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

Revision as of 09:43, 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}

Revision as of 09:42, 10 September 2018 (view source) MishaT (talk \| contribs) (→‎Math & Physics) ← Older edit		Revision as of 09:43, 10 September 2018 (view source) MishaT (talk \| contribs) (→‎Math & Physics) Newer edit →
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	Since average pressure is: <math>\bar P = \frac{\int P dx}{\int dx}</math>		Since average pressure is: <math>\bar P = \frac{\int P dx}{\int dx}</math>

−	:<math> \bar P = \frac{ \int \limits_{0}^{x_e/2} \left ( \frac{q B \mu}{2ky_eh} + P_{wf} \right ) dx}{\int dx}</math>	+	:<math> \bar P = \frac{ \int \limits_{0}^{x_e/2} \left ( \frac{q B \mu}{2ky_eh} x + P_{wf} \right ) dx}{\int dx}</math>