Page 461 - Petrophysics

LINEAR FLOW THROUGH FRACTURES AND CHANNELS 429

Equating this expression with Darcy’s law (Equation 7.33) and solving
for the actual permeability of the fracture (in Darcy) yields:

kf = 8.444 x 106(1 - Swcf)$fW: (7.40)

EXAMPLE

A cubic block of a carbonate rock with an intercrystalline-intergranular
porosity system has a matrix porosity of 19%. The permeability of the
matrix is 1 mD. Calculate:

(a) the permeability of the fracture if each square foot contains one
fracture in the direction of fluid flow, and
(b) the flow rate in field units through the fracture and the fracture-
matrix system.

The width of the fracture is 2.5 x lop3 in., the viscosity of the flowing
fluid is 1.5 cP, and Ap across this block is 10psia.

SOLUTION

(a) The permeability of a fracture is estimated from Equation 7.34, where
wf = 2.5 x lop3 x 2.54 = 6.35 x cm:

kf = 8.444 x lo6 x (6.35 x = 340.5Darcy

It is obvious from this extremely high value of permeability that
fractures contribute substantially to the recovery of oil from tight
formations that otherwise would be noncommercial. This contri-
bution is actually even higher as one square foot of carbonate rock
is generally likely to contain more than one fracture.
(b) The flow rate through the fracture only can be estimated from Darcy’s
law (Equation 7.6), where L = 1 ft, Ap = 10 psia, k = 340.5 Darcy,
p = 1.5 cP, and Af = 0.0025 x 1 = 2.08 x ft2. Thus:

340.5 x 2.08 x lop4 x 10
q = 1.127 = 0.533 bbl/Day
1.5 x 1

The flow rate through the matrix only is also obtained from
Equation 7.6, where the permeability of the matrix is 1mD and
A, = Af = 1 - 2.08 x lo-* 1 ft2.ThuS:

1xlxlO
q = 1.127 x IOp3 = 0.0075 bbl/Day
1.5 x 1