Q1: For a homogeneous earth dam 50 m high having 2 m free broad, a flow net was constructed and the results were : Number of potential drops = 2.4 Number of flow channels = 0.4. If coefficiency of permeability of the dam mateiral is 3 x 10^-3 cm³/sec, the discharge per metre length of dam, is

ANS:D - 24 x 10^-5 m³/sec

Number of potential drops = 2.4,
Number of flow channels = 0.4.
Coefficient of permeability of the dam material = 3 x 10^-3 cm^3/sec.
Height of the dam = 50 m.
Freeboard of the dam = 2 m.

Calculations:
Head loss, h = 50 m - 2 m = 48 m.
Discharge per meter length of the dam, Q = kh * (Nf/Nd).
= (3 x 10^-3 cm^3/sec) * 48 m * (0.4/2.4).
= 24 x 10^-5 m^3/sec.
Therefore, the discharge per meter length of the dam is 24 x 10^-5 m^3/sec.
K = 3*10^-3cm3/sec.
(0.003/100) = 0.00003 m3/sec, OR.
3*10^-5m3/sec.

H = (50 -2 ) 48 m.
Nf = .4,
Nd = 2.4,

Then formula;
Q = kh* (Nf/Nd).

Q = .00003 * 48 * (.4/2.4)
Q = 0.00024 OR 24*10^-5 m3/sec. Given data,
K = 3*10^-3cm3/sec
(0.003/100) = 0.00003 m3/sec, OR
3*10^-5m3/sec.

H = (50 -2 ) 48 m.
Nf = .4
Nd = 2.4
Than formula,
Q = kh* (Nf / Nd)

Q = .00003 * 48 * (.4 / 2.4
Q = 0.00024 OR 24*10^-5 m3/sec. The discharge PER METER is asked, so the calculated answer must be divided by 100
since 1m=100cm.