Page 35 - A Practical Companion to Reservoir Stimulation

PRACTICAL COMPANION TO RESERVOIR STIMULATION

EXAMPLE B-5
and therefore the minimum wellhead pressure should be
Fracture Initiation Pressure
5300 - 4167 = 1133 psi. This pressure would require a sig-
nificant addition to account for the friction pressure drop.
Calculate the fracture initiation pressure for a formation with For example, pumping a crosslinked fluid at 30 BPM down a
CJH,,~,;,, = 5000 psi, CJ~,,~~,~ 6500 psi, p = 3700 psi and T = 3.5-in. tubing may result in a friction pressure drop equal to
=
500 psi (tensile stress). If this well is 10,000 ft deep, and if a 150 psi/lOOO ft and thus add 1500 psi to the wellhead pres-
water-base fracturing fluid is used (assume p = 60 lb/ft’), what sure requirement.
would be the minimum wellhead pressure to initiate the The breakdown pressure calculated from Eq. 2-50 is an
fracture (i.e., assume zero friction pressure drop in the well). upper limit. For a lower limit, Eqs. 2-52 and 2-53 can be used.

Solution (Ref. Section 2-4.7) Thus, assuming that Biot’s constant, a, is 0.7 and the
Poisson’s ratio, v, is 0.25, then from Eq. 2-53,
Equation 2-50 provides the expression for the fracture initia-
tion pressure: 0.7 ( 1 - 0.5)
77= = 0.233, (B-20)
Phrrnkdo,,,,, = 3 (5000) - 6500 - 3700 + 500 2 (1 - 0.25)
= 5300psi. (B- 18) and
This value is bottomhole. Thus, the hydrostatic pressure
head must be subtracted in order to calculate the wellhead 3 (5000) - 6500 - 2 (0.233) (3700) + 500
pressure requirement:
2(1 - 0.233)
(60) ( 10,000)
p H
- _- = 4743psi. (B-21)
APhvdrosrur ic - -
144 144
-
- 4167 psi, (B-19)

B-6