Page 119 - A Practical Companion to Reservoir Stimulation

PRACTICAL COMPANION TO RESERVOIR STIMULATION

EXAMPLE G-6 EXAMPLE G-7
Calculation of Acid Dissolving Power Fluid Volume Requirements

Estimate the acid dissolving power in gallons of rock per Calculate the fluid volume requirements for the acidizing of a
gallons of acid fora limestone. Use density of limestone as 2.7 sandstone assuming that kaolinite clay is to be removed with
g/cm3. The acid is 28% by weight. mud acid. The weight fraction of the kaolinite is 5%. The
dissolving power of acid is 0.05 gal of rocWga1 of acid. The
Solution (Ref. Section 14-7) porosity is 0.25, and the well radius is 0.328 ft. Do this for
The stoichiometric relationship of the reaction between HCl depths of damage from 1 to 5 ft.
and limestone (CaC03) is:
Soluti6n (Ref. Section 14-7)
CaC03 +2HC1 + CaCl, +H20+C02. From Eq. 14-8, the volume of acid required per foot of forma-
100 73 111 18 44 tion thickness can be calculated.

Thus, for 100 g of CaC03, there is a need for 73 g of HC1. (3.14)( 1 - 0.25)( 0.05)( r,’ - 0.328’)
One gallon of 28% HCI has a volume of 3785 cm3 and yc,d = 7.48
weighs approximately 3785 g. Therefore, it contains 1060 g (0.05)
of HCl, which can dissolve (1060) (100/73) = 1452 g of = 17.6 (r,’ - 0.328’). ((3-12)
CaC03. Since the density is 2.7 g/cm’, the volume of lime-
stone that can be dissolved is 538 cm3, or 0.142 gal. Thus, the Figure G-4 is a plot of the volume of acid required to
dissolving power of acid is 0.142 gal of rock/gal of acid. remove the range of damage.

G-8