DHARM

198 GEOTECHNICAL ENGINEERING

or S = ()92^22 +− m = 4.77 m 30 92
The quantity of seepage per metre unit length of the dam
q = k. S
= 3 × 10–2 × 10–3 × 4.77 m^3 /s
= 14.31 × 10–5 m^3 /S
= 143.1 ml/s.
Example 6.8: For a homogeneous earth dam 32 m high and 2 m free board, a flow net was
constructed with four flow channels. The number of potential drops was 20. The dam has a
horizontal filter at the base near the toe. The coefficient of permeability of the soil was 9 × 10–

(^2) mm/s. Determine the anticipated seepage, if the length of the dam is 100 metres.
Head of water, H = (32 – 2) m = 30 m
Permeability of the soil = 9 × 10–2 mm/s
Number of flow channels = 4
Number of head drops = 20
Discharge per metre length = k. H. n
n
f
d

910
10
2
3
× −
× 30 ×
4
20 m
(^3) /s
Seepage anticipated for the entire length of the dam

100 9 10 30 4
1000 20
××^2 × ×
×
−
m^3 /s
= 0.054 m^3 /s = 541 ml/s.
Example 6.9: An earth dam is built on an impervious foundation with a horizontal filter at
the base near the toe. The permeability of the soil in the horizontal and vertical directions are
3 × 10–2 mm/s and 1 × 10–2 mm/s respectively. The full reservoir level is 30 m above the filter.
A flow net constructed for the transformed section of the dam, consists of 4 flow channels and
16 head drops. Estimate the seepage loss per metre length of the dam.
kh = 3 × 10–2 mm/s kv = 1 × 10–2 mm/s
kh = 3 × 10–5 m/s kv = 10–5 m/s
H = 30 m
Equivalent permeability ke = kkhv
= 310 110×××−−^55 m/s
= 1.732 × 10–5 m/s
Seepage loss per metre length of the dam
= ke. H. n
n
f
d
= 0.433 × 10–5 × 30 ×
4
16
m^3 /s
= 1.299 × 10–4 m^3 /s
= 129.9 ml/s.