p0796

# p0796 - Fr 1 = u 1 √ gy 1 = y ± 2 g y y 1 1 gy 1 = y ±...

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900 CHAPTER7 . CONSERVATIONOFENERGY 7.96 Chapter 7, Problem 96 Figure 7.59: Open-channel flow over a bump. 7.96(a): Because the flow is steady and incompressible, the mass-flow rate, and hence the volume-flow rate, is constant. So, since the cross-sectional area in a rectangular channel is A = by , Q = u 0 by 0 = u 1 by 1 = u 0 = y 1 y 0 u 1 Since there is no head loss ahead of the hydraulic jump, energy conservation tells us u 2 0 2 g + y 0 = u 2 1 2 g + y 1 Then ,appea l ingtomassconse rva t iontoe l im ina te u 0 in favor of u 1 ,wef ind X y 1 y 0 ~ 2 u 2 1 2 g + y 0 = u 2 1 2 g + y 1 = 1 X y 1 y 0 ~ 2 u 2 1 2 g = y 0 y 1 Rearranging terms a little, there follows u 2 1 = 2 g ( y 0 y 1 ) ( y 2 0 y 2 1 ) /y 2 0 = 2 gy 2 0 y 0 + y 1 Therefore, the velocities u 0 and u 1 are u 0 = y 1 ± 2 g y 0 + y 1 , u 1

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Unformatted text preview: Fr 1 = u 1 √ gy 1 = y ± 2 g y + y 1 1 gy 1 = y ± 2 y 1 ( y + y 1 ) 7.96. CHAPTER 7, PROBLEM 96 901 7.96(b): For the given conditions ( y = 3 m, y 1 = 1 . 1 m), we have Fr 1 = (3 m) ± 2 (1 . 1 m)(3 m + 1 . 1 m) = 2 . 00 7.96(c): From the hydraulic-jump relations, we have y 2 y 1 = 1 2 } ² 1 + 8 Fr 2 1 − 1 ] = 1 2 } ² 1 + 8(2 . 00) 2 − 1 ] = 2 . 37 Therefore, the depth after the hydraulic jump is y 2 = 2 . 61 m...
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p0796 - Fr 1 = u 1 √ gy 1 = y ± 2 g y y 1 1 gy 1 = y ±...

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