Page 170 - Fundamentals of Reservoir Engineering

DARCY'S LAW AND APPLICATIONS 108

equ. (4.14) is, say, 200 reservoir cc/sec, then the left hand side of equ. (4.15) must
also have the numerical value of 200, even though q in the latter is in stb/d, i.e.

conversion
q(stb / d) × = q(r.cc / sec)
factor

which is satisfied by

r.cc / sec
q(stb / d) = q(r.cc / sec)
stb / d

This preserves the balance on the left hand side of both equations. The conversion
factor can be expanded as

r.cc / sec r.cc / sec rb / d
= =
stb/ d rb/ d stb / d

Applying this method throughout, then

2
D 2 cm atm
kmD × A ft 2
stb r.cc / sec rb / d mD ft dp psi psi
q =− ×
d rb / d stb / d µ (cp) dl ft cm (4.16)
ft
D 1 cm atm 1
and since = ; = 30.48 and = ; equ.(4.16)
mD 1000 ft psi 14.7

can be evaluated as

kA dp
×
q =− 1.127 10 − 3 (stb / d) (4.17)
µ o dl

EXERCISE 4.1 UNITS CONVERSION

2
2
1) What is the conversion factor between k, expressed in Darcies, and in cm and metre ,
respectively.
2) Convert the full equation for the linear flow of an incompressible fluid, which in Darcy
units is

kA dp ρ g dz
q =− +
µ dl 1.0133 10 6 dl
×
to field units.