Page 136 - A Practical Companion to Reservoir Stimulation

DESIGN AND PERFORMANCE OF ACID FRACTURES

EXAMPLE 1-4
Then, from Eq. 18-23 and rearrangement,
Performance of an Acid Fracture
(2) (0.16) (SO) ( 1.2 x ( 100’) ( 1.6 x lo7)
Calculate the oil and gas cumulative production after 10 days Gp = (0.376) (0.95) (640)
for the two fractures described in Table 1-2. In both cases, the
effective stress is 3000 psi and the formation strength is = 1.3 x lo6 MSCF. (1-14)
60,000 psi.

Solution (Ref. Section 18-5)
Oil Well Gas Well
A. Oil well
The dimensionless time at 10 days is k =lmd k = 0.2 md
(0.000264) ( 1) ( 10 x 24) I h =28ft Ih = 50ft I
tD.r,f = = 3.35, ’ (I- 11) I 4 I4 I
(0.21) (1) (9 x 10-6) (1002) = 0.21 = 0.16
leading to el, = 2.5 from Fig. 18-13. ,Ll =Icp ,Ll = 0.025cp
Then, from Eq. 18-22 and rearrangement, ct = 9 x psi-’ Z = 0.95

(2.5) (0.21) (28) (9x (100’) (2000) Xf = 100ft T = 640”R
Np =
(3.73 x 10-2) (1.15) B = 1.15 resbbl/STB c, = 1.2 x psi-’
= 62,000 STB. (1-12) I p,-pwf = 2000 psi I p,’-p$f= 1.6 x lo7 psi2 I
Xf = 100ft
B. Gas well Table I-2-Well and reservoir data for Example 1-4.
The dimensionless time at 10 days is

(0.000264) (0.2) ( 10 x 24)
tnrf =
(0.16) (0.025) (1.2 x 10-4) (1002)

= 2.64, (1-13)
and from Fig. 18- 13, QD = 2.

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