Rock mechanics for hydraulic fracturing

Quick refresher

The stresses in the reservoir are described by three principal stresses: the vertical $\sigma_v$ (the weight of the rocks) and the horizontal $\sigma_H$ and $\sigma_h$ . A fracture opens against the smallest of them, usually $\sigma_h$ , and extends along $\sigma_H$ .
The minimum horizontal stress is made up of a share of the effective vertical stress (the part of the weight of the rock carried by the rock skeleton, not by the pore fluid), a part of the pore pressure, and a tectonic addition proportional to the stiffness of the layer. So it is different in different layers, and we calibrate the stress model against the fracture closure pressure.
The wall of the well concentrates stress. The hoop stress on it (the stress along the circumference) changes around the circumference from $3\sigma_h - \sigma_H - p_w$ to $3\sigma_H - \sigma_h - p_w$ . The breakdown pressure of the wall depends on the tensile strength and on whether fluid penetrates the rock. So $\sigma_h$ is measured not by the breakdown pressure, but by the closure pressure.
After injection stops, the pressure falls from the instantaneous shut-in pressure (ISIP) to the closure pressure. The difference is the net pressure at the end of injection. We find the closure from the point where the record leaves a straight line on a plot against the $G$ function.
The height of a fracture is limited by the stress contrast between the reservoir and the barriers. While the net pressure is clearly smaller than the contrast, the fracture hardly leaves the reservoir. When the net pressure approaches the contrast, the fracture grows in height without restraint. Depletion of the reservoir lowers $\sigma_h$ in it and improves containment (keeping the fracture inside the reservoir).

Key relationships

\sigma_h = \dfrac{\nu}{1-\nu}(\sigma_v - \alpha p_p) + \alpha p_p + \dfrac{E\,(\varepsilon_h + \nu\varepsilon_H)}{1-\nu^2}

poroelastic model with tectonic strains

\sigma_\theta = \sigma_H + \sigma_h - 2(\sigma_H - \sigma_h)\cos 2\theta - p_w

hoop stress at the wall of a vertical well; $\theta$ is measured from the direction of $\sigma_H$

p_b = \dfrac{3\sigma_h - \sigma_H + T - A\,p_p}{2 - A},\quad A = \dfrac{\alpha(1-2\nu)}{1-\nu}

breakdown pressure of the wall when fluid penetrates the rock

p_{ISIP} = p_c + p_{net}

instantaneous shut-in pressure, closure pressure and net pressure

h_f = \dfrac{h}{\cos\left(\pi p_{net}/(2\Delta\sigma)\right)}

equilibrium height of a fracture in a reservoir of thickness $h$ between barriers with a stress contrast $\Delta\sigma$

Common pitfalls

Taking the breakdown pressure of the wall or the fracture extension pressure as the minimum stress. Both are above the closure pressure.
Calculating $\sigma_h$ from a single Poisson's ratio, without a tectonic addition and without calibration against a measurement.
Thinking that shale will always hold a fracture in. Under tectonic compression a stiff sandstone carries more stress, and a fracture breaks through a thin barrier.
Reading wellhead pressures without converting them to the bottom of the well, that is, without the hydrostatic head, friction in the pipes, and losses in the perforations and near the wellbore.
Forgetting that depletion and injection change the stresses. A profile measured at the start of development may be different years later.

After this lesson you can

describe the stress state of a reservoir by three principal stresses, estimate the minimum horizontal stress with a poroelastic model that has a Biot coefficient and a tectonic strain, and calibrate it against a measurement
calculate the stress concentration at the wall of the well, and the breakdown pressure of the wall without and with fluid penetrating the rock
find the fracture closure pressure from the pressure decline after a calibration injection, and the fracture extension pressure (the pressure at which the fracture keeps growing; also called the propagation pressure) from a step-rate test (injection at steadily higher rates)
explain what the net pressure is, how it is related to the width of the fracture, and what its behaviour during injection tells us
estimate the growth of a fracture in height for a given stress contrast, and explain how depletion (the fall of reservoir pressure), the thickness of the barriers and the stiffness of the layers affect it
predict the orientation of a fracture in vertical, deviated and horizontal wells, and explain why the orientation of the perforations matters

Three principal stresses in the reservoir

The state of stress of a rock at any point is described by three principal stresses: the normal stresses on three mutually perpendicular planes on which there are no shear stresses. In a quiet sedimentary basin one of them is vertical, $\sigma_v$ , and the other two are horizontal: the largest $\sigma_H$ and the smallest $\sigma_h$ . Why a fracture opens against the smallest stress, and how the three stresses are used to tell the tectonic regimes apart, is explained in the lesson Fracture gradient and leak-off testing. Here we look at where their values come from and how they are measured.

Vertical stress $\sigma_v$: The weight of the overlying rocks, that is, the overburden pressure. We find it by integrating the density log (a record of rock density against depth) from the surface (Pore pressure and overburden pressure). On average it is 22–25 kPa per metre of depth, and less for young, uncompacted sediments and under the deep sea. Of the three stresses, it is the one we know most reliably.
Minimum horizontal stress $\sigma_h$: The main quantity for hydraulic fracturing. We measure it: it is the closure pressure of the fracture in a calibration injection and in an extended leak-off test (a pressure test of the borehole wall, carried on far enough to show the closing of the fracture). Between measurements we carry it along the section with a model, described in the next section.
Maximum horizontal stress $\sigma_H$: It cannot be measured directly. We estimate it from damage to the wall of the well (breakouts and tensile fractures; see the section on the stresses at the wall), from the fracture reopening pressure, and from models. Its upper limit is set by the strength of faults (advanced section).

We learn the direction of the stresses from the well itself. Breakouts are places where the rock has failed under compression. They elongate the cross-section of the hole in the direction of $\sigma_h$ . The thin tensile fractures that drilling induces, on the contrary, are oriented along $\sigma_H$ . Both can be seen on borehole images and on caliper logs (measurements of the hole diameter) in two mutually perpendicular planes. The direction of $\sigma_H$ is also often shown by shear-wave anisotropy (shear waves travel at different speeds in different directions) in dipole sonic logging (Sonic log). Within a field the direction is usually consistent, so hydraulic fractures in neighbouring wells are almost parallel.

Where stresses behave unusually

Rock salt flows like a very viscous liquid, so the stresses in it are almost the same in all directions and close to the overburden pressure. A salt seal stops the growth of a fracture upwards: the stress in it is many megapascals higher than in the reservoir below it.
Erosion. If part of the section has been eroded away, the vertical stress has decreased, but the horizontal stresses have largely stayed. The rock “remembers” its former load. Near the surface the smallest stress may be the vertical one, and then fractures are horizontal.
Faults, folds and salt domes rotate the principal stresses and change their size. In regions of active mountain building the principal stresses can be tilted.
Field development changes the stresses. Depletion (the fall of reservoir pressure as fluid is produced) lowers the horizontal stresses in the reservoir. Water injection raises them, because the pressure rises. And cooling of the rock near injection wells lowers them.

Minimum horizontal stress: the poroelastic model

Where does the horizontal stress in the rock come from? The weight of the overlying layers squeezes a layer from above, and the layer tries to expand sideways, like a rubber eraser under a finger. The neighbouring rocks do not let it, and they push on it from the sides. In its simplest form this idea is already used in the Eaton method (estimating the fracture pressure): the horizontal effective stress is $\nu/(1-\nu)$ of the vertical one. For hydraulic fracturing this is not enough, and we refine the model with two additions.

Biot coefficient $\alpha$ . The pore pressure does not take the whole load off the rock skeleton (the framework of grains and cement), only the part $\alpha p_p$ . In a cemented rock part of the load passes through the cement, which the fluid pressure does not act on. For loose rocks $\alpha$ is close to one, and for tight cemented sandstones it is 0.6–0.9 (where the Biot coefficient comes from).
Tectonic strain. The Earth's crust is not only loaded by the weight of the rocks. It is also squeezed or stretched horizontally by the movement of plates. We describe this strain by two small numbers, $\varepsilon_h$ and $\varepsilon_H$ : the shortening along the directions of $\sigma_h$ and $\sigma_H$ . The same strain creates more stress in a stiff layer than in a soft one. The addition is proportional to Young's modulus (the stiffness of the rock).

Horizontal stresses of a poroelastic layer with tectonic strains

\sigma_h = \frac{\nu}{1-\nu}\left(\sigma_v - \alpha p_p\right) + \alpha p_p + \frac{E}{1-\nu^2}\left(\varepsilon_h + \nu\,\varepsilon_H\right)

where:

$\sigma_h$ minimum horizontal stress (total), MPa
$\sigma_v$ vertical stress (overburden pressure), MPa
$p_p$ pore pressure, MPa
$\nu$ static Poisson's ratio of the layer, —
$E$ static Young's modulus of the layer, MPa
$\alpha$ Biot coefficient, from 0 to 1, —
$\varepsilon_h,\ \varepsilon_H$ tectonic strains (shortenings) along the directions of $\sigma_h$ and $\sigma_H$ ; positive in compression, —

Units: any identical units of stress and modulus (MPa or psi); the strains are dimensionless

Holds when

The layer is linearly elastic and isotropic, and wide compared with its thickness. The vertical stress is a principal stress.
The strains are measured from the state in which the horizontal stresses were created only by the weight of the rocks. $\varepsilon_h$ and $\varepsilon_H$ are the same for all layers of the section, but the stresses they cause are different in each layer.
The moduli are static: measured by slowly loading a core sample (a cylinder of rock taken from the well), or converted from dynamic moduli (found from sound speeds) with a relationship that has been checked on core (Sonic log).
$\sigma_H$ is found with the same formula by swapping $\varepsilon_h$ and $\varepsilon_H$ . At $\varepsilon_h = \varepsilon_H = 0$ the two horizontal stresses are equal, and at $\alpha = 1$ we get the Eaton formula.

The first two terms are the contribution of the weight of the rocks: the share $\nu/(1-\nu)$ of the effective vertical stress, plus the part of the pore pressure that is carried by the fluid. The third term is the tectonic addition. $E/(1-\nu^2)$ shows how much stress a unit strain creates in the layer when the vertical stress is held constant (the layer is free to deform vertically). We cannot measure the strains directly. We choose them so that the model matches the measured closure pressures, and we take the ratio $\varepsilon_H/\varepsilon_h$ from regional data or from estimates of $\sigma_H$ .

Change of the horizontal stress when the pore pressure changes

\Delta\sigma_h = A\,\Delta p_p,\qquad A = \frac{\alpha\,(1 - 2\nu)}{1 - \nu}

where:

$\Delta\sigma_h$ change of the minimum horizontal stress, MPa
$\Delta p_p$ change of the pore pressure (negative in depletion), MPa
$A$ poroelastic stress coefficient; the literature also uses a value half as large, $\eta = A/2$ , —
$\alpha,\ \nu$ Biot coefficient and Poisson's ratio of the reservoir, —

Units: any identical units of pressure

Holds when

The pressure changes over the whole wide reservoir, and the reservoir deforms only vertically (uniaxial strain).
The rock is elastic, and the overburden pressure does not change during development.
The ratios $\Delta\sigma_h/\Delta p_p$ measured in fields are scattered, and they can differ noticeably from the calculated one. We check the ratio by repeated measurements of the closure pressure.

At $\alpha = 1$ this is $(1-2\nu)/(1-\nu) = 1 - K$ , where $K = \nu/(1-\nu)$ is the lateral earth pressure coefficient (the ratio of the horizontal to the vertical effective stress). This is the same factor with which, in the lesson Fracture gradient and leak-off testing, the fracture pressure falls with depletion. Usually $A$ is 0.5–0.8: for every 10 MPa of pressure drop, the $\sigma_h$ of the reservoir falls by 5–8 MPa. The shales above and below it hardly deplete, and the stress contrast grows.

CalculatedThe stresses in the interval of the example below: a sandstone 24 m thick between shales. The minimum horizontal stress by the poroelastic model with a calibrated tectonic strain (solid line) and without it (dashed line), and the overburden and pore pressures. The vertical dashed line is the pressure in the fracture at the end of the calibration injection (44.4 MPa).

Worked exampleCalibrating the stress model against the closure pressure

Given (assumed data)

Middle of the reservoir: 2900 m; $\sigma_v = 65.54$ MPa (average gradient 22.6 kPa/m); $p_p = 30.5$ MPa
Sandstone (static properties): $\nu = 0.22$ ; $E = 22$ GPa; $\alpha = 0.80$
Shales above and below the reservoir: $\nu = 0.28$ ; $E = 12$ GPa; $\alpha = 1.0$
Closure pressure in the sandstone from the calibration injection: 41.8 MPa (the example in the section on injection)
Ratio $\varepsilon_H/\varepsilon_h$ from regional data: 3

Solution

Without tectonics: $\sigma_h = 0.2821\cdot(65.54 - 0.8\cdot30.5) + 0.8\cdot30.5 = 0.2821\cdot41.14 + 24.4 = 36.0$ MPa, which is 5.8 MPa less than the measured value.
The tectonic addition $E'(\varepsilon_h + \nu\varepsilon_H) = E'\varepsilon_h(1 + 3\nu)$ makes up the shortfall, where $E' = E/(1-\nu^2) = 23{,}119$ MPa. From this $\varepsilon_h = 5.8/(23{,}119\cdot1.66) = 0.151\cdot10^{-3}$ and $\varepsilon_H = 0.453\cdot10^{-3}$ .
The maximum horizontal stress: $\sigma_H = 36.0 + 23{,}119\cdot(0.453 + 0.22\cdot0.151)\cdot10^{-3} = 47.2$ MPa. The regime is extensional (normal faulting): $\sigma_v > \sigma_H > \sigma_h$ , so the fracture is vertical and extends along $\sigma_H$ .
Shale without tectonics: $\sigma_h = 0.3889\cdot(65.54 - 30.5) + 30.5 = 44.1$ MPa. The tectonic addition in it is $13{,}021\cdot(0.151 + 0.28\cdot0.453)\cdot10^{-3} = 3.62$ MPa, so the total is 47.7 MPa.
The stress contrast between shale and sandstone: $47.7 - 41.8 = 5.9$ MPa. Without tectonics it would have been $44.1 - 36.0 = 8.1$ MPa.

Answer: A tectonic shortening of about

0.151\cdot10^{-3}

along

\sigma_h

explains the measurement.

\sigma_H \approx 47.2

MPa, and the stress contrast between the shale and the sandstone is about 5.9 MPa.

What it means. A calculation with a single Poisson's ratio would have underestimated the $\sigma_h$ of the sandstone by almost 6 MPa. It would have overstated the contrast with the shale by 2.2 MPa, which means that it would have overrated how well the shale keeps the fracture from growing in height. The stiff sandstone took more of the tectonic compression than the soft shale. We cannot find two strains from one measurement. The ratio 3 is taken from regional data, and $\sigma_H$ should be checked against the breakouts in the wellbore. The data in this example are assumed for illustration.

Stresses at the wall of the well and the breakdown pressure

When we drill a well, we remove a cylinder from loaded rock. The load that it carried is redistributed onto the rock around it. The stresses at the wall “flow round” the hole and become concentrated, like the streamlines of water around a bridge pier. For an elastic rock and a vertical well, this picture is described by the Kirsch solution. We are interested first of all in the hoop (tangential) stress on the wall itself. It decides where the rock will break in compression and where it will split in tension.

Stresses on the wall of a vertical well (the Kirsch solution)

\sigma_\theta = \sigma_H + \sigma_h - 2\,(\sigma_H - \sigma_h)\cos 2\theta - p_w,\qquad \sigma_r = p_w

where:

$\sigma_\theta$ hoop (tangential) total stress on the wall, MPa
$\sigma_r$ radial stress on the wall, equal to the pressure in the well, MPa
$\theta$ angle on the wall, measured from the direction of $\sigma_H$ , rad
$\sigma_H,\ \sigma_h$ maximum and minimum horizontal stresses far from the well, MPa
$p_w$ pressure in the well at this depth, MPa

Units: any identical units of stress; compression is positive

Holds when

The rock is linearly elastic, homogeneous and isotropic. The well is vertical, and the vertical stress is a principal stress.
The pore pressure at the wall is the same as the formation pressure. Either a filter cake (a thin layer of solids deposited on the wall) separates the wall from the well, or the pressure changes so fast that the fluid has no time to enter the rock (the effect of filtration is below).
The temperature of the wall equals the formation temperature.
Plane strain along the axis of the well (the rock does not stretch along it). With distance from the wall the disturbance dies out quickly, and already a few radii from the axis the stresses hardly differ from $\sigma_H$ and $\sigma_h$ (advanced section).

The hoop stress is largest at $\theta = 90°$ , at the points of the wall that lie in the direction of $\sigma_h$ . There it is $3\sigma_H - \sigma_h - p_w$ , and there the rock fails in compression and forms breakouts. It is smallest at $\theta = 0$ and $180°$ , in the direction of $\sigma_H$ : $3\sigma_h - \sigma_H - p_w$ . A rise of the pressure in the well decreases the hoop stress everywhere. When the stress in the direction of $\sigma_H$ becomes tensile and exceeds the tensile strength, the wall splits. So the fracture starts in the plane in which it is going to grow, along $\sigma_H$ .

CalculatedEffective hoop stress

\sigma_\theta - p_p

on the wall of a well in the sandstone of the examples at three pressures in the well: equal to the pore pressure, equal to the closure pressure, and equal to the breakdown pressure without penetration of fluid. The dashed line is the tensile strength

-T

. At the breakdown pressure the curve touches it in the direction of

\sigma_H

The breakdown pressure of the wall without penetration of fluid into the rock, $p_b = 3\sigma_h - \sigma_H - p_p + T$ , was derived from this solution in the lesson on the fracture gradient. But in hydraulic fracturing the fluid often enters the pores at the wall. This happens through perforation tunnels (the channels that perforating guns shoot through the casing into the rock), where there is no filter cake, and when a low-viscosity fluid is injected slowly into a permeable rock. Then the pore pressure at the wall rises together with the pressure in the well, and the rock splits earlier: the pressure in the pores pushes the grains apart from inside. A poroelastic effect offsets this in part. The rock at the wall, “inflated” by the pressure, tries to expand, its neighbours prevent it, and the hoop compression on the wall grows by $A\,(p_w - p_p)$ , where $A$ is the same poroelastic coefficient as in the previous section.

Breakdown pressure of the wall when fluid penetrates the rock

p_b = \frac{3\sigma_h - \sigma_H + T - A\,p_p}{2 - A},\qquad A = \frac{\alpha\,(1 - 2\nu)}{1 - \nu}

where:

$p_b$ breakdown pressure of the wall at the bottom of the well, MPa
$\sigma_h,\ \sigma_H$ minimum and maximum horizontal stresses, MPa
$p_p$ formation (pore) pressure before injection, MPa
$T$ tensile strength of the rock; zero for a wall that already has fractures, and after the first breakdown, MPa
$A$ poroelastic stress coefficient, from 0 to 1, —

Units: any identical units of pressure

Holds when

The same as for the Kirsch solution, but the pore pressure at the wall equals the pressure in the well, and the change of the pore pressure around the well is the same in all directions (axisymmetric). The wall splits when the effective hoop stress $\sigma_\theta - p_w$ reaches $-T$ .
The loading is slow: at every moment the rock has time to come to equilibrium with the pressure in the pores.
The real breakdown pressure lies between this value and the value without penetration. Which one it is closer to depends on the injection rate, the viscosity of the fluid, the filter cake and the permeability.

At $A = 1$ the formula coincides with the formula without penetration. At $A \to 0$ it gives $\sigma_h - (\sigma_H - \sigma_h)/2 + T/2$ , which is much less. So the breakdown pressure is a poor measure of stresses: it depends on a strength that is hard to measure, and on how the injection was run. To estimate $\sigma_h$ we use the fracture closure pressure.

Worked exampleBreakdown pressure of the wall in the sandstone

Given (assumed data)

Stresses (example above): $\sigma_h = 41.8$ MPa; $\sigma_H = 47.2$ MPa; $p_p = 30.5$ MPa
Sandstone: $T = 4$ MPa; $\nu = 0.22$ ; $\alpha = 0.80$
Depth: 2900 m

Solution

Without penetration: $p_b = 3\cdot41.8 - 47.2 - 30.5 + 4 = 51.7$ MPa.
The poroelastic coefficient: $A = 0.8\cdot(1 - 0.44)/0.78 = 0.574$ .
With penetration: $p_b = (3\cdot41.8 - 47.2 + 4 - 0.574\cdot30.5)/(2 - 0.574) = 64.69/1.426 = 45.4$ MPa.
Reopening of a fracture on the wall ( $T = 0$ , no penetration): $3\cdot41.8 - 47.2 - 30.5 = 47.7$ MPa.
In equivalent densities (the density of a fluid column that gives this pressure at the depth, $p/(0.00981\cdot2900)$ ): closure 1.47, breakdown 1.60…1.82, reopening 1.68 g/cm³.

Answer: The wall will split at 45.4…51.7 MPa at the bottom of the well, depending on whether fluid penetrates the rock. The fracture will then close at 41.8 MPa.

What it means. The breakdown pressure is 3.6…9.9 MPa above the closure pressure, and the pumps must supply it, once the hydrostatic head and the friction in the pipes are taken into account. If the wall already has fractures and the perforations are oriented along $\sigma_H$ , the tensile strength plays no part, and the pressure at the start of injection is closer to 47.7 MPa. The spread of 6 MPa between the two limiting cases is one more reason not to judge $\sigma_h$ by the breakdown pressure. The data in this example are assumed for illustration.

Calibration injection: closure pressure and net pressure

Before the main treatment, a small volume of fluid without proppant is pumped into the reservoir, from a few to a few tens of cubic metres, and the pressure is recorded during injection and after shut-in (when pumping stops). This is a calibration injection, also called a mini-frac or a diagnostic fracture injection test (DFIT). It is best to record the pressure with a bottomhole gauge. A reading at the wellhead must be converted to the bottom of the well. This needs the hydrostatic head (the weight of the fluid column) and the friction losses in the pipes, and during injection we do not know these accurately.

What happens during injection and after shut-in

1
Pressure rise
The closed system is compressed, and some of the fluid goes into the pores. The pressure rises almost in a straight line.
2
Breakdown of the wall
The wall splits, and the pressure falls. If a fracture already exists (a repeat cycle, perforations oriented along $\sigma_H$ ), there is no peak: the fracture simply opens.
3
Propagation
The fracture grows at an almost constant pressure, which is the sum of the closure pressure, the net pressure and the losses near the wellbore.
4
Shut-in
The friction losses vanish at once. The pressure right after shut-in is the ISIP (instantaneous shut-in pressure).
5
Closing
The fluid leaks off (escapes through the walls of the fracture into the rock), and the fracture narrows. When its walls come together, the pressure passes the closure pressure, and the way it declines changes.
6
After closure
The pressure slowly approaches the reservoir pressure. From this part of the record we estimate the reservoir pressure and the flow capacity of the reservoir.

Bottomhole pressure during injection and after shut-in

p_{wf} = p_c + p_{net} + \Delta p_{\text{perf}} + \Delta p_{\text{nw}},\qquad p_{ISIP} = p_c + p_{net}

where:

$p_{wf}$ bottomhole pressure during injection, MPa
$p_c$ fracture closure pressure; in a uniform reservoir it equals $\sigma_h$ , MPa
$p_{net}$ net pressure: the excess of the pressure in the fracture over the closure pressure, MPa
$\Delta p_{\text{perf}}$ pressure losses in the perforations; they grow as the square of the rate, MPa
$\Delta p_{\text{nw}}$ losses in the near-wellbore zone: in the tortuous, narrowed start of the fracture (tortuosity), MPa
$p_{ISIP}$ instantaneous bottomhole pressure after shut-in, MPa

Units: any identical units of pressure; all pressures are at the depth of the reservoir

Holds when

The wellhead pressure is converted to the bottom of the well by adding the hydrostatic head of the column of fluid and subtracting the friction losses in the pipes.
At the moment of shut-in, the friction losses in the pipes, in the perforations and near the wellbore vanish, and the net pressure has not yet had time to change.

The net pressure keeps the fracture open to the required width, pushes the fluid along it, and splits the rock at the tip. We separate the near-wellbore losses from the net pressure by lowering the rate in steps before shut-in. The losses in the perforations are proportional to the square of the rate. In the tortuous start of the fracture they grow with the rate more slowly, roughly as a square root.

How do we find the moment of closure on the pressure decline curve? While the fracture is open, fluid leaks off into the reservoir through its whole surface, and the net pressure is proportional to the width of the fracture and so to the volume of fluid in it. It is convenient to express the volume that has left since shut-in through the $G$ function. This is a dimensionless time that takes into account that different parts of the fracture opened at different moments and leak off differently.

The G function (for a fracture whose area grew in proportion to time)

G(\Delta t_D) = \frac{16}{3\pi}\left[(1 + \Delta t_D)^{1.5} - \Delta t_D^{1.5} - 1\right],\qquad \Delta t_D = \frac{\Delta t}{t_p}

where:

$G$ Nolte's function: the dimensionless time of the pressure decline, —
$\Delta t$ time after the end of injection, min
$t_p$ duration of injection, min

Units: dimensionless; $\Delta t$ and $t_p$ in the same units

Holds when

The injection is at a constant rate. After shut-in the area of the fracture does not change.
Leak-off from the fracture into the reservoir follows Carter's law: the leak-off rate is inversely proportional to the square root of the time since the given part of the wall opened.
The stiffness of the fracture is constant: the net pressure is proportional to the volume of fluid in it.
There is also a second limiting form of the function, for a fracture whose area grew as the square root of time. The two forms differ by no more than about 20% (the later the closure, the smaller the difference), and the choice does not affect the closure pressure that is found.

Under these conditions, before closure the pressure falls along a straight line in the coordinates $p$ against $G$ , and the derivative $G\,\dd p/\dd G$ lies on a straight line through the origin. When the walls of the fracture come together, the leak-off area and the stiffness change, and the derivative leaves the line downwards. This is the closure. The derivation is in the advanced section.

CalculatedThe pressure decline after the calibration injection of the example below, in the coordinates of the

G

function. Top: the bottomhole pressure and the straight line before closure. Bottom: the derivative

G\,\dd p/\dd G

and the straight line through the origin. The closure is where the derivative leaves the line:

G_c \approx 4.27

p_c = 41.8

MPa. The record is built from a simple model of the decline and is assumed for illustration.

Worked exampleClosure pressure and net pressure from a calibration injection

Given (assumed data)

Injection: 8 min at a constant rate, then shut-in
Instantaneous shut-in pressure at the bottom: 44.4 MPa
The derivative leaves the straight line (from the plot): 40 min after shut-in, the pressure is 41.8 MPa
Depth of the middle of the reservoir: 2900 m

Solution

Dimensionless time of closure: $\Delta t_D = 40/8 = 5$ .
The $G$ function at closure: $G_c = 1.698\cdot(6^{1.5} - 5^{1.5} - 1) = 1.698\cdot(14.697 - 11.180 - 1) = 4.27$ .
The slope of the line before closure: $m = (44.4 - 41.8)/4.27 = 0.609$ MPa per unit of $G$ .
The net pressure at the end of injection: $p_{net} = 44.4 - 41.8 = 2.6$ MPa.
The closure pressure gradient: $41.8/2900 = 14.41$ kPa/m (0.637 psi/ft), and the equivalent density is $41.8/(0.00981\cdot2900) = 1.47$ g/cm³.
With the second limiting form of the $G$ function, the closure would fall at $G_c = 4.06$ , which is 5% less. The closure pressure read from the curve is the same.

Answer: The closure pressure is 41.8 MPa (14.41 kPa/m). This is the minimum stress in the sandstone. The net pressure at the end of injection is 2.6 MPa.

What it means. The closure pressure was used to calibrate the stress model (the example in the previous section). It helps to remember the order of the pressures in one and the same well. The closure, 41.8 MPa, is below the ISIP by the net pressure. The pressure during injection is above the ISIP by the losses near the wellbore. And the breakdown pressure of the wall is above them all. The time to closure tells how fast the fluid leaks off into the reservoir: the longer the fracture stays open, the higher the share of the pumped fluid that stays in it. These data go into the design of the main treatment (Fracture design: size and conductivity). The data in this example are assumed for illustration.

Step-rate test

In injection wells it matters at what pressure the reservoir starts to take water through a growing fracture. For this we run a step-rate test. We raise the rate in steps of equal duration and record the bottomhole pressure at the end of each step. While there is no fracture, the water goes into the pores, and the pressure rises steeply with the rate, almost in a straight line, as Darcy's law says. When the pressure reaches the fracture extension pressure, the reservoir starts to take water through the fracture, and each extra cubic metre per day adds only a little pressure. The crossing of the two lines gives the fracture extension pressure (also called the fracture propagation pressure).

Step-rate test in an injection well (reservoir pressure 18.0 MPa)

Rate, m³/d: 60
Bottomhole pressure, MPa: 19.92

Rate, m³/d: 120
Bottomhole pressure, MPa: 21.80

Rate, m³/d: 180
Bottomhole pressure, MPa: 23.73

Rate, m³/d: 240
Bottomhole pressure, MPa: 25.62

Rate, m³/d: 300
Bottomhole pressure, MPa: 27.55

Rate, m³/d: 360
Bottomhole pressure, MPa: 28.41

Rate, m³/d: 420
Bottomhole pressure, MPa: 28.74

Rate, m³/d: 480
Bottomhole pressure, MPa: 29.08

Rate, m³/d: 540
Bottomhole pressure, MPa: 29.40

Rate, m³/d	Bottomhole pressure, MPa
60	19.92
120	21.80
180	23.73
240	25.62
300	27.55
360	28.41
420	28.74
480	29.08
540	29.40

The data are assumed for illustration. The pressure is at the end of a 30-minute step; the first five steps are injection into the pores.

CalculatedThe step-rate test from the table: a straight line through the first five steps (injection into the pores), a straight line through the last four (injection through the fracture), and their crossing, the fracture extension pressure.

Worked exampleFracture extension pressure from a step-rate test

Given (assumed data)

Test data: the table above
Closure pressure from the pressure decline after the test: 27.5 MPa

Solution

A straight line through the first five steps (a least-squares fit): $p = 18.00 + 0.0318\,q$ . The intercept on the pressure axis coincided with the reservoir pressure: the steps stabilised in the same way.
A straight line through the last four: $p = 26.425 + 0.00552\,q$ .
The crossing: $q^* = (26.425 - 18.00)/(0.0318 - 0.00552) = 321$ m³/d. The extension pressure is $p^* = 28.2$ MPa.
Injectivity (how much water the well takes per unit of pressure): before the fracture $1/0.0318 = 31.4$ m³/(d·MPa). With the fracture it is 181, which is 5.8 times larger.
The extension pressure is above the closure pressure by 0.7 MPa.

Answer: The fracture grows at a bottomhole pressure above 28.2 MPa, that is, at an injection rate above about 321 m³/d.

What it means. To inject into the pores, we keep the bottomhole pressure below 28.2 MPa with a margin, after converting it to the pressure at the wellhead. Above that pressure the well takes water more readily, but through a fracture that extends along $\sigma_H$ . If a producing well stands in that direction, the water may break through to it too early. Sometimes injection above the opening pressure is done on purpose, in tight reservoirs, and then the well pattern is matched to the direction of the fractures (Why water is injected: pressure maintenance and oil displacement). The extension pressure is an upper estimate of the closure pressure. The difference is the net pressure and the losses near the wellbore. The data in this example are assumed for illustration.

Net pressure and the width of the fracture

The walls of the fracture are squeezed by the stress $\sigma_h$ and pushed apart by the fluid pressure. Only the excess of one over the other, the net pressure, creates width. The rock is elastic, so the width of the fracture is proportional to the net pressure and to its smallest dimension. For a fracture that is long and limited in height, this dimension is the height.

Maximum width of a fracture at a given net pressure

w_{\max} = \frac{2\,p_{net}\,h_f}{E'},\qquad E' = \frac{E}{1 - \nu^2}

where:

$w_{\max}$ the largest width of the fracture, at the middle of its vertical section, m
$p_{net}$ net pressure, MPa
$h_f$ height of the fracture, m
$E'$ plane-strain modulus of the rock, MPa

Units: consistent: $p_{net}$ and $E'$ in the same units, $w$ in the units of $h_f$

Holds when

The fracture is much longer than its height, and its vertical section is an ellipse (plane strain).
The net pressure is the same over the height. The rock is homogeneous and elastic.
This is an order-of-magnitude estimate. In reality the pressure falls towards the tip of the fracture, and layers of different stiffness open differently.

With a modulus of tens of gigapascals and a net pressure of a few megapascals, the width of a fracture during injection comes to millimetres. Soft rocks open wider, and stiff rocks open narrower. In tight, stiff sandstones the fracture is narrow and long, and the proppant enters it with difficulty. Fracture growth models combine this relation with the flow of fluid and the leak-off into the reservoir, and they calculate the width and the length (Fracture design: size and conductivity).

How the net pressure changes during injection depends on where the fracture is growing. If the height is limited by barriers and the fracture grows in length, the fluid has to be pushed along a longer and longer narrow channel, and the net pressure rises slowly. If the fracture grows both in height and in length without restraint, the net pressure falls. So the engineer who runs the injection reads from the plot of the net pressure against time, on log–log axes, what is happening in the reservoir.

How to read the net pressure during injection

Behaviour of $\log p_{net}$ against $\log t$: Slow rise, slope 0.1–0.25
What is probably happening: the fracture grows in length, the height is contained
What to do: continue as planned

Behaviour of $\log p_{net}$ against $\log t$: Almost constant
What is probably happening: moderate growth in height, or fluid going into natural fractures
What to do: watch: the height may be growing

Behaviour of $\log p_{net}$ against $\log t$: Sharp rise, slope about 1 or more
What is probably happening: the proppant has stopped at the tip, and the fracture is only widening
What to do: get ready for a screen-out (the fracture stops growing; sometimes this is the aim)

Behaviour of $\log p_{net}$ against $\log t$: Fall
What is probably happening: breakthrough of a barrier, growth into a layer with lower stress
What to do: check whether the fracture has gone into water or gas

Behaviour of $\log p_{net}$ against $\log t$	What is probably happening	What to do
Slow rise, slope 0.1–0.25	the fracture grows in length, the height is contained	continue as planned
Almost constant	moderate growth in height, or fluid going into natural fractures	watch: the height may be growing
Sharp rise, slope about 1 or more	the proppant has stopped at the tip, and the fracture is only widening	get ready for a screen-out (the fracture stops growing; sometimes this is the aim)
Fall	breakthrough of a barrier, growth into a layer with lower stress	check whether the fracture has gone into water or gas

These are qualitative signs; the slopes depend on the properties of the fluid and on leak-off. The net pressure is calculated from the bottomhole pressure minus the closure pressure and the losses near the wellbore.

Growth of a fracture in height

The height of a fracture is set not by the boundaries of the productive layer, but by how hard it is for the fracture to enter the neighbouring layers. The main thing is the stress contrast. Suppose $\sigma_h$ in the shale above the reservoir is a few megapascals higher than in the reservoir. For the fracture to open in the shale, the pressure must be above this stress. But a fracture is not a slit that is open only where the pressure exceeds the stress. The pressure on its whole surface creates a stress concentration at the tips, and the tips go into the barrier. How far they go depends on the ratio of the net pressure to the contrast.

Equilibrium height of a fracture between two barriers

h_f = \frac{h}{\cos\left(\dfrac{\pi\,p_{net}}{2\,\Delta\sigma}\right)},\qquad p_{net} < \Delta\sigma

where:

$h_f$ total height of the fracture, m
$h$ thickness of the reservoir between the barriers, m
$p_{net}$ net pressure in the fracture, MPa
$\Delta\sigma$ stress contrast: the excess of $\sigma_h$ of the barriers over $\sigma_h$ of the reservoir, MPa

Units: $p_{net}$ and $\Delta\sigma$ in the same units; $h_f$ in the units of $h$ ; the angle is in radians

Holds when

The fracture is symmetric about the middle of the reservoir. The barriers are identical and thick enough, with no layers of low stress behind them.
The pressure is the same over the height of the fracture (the weight of the fluid and the friction losses are not taken into account).
The fracture toughness of the rock (its resistance to the growth of a crack) is not taken into account (with it, the fracture goes into the barriers a little less). The elastic moduli of the layers are the same.

While the net pressure is not more than about half of the contrast, the fracture goes into the barriers by only fractions of the reservoir thickness. After that the growth speeds up, and at $p_{net} \to \Delta\sigma$ the height in the model is unlimited: the barrier stops holding. From this comes a practical rule: the net pressure of the main injection must stay clearly below the stress contrast.

CalculatedRatio of the fracture height to the reservoir thickness against the ratio of the net pressure to the stress contrast (symmetric thick barriers, fracture toughness not taken into account). The points are the cases of the example below: the calibration injection, the main injection with the growth of the net pressure, and the same injection after depletion of the reservoir.

Worked exampleWill the shale hold the fracture?

Given (assumed data)

Reservoir and barriers: thickness 24 m; stress contrast 5.9 MPa (the example of calibrating the model)
Calibration injection: net pressure 2.6 MPa; $E' = 23.1$ GPa
Main injection: a viscous fluid; the net pressure may rise to 4.5 MPa
After some years: the reservoir is depleted by 8 MPa; $A = 0.574$ ; the pressure in the shales is as before

Solution

Calibration injection: $h_f = 24/\cos(\pi\cdot2.6/(2\cdot5.9)) = 24/\cos 0.692 = 31.2$ m, which is 3.6 m into each barrier.
The width in the middle: $w = 2\cdot2.6\cdot31.2\cdot10^3/23{,}100 = 7.0$ mm (the factor $10^3$ converts metres to millimetres).
The main injection: $h_f = 24/\cos(\pi\cdot4.5/11.8) = 24/\cos 1.198 = 66$ m. The fracture goes into each barrier by 21 m.
Depletion by 8 MPa lowers the $\sigma_h$ of the sandstone by $0.574\cdot8 = 4.59$ MPa. The contrast becomes $5.9 + 4.59 = 10.49$ MPa, and at the same 4.5 MPa $h_f = 30.7$ m.

Answer: At a net pressure of 2.6 MPa the fracture hardly leaves the reservoir (31.2 m for a thickness of 24 m). At 4.5 MPa its height is about 66 m. In the depleted reservoir the same 4.5 MPa gives only 30.7 m.

What it means. The height grows far faster than in proportion to the net pressure: an increase of $p_{net}$ by 1.9 MPa doubled the height. Suppose a water-bearing sandstone with low stress lies 20 m above the reservoir. The main injection will break through into it, and a model with thick barriers no longer applies here. So we limit the net pressure, through the pumping rate and the viscosity of the fluid, and we watch it during injection. Depletion of the reservoir improves containment: fractures made in the late years of development are held in the reservoir better than those made at the start. A model without fracture toughness and without the fall of pressure towards the tip of the fracture somewhat overstates the height, and a model without low-stress layers behind the barrier understates it. The data in this example are assumed for illustration.

Thickness of the barrier. A fracture goes straight through a thin shale layer of high stress if a layer of low stress lies behind it. Only a thick barrier holds it.
Stiffness of the layers. A stiff layer opens less, and a soft one opens more. But by itself a change of modulus does not stop a fracture: the stress contrast decides.
Weight of the fluid. The pressure in the fracture grows downwards with the gradient of the fluid, about 10–12 kPa/m. The $\sigma_h$ in a uniform reservoir usually grows with depth faster, by 12–16 kPa/m, so the net pressure at the top of the fracture is greater than at the bottom. If the stress grows with depth more slowly than the weight of the fluid, a heavy fluid pulls the fracture down. If it grows clearly faster, the fracture grows upwards more readily.
Leak-off and layer boundaries. A permeable layer that takes a lot of fluid slows the growth. Weak boundaries between layers can open and blunt the tip of the fracture.

We measure the height of the fracture that was made in several ways: by temperature logging before and after injection (cold fluid cools the interval that the fracture went into), by gamma-ray logging after injecting proppant with a radioactive tracer, by recording microseismic events (tiny earthquakes that the fracture causes) in neighbouring wells, and by tiltmeters (instruments that measure the tilt of the ground). The first two methods see the fracture only at the wellbore. The created, propped and working heights differ: the working height is the part through which fluid really flows into the well.

Orientation of the fracture, perforation and the well trajectory

Far from the well, a fracture grows perpendicular to the smallest stress. In the extensional (normal-faulting) and strike-slip regimes this is a vertical plane along $\sigma_H$ . In the compressional (thrust) regime, and at shallow depth where $\sigma_v$ becomes the smallest stress, it is a horizontal plane. But right at the wellbore the fracture starts where the perforations and the stress concentration allow it. If these directions differ, it has to turn.

Perforations not in the direction of the fracture. The fracture starts at the hole and turns towards the favourable plane. At the turn its walls are squeezed not by $\sigma_h$ but by a larger stress, so the width is small and the path is winding. The pressure losses near the wellbore grow, and the proppant may get stuck at the entrance. This causes a premature screen-out (the proppant blocks the fracture and stops its growth). It helps to use perforations with 60° phasing (the angle between perforations around the casing), so that at least one hole is no more than 30° from the favourable plane, or perforations oriented along $\sigma_H$ .
A deviated well (one tilted away from the vertical). If the wellbore lies in the plane of the fracture, the fracture is opened over a long length. If not, the fracture crosses the wellbore over a short stretch, and several parallel fractures may form near the wellbore. Sometimes the well is made to pass through the reservoir vertically, with an S-shaped profile.
A horizontal well drilled along $\sigma_h$ gets transverse fractures (across the well). This is how multistage hydraulic fracturing works. Along $\sigma_H$ the well gets one longitudinal fracture along the wellbore (Horizontal wells).
Stresses change during development. Depletion around producing wells lowers the horizontal stresses unevenly and may rotate them. A repeat fracture in an old well sometimes grows in a different direction. Neighbouring fractures of a multistage treatment raise the stress next to each other (the “stress shadow” effect) and deflect each other.
Development patterns. The fractures of injection wells extend along $\sigma_H$ , so the rows of wells are placed along this direction. Then the water goes from the fracture to the producing wells, and not along the fracture to them.

Terms and symbols

Principal stresses $\sigma_v$ , $\sigma_H$ , $\sigma_h$: Three mutually perpendicular normal stresses, on whose planes there are no shear stresses. In a quiet basin they are the vertical one and two horizontal ones.
Biot coefficient $\alpha$: The share of the pore pressure that takes the load off the rock skeleton. The effective stress is $\sigma - \alpha p_p$ .
Tectonic strain $\varepsilon_h$ , $\varepsilon_H$: Horizontal shortening of the crust that creates in each layer an addition to the stress proportional to its stiffness. It is the calibration parameter of the stress model.
Poroelastic coefficient $A$: It equals $\alpha(1-2\nu)/(1-\nu)$ . It is the change of $\sigma_h$ per unit change of the pore pressure in uniaxial strain, and also the correction to the hoop stress at the wall when fluid penetrates the rock.
Breakdown pressure $p_b$: The pressure in the well at which the wall splits. It depends on the stresses, the tensile strength and the penetration of fluid.
Closure pressure $p_c$: The pressure at which the walls of the fracture come together. It is a measure of the minimum stress.
Net pressure $p_{net}$: The excess of the pressure in the fracture over the closure pressure. It creates the width of the fracture.
$G$ function: The dimensionless time of the pressure decline after injection, in which the pressure falls along a straight line before closure.
Stress contrast $\Delta\sigma$: The excess of $\sigma_h$ of the neighbouring layers over $\sigma_h$ of the reservoir. It is the main factor in holding the fracture in height.

Common mistakes and misconceptions

Wrong

“The breakdown pressure on the record is the minimum stress.”

In fact

The breakdown pressure is above the minimum stress by an amount that depends on $\sigma_H$ , the tensile strength and the penetration of fluid. In the lesson's example the difference is 3.6–9.9 MPa. The measure of $\sigma_h$ is the closure pressure.

Wrong

“ISIP is equal to the closure pressure.”

In fact

ISIP is above the closure pressure by the net pressure, which at the end of injection keeps the fracture open: in the lesson's example, 44.4 against 41.8 MPa. We find the closure from the further decline of the pressure.

Wrong

“Since shale has a higher Poisson's ratio, it will always hold the fracture.”

In fact

Tectonic compression loads a stiff sandstone more than a soft shale, and it reduces the contrast: in the lesson's example, from 8.1 to 5.9 MPa. The thickness of the barrier and what lies behind it also decide.

Wrong

“A step-rate test shows the closure pressure.”

In fact

The bend on the plot of pressure against rate is the fracture extension pressure. It is above the closure pressure by the net pressure and the losses near the wellbore. We find the closure from the pressure decline or from the flowback after the test.

Wrong

“A stress profile measured at the first hydraulic fracturing is good for the whole life of development.”

In fact

Depletion lowers the $\sigma_h$ of the reservoir by about $A$ times the pressure drop. A rise of pressure from water injection raises it, and cooling of the rock by cold water lowers it. Before repeat treatments we measure the stresses again.

In depth: advanced detail

The stress field around the wellbore and the poroelastic correction

The Kirsch solution gives the stresses not only at the wall but also in the surrounding rock. Here are the formulas in polar coordinates with the origin on the axis of the well, with the radius of the well $r_w$ and the angle $\theta$ from the direction of $\sigma_H$ :

The Kirsch solution for a vertical well

\begin{aligned} \sigma_r &= \tfrac{\sigma_H+\sigma_h}{2}\left(1-\tfrac{r_w^2}{r^2}\right) + \tfrac{\sigma_H-\sigma_h}{2}\left(1-\tfrac{4r_w^2}{r^2}+\tfrac{3r_w^4}{r^4}\right)\cos 2\theta + p_w\tfrac{r_w^2}{r^2}\\ \sigma_\theta &= \tfrac{\sigma_H+\sigma_h}{2}\left(1+\tfrac{r_w^2}{r^2}\right) - \tfrac{\sigma_H-\sigma_h}{2}\left(1+\tfrac{3r_w^4}{r^4}\right)\cos 2\theta - p_w\tfrac{r_w^2}{r^2}\\ \tau_{r\theta} &= -\tfrac{\sigma_H-\sigma_h}{2}\left(1+\tfrac{2r_w^2}{r^2}-\tfrac{3r_w^4}{r^4}\right)\sin 2\theta \end{aligned}

where:

$\sigma_r,\ \sigma_\theta,\ \tau_{r\theta}$ radial, hoop and shear stresses at a distance $r$ from the axis, MPa
$r_w,\ r$ radius of the well and distance from the axis, $r \ge r_w$ , m
$\theta$ angle from the direction of $\sigma_H$ , rad

Units: any identical units of stress; compression is positive

Holds when

The same as for the formula for the wall. At $r = r_w$ we get $\sigma_r = p_w$ , the formula for $\sigma_\theta$ at the wall, and $\tau_{r\theta} = 0$ .

The disturbance dies out as $(r_w/r)^2$ and $(r_w/r)^4$ . For example, in the direction of $\sigma_H$ for the sandstone of the examples at $p_w = p_p$ , the hoop stress is 47.7 MPa at the wall, 43.3 MPa at a distance of three radii from the axis, and 41.8 MPa far from the well. A fracture that has gone a few radii away from the wellbore “sees” only $\sigma_h$ .

Where the poroelastic correction comes from. Let the pore pressure around the well change by $\Delta p(r)$ , equally in all directions and only near the well. The solution of the elasticity problem for such an axisymmetric addition (with the same equations as for thermal expansion, where the pressure in the pores plays the role of temperature) gives at the wall, where the radial stress does not change, $\Delta\sigma_\theta = A\,\Delta p(r_w)$ . It does not depend on how far into the reservoir the change has reached. During injection $\Delta p > 0$ , and the hoop compression grows. During production $\Delta p < 0$ , and it falls. The formula for the breakdown pressure with penetration is built on this, and so is the calculation of the critical drawdown in the lesson Sand production: types, prediction and cost of the problem.

Limits of stress: friction on faults

The Earth's crust is cut by faults and fractures of different orientation. If the ratio of the largest effective stress to the smallest becomes too large, sliding starts on well-oriented faults, and the stresses are released. So this ratio is limited by friction on faults:

Limiting ratio of effective stresses with friction on faults

\frac{\sigma_1 - p_p}{\sigma_3 - p_p} \le \left(\sqrt{\mu^2 + 1} + \mu\right)^2

where:

$\sigma_1,\ \sigma_3$ largest and smallest principal stresses, MPa
$p_p$ pore pressure, MPa
$\mu$ coefficient of friction on the fault; for most rocks 0.6–1.0, —

Units: any identical units of pressure

Holds when

There are faults of every orientation in the rock mass, with no cohesion. The effective stress is the simple one of Terzaghi (stress minus pore pressure; the Biot coefficient for faults is taken as 1).

At $\mu = 0.6$ the right-hand side is 3.12. In the normal-faulting regime $\sigma_1 = \sigma_v$ and $\sigma_3 = \sigma_h$ , and the minimum horizontal stress cannot be less than $p_p + (\sigma_v - p_p)/3.12$ . That is, the effective horizontal stress is not less than 0.32 of the vertical one, about a third. This is the physical meaning of the “one third” rule from the lesson Fracture gradient and leak-off testing: it describes rocks in a state of limiting friction in areas of extension. For the reservoir of the examples the lower limit is 41.7 MPa. The model without tectonics gave 36.0 MPa, which is below the limit, and the measured 41.8 MPa lies right at the boundary: the reservoir is in an area where faults are close to sliding. Such checks rule out impossible combinations of stresses, especially when we estimate $\sigma_H$ .

Where the G function comes from

Let injection at a constant rate last for a time $t_p$ , and let the area of the fracture grow in proportion to time. Each part of the wall that opened at the moment $\tau$ loses fluid at the rate $C_L/\sqrt{t - \tau}$ (Carter's law; $C_L$ is the leak-off coefficient). If the area does not change after shut-in, the volume of fluid that has gone through both walls by the time $\Delta t$ after shut-in is found by integrating over the area. It is proportional to $(1 + \Delta t_D)^{1.5} - \Delta t_D^{1.5} - 1$ , where $\Delta t_D = \Delta t/t_p$ . The factor $16/(3\pi)$ is the normalisation adopted for the $G$ function.

While the fracture is open and its stiffness is constant, the net pressure is proportional to the volume of fluid in it. That is, the fall of pressure is proportional to the volume that has left, and so to the $G$ function. This gives the straight line of $p$ against $G$ before closure. The derivative $G\,\dd p/\dd G$ at constant $\dd p/\dd G$ is a straight line through the origin. This line is convenient to watch, because a deviation from a straight line through zero is easier to see than a bend in the pressure curve itself. If the area of the fracture grew as the square root of time, the integral gives another form of the function, the second limiting one, which was used in the lesson's example for comparison. From the time to closure and the slope of the line we estimate the leak-off coefficient and the share of the fluid that stays in the fracture at the end of injection. These are the initial data for designing the main treatment.

Connections to other topics

Fracture gradient and leak-off testing — the smallest stress, the tectonic regimes, the breakdown pressure of the wall without penetration, and tests under the casing shoe.
Hydraulic fracturing: purpose and well selection — what a fracture gives, its half-length and conductivity, and the choice of wells.
Pore pressure and overburden pressure — overburden pressure from the density log, and effective stress.
Rock compressibility and the effect of stress on core properties — the Biot coefficient, uniaxial strain, and the growth of effective stress with depletion.
Sonic log — dynamic elastic moduli from the interval transit times of compressional and shear waves.
Inflow to a hydraulically fractured well — inflow to a well with a fracture, and the duration of the flow regimes.
Fracture design: size and conductivity — fracture growth models, the leak-off coefficient, proppant and optimal dimensions.
Sand production: types, prediction and cost of the problem — the same stresses at the wall, but in production and not in injection: rock strength and critical drawdown.

Check yourself

1
On the record of a calibration injection, the bottomhole pressure peak is 53 MPa, the pressure during injection is 46 MPa, the ISIP is 44 MPa, and the closure by the $G$ function is 41 MPa. Which value should we take as the minimum horizontal stress?
1. A.46 MPa, the extension pressure
2. B.44 MPa, the ISIP
3. C.41 MPa, the closure pressure
4. D.53 MPa, the breakdown pressure
Show the answer
Correct answer: C. 41 MPa, the closure pressure
The breakdown pressure includes the stress concentration and the tensile strength. The pressure during injection includes the net pressure and the losses near the wellbore. The ISIP includes the net pressure at the end of injection. Only the closure pressure, at which the walls of the fracture come together, is equal to the stress that squeezes them, $\sigma_h$ . The net pressure at the end of injection is 3 MPa.
2
In a reservoir $\sigma_h = 35$ MPa, $\sigma_H = 40$ MPa, the pore pressure is 24 MPa, the tensile strength is 3 MPa, and $A = 0.6$ . Find the breakdown pressure of the wall without and with penetration of fluid.
Show the answer
Without penetration: $3\cdot35 - 40 - 24 + 3 = 44$ MPa. With penetration: $(105 - 40 + 3 - 0.6\cdot24)/(2 - 0.6) = 53.6/1.4 \approx 38.3$ MPa. The difference is almost 6 MPa: the breakdown pressure depends on how the injection is run, so it is a poor measure of $\sigma_h$ .
3
A reservoir is depleted by 12 MPa, and $A = 0.55$ . The pressure in the shales above and below it has not changed, and the stress contrast before development was 3 MPa. How does the containment of the fracture in height change?
1. A.It gets better: the contrast grows to about 9.6 MPa
2. B.It gets better, but the contrast grows by the whole 12 MPa
3. C.It does not change: stresses do not depend on the pore pressure
4. D.It gets worse: a depleted reservoir is weaker
Show the answer
Correct answer: A. It gets better: the contrast grows to about 9.6 MPa
The $\sigma_h$ of the reservoir will fall by $0.55\cdot12 = 6.6$ MPa, and in the shales it stays as before. The contrast becomes $3 + 6.6 = 9.6$ MPa, and at the same net pressure the fracture will go into the barriers much less. The stress falls not by the whole drop of the pore pressure, but by the share $A$ of it.
4
A reservoir 30 m thick lies between thick shales, and the stress contrast is 4 MPa. Estimate the height of the fracture at a net pressure of 2 MPa. What will happen if the net pressure rises to 3.6 MPa?
Show the answer
$h_f = 30/\cos(\pi\cdot2/8) = 30/\cos 45° \approx 42$ m, that is, 6 m into each barrier. At 3.6 MPa: $30/\cos(\pi\cdot3.6/8) = 30/\cos 81° \approx 192$ m. The ratio $p_{net}/\Delta\sigma = 0.9$ , and the barrier hardly holds at all. We must limit the net pressure through the pumping rate and the viscosity of the fluid, and watch it during injection.
5
We want a horizontal well for multistage hydraulic fracturing to get transverse fractures. In what direction should we drill it?
1. A.Along $\sigma_h$
2. B.The direction does not matter
3. C.At 45° to $\sigma_H$
4. D.Along $\sigma_H$
Show the answer
Correct answer: A. Along $\sigma_h$
A vertical fracture grows in the plane perpendicular to $\sigma_h$ , that is, along $\sigma_H$ . For it to cross the wellbore transversely, the wellbore must run along $\sigma_h$ . If the wellbore runs along $\sigma_H$ , the fracture will be longitudinal: one fracture along the whole wellbore.
6
Why are all the steps of a step-rate test made equal in time, and what does the bend on the plot of pressure against rate give?
Show the answer
With equal steps, transient flow distorts the pressure of every step in the same way, and the points before the fracture appears lie on a straight line. The bend is the fracture extension pressure: above it the reservoir takes fluid through a growing fracture. It is above the closure pressure by the net pressure and the losses near the wellbore, so we find the closure from the pressure decline or from the flowback after the test.

Updated: 28 October 2026