Question

$\text{The discharge over a}{90}^{\circ }\text{V-notch is given as}Q=1.37H^{\frac{5}{2}}{\text{, where Q is in m}}^3\text{/s and H in m, the}{C}_d\text{of the notch is}$

• A. 0.611
• B. 0.580
• C. 0.464
• D. 0.710

Answer 1

The correct option is B 0.580

Flow over Triangular weir or V-Notch


$Q=\frac{8}{15}{C}_d\sqrt{2g}\tan \frac{\theta }{2}{H}^{\frac{5}{2}}...\left(i\right)$

Equation of discharge given in question is:

$Q=1.37H^{\frac{5}{2}}...\left(ii\right)$

Compare equation (i) and (ii)

$\Rightarrow \frac{8}{15}{C}_d\sqrt{2g}\tan \left(\frac{\theta }{2}\right)=1.37$

$\text{Given}:{90}^{\circ }\text{V-notch}\Rightarrow \theta ={90}^{\circ }$

$\Rightarrow \frac{8}{15}{C}_d\sqrt{2g}\tan {45}^{\circ }=1.37$

$C_d=\frac{1.37\times 15}{8\sqrt{2g}}=\frac{1.37\times 15}{8\times 4.429}=\frac{20.55}{35.432}=0.58$