Page 70 - A Practical Companion to Reservoir Stimulation

PRACTICAL COMPANION TO RESERVOIR STIMULATION

EXAMPLE E-6
If the efficiency were 0.15, then
Calculation of Total Fluid Volume Requirements
and Pad Portion 8
KL = - (0.15) + 3.14 (1 - 0.15) = 3.07, (E-25)
3
Assuming that the average fracture width is 0.24 in. (from and the quadratic equation becomes
Example E-5) and using the variables in Table E-4, calculate
the total fluid volume requirements and the pad portion. t - 10.3h - 17.8 = 0, (E-26)
Repeat the calculation for a very low (0.15) and a very high resulting in a total pumping time equal to 139 min.
(0.75) efficiency. The total volume required is then

Solution (Ref. Sections 8-2.4 and 8-2.6) = (40) (42) ( 139) = 2.34 x 10' gal. (E-27
The first variable to calculate is the multiplier to the fluid loss The pad volume would be significantly different.
coefficient, effective during pumping. This is given by Eq.
8-20. :.::I
= 2.34 x lo" [: z
8
= - (0.35) + 3.14 (1 - 0.35) = 2.97. (E-19)
KL 3 = 1.73 x 10' gal (E-28
The leakoff area, A, is then representing 74% of the total volume and requiring I03 min
of pumping.
A, = 4 (100) (1000) = 4 x 105ft2. (E-20) If the efficiency were 0.75, then
Then, from Eq. 8- 18 (in the textbook, the leakoff term must be 8
divided by 2), KL = - (0.75) + 3.14 (1 - 0.75) = 2.79, (E-29)
3
and the quadratic equation becomes

t - 9.3h - 17.8 = 0. (E-30)
The required time is then 119 min, and the total fluid
volume is
Equation E-2 1 reduces to V, = (1 19) (40) (42) = 2.0 x 10' gal. (E-31)

t - 9.94; - 17.8 = 0, (E-22) The pad volume is
which is a quadratic equation, and its real solution yields %u, = 2.0 x lo5 (i :.:)
t= 131 min(2hrand I1 min). = 2.8 x 104 gal (E-32)
This result is an inverse calculation. It calculates the time
required to generate a fracture volume while the penalty of requiring only 17 min of pumping and representing 14% of
leakoff is in effect. total fluid demand.
The total volume required to generate the fracture is then

V, = (40) (42) (131) = 2.2 x 10' gal. (E-23) gj = 40BPM
From Eq. 8-32 the pad volume may then be calculated: TJ = 0.35
I w = 0.24 in. I

Xf = l000ft
= 1.06 x 105gal, (E-24) rp = 0.7 = (h,/hf)

representing 48% of the total volume. At 40 BPM it would hf = 100 ft
require 63 min of pumping.
Table E-4-Variables for total fluid volume calculation for
Example E-6.

E- 10