Page 59 - A Practical Companion to Reservoir Stimulation

FRACTURE CALIBRATION TREATMENTS

EXAMPLE D-9 The leakoff coefficient can then be found from Eq. 7-96:
Impact of Fracture Height on Leakoff Coefficient (2) (465) (3.8 x lo-’)
and Fracture Length Calculation c, =
(3.14) (501 140) 65
Repeat the calculation in Section 7-4.5.1 but use h, = 140 ft
(instead of 70 ft). Use the same well and pressure data as in = 5.3 x 10~’ft/min,”’ (D-30)
Tables 7-4 and 7-5.
which is four times larger than the value in Section 7-4.5.1.
Solution (Ref. Sections 7-4.5 and 7-4.5.1) The fracture half-length can then be calculated from
The fracture compliance can be calculated from Eq. 7-23a, and Eq. 747 using q = 0.46, V, = 2850 ft3 and g,, = 1.52 (from
thus, Table.7-3, a = 0.6). Thus,
(3.14) (0.74) (140) (1 - 0.46) (2850)
ci = = 3.8 x IO-’ft/psi. (D-29) xi =
(2) (4.3 x loh) (2 x 140) (2 x 1.52) (5.3 x lo-’) (50/ 140) 65

This compares with 1.9 x ft/psi for the example in
Section 7-4.5.1. = 165ft, (D-31)
which is one-fourth of the value calculated in Section 7-4.5.1.

D-9