Question

Discharge (Q) over a triangular notch or weir is given by - 
1. \({c_d}\sqrt {2g} \tan \frac{\theta }{2} \cdot {H^{\frac{5}{2}}}\)
2. \(\frac{1}{2}{c_d}\sqrt {2g} \tan \frac{\theta }{2} \cdot {H^{\frac{5}{2}}}\)
3. \(\frac{8}{15}{c_d}\sqrt {2g} \tan \frac{\theta }{2} \cdot {H^{\frac{5}{2}}}\)
4. \(\frac{2}{3}{c_d}\sqrt {2g} \tan \frac{\theta }{2} \cdot {H^{\frac{5}{2}}}\)

Answer 1

Correct Answer - Option 3 : \(\frac{8}{15}{c_d}\sqrt {2g} \tan \frac{\theta }{2} \cdot {H^{\frac{5}{2}}}\)

Discharge over a rectangular weir:

\(Q = \frac{2}{3}{C_d}.L\sqrt {2g} {\left( H \right)^{\frac{3}{2}}}\)

Where H: - still water head

Flow over a triangular weir (V-weir):

\(Q = \frac{8}{{15}}{C_d}\sqrt {2g} \;{H^{\frac{5}{2}}}\tan \left( {\frac{\theta }{2}} \right)\)

θ: Included angel of Notch.

Flow over a trapezoidal weir (or) Notch:

\(Q = \frac{2}{3}{C_{{d_1}}}\sqrt {2g} \;LH_1^{\frac{3}{2}} + \frac{8}{{15}}{C_{{d_2}}}\sqrt {2g} .\tan \frac{\theta }{2}.{H^{\frac{5}{2}}}\)

Where, \(\left( {\frac{\theta }{2}} \right)\) : weir angle of inclination with the vertical.

\({C_{{d_1}}}\) = Coefficient of discharge for rectangular portion.

\({C_{{d_2}}}\) = Coefficient of discharge for the triangular portion.