148 PermeabilityA9.5 To solve such network problems, we have to set up and solve a
series of simultaneous equations for the unknown nodal heads. (Because
the network is horizontal, we do not need to know the elevation of
the network above the datum, i.e. the solution is independent of the
elevation head.) We do this by writing an equation for the net flow
at each node in terms of the heads at the adjacent nodes and the
conductances of the connecting channels, using the fact that the net flow
at each node is zero and that the flow in each fracture segment is given
by the product of the conductance of the fracture and the head loss
along it.
The first step is to establish a matrix of apertures for the network
segments linking the various nodes, as follows.
In this case, all the apertures are equal, but in the next question, Q9.6,
we will be solving the same network for the case when the fracture
aperture values are different between segments. Note that this matrix is
symmetrical, and we have suppressed the values in the lower left half of
the matrix.
Using this matrix, we now compute the associated matrix of conduct-
ances, using the formula qj = (ge:j)/(12vLi,j), where c;,,, e;,, and Li,j
represent the conductance, aperture and length of the fracture segment
between node i and node j, and where g and v are the acceleration
due to gravity and the kinematic viscosity of water, respectively. The
numerical values of g and v used here are 9.8 m/s2 and 1 x m2/s.
As the length of the channels is given in m, to simplify the handling of
the units, it is convenient to convert the aperture values from mm to m
as well. Once this has been done, and remembering that the block has
a height of 1 m, it is found that the values are as below with units of
conductance in m2/s.
We now determine an equation for each nodal head in the interior of
the network, using the heads of the adjacent nodes. The fundamental