Fracture gradient and leak-off testing

Quick refresher

A fracture opens when the pressure in the well exceeds the smallest principal stress, and it grows perpendicular to it. In most sedimentary basins (large areas where thick layers of sediment have piled up) the smallest stress is horizontal, and the fractures are vertical.
Fracture pressure grows with the pore pressure and the overburden pressure (the pressure from the weight of all the rock above): $p_f = p_p + K(\sigma_v - p_p)$ . The methods of estimation differ in how they find the coefficient $K$ . The most reliable way is calibration against tests in offset wells.
When a reservoir is depleted, the fracture pressure falls, although more slowly than the pore pressure. Under deep sea it is low, because the overburden pressure is small.
A test under the casing shoe checks the cement and the rock a few metres below the shoe. On the pressure–volume plot the points first follow a straight line and then bend away from it: this is the start of leak-off. The LOT is stopped there, and the FIT is stopped at a pressure set in advance.
The pressure under the shoe at leak-off equals the wellhead pressure plus the pressure of the mud column down to the shoe. From it we get the equivalent mud weight, the permitted mud density (with a margin) and the MAASP.
We choose the casing strings from the bottom up: the shoe must withstand the mud that is needed at the bottom of the next interval, with a margin for a kick.

Key relationships

p_f = p_p + K\,(\sigma_v - p_p)

fracture pressure from the effective stress; in the Eaton method $K = \nu/(1-\nu)$

\rho_{LOT} = \rho_{\text{test}} + \dfrac{p_{LO}}{g\,h_{\text{shoe}}}

equivalent mud weight from the test: the wellhead pressure $p_{LO}$ and the mud density during the test $\rho_{\text{test}}$

\text{MAASP} = (\rho_{LOT} - \rho_m)\,g\,h_{\text{shoe}}

permitted pressure at the wellhead in the annulus with the current mud $\rho_m$

\rho_{ECD} = \rho_m + \dfrac{\Delta p_{\text{fr}}}{g\,h}

equivalent circulating density at depth $h$

Common pitfalls

Thinking that a test under the shoe describes the whole future open hole. Weaker rocks may lie below: depleted sands and fractured zones.
Forgetting to add the pressure of the mud column down to the shoe to the wellhead pressure, or taking a mud density other than the one that was in the well during the test.
Comparing the static mud density with the leak-off pressure, without counting the ECD and the pressure surges when pipe is run in.
Not recalculating the MAASP after the mud is made heavier.
Pushing a test as far as rupture without need. Rock that has been ruptured below the shoe holds less pressure afterwards.

After this lesson you can

explain why a fracture opens against the smallest principal stress (the smallest of the three main stresses in the rock; stress is force per unit area), and tell apart the leak-off pressure, the breakdown pressure, and the fracture propagation and closure pressures
estimate the fracture pressure before drilling from the effective stress, by the Eaton, the Matthews and Kelly and the “one third” methods, and explain how it is affected by the pore pressure, by the depletion of a reservoir and by the depth of the sea
describe the leak-off test (LOT) and the formation integrity test (FIT), and find the leak-off pressure from the pressure–volume plot
calculate from the test results the equivalent mud weight at the shoe, the permitted mud density and the maximum allowable annular surface pressure (MAASP), also taking account of the equivalent circulating density (ECD)
explain how the casing setting depths are chosen from the curves of the pore pressure and the fracture pressure

The upper limit of the pressure in the well

The pressure in the open hole must lie in a window. It must be above the pore pressure of the permeable beds and above the pressure at which shales cave in, but below the pressure at which the rock ruptures. If the pressure goes above the window, a fracture opens in the rock, and the mud escapes into it. This is lost circulation. The level in the well falls, the pressure of the column on all the beds decreases, and the well may start to kick. In the worst case, fluid from a bed with high pressure flows along the hole into the ruptured bed. This is called a crossflow between formations, or an underground blowout (see the lesson on well control).

Fracture pressure, like pore pressure, is conveniently expressed as a gradient (pressure per metre of depth) or as an equivalent mud weight. Then we can compare it directly with the density of the mud. In drilling, the fracture gradient usually means the average gradient from the rotary table (the point on the rig floor from which depths are measured): the fracture pressure divided by the true vertical depth (see the lesson on pore and overburden pressure).

Overburden pressure and the stresses in the rock grow with depth. So in an interval with normal pore pressure the fracture pressure also grows with depth, and the weakest place in the open hole is usually at the top, just under the shoe of the last casing string. That is where we check it with a leak-off test. The exceptions are depleted sandstones, fractured and cavernous rocks, and loose sands. They may be weaker than the rock under the shoe.

Stresses in the rock and the direction of the fracture

The state of stress of a rock at any point is described by three principal stresses that are at right angles to each other. In a tectonically quiet sedimentary basin one of them is vertical: this is the overburden pressure $\sigma_v$ . The other two are horizontal: the largest, $\sigma_H$ , and the smallest, $\sigma_h$ .

The pressure of the fluid in a fracture pushes its walls apart, and the stress perpendicular to the fracture presses them together. So a fracture opens and grows in the plane that is easiest to open, which is perpendicular to the smallest principal stress. For a fracture to grow, the pressure in it must stay above this stress. When the pressure falls below it, the fracture closes.

Which stress is the smallest, and what the fracture is like

Tectonic regime: Extension (normal faults: the rock is pulled apart)
Relation of the stresses: $\sigma_v > \sigma_H > \sigma_h$
Fracture: vertical
Fracture pressure is close to: $\sigma_h$ , markedly lower than the overburden pressure

Tectonic regime: Strike-slip (blocks slide past each other)
Relation of the stresses: $\sigma_H > \sigma_v > \sigma_h$
Fracture: vertical
Fracture pressure is close to: $\sigma_h$

Tectonic regime: Compression (thrust faults: the rock is squeezed together)
Relation of the stresses: $\sigma_H > \sigma_h > \sigma_v$
Fracture: horizontal
Fracture pressure is close to: $\sigma_v$ , the overburden pressure

Tectonic regime	Relation of the stresses	Fracture	Fracture pressure is close to
Extension (normal faults: the rock is pulled apart)	$\sigma_v > \sigma_H > \sigma_h$	vertical	$\sigma_h$ , markedly lower than the overburden pressure
Strike-slip (blocks slide past each other)	$\sigma_H > \sigma_v > \sigma_h$	vertical	$\sigma_h$
Compression (thrust faults: the rock is squeezed together)	$\sigma_H > \sigma_h > \sigma_v$	horizontal	$\sigma_v$ , the overburden pressure

In most sedimentary basins the smallest stress is horizontal. We judge the direction and the relation of the stresses from the faults of the area, from breakouts (broken-off places in the wall of the hole) and fractures on borehole images (pictures of the wall of the hole), and from the fracture closure pressure in tests.

The horizontal stresses in a sedimentary basin are created first of all by the weight of the rocks. The rock skeleton, squeezed from above, tends to expand sideways, but the neighbouring rocks do not let it. The skeleton (the solid framework of grains) carries only the part of the load that is called the effective stress (see effective stress). So the horizontal effective stress is linked with the vertical effective stress: $\sigma_h - p_p = K\,(\sigma_v - p_p)$ . The coefficient $K$ is the horizontal effective stress divided by the vertical one. It shows how strongly the rock pushes sideways when it is squeezed from above. This is the lateral earth pressure coefficient. (Russian-language rock mechanics calls it the coefficient of lateral thrust.) Rock salt behaves like a very viscous liquid. The stresses in it are almost equal in all directions ( $K \approx 1$ ) and are close to the overburden pressure.

Estimating the fracture pressure before drilling

We need to know the fracture pressure as early as the design of the well, when there are no tests under its shoes yet. All the usual methods of estimation start from one idea. They take the fracture pressure equal to the minimum horizontal stress, and they express it through the pore pressure and the vertical effective stress.

Fracture pressure from the effective stress

p_f = p_p + K\,(\sigma_v - p_p)

where:

$p_f$ fracture pressure (the minimum horizontal stress $\sigma_h$ ), MPa
$p_p$ pore pressure at this depth, MPa
$\sigma_v$ overburden pressure at this depth, MPa
$K$ ratio of the effective stresses $(\sigma_h - p_p)/(\sigma_v - p_p)$ , the lateral earth pressure coefficient, —

Units: any identical units of pressure at one depth, so the formula holds both for gradients and for equivalent mud weights

Holds when

the smallest principal stress is horizontal (extension or strike-slip);
the fracture pressure is taken equal to the smallest stress: the fracture already exists, or the tensile strength of the rock (how much pulling it can take before it breaks) is small. This is the fracture propagation pressure (also called the fracture extension pressure) and a cautious estimate of the leak-off pressure;
$K$ is fitted for an area. Tectonic stresses, salt and temperature change it.

It is convenient to rewrite the formula as $p_f = K\sigma_v + (1 - K)\,p_p$ . A rise of the pore pressure by $\Delta p$ raises the fracture pressure only by $(1 - K)\Delta p$ . So in a zone of overpressure (abnormally high formation pressure) the window between the pore pressure and the fracture pressure narrows. When a reservoir is depleted, the fracture pressure falls, although more slowly than the pore pressure.

Methods of estimation: how

K

is found

Method: “One third” (Hubbert and Willis)
How $K$ is found: $K \approx 1/3$ ; 1/2 is taken as an upper estimate
Features: young loose sediments in areas of extension; a cautious lower estimate

Method: Matthews and Kelly
How $K$ is found: from regional plots of $K$ against depth (in an overpressure zone, against the equivalent depth with the same effective stress)
Features: empirical, tied to the area where the plots were built

Method: Eaton
How $K$ is found: $K = \nu/(1-\nu)$ , with a “Poisson's ratio” $\nu$ from a plot or from calibration against depth
Features: the most widely used; $\nu$ is a calibration parameter

Method: Calibration against offset wells
How $K$ is found: $K$ is chosen so that the calculation matches the tests under the shoes
Features: the most reliable way, if there are enough tests

Method	How $K$ is found	Features
“One third” (Hubbert and Willis)	$K \approx 1/3$ ; 1/2 is taken as an upper estimate	young loose sediments in areas of extension; a cautious lower estimate
Matthews and Kelly	from regional plots of $K$ against depth (in an overpressure zone, against the equivalent depth with the same effective stress)	empirical, tied to the area where the plots were built
Eaton	$K = \nu/(1-\nu)$ , with a “Poisson's ratio” $\nu$ from a plot or from calibration against depth	the most widely used; $\nu$ is a calibration parameter
Calibration against offset wells	$K$ is chosen so that the calculation matches the tests under the shoes	the most reliable way, if there are enough tests

Fracture pressure by the Eaton method

p_f = p_p + \frac{\nu}{1-\nu}\,(\sigma_v - p_p)

where:

$p_f$ fracture pressure, MPa
$p_p$ pore pressure (measured or forecast), MPa
$\sigma_v$ overburden pressure, MPa
$\nu$ Poisson's ratio in the sense of the method: a parameter fitted to the tests in the area (the real Poisson's ratio shows how much a rock expands sideways when it is squeezed along its axis), —

Units: any identical units of pressure at one depth; the same for gradients and for equivalent mud weights

Holds when

the horizontal stresses are created by the weight of the rocks, with the rock prevented from expanding sideways and with no tectonic additions. The rock is elastic and isotropic (the same in all directions);
the fracture pressure equals the smallest horizontal stress (the tensile strength and the concentration of stress at the wall are not counted);
$\nu$ usually grows with depth: in young shaly sections from about 0.25 at the surface to 0.4–0.5 at depth. As $\nu \to 0.5$ the ratio tends to one, and the fracture pressure tends to the overburden pressure.

The ratio $\nu/(1-\nu)$ is the coefficient $K$ for an elastic rock that is not allowed to expand sideways (the derivation is in the advanced section). At $\nu = 0.25$ it gives 1/3, and at $\nu = 0.4$ it gives 2/3.

Worked exampleForecast of the fracture pressure for a planned well

Given (assumed data)

True vertical depth: 2400 m
Overburden pressure from the density log of a neighbouring well: an equivalent mud weight of 2.18 g/cm³
Pore pressure from the forecast: an equivalent mud weight of 1.07 g/cm³
Parameter $\nu$ fitted to the tests in the area: 0.40

Solution

The coefficient: $K = \dfrac{0.40}{1 - 0.40} = 0.667$ .
The equivalent mud weight of the fracture pressure: $\rho_f = 1.07 + 0.667\cdot(2.18 - 1.07) = 1.81$ g/cm³. The pressure is $0.00981\cdot 1.81\cdot 2400 = 42.6$ MPa, and the gradient is 17.8 kPa/m (0.785 psi/ft, 15.1 ppg).
For comparison, by the “one third” rule: $\rho_f = 1.07 + (2.18 - 1.07)/3 = 1.44$ g/cm³ ( $33.9$ MPa), which is 0.37 g/cm³ less.
If the sandstone at this depth has been depleted by production to an equivalent mud weight of 0.80 g/cm³: $\rho_f = 0.80 + 0.667\cdot(2.18 - 0.80) = 1.72$ g/cm³. The pore pressure fell by 6.4 MPa, and the fracture pressure fell by 2.1 MPa.

Answer: The fracture pressure at 2400 m is about 42.6 MPa (1.81 g/cm³). In the depleted sandstone it is about 1.72 g/cm³.

What it means. The “one third” rule would give a much narrower window. The choice of the method, and the calibration of $\nu$ against tests in the area, matter more here than the precision of the arithmetic. By this model, depletion lowers the fracture pressure by $(1 - K)$ of the fall of the pore pressure, which is three times more slowly. On real fields the fall is often larger, half of the fall of the pore pressure or more, so we check the model by tests. Even by this model, the drop is enough for the mud that the neighbouring shales need to start leaking into the depleted reservoir when we drill through it. The data in this example are assumed for illustration.

The test under the shoe: LOT and FIT

After a casing string has been run and cemented, and before the next interval is drilled, we check the whole system. It includes the casing with its threaded connections, the cement at the shoe, the wellhead (the equipment at the top of the well), the blowout preventer (BOP) equipment (valves that can close the well), and the rock below the shoe. The test answers two questions. Does the cement at the shoe leak? What pressure can the rock take at the top of the next open hole, which is usually its weakest part? In Russian-language practice such tests are also called pressure testing of the open hole under the shoe.

How the test is done (in general terms)

1
Drilling out
We drill out the cement plug and the shoe, and go 3–5 m into new rock.
2
Preparation
We circulate until the density of the mud is uniform, and we pull the bit back into the shoe, so that no pipe stays in the open hole.
3
Shutting in
We close the blowout preventer. We build the pressure with a pump unit that has an accurate gauge, and we keep track of the volume pumped.
4
Pumping
We pump the mud at a small, constant rate, some tens of litres per minute, or in equal portions. We record the pressure and the volume, and we draw the plot as the test goes on.
5
Stopping
In a leak-off test, we stop as soon as the points depart from the straight line. In a test to a set pressure, we stop at that pressure. We watch the pressure in the shut-in well for a few minutes.
6
Bleeding off
We release the pressure and measure the volume that comes back. If it is clearly less than the volume pumped, part of the mud has gone into the rock.

Kinds of test

Test: Leak-off test (LOT)
When it is stopped: at the first departure from the straight line
What it gives: the leak-off pressure: the strength of the rock under the shoe

Test: Formation integrity test (FIT, limit test)
When it is stopped: at a pressure chosen in advance
What it gives: a confirmation that the rock will withstand the pressure that is needed; its limit stays unknown

Test: Formation breakdown test
When it is stopped: after the rupture and the propagation of the fracture
What it gives: the breakdown and propagation pressures; the rock under the shoe is weakened

Test: Extended leak-off test (extended LOT)
When it is stopped: after several cycles of pumping and fall-off
What it gives: the fracture closure pressure: the smallest stress

Test	When it is stopped	What it gives
Leak-off test (LOT)	at the first departure from the straight line	the leak-off pressure: the strength of the rock under the shoe
Formation integrity test (FIT, limit test)	at a pressure chosen in advance	a confirmation that the rock will withstand the pressure that is needed; its limit stays unknown
Formation breakdown test	after the rupture and the propagation of the fracture	the breakdown and propagation pressures; the rock under the shoe is weakened
Extended leak-off test (extended LOT)	after several cycles of pumping and fall-off	the fracture closure pressure: the smallest stress

In drilling we usually keep to the LOT or the FIT. A breakdown test and an extended test are run when data on stresses are needed, for example for hydraulic fracturing or for calculations of the stability of the hole.

We choose the FIT where the fracture pressure is known from offset wells. The test confirms that the shoe will withstand the density of the mud that is needed for the next interval, with a margin, and it does not load the rock more than necessary. A LOT gives more information, and it is needed in exploration wells (wells drilled to find new fields) and in new areas.

How to read the pressure–volume plot

While the rock is intact, pumping into the shut-in well only compresses the mud and stretches the casing a little. The pressure grows in proportion to the volume pumped, and the points lie on a straight line. Its slope depends on the volume of the mud in the well and its compressibility (how much the mud shrinks under pressure; a check of the slope is in the advanced section). When the rock at the wall begins to take mud (a fracture opens, or mud goes into the pores and micro-fractures), part of the pumped volume leaves the well, and the pressure begins to lag behind the straight line.

CalculatedThe test under the shoe from the example below: the points are measurements of the wellhead pressure, and the dashed line is the straight line through the points from 40 to 360 L. At 400 L the pressure is 0.5 MPa below the line: this is the start of leak-off, and an extra portion of 20 L confirms the departure.

Test data (example)

Pumped, L: 0
Wellhead pressure, MPa: 0

Pumped, L: 40
Wellhead pressure, MPa: 0.85

Pumped, L: 80
Wellhead pressure, MPa: 1.80

Pumped, L: 120
Wellhead pressure, MPa: 2.70

Pumped, L: 160
Wellhead pressure, MPa: 3.62

Pumped, L: 200
Wellhead pressure, MPa: 4.50

Pumped, L: 240
Wellhead pressure, MPa: 5.42

Pumped, L: 280
Wellhead pressure, MPa: 6.30

Pumped, L: 320
Wellhead pressure, MPa: 7.22

Pumped, L: 360
Wellhead pressure, MPa: 8.08

Pumped, L: 400
Wellhead pressure, MPa: 8.50

Pumped, L: 420
Wellhead pressure, MPa: 8.62

Pumped, L	Wellhead pressure, MPa
0	0
40	0.85
80	1.80
120	2.70
160	3.62
200	4.50
240	5.42
280	6.30
320	7.22
360	8.08
400	8.50
420	8.62

The data are assumed. The pressure is gauge pressure, read on the gauge at the wellhead. The well holds mud of 1.15 g/cm³, and the shoe is at 1850 m true vertical depth.

Leak-off pressure: The pressure at the point where the plot departs from the straight line: the mud has begun to go into the rock. We calculate the permitted density and the MAASP from it. To keep a margin, some companies take the last point on the straight line instead.
Breakdown pressure: The highest pressure reached as pumping goes on: the rock at the wall splits, and the pressure drops sharply.
Propagation pressure: The nearly constant pressure at which the fracture grows when pumping goes on. It is a little above the smallest stress.
Instantaneous shut-in pressure (ISIP): The pressure right after the pumps are stopped, when the friction losses have gone. It is close to the smallest stress.
Closure pressure: The pressure at which the walls of the fracture close as the pressure falls. It is the best estimate of the minimum horizontal stress.

The shape of the plot depends on the rock. In ductile shales (shales that squash without breaking) the curve bends smoothly, and after the departure the pressure grows more and more slowly, with no sharp peak. In brittle dense rocks (rocks that break suddenly) the pressure grows almost along a straight line up to a sudden rupture, and then it falls sharply. There is almost no warning. So we pump in small portions and watch the slope closely.

What the results give: permitted density, ECD and MAASP

The wellhead pressure at leak-off is only a part of the pressure on the rock under the shoe: to it we add the pressure of the mud column down to the shoe. We express the sum as an equivalent mud weight, to compare it with the density of the mud.

Equivalent mud weight of the leak-off pressure under the shoe

\rho_{LOT} = \rho_{\text{test}} + \frac{p_{LO}}{g\,h_{\text{shoe}}}

where:

$\rho_{LOT}$ equivalent mud weight of the leak-off pressure under the shoe, kg/m³
$\rho_{\text{test}}$ density of the mud in the well during the test (not to be confused with the density of the current mud $\rho_m$ in the MAASP formula), kg/m³
$p_{LO}$ leak-off pressure at the wellhead (gauge), Pa
$h_{\text{shoe}}$ true vertical depth of the shoe from the rotary table, m
$g$ acceleration due to gravity, m/s²

Units: SI

Holds when

the mud in the well during the test is at rest and uniform in density;
the pressure under the shoe is $p_{LO} + \rho_{\text{test}} g h_{\text{shoe}}$ , the sum of the wellhead pressure and the hydrostatic pressure. There are almost no friction losses at slow pumping;
all the pressures are gauge pressures (pressures above the atmosphere), like the reading of the gauge at the wellhead. The absolute pressure is greater by the atmospheric pressure (about 0.1 MPa), but we calculate the equivalent mud weight from the gauge pressure;
the result refers to the rock under the shoe, and not to the whole open hole.

A mud of this density would create the leak-off pressure at the shoe with the wellhead open. The permitted mud density for the next interval is lower than it by a margin set by the design, for example 0.05–0.06 g/cm³ (about 0.5 ppg).

In other unit systems (2)

Metric units

\rho_{LOT}\,[\text{g/cm}^3] = \rho_{\text{test}} + \frac{p_{LO}\,[\text{MPa}]}{0.00981\; h_{\text{shoe}}\,[\text{m}]}

densities in g/cm³, pressure in MPa, depth in m.

0.00981 MPa/m is the pressure of a column of liquid of density 1 g/cm³ and height 1 m.

Oilfield units

\rho_{LOT}\,[\text{ppg}] = \rho_{\text{test}} + \frac{p_{LO}\,[\text{psi}]}{0.052\; h_{\text{shoe}}\,[\text{ft}]}

densities in ppg, pressure in psi, depth in feet.

0.052 psi/ft per ppg comes from the oilfield formula for hydrostatic pressure.

While we drill, the pressure at the shoe is larger than the hydrostatic pressure. The friction pressure losses in the annulus above the shoe add to it, and so do the pressure surges (short pressure peaks) when pipe is run in (see Circulating pressure losses). So we compare the equivalent circulating density (ECD) at the depth of the shoe with the leak-off pressure.

Equivalent circulating density at the depth of the shoe

\rho_{ECD} = \rho_m + \frac{\Delta p_{\text{fr}}}{g\,h_{\text{shoe}}}

where:

$\rho_{ECD}$ equivalent circulating density at the depth of the shoe, kg/m³
$\Delta p_{\text{fr}}$ friction pressure loss in the annulus from the shoe to the surface, Pa
$\rho_m,\ g,\ h_{\text{shoe}}$ mud density, acceleration due to gravity, true vertical depth of the shoe, kg/m³, m/s², m

Units: SI; in metric units $g h$ is replaced by $0.00981\,h$ (g/cm³, MPa, m)

Holds when

to check the shoe we take only the losses above the shoe. For the bottom of the hole, we take the losses in the whole annulus and the depth of the bottom.

The ECD must stay below $\rho_{LOT}$ with a margin for the pressure surges when pipe is run in and for the inaccuracy of the loss calculation.

If the well is shut in after a kick, the pressure at the wellhead in the annulus is passed downwards and adds to the pressure of the mud column. The limit for this pressure is called the maximum allowable annular surface pressure (MAASP).

Maximum allowable annular surface pressure (MAASP)

\text{MAASP} = (\rho_{LOT} - \rho_m)\,g\,h_{\text{shoe}}

where:

$\rho_{LOT}$ equivalent mud weight of the leak-off pressure under the shoe, kg/m³
$\rho_m$ density of the mud that fills the annulus now, kg/m³
$g,\ h_{\text{shoe}}$ acceleration due to gravity, true vertical depth of the shoe, m/s², m

Units: SI; in metric units $\text{MAASP}\,[\text{MPa}] = 0.00981\,(\rho_{LOT} - \rho_m)\,h_{\text{shoe}}$ (g/cm³, m)

Holds when

the annulus from the wellhead to the shoe is filled with mud of density $\rho_m$ . If there is gas in it, or mud of another density, we calculate the pressure at the shoe from the real column;
the weakest place is the rock under the shoe. If there is a weaker bed in the open hole, it sets the limit.

The MAASP gets smaller when the mud is made heavier: the same margin to the leak-off pressure is now “eaten up” by a heavier column. So we recalculate it after every change of density, and we keep it in front of the driller.

Worked exampleResults of the test under the shoe

Given (assumed data)

Test data: the table and the plot in the previous section
True vertical depth of the shoe: 1850 m
Mud density during the test: 1.15 g/cm³
Mud for the next interval: 1.21 g/cm³; the friction losses in the annulus above the shoe are 1.1 MPa
Margin on density set by the design: 0.06 g/cm³

Solution

The straight line through the points from 40 to 360 L has a slope of 0.0226 MPa/L, which is about 0.90 MPa per 40 L. At 400 L the line gives 9.01 MPa, but 8.50 MPa was measured: a departure of 0.51 MPa. The leak-off pressure at the wellhead is $p_{LO} = 8.5$ MPa.
The pressure under the shoe (gauge pressure, like the reading of the gauge): $8.5 + 0.00981\cdot 1.15\cdot 1850 = 8.5 + 20.87 = 29.37$ MPa. The gradient is 15.88 kPa/m (0.702 psi/ft).
Equivalent mud weight: $\rho_{LOT} = 1.15 + \dfrac{8.5}{0.00981\cdot 1850} = 1.618$ g/cm³ (13.5 ppg). The permitted mud density with the margin of 0.06 g/cm³ is $1.618 - 0.06 = 1.558$ g/cm³. We adjust the density in steps of 0.01 g/cm³, so we round the upper limit down: 1.55 g/cm³.
ECD at the shoe when drilling with mud of 1.21 g/cm³: $1.21 + \dfrac{1.1}{0.00981\cdot 1850} = 1.27$ g/cm³. The margin to the leak-off pressure is 0.35 g/cm³.
MAASP with mud of 1.21 g/cm³: $0.00981\cdot(1.618 - 1.21)\cdot 1850 = 7.4$ MPa (1075 psi). If the mud is made heavier, to 1.25 g/cm³, the MAASP falls to 6.7 MPa.

Answer: The rock under the shoe begins to take mud at 29.37 MPa, an equivalent mud weight of 1.618 g/cm³. The mud density in the next interval must not exceed 1.55 g/cm³. The MAASP with mud of 1.21 g/cm³ is 7.4 MPa.

What it means. This is the same shoe as in the example of choosing the mud density in the lesson on pore pressure: mud of 1.21 g/cm³ with an ECD of 1.27 g/cm³ fits the window with a margin. If the forecast called for a mud heavier than 1.55 g/cm³, we would have to run one more casing string. If we take the last point on the straight line (8.08 MPa) as the leak-off pressure, the equivalent mud weight is 1.595 g/cm³, which is a cautious estimate. The data in this example are assumed for illustration.

How pressures set the casing setting depths

Each interval of the open hole is drilled with one mud. Its density must be above the pore pressure (with a margin) at the point of the interval where the pore pressure is highest, and below the fracture pressure (with a margin) at the weakest point, usually under the shoe. If one mud is not enough for the whole interval, we split the interval by running a casing string. It is convenient to choose the depths of the shoes from the bottom up, from the curves of the pore pressure and the fracture pressure in equivalent mud weights.

Choosing the shoes from the bottom up

1
Mud at the bottom
At the planned bottom we find the mud density that is needed: the equivalent mud weight of the pore pressure plus a margin for swabbing (the drop of pressure when the string is pulled).
2
Requirement for the shoe
We add a margin for a kick: the rock under the shoe must withstand not less than this sum.
3
Depth of the shoe
On the curve of the fracture pressure we find the depth at which it reaches the required value. We set the shoe no shallower than that.
4
Next string
For the new depth of the shoe we repeat the same: which mud is needed to drill down to it, and where the rock will withstand it with a margin.
5
Checking
We check the other limits: the overbalance on beds with normal pressure (differential sticking, where the pressure difference holds the pipe against the wall), the kick tolerance, unstable shales, the protection of fresh water, and the standard sizes of casing and bits.

CalculatedAn illustrative forecast for a planned well: the equivalent mud weights of the pore pressure (the top of the overpressure at 2600 m) and of the fracture pressure by the Eaton method. The vertical dashed line is the requirement for the shoe for drilling to 4000 m (1.70 g/cm³). It crosses the fracture pressure curve at a depth of about 1920 m.

Forecast by depth (for the example below)

Depth, m: 1800
Pore pressure, g/cm³: 1.05
Fracture pressure, g/cm³: 1.683

Depth, m: 1900
Pore pressure, g/cm³: 1.05
Fracture pressure, g/cm³: 1.697

Depth, m: 2000
Pore pressure, g/cm³: 1.05
Fracture pressure, g/cm³: 1.711

Depth, m: 2100
Pore pressure, g/cm³: 1.05
Fracture pressure, g/cm³: 1.724

Depth, m: 2600
Pore pressure, g/cm³: 1.05
Fracture pressure, g/cm³: 1.784

Depth, m: 4000
Pore pressure, g/cm³: 1.60
Fracture pressure, g/cm³: 2.099

Depth, m	Pore pressure, g/cm³	Fracture pressure, g/cm³
1800	1.05	1.683
1900	1.05	1.697
2000	1.05	1.711
2100	1.05	1.724
2600	1.05	1.784
4000	1.60	2.099

Equivalent mud weights relative to the rotary table. The data are assumed.

Worked exampleWhere the shoe of an intermediate string may stand

Given (assumed data)

Planned bottom: 4000 m true vertical, pore pressure 1.60 g/cm³
Margin for swabbing: 0.04 g/cm³
Margin for a kick: 0.06 g/cm³
Forecast of pressures by depth: the table above

Solution

Mud at the bottom: $1.60 + 0.04 = 1.64$ g/cm³. The requirement for the rock under the shoe: $1.64 + 0.06 = 1.70$ g/cm³.
From the table the fracture pressure is 1.697 g/cm³ at 1900 m and 1.711 g/cm³ at 2000 m. By interpolation, $1900 + \dfrac{1.70 - 1.697}{1.711 - 1.697}\cdot 100 = 1921$ m. The shoe must be no shallower than this depth.
Check of the overbalance: if the shoe is set just below this depth, the mud of 1.64 g/cm³ will press on the sandstones with normal pressure at the top of the overpressure (2600 m) with an overbalance of $0.00981\cdot(1.64 - 1.05)\cdot 2600 = 15.0$ MPa.
A shoe at the top of the overpressure, at 2600 m: the fracture pressure there is 1.784 g/cm³, which is 0.084 g/cm³ above the requirement.

Answer: By the fracture pressure, the shoe of the intermediate string must be no shallower than 1921 m. Allowing for the overbalance on the beds with normal pressure, it is reasonable to run it down to the top of the overpressure, about 2600 m.

What it means. The calculation from the bottom up gives only the minimum depth of the shoe. An overbalance of about 15 MPa on sandstones with normal pressure threatens differential sticking and lost circulation. So the interval with normal pressure is usually cased off entirely, and the overpressure zone is drilled separately, with a heavier mud. This is why it is so important to know the depth of the top of the overpressure (Abnormal formation pressures: origin, prediction and recognition). To drill down to 2600 m with a mud of about 1.09 g/cm³, we need a shoe that withstands 1.15 g/cm³. By the forecast, the rock withstands this at any depth, so we choose the depth of the surface casing (the string that shuts off fresh water and carries the blowout preventers) for other reasons. The data in this example are assumed for illustration.

Terms and symbols

Fracture pressure, $p_f$: The pressure in the well at which the rock ruptures and begins to take mud. As an equivalent mud weight, it is the fracture gradient.
Smallest principal stress: The stress against which a fracture opens. In most basins it is the minimum horizontal stress $\sigma_h$ .
Lateral earth pressure coefficient $K$: The ratio of the horizontal effective stress to the vertical one. In the Eaton method $K = \nu/(1-\nu)$ .
LOT: Leak-off test: pumping into the shut-in well is stopped at the first departure of the plot from the straight line.
FIT: Formation integrity test: a test to a set pressure. It confirms that the rock withstands the pressure that is needed, without taking it to leak-off.
Leak-off pressure: The pressure at which the pressure–volume plot departs from the straight line.
Closure pressure: The pressure at which the walls of the fracture close as the pressure falls. It is an estimate of the smallest stress.
MAASP: The maximum allowable annular surface pressure: $(\rho_{LOT} - \rho_m)\,g\,h_{\text{shoe}}$ . It gets smaller when the mud is made heavier.

Common mistakes and misconceptions

Wrong

“The test under the shoe gave 1.62 g/cm³, so the whole next interval will withstand such a mud.”

In fact

The test describes a few metres of rock under the shoe. Weaker rocks may lie below: depleted sandstones, fractured and cavernous zones.

Wrong

“The leak-off pressure is what the gauge at the wellhead showed.”

In fact

The rock is pressed by the sum of the wellhead pressure and the pressure of the mud column down to the shoe. The equivalent mud weight is the density of the mud during the test plus $p_{LO}/(g\,h_{\text{shoe}})$ .

Wrong

“The higher the pressure to which the test was taken, the more reliable the shoe.”

In fact

After a rupture the rock loses its tensile strength, and the fracture opens again at a lower pressure. So the LOT is stopped at the first departure, and where the limit is known we keep to the FIT.

Wrong

“It is enough for the mud density to be lower than the equivalent mud weight from the test.”

In fact

During circulation the ECD presses on the shoe, and when pipe is run in the pressure surges press on it too. We choose the density with a margin, and we check the ECD at the shoe separately.

Wrong

“The MAASP was calculated after the test, and it holds to the end of the interval.”

In fact

The MAASP depends on the density of the mud in the annulus and gets smaller with every increase of density. We recalculate it after every change of density.

In depth: advanced detail

Why the breakdown pressure is higher than the smallest stress

A round hole in a loaded rock concentrates the stresses. At the wall of a vertical well the hoop (tangential) stress, the stress along the circle of the wall, changes around the circumference, and the rock ruptures first at the point where this stress is smallest. For an elastic rock, and a wall that the mudcake (the thin layer of solids that the mud leaves on the wall) separates from the mud, the solution for a hole in a plate (the Kirsch solution) gives the breakdown pressure:

Breakdown pressure of a vertical well

p_b = 3\sigma_h - \sigma_H - p_p + T

where:

$p_b$ breakdown pressure, MPa
$\sigma_h,\ \sigma_H$ smallest and largest horizontal stresses (total), MPa
$p_p$ pore pressure, MPa
$T$ tensile strength of the rock; zero for a fractured rock, MPa

Units: any identical units of pressure

Holds when

a vertical well, with the vertical stress as one of the principal stresses; the rock is elastic and uniform;
the mud does not penetrate the rock before the rupture. If it does, the breakdown pressure is lower;
the temperature of the wall equals the temperature of the formation. Cooling of the wall by the mud lowers the breakdown pressure.

With equal horizontal stresses, $p_b = 2\sigma_h - p_p + T$ . For example, at $\sigma_h = 30$ MPa, $p_p = 20$ MPa and $T = 0$ the wall holds 40 MPa, although a pressure only a little above 30 MPa already propagates the fracture. After the first rupture $T$ disappears, and the shoe holds less. If $\sigma_H$ is much larger than $\sigma_h$ , the formula gives a pressure below $\sigma_h$ . A fracture at the wall appears earlier, but it cannot go deep until the pressure exceeds $\sigma_h$ .

For inclined wells the concentration of stress depends on the orientation of the hole relative to the principal stresses. When the vertical stress is the largest, the pressure at which rupture starts usually falls as the inclination angle grows. Such calculations are the subject of the geomechanics of the wellbore (Rock mechanics for hydraulic fracturing).

Where the ratio ν/(1 − ν) comes from

For an elastic isotropic rock, Hooke's law (stress is proportional to strain) in effective stresses gives the horizontal strain (the relative stretching sideways) $\varepsilon_h = [\sigma'_h - \nu(\sigma'_H + \sigma'_v)]/E$ , where $E$ is Young's modulus (the stiffness of the rock). If the horizontal stresses are equal ( $\sigma'_H = \sigma'_h$ ), and the rock cannot expand sideways ( $\varepsilon_h = 0$ ), then $\sigma'_h(1 - \nu) = \nu\,\sigma'_v$ , which gives $\sigma'_h = \dfrac{\nu}{1-\nu}\,\sigma'_v$ . In Russian-language rock mechanics this relation has long been used to estimate the initial stresses in a rock mass (the hypothesis of A. N. Dinnik).

The conditions of the derivation hold only roughly: sediments compact irreversibly as they sink, shales creep (they deform slowly under load), and tectonics adds horizontal stresses. So in the Eaton method we fit $\nu$ from tests. The same derivation explains the fall of the fracture pressure in depletion. When the pore pressure falls by $\Delta p$ and the rock cannot expand sideways, $\sigma_h$ falls by about $(1 - K)\,\Delta p$ . Ratios measured on fields can be larger or smaller, so we check them by tests.

Checking the test by the slope of the straight line

We can estimate the slope of the straight part in advance. The pumped volume $\Delta V$ goes into compressing the mud and expanding the casing: $\Delta V \approx V\,c_{\text{eff}}\,\Delta p$ , where $V$ is the volume of mud in the closed system, and $c_{\text{eff}}$ is the effective compressibility (for water about $4.5\cdot 10^{-4}$ MPa⁻¹, and more if we count the expansion of the casing and the lines). So the expected slope is $\Delta p/\Delta V \approx 1/(V\,c_{\text{eff}})$ .

For example, at $V$ = 65 m³ and $c_{\text{eff}} = 6.5\cdot 10^{-4}$ MPa⁻¹ the slope is about 0.024 MPa/L, which is about 0.95 MPa per 40 L. In the lesson's example the slope is 0.0226 MPa/L: the test agrees with the volume of the well. If the slope is clearly smaller than expected, there is air or gas in the system, or a leak. Such a test is not fit for calculation.

Connections to other topics

Pore pressure and overburden pressure — overburden pressure, effective stress and equivalent mud weight.
Abnormal formation pressures: origin, prediction and recognition — where the pore pressure comes from, on which the fracture pressure depends, and where overpressure begins.
Well control: primary barrier and kick indicators — lost circulation as a cause of a kick, and the role of the shoe when the well is shut in.
Well kill methods — the MAASP during killing, and the pressure at the shoe while gas is circulated out.
Circulating pressure losses — where the ECD comes from and how to calculate it.
Casing design — how the depths of the shoes and the pressures pass into the design of the casing strings.
Rock mechanics for hydraulic fracturing — stresses in the formation, and the fracture pressure and closure pressure in hydraulic fracturing.

Check yourself

1
In an area of normal faulting, where $\sigma_v > \sigma_H > \sigma_h$ , what is the fracture that forms during lost circulation like?
1. A.Horizontal: the overburden pressure opens it
2. B.The direction of the fracture does not depend on the stresses
3. C.Vertical, perpendicular to the largest horizontal stress
4. D.Vertical, perpendicular to the smallest horizontal stress
Show the answer
Correct answer: D. Vertical, perpendicular to the smallest horizontal stress
A fracture opens against the smallest principal stress, which here is $\sigma_h$ . Its plane is vertical and perpendicular to $\sigma_h$ , that is, it is oriented along $\sigma_H$ .
2
The shoe is at 2200 m true vertical depth, the mud during the test is 1.20 g/cm³, and the leak-off pressure at the wellhead is 7.0 MPa. Find the equivalent mud weight and the MAASP after the mud is made heavier, to 1.30 g/cm³.
Show the answer
$\rho_{LOT} = 1.20 + 7.0/(0.00981\cdot 2200) \approx 1.20 + 0.324 = 1.524$ g/cm³. The MAASP at 1.30 g/cm³ is $0.00981\cdot(1.524 - 1.30)\cdot 2200 \approx 4.8$ MPa. This is less than the 7.0 MPa with mud of 1.20 g/cm³.
3
Why is a leak-off test stopped at the first departure from the straight line, and not taken to rupture?
1. A.After rupture the rock under the shoe loses its tensile strength and opens again at a lower pressure
2. B.A departure from the straight line always means a leak in the equipment
3. C.To save mud
4. D.A rupture cannot be recorded with a gauge
Show the answer
Correct answer: A. After rupture the rock under the shoe loses its tensile strength and opens again at a lower pressure
An intact wall holds more thanks to its tensile strength and the concentration of stresses. After a rupture the fracture opens at a pressure close to the smallest stress, and the shoe becomes weaker. So the test is stopped at the start of leak-off.
4
The pore pressure in a sandstone fell by 6 MPa during production. How will the fracture pressure change if $K = 0.6$ ?
Show the answer
By $p_f = K\sigma_v + (1 - K)\,p_p$ , at an unchanged overburden pressure, the fracture pressure falls by $(1 - 0.6)\cdot 6 = 2.4$ MPa. The window narrows from above. The mud that is needed for neighbouring shales with their former pressure may begin to be lost into the depleted sandstone, and the overbalance on the sandstone itself grows, so the risk of differential sticking is higher.
5
The equivalent mud weight from the test under the shoe is 1.55 g/cm³, the mud is 1.45 g/cm³, and the friction losses in the annulus above the shoe at a depth of 2000 m are 2.5 MPa. Is circulation safe?
1. A.Yes: friction losses do not act on the shoe
2. B.No: the ECD at the shoe is about 1.58 g/cm³ and exceeds 1.55
3. C.We cannot say without the bit diameter
4. D.Yes: the mud density is less than 1.55 g/cm³
Show the answer
Correct answer: B. No: the ECD at the shoe is about 1.58 g/cm³ and exceeds 1.55
$\rho_{ECD} = 1.45 + 2.5/(0.00981\cdot 2000) \approx 1.45 + 0.127 = 1.58$ g/cm³, which is above the leak-off pressure. With such a circulation rate the rock under the shoe will begin to take mud. We must reduce the pump rate, make the mud lighter or change its properties.
6
By the forecast, a mud of 1.50 g/cm³ is needed to drill to the bottom, and the margin for a kick is 0.06 g/cm³. At what depths can the shoe stand, if the fracture pressure grows with depth and is 1.56 g/cm³ at 1400 m?
Show the answer
The rock under the shoe must withstand $1.50 + 0.06 = 1.56$ g/cm³. The fracture pressure grows with depth, so the shoe must stand no shallower than 1400 m. Deeper is still more reliable, if that does not contradict other limits (the overbalance on beds with normal pressure, the kick tolerance).

Updated: 19 October 2026