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I. Design and Performance of Acid Fractures
EXAMPLE 1-1 EXAMPLE 1-2
Acid-Fracture Penetration Calculation of Effective Acid-Fracture Conductivity
Assuming that the velocity along an acid fracture, u,, is given Estimate the effective conductivity of an acid fracture if the
by closure pressure is 7000 psi, the rock embedment strength is
9; 60,000 psi and the fracture width is 0.1 in. Repeat this
u, = - ‘ (1-1) calculation for two bottomhole pressures, 1000 psi and 4000
2 I- wh’
psi, respectively.
where qi is the injection rate, and is equal to I for the KGD
model and 7d4 for the PKN model, show a relationship be- Solution (Ref. Section 18-4)
tween acid fracture penetration, +,, and treatment variables. Case I (pK , = 1000 psi)
Using the data in Table I- 1, calculate the acid penetration. The
is
width of the fracture is already generated through the injection The effective stress, o’, approximately 6000 psi (7000 -
of a polymer pad. 1000, using Biot’s constant equal to 1).
Therefore, from Eqs. 18-19, 3-158 and 3-155,
Solution (Ref. Section 18-2.1) cI = 1.77 x 108(0.1)2.47 = 6 x lo5, (1-4)
The acid penetration is given by Eq. 18-17, and combination
with Eq. 1-1 results in
c2 = (3.8 - 0.28 In 60,000) x lo-’
9i
Xf, = - -. W (1-2) = 7.19 x (1-5)
8rDeJ:f
and
From Table 1-1 and by converting units into a consistent set,
5 -(7.19 x 10-J)(6000)
wk, = 6x 10 e
(20) (2.648~ lo3 cm’//s/BPM)
Xfil = = 8027 md - in., (1-6)
(70) (30.48cm/ft)
(0.20) (2.54 cm/ in.) I or 670 md-ft.
If pllf = 4000 psi, then Eq. 1-6 changes to
(8) (I) (4 x = 4.27 x lo3 cm, (1-3)
5 -(7.19 x 10-‘)(3000)
wkr = 6x 10 e
or 140 ft.
Obviously, the two controlling variables are the injection = 6.94 x lo4 md - in., (1-7)
rate (usually qi/h = 0.2 to 0.4 BPM/ft) and the effective
diffusion coefficient, DeF High-efficiency acids with low Defl or 5800 md-ft, denoting the significance of the drawdown
would result in long acid-fracture penetrations. pressure and the resulting effective stress on the conductivity
of an acid fracture.
9, = 20BPM
~ ~ ~
w = 0.20 in.
h = 70ft
I ~ ~ = f 4x 10-4cm2/s I
i
I r = 1 (KGDmodel) I
Table I-1-Treatment variables for Example 1-1.
I- 1