Page 212 - Introduction to Petroleum Engineering

HYDRAULIC FRACTURING 199
unfractured cases multiplied by a scaling factor.” In addition, relative conductivity
“expresses the ability of the fracture to conduct fluid relative to that of the formation. It
is the ratio of two products—fracture permeability times fracture width divided by
formation permeability times the width of the drained area ( drainage radius).” Relative
conductivity should be dimensionless. For convenience, the authors replaced drainage
radius with A where A is the spacing in acres. The constant 40 in Equation 10.18 is a
scaling factor. The magnitude of the scaling factor 40/A is 1 when A is 40 acres.
A key to using Figure 10.4 is strict compliance with the units specified earlier. Use
of Figure 10.4 is illustrated with the following examples. We begin by calculating
relative conductivity for a specified well pattern and formation.

Example 10.9 Relative Conductivity
Find the relative conductivity for a fracture in a 40‐acre pattern. The formation
permeability is 2 md, the fracture permeability is 150 darcies, and the fracture
width is 0.3 in.
Answer
Substitute the given values into the definition of relative conductivity using the
specified units:

.
wk 40 (03. in )(150000 md) 40
Relativeconductivity = f = = 22500
k A (2 md) 40 acrres

The productivity index ratios for two different well radii are now compared.

Example 10.10 Productivity Index Ratio
A. Find the ratio of productivity indices for a reservoir with a 120‐ft fracture
half‐length using the relative conductivity and other input from the
previous example. The well radius r is 3 in.
w
b. Repeat the preceding example with well radius r equal to 4 in.
w
Answer
A. The ratio R = L L = 120 /660 = 018. For relative conductivity of
.
/
f q
22 500, the reading on the vertical axis of Figure 10.4 is about 4.0,
so the ratio of productivity indices is 4.0. In other words, produc-
tivity after fracking is four times greater than the productivity of
the unfracked formation. Next, the productivity scaling term
025 )) =
/
.
.
t/
.
f
)
(
/
.
.
/
7 13 ln( 0 472 Lr w ) = 713 ln( ( 0 472 660 ) (. ft 1 00.
q
b. For L L = 018 and relative conductivity of 22 500, the reading on the
/
.
f q
vertical axis of Figure 10.4 is still about 4.0. However, the productivity
.
.
)
scaling term is slightly different: 7 13. / ln((. )( f 033ft) = 1 04.
0 472 660 )t/(
So the ratio of productivity indices is 4 01 04. / . = 38.
.