Rock permeability · MunaiHub

Quick refresher

Porosity $\phi$ is the capacity of the rock (how much fluid it can hold). Permeability $k$ is its conductivity (how easily fluid moves). The two are linked, but not in a universal way.
Darcy's law: the flow rate is directly proportional to the permeability, the cross-section area and the pressure difference, and inversely proportional to the viscosity and the length of the path.
1 darcy ≈ $0.987\cdot 10^{-12}$ m² ≈ 1 µm². In practice engineers use the millidarcy (mD): 1 D = 1000 mD.
In radial flow (flow towards the well from all sides) most of the pressure difference falls in the zone near the well, so the state of that zone has such a strong effect on the rate.
Permeability depends on direction: in layered rocks the vertical permeability is usually lower than the horizontal one. We average layers that work in parallel (flow along the layers) arithmetically, weighted by thickness. We average layers that work in series (flow across the layers) harmonically.

Key relationships

q = \dfrac{k A}{\mu}\cdot\dfrac{p_1 - p_2}{L}

linear flow, any consistent system of units

q = \dfrac{2\pi k h\,(p_e - p_w)}{\mu\,\ln(r_e/r_w)}

steady-state radial flow to a well (nothing changes with time; at reservoir conditions)

1\ \text{D} = 1000\ \text{mD} = 0.9869\cdot 10^{-12}\ \text{m}^2

unit conversion

Common pitfalls

Mixing up the millidarcy and the darcy: an error of a factor of a thousand.
Assuming that high porosity guarantees high permeability.
Using the gas permeability of core without a correction for gas slippage (the Klinkenberg effect: gas slips along the pore walls, so the gas permeability reads too high).
Averaging the permeability of layers arithmetically when the flow goes across the layers.

After this lesson you can

explain how permeability differs from porosity, and why a porous rock can be almost impermeable
write Darcy's law for linear and radial flow, and list the conditions under which it holds
convert permeability between darcies, millidarcies and square metres, and recognise the constants in oilfield formulas
calculate permeability from a core test, and estimate the rate of a well with the radial flow formula
choose the way to average the permeability of a layered reservoir according to the direction of flow

Porosity and permeability: capacity and conductivity

A reservoir rock consists of grains and the voids between them. Porosity ( $\phi$ ) shows what share of the rock volume is voids, that is, how much fluid the rock can hold. Permeability shows something different: how easily fluid can move through these voids.

The movement of fluid is limited not by the pores themselves but by the narrowest places between them, the pore throats. A rock may contain many voids, but if they are joined by thin channels, or not joined at all, fluid will hardly pass through it.

Porosity and permeability behave differently

Rock: Well-sorted coarse-grained sandstone
Porosity: high
Permeability: high
Why: coarse grains leave wide channels between the pores

Rock: Clay, mudstone
Porosity: can be high
Permeability: very low
Why: the pores are very small, the channels are narrow, and the water is bound to the surface of the clay particles

Rock: Limestone with isolated vugs
Porosity: high
Permeability: low
Why: the vugs (cavities) hardly connect with each other

Rock: Fractured tight rock
Porosity: low
Permeability: can be high
Why: fractures take a small volume but form a network that conducts well

Rock	Porosity	Permeability	Why
Well-sorted coarse-grained sandstone	high	high	coarse grains leave wide channels between the pores
Clay, mudstone	can be high	very low	the pores are very small, the channels are narrow, and the water is bound to the surface of the clay particles
Limestone with isolated vugs	high	low	the vugs (cavities) hardly connect with each other
Fractured tight rock	low	can be high	fractures take a small volume but form a network that conducts well

A qualitative comparison; the actual values depend on the rock and the conditions.

Darcy's law

Darcy's law gives a quantitative description of permeability. Take a rock sample of length $L$ and cross-section area $A$ , completely filled with one fluid, and pump fluid through it so that the pressure at the inlet is $p_1$ and at the outlet $p_2$ . The experiment shows that for slow flow the flow rate $q$ is proportional to the pressure difference and to the area, and inversely proportional to the viscosity of the fluid and the length of the sample.

SchematicScheme of Darcy's experiment: a rock sample of length

L

and cross-section

A

, the pressures at the inlet and the outlet, the flow rate

q

Darcy's law for linear flow

q = \frac{k A}{\mu}\cdot\frac{p_1 - p_2}{L}

where:

$q$ volumetric flow rate of the fluid at the conditions of the test (in the reservoir, at reservoir conditions), m³/s
$k$ permeability of the rock, m²
$A$ area of the cross-section perpendicular to the flow, m²
$\mu$ dynamic viscosity of the fluid, Pa·s
$p_1,\ p_2$ pressure at the inlet and at the outlet, Pa
$L$ length of the flow path, m

Units: SI (any consistent system of units)

Holds when

Steady-state flow: the pressures and the flow rate do not change with time.
Laminar (slow) flow, with inertial forces negligibly small.
The pores are filled with one fluid (single-phase flow).
The fluid is incompressible and does not react with the rock.
The flow is horizontal. For inclined flow, instead of the pressure difference we take the potential difference, which allows for the weight of the column of fluid.

Permeability is the coefficient of proportionality between the “driving force” (the pressure gradient, that is, how fast the pressure changes along the path) and the flow, with the effect of the fluid removed. That is why $k$ is a property of the rock, not of the fluid: single-phase flow of water or oil through the same rock gives the same value.

In other unit systems (3)

Darcy units

q\,[\text{cm}^3/\text{s}] = \frac{k\,[\text{D}]\;A\,[\text{cm}^2]\;\Delta p\,[\text{atm}]}{\mu\,[\text{cP}]\;L\,[\text{cm}]}

$k$ in darcies, $A$ in cm², $\Delta p$ in physical atmospheres, $\mu$ in centipoise (cP = mPa·s), $L$ in cm, $q$ in cm³/s.

There is no constant: this is how the unit “darcy” is defined.

Oilfield units

q\,[\text{bbl/d}] = 1.127\cdot 10^{-3}\,\frac{k\,[\text{mD}]\;A\,[\text{ft}^2]\;\Delta p\,[\text{psi}]}{\mu\,[\text{cP}]\;L\,[\text{ft}]}

$k$ in mD, $A$ in square feet, $\Delta p$ in psi, $\mu$ in cP, $L$ in feet, $q$ in barrels per day.

$1.127\cdot10^{-3}$ comes from converting the Darcy units: mD → D, ft² → cm², psi → atm, ft → cm, cm³/s → bbl/d.

Metric oilfield units

q\,[\text{m}^3/\text{d}] = 0.0853\,\frac{k\,[\text{mD}]\;A\,[\text{m}^2]\;\Delta p\,[\text{MPa}]}{\mu\,[\text{mPa·s}]\;L\,[\text{m}]}

$k$ in mD, $A$ in m², $\Delta p$ in MPa, $\mu$ in mPa·s, $L$ in m, $q$ in m³/d.

$0.0853 = 0.9869\cdot10^{-15}\cdot10^{6}/10^{-3}\cdot 86{,}400$ (the conversion of mD → m², MPa → Pa, mPa·s → Pa·s, and of seconds to days).

Units of permeability

In SI (the international system of units), permeability is measured in square metres: in essence it is the typical “cross-section area” of the channels that carry the flow. A rock with a permeability of 1 darcy (D) lets through 1 cm³/s of a fluid with a viscosity of 1 cP (centipoise; water is about 1 cP) across a sample with a cross-section of 1 cm² and a length of 1 cm at a pressure difference of 1 atm. In square metres this is a very small quantity, so in practice engineers use the darcy and the millidarcy (mD).

Conversion of permeability units

Unit: 1 D
In m²: $0.9869\cdot10^{-12}$
Note: ≈ 1 µm² (square micrometre)

Unit: 1 mD
In m²: $0.9869\cdot10^{-15}$
Note: the main oilfield unit

Unit: 1 µD
In m²: $0.9869\cdot10^{-18}$
Note: tight gas reservoirs, seals

Unit: 1 nD
In m²: $0.9869\cdot10^{-21}$
Note: shales, mudstones

Unit	In m²	Note
1 D	$0.9869\cdot10^{-12}$	≈ 1 µm² (square micrometre)
1 mD	$0.9869\cdot10^{-15}$	the main oilfield unit
1 µD	$0.9869\cdot10^{-18}$	tight gas reservoirs, seals
1 nD	$0.9869\cdot10^{-21}$	shales, mudstones

In Russian-language practice µm² and 10⁻³ µm² are also used; 1 mD ≈ 10⁻³ µm² with an error of about 1.3%.

Approximate ranges of permeability

Rocks: Seals (rocks that stop oil and gas escaping): clays, salts, tight mudstones
Order of permeability: nanodarcies to microdarcies

Rocks: Tight (low-permeability) gas reservoirs
Order of permeability: less than 0.1 mD

Rocks: Typical clastic (made of grains of broken rock) and carbonate (limestone and dolomite) reservoirs
Order of permeability: from a few to hundreds of mD

Rocks: High-permeability sands and fractured rocks
Order of permeability: from hundreds of mD to several darcies

Rocks	Order of permeability
Seals (rocks that stop oil and gas escaping): clays, salts, tight mudstones	nanodarcies to microdarcies
Tight (low-permeability) gas reservoirs	less than 0.1 mD
Typical clastic (made of grains of broken rock) and carbonate (limestone and dolomite) reservoirs	from a few to hundreds of mD
High-permeability sands and fractured rocks	from hundreds of mD to several darcies

Illustrative orders of magnitude for orientation, not a classification: the class boundaries differ in different regulations.

How permeability is measured

Permeability cannot be seen directly: it is calculated from the flow and the pressure difference. Different methods cover very different volumes of rock, so they give different but equally valid numbers.

On core, the most common method is the steady-state method. We pump gas or liquid at a constant rate through a sample in a core holder (a cell that grips the sample), under a set confining pressure (a pressure that squeezes the sample from all sides), and measure the pressure difference. For very tight rocks the unsteady-state methods are more convenient. They find permeability, down to fractions of a microdarcy, from the rate at which the pressure falls or evens out in a “chamber and sample” system.

Sources of permeability data

Method: Test on core (a sample)
Volume of rock: a few cubic centimetres
What it gives: permeability of the sample in the chosen direction
Limitations: small volume; test conditions differ from reservoir conditions; fractures and large cavities often do not get into the sample

Method: Wireline formation tester (a tool on a cable that presses a probe against the wall and draws a little fluid)
Volume of rock: the first metres around the probe
What it gives: the local mobility of the fluid $k/\mu$
Limitations: small radius of investigation, effect of the invaded zone

Method: Well testing
Volume of rock: hundreds of metres around the well
What it gives: the mean effective permeability for the current fluid, and the skin factor (a number that shows how much the zone around the well helps or hinders flow)
Limitations: needs an interpretation model; the result is the product $kh$

Method: Well logs (estimates from correlations)
Volume of rock: along the whole wellbore
What it gives: a continuous but indirect estimate
Limitations: needs calibration with core

Method	Volume of rock	What it gives	Limitations
Test on core (a sample)	a few cubic centimetres	permeability of the sample in the chosen direction	small volume; test conditions differ from reservoir conditions; fractures and large cavities often do not get into the sample
Wireline formation tester (a tool on a cable that presses a probe against the wall and draws a little fluid)	the first metres around the probe	the local mobility of the fluid $k/\mu$	small radius of investigation, effect of the invaded zone
Well testing	hundreds of metres around the well	the mean effective permeability for the current fluid, and the skin factor (a number that shows how much the zone around the well helps or hinders flow)	needs an interpretation model; the result is the product $kh$
Well logs (estimates from correlations)	along the whole wellbore	a continuous but indirect estimate	needs calibration with core

Worked examplePermeability from a core test

Given (assumed data)

Sample length $L$: 5.0 cm
Sample diameter $d$: 3.8 cm
Brine viscosity $\mu$: 1.0 cP
Steady flow rate $q$: 0.50 cm³/s
Pressure difference $\Delta p$: 2.0 atm

Solution

Cross-section area: $A = \pi d^2/4 = \pi\cdot 3.8^2/4 = 11.34$ cm².
We find the permeability from Darcy's law in darcy units: $k = \dfrac{q\,\mu\,L}{A\,\Delta p} = \dfrac{0.50\cdot 1.0\cdot 5.0}{11.34\cdot 2.0} = 0.110$ D.
Converting: $k = 110$ mD $= 1.088\cdot 10^{-13}$ m².

Answer: The permeability of the sample to brine is about 110 mD.

What it means. This is the permeability of a good reservoir rock. It refers to a small sample and to the direction in which it was drilled (usually horizontal). For the reservoir as a whole it is compared with other samples and with the results of well tests. The data in this example are assumed for illustration.

Radial flow to a well

In a reservoir the fluid goes to a well not in a straight line but from all sides, in a converging flow. The area through which it passes, the side surface of a cylinder $2\pi r h$ , gets smaller as the well is approached, so the flow velocity (the rate divided by the area) and the pressure gradient are greatest at the well wall.

Steady-state radial flow (the Dupuit equation)

q = \frac{2\pi k h\,(p_e - p_w)}{\mu\,\ln(r_e/r_w)}

where:

$q$ rate at reservoir conditions; the surface rate is $q/B$ , where $B$ is the formation volume factor (the ratio of the volume of the oil in the reservoir to its volume at the surface), m³/s
$k$ permeability of the reservoir (horizontal), m²
$h$ net thickness of the reservoir, m
$p_e$ pressure at the outer boundary of the drainage area, Pa
$p_w$ pressure at the well wall (bottomhole), Pa
$r_e$ radius of the drainage area (the area that feeds the well), m
$r_w$ radius of the well, m
$\mu$ viscosity of the fluid at reservoir conditions, Pa·s

Units: SI

Holds when

Steady state: the pressure at the outer boundary $r_e$ is kept constant.
A uniform isotropic reservoir (the same in all directions) of constant thickness; the well penetrates the whole thickness.
Single-phase flow of a slightly compressible fluid (one whose volume hardly changes with pressure), laminar flow.
No damage to or improvement of the near-wellbore zone, the rock close to the well (the skin factor is zero).

The rate is proportional to the transmissibility $kh/\mu$ (permeability times thickness, divided by viscosity) and to the drawdown $p_e - p_w$ (the pressure difference between the reservoir and the well). The sizes of the drainage area and of the well enter only through a logarithm, so they have a weak effect.

In other unit systems (2)

Oilfield units

q\,[\text{STB/d}] = \frac{7.08\cdot10^{-3}\,k\,h\,(p_e - p_w)}{\mu\,B\,\ln(r_e/r_w)}

$k$ in mD, $h$ in ft, $p$ in psi, $\mu$ in cP, $B$ in rb/STB, $q$ in barrels of oil at standard conditions per day.

$7.08\cdot10^{-3} = 2\pi\cdot 1.127\cdot10^{-3}$ .

Metric oilfield units

q\,[\text{m}^3/\text{d}] = \frac{0.536\,k\,h\,(p_e - p_w)}{\mu\,B\,\ln(r_e/r_w)}

$k$ in mD, $h$ in m, $p$ in MPa, $\mu$ in mPa·s, $B$ in m³/m³, $q$ in m³/d at standard conditions.

$0.536 = 2\pi\cdot 0.0853$ .

CalculatedCalculated pressure distribution around the well for the data of the example below: the pressure grows with the distance following a logarithmic law.

Worked exampleWell rate and the distribution of the pressure difference

Given (assumed data)

Permeability $k$: 50 mD
Net thickness $h$: 12 m
Drawdown $p_e - p_w$: 5.0 MPa
Oil viscosity in the reservoir $\mu$: 1.5 mPa·s
Formation volume factor $B$: 1.2 m³/m³
Radii $r_e$ and $r_w$: 300 m and 0.1 m

Solution

Logarithm of the ratio of the radii: $\ln(r_e/r_w) = \ln 3000 = 8.006$ .
Rate at standard conditions by the metric oilfield form: $q = \dfrac{0.536\cdot 50\cdot 12\cdot 5.0}{1.5\cdot 1.2\cdot 8.006} = 112$ m³/d.
The share of the drawdown that is used within 10 m of the well: $\dfrac{\ln(10/0.1)}{\ln(300/0.1)} = 0.575$ , that is about 58%.

Answer: The well can give about 112 m³/d of oil, and 58% of the pressure difference is used up in the first 10 m around it.

What it means. The rate is an estimate for an ideal well without a skin factor. The main practical conclusion is in the second number. Because more than half of the drawdown is “spent” right at the well, damage to the near-wellbore zone by drilling mud changes the rate much more than a change of the permeability far out in the reservoir does. So does the opposite, cleaning the zone with acid or by hydraulic fracturing. The data in this example are assumed for illustration.

Anisotropy and averaging over layers

Permeability depends on direction. This is anisotropy. In layered sedimentary rocks the grains and thin beds are elongated along the bedding (the layers of the rock), so the horizontal permeability $k_h$ is usually greater than the vertical permeability $k_v$ . Thin clay layers can reduce the vertical permeability by orders of magnitude and hardly change the horizontal one.

When a reservoir consists of layers with different permeability, the way of averaging depends on how the fluid flows relative to the layers.

Mean permeability of layers working in parallel

\bar{k} = \frac{\sum_i k_i h_i}{\sum_i h_i}

where:

$k_i$ permeability of layer $i$ , mD
$h_i$ thickness of layer $i$ , m

Units: any, the same for all layers

Holds when

The flow goes along the layers, and the pressure difference is the same for all the layers.
There is no crossflow between the layers (fluid moving from one layer to another), or it does not change the picture.

The arithmetic mean weighted by thickness: the layers add their conductivities $k_i h_i$ , like pipes connected in parallel.

Mean permeability of layers working in series

\bar{k} = \frac{\sum_i L_i}{\sum_i L_i / k_i}

where:

$k_i$ permeability of layer $i$ , mD
$L_i$ length of the path in layer $i$ (for vertical flow, the thickness of the layer), m

Units: any, the same for all layers

Holds when

The flow crosses the layers one after another, and the flow rate through all the layers is the same.

The harmonic mean: the resistances $L_i/k_i$ add up, so the result is set by the worst layer.

Worked exampleHorizontal and vertical permeability of a three-layer reservoir

Given (assumed data)

Layer 1: $h = 2$ m, $k = 500$ mD
Layer 2: $h = 5$ m, $k = 50$ mD
Layer 3: $h = 3$ m, $k = 5$ mD

Solution

Along the layers (horizontal flow): $\bar k_h = \dfrac{500\cdot2 + 50\cdot5 + 5\cdot3}{2 + 5 + 3} = 126.5$ mD.
Across the layers (vertical flow): $\bar k_v = \dfrac{2 + 5 + 3}{2/500 + 5/50 + 3/5} = 14.2$ mD.
The ratio $\bar k_v/\bar k_h = 0.11$ .

Answer: The horizontal permeability is about 126.5 mD, and the vertical permeability is about 14.2 mD.

What it means. The horizontal conductivity comes almost entirely from the thin high-permeability layer, and the vertical one is limited by the tightest layer. This difference matters for coning (the cone-shaped rise of water or gas towards the well), for horizontal wells and for gravity drainage (oil draining downward under its own weight). The data in this example are assumed for illustration.

What permeability depends on

Grain size. Other things being equal, permeability grows roughly in proportion to the square of the grain size: the larger the grains, the wider the channels.
Sorting. In a poorly sorted rock, small grains fill the spaces between large ones and narrow the channels.
Cement and clay minerals. Cement is the mineral that glues the grains together. Calcite, quartz and clays deposited in the pores block the channels. Fibrous clays (for example, illite) reduce permeability especially sharply.
Compaction and effective stress. Effective stress is the part of the load that the rock grains carry. With depth, and as the reservoir pressure falls, the rock is compacted and the channels narrow. In the laboratory, we measure permeability under a load close to that in the reservoir.
Fractures. Even a few open fractures can increase the permeability of a reservoir many times over in one direction.
Fluid chemistry. Fresh or incompatible water can make clay particles swell and move, and then the permeability to water turns out noticeably lower than that to gas or oil.

Terms and symbols

Absolute permeability: Permeability when one fluid fills the pores completely; a property of the rock.
Effective permeability: Permeability to one phase when the pores hold several fluids; it is lower than the absolute permeability.
Relative permeability: The ratio of the effective permeability to a base value (usually the absolute permeability); a number from 0 to 1 that depends on saturation.
Transmissibility: The quantity $kh/\mu$ , which sets the rate of a well at a given drawdown.
Anisotropy: The dependence of permeability on direction; it is usually described by the ratio $k_v/k_h$ .
$k$ , mD: The symbol and the main unit of permeability; 1 mD ≈ $0.987\cdot10^{-15}$ m².

Common mistakes and misconceptions

Wrong

“Porosity is high, so the reservoir is good.”

In fact

Porosity tells you how much hydrocarbon is in place; permeability tells you whether it can be taken out. A clay with high porosity is practically impermeable.

Wrong

“Permeability to water and to gas must be the same.”

In fact

For single-phase flow of liquids, yes. Gas “slips” along the walls of the channels, so its permeability at low pressure comes out too high. The Klinkenberg correction fixes this.

Wrong

“The mean permeability of a reservoir is the arithmetic mean over the samples.”

In fact

The way of averaging depends on the direction of flow and on how the layers are arranged. For vertical flow you need the harmonic mean.

Wrong

“If the rate is half the calculated one, the permeability is half as large.”

In fact

More often the near-wellbore zone (the skin factor) is to blame: it takes most of the drawdown. Well tests separate the permeability from the skin.

In depth: advanced detail

Derivation of the radial flow formula

Write Darcy's law for a cylindrical surface of radius $r$ through which the whole flow $q$ passes: $q = \dfrac{k\,(2\pi r h)}{\mu}\,\dfrac{\dd p}{\dd r}$ . The sign is plus because the pressure grows with the distance from the well, while the flow is directed towards it.

Separate the variables, $\dd p = \dfrac{q\mu}{2\pi k h}\,\dfrac{\dd r}{r}$ , and integrate from the well wall ( $r_w$ , $p_w$ ) to the outer boundary ( $r_e$ , $p_e$ ): $p_e - p_w = \dfrac{q\mu}{2\pi k h}\ln\dfrac{r_e}{r_w}$ . This gives the Dupuit equation. The integral from $r_w$ to any $r$ gives the logarithmic pressure profile $p(r) = p_w + \dfrac{q\mu}{2\pi k h}\ln\dfrac{r}{r_w}$ , the one shown on the graph.

Gas permeability and the Klinkenberg correction

Gas is compressible, so along the sample its volumetric flow rate grows as the pressure falls, and Darcy's law for gas is written with the squares of the pressures. For steady flow of an ideal gas at constant temperature, the flow rate, converted to the pressure at which it is measured at the outlet, is:

Darcy's law for linear flow of gas

q_b = \frac{k A\,(p_1^2 - p_2^2)}{2\,\mu\,L\,p_b}

where:

$q_b$ volumetric flow rate of the gas at the measurement pressure $p_b$ , m³/s
$p_1,\ p_2$ absolute pressure at the inlet and the outlet of the sample, Pa
$p_b$ absolute pressure at which the flow rate is measured (usually close to atmospheric), Pa
$k,\ A,\ \mu,\ L$ permeability, area, gas viscosity and length of the sample, as in Darcy's law, m², m², Pa·s, m

Units: SI (any consistent system)

Holds when

Steady isothermal flow of an ideal gas.
Laminar flow without slippage (the Klinkenberg correction is added separately).
All pressures are absolute.

We use this formula to calculate the gas permeability of a sample from the measured $q_b$ , $p_1$ and $p_2$ . The mean pressure in the sample $\bar p = (p_1 + p_2)/2$ is needed for the Klinkenberg correction below.

When the mean free path of the gas molecules (the average distance a molecule travels between collisions) is comparable with the size of the pore channels, the gas does not “stick” to the walls but slips along them. So the gas permeability measured at low pressure comes out higher than the liquid permeability. The effect is stronger in tight rocks and at a low mean pressure in the test.

Klinkenberg correction

k_g = k_\infty\left(1 + \frac{b}{\bar p}\right)

where:

$k_g$ permeability measured with gas, mD
$k_\infty$ equivalent liquid permeability (the limit at infinite pressure), mD
$\bar p$ mean pressure in the sample, $(p_1 + p_2)/2$ , absolute, atm or MPa
$b$ slip coefficient, which depends on the gas and on the size of the channels, the same units of pressure

Units: any consistent units

Holds when

Single-phase gas flow at low velocities.
The pressure is absolute, not gauge.

Plot $k_g$ against $1/\bar p$ from tests at several pressures, and the points lie on a straight line. Its intercept on the vertical axis (at $1/\bar p \to 0$ ) is the liquid permeability $k_\infty$ .

Worked exampleExtrapolation from two measurements

Given (assumed data)

Test 1: $\bar p = 1.5$ atm, $k_g = 12.6$ mD
Test 2: $\bar p = 4.0$ atm, $k_g = 10.8$ mD

Solution

Values on the horizontal axis: $1/\bar p_1 = 0.667$ atm⁻¹, $1/\bar p_2 = 0.250$ atm⁻¹.
Slope of the line: $s = \dfrac{12.6 - 10.8}{0.667 - 0.250} = 4.32$ mD·atm.
Intercept on the axis: $k_\infty = 10.8 - 4.32\cdot0.250 = 9.7$ mD; the coefficient is $b = s/k_\infty = 0.44$ atm.

Answer: The liquid permeability of the sample is about 9.7 mD, which is 10% lower than the gas permeability at 4 atm.

What it means. In practice we take more than two measurements and check that the points really lie on a straight line. The data in this example are assumed for illustration.

When Darcy's law stops working

At high flow velocities, above all near gas wells, inertial losses add to the viscous resistance: the fluid is constantly speeding up and slowing down in the winding channels. The Forchheimer equation describes these losses.

The Forchheimer equation

-\frac{\dd p}{\dd x} = \frac{\mu}{k}\,v + \beta\,\rho\,v^2

where:

$v$ flow velocity (the flow rate divided by the area), m/s
$\beta$ inertial resistance coefficient of the rock, m⁻¹
$\rho$ density of the fluid, kg/m³
$\mu,\ k$ viscosity and permeability, as in Darcy's law, Pa·s, m²

Units: SI

Holds when

Single-phase flow; $\beta$ is found from a test or from correlations with permeability and porosity.

At small $v$ the second term disappears and Darcy's law remains. In a well the additional losses show up as a skin factor that grows with the rate ( $Dq$ ). We separate it out by tests at several flow rates.

The link between permeability, porosity and grain size

The Kozeny–Carman model represents the pore space as a bundle of winding capillaries and links permeability with porosity and specific surface (the surface area of the grains per unit volume). For a packing of identical spherical grains of diameter $d$ it is often written like this:

The Kozeny–Carman equation (spherical grains)

k = \frac{d^2\,\phi^3}{180\,(1 - \phi)^2}

where:

$d$ diameter of the grains, m
$\phi$ porosity (fraction), —
$k$ permeability, m²

Units: SI ( $k$ comes out in the same square units as $d^2$ )

Holds when

An unconsolidated (not cemented) packing of spherical grains of one size.
The coefficient 180 allows empirically for the winding and shape of the channels; in the Blake–Kozeny variant (and in the Ergun equation) 150 is used.

The formula explains the main patterns: the strong dependence on grain size ( $k \propto d^2$ ) and on porosity. For real cemented rocks it gives only the order of magnitude, so in practice we build the “porosity–permeability” relations from core for each reservoir.

Estimating permeability without core: correlations and well logs

A well log does not measure permeability directly, so we estimate it from relations with what we can measure. One of the best known is the dependence on porosity and irreducible water saturation (the water that cannot be pushed out of the pores): the finer the grains and the narrower the channels, the more water is held by capillary forces, and the lower the permeability.

The Timur correlation (permeability from porosity and irreducible water saturation)

k \approx 10^{4}\,\frac{\phi^{4.5}}{S_{wi}^{2}}

where:

$k$ permeability, mD
$\phi$ porosity, fraction
$S_{wi}$ irreducible (residual) water saturation, fraction

Units: empirical: $k$ in mD, $\phi$ and $S_{wi}$ as fractions

Holds when

An empirical relation obtained for sandstones; the constants are fitted to the core of a particular reservoir.
$S_{wi}$ must really be irreducible (the reservoir is above the transition zone).

The formula explains why permeability can be estimated from nuclear magnetic resonance logging (a well-logging method that measures the fluids in the pores) data on bound water. Without calibration to core, the error easily reaches an order of magnitude (a factor of 10).

Permeability of a fracture and of a fractured rock

Flow between two smooth parallel walls separated by a gap of width $w$ is described by an exact solution, and comparison with Darcy's law gives the permeability of the fracture itself, $k_f = w^2/12$ . The permeability grows as the square of the aperture (the width of the gap), and the flow rate through the fracture as the cube, so even thin fractures conduct a great deal.

Worked exampleFractures with almost no porosity but with noticeable permeability

Given (assumed data)

Fracture aperture $w$: 0.1 mm
Spacing between parallel fractures $s$: 1 m
Matrix permeability (of the solid rock between the fractures): negligibly small

Solution

Permeability of the fracture itself: $k_f = w^2/12 = (10^{-4})^2/12 = 8.333\cdot 10^{-10}$ m², that is about 844 D.
The fractures take a share of the volume $w/s = 0.0001$ , that is 0.01%: the porosity practically does not change.
Permeability of the rock along the fractures: $k \approx k_f\,w/s = 84$ mD.

Answer: A set of fractures with an aperture of 0.1 mm at every metre gives the rock about 84 mD along the fractures.

What it means. A rock with almost zero matrix permeability behaves like a good reservoir along the fractures, and stays almost impermeable across them. The aperture of fractures decreases under load, so their conductivity is sensitive to the fall of the reservoir pressure. The data in this example are assumed for illustration.

Connections to other topics

Porosity — the capacity of the pore space and the ways of finding it.
Relative permeability — how permeability is shared between oil, gas and water when they flow together.
Skin factor — why the rate is often decided not by the reservoir but by the near-wellbore zone.
Pressure build-up analysis — how to find the permeability of the reservoir from the pressure build-up in a well.

Check yourself

1
A clay sample has a porosity of 25%, and a sandstone sample has a porosity of 18%. Which of them is probably more permeable?
1. A.The sandstone
2. B.The clay: it has higher porosity
3. C.The same: permeability is determined by porosity
4. D.Nothing can be said
Show the answer
Correct answer: A. The sandstone
Permeability is determined by the size and connectedness of the pore channels, not by the share of voids. In clay the pores are very small, so with high porosity it is almost impermeable. A sandstone with lower porosity but wide channels passes fluid much better.
2
Water was pumped through a sample, and then oil with twice the viscosity at the same pressure difference. How did the flow rate and the calculated permeability change?
1. A.Nothing changed
2. B.The flow rate is the same, the permeability is half
3. C.The flow rate is half, the permeability is the same
4. D.Both fell by half
Show the answer
Correct answer: C. The flow rate is half, the permeability is the same
By Darcy's law the flow rate is inversely proportional to the viscosity, and permeability is a property of the rock from which the viscosity has already been removed. For single-phase flow of both fluids $k$ comes out the same.
3
A reservoir consists of two layers of equal thickness: 200 mD and 2 mD. Which mean permeability governs the flow along the layers, and which across them?
Show the answer
Along the layers it is the arithmetic mean: $(200 + 2)/2 = 101$ mD. Across the layers it is the harmonic mean: $2/(1/200 + 1/2) \approx 3.96$ mD. Vertical flow is almost entirely limited by the tight layer.
4
Why can damage to the near-wellbore zone with a radius of only a few metres noticeably reduce the rate of a well?
Show the answer
In radial flow the pressure changes with the logarithm of the distance, so a large part of the drawdown is spent near the well (in the lesson's example, more than half within 10 m). A reduction of permeability exactly there greatly increases the resistance of the whole system.
5
The gas permeability of a sample, measured at a mean pressure of 2 atm, is 5 mD. Can this value be used at once to calculate oil inflow?
1. A.No: gas permeability is always lower than liquid permeability
2. B.Yes, permeability does not depend on the fluid
3. C.No: the Klinkenberg correction to liquid permeability is needed first
Show the answer
Correct answer: C. No: the Klinkenberg correction to liquid permeability is needed first
Because of gas slippage at low pressure, the measured gas permeability is too high, especially in tight rocks. We get the equivalent liquid permeability by extrapolating $k_g$ (extending the straight line) to $1/\bar p \to 0$ .

Updated: 22 September 2026