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GeoscienceStructural geology Intermediate 27 min read

Fractures and naturally fractured reservoirs

A fracture is a surface in rock where the rock is no longer continuous. The walls of the fracture have moved apart, or have slid along each other, by a distance that is small compared with the size of the fracture itself. A joint (extension fracture) has opened at right angles to its surface, with no noticeable slip along it. A shear fracture has slipped along its surface. When the slip is noticeable, we call it a fault (Faults). A naturally fractured reservoir is a layer in which open natural fractures clearly affect the flow of fluids and the production.

Why it matters

Fractures can make an almost impermeable rock productive. They can also make a good layer produce mostly water within months: the water we inject runs to the producing well along the fractures and bypasses the oil in the matrix (the rock between the fractures). The direction of fractures decides how we orient horizontal wells and rows of injection wells. Their opening decides how sensitive flow rates are to a fall in pressure. Their density decides whether we need hydraulic fracturing (cracking the rock on purpose by pumping in fluid) and what model of the layer to build. Fractures swallow drilling fluid (mud), and they reveal the present-day stresses at the wellbore. To predict fractures where there are no wells, we need to understand how and why they form.

Quick refresher

  • A joint opens in the direction of the least principal stress σ3\sigma_3 (the weakest of the three main pushes on the rock) and lies in the plane of σ1\sigma_1 and σ2\sigma_2 (the other two). Shear fractures form conjugate pairs (two sets at an angle to each other) at about 30° to σ1\sigma_1.
  • At depth all stresses are compressive. A joint forms only when two things hold. The pore pressure must exceed σ3\sigma_3 by the tensile strength of the rock TT (the pull the rock can bear before it breaks). And the differential stress (the difference between the greatest and the least stress) must be not more than 4T4T. This links joints with abnormally high pressure, and with the uplift and cooling of rocks.
  • Fractures group into sets. We show their orientation by poles on a stereographic projection (a circular chart on which each fracture plane becomes a point, its pole). A well crosses a set less often the closer its direction is to the set. We divide the number of fractures along the well by cos⁡γ\cos\gamma. This is the Terzaghi correction.
  • In layered sequences, joints are often limited to one mechanical layer (a bed or group of beds that breaks as one unit), and their spacing grows with its thickness. In a well-developed set the spacing is close to the thickness.
  • The flow through a fracture is proportional to the cube of its aperture (opening). The permeability of a set is k=∑w3/(12L)k = \sum w^3/(12L), and the widest fractures decide it. So it falls sharply when fractures close under load.
  • By the role of fractures, we tell four types of reservoir apart. In one, fractures give both storage and permeability. In another, they only create anisotropy (permeability that differs with direction). Fractures speed up production, but they also bring early water breakthrough, mud losses, and a fall in flow rates when pressure falls.
Key relationships
σ3−p=−T,σ1−σ3≤4T\sigma_3 - p = -T,\qquad \sigma_1 - \sigma_3 \le 4T

conditions for a joint to form by the Griffith criterion

P10=Nl cos⁡γ,s=1P10P_{10} = \frac{N}{l\,\cos\gamma},\qquad s = \frac{1}{P_{10}}

true linear intensity and spacing of a set from the number NN of its fractures on a stretch of well of length ll; γ\gamma is the angle between the well and the normal to the fractures

k=∑wi312 L,kf=w212k = \frac{\sum w_i^3}{12\,L},\qquad k_f = \frac{w^2}{12}

permeability of the rock along a fracture set counted over a stretch of length LL across it, and the permeability of one fracture

ΔV=σn′VmKniVm+σn′\Delta V = \frac{\sigma_n' V_m}{K_{ni} V_m + \sigma_n'}

hyperbolic law of closure of a fracture under effective normal stress

Common pitfalls
  • Believing that a vertical well sees all fractures: it almost misses steep sets.
  • Taking light, high-resistivity sinusoids (wave-shaped traces) on an image in water-based mud for open fractures, and all dark ones for conductive fractures.
  • Estimating the permeability of fractures from the average aperture: because of the cube, the contribution of the wide fractures is out of proportion.
  • Transferring fracture density from outcrops (places where rock shows at the surface) to the layer with no correction: near the surface, unloading and weathering fractures are added to the tectonic ones.
  • Taking the permeability of fractures as constant: when reservoir pressure falls, fractures close.

After this lesson you can

  • tell joints, shear fractures and hybrid fractures apart, and explain with the Mohr circle (a diagram of the stresses on all planes through a point) at what stresses and pore pressure (the pressure of the fluid in the pores) each one forms
  • link fracture sets to their origin (regional stresses, folds, faults, uplift and cooling) and predict their orientation
  • describe fracturing by linear intensity and spacing, apply the Terzaghi correction, and take into account the link between spacing and the thickness of a mechanical layer
  • calculate the permeability of a fracture set with the cubic law, and explain why the widest fractures decide it and why it is sensitive to stress
  • assign a reservoir to a type by the role of its fractures, and draw the consequences for development: anisotropy, early water breakthrough, mud losses, and the loss of permeability when pressure falls
  • read fractures on core and borehole images, and tell natural open fractures from healed ones and from those induced by drilling

Fractures, veins and faults: what is called what

Almost every core and every outcrop is cut by fractures, and almost every layer has them. The question is not whether there are fractures. It is whether they affect the flow of fluids. To answer it, we first have to understand what kind of break we have. In fracture mechanics, fractures are divided by how their walls moved. Opening is mode I. Sliding along the surface is mode II (across the edge of the fracture) or mode III (along the edge). For a reservoir it also matters what fills the break, and what has happened to the grains at its edges.

Breaks in rock and their role for flow
How the walls moved
opened apart (mode I)
Name
joint (extension fracture); if filled with minerals, a vein
Effect on flow
open: a channel along the fracture; healed: no channel, sometimes a barrier
How the walls moved
slid along the surface (modes II and III)
Name
shear fracture
Effect on flow
from a channel to a barrier, depending on roughness and fill
How the walls moved
opened and slid
Name
hybrid fracture
Effect on flow
more often a channel
How the walls moved
moved closer, the rock dissolved or compacted
Name
stylolite (a jagged dissolution seam), compaction band
Effect on flow
a barrier across, sometimes a channel along (stylolites)
How the walls moved
slid, crushing the grains
Name
cataclastic band (deformation band)
Effect on flow
a barrier across (fault zone)

The boundaries are conventional. A shear fracture with noticeable slip is called a fault, and in reports the word “fractures” often stands for all breaks at once.

What we can read on the walls of a fracture in core:

  • Plumose structure (a feather-like pattern): fan-shaped grooves on a joint. They spread from the place where the fracture started, in the direction in which it grew, and arc-shaped lines mark stops of its front.
  • Slickensides and slickenlines (polished surfaces and the grooves on them): a sign of a shear fracture. They show the direction of the last slip (parts of a fault).
  • Druses: crystals that grew freely on the wall. Their growth needs an open cavity filled with mineral-rich water. So the fracture was open at least at some time.
  • Fibrous fill across a vein: the fracture opened a little at a time, cracking and healing again and again (crack-seal).
  • Mineral bridges between the walls of a partly healed fracture. They work as natural proppant (grains that hold a fracture open) and keep the fracture open as the load grows (see the section on fracture closure).

Why rock cracks: opening and shear

Intact rock breaks when the Mohr circle of effective stresses touches the failure envelope (Mohr circle). Effective stresses are the stresses carried by the rock itself, after the pore pressure is subtracted. The failure envelope is the limit line beyond which the rock breaks. Under compressive normal stress the envelope is close to the Coulomb straight line (Mohr–Coulomb criterion). In the tensile region it bends towards the point σn′=−T\sigma_n' = -T, where TT is the tensile strength of the rock, usually a few megapascals. A handy model of this part of the envelope is the Griffith parabola.

Griffith envelope and the condition for a joint to form
τ2=4T (σn′+T);joint:  σ3′=−T,σ1−σ3≤4T\tau^2 = 4T\,(\sigma_n' + T);\qquad \text{joint:}\ \ \sigma_3' = -T,\quad \sigma_1 - \sigma_3 \le 4T
where:
  • τ\taushear stress on a plane, MPa
  • σn′\sigma_n'effective normal stress on the plane, σn−p\sigma_n - p; compression is positive and tension is negative, MPa
  • σ1′, σ3′\sigma_1',\ \sigma_3'greatest and least effective principal stress, σ1−p\sigma_1 - p and σ3−p\sigma_3 - p, MPa
  • pppore pressure, MPa
  • TTtensile strength of the rock (a positive number), MPa

Units: MPa; effective stresses use the full pore pressure, as in failure criteria

Holds when
  • The rock is intact, with no weak planes. On existing fractures and faults, slip starts earlier, by the law of friction (slip on a fracture).
  • The parabola describes the tensile and weakly compressive part of the envelope. At high compressive stress, the real strength is closer to the Coulomb line, and the boundary between hybrid and shear fractures moves a little.
  • The differential stress σ1−σ3\sigma_1 - \sigma_3 does not depend on pore pressure.

When the circle touches the envelope at its tip, it gives a joint at right angles to σ3\sigma_3. This is possible only while the radius of the circle is no larger than the radius of curvature of the parabola at its tip, 2T2T, that is, while σ1−σ3≤4T\sigma_1 - \sigma_3 \le 4T. With a larger differential stress, the circle touches the parabola to the left of zero normal stress. Then hybrid fractures form, with both opening and shear, at an acute angle of up to 22.5° to σ1\sigma_1. When σ1−σ3>42 T≈5.7 T\sigma_1 - \sigma_3 > 4\sqrt2\,T \approx 5.7\,T, the point of contact moves into the compressive region, and shear fractures form. There the envelope is close to the Coulomb line, and with a friction coefficient of about 0.6 the fractures run at about 30° to σ1\sigma_1. The derivation is in the advanced section.

From this comes the main rule of orientation. A joint opens in the direction of least compression, so it is at right angles to σ3\sigma_3 and lies in the plane of σ1\sigma_1 and σ2\sigma_2. In an extension setting, where σ1\sigma_1 is vertical, joints are vertical and run along the greatest horizontal stress σH\sigma_H. In compression, where σ3\sigma_3 is vertical, they are horizontal. Shear fractures form in pairs, as conjugate sets, symmetric about σ1\sigma_1. The acute angle between them, about 60°, is cut in half by the direction of σ1\sigma_1 (Anderson angles).

Orientation of fractures relative to the principal stressesσ₃σ₃σ₁jointsσ₁hybrid≈ 15°σ₁shear≈ 30°

Three square blocks of rock in a row. Above and below each block are vertical arrows of sigma one, which squeeze the block. To the left and right of the row of blocks are horizontal arrows of sigma three. The left block has several parallel vertical fractures and the label “joints”. The middle block has two sets of steep fractures leaning in opposite directions at about 15 degrees to the vertical, crossing each other, and the label “hybrid”. The right block has two sets of fractures at about 30 degrees to the vertical, crossing each other, and the label “shear”.

SchematicOrientation of new fractures relative to the principal stresses in a vertical section in extension, with σ1\sigma_1 vertical (schematic). Joints run along σ1\sigma_1, at right angles to σ3\sigma_3. Hybrid fractures form conjugate pairs at a small angle to σ1\sigma_1 (about 15° in case 2 of the example below). Shear fractures form conjugate pairs at about 30° to σ1\sigma_1 at a friction coefficient of 0.6.

At depth all the principal stresses are compressive, so a joint needs a “helper”: pore pressure. It moves the Mohr circle to the left without changing its size. When pp exceeds σ3\sigma_3 by TT, the rock breaks in the same way as in hydraulic fracturing in a well (fracture pressure). Such natural hydraulic fracturing happens where the pore pressure is abnormally high (Abnormal formation pressures: origin, prediction and recognition): in quickly sinking shales, in source rocks while they are making oil and gas, and under reliable seals (rocks that hold fluids in). The second way is to reduce the stresses. When rock is uplifted and eroded, it cools and shrinks, and the removal of load from above also lowers the horizontal stresses. The smallest of them can become tensile, and unloading joints open in the rock.

Worked exampleWhat pressure is needed to open joints at depth
Given (assumed data)
Depth
2800 m
Vertical stress σv=σ1\sigma_v = \sigma_1
gradient 23.5 kPa/m
Minimum horizontal stress σh=σ3\sigma_h = \sigma_3
case 1: 51.0 MPa; case 2: 42.5 MPa
Tensile strength of the rock TT
5 MPa (a tight cemented sandstone; only the order of magnitude is meant)
Hydrostatic pressure
gradient 10.2 kPa/m
Solution
  1. Stresses: σ1=0.0235⋅2800=65.8\sigma_1 = 0.0235\cdot2800 = 65.8 MPa, and the hydrostatic pressure is 28.56 MPa. The Griffith limits are 4T=204T = 20 MPa and 42 T=28.34\sqrt2\,T = 28.3 MPa.
  2. Case 1: σ1−σ3=14.8\sigma_1 - \sigma_3 = 14.8 MPa, which is less than 4T4T, so a joint is possible. It will open at p=σ3+T=51.0+5=56.0p = \sigma_3 + T = 51.0 + 5 = 56.0 MPa. This is a gradient of 20.0 kPa/m, which is 1.96 times the hydrostatic one.
  3. Case 2: σ1−σ3=23.3\sigma_1 - \sigma_3 = 23.3 MPa, which is between 4T4T and 42 T4\sqrt2\,T, so the failure is hybrid. The radius of the Mohr circle is r=11.65r = 11.65 MPa. The circle touches the parabola when its centre, in effective stresses, equals r2/(4T)=6.79r^2/(4T) = 6.79 MPa. That is at p=(65.8+42.5)/2−6.79=47.4p = (65.8 + 42.5)/2 - 6.79 = 47.4 MPa, a gradient of 16.9 kPa/m, which is 1.66 times the hydrostatic one. The fractures will run at about 15° to the vertical σ1\sigma_1.
  4. Check: suppose the rock has existing planes with no cohesion (no bond between their walls), well oriented for slip. They start to slip when the ratio (σ1−p)/(σ3−p)(\sigma_1 - p)/(\sigma_3 - p) reaches (μ2+1+μ)2=3.12(\sqrt{\mu^2 + 1} + \mu)^2 = 3.12 at μ=0.6\mu = 0.6 (frictional limit on faults). In case 1 this happens at p=(3.12⋅51.0−65.8)/(3.12−1)=44.0p = (3.12\cdot 51.0 - 65.8)/(3.12 - 1) = 44.0 MPa, earlier than the joints open.
Answer: At hydrostatic pressure no new fractures form in either case. In case 1, vertical joints along σH\sigma_H open at a pore pressure of about 56 MPa, almost twice the hydrostatic one. In case 2, hybrid fractures appear at 47 MPa.

What it means. Joints at depth are evidence that the pore pressure once exceeded the least stress, or that the stresses fell during uplift. The calculation is simplified, and it errs in one direction: as pore pressure in a layer rises, the horizontal stress rises too (poroelastic model). So to break the rock, an even higher pressure than the calculated one is needed. Existing weak planes, on the contrary, set an upper limit on the pressure. In case 2, well-oriented planes with no cohesion would slip already at 31.5 MPa, long before new fractures form, and the pressure would be released along them. The data in this example are assumed for illustration.

Mohr circles of the example and the Griffith envelope−5051015202530350510152025Effective normal stress σₙ′, MPaShear stress τ, MPa1, hydrostatic1, joint2, hybridGriffith envelope

Graph with the same scale on both axes. The horizontal axis is the effective normal stress from minus 8 to 38 MPa. The vertical axis is the shear stress from 0 to 26 MPa. The Griffith parabola starts at the point minus 5 on the horizontal axis and rises to the right, passing through 10 MPa of shear stress at zero normal stress. On the right is a small dashed semicircle for case 1 at hydrostatic pressure, from 22 to 37 MPa, far under the parabola. The same semicircle, moved left until it touches the tip of the parabola at minus 5, is the moment a joint forms. The larger semicircle of case 2, from minus 4.9 to 18.4 MPa, touches the parabola at a point near minus 3.2 MPa and 6 MPa of shear stress: hybrid fractures.

CalculatedMohr circles in effective stresses for the example, and the Griffith parabola for T=5T = 5 MPa. Case 1 is shown at hydrostatic pressure (far from the envelope) and at the moment a joint forms, when the circle touches the tip of the parabola. Case 2 is shown at the moment hybrid fractures form, when the contact is to the left of zero normal stress.

Where fractures form: stresses, folds, faults, uplift

The mechanics is one, but the settings in which rock reaches failure are different, and each leaves its own pattern of fractures. Once we know the pattern, we can predict fractures between wells.

Origin of fractures and their pattern
Cause
Regional stresses
Orientation
vertical sets along σH\sigma_H over large areas
Where they are concentrated
in all brittle layers; the intensity depends on the layer
Cause
Bending of folds
Orientation
along the hinge (the line of sharpest bending) in the outer arc of the crest, and also across the hinge
Where they are concentrated
in crests and on steep limbs (stretching at the crest)
Cause
Faults
Orientation
parallel to the fault and at an acute angle to it
Where they are concentrated
in the damage zone; the intensity falls with distance from the fault (fault zone)
Cause
Uplift, erosion, cooling
Orientation
vertical, often along the present-day σH\sigma_H; near the surface, parallel to it
Where they are concentrated
in uplifted blocks; near the surface, unloading and weathering fractures
Cause
Instability of the edge of a carbonate platform
Orientation
steep, parallel and perpendicular to the edge
Where they are concentrated
in the strong rocks of the outer platform and the upper slope
Cause
Cooling of magma
Orientation
columnar, at right angles to the contacts
Where they are concentrated
in lava flows, sills and dykes (sheets of solidified magma)

This is a simplification: one layer usually has several generations of fractures of different origin.

The intensity of fractures depends on the rock too. The stronger and stiffer it is, and the fewer pores and clay it has, the more fractures it gets for the same deformation. Dolomites, silicified limestones, cherts and cemented quartz sandstones crack more densely than porous limestones, weak sandstones and clays. In clays fractures either do not form or close up. So in a layered sequence the fractures are often confined to strong layers and stop at their boundaries with clays. Each such layer works as a separate mechanical layer. The alternation of layers by their mechanical properties is called mechanical stratigraphy.

In folds we predict fracturing from curvature. The stretching in the outer arc of a layer is proportional to its thickness and curvature (curvature and stretching), so there are more fractures in crests and where a fold bends sharply. Maps of the curvature of horizons (the mapped tops of layers) from seismic data are a common way to look for such zones. Near faults, the fracture density is greatest in the damage zone (the cracked rock around a fault) and falls quickly with distance. At the tips of faults and in relay ramps, where the slip passes from one segment to the next, fracturing is especially strong (relay ramps).

Sets, intensity and spacing of fractures

Fractures of one orientation and origin form a set. Systematic fractures are long, planar and almost parallel. Non-systematic fractures are short and curved and end against the systematic ones. We show the orientation of sets by poles on a stereographic projection: each set is a cluster of points (stereographic projection). For vertical sets a rose diagram of strikes (a circular histogram of directions) is handy too. The relative age of sets is found from their junctions: a younger fracture ends against an older one, making a T-shaped junction, or offsets it.

How many fractures there are in the rock is described by the fracture intensity. The easiest to measure is the linear intensity P10P_{10} (the density): the number of fractures per metre of a line of observation, such as a well, a core, or a line on an outcrop. One divided by it is the average spacing ss of the set, if the line is at right angles to the fractures. Some measures do not depend on the direction of the line: the total length of fracture traces per unit area (P21P_{21}, m/m²) and the total area of fractures in a unit volume (P32P_{32}, m²/m³). It is P32P_{32} that is usually set in fracture models.

True linear intensity of a set from observations in a well (Terzaghi correction)
P10=Nl cos⁡γ,s=1P10,cos⁡γ=∣n⋅w∣P_{10} = \frac{N}{l\,\cos\gamma},\qquad s = \frac{1}{P_{10}},\qquad \cos\gamma = |\mathbf n\cdot\mathbf w|
where:
  • NNnumber of fractures of the set crossed by the well in the interval, —
  • lllength of the interval along the well, m
  • γ\gammaangle between the well and the normal to the fractures of the set, °
  • n, w\mathbf n,\ \mathbf wunit vectors of the normal to the fractures and of the direction of the well, as in the vector calculation, —
  • P10P_{10}linear intensity along the normal to the set, m⁻¹
  • ssaverage spacing of the set, m

Units: metres and inverse metres; angles in degrees

Holds when
  • The fractures of the set are parallel and much longer than the diameter of the well. If the orientations scatter, we calculate the angle γ\gamma for each fracture separately and add up the weights 1/cos⁡γ1/\cos\gamma.
  • When the well is almost parallel to the fractures (γ\gamma close to 90°), the correction is unreliable: the fractures are crossed rarely, and one chance fracture gives a huge contribution. So the weight is limited, usually when the angle between the well and the plane of the fractures is less than 10–15°.
  • The correction cannot restore fractures that the well did not cross: a set parallel to the well may not be noticed at all.

A well at an angle to the normal of the set crosses 1/cos⁡γ1/\cos\gamma times fewer fractures than a line of the same length at right angles to them. Dividing by cos⁡γ\cos\gamma returns the true intensity. This gives a practical rule: to see steep fractures, we drill inclined and horizontal wells across their strike.

Worked exampleHow many fractures a well will cross
Given (assumed data)
Fracture set (from oriented core, whose compass direction is known, of a neighbouring well)
dip azimuth 110°, dip 78°
Vertical pilot well
over a 30 m interval of the layer the image showed 3 fractures of this set
Planned horizontal wells 800 m long
at azimuths 110°, 65° and 20°
Solution
  1. For a vertical well cos⁡γ=cos⁡78°=0.2079\cos\gamma = \cos 78° = 0.2079. The well crosses the fractures at only 12° to their plane, and the Terzaghi weight is 1/0.2079=4.811/0.2079 = 4.81.
  2. The linear intensity along the well is 3/30=0.1003/30 = 0.100 m⁻¹. The true one is 0.100/0.2079=0.4810.100/0.2079 = 0.481 m⁻¹, and the spacing of the set is s=1/0.481=2.08s = 1/0.481 = 2.08 m.
  3. A horizontal well at azimuth 110° runs in the direction of dip of the fractures: cos⁡γ=sin⁡78°=0.9781\cos\gamma = \sin 78° = 0.9781. It will cross about 800⋅0.9781⋅0.481=376800\cdot0.9781\cdot0.481 = 376 fractures.
  4. A well at azimuth 65°, at 45° to the strike: cos⁡γ=sin⁡78°⋅cos⁡45°=0.6917\cos\gamma = \sin 78°\cdot\cos 45° = 0.6917, about 266 fractures. A well at azimuth 20° runs along the strike: cos⁡γ=0.00\cos\gamma = 0.00, and it will hardly cross any fractures of this set.
Answer: The true spacing of the set is about 2.1 m. A horizontal well across the strike will cross about 376 fractures, a well at 45° to it about 266, and a well along the strike almost none.

What it means. Without the correction, the pilot well would understate the intensity almost five times: a spacing of 10 m instead of 2. A vertical well is a poor sampler of steep fractures. Three fractures in 30 m is too small a sample, and the true intensity may differ from the estimate several times over. An angle of 12° to the plane of the fractures is also near the limit beyond which the correction is not used. For a confident estimate we drill an inclined pilot well across the strike. A spacing of about 2 m agrees with the rule of the mechanical layer, if the strong dolomite beds in the layer are about 2 m thick. The well at azimuth 110° will open the most fractures and give the highest inflow. But it will also give the fastest water breakthrough, if the fractures link it to the water-bearing part of the layer. The data in this example are assumed for illustration.

Vertical and horizontal wells in a fractured layervertical well: 1 crossing in the layerhorizontal well across the fractures crosses every onefractures dip at 78°

Vertical section of a layer between its top and base, with grey shales above and below. The layer is cut by many parallel steep fractures tilted at 78 degrees. On the left a vertical well runs from top to bottom. In the layer it crosses one fracture, and the crossing point is marked. Lower down, the well bends, crosses the same fracture again, and runs horizontally to the right along the layer, crossing all the fractures one after another. The crossing points are marked along its whole length. Labels: the vertical well has one crossing in the layer, the horizontal well across the fractures crosses every one, the fractures dip at 78 degrees.

SchematicA vertical and a horizontal well in a layer with a steep fracture set (schematic, a section along the dip of the fractures). The vertical well is almost parallel to the fractures and crosses only a few of them. The horizontal well across the strike crosses every one.

In layered sequences, the joints of one set are most often limited by a mechanical layer, and their spacing grows with its thickness: in thin layers the fractures are dense, and in thick layers they are sparse. The explanation is mechanical. The stretching passes into the layer through its boundaries, and next to each fracture the stress in the layer is relieved over a distance of about the thickness of the layer. A new fracture between two neighbours can appear only if they are farther apart than about this thickness. As the strain (the amount of stretching) grows, the spacing falls to a limit, the state of saturation, in which the ratio of spacing to layer thickness is close to one (0.8–1.2 in observations). After that, the stretching goes into opening the existing fractures. The rule is not universal. In thick layers (thicker than one and a half to two metres), the spacing, according to a number of observations, hardly depends on thickness, and fractures of natural hydraulic fracturing do not follow it.

Fracture aperture and the cubic law

Fluid flows through a fracture between two walls. If the walls are smooth and parallel, an exact solution describes the flow. The flow rate through the fracture is then proportional to the cube of the aperture ww (the opening; in Russian-language literature the term “openness of a fracture” is more common). This is the cubic law. The permeability of the fracture itself is w2/12w^2/12 (permeability of a fracture). For a rock cut by a set, we multiply it by the share of volume taken by fractures, and get the sum of the cubes of the apertures.

The cubic law: flow rate through a fracture and permeability of a set
q=w3h12 μ⋅ΔpLf,k=∑iwi312 L,ϕf=∑iwiLq = \frac{w^3 h}{12\,\mu}\cdot\frac{\Delta p}{L_f},\qquad k = \frac{\sum_i w_i^3}{12\,L},\qquad \phi_f = \frac{\sum_i w_i}{L}
where:
  • qqflow rate through one fracture, m³/s
  • w, wiw,\ w_ihydraulic aperture of the fracture, m
  • hhsize of the fracture in its plane, across the flow, m
  • μ\muviscosity of the fluid, Pa·s
  • Δp, Lf\Delta p,\ L_fpressure drop over the length of the path along the fracture, Pa, m
  • kkpermeability of the rock along the fracture set, created by the fractures, m²
  • LLlength of the stretch across the set on which the fractures are counted, m
  • ϕf\phi_ffracture porosity, fraction

Units: SI; to convert permeability to millidarcies, divide by 0.9869⋅10−150.9869\cdot10^{-15} m²

Holds when
  • Laminar flow (smooth and orderly) between smooth parallel walls. For rough fractures with contact points, we use the hydraulic aperture, which is smaller than the average mechanical one.
  • The fractures of the set are parallel, and the flow is along them. Across the set the fractures hardly help, and the matrix decides the permeability in that direction.
  • The matrix adds its own permeability to that along the fractures: kalong≈km+kk_{\text{along}} \approx k_m + k.

A fracture twice as wide passes 8 times more fluid, and its own permeability is 4 times higher. So the conductivity of a set is decided by the few widest fractures, and a calculation from the average aperture greatly understates it. The porosity of fractures is tiny: usually not more than a few tenths of a per cent.

In other unit systems (1)
Practical form: apertures in micrometres
k [mD]=8.444⋅10−5 ∑wi3 [μm3]L [m]k\,[\text{mD}] = 8.444\cdot10^{-5}\,\frac{\sum w_i^3\,[\text{$\mu$m}^3]}{L\,[\text{m}]}

wiw_i in µm, LL in m, kk in mD.

8.444⋅10−5=10−18/(12⋅0.9869⋅10−15)8.444\cdot10^{-5} = 10^{-18}/(12\cdot 0.9869\cdot10^{-15}): conversion of µm³ to m³, division by 12 and conversion of m² to mD.

The aperture that we see on core or estimate from an image is the mechanical one: the average distance between the walls. Real walls are rough and touch each other at separate points, and the fluid goes round the contacts through winding channels. So the hydraulic aperture, the one that gives the measured flow rate when it is put into the cubic law, is smaller than the mechanical one. The rougher the walls and the larger the area of contacts, the smaller it is. The aperture on core is also distorted: when the core is brought up, the stresses are removed, and the fractures open wider than in the reservoir.

Worked examplePermeability of a fracture set, and the role of the widest fracture
Given (assumed data)
Horizontal interval of a well across the set
8.0 m; 10 fractures on it
Hydraulic apertures of the fractures (estimated from the image and tests)
30, 30, 40, 40, 50, 50, 60, 80, 100 and 250 µm
Matrix permeability
0.5 mD
Solution
  1. Average aperture: 730/10=73730/10 = 73 µm. Sum of cubes: ∑w3=17.785⋅106\sum w^3 = 17.785\cdot10^{6} µm³. Of this, the widest fracture accounts for 2503=15.625⋅106250^3 = 15.625\cdot10^{6} µm³, which is 87.9%.
  2. Permeability along the set: k=8.444⋅10−5⋅17.785⋅106/8.0=187.7k = 8.444\cdot10^{-5}\cdot 17.785\cdot10^{6}/8.0 = 187.7 mD. With the matrix, it is 188.2 mD. Across the set, the matrix permeability, 0.5 mD, remains.
  3. If we put the average aperture into all the fractures: k=8.444⋅10−5⋅10⋅733/8.0=41.1k = 8.444\cdot10^{-5}\cdot 10\cdot 73^3/8.0 = 41.1 mD. This is 4.57 times less.
  4. The hydraulic aperture of the set is the aperture that, the same for all fractures, gives the same sum of cubes: 17.785⋅106/103=121.2\sqrt[3]{17.785\cdot10^{6}/10} = 121.2 µm. Fracture porosity: 730⋅10−6/8.0=0.0091730\cdot10^{-6}/8.0 = 0.0091%.
Answer: Along the set, the rock conducts about 188 mD, hundreds of times more than the matrix. About 88% of this flow goes through one fracture out of ten.

What it means. The conductivity of fractured rock is decided by a few wide fractures. So an estimate from the average aperture or from the number of fractures understates it badly, and missing one wide fracture in the sample changes the result several times over. Fracture porosity is tiny, 0.0091%. The hydrocarbons in place in such a rock are in the matrix, and the fractures only conduct, as in Type II reservoirs (see the section on reservoir types). The anisotropy is huge: if there is no second set, the permeability along the fractures is several hundred times higher than across them. The calculation assumes smooth parallel walls. If we put mechanical apertures into it instead of hydraulic ones, the conductivity comes out too high. The data in this example are assumed for illustration.

Permeability along a fracture set by the cubic law103010030010000.01110010,000Hydraulic aperture w, µmPermeability k, mDspacing 0.2 mspacing 1 mspacing 5 m

Graph on logarithmic scales. The horizontal axis is the hydraulic aperture from 10 to 1000 micrometres. The vertical axis is the permeability from 0.01 to 100,000 millidarcies. There are three parallel straight lines with slope three: the upper one for a spacing of 0.2 metre, the middle one for 1 metre, the lower one for 5 metres. For example, at an aperture of 100 micrometres the permeability is about 420, 84 and 17 millidarcies respectively. The point of the example is at about 121 micrometres and 188 millidarcies, between the lines for spacings of 0.2 and 1 metre.

CalculatedPermeability of the rock along a set of identical parallel fractures against their hydraulic aperture, for spacings of 0.2, 1 and 5 m, by the cubic law (logarithmic scales). The point is the set of the example, replaced by fractures of the same hydraulic aperture at a spacing of 0.8 m.

Fractures under load: closure and critically stressed fractures

A fracture stays open because its uneven walls rest on each other at their bumps, and sometimes on minerals that have grown on the walls. As the effective normal stress grows, the bumps are crushed and the fracture closes. First this happens fast, then more and more slowly, because the contact area grows and the fracture becomes stiffer. This relation is often described by a hyperbolic law.

Closure of a fracture under load (hyperbolic law)
ΔV=σn′VmKniVm+σn′\Delta V = \frac{\sigma_n' V_m}{K_{ni} V_m + \sigma_n'}
where:
  • ΔV\Delta Vclosure of the fracture compared with the unloaded state, µm
  • σn′\sigma_n'effective normal stress on the fracture, MPa
  • VmV_mgreatest possible closure, µm
  • KniK_{ni}initial normal stiffness of the fracture, MPa/µm

Units: stresses in MPa, closure and VmV_m in µm, KniK_{ni} in MPa/µm

Holds when
  • An empirical relation. VmV_m and KniK_{ni} are fitted to tests on samples with a fracture, and they depend on the roughness and strength of the walls.
  • Closure is reversible only in part: under repeated load the fracture is stiffer, while slip along it can, on the contrary, open it through the bumps of the walls.
  • We take the change of the hydraulic aperture to equal the change of the mechanical one. For rough fractures, the hydraulic aperture can fall faster as the contacts grow.

At low stress the fracture is soft: closure grows almost in proportion to σn′/Kni\sigma_n'/K_{ni}. At high stress it approaches VmV_m, and the fracture almost stops closing. Because permeability is proportional to the cube of the aperture, even a small closure lowers it noticeably.

During development reservoir pressure falls, and the effective normal stress on the fractures grows. But not by the whole drop in pressure: the horizontal stress also falls with depletion (the fall of pressure as fluids are produced), by a share AA of the pressure drop (poroelastic coefficient). For vertical fractures at right angles to σh\sigma_h, the effective normal stress grows by (1−A) Δp(1 - A)\,\Delta p.

Worked exampleHow far the permeability of fractures falls when pressure drops
Given (assumed data)
Hydraulic aperture of the set at the start of development
121.2 µm (the example above) at an effective normal stress of 22 MPa
Closure parameters (assumed)
Vm=180V_m = 180 µm, Kni=0.08K_{ni} = 0.08 MPa/µm
Fall in reservoir pressure
18 MPa
Poroelastic coefficient of the horizontal stress
A=2/3A = 2/3 (Poisson's ratio 0.25, Biot coefficient 1)
Solution
  1. The horizontal stress falls by 23⋅18=12\tfrac23\cdot 18 = 12 MPa, and the effective normal stress on the fractures grows by 18−12=618 - 12 = 6 MPa, to 28 MPa.
  2. Closure at 22 MPa: 22⋅1800.08⋅180+22=108.8\dfrac{22\cdot 180}{0.08\cdot 180 + 22} = 108.8 µm. At 28 MPa: 28⋅18014.4+28=118.9\dfrac{28\cdot 180}{14.4 + 28} = 118.9 µm. The fractures close by a further 10.1 µm, to 111.1 µm.
  3. Permeability is proportional to the cube of the aperture: (111.1/121.2)3=0.77(111.1/121.2)^3 = 0.77, a loss of about 23%.
  4. If the horizontal stress did not fall, the effective stress would grow by the whole 18 MPa, to 40 MPa. The closure would be 132.4 µm, the aperture 121.2−(132.4−108.8)=97.6121.2 - (132.4 - 108.8) = 97.6 µm, and (97.6/121.2)3=0.52(97.6/121.2)^3 = 0.52, a loss of about 48%.
Answer: When pressure falls by 18 MPa, the permeability of the fracture set falls by about 23%. Without the fall of the horizontal stress, it falls almost by half, by 48%.

What it means. The aperture has fallen by only 8%, but permeability by 23%: the cube strengthens any closure. Expressed as the permeability modulus from the lesson on stresses in the reservoir, this is γ≈0.014\gamma \approx 0.014 MPa⁻¹, a value typical of fractured rocks (permeability under load). The model is simplified. The hyperbolic law describes mechanical closure, while the hydraulic aperture of rough fractures may fall faster, and then the loss is larger. Mineral bridges and slip along fractures, on the contrary, keep them open. The closure of fractures during depletion is only partly recovered when pressure rises. This is one of the arguments for maintaining pressure early in fractured layers. The data in this example are assumed for illustration.

Closure of fractures and permeability of the set under load01020304050600.00.51.01.52.0Effective normal stress σₙ′, MPaFraction of the value at 22 MPaaperturepermeabilityk = 0.77 at 28 MPa (σh falls)k = 0.52 at 40 MPa (σh unchanged)

Graph. The horizontal axis is the effective normal stress from 0 to 60 MPa. The vertical axis is the ratio to the value at 22 MPa, from 0 to 2.2. There are two falling curves that cross at one at 22 MPa. The aperture curve is gentle: about 1.9 at zero stress and about 0.7 at 60 MPa. The permeability curve is steeper. It enters the graph from above at about 10 MPa (at zero stress it would be about 6.8). It passes through 0.77 at 28 MPa, 0.52 at 40 MPa and about 0.34 at 60 MPa. Points are marked at 22, 28 and 40 MPa, and the values 0.77 and 0.52 are written below.

CalculatedAperture of the fractures and permeability of the set of the example, as fractions of their values at 22 MPa, against effective normal stress, by the hyperbolic law (Vm=180V_m = 180 µm, Kni=0.08K_{ni} = 0.08 MPa/µm). The points are the initial state and the states after a pressure fall of 18 MPa: with the horizontal stress falling (28 MPa) and without it (40 MPa).

Orientation decides which fractures are open. Fractures at right angles to the least stress σh\sigma_h are squeezed less than the others and are more open. So the permeability of a fractured layer is usually greatest along σH\sigma_H (fractures and the direction of stresses). Observations in deep wells suggest a second rule: the conductive fractures are often the critically stressed fractures. These are fractures on which the ratio of shear stress to effective normal stress is close to the coefficient of friction (how close a fracture is to slip). Small slips along them do not let the uneven walls close. The rule is not universal. In some rock masses, fractures inherited from earlier stages conduct, whatever the present stresses.

Fractured reservoirs: the role of fractures and the consequences for development

Almost every layer has fractures, but a reservoir is called fractured when open natural fractures clearly change its behaviour. A widely used classification splits such reservoirs into four types, by the share of storage and permeability that the fractures give.

Types of fractured reservoirs by the role of fractures
Type
I
Storage
mostly fractures and the vugs linked to them
Permeability
fractures
Behaviour
high initial rates and a fast decline; hydrocarbons in place are small
Type
II
Storage
matrix
Permeability
fractures
Behaviour
the matrix feeds the fractures; rates hold up if the matrix can give up its oil in time
Type
III
Storage
matrix
Permeability
matrix; fractures raise it
Behaviour
an ordinary reservoir with raised and directional permeability
Type
IV
Storage
matrix
Permeability
matrix; fractures add nothing
Behaviour
fractures create anisotropy, and healed ones create barriers

The boundaries between types are conventional, and one layer may change type across the area and as development goes on. In Russian-language practice, reservoirs are also divided by the type of voids: pore, fracture, vuggy and mixed, for example fracture–pore or pore–fracture.

  • Anisotropy. Permeability along the fractures can be tens and hundreds of times higher than across them. Wells interact more strongly along the fractures, the drainage area is elongated, and horizontal wells are better drilled across the fractures (direction of the well).
  • Early breakthrough of water and gas. Injected water, and water from the water-bearing zone, go along the fractures and leave the oil in the matrix. The water cut (the share of water in the produced fluid) rises early and fast. We place injection wells in rows along the strike of the main fracture set, so that water moves across the fractures. We do not put producers on the continuation of the fractures from the injectors (flood patterns).
  • Exchange between matrix and fractures. Oil leaves the matrix blocks for the fractures by capillary imbibition in water-wet rock (the rock soaks up water, like a sponge, and pushes the oil out), by the difference in densities, and by elastic expansion. The rate and completeness of recovery depend on the size of the blocks and on wettability. In models, such layers are described by two systems of pores, the matrix and the fractures (dual porosity).
  • Fall of permeability when pressure falls (the section above). So for fractured layers it is especially important when and how to maintain pressure.
  • Drilling. Open fractures take drilling mud, up to a complete loss of circulation, and when the pressure in the hole is lower than reservoir pressure, they give inflows. Mud losses are one of the most reliable signs of conducting fractures.
  • Well tests. Permeability from a test is often many times higher than from core. On the diagnostic plot (the plot used to recognise flow regimes), the pressure derivative of a layer with dual porosity makes a dip between two stretches of radial flow (pressure derivative).

In Kazakhstan, fractures are especially important in the subsalt carbonates of the Pre-Caspian Basin (subsalt tier). At Tengiz, most of the porosity is vugs, the primary porosity does not exceed 2–3%, and from the available data the permeability of the carbonates is controlled mainly by fractures. The fractures there are of several generations, from early ones that formed before the rocks were buried deep to late tectonic ones. Researchers rate differently which of them matter more for the present inflow. The general picture is this. The early fractures are steep and run parallel or perpendicular to the edge of the platform. There are most fractures in the strong framework rocks, the boundstones, of the outer platform and the upper slope, and they can be followed into the platform up to 2 km from the edge. In the inner part of the platform there are few fractures, and the inflow is set by the properties of the matrix. Some of the fractures have been widened by dissolution up to vugs. Such zones give both strong mud losses in drilling and highly productive wells.

At Kashagan, three generations of fractures have been identified from core. The first are early ones, linked to the instability of the edge of the buildup, and more frequent along it. The second are burial fractures, linked to stylolites. The third are thin fractures that formed while the accumulation was filling, when the fluid pressure had risen enough to break the seal. The large-scale permeability of the edge of the buildup is linked with the early fractures. Mud losses are used as an indirect measure of the conductivity of fractures. On an image, a high or low resistivity of a fracture does not by itself say whether it conducts.

Another example is the Triassic deposits of South Mangyshlak. The accumulations in the Middle Triassic carbonates are linked to fractured dolomites with vuggy porosity. The matrix of the dolomites is almost impermeable, with porosity of not more than 3–4%, and the voids were dissolved out (leached) along fracture sets that formed after the Early Miocene. At the Oymasha field the oil accumulation lies in fractured and weathered granites of the basement (the old crystalline rock beneath the sedimentary layers), productive down to 300 m below the top of the basement. The voids in such a reservoir were made mainly by fractures and weathering. In neighbouring areas the accumulations are controlled less by uplifts than by zones of fracturing and leaching that strike north-east and north-west. The north-east zones are linked with faults that have a considerable strike-slip component. So we can expect such accumulations outside uplifts too (strike-slip zones).

Fractures in core, on images and in other data

Core is the only source where we can touch a fracture: we see whether it is open or healed, what fills it, and whether its walls have slickenlines or a plumose pattern. But core is small, and it is not always oriented. We restore its orientation from the image of the same interval, or we take oriented core (coring). It is important to tell natural fractures from induced ones. Drilling and bringing up the core create their own. These are discs into which the core splits when it is drilled under high stress (core discing), fractures along the core axis, and breaks made when the core is pulled out and laid in boxes. A natural fracture is given away by mineral fill, traces of slip, altered walls, and an orientation that agrees with other fractures of the set.

Borehole images (borehole image log) give the orientation of every fracture from its sinusoid (attitude from an image) and a continuous picture along the whole well. On electrical images in water-based mud, an open fracture filled with mud filtrate (the liquid from the mud that has entered the rock) conducts current and looks like a dark sinusoid. A fracture healed with calcite, quartz or anhydrite looks like a light, high-resistivity one. Fractures with clay or pyrite fill, and thin shale beds, can be dark too. So we tell a fracture from a bed by its angle and direction of dip relative to the bedding, and by whether the sinusoid breaks off. In oil-based mud, open and healed fractures are equally high in resistivity on ordinary electrical images, and they are harder to tell apart. On acoustic images, open fractures show up by a weaker reflected signal.

Drilling itself creates fractures and breakouts on the wall of the hole, and this is valuable information about stresses (direction of stresses). In a vertical well, thin tensile fractures induced by drilling run along the hole in pairs on opposite sides, at the azimuth of σH\sigma_H. Breakouts (pieces of the wall that have broken off, seen as wide smeared bands) lie 90° from them, at the azimuth of σh\sigma_h. Natural fractures parallel to σH\sigma_H are more open than the rest.

Fractures, breakouts and drilling-induced fractures on an image of the borehole wallNESWN0 m0.5 m1 m1.5 mopen fracturehealed fracturebeddingdashes at 20 and 200°: drilling-induced fractures (direction of σH)smeared bands at 110 and 290°: breakouts (direction of σh)

An unrolled view of the wall of the hole. The horizontal axis is azimuth from 0 to 360 degrees, marked N, E, S, W, N. The vertical axis is depth downwards, about 1.6 metres. Several almost horizontal gentle sinusoids are bedding. One dark sinusoid of large swing, about a metre, with its lowest point at azimuth 110 degrees, is an open fracture. One light sinusoid of similar shape, with its lowest point near 250 degrees, is a healed fracture. At azimuths 20 and 200 degrees there are short dark vertical dashes in echelon: fractures induced by drilling. At azimuths 110 and 290 degrees there are wide smeared dark vertical bands: breakouts.

SchematicAn electrical image of a vertical well in water-based mud (schematic). Gentle sinusoids are bedding. The dark steep sinusoid is the open fracture of the example (dip azimuth 110°, dip 78°, lowest point at azimuth 110°). The light one is a healed fracture. Short vertical dark dashes at azimuths 20 and 200° are drilling-induced fractures (the direction of σH\sigma_H). Wide smeared bands at azimuths 110 and 290° are breakouts (the direction of σh\sigma_h).

Other observations reveal fractures too. One sign is mud losses and drops of the tool. Another sign is a flow profile from production logging with a spinner flowmeter (a tool that measures the flow in the well), when the inflow comes from separate narrow intervals. A third is permeability from a well test that is higher than the core permeability. A fourth is shear-wave anisotropy in dipole sonic logging (Sonic log). Between wells, seismic signs help: curvature of horizons, breaks in the continuity of reflections, and azimuthal anisotropy. No single sign proves that fractures conduct. Mud losses and inflow are checked, and images are calibrated against core and tests.

Order of studying fracturing
  1. 1
    Description

    From core and images, pick out the sets, their orientation, fill and aperture. Discard the fractures induced by drilling.

  2. 2
    Sampling correction

    Recalculate the intensity with the Terzaghi correction and estimate which sets the well may have missed.

  3. 3
    Origin

    Link the sets to stresses, folds, faults and mechanical layers. This is the basis of the prediction between wells.

  4. 4
    Conductivity

    Estimate it by the cubic law and check it against mud losses, the flow profile and well tests.

  5. 5
    Model and project

    Find the type of reservoir, the anisotropy and the sensitivity to pressure. Choose the direction of wells, the placement of injection wells and the model of the layer.

Terms and symbols

Joint
A fracture whose walls have moved apart at right angles to its surface. It is at right angles to σ3\sigma_3 (extension fracture).
Shear fracture
A fracture with slip of its walls along its surface. It forms in conjugate pairs at about 30° to σ1\sigma_1.
Hybrid fracture
A fracture with both opening and shear, at a small angle to σ1\sigma_1. It forms at a differential stress from 4T4T to 42 T4\sqrt2\,T.
Vein
A fracture filled with minerals that settled out of solution.
Fracture set
Fractures of one orientation and origin. On a stereographic projection it is a cluster of poles.
Mechanical layer
A layer inside which fractures are confined and at whose boundaries they end.
Linear intensity P10P_{10} and spacing ss
Linear intensity is the number of fractures per metre of a line of observation. Spacing is one divided by it, along the normal to the set.
Terzaghi correction
Recalculation of the number of fractures crossed by a well into the true intensity, by dividing by cos⁡γ\cos\gamma.
Mechanical and hydraulic aperture
The average distance between the walls of a fracture, and the aperture that gives the measured flow rate by the cubic law. The second is smaller.
Cubic law
The flow rate through a fracture is proportional to the cube of its aperture.
Critically stressed fracture
A fracture that is close to slip under the present stresses. Such fractures are often conductive.
Fractured reservoir of Type I–IV
A classification by the role of fractures: from fractures that give both storage and permeability to fractures that create only anisotropy.

Common mistakes and misconceptions

Wrong

“A vertical well met no fractures, so the layer is not fractured.”

In fact

A vertical well rarely crosses a steep set. With a spacing of 2 m and a hole diameter of 0.2 m, the chance of meeting at least one fracture is about 10%. We need inclined wells, the Terzaghi correction and other signs, such as mud losses and well tests.

Wrong

“The permeability of fractured rock can be estimated from the average aperture and the number of fractures.”

In fact

The flow rate is proportional to the cube of the aperture, and the widest fractures decide the conductivity. An estimate from the average aperture understates it several times over.

Wrong

“The more fractures on the image, the better the reservoir.”

In fact

What matters are open fractures that connect with each other. Healed fractures are barriers, and shale beds also give a dark sinusoid. We confirm conductivity from mud losses, inflow and tests.

Wrong

“The fractures in outcrops show how many there are in the layer at depth.”

In fact

Near the surface, unloading and weathering fractures are added. From outcrops we take the geometry of the sets, and we transfer the density to the layer with care.

Wrong

“The permeability of fractures does not change during development.”

In fact

As pressure falls, fractures close, and permeability falls as the cube of the aperture. In the example of the lesson, a pressure drop of 18 MPa takes away almost a quarter of it, and without the fall of the horizontal stress almost a half.

In depth: advanced detail

Where the limits 4T4T and 42 T4\sqrt2\,T come from

Write the Griffith envelope as σn′=τ2/(4T)−T\sigma_n' = \tau^2/(4T) - T. The square of the distance from the centre (c′,0)(c', 0) of the Mohr circle to the point of the envelope with τ2=u\tau^2 = u is (u/(4T)−T−c′)2+u(u/(4T) - T - c')^2 + u. Setting the derivative with respect to uu to zero, we get u=4T(c′−T)u = 4T(c' - T). So the point of the envelope nearest to the centre lies at σn′=c′−2T\sigma_n' = c' - 2T, and it is 2Tc′2\sqrt{T c'} from the centre. This is so as long as c′≥Tc' \ge T. For smaller c′c', the nearest point is the tip of the parabola (−T,0)(-T, 0).

A circle of radius rr touches the envelope when its radius equals this distance. If r≤2Tr \le 2T, the contact is at the tip: σ3′=c′−r=−T\sigma_3' = c' - r = -T. This is a joint at right angles to σ3\sigma_3. From this, σ1−σ3=2r≤4T\sigma_1 - \sigma_3 = 2r \le 4T. If r>2Tr > 2T, the contact comes at c′=r2/(4T)c' = r^2/(4T), at the point σn′=c′−2T\sigma_n' = c' - 2T. While this point lies in the tensile region (c′<2Tc' < 2T, that is r<22 Tr < 2\sqrt2\,T), the fracture is hybrid. The angle θ\theta between σ1\sigma_1 and the normal to the fracture is found from cos⁡2θ=−2T/r\cos 2\theta = -2T/r, and it changes from 90° to 67.5°. So the fracture itself deviates from σ1\sigma_1 by not more than 22.5°. When r>22 Tr > 2\sqrt2\,T, that is σ1−σ3>42 T\sigma_1 - \sigma_3 > 4\sqrt2\,T, the contact is in the compressive part of the envelope, and shear fractures form.

Where the cubic law comes from

Consider steady laminar flow of a viscous liquid between two parallel walls a distance ww apart. The xx axis is along the flow, and yy is across the fracture, from its middle. The balance of a layer of liquid of thickness  dy\dd y gives μ  d2u/ dy2= dp/ dx\mu\,\dd^2 u/\dd y^2 = \dd p/\dd x, and the sticking of the liquid to the walls gives u=0u = 0 at y=±w/2y = \pm w/2. The solution is a parabola: u(y)=12μ(− dp dx)(w24−y2)u(y) = \dfrac{1}{2\mu}\left(-\dfrac{\dd p}{\dd x}\right)\left(\dfrac{w^2}{4} - y^2\right).

The integral of the velocity across the width gives the flow rate per unit height of the fracture: q/h=w312μ(− dp dx)q/h = \dfrac{w^3}{12\mu}\left(-\dfrac{\dd p}{\dd x}\right). This is the cubic law. The average velocity in the fracture is w212μ(− dp dx)\dfrac{w^2}{12\mu}\left(-\dfrac{\dd p}{\dd x}\right). Comparing it with Darcy's law, we get the permeability of the fracture itself, kf=w2/12k_f = w^2/12. Suppose fractures with apertures wiw_i lie on a stretch of length LL across the set. Their total flow rate per unit height, referred to the whole cross-section LL, gives k=∑wi3/(12L)k = \sum w_i^3/(12L).

Connections to other topics

Check yourself

  1. 1

    In an area with an extension regime (σ1\sigma_1 vertical, σH\sigma_H pointing from north to south), joints have formed at depth. How are they oriented?

    1. A.Inclined, at 30° to the vertical
    2. B.Vertical, striking west–east
    3. C.Vertical, striking north–south
    4. D.Horizontal
    Show the answer

    Correct answer: C. Vertical, striking north–south

    A joint is at right angles to the least stress. Here it is σh\sigma_h, which points from west to east. A joint contains σ1\sigma_1 and σ2\sigma_2: the vertical and the north–south direction. So the joints are vertical and strike north–south. At depth they need a pore pressure that exceeds σh\sigma_h by the tensile strength of the rock.

  2. 2

    A vertical well crossed 4 fractures of one set over a 40 m interval; the fractures dip at 80°. What is the true spacing of the set, and how many fractures will a 500 m horizontal well cross, drilled in the direction of their dip?

    Show the answer

    For a vertical well cos⁡γ=cos⁡80°≈0.1736\cos\gamma = \cos 80° \approx 0.1736. The linear intensity along the well is 0.1 m⁻¹, and the true one is 0.1/0.1736≈0.5760.1/0.1736 \approx 0.576 m⁻¹, a spacing of about 1.74 m. A horizontal well in the direction of dip has cos⁡γ=sin⁡80°≈0.985\cos\gamma = \sin 80° \approx 0.985. It will cross about 500⋅0.985⋅0.576≈284500\cdot0.985\cdot0.576 \approx 284 fractures. The estimate is uncertain: the vertical well ran at only 10° to the fractures, at the limit of the correction, and a sample of four fractures is small.

  3. 3

    Two fractures have a hydraulic aperture of 50 and 100 µm. How many times more fluid does the second pass at the same pressure gradient?

    1. A.8 times
    2. B.4 times
    3. C.16 times
    4. D.2 times
    Show the answer

    Correct answer: A. 8 times

    The flow rate through a fracture is proportional to the cube of the aperture: (100/50)3=8(100/50)^3 = 8. The permeability of the fracture itself, w2/12w^2/12, is 4 times higher, and the area of its cross-section is 2 times larger.

  4. 4

    A limestone matrix has porosity of 12% and permeability of 0.3 mD. Wells give commercial rates only where they open fractures, and well tests show permeability hundreds of times above that of core. What type of reservoir is this by the role of fractures?

    1. A.II: storage is in the matrix, permeability is given by fractures
    2. B.IV: fractures do not affect permeability
    3. C.III: fractures only raise a permeability that is already high enough
    4. D.I: fractures give both storage and permeability
    Show the answer

    Correct answer: A. II: storage is in the matrix, permeability is given by fractures

    With a porosity of 12%, the hydrocarbons in place are almost all in the matrix. Its permeability of 0.3 mD is too low for commercial rates, so the fractures conduct. This is Type II. For it, the exchange between matrix and fractures and the size of the blocks are important, and the risk is a fast water breakthrough along the fractures.

  5. 5

    Why, in a fractured layer, are injection wells placed in rows along the strike of the main fracture set, and not on the continuation of the fractures from the producers?

    Show the answer

    Water goes first of all along the fractures. If a producer stands on the continuation of the fractures from an injector, water quickly breaks through to it and leaves the oil in the matrix. A row of injectors along the strike of the fractures works like a line source (a long, continuous source of water). The fractures between them quickly fill with water, and the front moves across the fractures towards the rows of producers, sweeping the matrix blocks between them. Such a layout improves sweep, although the rate at which oil leaves the blocks is still set by imbibition and block size.

  6. 6

    On an electrical image of a well drilled with water-based mud, there are two steep sinusoids of the same direction: a dark continuous one and a light one. Which of them is more likely an open fracture?

    1. A.Open fractures cannot be seen on an image
    2. B.Both are equally likely
    3. C.The dark one: it is filled with conductive mud filtrate
    4. D.The light one: an open fracture has high resistivity
    Show the answer

    Correct answer: C. The dark one: it is filled with conductive mud filtrate

    In water-based mud an open fracture is filled with conductive filtrate and looks dark. A light, high-resistivity sinusoid is more likely a fracture healed with calcite, quartz or anhydrite. But fractures with clay fill can be dark too, so we check the conclusion against core, mud losses and inflow. In oil-based mud both would look light.

Updated: 15 October 2026