Page 74 - A Practical Companion to Reservoir Stimulation

PRACTICAL COMPANION TO RESERVOIR STIMULATION

EXAMPLE E-9
Thus, at x, = 1000 ft,
Net Pressure Determination
I’
Using the data in Table E-5, calculate the net pressure in the Ap, (psi) = 0.0254 (2.5 x lo6)’ (40) (100) (1000) ‘
fracture for the PKN model when the fracture half-length is ( 1 - 0.25).’ 904
1000 ft. Plot the treating pressure for both the PKN and the
KGD models for fracture half-lengths up to 2000 ft. = 990psi. (E-44)

Solution (Ref. Sections 8-2.7 and 7-2.3.1) For the PKN model, the net pressure (for any .x,) is
Equation 8-5 1 relates the net pressure at the wellbore for the I
PKN model while Eq. 8-52 describes the net pressure for the Ap, = 175.~,‘; (E-45)
KGD model. These relationships are not exact.
To reach an exact relationship, Eq. 7- 13 is adjusted in terms whereas for the KGD model, the relationship is
of G (and remembering that d = h, for PKN); i.e.,a I
Apf = 98091;’. (E-46)
(E-40) Figure E-7 is a graph of the treating pressure for the two
models, remembering that the treating pressure is
and then after combination with Eq. 8-19, the net pressure Ap,
can be obtained: PI = Apf + 6000. (E-47)
(Note that it is unlikely that the KGD model would be
(E-41) applicable for such long fracture lengths, hence the dashed
line.)
For the units in Table E-5, Eq. E-41 becomes

I G = 2.5 x lo6 psi
Ayl (psi) = 0.0254 (E-42) I v = 0.25

For the KGD model, the equation is
Apl (psi) = 0.050 1 G”q’P 1: 100 cp
TI
r
(1 - v)”h/x; (E-43) CTH,~,” = 6000 psi
Table Ed-Variables for net pressure determination in
Example E-9.

E-14