Civil Engineering Reference
In-Depth Information
Example 2.10:
Seepage loss through dam (iii)
A dam has the same details as in Example
2.8,
except that there is no filter drain at the
toe.
Solution:
flow net it is also seen that a
+
Δ
a
=
22.4 m. Now
α
=
45°, and hence (according to
Casagrande):
∆
a
=
0 34
.
taken from Fig
(
.
2 24
.
b
).
a
+
∆
a
Hence
Δ
a
=
7.6 m.
4
18
−
7
Total seepage loss
=
300 5 8
× × × × × × ×
.
35 60 60 24 10
=
117
m day
3
/
Fig. 2.28
Example
2.10.
2.15.3 Permeability of sedimentary deposits
A sedimentary deposit may consist of several different soils and it is often necessary to determine the
average values of permeability in two directions, one parallel to the bedding planes and the other at right
angles to them.
Let there be n layers of thicknesses H
1
, H
2
, H
3
, . . . H
n
.
Let the total thickness of the layers be H.
Let k
1
, k
2
, k
3
, . . . k
n
be the respective coefficients of permeability for each individual layer.
Let the average permeability for the whole deposit be k
x
for flow parallel to the bedding planes and k
z
for flow perpendicular to this direction.
Consider flow parallel to the bedding planes:
Total flow q Ak i
x
= =
where A
=
total area and i
=
i = hydraulic gradient.
