14%, 4 samples (the core was only partly recovered)
For the calculation of oil in place: area 2.0 km², $S_w = 0.25$ , $B_o = 1.20$ m³/m³, stock-tank oil density 850 kg/m³

Solution

Average over samples: $\bar\phi_{\text{plug}} = \dfrac{8\cdot 0.25 + 4\cdot 0.14}{12} = 0.213$ .
Average over thickness: $\bar\phi = \dfrac{0.25\cdot 2.0 + 0.14\cdot 8.0}{2.0 + 8.0} = 0.162$ .
Oil in place with correct averaging: $N = \dfrac{2.0\cdot 10^6\cdot 10\cdot 0.162\cdot (1 - 0.25)}{1.20} = 2.025\cdot 10^6$ m³, or 1.72 million t.
With the average over samples it would come out at 2.67 million m³, 32% more.

Answer: The average porosity of the reservoir is 16.2%, not 21.3%. The oil in place is about 2.0 million m³ of oil (1.7 million t).

What it means. The average over samples overestimates the pore volume by almost a third, only because the good layer was sampled more densely. A log is recorded continuously, so it does not make this error. This is why core is used mainly to calibrate logs, not for direct averaging. The data in this example are assumed for illustration.

The volumetric method as a whole is covered in the topics Volumetric estimation of hydrocarbons in place and Probabilistic volumetrics by Monte Carlo simulation. It includes maps of thickness and porosity, and a probabilistic estimate (an estimate that gives a range of outcomes, not a single number).

Terms and symbols

Porosity $\phi$ ( $K_{\text{p}}$ in Russian-language notation): The fraction of the rock volume taken up by voids. It has no units, and reports give it as a percentage.
Open porosity: The fraction of the rock volume taken up by connected pores. English-language literature calls it effective porosity.
Total porosity: The fraction of the rock volume taken up by all voids, including isolated ones.
Secondary porosity: Voids that appeared after the rock formed: dissolution vugs, fractures and dolomitization pores.
Grain (matrix) density $\rho_{ma}$: The density of the solid part of the rock, without the pores: quartz 2.65 g/cm³, calcite 2.71, dolomite 2.87.
Net thickness: The total thickness of the reservoir layers picked out by cut-off values. Its ratio to the gross thickness is the net-to-gross ratio.

Common mistakes and misconceptions

Wrong

“Porosity is 20%, so I put 20 into the formula.”

In fact

Formulas take porosity as a fraction: 0.20. Otherwise the result is 100 times too large.

Wrong

“The helium porosimeter showed 18%, so that is the total porosity.”

In fact

Gas does not enter isolated pores, so the porosimeter gives the open porosity. The total porosity is measured on a crushed sample.

Wrong

“The effective porosity in an English-language report is the same as the effective porosity in Russian-language calculations of hydrocarbons in place.”

In fact

English-language effective porosity is the open porosity (from core) or the total porosity minus clay-bound water (from logs). Russian-language effective porosity also leaves out all the residual water.

Wrong

“The average porosity of a reservoir is the average over all the core samples.”

In fact

Average with thickness as the weight, and over reservoir rock only. Uneven sampling skews a simple average.

Wrong

“The density log gives porosity regardless of the rock.”

In fact

The result depends on the assumed matrix density: the same reading of 2.32 g/cm³ gives 20% for sandstone and 29% for dolomite.

In depth: advanced detail

Deriving the gas porosimeter formula: why gauge pressures can be used

Before the valve is opened, the reference chamber (volume $V_r$ ) holds gas at absolute pressure $p_1$ (pressure measured from vacuum). The free volume of the chamber with the sample, $V_c - V_{gr}$ , holds gas at absolute pressure $p_0$ . After the valve is opened, the gas fills the whole volume $V_r + V_c - V_{gr}$ at pressure $p_2$ . For an ideal gas at constant temperature, the amount of gas is proportional to the product $pV$ , and it does not change:

$p_1 V_r + p_0\,(V_c - V_{gr}) = p_2\,(V_r + V_c - V_{gr})$ , so $(V_c - V_{gr})(p_2 - p_0) = V_r\,(p_1 - p_2)$ and $V_{gr} = V_c - V_r\,\dfrac{p_1 - p_2}{p_2 - p_0}$ .

Only differences between pressures enter the result. So shifting all the pressures by the atmospheric pressure changes nothing. If the chamber with the sample started at atmospheric pressure, you can work with the gauge readings and put $p_0 = 0$ . If the chamber was pumped down to a vacuum beforehand, or its pressure differed from atmospheric, you must put $p_0$ in explicitly, on the absolute scale. If the gas is noticeably non-ideal, replace each product $pV$ by $pV/z$ , where $z$ is the gas deviation factor (it shows how far the real gas departs from an ideal one).

Compaction with depth

Compaction is normal when the pore water has time to escape. Then porosity decreases with depth roughly exponentially:

Exponential trend of normal compaction

\phi(z) = \phi_0\,\e^{-c\,z}

where:

$\phi(z)$ porosity at burial depth $z$ , fraction
$\phi_0$ porosity of the sediment at the surface: about 0.6 for clay-rich muds and 0.4–0.5 for sands, fraction
$c$ compaction coefficient. It is fitted to the data of a particular basin, and it is larger for clays than for sands, km⁻¹
$z$ burial depth, km

Units: $c$ and $z$ in matching inverse units (for example, km⁻¹ and km)

Holds when

Normal compaction: the pore pressure is close to hydrostatic (the pressure of a column of water).
One lithology only. Cementation and dissolution do not disturb the trend.

The trend describes mechanical compaction. If the porosity of clays at some depth is clearly above the trend, the water has not had time to escape, and the pore pressure is probably abnormally high. This is used to predict pressure before drilling (see Abnormal formation pressures: origin, prediction and recognition).

Connections to other topics

Rock permeability — why high porosity does not by itself mean that fluid can flow.
Fluid saturation — what share of the pore volume is filled with oil, gas and water.
Core analysis — how core is sampled and prepared, and the full range of laboratory tests on it.
Porosity from the density log — calculating porosity from the density log, and the corrections to it.
Rock compressibility and the effect of stress on core properties — how porosity changes under load and as the reservoir pressure declines.
Volumetric estimation of hydrocarbons in place — calculating hydrocarbons in place by the volumetric method from maps of thickness, porosity and saturation.

Check yourself

1
A sample has a bulk volume of 50.0 cm³. A helium porosimeter gives a grain volume of 41.0 cm³. What is the porosity, and which kind of porosity is it?
1. A.82%, total
2. B.18%, total
3. C.18%, open
4. D.22%, open
Show the answer
Correct answer: C. 18%, open
$\phi = (50.0 - 41.0)/50.0 = 0.18$ . Helium does not enter isolated pores. They are counted in the “grain volume”, so the result is the open porosity.
2
Two samples are packed in the same way (cubic packing), one from spheres 1 mm in diameter and one from spheres 0.1 mm in diameter. How do their porosities compare?
1. A.The small spheres give more
2. B.The small spheres give 100 times less
3. C.The small spheres give 10 times less
4. D.They are equal: about 47.6%
Show the answer
Correct answer: D. They are equal: about 47.6%
The radius cancels out: $\phi = 1 - \pi/6$ at any sphere size. Grain size changes the width of the channels, and so the permeability. Porosity is set by sorting, packing, compaction and cement.
3
Which method gives the total porosity rather than the open porosity?
1. A.Measuring the volume of the solid phase of a crushed sample
2. B.Saturating a dry sample with brine
3. C.A helium porosimeter on an intact sample
4. D.Injecting helium into the pores of a sample in a core holder
Show the answer
Correct answer: A. Measuring the volume of the solid phase of a crushed sample
Crushing opens the isolated pores, and the volume of the solid phase of the powder no longer includes them. The other methods work with a gas or a liquid, which reach only the connected pores.
4
The density log in a limestone reads 2.40 g/cm³, and the mud filtrate is fresh (1.00 g/cm³). What porosity do you get, and what happens if you wrongly take the framework to be quartz?
Show the answer
For calcite: $\phi_D = (2.71 - 2.40)/(2.71 - 1.00) = 0.181$ , about 18%. With a quartz framework: $(2.65 - 2.40)/(2.65 - 1.00) = 0.152$ , about 15%. The wrong lithology would make the porosity 3 percentage points too low, and the pore volume and the hydrocarbons in place about 16% too low.
5
Why can the porosity from the sonic log in a vuggy carbonate be noticeably lower than from the density and neutron logs?
Show the answer
A sound (elastic) wave travels by the fastest path, through the framework and the intergranular pores, and hardly notices large vugs and fractures. The density and neutron methods respond to the whole volume of voids. The difference between the estimates points to secondary porosity.
6
The average over core samples gives a reservoir porosity of 21%. The log, averaged over thickness, gives 16%. What is the most likely cause of the difference, and which figure should you trust when calculating hydrocarbons in place?
Show the answer
Most likely the samples were taken unevenly. Good layers were sampled more densely, and tight or crumbled ones less often, so the simple average is pushed upwards. For hydrocarbons in place, use the thickness-weighted average from the calibrated log. Also check that core and logs measured the same kind of porosity, and that the confining stress on the samples was taken into account.

Updated: 19 October 2026