Difference between revisions of "6/π stimulated well potential"

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From Mass conservation:
 
From Mass conservation:
  
−
:<math>\frac{d(\rho q)}{2 dx}=y_e h \phi \frac{d\rho}{dt}</math>
+
:<math>\frac{d(\rho q)}{2 dx}=y_e h \phi \frac{d\rho}{dt}</math> (1)
  
 
From [[Darcy's law]]:
 
From [[Darcy's law]]:
  
−
:<math>\frac{q}{2}=\frac{kA}{\mu}\ \frac{dP}{dx}</math>
+
:<math>\frac{q}{2}=\frac{kA}{\mu}\ \frac{dP}{dx}</math> (2)
  
 
:<math> A =y_e*h</math>
 
:<math> A =y_e*h</math>
 +
 +
(2) →(1):
  
 
:<math>dP=\frac{q \mu}{2ky_eh} dx</math>
 
:<math>dP=\frac{q \mu}{2ky_eh} dx</math>

Revision as of 09:09, 12 September 2018

Brief

Stimulated well drainage

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

Math & Physics

Pseudo steady state flow boundary conditions:

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

From Mass conservation:

\frac{d(\rho q)}{2 dx}=y_e h \phi \frac{d\rho}{dt} (1)

From Darcy's law:

\frac{q}{2}=\frac{kA}{\mu}\ \frac{dP}{dx} (2)
A =y_e*h

(2) →(1):

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

Integration gives: P-P_{wf}=\frac{q \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 \mu}{2ky_eh} x + P_{wf} \right ) dx}{\int dx} = \left. \frac{q \mu}{2ky_eh} \frac{x}{2} \right|_{x=0}^{x=x_e/2} + P_{wf} = \frac{q \mu x_e}{8ky_eh} + P_{wf}
J_D=\frac{q \mu}{2 \pi k h} \frac{1}{( \bar P - P_{wf})} =\frac{q \mu}{2 \pi k h} \frac{8ky_eh}{q \mu x_e} = \frac{4y_e}{\pi x_e}=\frac{4}{\pi}

See also

optiFrac
fracDesign
Production Potential

Nomenclature

A = cross-sectional area, cm2
h = thickness, m
k = permeability, d
P = pressure, atm
P_i = initial pressure, atm
\bar P = average pressure, atm
q = flow rate, cm3/sec
x = length, m
x_e = drinage area length, m
y_e = drinage area width, m

Greek symbols

\mu = oil viscosity, cp
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