6/π stimulated well potential

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:

$\left. \frac{dP}{dx} \right|_{x=x_e/2} = \left. \frac{dP}{dx} \right|_{x=-x_e/2} = 0$

$\frac{dP}{dx} =const\ for \ \forall x$

From Diffusivity Equation:

$\frac{d^2P}{dx^2}=\frac{\phi \mu c}{k} \frac{dP}{dt}$ ( 1 )

From Material Balance:

$q/2 =\frac{dV}{dt}$

$V =y_e*h*x_e/2*\phi$

$c=\frac{1}{V} \frac{dV}{dP}$

$\frac{dP}{dt} = \frac{q}{2 c y_e h \phi} \frac{2}{x_e}$ ( 2 )

( 2 ) - > ( 1 ) :

$\frac{d^2P}{dx^2}=\frac{q \mu}{2 k y_e h} \frac{2}{x_e}$ ( 3 )

Integrating ( 3 ):

$\frac{dP}{dx}=\frac{q \mu}{2 k y_e h} \frac{2}{x_e} x + c_1$

$c_1$ must satisfy boundary condition:

$c_1 = \frac{q \mu}{2 k y_e h}$

$\frac{dP}{dx}=\frac{q \mu}{k x_e y_e h} \left ( x- \frac{x_e}{2} \right )$ ( 4 )

Integrating ( 4 ):

$P - P_{wf} = \frac{q \mu}{k x_e y_e h} \left ( \frac{x^2}{2} - \frac{x x_e}{2} \right )$ ( 5 )

Since average pressure is: $\bar P = \frac{\int P dx}{\int dx}$:

$\bar P = \frac{ \int \limits_{0}^{x_e/2} \left ( \frac{q \mu}{k x_e y_e h} \left ( \frac{x^2}{2} - \frac{x x_e}{2} \right ) + P_{wf} \right ) dx}{\int \limits_{0}^{x_e/2}dx} = \frac{q \mu}{2 k x_e y_e h} \left. \frac{\frac{x^3}{3} - x_e \frac{x^2}{2}}{x} \right|_{x=0}^{x=x_e/2} + P_{wf}$

$\bar P - P_{wf} = \frac{q \mu}{2 k x_e y_e h} \frac{\frac{1}{3} \frac{x_e^3}{8} -\frac{1}{2} \frac{x_e^3}{4}}{\frac{x_e}{2}} = \frac{q \mu}{12 k h} \frac{x_e}{y_e}$

$J_D=\frac{q \mu}{2 \pi k h} \frac{1}{( \bar P - P_{wf})} =\frac{q \mu}{2 \pi k h} \frac{12 k h}{q \mu} \frac{y_e}{x_e} = \frac{6 y_e}{\pi x_e}=\frac{6}{\pi}$

Diff eq

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 ):

$\frac{d}{dx} \left ( \frac{\rho k y_e h}{\mu} \frac{dP}{dx} \right )=y_e h \phi \frac{d\rho}{dt}$ ( 3 )

$\frac{d}{dx} \left ( \frac{k \rho}{\mu} \frac{dP}{dx} \right )=\phi \frac{d\rho}{dt}$ ( 4 )

$c=\frac{1}{\rho} \frac{d \rho}{dP}$ ( 5 )

( 5 ) -> ( 4 ):

$\frac{d}{dx} \left ( \frac{k \rho}{\mu} \frac{dP}{dx} \right )=\phi c \rho \frac{dP}{dt}$ ( 6 )

$\frac{d}{dx} \left ( \frac{k}{\mu} \right ) \rho \frac{dP}{dx} + \frac{k}{\mu} \left ( \frac{d \rho}{dx} \right ) \frac{dP}{dx} + \frac{k \rho}{\mu} \frac{d^2P}{dx^2}=\phi c \rho \frac{dP}{dt}$ ( 7 )

Assumption that viscosity is constant cancels out first term in left hand side of (7):

$\frac{k}{\mu} \left ( \frac{d \rho}{dx} \right ) \frac{dP}{dx} + \frac{k \rho}{\mu} \frac{d^2P}{dx^2}=\phi c \rho \frac{dP}{dt}$ ( 8 )

$\frac{d \rho}{dx} = c \rho \frac{d P}{dx}$ ( 9 )

( 9 ) -> ( 8 ):

$\frac{k}{\mu} c \rho \left ( \frac{dP}{dx} \right )^2+ \frac{k \rho}{\mu} \frac{d^2P}{dx^2}=\phi c \rho \frac{dP}{dt}$ ( 10 )

Term $\left ( \frac{dP}{dx} \right )^2$ in (10) is second order of magnitude low and can be cancelled out, which yields:

$\frac{d^2P}{dx^2}=\frac{\phi c \mu}{k} \frac{dP}{dt}$ ( 11 )

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