Difference between revisions of "4/π stimulated well potential"

Revision as of 09:56, 10 September 2018

Stimulated well drainage

4/π is the maximum possible stimulation potential for steady state linear flow in a square well spacing.

Steady state flow boundary conditions:

P |_{x=x_e/2} = P |_{x=-x_e/2} = P_i

\frac{dp}{dt} =0\ for \ \forall x

\frac{q}{2}=\frac{kA}{B \mu}\ \frac{dP}{dx}

A =y_e*h

dP=\frac{q B \mu}{2ky_eh} dx

Integration gives: $P-P_{wf}=\frac{q B \mu}{2ky_eh} x$

Since average pressure is: $\bar P = \frac{\int P dx}{\int dx}$

\bar P = \frac{ \int \limits_{0}^{x_e/2} \left ( \frac{q B \mu}{2ky_eh} x + P_{wf} \right ) dx}{\int dx} = \left. \frac{q B \mu}{2ky_eh} \frac{x}{2} \right|_{x=0}^{x=x_e/2} + P_{wf} = \frac{q B \mu x_e}{8ky_eh} + P_{wf}

$JD=\frac{q B \mu}{2 \pi k h} \frac{1}{( \bar P - P_{wf}} =$

@@ Line 26: / Line 26: @@
 :<math> \bar P = \frac{ \int \limits_{0}^{x_e/2} \left ( \frac{q B \mu}{2ky_eh} x + P_{wf} \right ) dx}{\int dx} = \left. \frac{q B \mu}{2ky_eh} \frac{x}{2} \right|_{x=0}^{x=x_e/2} + P_{wf} = \frac{q B \mu x_e}{8ky_eh} + P_{wf}</math>
-<math>JD=\frac{q B \mu}{2 \pi k h ( /bar P - P_{wf}} = </math>
+<math>JD=\frac{q B \mu}{2 \pi k h} \frac{1}{( \bar P - P_{wf}} = </math>
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