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Sequential Methods for Coupled Geomechanics and Multiphase Flow

Sequential Methods for Coupled Geomechanics and Multiphase Flow

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36 CHAPTER 3. STABILITY OF THE DRAINED AND UNDRAINED SPLITS<br />

3.3.2 Drained split<br />

In the drained split, the solution is obtained sequentially by first solving the mechanics<br />

problem, <strong>and</strong> then the flow problem. The pressure field is frozen when the mechanical<br />

problem is solved. The drained-split approximation of the operator A can be written as<br />

⎡<br />

⎤<br />

⎡<br />

⎣ un<br />

pn ⎦ Au dr<br />

−→ ⎣ un+1<br />

pn ⎤<br />

⎦ Ap<br />

dr<br />

−→<br />

⎡<br />

⎣ un+1<br />

pn+1 ⎤<br />

⎦, where<br />

⎧<br />

⎪⎨ Au dr : Div σ + ρbg = 0, δp = 0,<br />

⎪⎩ A p<br />

dr : ˙m + Div w = ρf,0f, ˙ε : prescribed,<br />

(3.32)<br />

where the superscript n indicates the time level. One solves the mechanical problem with<br />

no pressure change, thus allowing the fluid to drain. Then, the fluid flow problem is solved<br />

with a frozen displacement field.<br />

3.3.3 Undrained split<br />

In contrast to the drained split, the undrained split uses a different pressure predictor <strong>for</strong><br />

the mechanical problem, which is computed by imposing that the fluid mass in each grid<br />

block remains constant during the mechanical step (δm = 0). The original operator A is<br />

split as follows:<br />

⎡<br />

⎤<br />

⎡<br />

⎣ un<br />

pn ⎦ Au ud<br />

−→ ⎣ un+1<br />

p ∗<br />

⎤<br />

⎦ Ap<br />

ud<br />

−→<br />

⎡<br />

⎣ un+1<br />

pn+1 ⎤<br />

⎦, where<br />

⎧<br />

⎪⎨ Au ud : Div σ + ρbg = 0, δm = 0,<br />

⎪⎩ A p<br />

ud : ˙m + Div w = ρf,0f, ˙ε : prescribed.<br />

(3.33)<br />

The undrained strategy allows the pressure to change locally when the mechanical prob-<br />

lem is solved. From Equation 3.3, the undrained condition (δm = 0) yields<br />

<strong>and</strong> the pressure is updated locally in each element using<br />

0 = bδεv + 1<br />

δp, (3.34)<br />

M<br />

p ∗ = p n − bM(ε n+1<br />

v − ε n v). (3.35)

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