Page 127 - A Practical Companion to Reservoir Stimulation

PRACTICAL COMPANION TO RESERVOIR STIMULATION

EXAMPLE H-6
C. Concentration of diverter
Design of a Matrix lkeatment
Since the permeability of the high-permeability zone is twice
Table H-2 contains the variables for a carbonate formation to the permeability of the other zone (since hl = hz),
be acidized. Calculate the maximum allowable injection rate hlkl +2klhI = kh, (H-21)
and the volume of acid required to remove all of the skin effect.
Do the above assuming a single layer. where hl = hZ = 75/2.
If the formation is considered a two-layer system, with one Therefore,
permeability twice as large as the other (assume equal heights),
calculate the concentration of diverter required in the high-
permeability layer to accept 40% of the total (maximum) leading to kl = 6.7 md, and thus k2 = 13.4 md.
injection rate within 3 min of pumping. The acid is 28% HCI. From Eq. H- 13,
Solution (Ref. Sections 13-3,14-6 and 15-5) r 71

A. Muximum injection rate
From Eq. 14-6,
and therefore (by appropriate unit conversions),
qin,u, (BPM 1 =
(2) (3.14) (13.4) (9.86 x 10-I6) (37.5)
(4.917 x (10) (75) [(0.7) (8200) - 200- 43001
(0.4) (0.47) (2.65 x
(0.7) (1) (8 + 4)
= 0.55 BPM. (H-17) (0.3048) (1240) (6.9 x 10') I' (H-24)
- 1,
(0.7) ( lo-') (8 + 4)
B. Volutne of acid leading to K2 = 1.5 x IO-'s-'.
The acid capacity number for such a treatment has been From Eq. H-14,
calculated by Eq. G-7 (in Example G-3) and is equal to 0.02 17.
+
The Peclet number can be calculated from Eq. (3-3 with K, A2p[h(c,/rw) sI2
appropriate unit conversions using the maximum injection CJi), = 8 2 (kh) * K, Ap , (H-25)
rate calculated from Eq. H-17.
and therefore.
(0.55) (2.648 x IO'cm'/s/BPM) (BPM)
NP, = (1.5 x lo-') (0.8),
( cm2/s) (75) (30.48 cm/ft) ft
"" [ (8) (3.14)2 ( 13.4), (9.86 x 10-l6)* (373,.
=
= 6.4 x lo4. (H-18)
(0.7) (lo-') (8 + 4)* 1
From Eq. 13-13 and rearrangement, we can solve for the (0.3048)' (6.15 x 10l6) (1240) (6.9 x 10')
volume of acid required to result in As = 4. Thus,
= 0.001, (H-26)

(H- 1 9) or 0.1 % by weight.

and after substitution (remembering that d = 2),

(3.14) (22.86) (0.2) (0.1)*
V=
(0.0217) (6.4~ 104)-1/3 (1.7 x 104)

(e(2)(4) 1) = 46.2 m', (H-20)
-
or 12,200 gal, or approximately 160 gal/ft.

H-6