written 2.6 years ago by
RakeshBhuse
• 3.2k
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•
modified 2.6 years ago
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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)
$\begin {aligned}
E_{\min } &=\frac{3 h_{c}}{2}\\
&=\frac{3 \times 0.71}{2}\\
&=1.065 \mathrm{~m}
\end{aligned}$