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

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:<math> \bar P = \frac{ \int \limits_{0}^{x_e/2} \left ( \frac{q B \mu}{2ky_eh} x + P_{wf} \right ) dx}{\int dx} = \frac{q B \mu}{2ky_eh} \frac{x}{2} | + P_{wf}</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} = \frac{q B \mu}{2ky_eh} \frac{x}{2} | + P_{wf}</math>
  
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:<math> \mathop{|}_0^1 </math>
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:<math> \mathop{|}_0^1 |_0^1</math>
  
 
[[Category:Technology]]
 
[[Category:Technology]]

Revision as of 09:50, 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} \left ( \frac{q B \mu}{2ky_eh} x + P_{wf} \right ) dx}{\int dx} = \frac{q B \mu}{2ky_eh} \frac{x}{2} | + P_{wf}
\mathop{|}_0^1 |_0^1
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