Engineering Rock Mechanics

Flow through discontinuity networks 155

A

R 21.5m

I25m I

G-2

VE

111 VI1

5 5 Pch,.PdY, 0 0 5

L 25 25 18 10 30 10 0

H 30 30 Hc H, 30 10 5

Conductance = c = ge3 = 9.81 x (5E-4)3 = 1.022E-6

1TL 12x 1E-6xL L

AC BC CD ED DF DG

L(m) 21.5 7.0 26.0 20.0 21.0 32.0

C 4.75 14.6 3.93 5.1 1 4.87 3.19

(x E-6)

At C: Hc = 24.94 + 0.169 H,

At D: H, = 12.75 + 0.230 H,

Hence Hc = 28.19111 H, = 19.23 m

P, = 99.96 kN/m3 P, = 90.55 kN/m2

Figure 9.6 Example of network flow calculations.

the resulting set of simultaneous equations, and finally computing the flow
through each of the individual channels. An example of this calculation is
gven in Fig. 9.6.
Figure 9.7 illustrates the results of an analysis of a simulated discon-
tinuity array. The numbers on the diagram indicate the total head at
each node. From these it can be seen that the boundary conditions are such
that overall flow is from left to right across the network however, local
idiosyncratic flows can be in an opposing direction, as the figure
demonstrates.
Obviously, for more complex networks, the use of a suitable computer-
based numerical solution is necessary. It should be noted that the analysis
presented here is for an essentially two-dimensional network the analysis
cannot be simply extended into three dimensions because two dis-
continuity planes will meet along an intersection line, along which the
hydraulic head may be changing. However, commercial computer
programs are available for studying fluid flow through three-dimensional
fracture networks (noting that the word ’fracture’ is used instead of
‘discontinuity’ in hydrogeological literature).