4/π stimulated well potential

Brief

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

Math & Physics $P |_{x=x_e/2} = P |_{x=-x_e/2} = P_i$ $\frac{dP}{dt} =0\ for \ \forall x$

From Darcy's law: $\frac{q}{2}=\frac{kA}{\mu}\ \frac{dP}{dx}$ $A =y_e*h$ $dP=\frac{q \mu}{2ky_eh} dx$

Integration gives: $P-P_{wf}=\frac{q \mu}{2ky_eh} x$

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

Nomenclature $A$ = cross-sectional area, cm2 $h$ = thickness, m $J_D$ = dimensionless productivity index, dimensionless $k$ = 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 $\mu$ =viscosity, cp