Dynamic Wellbore Stability Analysis Under Tripping Operations

Abstract

The loadings which act on the wellbore are more frequently dynamic than static, such as the surge/swab pressure caused by tripping operations. The changing rate of loading could induce a change in wellbore stress and result in wellbore instability. The conventional Kirsch solution to calculate wellbore stress is only applicable to the steady state without considering the coupled deformation–diffusion effect. To account for these deficiencies, this paper introduces a coupled poroelastodynamics model to obtain wellbore stress distribution under dynamic loading. The model is solved by the implicit finite-difference method with the surge/swab pressure caused by tripping operations taken as the specific dynamic loading source. Then, failure criteria are applied to analyze dynamic wellbore stability and the hemisphere plots of minimum mud density (MMD) to avoid wellbore collapse are generated. The effect of in-situ stress regimes, failure criteria, and permeable properties on the instability of the borehole has been investigated, and the applicability and accuracy of the conventional failure criteria for dynamic loading conditions have been studied by the extensive triaxial compression tests performed on Bedford limestone under different strain rates. Furthermore, two specific cases have been analyzed and results show that compared with the poroelastodynamics model, the conventional method either overestimates or underestimates the MMD, the difference of which can be as significant as 0.11 g/cm³. Not limited to the tripping operations, the developed poroelastodynamics model can be applied to any specific dynamic loading conditions with the fluid inertia neglected.

Appendix

1.1 A1 Numerical Solution Strategy of the Poroelastodynamics Model

For the governing equations of the poroelastodynamics model, the numerical solving method is preferred to be used. The reason is, on one hand, the pore pressure term in the displacement equation makes it difficult to find the analytical solution. On the other hand, the analytical solution is not suitable, since the dynamic surge/swab pressure changes arbitrarily which cannot be simulated simply by any specific function. To avoid the redundant unit conversion, non-dimensional quantities should be introduced to modify Eq. (12).

• Non-dimensional distance: $\xi =\frac{r}{{r}_{w}}$

• Non-dimensional displacement: $\phi =\frac{2G}{{p}_{0}}\cdot \frac{{u}_{r}}{{r}_{w}}$

• Non-dimensional pore pressure: $\chi =\frac{p}{{p}_{0}}$

• Non-dimensional time: $\tau =\frac{ct}{{r}_{w}}$

• Non-dimensional radial stress: ${S}_{r}=\frac{{\sigma }_{r}}{{p}_{0}}$

Introducing these non-dimensional quantities into Eq. (12):

$$\left\{ \begin{gathered} \frac{{\partial ^{2} \phi }}{{\partial \xi ^{2} }} + \frac{1}{\xi }\frac{{\partial \phi }}{{\partial \xi }} - \frac{\phi }{{\xi ^{2} }} - \alpha \frac{{1 - 2v}}{{1 - v}}\frac{{\partial \chi }}{{\partial \xi }} = \frac{{\partial ^{2} \phi }}{{\partial \tau ^{2} }} \hfill \\ \frac{{\partial ^{2} \chi }}{{\partial \xi ^{2} }} + \frac{1}{\xi }\frac{{\partial \chi }}{{\partial \xi }} = \frac{{c\dot{r}_{w} }}{{c_{v} }}\frac{\chi }{{\partial \tau }} \hfill \\ \end{gathered} \right..$$

(17)

The implicit difference method is used to differentiate the diffusion equation:

$$\left\{ \begin{gathered} \frac{{\partial ^{2} \chi }}{{\partial \xi ^{2} }} + \frac{1}{\xi }\frac{{\partial \chi }}{{\partial \xi }} = \frac{{c \cdot r_{w} }}{{c_{v} }}\frac{{\partial \chi }}{{\partial \tau }} \hfill \\ \chi \left( {1,\tau } \right) = f\left( {g\frac{{r_{w} }}{c}\tau } \right) \hfill \\ \chi \left( {\infty ,\tau } \right) = 0 \hfill \\ \chi \left( {\xi ,0} \right) = 0 \hfill \\ \end{gathered} \right..$$

(18)

Since the governing equation is in cylindrical coordinates, it is better to transform the radial direction into Cartesian coordinates. The way is as follows:

$$\xi ={\xi }_{w}{e}^{x}$$

(19)

Then, the diffusion equation in Eq. (18) becomes

$$\frac{{\partial }^{2}\chi }{\partial {x}^{2}}={\xi }^{2}\frac{c\cdot {r}_{w}}{{c}_{v}}\frac{\partial \chi }{\partial \tau }={\xi }_{w}{e}^{2x}\frac{c\cdot {r}_{w}}{{c}_{v}}\frac{\partial \chi }{\partial \tau }$$

(20)

Therefore, the implicit finite differential formula of Eq. (20) is

$$\frac{{\chi }_{i-1}^{n+1}-2{\chi }_{i}^{n+1}+{\chi }_{i+1}^{n+1}}{{\left({\Delta }x\right)}^{2}}=\left({\xi }_{w}{e}^{2ix}\frac{c\cdot {r}_{w}}{{c}_{v}}\right)\frac{{\chi }_{i}^{n+1}-{\chi }_{i}^{n}}{{\Delta }t}$$

(21)

Through some derivations, we have

$$\left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} { - l_{1} } & 1 & {} \\ 1 & { - l_{2} } & 1 \\ \end{array} } & \cdots & {} \\ \vdots & \ddots & \vdots \\ {} & \cdots & {\begin{array}{*{20}c} 1 & { - l_{{m - 2}} } & 1 \\ {} & 1 & { - l_{{m - 1}} } \\ \end{array} } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {\chi _{1}^{{n + 1}} } \\ {\chi _{2}^{{n + 1}} } \\ \vdots \\ \end{array} } \\ {\chi _{{m - 2}}^{{n + 1}} } \\ {\chi _{{m - 1}}^{{n + 1}} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {d_{1} - f[\omega (n + 1)]} \\ {d_{2} } \\ \vdots \\ \end{array} } \\ {d_{{m - 2}} } \\ {d_{{m - 1}} } \\ \end{array} } \right]$$

(22)

Implicit finite-difference method is also used to differentiate the displacement equation:

$$\left\{ \begin{gathered} \frac{{\partial ^{2} \phi }}{{\partial \xi ^{2} }} + \frac{1}{\xi }\frac{{\partial \phi }}{{\partial \xi }} - \frac{\phi }{{\xi ^{2} }} - \alpha \frac{{1 - 2v}}{{1 - v}}\frac{{\partial \chi }}{{\partial \xi }} = \frac{{\partial ^{2} \phi }}{{\partial \tau ^{2} }} \hfill \\ f\left[ {\omega (n + 1)\Delta \tau } \right] = \frac{{1 - v}}{{1 - 2v}}\frac{{\partial \phi }}{{e^{x} \partial x}}\left| {(x = 0)} \right. + \frac{v}{{1 - 2v}}\frac{\phi }{{e^{x} }}\left| {(x = 0)} \right. - \alpha \chi \hfill \\ 0 = \frac{1}{{e^{x} }}\left| {(x = 0)} \right.\left( {\frac{{1 - v}}{{1 - 2v}}\frac{{\partial \phi }}{{\partial x}} + \frac{v}{{1 - 2v}}\phi } \right) - \alpha \chi \hfill \\ \phi \left( {x,0} \right) = 0 \hfill \\ \end{gathered} \right..$$

(23)

Using Eq. (19), the displacement equation in Eq. (23) could be converted into

$$\frac{1}{{e}^{2}}\frac{{\partial }^{2}\phi }{\partial {x}^{2}}-\frac{\phi }{{e}^{2x}}-\alpha \frac{1-2v}{1-v}\frac{\partial \chi }{{e}^{x}\partial x}=\frac{{\partial }^{2}\phi }{\partial {\tau }^{2}}$$

(24)

Therefore, the implicit finite differential formula of Eq. (24) is

$$\frac{1}{{e}^{2i\cdot {\Delta }x}}\frac{{\phi }_{i-1}^{n+1}-2{\phi }_{i}^{n+1}+{\phi }_{i+1}^{n+1}}{{\left({\Delta }x\right)}^{2}}-\frac{{\phi }_{i}^{n+1}}{{e}^{2i\cdot {\Delta }x}}-\alpha \frac{1-2v}{1-v}\frac{{\chi }_{i+1}^{n+1}-{\chi }_{i+1}^{n+1}}{{e}^{i\cdot \varDelta x}\cdot \varDelta x}=\frac{{\phi }_{i}^{n-1}-2{\phi }_{i}^{n}+{\phi }_{i}^{n+1}}{{\left({\Delta }\tau \right)}^{2}}$$

(25)

$$\left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {B - L_{1} } & 1 & {} \\ 1 & { - L_{2} } & 1 \\ \end{array} } & \cdots & {} \\ \vdots & \ddots & \vdots \\ {} & \cdots & {\begin{array}{*{20}c} 1 & { - L_{{m - 2}} } & 1 \\ {} & 1 & {C_{n} - L_{{m - 1}} } \\ \end{array} } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {\phi _{1}^{{n + 1}} } \\ {\phi _{2}^{{n + 1}} } \\ \vdots \\ \end{array} } \\ {\phi _{{m - 2}}^{{n + 1}} } \\ {\phi _{{m - 1}}^{{n + 1}} } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {E_{1} - A_{n} } \\ {E_{1} } \\ \vdots \\ \end{array} } \\ {E_{{m - 2}} } \\ {E_{{m - 1}} - D_{n} } \\ \end{array} } \right]$$

(26)

$${L}_{i}=\frac{1+\frac{2{(\varDelta t)}^{2}}{{e}^{2i\cdot \varDelta x}{(\varDelta x)}^{2}}-\frac{{\left(\varDelta t\right)}^{2}}{{e}^{2i\cdot \varDelta x}}}{\frac{{(\varDelta t)}^{2}}{{e}^{2i\cdot \varDelta x}{(\varDelta x)}^{2}}}$$

$${E}_{i}=\frac{1}{\frac{{(\varDelta t)}^{2}}{{e}^{2i\cdot \varDelta x}{(\varDelta x)}^{2}}}\left[{\phi }_{i}^{n-1}-2{\phi }_{i}^{n}+\alpha \frac{1-2v}{1-v}\frac{{(\varDelta t)}^{2}}{{e}^{2i\cdot \varDelta x}\varDelta x}\left({\chi }_{i+1}^{n+1}-{\chi }_{i}^{n+1}\right)\right]$$

$${A}_{n}=\frac{(1+\alpha )sin\left[\omega (n+1)\tau \right]}{\frac{v}{1-2v}-\frac{1-v}{1-2v}\frac{1}{\varDelta x}}, B=\frac{-\frac{1-v}{1-2v}\frac{1}{\varDelta x}}{\frac{v}{1-2v}-\frac{1-v}{1-2v}\frac{1}{\varDelta x}},$$

$${C}_{n}=\frac{\frac{1-v}{1-2v}\frac{1}{\varDelta x}}{\frac{v}{1-2v}+\frac{1-v}{1-2v}\frac{1}{\varDelta x}}, {D}_{n}=\frac{\alpha {\chi }_{m}^{n+1}}{\frac{v}{1-2v}+\frac{1-v}{1-2v}\frac{1}{\varDelta x}}.$$

The tridiagonal matrix algorithm (TDMA), also known as the Thomas algorithm, is a simplified form of Gaussian elimination that can be used to solve Eqs. (22) and (26)

After rearrangement, we have

1.2 A2 Conventional Method of Calculating Wellbore Stress

The in-situ stress of the virgin formation for a deviated well is as follows (Fig. 24):

$$\left[\begin{array}{c}{\sigma }_{x}^{0}\\ {\sigma }_{y}^{0}\\ {\sigma }_{z}^{0}\end{array}\right]=\left[\begin{array}{ccc}{cos}^{2}a\cdot {cos}^{2}i& {sin}^{2}a\cdot {cos}^{2}i& {sin}^{2}i\\ {sin}^{2}a& {cos}^{2}a& 0\\ {cos}^{2}a\cdot {sin}^{2}i& {sin}^{2}a\cdot {sin}^{2}i& {cos}^{2}i\end{array}\right]\left[\begin{array}{c}{\sigma }_{H}\\ {\sigma }_{h}\\ {\sigma }_{v}\end{array}\right]$$

$$\left[\begin{array}{c}{\sigma }_{xy}\\ {\sigma }_{xz}\\ {\sigma }_{yz}\end{array}\right]=\frac{1}{2}\left[\begin{array}{ccc}-sin2a\cdot cosi& sin2a\cdot cosi& 0\\ {cos}^{2}a\cdot sin2i& {sin}^{2}a\cdot sin2i& -sin2i\\ -sin2a\cdot sini& sin2a\cdot sini& 0\end{array}\right]\left[\begin{array}{c}{\sigma }_{H}\\ {\sigma }_{h}\\ {\sigma }_{v}\end{array}\right]$$

(27)

Under static loading conditions, the conventional solution for impermeable wellbore is provided by Fjar et al. (2008). It is very well-known and not include here. The conventional solution for permeable wellbore is (Chen 2008; Ma 2015)

$${\sigma }_{r}=\left(\frac{{\sigma }_{x}^{0}+{\sigma }_{y}^{0}}{2}\right)\left(1-\frac{{r}_{w}^{2}}{{r}^{2}}\right)+\left(\frac{{\sigma }_{x}^{0}-{\sigma }_{y}^{0}}{2}\right)\left(1+\frac{{3r}_{w}^{4}}{{r}^{4}}-\frac{{4r}_{w}^{2}}{{r}^{2}}\right)cos2\theta +{\sigma }_{xy}\left(1+\frac{{3r}_{w}^{4}}{{r}^{4}}-\frac{{4r}_{w}^{2}}{{r}^{2}}\right)sin2\theta +{P}_{w}\frac{{r}_{w}^{2}}{{r}^{2}}+\left[\frac{\alpha \left(1-2v\right)}{2(1-v)}\left(1-\frac{{r}_{w}^{2}}{{r}^{2}}\right)-\varnothing \right]\left({p}_{i}-{p}_{p}\right)-\alpha {p}_{p}$$

$$\begin{aligned} \sigma _{\theta } = & \left( {\frac{{\sigma _{x}^{0} + \sigma _{y}^{0} }}{2}} \right)\left( {1 + \frac{{r_{w}^{2} }}{{r^{2} }}} \right) - \left( {\frac{{\sigma _{x}^{0} - \sigma _{y}^{0} }}{2}} \right)\left( {1 + \frac{{3r_{w}^{4} }}{{r^{4} }}} \right)\cos 2\theta - \sigma _{{xy}} \left( {1 + \frac{{3r_{w}^{4} }}{{r^{4} }}} \right)\sin 2\theta \\ & - P_{w} \frac{{r_{w}^{2} }}{{r^{2} }} + \left[ {\frac{{\alpha \left( {1 - 2v} \right)}}{{2(1 - v)}}\left( {1 - \frac{{r_{w}^{2} }}{{r^{2} }}} \right) - \emptyset } \right]\left( {p_{i} - p_{p} } \right) - \alpha p_{p} \\ \end{aligned}$$

(28)

$${\sigma }_{z}={\sigma }_{z}^{0}-v\left[\frac{2{r}_{w}^{2}}{{r}^{2}}\left({\sigma }_{H}-{\sigma }_{h}\right)cos2\theta +4{\sigma }_{xy}\frac{{r}_{w}^{2}}{{r}^{2}}sin2\theta \right]+\left[\frac{\alpha \left(1-2v\right)}{2(1-v)}\left(1-\frac{{r}_{w}^{2}}{{r}^{2}}\right)-\varnothing \right]\left({p}_{i}-{p}_{p}\right)-\alpha {p}_{p}$$

$${\sigma }_{r\theta }=\left(\frac{{\sigma }_{x}^{0}-{\sigma }_{y}^{0}}{2}\right)\left(1-\frac{{3r}_{w}^{4}}{{r}^{4}}+\frac{{2r}_{w}^{2}}{{r}^{2}}\right)sin2\theta +{\sigma }_{xy}\left(1-\frac{{3r}_{w}^{4}}{{r}^{4}}+\frac{{2r}_{w}^{2}}{{r}^{2}}\right)cos2\theta$$

$${\sigma }_{\theta z}=\left({-\sigma }_{xz}sin\theta +{\sigma }_{yz}cos\theta \right)\left(1+\frac{{r}_{w}^{2}}{{r}^{2}}\right)$$

$${\sigma }_{rz}=\left({\sigma }_{xz}cos\theta +{\sigma }_{yz}sin\theta \right)\left(1-\frac{{r}_{w}^{2}}{{r}^{2}}\right),$$

where ${p}_{0}$ is the static wellbore pressure, and $\varnothing$ is the porosity.