Page 169 - Fundamentals of Reservoir Engineering

DARCY'S LAW AND APPLICATIONS 107

kdΦ k dp dz
ρ
u = = − + ρ g (4.12)
µ dl µ dl dl
in absolute units, while

kdp ρ g dz
u =− + (4.13)
µ dl 1.0133 10 6 dl
×
6
2
in Darcy units. The constant 1.0133 × 10 is the number of dyne/cm in one
atmosphere and is required because both ρ and g have the same units in both the cgs
and Darcy systems, and yet, the second term within the parenthesis of equ. (4.13)
must have the same units as the first, namely, atm/cm.

In spite of this obvious drawback, reservoir engineers tend to work theoretically using
equations expressed in Darcy units. This practice will generally be adhered to in this
text and, in the remaining chapters, the majority of the theoretical arguments will be
developed with equations expressed in these units.

When dealing with the more practical aspects of reservoir engineering, such as well
test analysis described in Chapters 7 and 8, it is conventional to switch to what are
called practical, or field units. The word practical is applied to such systems because all
the units employed are of a convenient magnitude. There are no rules governing field
units which therefore vary between countries and companies. The set of such units
presented in table 4.1 is, however, probably the most widely accepted in the industry at
the time of writing this book.

Because of the wide variation in unit systems employed by the industry, it is very
important that reservoir engineers should be adept at converting equations expressed
in Darcy units to the equivalent form in field units, or for that matter, any other set of
units. There is a systematic approach in making such conversions which, if rigorously
applied, will exclude the possibility of error. Consider, as an example, the conversion of
equ. (4.11) from Darcy to field units. Since

2
q = u(cm/sec) × A(cm )

the equation can be expressed in more practical form, in Darcy units, as
2
k(D) A(cm ) dp
q(cc / sec) =− (atm/ cm) (4.14)
µ (cp) dl

which, when converted to field units will have the form

2
k(mD) A(ft ) dp
q(std/ d) =− (constant) (psi/ ft) (4.15)
µ (cp) dl

in which the same symbols are used in both equations.

Making the conversion amounts to evaluating the constant in equ. (4.15) and this can
be achieved simply by remembering that equations must balance. Thus, if q in