Page 33 - A Practical Companion to Reservoir Stimulation

PRACTICAL COMPANION TO RESERVOIR STIMULATION

EXAMPLE B-3 EXAMPLE B-4

Calculation of Horizontal Stresses Critical Depth for Horizontal
vs. Vertical Hydraulic Fractures
Calculate the minimum horizontal stress at 10,000 ft depth
for a formation with the properties used in Example B-2. Assuming that the Poisson’s relationship is in effect and
Assume that the Poisson’s ratio is 0.25. Will the absolute assuming that the horizontal stresses are ‘‘locked’’ in place,
minimum stress increase or decrease during reservoir deple- what would be the critical depth above which horizontal
tion? fractures would be generated if 2000 ft of overburden were
removed by some geologic means? Use the formation vari-
Solution (Ref. Section 2-4.1) ables given in Example B-2.
Equation 2-41 a provides the absolute minimum horizontal
stress as a function of the vertical stress. The vertical stress is Solution (Ref. Section 2-4.1)
equal to the weight of the overburden. Furthermore, the pore The solution to this problem is outlined in the graphical
pressure was assumed to be the hydrostatic value. Thus, at construction shown in Fig. 2-20. Eq. 2-4 1 a can be rewritten
10,000 ft, as a function of depth, and with substitution of the calculated
variables in Example B-2, it becomes
0.25
- [I 1,483 - (0.72) (3507)l
OH, miri - (B-17)
(I - 0.25)
Figure B-2 is a graph of the vertical and minimum hori-
+ (0.72) (3507) = 55 1 1 psi. (B- 13) zontal stresses. The “original” vertical stress is shown as a
The effective minimum stress is then dashed line, whereas the current vertical stress, relieved by
2000 ft of overburden, is shown as a solid line, starting at
(5~,.,jr, = 551 1 - (0.72) (3507) = 2986 psi. (B-14) H = -2000 ft. The slope of this curve is also 165/144 =
1.146, as for the “original” vertical stress.
During depletion a rearrangement of Eq. 2-41a can be The intercept happens at 3847 ft from the original ground
instructive: surface or at 1847 ft from the current ground surface. Thus,
depths shallower than this will likely develop a horizontal
hydraulic fracture, and depths below this level will develop a
vertical hydraulic fracture.
As the pore pressure decreases, the C J decreases. For a ~ ~ ~ ~ ~
~
~
~
1 000-psi depletion in the reservoir pressure, the minimum
horizontal stress for this formation will decrease by
A~~~,,,jJl (0.72) (1000) [ 111:;5] = 480psi. (B-16)
=
~

B-4