The discharge of water through a rectangular channel of width 8m is 15 m3/sec when depth of flow of water is 1.2m Find i) Specific energy of flowing water, ii) Critical depth, iii) Critical velocity , iv) Value of minimum Specific energy.

Solution :

Given :

Discharge,$\quad Q=15 \mathrm{~m}^{3} / \mathrm{s} $

Width of channel,$\quad b=8 \mathrm{~m} $

Depth, $\quad h=1.2 \mathrm{~m}$

Discharge per unit width,

$\begin{aligned}q=\frac{Q}{b}=\frac{15}{8}=1.875 \mathrm{~m}^{2} / \mathrm{s} \end{aligned}$

Velocity of flow,

$\begin{aligned} V &=\frac{Q}{\text { Area }}\\ &=\frac{15}{b \times h}\\ &=\frac{15.0}{8 \times 1.2}\\ &=1.5625 \mathrm{~m} / \mathrm{s} \end{aligned}$

1. Specific energy (E)

$\begin {aligned} E &=h+\frac{V^{2}}{2 g}\\ &=1.2+\frac{1.5625^{2}}{8 \times 9.81}\\ &=1.20+0.124\\ &=1.324 \mathrm{~m} \end{aligned}$

2. Critical depth (hc)

$\begin{aligned} h_{c} &=\left(\frac{q^{2}}{g}\right)^{1 / 3}\\ &= \left(\frac{1.875^{2}}{9.81}\right)^{1 / 3}\\ &=0.71 \mathrm{~m} \end{aligned}$

3. Critical velocity (Vc)

$\begin{aligned} V_{c} &=\sqrt{g \times h_{c}}\\ &=\sqrt{9.81 \times 0.71} \\ &=2.639 \mathrm{~m} / \mathrm{s} \end{aligned}$

4. Minimum specific energy (E min)

$\begin {aligned} E_{\min } &=\frac{3 h_{c}}{2}\\ &=\frac{3 \times 0.71}{2}\\ &=1.065 \mathrm{~m} \end{aligned}$