Page 225 - Fundamentals of Reservoir Engineering

OILWELL TESTING 163

D cm atm
2kmD × h ft × (p − p ) psi
π
i
wf
p D = mD ft psi fieldunits
qstb / d rb / d rcc / sec µ

stb/ d rb/ d

π
2 (1/1000) × (30.48) × (1/14.7) kh
= (p − p )
i
wf
B(1.84) qµ
o
kh
p D = 7.08 × 10 − 3 (p − p ) (7.22)
i
wf
qB
µ
o
The reasons for using dimensionless variables in pressure analysis are as follows.
a) The variables lead to both a simplification and generality in the mathematics. The latter
is probably the more important and implies that if the radial flow of any fluid can be
described by the differential equ. (7.18) then the solutions will be identical irrespective
of the nature of the fluid. In this current chapter equ. (7.18) is being applied to a fluid of
small and constant compressibility for which the solutions are the p D (r D,t D ) functions.
In Chapter 8, however, it will be shown that an equation identical in form to equ. (7.18)
can be applied to the flow of a real gas. In this case the solutions are for m D (r D, t D )
functions which are dimensionless real gas pseudo pressures. Nevertheless, solutions
of equ. (7.18) expressed as p D functions will have the same form as solutions in terms
of the m D functions.

b) Since the variables are dimensionless then equations expressed in terms of them are
invariant in form, irrespective of the units system used. The same holds true, of course,
for dimensionless plots of p D as a function of t D. The scales have the same numerical
value whether Darcy, field or Sl units are employed. This latter point will be referred to
again in connection with the Matthews, Brons and Hazebroek plots presented in
sec. 7.6. Thus suppose, for instance, a value of p D (t D) = 35.71 is determined as the
result of solving an equation or reading a chart for a certain value of t D. Then if the
reservoir parameters, fluid properties and rate are

= 3500 psi (238.1 atm) = 1.2 rb/stb
p i B o
k = 150 mD (.15D) µ = 3 cp
h = 20 ft (609.6 cm) q = 100 stb/d (220.8 rcc/sec)

S = 3

then equ. (7.22) (field units) can be used to determine p wf in psi as

−
7.08 × 10 − 3 × 150 × 20 (3500 p wf ) = 35.71 S
+
100 × 3 × 1.2

0.059(3500 p ) = 38.71
−
wf
p wf = 2844psi