Page 93 - A Practical Companion to Reservoir Stimulation

PRACTICAL COMPANION TO RESERVOIR STIMULATION

EXAMPLE F-3
md-ft, then while r',.D = 0.5 from Eq. 11-3, r',, increases
Relative Importance of Increasing Fracture Length threefold. Since r', = rw,e-sj, where sris the fracture skin effect,
or Conductivity increasing the fracture half-length would result in a major
increase in production. Thus, the conclusion for a very low
Assume that two reservoirs have permeabilities equal to 0.05 permeability reservoir is that a longer fracture length is
and md, respectively. Investigate the importance Of extremely important, whereas k,w is relatively less important.
creating fractures of 500- and 1500-ft half-lengths with kjw For the 5-md reservoir, x,= 500 ft and k,p = 500 md-ft.
products of 500 and I500 md-ft.
(3.1 4) (5) (500)
Solution (Ref. Section 11-2) a= 2 (500) = 8,
The relative capacity parameter, a, as given by Eq. 1 1 - 1 can be
calculated. For the 0.05-md reservoir and for xf = 500 ft and and thus as shown by Eq. 11-6, the effective wellbore radius
kjw = 500 md-ft, is not affected by increasing the fracture length. Thus, a longer
fracture would not result in any appreciable benefit. However,
(3.14)(0.05)(500) doubling the k,w product would result in the doubling of r:.
a= = 0.08. (F-7)
2 (500) The conclusion for high-permeability reservoirs, resulting in
low-conductivity fractures, is that the k,(w product and its
If the k,w product is 1500 md-ft, then a = 0.026 (ie., an maximization are crucial. Using a high-permeability prop-
even more conductive fracture). In both cases, the dimension- pant-one that can sustain its integrity under stress for a long
less effective wellbore radius is equal to 0.5. Thus, in a low- period of time-and especially low-damage fracturing fluids
permeability reservoir, even a low kjw product (lackluster is essential to the success of hydraulic fracture treatments in
fracture) can result in a major increase in production. If, on high-permeability reservoirs. Fracture length in this case is
the other hand, xf increases to 1500 md-ft with kjw = 1500 relatively less important.

F-2