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

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(Math & Physics)
(Math & Physics)
Line 24: Line 24:
 
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}\frac{q B \mu}{2ky_eh} + P_{wf} dx}{\int dx}</math>
+
:<math> \bar P = \frac{ \int \limits_{0}^{x_e/2}\frac{q B \mu}{2ky_eh} + P_{wf} dx}{\int dx}</math>
  
  

Revision as of 09:42, 10 September 2018

Brief

Stimulated well drainage

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

Math & Physics

Steady state flow boundary conditions:

P |_{x=x_e/2} = P |_{x=-x_e/2} = P_i
 \frac{dp}{dt} =0\ for \ \forall x

From Darcy's law:

\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}\frac{q B \mu}{2ky_eh} + P_{wf} dx}{\int dx}