Page 58 - A Practical Companion to Reservoir Stimulation

PRACTICAL COMPANION TO RESERVOIR STIMULATION

EXAMPLE D-7 EXAMPLE D-8

Calculation of Fluid Efficiency from Closure Time Closure Time with Proppant

If the pumping time and the closure time of a calibration If 300,000 Ib of proppant are pumped in a 500.000-gal total
treatment are 30 and 15 min, respectively, what is the fluid injection treatment, what should the closure time be? Use the
efficiency? What would the closure time be for an efficiency treatment variables in Table D-5.
equal to 0.75?
Solution (Ref. Section 7-4.1)
Solution (Ref. Section 7-4.1) From Eq. 7-78,
The dimensionless closure time AtD is given by
300,000
15 !JfJVpr”/’ =
At,> = - = 0.5. (D-25) (165) ( 1 - 0.37) (500,000 I 7.48)
30
Thus, from Fig. 7-18 (between the two bounds?, 11 G 0.3. = 0.043. (D-26)
If q = 0.75, then from Fig. 7- 18, At, = 7, and therefore From Eq. 7-85,
At = (7) (30) = 2 10 min.
17 = 0.043 (1 - 0.4) + 0.4 = 0.426. (D-27)
Since the injection rate is 30 BPM, then
500,000 142
f, = = 397min. (D-28)
30
With an efficiency equal to 0.426 from Fig. 7-18,
At,, G 1.2, and therefore At = ( 1.2) (397) = 476 min (8 hr).

gi = 30BPM
pprop = 165 Ib/ft3
1
I r~ ‘ = 0.4 (from a calibration treatment)

Table Dd-Treatment variables for Example D-8.

D-8