Deformation in depleting reservoirs¶

Subsidence¶

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Source: Wikimedia Commons -- Public Domain

Geertsma (1973) dispacement solution¶

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$$ \begin{align} u_z(r,0) &= -\frac{1}{\pi} c_m(1-\nu) \frac{D}{\left(r^2 + D^2\right)^{3/2}} \Delta P_p V \\ u_r(r,0) &= +\frac{1}{\pi} c_m(1-\nu) \frac{D}{\left(r^2 + D^2\right)^{3/2}} \Delta P_p V \end{align} $$

© Cambridge University Press Zoback, Reservoir Geomechanics (Fig. 12.17, pp. 414)

$$ \begin{align} \frac{u_z(r,0)}{\Delta H} = A(\rho, \eta) & \qquad \rho = r/R \\ \frac{u_r(r,0)}{\Delta H} = B(\rho, \eta) & \qquad \eta = D/R \end{align} $$

where

$$ A = \begin{cases} -\frac{k\eta}{4\sqrt{\rho}} F_0(m) - \frac{1}{2} \Lambda_0(p,k)+1 &\quad \mbox{for} \quad (p < 1) \\ -\frac{k\eta}{4} F_0(m) + \frac{1}{2} &\quad \mbox{for} \quad (p = 1) \\ -\frac{k\eta}{4\sqrt{\rho}} F_0(m) + \frac{1}{2} \Lambda_0(p,k) &\quad \mbox{for} \quad (p > 1) \\ \end{cases} $$$$ B = \frac{1}{k\sqrt{\rho}} \left( \left(1-\frac{1}{2} k^2 \right) F_0(m) - E_0(m) \right) $$

with

$$ m = k^2 = \frac{\rho}{\left(1-\rho\right)^2 + \eta^2} \quad \mbox{and} \quad p = \frac{k^2\left(\left(1-\rho\right)^2 + \eta^2\right)}{\left(1-\rho\right)^2 + k^2} $$

Subsidence and horizontal dispacement¶

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© Cambridge University Press Zoback, Reservoir Geomechanics (Fig. 12.17, pp. 414)

Leeville case study¶

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© Cambridge University Press Zoback, Reservoir Geomechanics (Fig. 12.18, pp. 416)