Page 57 - A Practical Companion to Reservoir Stimulation

FRACTURE CALIBRA TION TREATMENTS

EXAMPLE D-5 EXAMPLE D-6
Calculation of Distance to Restricted Fracture Comparison of Fluid Loss
During Pumping and During Closure
Assume that after 30 min of pumping the pressure starts
increasing at 200 psi/min. If E’ = 3 x lo6 psi, qi = 40 BPM If the pumping time and the closure time is the same (30 min),
and h = 50 ft, calculate the maximum distance to the point of calculate the total fluid lost during pumping and closure. Use
restricted fracture. CL = ft/min’I2, x,= 1000 ft, h = 100 ft and hf= 200 ft.
In both cases use an average between the upper and lower
Solution (Ref. Section 7-3.3) bounds.

Since dpJdt = m = 200 psi/min, then from Eq. 7-57, Solution (Ref. Section 7-4.1)
2E%,
x/ = -. (D-19) From &. 7-66 and remembering that A,, = 4xf h$
rhj m V,, = (2) (lo-’) (4) (1000) (100) (30)”’

For the upper bound of the xf value, q = 1, and therefore,
(4/3) + (r/2)
(2) (3 x lo6) (40 x 5.615) = 6360ft’ = 1133bbl. (D-21)
L
Xf = (3.14) (50’) (200) = 860ft. (D-20) From Eq. 7-70 and since At, = 1,

From Eq. 7-7 1,
1.33 + 1.57
g” = = 1.45. (D-23)
9
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Then, finally, from Eq. 7-69 (and remembering that
r,,Af= A,,),

V,,T = (2) (10-j) (4) (1000) (100) (30)”’

(2.5 I - 1.45) = 4645 ft’ = 827 bbl. (D-24)
Invariably, the rate of fluid loss during pumping is larger
than during closure.

D-7