Reservoir heterogeneity and permeability averaging

Quick refresher

Flow is proportional to permeability, not to its logarithm. A layer of high permeability can carry most of the flow, even though it looks modest on a logarithmic plot.
Layers that work in parallel — flow along the layers, including radial inflow to a well — are averaged arithmetically, with the thicknesses as weights. Layers that work in series (the flow has to cross one after another) are averaged harmonically. For ring-shaped zones around a well, the weight is the logarithm of the ratio of the radii. The zone nearest the well weighs the most.
The equivalent permeability of a random mosaic lies between the harmonic and the arithmetic mean. For equidimensional patches (about as long as they are wide) it is close to the geometric mean. For log-normal permeability (permeability whose logarithm follows the bell-shaped normal curve), $\bar k_{\text{A}} = k_Ge^{\sigma^2/2}$ and $\bar k_{\text{H}} = k_Ge^{-\sigma^2/2}$ . Their ratio $e^{\sigma^2}$ grows quickly with the spread.
The ratio $k_v/k_h$ of a unit (a group of neighbouring layers) is set not by core samples but by the thin layers in it. Continuous shale layers cut the vertical permeability by orders of magnitude, and broken ones cut it by much less. The number $R_L$ shows whether the layers communicate over the distance between wells.
The Dykstra–Parsons coefficient $V_{DP}$ and the Lorenz coefficient $L_c$ measure the spread of permeability, but not where the permeability sits. For log-normal permeability both can be written with $\sigma_{\ln k}$ .
The correlation length is the distance over which values are still similar to each other. If it is longer than the distance between wells, the reservoir behaves as a layered one. If it is much shorter, the reservoir behaves as a mosaic with one equivalent permeability.

Key relationships

\bar k = \dfrac{\ln(r_e/r_w)}{\sum_j \ln(r_j/r_{j-1})/k_j}

ring-shaped zones around a well, steady-state flow

k_\omega = k_G\,e^{\omega\sigma^2/2}

power means of log-normal permeability: $\omega = 1$ , $0$ , $-1$ give the arithmetic, geometric and harmonic mean

V_{DP} = 1 - e^{-\sigma_{\ln k}}

Dykstra–Parsons coefficient of log-normal permeability

L_c = 2\,\Phi\!\left(\sigma_{\ln k}/\sqrt{2}\right) - 1

Lorenz coefficient of log-normal permeability

R_L = \dfrac{L}{H}\sqrt{\dfrac{k_v}{k_h}}

effective aspect ratio (length to thickness)

Common pitfalls

Judging the permeability contrast from a logarithmic plot. Flow follows the permeability itself.
Averaging the ring zones around a well by area. The zone nearest the well weighs far more.
Assuming that equal $V_{DP}$ means equal behaviour of two reservoirs. What matters is whether the layers continue from well to well.
Taking the $k_v/k_h$ of a reservoir from a couple of core samples. The vertical permeability of a unit is set by thin layers that the samples do not contain.
Trusting the second decimal of a $V_{DP}$ calculated from about ten samples.
Confusing the equivalent permeability of a heterogeneous reservoir with the effective (phase) permeability in multiphase flow.

After this lesson you can

describe heterogeneity at scales from the pores to the whole accumulation, and tell layered, mosaic and correlated reservoir structure apart
choose the average permeability for linear and for radial flow from the arrangement of the layers and zones, and estimate how far the different averages differ when permeability is log-normally distributed
explain what the ratio $k_v/k_h$ (vertical to horizontal permeability) of a unit of layers depends on, and use it to judge whether the layers communicate (pass fluid to each other) over the distance between wells
calculate the Dykstra–Parsons and Lorenz coefficients for a layered section, link them to the spread of the logarithm of permeability, and say what they do not show
explain what the correlation length is and how comparing it with the distance between wells decides the reservoir model

What heterogeneity is and at which scales we see it

There is no such thing as a uniform reservoir. Sediment settles in layers, and each layer has its own grain size and sorting (how alike the grains are in size). Later, cement partly fills the pores, clay minerals grow in the flow channels, and tectonic movements add fractures (what permeability depends on). The porosity of one reservoir (the share of pore space) usually changes by a factor of 1.5 to 2. Permeability changes by orders of magnitude (factors of ten). So how a reservoir behaves in production and injection is decided first of all by the unevenness of its permeability.

Scales of heterogeneity

Scale: Pores and grains
Size: micrometres to millimetres
What varies: size of the flow channels, clay minerals, cement
How we see it: thin sections (rock slices under a microscope), tomography (X-ray scanning), mercury injection (MICP: mercury is forced into the pores)

Scale: Laminae (very thin layers) and layers
Size: centimetres to metres
What varies: bedding, cross-bedding (thin layers lying at an angle), thin shale layers
How we see it: core, probe permeameter (measures permeability at a point of the core), well logs, borehole image logs (pictures of the borehole wall)

Scale: Units and sandstone bodies
Size: metres to hundreds of metres
What varies: pinch-outs (layers that thin out to nothing), channels, lenses
How we see it: well correlation (matching the same layers from well to well), well tests

Scale: Accumulation
Size: hundreds of metres to kilometres
What varies: blocks, faults, facies zones (zones laid down in different environments)
How we see it: seismic surveys, production history

Scale	Size	What varies	How we see it
Pores and grains	micrometres to millimetres	size of the flow channels, clay minerals, cement	thin sections (rock slices under a microscope), tomography (X-ray scanning), mercury injection (MICP: mercury is forced into the pores)
Laminae (very thin layers) and layers	centimetres to metres	bedding, cross-bedding (thin layers lying at an angle), thin shale layers	core, probe permeameter (measures permeability at a point of the core), well logs, borehole image logs (pictures of the borehole wall)
Units and sandstone bodies	metres to hundreds of metres	pinch-outs (layers that thin out to nothing), channels, lenses	well correlation (matching the same layers from well to well), well tests
Accumulation	hundreds of metres to kilometres	blocks, faults, facies zones (zones laid down in different environments)	seismic surveys, production history

The borders between scales are approximate. A property measured at one scale cannot be carried over to another without re-calculation (core, logs and tests).

To describe heterogeneity, we answer three questions. How much does permeability change? This is the distribution and the measures of heterogeneity. How are the patches of different permeability arranged relative to the flow: as continuous layers, as lenses or as a random mosaic? The correlation length matters here. In which direction does it change? This is anisotropy, above all the ratio of vertical to horizontal permeability. Two reservoirs with the same set of permeabilities can behave quite differently if they differ in the second and the third.

Worked exampleA thin high-permeability layer

Given (assumed data)

Unit: 24 m thick: a layer of 1.2 m with a permeability of 1500 mD, and the other 22.8 m with about 60 mD
Porosity: almost the same throughout the unit

Solution

The flow capacity of the layer (permeability times thickness) is $kh = 1500\cdot1.2 = 1800$ mD·m. For the rest of the unit it is $60\cdot22.8 = 1368$ mD·m.
The share of the flow along the unit that goes through the layer is $1800/(1800 + 1368) = 0.568$ . That is 57% of the flow through 5% of the thickness.
The average permeability of the unit along the layers is $(1800 + 1368)/24 = 132$ mD. This is 2.2 times higher than in the main rock.
On a logarithmic scale the layer differs from the rest of the unit by $\log_{10} 1500 - \log_{10} 60 = 1.40$ orders of magnitude (powers of ten). With the same porosity and the same pressure drop, the fluid moves in it $1500/60 = 25$ times faster.

Answer: A layer with 5% of the thickness carries 57% of the flow.

What it means. On a logarithmic plot of permeability against depth, such a layer looks like a noticeable but not exceptional peak, while the flow follows the permeability itself. For the rate of a well the layer is a gift: it gives more than half of the inflow. For waterflooding it is a threat. If the layers do not communicate, the water will reach the producing wells along the layer many times faster than through the main rock, and after that it will keep circulating along the path it has already flushed (Why water is injected: pressure maintenance and oil displacement; how such a reservoir is flooded is explained in the lesson Displacement in heterogeneous reservoirs). Thin high-permeability layers are easy to miss. Loose rock is washed out when the core is cut, and a well log averages the layer with the rocks next to it. The data in this example are assumed for illustration.

Layers and zones: linear and radial flow

The rules for layers are already familiar. Layers that work in parallel add up their flow capacities $k_ih_i$ , and the arithmetic mean, weighted by thickness, describes the flow along them. Layers that work in series add up their resistances $L_i/k_i$ , and the harmonic mean describes the flow across them (averaging over layers, the three averages for permeability). Here we look at what changes when the flow towards the well is radial (it converges on the well from all sides).

Layers in radial inflow. Suppose a well penetrates several layers, and all of them have the same pressure at the bottom of the well and the same pressure at the outer boundary. Each layer gives an inflow proportional to its flow capacity $k_ih_i$ . The sum of these inflows equals the inflow of a uniform reservoir with permeability $\sum k_ih_i/\sum h_i$ — the same arithmetic mean as in linear flow. If the layers communicate vertically, this changes nothing. The pressure distribution in them is the same, so there is nothing to flow between them. Crossflow (flow from one layer into another) appears when the pressures at the layer boundaries differ. It also appears when the layers have different skin factors (a skin factor is a number that shows how much the zone right around the well helps or hinders flow). This happens, for example, when the layers are depleted by different amounts. A transient well test (a test in which the pressure is still changing with time) in a layered reservoir also finds the total flow capacity $\sum k_ih_i$ (Pressure build-up analysis).

Ring-shaped zones around the well work in series: all the inflow passes through each of them. But the area the flow passes through grows with the distance from the well. So the resistance of a zone is proportional not to its width but to the logarithm of the ratio of its radii.

Average permeability of ring zones in radial inflow

\bar k = \frac{\ln(r_e/r_w)}{\sum_{j=1}^{n} \ln(r_j/r_{j-1})/k_j}

where:

$\bar k$ permeability of a uniform reservoir that gives the same inflow, mD
$k_j$ permeability of zone $j$ , the ring between the radii $r_{j-1}$ and $r_j$ , mD
$r_j$ outer radius of zone $j$ ; $r_0 = r_w$ , $r_n = r_e$ , m
$r_w,\ r_e$ radius of the well and radius of the outer boundary of the drainage area (the part of the reservoir that feeds the well), m

Units: permeabilities in any identical units; radii in identical units (only the ratios enter)

Holds when

The flow is steady-state and single-phase. The zones are coaxial cylinders (they share one axis, the well) across the whole thickness of the reservoir, and each zone is uniform.
The pressure at the outer boundary is constant.
With one inner zone, the formula is equivalent to the Hawkins skin factor: $\bar k = k\ln(r_e/r_w)/[\ln(r_e/r_w) + S]$ (Skin factor).

Rings with the same ratio of outer to inner radius weigh the same: the zone from 0.1 to 1 m matters for the inflow as much as the zone from 100 to 1000 m. So the permeability near the well controls the inflow much more strongly than its share of the reservoir volume would suggest.

Worked exampleGood rock near the well or far from it

Given (assumed data)

Well: radius 0.1 m; the outer boundary of the drainage area is at 500 m
Case A: out to 100 m from the well the rock is 200 mD, beyond that 20 mD
Case B: the other way round: 20 mD out to 100 m, 200 mD beyond

Solution

Logarithms: $\ln(100/0.1) = 6.908$ , $\ln(500/100) = 1.609$ , $\ln(500/0.1) = 8.517$ .
Case A: $\bar k = 8.517/(6.908/200 + 1.609/20) = 8.517/0.1150 = 74.1$ mD.
Case B: $\bar k = 8.517/(6.908/20 + 1.609/200) = 8.517/0.3534 = 24.1$ mD. The inflow in Case A is 3.1 times larger.
The inner zone takes only $(100/500)^2 = 0.04$ of the area. An average by area would give $0.04\cdot200 + 0.96\cdot20 = 27.2$ mD for Case A and $0.04\cdot20 + 0.96\cdot200 = 192.8$ mD for Case B. That is a completely different picture.

Answer: With the same volumes of rock, their arrangement decides the inflow: 74.1 mD if the good rock is near the well, and 24.1 mD if it begins beyond 100 m.

What it means. Near the well the flow converges, so the flow velocity and the pressure gradient (how fast pressure changes with distance) are highest there. The near zone therefore “weighs” far more than the far zone. In Case B almost the whole pressure drop — 98% — is spent in the tight zone of 100 m radius (see the figure). This has practical consequences. A well test sees the permeability of the near zone at the start of the record. If the rock near the well is worse than the rock farther out, a treatment of the near-wellbore zone (well stimulation) or hydraulic fracturing helps a lot. Hydraulic fracturing means cracking the rock with fluid under high pressure to make a path for flow. And we cannot carry the permeability found in one well over to the whole accumulation without knowing where that well is placed. The data in this example are assumed for illustration.

CalculatedThe share of the drawdown (the whole pressure drop from the outer boundary to the well) that is spent between the wall of the well and the distance

r

: Cases A and B from the example and a uniform reservoir (steady-state flow,

r_w

= 0.1 m,

r_e

= 500 m; the distance axis is logarithmic).

Random mosaic and log-normal permeability

Suppose there are no continuous layers. Instead, patches of different permeability are mixed at random, and they are small compared with the distance between wells. The flow then picks its way through the better patches, but it cannot avoid the poor ones completely. The equivalent permeability of such a mosaic lies between the harmonic and the arithmetic mean. These are strict limits for any arrangement of the patches. If the patches are not stretched in any direction, the equivalent permeability is close to the geometric mean.

For log-normally distributed permeability (permeability whose natural logarithm follows the normal, bell-shaped curve), all the averages can be written with the geometric mean $k_G$ and the standard deviation (the usual measure of spread) of the natural logarithm, $\sigma = \sigma_{\ln k}$ .

Means of log-normally distributed permeability

\bar k_{\text{A}} = k_G\,e^{\sigma^2/2},\qquad \bar k_{\text{H}} = k_G\,e^{-\sigma^2/2},\qquad \frac{\bar k_{\text{A}}}{\bar k_{\text{H}}} = e^{\sigma^2}

where:

$\bar k_{\text{A}},\ \bar k_{\text{H}}$ arithmetic and harmonic means, mD
$k_G$ geometric mean, which is also the median (the middle value) of the log-normal distribution, mD
$\sigma$ standard deviation of the natural logarithm of permeability, —

Units: permeabilities in any identical units; $\sigma$ is dimensionless

Holds when

Permeability is log-normally distributed. The means refer to the whole population (a large sample, with volumes as weights).
From about ten samples the formulas give only a rough guide, because small samples represent the tails of the distribution poorly.

The ratio of the two extreme means grows very fast with the spread: at $\sigma = 1$ it is about 2.7, at 1.5 about 9.5, and at 2 about 55. The more uneven the reservoir, the more important it is to know how the patches of different permeability are arranged in it.

Where the equivalent permeability falls between the limits depends on the dimension of the flow. In a chain of patches (one-dimensional flow), the harmonic mean works. For two-dimensional flow in an isotropic random medium (one with the same properties in all directions) with log-normal permeability, the equivalent permeability is exactly equal to the geometric mean. In a three-dimensional medium the flow has more ways round the poor patches, so the equivalent permeability is higher — about $k_Ge^{\sigma^2/6}$ . This is an estimate, and we check it by calculation on a model.

Equivalent permeability of a log-normal rock for different arrangements

Arrangement relative to the flow: Continuous layers, flow along them
Equivalent permeability: $k_Ge^{\sigma^2/2}$ — the arithmetic mean
Exponent $\omega$: 1

Arrangement relative to the flow: Random mosaic, three-dimensional flow
Equivalent permeability: about $k_Ge^{\sigma^2/6}$
Exponent $\omega$: about 1/3

Arrangement relative to the flow: Random mosaic, two-dimensional flow
Equivalent permeability: $k_G$ — the geometric mean
Exponent $\omega$: 0

Arrangement relative to the flow: Layers across the flow, a chain of patches
Equivalent permeability: $k_Ge^{-\sigma^2/2}$ — the harmonic mean
Exponent $\omega$: −1

Arrangement relative to the flow	Equivalent permeability	Exponent $\omega$
Continuous layers, flow along them	$k_Ge^{\sigma^2/2}$ — the arithmetic mean	1
Random mosaic, three-dimensional flow	about $k_Ge^{\sigma^2/6}$	about 1/3
Random mosaic, two-dimensional flow	$k_G$ — the geometric mean	0
Layers across the flow, a chain of patches	$k_Ge^{-\sigma^2/2}$ — the harmonic mean	−1

$\omega$ is the exponent of the power mean that equals the equivalent permeability (see the advanced section). For lenses stretched along the flow it is closer to 1, and across the flow closer to −1.

Worked exampleOne rock, three arrangements

Given (assumed data)

Permeability: log-normal; median (geometric mean) $k_G$ = 40 mD, $V_{DP}$ = 0.70 (the Dykstra–Parsons coefficient, explained below)

Solution

Spread of the logarithm: $\sigma = -\ln(1 - 0.70) = -\ln 0.30 = 1.204$ ; $\sigma^2 = 1.450$ .
Continuous layers, flow along them: $40\cdot e^{1.450/2} = 40\cdot e^{0.725} = 82.6$ mD.
The same layers, flow across them: $40\cdot e^{-0.725} = 19.4$ mD.
Random mosaic: 40 mD in two-dimensional flow, and about $40\cdot e^{1.450/6} = 40\cdot e^{0.242} = 51.0$ mD in three-dimensional flow.

Answer: From 19.4 to 82.6 mD — a factor of 4.3 — depending on how the same patches of rock are arranged.

What it means. The permeability distribution is the same, but the equivalent permeability differs by a factor of four. To calculate the inflow, it is not enough to know the histogram of core permeability. We must also know whether the layers continue across the area, and in which direction the flow goes. Comparing the core averages with the permeability from a well test shows which picture is closer to reality (core, logs and tests). The data in this example are assumed for illustration.

Anisotropy: vertical and horizontal permeability

In sedimentary rocks the vertical permeability $k_v$ is usually lower than the horizontal permeability $k_h$ (permeability anisotropy). For a single sample of clean sandstone, the ratio $k_v/k_h$ is most often between a few tenths and one. Fine laminae (very thin layers) and flattened grains lower it. For a unit several metres thick, the ratio is set by the shale and tight thin layers, and it can be as low as hundredths or thousandths. So $k_v/k_h$ is a property of the scale, not of the rock.

How much the thin layers lower $k_v$ depends on how far they extend. A continuous thin layer works in series with the sandstone, so the vertical permeability of the unit is the harmonic mean. A layer only a few centimetres thick can make it hundreds of times smaller (the example with the missed thin layer). The flow goes round broken thin layers, which makes its path longer. The more shale the unit holds, and the longer the thin layers are compared with the gaps between them, the more $k_v$ falls. The harmonic mean of the samples is a lower limit if all the thin layers got into the samples. If thin layers were missed, it is far too high. We estimate $k_v/k_h$ from the following data.

Samples drilled vertically and horizontally, and whole (full-diameter) core. This is the scale of centimetres and decimetres (Core analysis).
A formation tester with several probes at different heights along the wellbore: the pressure response at a neighbouring probe gives $k_v$ of an interval of about a metre.
Well tests in wells that penetrate the reservoir only partly, where the flow near the well is spherical (it converges on one point), and in horizontal wells (Partial penetration and well deviation, Horizontal wells).
Production history: the breakthrough time of water and gas cones (water or gas drawn towards the well from below or above) and the growth of the gas–oil ratio (Water and gas coning).

The ratio $k_v/k_h$ affects the critical (water-free) rates and the breakthrough time of cones, the productivity of horizontal wells, and the pseudo-skin of partial penetration (the extra skin that appears when the well opens only part of the pay thickness). It also affects the speed at which gravity separates the phases, and whether the layers communicate during waterflooding. We can check the last point with a dimensionless number (a number without units).

Effective aspect ratio (length to thickness)

R_L = \frac{L}{H}\sqrt{\frac{k_v}{k_h}} = \frac{L}{I\,H}

where:

$R_L$ effective aspect ratio — the ratio of length to thickness, corrected for anisotropy, —
$L$ distance along the reservoir, for example between an injection well and a producing well, m
$H$ thickness of the unit, m
$k_v,\ k_h$ vertical and horizontal permeability of the unit, mD
$I$ anisotropy index $\sqrt{k_h/k_v}$ , as in the lesson Horizontal wells, —

Units: any identical units of length and of permeability

Holds when

$k_v$ and $k_h$ are the equivalent permeabilities of the unit in the two directions.
This is an estimate of the order of magnitude: the number compares scales and is not an exact criterion.

If we stretch the vertical coordinate by a factor $I$ , the anisotropic reservoir becomes an isotropic one of thickness $IH$ . Then $R_L^2$ is the time for pressure to even out along the reservoir over the distance $L$ , divided by the time for it to even out across the thickness. When $R_L$ is much larger than one, the fluids have time to redistribute vertically while they move between the wells, so the layers work as communicating layers. When $R_L$ is clearly smaller than one, the crossflow between the layers during that time is small, and the layers work almost independently.

Worked exampleDo the layers communicate between the wells?

Given (assumed data)

Unit: 20 m thick; the distance between the injection well and the producing well is 400 m
Case 1: broken shale layers: $k_v/k_h$ = 0.05
Case 2: one continuous shale layer in the middle: $k_v/k_h$ = 0.001

Solution

Case 1: $I = \sqrt{1/0.05} = 4.472$ ; $R_L = 400/(4.472\cdot20) = 4.47$ . The “stretched” thickness is $IH = 89$ m: the unit behaves as an isotropic one about 89 m thick over a length of 400 m.
Case 2: $I = \sqrt{1/0.001} = 31.62$ ; $R_L = 400/(31.62\cdot20) = 0.63$ ; $IH = 632$ m, which is more than the distance between the wells.

Answer:

R_L

is about 4.47 in Case 1 and about 0.63 in Case 2.

What it means. In Case 1 the pressure evens out across the thickness about $R_L^2 \approx 20$ times faster than over the distance between the wells. The layers clearly communicate: the injected water will flow between them and, under gravity, sink to the bottom of the unit. In Case 2 the vertical evening-out is slower than the horizontal one. The top and the bottom of the unit communicate only weakly: there is some crossflow through the shale layer, but it does not have time to even them out. To calculate the waterflood we then need a model with separate layers and crossflow between them. What this means for the share of the reservoir the water sweeps is explained in the lesson Displacement in heterogeneous reservoirs. The data in this example are assumed for illustration.

Measures of heterogeneity: the Dykstra–Parsons and Lorenz coefficients

There are several ways to describe the spread of permeability by a single number. The coefficient of variation $C_v$ (the standard deviation divided by the mean) and the standard deviation of the logarithm $\sigma_{\ln k}$ are covered in the lessons Descriptive statistics in geology and Distributions and sample size. In calculations of waterflooding in layered reservoirs, two more are traditionally used. The Dykstra–Parsons coefficient $V_{DP}$ is read from a probability plot of permeability (definition). The Lorenz coefficient $L_c$ is twice the area between the diagonal and the curve of “share of flow capacity ( $kh$ ) against share of storage capacity ( $\phi h$ )” (construction). Both are zero for a uniform reservoir, and both grow towards one as the reservoir becomes more heterogeneous.

How to find

V_{DP}

of a layered section

1
Divide the section into layers
From core and logs, pick out layers with thicknesses $h_i$ and permeabilities $k_i$ . If the data are core samples, give each sample the thickness it represents.
2
Sort
Arrange the layers in order of falling permeability. For each layer, find the share of the section thickness that has higher permeability (counted up to the middle of the layer).
3
Draw the probability plot
Plot $\ln k$ against the quantile of the normal distribution that belongs to this share (the position on the normal curve that matches the share), and draw a straight line through the central points.
4
Read the result
$k_{50}$ is the permeability at the 50% share, and $k_{84.1}$ is the permeability at the 84.1% share. Then $V_{DP} = (k_{50} - k_{84.1})/k_{50}$ . For a log-normal section the same value is $1 - e^{-\sigma}$ , where $\sigma$ is the standard deviation of $\ln k$ , weighted by thickness.

Lorenz coefficient of log-normal permeability

L_c = 2\,\Phi\!\left(\frac{\sigma_{\ln k}}{\sqrt{2}}\right) - 1

where:

$L_c$ Lorenz coefficient, —
$\Phi$ the standard normal distribution function (the share of a standard bell-shaped distribution that lies below the given value), —
$\sigma_{\ln k}$ standard deviation of the natural logarithm of permeability, —

Units: dimensionless

Holds when

Permeability is log-normal and the porosity of all the layers is the same. Otherwise the layers are sorted by $k/\phi$ and weighted by $\phi h$ , and the formula is only a rough guide.
There are many layers, so the Lorenz curve is smooth.

Both coefficients are fixed by $\sigma_{\ln k}$ alone, so each one gives the other: $V_{DP} = 1 - e^{-\sigma}$ and $L_c = 2\Phi(\sigma/\sqrt{2}) - 1$ , which is the same as $\operatorname{erf}(\sigma/2)$ (erf is the error function). The Lorenz curve of such a reservoir is $F = \Phi\bigl(\Phi^{-1}(C) + \sigma\bigr)$ . The derivation is in the advanced section.

Measures of heterogeneity of log-normal permeability

$V_{DP}$: 0.3
$\sigma_{\ln k}$: 0.36
$L_c$: 0.20
$\bar k_{\text{A}}/\bar k_{\text{H}}$: 1.14

$V_{DP}$: 0.5
$\sigma_{\ln k}$: 0.69
$L_c$: 0.38
$\bar k_{\text{A}}/\bar k_{\text{H}}$: 1.62

$V_{DP}$: 0.6
$\sigma_{\ln k}$: 0.92
$L_c$: 0.48
$\bar k_{\text{A}}/\bar k_{\text{H}}$: 2.32

$V_{DP}$: 0.7
$\sigma_{\ln k}$: 1.20
$L_c$: 0.61
$\bar k_{\text{A}}/\bar k_{\text{H}}$: 4.26

$V_{DP}$: 0.8
$\sigma_{\ln k}$: 1.61
$L_c$: 0.74
$\bar k_{\text{A}}/\bar k_{\text{H}}$: 13.3

$V_{DP}$: 0.9
$\sigma_{\ln k}$: 2.30
$L_c$: 0.90
$\bar k_{\text{A}}/\bar k_{\text{H}}$: 201

$V_{DP}$	$\sigma_{\ln k}$	$L_c$	$\bar k_{\text{A}}/\bar k_{\text{H}}$
0.3	0.36	0.20	1.14
0.5	0.69	0.38	1.62
0.6	0.92	0.48	2.32
0.7	1.20	0.61	4.26
0.8	1.61	0.74	13.3
0.9	2.30	0.90	201

For most reservoirs $V_{DP}$ is between 0.5 and 0.9. The scale is squeezed at the upper end: from 0.8 to 0.9 the ratio of the two extreme means grows fifteen-fold.

Layers of the section (from top to bottom)

Layer: 1
$h$ , m: 1.5
$k$ , mD: 310
$\phi$: 0.23

Layer: 2
$h$ , m: 2.0
$k$ , mD: 45
$\phi$: 0.18

Layer: 3
$h$ , m: 3.0
$k$ , mD: 120
$\phi$: 0.21

Layer: 4
$h$ , m: 1.0
$k$ , mD: 900
$\phi$: 0.25

Layer: 5
$h$ , m: 2.5
$k$ , mD: 18
$\phi$: 0.15

Layer: 6
$h$ , m: 2.0
$k$ , mD: 75
$\phi$: 0.19

Layer: 7
$h$ , m: 3.0
$k$ , mD: 8
$\phi$: 0.13

Layer: 8
$h$ , m: 1.0
$k$ , mD: 200
$\phi$: 0.22

Layer	$h$ , m	$k$ , mD	$\phi$
1	1.5	310	0.23
2	2.0	45	0.18
3	3.0	120	0.21
4	1.0	900	0.25
5	2.5	18	0.15
6	2.0	75	0.19
7	3.0	8	0.13
8	1.0	200	0.22

Total thickness 16.0 m. The data are assumed for illustration.

Worked exampleMeasures of heterogeneity of eight layers

Given (assumed data)

Section: eight layers with a total thickness of 16.0 m (table above)
Weights: the layer thicknesses: the layers make up the whole section

Solution

Natural logarithms of the layer permeabilities, in order: 5.737; 3.807; 4.787; 6.802; 2.890; 4.317; 2.079; 5.298.
Mean weighted by thickness: $m = \sum h_i\ln k_i/\sum h_i = 64.78/16.0 = 4.049$ . The geometric mean is $k_G = e^{4.049} \approx 57$ mD.
Spread: $\sigma^2 = \sum h_i(\ln k_i - m)^2/\sum h_i = 30.31/16.0 = 1.894$ ; $\sigma = 1.376$ .
Dykstra–Parsons coefficient: $V_{DP} = 1 - e^{-1.376} = 0.747$ .
Lorenz coefficient of the log-normal model: $L_c = 2\Phi(1.376/1.414) - 1 = 2\Phi(0.973) - 1 \approx 0.67$ .
The Lorenz curve of the layers themselves (table and figure below) gives $L_c \approx 0.56$ . With the same porosity in all the layers it would be 0.64.
Means weighted by thickness: arithmetic 139.6 mD, harmonic 25.8 mD. Their ratio is 5.4, against $e^{\sigma^2} = 6.6$ for the log-normal model.

Answer:

V_{DP} \approx 0.747

;

L_c \approx 0.56

from the layers and 0.67 from the log-normal model.

What it means. The section is clearly heterogeneous: the three best layers, with 28% of the storage capacity, give 70% of the flow capacity. The direct calculation gave a smaller Lorenz coefficient than the log-normal model, for two reasons. First, the best layers here are also the most porous. They not only conduct flow but also hold more, and this softens the contrast: with equal porosity the result would be 0.64. The rest comes from the small number of layers — a broken line through eight points cuts off the bend of the curve — and from the fact that the section is not strictly log-normal. The practical conclusion is that we can compare $V_{DP}$ and $L_c$ of different reservoirs only if they were found in the same way and from a comparable number of layers. The data in this example are assumed for illustration.

Lorenz curve of the layers

Layer: 4
$k/\phi$ , mD: 3600
$C$: 0.085
$F$: 0.403

Layer: 1
$k/\phi$ , mD: 1348
$C$: 0.202
$F$: 0.611

Layer: 8
$k/\phi$ , mD: 909
$C$: 0.276
$F$: 0.701

Layer: 3
$k/\phi$ , mD: 571
$C$: 0.490
$F$: 0.862

Layer: 6
$k/\phi$ , mD: 395
$C$: 0.619
$F$: 0.929

Layer: 2
$k/\phi$ , mD: 250
$C$: 0.741
$F$: 0.969

Layer: 5
$k/\phi$ , mD: 120
$C$: 0.868
$F$: 0.989

Layer: 7
$k/\phi$ , mD: 62
$C$: 1.000
$F$: 1.000

Layer	$k/\phi$ , mD	$C$	$F$
4	3600	0.085	0.403
1	1348	0.202	0.611
8	909	0.276	0.701
3	571	0.490	0.862
6	395	0.619	0.929
2	250	0.741	0.969
5	120	0.868	0.989
7	62	1.000	1.000

The layers are in order of falling $k/\phi$ . $C$ is the cumulative share of the storage capacity $\phi h$ , and $F$ is the cumulative share of the flow capacity $kh$ up to and including this layer.

CalculatedLorenz curves of log-normal permeability for

V_{DP}

= 0.5, 0.75 and 0.9, and the points of the eight layers of the example (their

V_{DP} \approx 0.75

). The dashed diagonal is a uniform reservoir. The points lie below the curve with the same

V_{DP}

: the best layers are also the most porous, and a broken line through eight points cuts off the bend.

Spatial correlation and the correlation length

The arrangement of patches of different permeability is described by spatial correlation: the values at neighbouring points are similar, and the similarity falls with distance. The distance at which it practically disappears is called the correlation length. In geostatistics (statistics for data spread in space) the same distance is called the range.

Correlation lengths differ with direction. Vertically they are short, from centimetres to metres, about the thickness of laminae and layers. Horizontally they run from tens of metres in channel sandstones (sandstones laid down in river channels) to kilometres in reservoirs laid down on a wide shallow sea shelf. The ratio of the horizontal length to the vertical one is tens or hundreds. This is why thin layers are stretched along the bedding (the layering of the rock), and why the anisotropy of a unit follows from the same correlation.

CalculatedTwo sections 300 m long and 16 m thick (the vertical scale is exaggerated), built from the same set of permeability values (the lesson's model: geometric mean 40 mD,

\sigma_{\ln k}

= 1.2). At the top the correlation lengths are about 15 m horizontally and 3 m vertically: a fine mosaic. At the bottom they are about 360 m and 4.5 m: stretched bodies whose high-permeability bands run for hundreds of metres.

Comparing the correlation length $\lambda$ with the distance between wells $L$ suggests which reservoir model to use.

$\lambda$ is much larger than $L$ . The thin layers can be followed from well to well, so the reservoir is layered. Each layer has its own permeability, the arithmetic mean describes the flow along the layers, and the waterflood goes layer by layer (Displacement in heterogeneous reservoirs).
$\lambda$ is much smaller than $L$ . Tens of independent patches fit between the wells, and their differences average out. The reservoir then behaves as a uniform one with a single equivalent permeability. For equidimensional patches (about as long as they are wide) it is close to the geometric mean. For patches stretched along the bedding, with flow along the reservoir, it is higher and closer to the arithmetic mean (see the table in the section on the random mosaic).
$\lambda$ is comparable to $L$ . This is the hardest case. Connectivity decides: whether the high-permeability bodies join into continuous paths between the wells. We describe such reservoirs with geostatistical models and test them by calculation on a reservoir simulation model (a computer model of fluid flow in the reservoir).

We measure the vertical correlation length from core and logs. The horizontal one is almost never measured, because the wells are too far apart. We take it from geology: from the depositional environments (the settings where the sediment was laid down), from outcrop analogues (exposed rocks of the same kind that we can walk along), from seismic data and from the correlation of well sections (Well correlation). We then refine it from interference tests (tests that show how wells affect each other) and from production history. The variogram describes correlation in numbers (Spatial correlation and the variogram).

Correlation also affects the accuracy of estimates. Samples taken every 0.3 m, when the vertical correlation length is about a metre, are not independent. Thirty such samples from nine metres of section carry less information than thirty independent measurements. The error of the means and of the measures of heterogeneity is then larger than the formulas for an independent sample give (see the advanced section).

How heterogeneity is taken into account in calculations

A uniform reservoir with an average permeability is fine when heterogeneity does not affect the flow at the scale of the problem: the patches of different permeability are small compared with the distance between wells and mixed without order. We choose the average according to the geometry of the flow.
A layered model (“layer cake”): layers that can be followed between wells, each with its own permeability and porosity and linked to its neighbours through $k_v$ . It is convenient for analytical waterflood calculations (made with formulas, not with a computer simulation) (Displacement in heterogeneous reservoirs).
Flow units: layers are grouped together when they have a similar link between porosity and permeability and behave similarly in flow.
A geostatistical model: many equally likely versions of the property distribution, all with the given histogram and correlation lengths. The differences between them show the uncertainty of the forecast (Spatial correlation and the variogram).
Moving to the cells of the reservoir simulation model: the detailed model is coarsened (upscaled to larger cells). For each cell we find the equivalent permeability in each direction (Permeability upscaling for single-phase flow), and for the flow between cells we find the transmissibility, the ability of the link between two cells to pass flow (Interblock flow and transmissibility).
Checking against the behaviour of the reservoir: permeability from well tests, inflow and injectivity profiles (how much each layer gives or takes), pressure measurements in the separate layers, and the water breakthrough time. A mismatch with the model points to heterogeneity that the model leaves out (History matching).

Terms and symbols

Reservoir heterogeneity: The variation of the rock properties from point to point and with direction.
Equivalent permeability: The permeability of a uniform reservoir that gives the same flow as the heterogeneous one under the same boundary conditions. English-language literature sometimes calls it effective permeability too. Do not confuse it with the effective (phase) permeability in multiphase flow.
Layered and mosaic reservoir: In the first, thin layers can be followed between wells. In the second, patches of different permeability are mixed finely and at random.
$k_v/k_h$: The ratio of vertical to horizontal permeability. It depends on the scale and is set by the thin layers.
$R_L$: The effective aspect ratio (length to thickness). It shows whether the layers have time to exchange fluid over the distance between wells.
$V_{DP}$ , $L_c$: The Dykstra–Parsons and Lorenz coefficients: measures of the spread of permeability, from 0 (a uniform reservoir) to 1.
Correlation length: The distance over which the values of a property are still similar to each other. It differs with direction.
Power mean: The mean $(\sum w_ik_i^{\omega}/\sum w_i)^{1/\omega}$ . For $\omega$ = 1, 0 and −1 it is the arithmetic, geometric and harmonic mean.

Common mistakes and misconceptions

Wrong

“On the logarithmic plot all the layers look close together, so the reservoir is almost uniform.”

In fact

Flow follows the permeability itself, not its logarithm. A layer that differs by one and a half orders of magnitude and is only 5% of the thickness of the unit can carry more than half of the flow.

Wrong

“The average permeability around a well is the average of the zones by area.”

In fact

In radial inflow the zones work in series with logarithmic weights. The zone nearest the well weighs much more than its share of the area.

Wrong

“Two reservoirs have $V_{DP}$ = 0.75, so the waterflood will go the same way.”

In fact

The coefficient describes the set of permeabilities, but not their arrangement. In a reservoir with continuous layers, the water moves layer by layer and breaks through early. In a fine random mosaic, the heterogeneity largely averages out.

Wrong

“The $k_v/k_h$ of the reservoir is 0.5: pairs of vertical and horizontal samples showed it.”

In fact

Samples describe the scale of centimetres. The vertical permeability of a unit is set by thin shale layers, which the samples usually do not contain. At the scale of the reservoir the ratio can be hundreds of times smaller.

Wrong

“The geometric mean is the universal average for a heterogeneous reservoir.”

In fact

It is exact only for two-dimensional flow in a random isotropic mosaic. Along continuous layers the arithmetic mean works, across them the harmonic mean, and in a three-dimensional mosaic the equivalent permeability is higher than the geometric mean.

In depth: advanced detail

Where the logarithmic weights come from

In steady-state flow the same rate $q$ passes through every zone. In zone $j$ the Dupuit equation (the basic formula for steady radial flow, see radial flow) gives the pressure drop $p_j - p_{j-1} = \dfrac{q\mu}{2\pi k_jh}\ln\dfrac{r_j}{r_{j-1}}$ . Adding the pressure drops of all the zones, we get $p_e - p_w = \dfrac{q\mu}{2\pi h}\sum_j \dfrac{\ln(r_j/r_{j-1})}{k_j}$ . A uniform reservoir with permeability $\bar k$ gives the same rate at $p_e - p_w = \dfrac{q\mu}{2\pi \bar k h}\ln\dfrac{r_e}{r_w}$ . Equating the two, we get the formula of the main section.

With one inner zone of permeability $k_s$ and radius $r_s$ in a reservoir of permeability $k$ , the sum equals $[\ln(r_e/r_w) + S]/k$ , where $S = (k/k_s - 1)\ln(r_s/r_w)$ is the Hawkins skin factor. The average permeability and the skin are two ways of saying the same thing. The first is convenient when the altered zone is large, and the second when it is thin.

The power mean and the dimension of the flow

The three means are special cases of one family.

Power mean of permeability

k_\omega = \left(\frac{\sum_i w_ik_i^{\,\omega}}{\sum_i w_i}\right)^{1/\omega}

where:

$k_\omega$ power mean with exponent $\omega$ , mD
$k_i,\ w_i$ permeability of a patch and its weight (volume or thickness), mD, —
$\omega$ exponent: 1 gives the arithmetic mean and −1 the harmonic mean; as $\omega \to 0$ we get the geometric mean, —

Units: permeabilities in any identical units

Holds when

The exponent $\omega$ is an empirical parameter of how the patches are arranged. We fit it by calculating the flow on a detailed model.
For log-normal permeability, $k_\omega = k_Ge^{\omega\sigma^2/2}$ .

The last equality follows from the fact that for a log-normal quantity the mean of $k^{\omega}$ is $e^{\omega\mu + \omega^2\sigma^2/2}$ , where $\mu$ is the mean of $\ln k$ . The larger $\omega$ is, the more the high-permeability patches decide the mean.

For an isotropic random log-normal medium, the exponent depends on the dimension of the flow $d$ roughly as $\omega \approx 1 - 2/d$ . It is $-1$ for a chain of patches, $0$ for planar flow (here the result is exact) and about $1/3$ for three-dimensional flow, which gives $k_Ge^{\sigma^2/6}$ . If the patches are stretched, the exponent for flow along the long axis grows towards one, and across it falls towards $-1$ . This is how correlation produces anisotropy of the equivalent permeability. Fitting the exponent, and the other ways of finding the equivalent permeability of a model cell, are the subject of the lesson Permeability upscaling for single-phase flow.

The Lorenz curve of log-normal permeability

Let the porosity be the same everywhere, and let $\ln k$ be normally distributed with mean $\mu$ and standard deviation $\sigma$ . Take all the layers with permeability above a threshold $k^*$ and write $z^* = (\ln k^* - \mu)/\sigma$ . Their share of the storage capacity is $C = 1 - \Phi(z^*)$ . Their share of the flow capacity is their share of the sum of $k$ ; for a log-normal quantity it equals $1 - \Phi(z^* - \sigma)$ . Eliminating $z^*$ , we get the curve $F = \Phi\bigl(\Phi^{-1}(C) + \sigma\bigr)$ .

The area under the curve equals the probability that one standard normal quantity is smaller than another one increased by $\sigma$ . This is $\Phi(\sigma/\sqrt{2})$ . Hence $L_c = 2\Phi(\sigma/\sqrt{2}) - 1$ . As $\sigma \to 0$ the curve merges with the diagonal. As $\sigma \to \infty$ it presses against the left and top sides of the square, and $L_c \to 1$ .

How accurate the measures of heterogeneity are from a small sample

Worked exampleAccuracy of

V_{DP}

from ten and from fifty samples

Given (assumed data)

Reservoir: log-normal permeability, $V_{DP}$ = 0.70, which means $\sigma$ = 1.204
Samples: 10 and 50 independent samples
Quantiles of the χ² distribution (from tables): 9 degrees of freedom: 3.325 (5%) and 16.919 (95%); 49 degrees of freedom: 33.93 and 66.34

Solution

The sample standard deviation of the logarithm, $s$ , is linked to the true one: the quantity $(n - 1)s^2/\sigma^2$ follows the χ² distribution with $n - 1$ degrees of freedom. So in 90% of samples $s$ lies between $\sigma\sqrt{\chi^2_{5\%}/(n - 1)}$ and $\sigma\sqrt{\chi^2_{95\%}/(n - 1)}$ .
10 samples: $s$ from $1.204\sqrt{3.325/9} = 0.732$ to $1.204\sqrt{16.919/9} = 1.651$ ; $V_{DP} = 1 - e^{-s}$ from 0.52 to 0.81.
50 samples: $s$ from $1.204\sqrt{33.93/49} = 1.002$ to $1.204\sqrt{66.34/49} = 1.401$ ; $V_{DP}$ from 0.63 to 0.75.

Answer: From ten independent samples, the

V_{DP}

of a reservoir whose true value is 0.70 will come out between 0.52 and 0.81 in 90% of cases. From fifty samples it will come out between 0.63 and 0.75.

What it means. The second decimal of a $V_{DP}$ calculated from about ten samples means nothing. Fifty samples make the interval about two and a half times narrower. If the samples are taken more often than the vertical correlation length, they are not independent, and the true interval is wider than the calculated one. The data in this example are assumed for illustration.

Connections to other topics

Rock permeability — Darcy's law and the averaging of layer permeability in linear flow.
Descriptive statistics in geology — the three means, the coefficient of variation and the construction of the Lorenz curve.
Distributions and sample size — the log-normal distribution, the probability plot and the definition of $V_{DP}$ .
Core analysis — how to compare core, logs and well tests, which work at different scales.
Skin factor — the altered zone around a well as a special case of ring zones.
Water and gas coning — why the ratio $k_v/k_h$ matters so much for cones.
Why water is injected: pressure maintenance and oil displacement — why heterogeneity decides how much of the reservoir the waterflood sweeps.
Displacement in heterogeneous reservoirs — calculation of displacement in layered reservoirs.
Spatial correlation and the variogram — a numerical description of spatial correlation.
Permeability upscaling for single-phase flow — equivalent permeability of the cells of the reservoir simulation model.

Check yourself

1
A well with no skin factor penetrates two layers with a common pressure at the outer boundary: 4 m of 150 mD and 10 m of 15 mD. What share of the inflow does each layer give, and which permeability decides the inflow of the well?
Show the answer
The layers work in parallel, and the inflow of each is proportional to $kh$ : 600 and 150 mD·m. The first layer gives $600/750 = 80$ % of the inflow with 29% of the thickness. The inflow of the well is decided by the arithmetic mean weighted by thickness: $750/14 \approx 53.6$ mD. Whether the layers communicate vertically does not matter: the pressure in them is distributed in the same way, so there is nothing to flow between them.
2
A well of radius 0.1 m is surrounded by a zone of radius 2 m whose permeability is reduced to 10 mD. The reservoir is 100 mD, and the outer boundary of the drainage area is at 300 m. Find the average permeability for steady-state flow and compare it with the calculation through the skin factor.
Show the answer
$\bar k = \ln 3000/(\ln 20/10 + \ln 150/100) = 8.006/(0.2996 + 0.0501) = 8.006/0.3497 \approx 22.9$ mD. Through the skin: $S = (100/10 - 1)\ln 20 \approx 27.0$ ; $\bar k = 100\cdot8.006/(8.006 + 27.0) \approx 22.9$ mD, the same. A zone of only 2 m radius has reduced the inflow more than fourfold.
3
Two reservoirs have $V_{DP}$ = 0.75. In the first, thin layers can be followed between all the wells. In the second, patches of different permeability are mixed finely and at random. Which statement is true?
1. A. $V_{DP}$ is not defined for a mosaic reservoir
2. B.In the second reservoir the water will break through earlier
3. C.In the first, the arithmetic mean describes the flow along the reservoir and the water will go along the best layers. In the second, the equivalent permeability is closer to the geometric mean, and the heterogeneity largely averages out
4. D.They will behave the same way: the heterogeneity is the same
Show the answer
Correct answer: C. In the first, the arithmetic mean describes the flow along the reservoir and the water will go along the best layers. In the second, the equivalent permeability is closer to the geometric mean, and the heterogeneity largely averages out
$V_{DP}$ depends only on the set of permeabilities, while the behaviour of the reservoir also depends on how the patches are arranged. Continuous thin layers give layer-by-layer flow and an early water breakthrough along the best of them. A fine random mosaic averages out over the distance between the wells.
4
The logarithm of the permeability of a reservoir is normally distributed, with $\sigma_{\ln k} = 1.5$ . How many times is the arithmetic mean larger than the harmonic mean? What are $V_{DP}$ and $L_c$ ?
Show the answer
$\bar k_{\text{A}}/\bar k_{\text{H}} = e^{1.5^2} = e^{2.25} \approx 9.5$ . $V_{DP} = 1 - e^{-1.5} \approx 0.78$ . $L_c = 2\Phi(1.5/1.414) - 1 = 2\Phi(1.06) - 1 \approx 2\cdot0.855 - 1 \approx 0.71$ .
5
A unit is 30 m thick, the distance between the injection and producing wells is 600 m, and $k_v/k_h$ = 0.0025. What is $R_L$ , and what does it tell us?
Show the answer
$R_L = (600/30)\sqrt{0.0025} = 20\cdot0.05 = 1.0$ . The pressure evens out across the thickness at about the same speed as over the distance between the wells, so the layers communicate partly. Neither a model of independent layers nor an assumption of instant vertical equalisation fits here. We need a calculation with crossflow between the layers.
6
The horizontal correlation length of permeability is about 20 m, and the distance between wells is 500 m. Which reservoir model is it sensible to use to estimate the inflow?
1. A.A uniform one with a single equivalent permeability, between the geometric and the arithmetic mean
2. B.A layered one: each layer has its own permeability, and the mean is arithmetic
3. C.The harmonic mean of all the samples
4. D.The highest permeability from core
Show the answer
Correct answer: A. A uniform one with a single equivalent permeability, between the geometric and the arithmetic mean
Tens of independent patches fit between the wells, and their differences average out: the reservoir behaves as a fine mosaic. Where exactly between the means the equivalent permeability falls depends on the shape of the patches. For equidimensional patches it is close to the geometric mean (a little higher in three-dimensional flow). For patches stretched along the bedding, with flow along the reservoir, it is closer to the arithmetic mean. A layered model is needed when the thin layers can be followed from well to well, that is, when the correlation length is greater than the distance between wells.

Updated: 19 October 2026