The discharge of water through a rectangular channel of width 8m, is $m^3/3$, when depth of flow of water is 1.2m calculate

1] specific energy of the flowing water.

2] critical depth and critical velocity.

3] value of minimum specific energy.

Given:

Discharge, Q = $15m^3/s$

width b = 8m

depth, h = 1.2m

$\therefore$ discharge per unit width,

$q = \frac{Q}{b}$

$= \frac{15}{8}$

$= 1.875 m^2/s$

Velocity of flow, $v = \frac{q}{Area}$

$= \frac{15}{b \times h}$

$\frac{15}{0 \times 1.2} = 1.5625$

$\therefore$ V = 1.5625 m/s

1] Specific energy (E)

$E = h + \frac{v^2}{2g}$

$= 1.2 + \frac{1.5625^2}{8 \times 9.81}$

= 1.2 + 0.124

[ E = 1.324m]

2] Critical path (hc)

$hc = (\frac{q^2}{g})^1/3$

$= (\frac{1.875^2}{9.81})^1/3$

= 0.71 m

Critical velocity (vc)

$Vc = \sqrt{9 \times hc}$

$\sqrt{9.81 \times 0.71}$

[Vc = 2.639 m/s]

3] Minimum specific energy (E min)

$E_{min} = \frac{3hc}{2}$

= $\frac{3 \times 0.71}{2}$

[ $E_{min}$ = 1.065m]