Question 12.16: Water in the channel in Fig. 12–37 is 2 m deep, and the dept......

Fluid Mechanics [1012629]

Water in the channel in Fig. 12–37 is 2 m deep, and the depth from the bottom of the triangular weir to the bottom of the channel is 1.75 m. If the discharge coefficient is $C_d$ = 0.57, determine the volumetric flow in the channel.

fig 12-37

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Fluid Description. The water level in the channel is assumed to be constant, so we have steady flow. Also, water is considered to be an ideal fluid.
Analysis. Here H = 2 m – 1.75 m = 0.25 m. Applying Eq. 12–35, the flow is therefore

Q_{actual}  =  C_d \frac{8}{15} \sqrt{2g} H^{5/2} \tan  \frac{\theta}{2}

=  (0.57) (\frac{8}{15}) \sqrt{2(9.81  m/s^2)} (0.25  m)^{5/2}  \tan  30°  =  0.0243  m^3/s

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