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

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<math>J_D=\frac{q B \mu}{2 \pi k h} \frac{1}{( \bar P - P_{wf})} =\frac{q B \mu}{2 \pi k h} \frac{8ky_eh}{q B \mu x_e} = \frac{4y_e}{\pi x_e}=\frac{4}{\pi}</math>
 
<math>J_D=\frac{q B \mu}{2 \pi k h} \frac{1}{( \bar P - P_{wf})} =\frac{q B \mu}{2 \pi k h} \frac{8ky_eh}{q B \mu x_e} = \frac{4y_e}{\pi x_e}=\frac{4}{\pi}</math>
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==See also==
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[[:Category:optiFrac | optiFrac]]
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[[:Category:fracDesign | fracDesign]]
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[[Category:Technology]]
 
[[Category:Technology]]

Revision as of 10:00, 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} = \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}

J_D=\frac{q B \mu}{2 \pi k h} \frac{1}{( \bar P - P_{wf})} =\frac{q B \mu}{2 \pi k h} \frac{8ky_eh}{q B \mu x_e} = \frac{4y_e}{\pi x_e}=\frac{4}{\pi}

See also

optiFrac fracDesign

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