Wellbore breakouts are a type of rock failure around the wellbore wall and occur when the stress anisotropy surpasses the shear strength limit of the rock. Maximum anisotropy is found at and , for which
(6.7) |
Hence, replacing and into a shear failure equation (Eq. ) permits finding the mud pressure that would produce a tiny shear failure (or breakout) at and :
(6.8) |
Hence,
(6.9) |
Mud pressure would extend rock failure and breakouts further in the neighborhood of and (See Fig. 6.10). Thus, is the lowest mud pressure before initiation of breakouts.
PROBLEM 6.1: Calculate the minimum mud weight (ppg) in a vertical wellbore for avoiding shear failure (breakouts) in a site onshore at 7,000 ft of depth where 4,300 psi and 6,300 psi and with hydrostatic pore pressure.
The rock mechanical properties are 3,500 psi, 0.6, and = 800 psi.
SOLUTION
Hydrostatic pore pressure results in:
The effective horizontal stresses are:
The friction angle is , and therefore, the friction coefficient is
Thus, the minimum mud pressure for avoiding shear failure (breakouts) is
This pressure can be achieved with an equivalent circulation density of
For a given set of problem variables (far field stress, pore pressure, and mud pressure), we can calculate the required strength of the rock to have a stable wellbore. Let us consider the example of Fig. 6.11 that shows the required to resist shear failure assuming the friction angle is . For example, if the rock had a MPa, one may expect a wide breakout in Fig. 6.11.
Alternatively, you could solve the previous problem analytically. The procedure consists in setting shear failure at the point in the wellbore at an angle from or . Hence, at a point on the wellbore wall at :
(6.10) |
Say hoop stress reaches the maximum principal stress anisotropy allowed by the Mohr-Coulomb shear failure criterion ( ) where the breakout begins (rock about to fail - Fig. 6.12), then
(6.11) |
which after some algebraic manipulations results in:
The breakout angle is
(6.13) |
The procedure assumes the rock in the breakout (likely already gone) is still resisting hoop stresses and therefore it is not accurate for large breakouts ( ).
You could also calculate the wellbore pressure for a predetermined breakout angle by rearranging Eq. 6.12 to
(6.14) |
PROBLEM 6.2: Calculate the breakout angle in a vertical wellbore for a mud weight of 10 ppg in a site onshore at 7,000 ft of depth where 4,300 psi and 6,300 psi and with hydrostatic pore pressure.
The rock mechanical properties are 3,500 psi, 0.6, and = 800 psi.
SOLUTION
The problem variables are the same of Problem 6.1.
For a 10 ppg mud, the resulting mud pressure is
Hence, the expected wellbore breakout angle is
Breakouts and tensile induced fractures (Section 6.4) can be identified and measured with borehole imaging tools (Fig. 6.13). Breakouts appear as wide bands of longer travel time or higher electrical resistivity in borehole images. Tensile fractures appear as narrow stripes of longer travel time or higher electrical resistivity. Borehole images also permit identifying the direction of the stresses that caused such breakouts or tensile induced fractures. For example, the azimuth of breakouts coincides with the direction of in vertical wells.
Breakouts can also be detected from caliper measurements. Caliper tools permit measuring directly the size and shape of the borehole (https://petrowiki.org/Openhole_caliper_logs). Thus, the caliper log is extremely useful to measure breakouts and extended breakouts (washouts). For the same mud density, the caliper log reflects changes of rock properties along the well and correlate with other well logging measurements.
Breakouts are a consequence of stress anisotropy in the plane perpendicular to the wellbore. Hence, knowing the size and orientation of breakouts permits measuring and calculating the direction and magnitude of stresses that caused such breakouts. This technique is very useful for measuring orientation of horizontal stresses. In addition, if we know the rock properties and , then it is possible to calculate the maximum principal stress in the plane perpendicular to the wellbore. For example, for a vertical wellbore the total maximum horizontal stress would be
(6.15) |