Subsections

6.6 Mechanical stability of deviated wellbores

The following subsections present a guide for calculating stresses at the wall of deviated wellbores and identifying stress magnitudes and locations for shear failure (breakouts) and tensile fractures.

6.6.1 Wellbore orientation

At any point along the trajectory of a deviated wellbore, the tangent orientation permits defining wellbore azimuth $\delta$ and deviation $\varphi$ (Fig. 6.19). Azimuth $\delta$ is the angle between the projection of the trajectory on a horizontal plane and the North. Deviation $\varphi$ is the angle between a vertical line and the trajectory line at the point of consideration. These two variables can be plotted in a half-hemisphere projection plot (stereonet). Notice that a point in this plot represents just one point along a wellbore trajectory. Fig. 6.19 shows an example of the full trajectory of a wellbore.

Figure 6.19: (a) Convention for plotting the orientation of a deviated wellbore on a lower hemisphere projection. (b) The example shows the deviation survey for a real wellbore (deviation range amplified to highlight small deviations). Notice that it starts slightly deviated on surface and then turns into vertical direction at depth.
\includegraphics[scale=0.45]{.././Figures/split/7-DevSurvey.pdf}

A given state of stress will result in different mud windows and locations of rock failure depending on the wellbore orientation. Breakouts and tensile fractures will depend on the stresses on the plane perpendicular to the wellbore. Figure 6.20 shows an example for normal faulting with generic stress magnitude values.

Figure 6.20: Example of expected location of wellbore breakouts and tensile fractures in a vertical and horizontal wellbores according to the state of stress (normal faulting with $S_{Hmax}$ in N-S direction here).
Image 7-DevWellFailure


PROBLEM 6.4: Consider a place where vertical stress $S_v$ is a principal stress and the maximum horizontal stress acts in E-W direction.

  1. Find the planes with maximum stress anisotropy for normal faulting, strike-slip, and reverse faulting stress regimes.
  2. Plot the orientation of wellbores in those planes of maximum stress anisotropy in a stereonet projection plot.
  3. Where around the wellbore would breakouts and tensile fractures occur in each case?

SOLUTION
The solution below shows just one of the possible solutions of two horizontal wells.

\includegraphics[scale=0.55]{.././Figures/split/8-ExampleDevWells.pdf}

6.6.2 Calculation of stresses on deviated wellbores

Let us define a coordinate system for a point along the trajectory of a deviated wellbore. The first element $x_b$ of the cartesian base goes from the center of a cross-section of the wellbore at a given depth to the deepest point around the cross-section (perpendicular to the axis). The second element of the base $y_b$ goes from the center to the side on a horizontal plane. The third element of the base $z_b$ goes along the direction of the wellbore.

Figure 6.21: The wellbore coordinate system.
\includegraphics[scale=0.65]{.././Figures/split/8-WellCoordSystem.pdf}

Based on the previous definition, it is possible to construct a transformation matrix $R_{GW}=f(\delta,\varphi)$ that links the geographical coordinate system and the wellbore coordinate system.

$\displaystyle \uuline{S}{}_W =
R_{GW} \uuline{S}{}_G R_{GW}^T$ (6.19)

Furthermore, the wellbore stresses can be calculated from the principal stress tensor according with:

$\displaystyle \uuline{S}{}_W =
R_{GW} R_{PG}^T \uuline{S}{}_P R_{PG} R_{GW}^T$ (6.20)

Where $\uuline{S}{}_P$ and $R_{PG}$ are the principal stress tensor and the corresponding change of coordinate matrix to the geographical coordinate system (Eq. 5.7). The tensor $\uuline{S}{}_W$ is composed by the following stresses:

\begin{displaymath}\uuline{S}{}_W =
\left[
\begin{array}{ccc}
S_{11} & S_{12} & ...
...2} & S_{23} \\
S_{31} & S_{32} & S_{33} \\
\end{array}\right]\end{displaymath} (6.21)

Stresses on the plane of the cross-section of the deviated wellbore at the wellbore wall $(\sigma_{rr}, \sigma_{\theta \theta}, \sigma_{zz}, \sigma_{\theta z})$ depend on far-field stresses $\sigma_{11} = S_{11} - P_p$, $\sigma_{22} = S_{22} - P_p$, $\sigma_{33} = S_{33} - P_p$, $\sigma_{12} = S_{12}$, $\sigma_{13} = S_{13}$, and $\sigma_{23} = S_{23}$. The Kirsch equations require additional far field shear terms $\sigma_{12}$, $\sigma_{13}$, and $\sigma_{23}$ in order to account for principal stresses not coinciding with the wellbore orientation. The solution of Kirsch equation for isotropic rock with far-field shear stresses is provided in Fig. 6.22.

Figure 6.22: Stresses around the wall of a deviated wellbore. Notice that principal stresses (directions $(x_1,x_2,x_3)$) may not be aligned with the wellbore trajectory.
\includegraphics[scale=0.65]{.././Figures/split/8-3.pdf}

Solving for the local principal stresses $(\sigma_{tmax}, \sigma_{rr}, \sigma_{tmin})$ on the wellbore wall permits checking for rock failure (tensile or shear). The local principal stresses may not be necessarily aligned with the wellbore axis leading to an angle $\omega$ (see Fig. 6.23). Because of such angle, tensile fractures in deviated wellbores can occur at an angle $\omega$ from the axis of the wellbore and appear as a series of short inclined (en-├ęchelon) fractures instead of a long tensile fracture parallel to the wellbore axis as in Fig. 6.16.

Figure 6.23: Principal stresses around the wall of a deviated wellbore. The hoop stress $\sigma _{\theta \theta }$ could be the least, the intermediate or the maximum principal stress depending on location (angle $\theta $).
Image 8-DevWellPS

6.6.3 Breakout analysis for deviated wellbores

Consider a place subjected to strike-slip stress regime with $S_{Hmax}$ oriented at an azimuth of 070$^{\circ }$ with known values of principal stresses (Fig 6.24). The maximum stress anisotropy lies in a plane that contains $S_1 = S_{Hmax}$ and $S_3 = S_{hmin}$, a plane perpendicular to the axis of a vertical wellbore. Hence, maximum stress amplification at the wellbore wall will happen for a vertical wellbore. The minimum stress anisotropy lies in a plane that contains $S_2 = S_{v}$ and $S_3 = S_{hmin}$, perpendicular to a horizontal wellbore drilled in direction of $S_{Hmax}$. Given a mud pressure and a fixed friction angle, we can calculate for a given wellbore orientation the stresses on the wellbore wall from equations in Fig. 6.23, and the required $UCS$ (using a shear failure criterion $\sigma_1 = UCS + q \: \sigma_3$) to avert shear failure. The plots in Fig. 6.24 are examples of this calculation. The maximum value of required $UCS$ corresponds to the wellbore direction with maximum stress anisotropy (vertical wellbore - red region), and the minimum value of required $UCS$ corresponds to the wellbore direction with minimum stress anisotropy (horizontal wellbore with $\delta = 070^{\circ}$ - blue region). Following the breakout concepts discussed before, we would expect breakouts at 160$^{\circ }$ and 340$^{\circ }$ of azimuth on the sides of a vertical wellbore. A horizontal wellbore drilled in the direction of $S_{hmin}$ would tend to develop breakouts on the top and bottom of the wellbore.

Figure 6.24: Stereonet plots to verify the rock strength required to avoid breakouts and the wellbore breakout angle for a given rock strength and wellbore pressure.
\includegraphics[scale=0.70]{.././Figures/split/8-BreakoutsDevWells.pdf}


PROBLEM 6.5: Consider a place with principal stresses $S_v = 70$ MPa, $S_{Hmax} = 67$ MPa (at 070$^{\circ }$), $S_{hmin} = 45$ MPa, $P_p = 32$ MPa, and $P_W = 32$ MPa. Calculate the required UCS using the Coulomb failure criterion (with $\mu_i = 0.8$) for all possible wellbore orientations. Plot results in a stereonet projection.

SOLUTION

\includegraphics[scale=0.70]{.././Figures/split/8-BreakoutsDevWells-EXNF.pdf}


PROBLEM 6.6: Consider a place with principal stresses $S_v = 70$ MPa, $S_{Hmax} = 105$ MPa (at 070$^{\circ }$), $S_{hmin} = 85$ MPa, $P_p = 32$ MPa, and $P_W = 32$ MPa. Calculate the required UCS using the Coulomb failure criterion (with $\mu_i = 0.8$) for all possible wellbore orientations. Plot results in a stereonet projection.

SOLUTION

\includegraphics[scale=0.70]{.././Figures/split/8-BreakoutsDevWells-EXRF.pdf}

6.6.4 Tensile fractures analysis for deviated wellbores

The procedure to find tensile failure is equivalent to the one used for shear failure, but using a tensile strength failure criterion. For example, consider a place subjected to normal faulting stress regime with $S_{Hmax}$ oriented at an azimuth of 070$^{\circ }$ and known values of principal stresses (Fig. 6.25). The maximum stress anisotropy lies in a plane that contains $S_1 = S_{v}$ and $S_3 = S_{hmin}$, perpendicular to a horizontal wellbore drilled in the direction of $S_{Hmax}$. For a given rock tensile strength, we can calculate the maximum mud pressure that the wellbore can bear without failing in tension. Fig. 6.25 shows an example of this calculation. The maximum possible $P_W$ corresponds to the wellbore direction with minimum stress anisotropy (blue region), and the minimum possible $P_W$ corresponds to the wellbore direction with maximum stress anisotropy (red region). For example, tensile fractures would tend to occur in the top and bottom of a horizontal wellbore drilled in the direction of $S_{Hmax}$.

Figure 6.25: Calculation of required $T_S$ as a function of wellbore orientation in order to prevent open-mode fractures with $P_W = 35$ MPa. Stresses and pore pressure: $S_v =$ 70 MPa, $S_{Hmax} =$ 55 MPa, $S_{hmin} =$ 45 MPa, and $P_p =$ 32 MPa, Notice that the stress regimes dictates the orientation of the least and most convenient drilling direction.
\includegraphics[scale=0.65]{.././Figures/split/8-TensfracsDevWells.pdf}


PROBLEM 6.7: Consider a place with principal stresses $S_v = 70$ MPa, $S_{Hmax} = 105$ MPa (at 070$^{\circ }$), $S_{hmin} = 55$ MPa, and $P_p = 32$ MPa. Calculate the maximum wellbore pressure $P_W$ at the limit of tensile strength with $T_s = 0$ MPa) for all possible wellbore orientations. Plot results in a stereonet projection.

SOLUTION
TBD


PROBLEM 6.8: Consider a place with principal stresses $S_v = 70$ MPa, $S_{Hmax} = 105$ MPa (at 070$^{\circ }$), $S_{hmin} = 85$ MPa, and $P_p = 32$ MPa. Calculate the maximum wellbore pressure $P_W$ at the limit of tensile strength with $T_s = 0$ MPa) for all possible wellbore orientations. Plot results in a stereonet projection.

SOLUTION
TBD