Question

The flow net below a dam consists of 25 equipotential lines and 9 flow lines. The available head is 6 m. Taking coefficient of permeability as $3.8\times {10}^{-6}m/s$, the seepage $\left(m^3/sperm\right)$ under the dam is

• A. $3.8\times {10}^{-6}$
• B. $45.6\times {10}^{-6}$
• C. $7.6\times {10}^{-6}$
• D. $8.21\times {10}^{-6}$

Answer 1

The correct option is C $7.6\times {10}^{-6}$

$N_f$ = Number of flow channels
or
$N_f$ = Number of flow lines -1

= 9-1 = 8

$N_d$ = Number of potential drops
or
$N_d$ = Number of equipotential lines - 1

= 25 - 1= 24

h = 6 m $k=3.8\times {10}^{-6}m/s$
$q=k\frac{N_f}{N_d}H$
$=3.8\times {10}^{-6}\times \frac{8}{24}\times 6=7.6\times {10}^{-6}{m}^3/sperm$