Page 96 - A Practical Companion to Reservoir Stimulation

EVALUATION OF TREATMENTS AND POSTFRACTURE PERFORMANCE

EXAMPLE F-7
Since a 1000-ft half-length fracture with kJw = 1000 md-ft
Posttreatment Pressure Transient Test Design
was intended, then
A fracture treatment was successful in an oil reservoir having
1 md permeability. The fracture design was intended to create (F-24)
a 1000-ft half-length with a kfw = 1000 md-ft. Using the data
in Table F-3, predict the length of time that the well should be and therefore the first foldout type curve can be used for the
definitely within the bilinear flow. What would be the pressure calculation.
drop at that time if the test flow rate is 500 STB/d? At CDJ= to be well within the bilinear flow regime,
the ratio fDAf/CD, must be equal to 100 (from the type curve).
Solution (Ref. Section 11-6) Thus,.
This exercise is to give a rough estimate of the required test
duration to provide meaningful and interpretable data. Often, tDxf = (ioo)(i x = iod (F-25)
it is necessary to assume plausible values of the variables to be From the definition of the dimensionless time (Eq. 11-10)
calculated in order to design a test of appropriate duration. The and rearrangement,
foldout curves at the end of Chapter 11 can be used for this
calculation. (lo- 3)(0.1)(2)(10- s)( 1000~)
At first, the dimensionless fracture wellbore storage coef- t= = 7.5 hr. (F-26)
ficient must be estimated. This is given by Eq. 11-22. (0.000264) ( 1 )
The dimensioned wellbore storage coefficient C is simply The dimensionless pressure at that time is 0.43 (from the
the well volume multiplied by the fluid compressibility within type curve). Thus, from Eq. 11-8 and rearrangement,
the tubing string. Assuming that only oil is produced, the oil
compressibility can be used. If gas is liberated in the tubing, Ap = (0.43) (14 1.2) (500) (1.1) (2)
which is often the case, then its impact must be taken into (1) (37)
account. For the purpose of this exercise, the oil is assumed to
have a very low bubblepoint pressure. Thus, = 1805 psi. (F-27)

c = K,C,, I
n( 2.44 1 )’ k = 1md
- (12,000)(6x 1
-
(4)( 144)(5.615)
= 4.17 x bbl /psi. (F-22)
1.1 resbbl/STB
From Eq. 11-22, I H = 12,000ft

(5.615)(4.17 x I
psi-’
CDJ = c, = 6~10-~
(2) (3.14) (0. I ) ( 10- 5, (37) ( 10002) 7
dtbg = 2.441 in.
= IX lo-? (F-23)

Table F-3-Posttreatment test data for Example F-7.

F-5