Water flows through a semi-circular weir 2 m deep and4 m wide at the top under a head of 1.50 m. Assuming C = 0.65, determine the discharge through the weir.
Please solve it something like the example below. However, the example is parabolic while the problem above is semi circular.
Please show step by step solution T.Y. :)
Extracted text: %3D Q = through the weir. 2 m Solution Q = 1 m By V 2 m Actua Actua 1.5 m 1.5-h
Extracted text: dQ = dA 2gh dA = 2x dh Express x in terms of h by squared property of parabola: 1. %D 1.5-h 2. x = 0.707 /1.5-h dA = 2(0.707 /1.5 -h )dh %D dQ = 2(0.707 /1.5 -h )dh 2gh %3D dQ = 6.263 /1.5-h h dh 1.5 Q = 6.263 V 1.5-hh dh By trigonometric substitution: h= 1.5 sin? 0 Vh = 1.2247 sin 0 Let %3D dh = 3 sin 0 cos 0 do %3D when h = 0, 0 = 0° %3D when h 1.5, 0= 90° = T/2 %3D top arge Q = 6.263 V1.5 -1.5 sin2 0(1.2247 sin 0)(3 sin @ cos ede) P/2 Q= 28.182 sin² 0 cos? Od0 By Walli's formula: 1(1) Q= 28.182 4(2) Q-5.5336 m/s (theoretical discharge) Actual discharge 3.597 m/s