Breakout angle determination

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. 6.10 that shows the required $UCS$ to resist shear failure assuming the friction angle is $\varphi = 30^{\circ}$. For example, if the rock had a $UCS \sim 25$ MPa, one may expect a $\sim 90^{\circ}$ wide breakout.

Figure 6.10: Example of required UCS with Mohr-Coulomb analysis to verify likelihood of wellbore breakouts.

Alternatively, you could solve the previous problem analytically. The procedure consists in setting shear failure at the point in the wellbore at an angle $(w_{BO}/2)$ from $\pi/2$ or $3\pi/2$. Hence, at a point on the wellbore wall at $\theta_B = \pi/2 - w_{BO}/2$:

$$\left\lbrace \begin{array}{rcl} \sigma_{\theta \theta} & = & ... ...\theta_B) \\ \sigma_{rr} & = & +(P_W - P_p) \end{array}\right.$$ (6.10)

Figure 6.11: Determination of breakout angle with Mohr-Coulomb failure criterion.

Say hoop stress reaches the maximum principal stress anisotropy allowed by the Mohr-Coulomb shear failure criterion ( $\sigma_1 = UCS + q \: \sigma_3$) where the breakout begins (rock about to fail - Fig. 6.11), then

$$\left[ -(P_W - P_p) + (\sigma_{Hmax} + \sigma_{hmin}) - 2(\sigma_{Hmax} - \sigma_{hmin}) \cos (2 \theta_B) \right] = UCS + q (P_W - P_p)$$ (6.11)

which after some algebraic manipulations results in:

$$2 \theta_B = \arccos \left[ \frac{ \sigma_{Hmax} + \sigma_{hmin} - UCS - (1+q)(P_W - P_p)}{2(\sigma_{Hmax} - \sigma_{hmin})} \right]$$ (6.12)

The breakout angle is

$$w_{BO} = \pi - 2 \theta_B$$ (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 ( $\sim w_{BO} > 60^{\circ}$).

You could also calculate the wellbore pressure for a predetermined breakout angle by rearranging Eq. 6.12

$$P_{WBO} = P_p + \frac{ (\sigma_{Hmax} + \sigma_{hmin}) - 2(\sigma_{Hmax} - \sigma_{hmin}) \cos (\pi - w_{BO}) - UCS}{1+q}$$ (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 7000 ft of depth where $S_{hmin} =$ 4300 psi and $S_{Hmax} =$ 6300 psi and with hydrostatic pore pressure. The rock mechanical properties are $UCS =$ 3500 psi, $\mu_i=$ 0.6, and $T_s$ = 800 psi.

SOLUTION
The problem variables are the same of problem 6.1. For a 10 ppg mud, the resulting mud pressure is

$$10$$    ppg$$\times \frac{0.44 \text{ psi/ft}} {8.3 \text{ ppg}} \times 7000 \text{ psi} = 3710 \text{ psi}$$

Hence, the expected wellbore breakout angle is

$$w_{BO} = 180^{\circ} - \arccos \left[ \frac{ 3220 \text{ psi} + 1... ...3220 \text{ psi} - 1220 \text{ psi})} \right] = 66 ^{\circ} \: \: \blacksquare$$