Integrating elemental flow across a triangular notch gives $Q=(8/15)C_d\sqrt{2g}\tan(\theta/2)H^{5/2}$. Here $\tan45^\circ=1$,so $Q=(8/15)(.6)\sqrt{19.62}(.49)^{5/2}=.2382263$ m³/s. Rounded as requested,**0.24 m³/s**.
More Fluid Mechanics questions
GATE 2024 AG Q55– A rectangular channel carries2 m³/s at 1 m/s. Depth is half the width. Manning n=.02. Find…
GATE 2023 AG Q24– Match1 Tensiometer,2 Piezometer,3 Lysimeter,4 Elbow meter,5 Pitot tube with a Consumptive…
GATE 2023 AG Q32– A best-hydraulic trapezoidal grassed waterway has side slope1H:1V, Q=1 m³/s, bed slope1%…
GATE 2023 AG Q57– Two cylindrical reservoirs A(diameter 8 m) and B(diameter 10 m) connect through a 30 m…
GATE 2022 AG Q20– The ratio of inertial forces to viscous forces is called .
GATE 2022 AG Q29– A wide rectangular earth channel carries 20 m³/s per metre width. Find critical depth in…