Page 152 - A Practical Companion to Reservoir Stimulation

PRACTICAL COMPANION TO RESERVOIR STIMULATION

EXAMPLE J-5 EXAMPLE 5-6

Vertical Fractured Wells Performance of a Single Hydraulic Fracture
vs. Openhole Horizontal Wells Connected Transversely to a Horizontal Well

Assume that vertical wells in a reservoir with the characteris- Calculate the production rate of a hydraulic fracture intercept-
tics in Table 5-6 are routinely fractured hydraulically. If one ing transversely a horizontal well. Compare with the perfor-
contemplates a 1500-ft horizontal well, what fracture length mance of the same fracture intercepted by a vertical well.
would be necessary in a vertical well to produce comparably? Table 5-7 contains the relevant well data.
Assume a very large conductivity fracture. What should the
fracture length be if /? = 0.25? Solution (Ref. Section 19-6)
First, the dimensionless fracture conductivity is calculated
Solution (Ref. Section 19-3) (Eq. 11-1 1):
From Eq. 19-2, a = 3028 ft. Then from Eq. 19-7 (also corrected
by substituting 2r,, with (/?+ 1) rw,), 1200 (5-17)
rlDxf. =
The dimensionless time, given by Eq. 1 1- 10, can be calculated
(2980) (750) for this problem as a function of real time:

(0.000264) ( 1) (24) t 7
tD.1.f = (0.15) (0.7) ( ( 12002) = 4.2 X 10-'t, (5-18)
= 114ft. (5- 16)
where t must be in days.
Since this is alargeconductivity fracture, rdSD=0.5 (seeFig. Using the large foldout type curve for FcD = 1 at the end
11-4), and therefore, xf= 228 ft. Thus, a fractured well would of Chapter 1 1, the production rate for a fractured vertical well
be far more attractive than a horizontal well since xf= 228 ft can be readily calculated.
can be achieved readily and much more economically. For example, if t = 30 days, then from Eq. 5-18, tD,f =
If p= 0.25, then the right-hand side of Eq. 19-7 is equal to 0.126, and the ratio fD.r/csf = I .26 x 10'. From the type
375 ft, and therefore, .xf = 750 ft. The latter is much more
difficult to accomplish in practice because reservoirs with
p= 0.25 are highly naturally fractured, and therefore, neither 31
FcD+m nor x,f = 800 ft is likely to be created.
This example demonstrates that an unfractured horizontal
well is not a likely substitute for fractured vertical wells in
almost all reservoirs that traditionally have undergone this
type of stimulation. However, in highly fissured reservoirs
that also experience routine screenouts and where it is not = 0.15
possible to propagate long fractures, a horizontal well could I kfw = 1200 md-ft I
be a viable alternative.

k = 0.5 md
p =3
I
I reH = rev= 2980 ft

I h = l00ft I I cof = 10-4 I
r, = 0.328 ft I I p,-pwf = 1500psi ~ I

Table J-6-Well and reservoir data for Example J-5. Table J-7-Well and reservoir data for Example J-6.

J- 12