Change is 120 ft and the velocity at entrance and

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change is 120 ft, and the velocity at entrance and exit are the same, so the equation simplifies to W ρ gQ = 120ft + 4.05 v 2 2g + 6.84 v 2 2g . For water, ρ g = 62.4 lb f ft 3 . Then
W = 120ft + 5.45 15.3 ft s 2 32.2 ft s 2 62.4 lb f ft 3 3 ft 3 s = 3x10 4 ft.lb f s . Now 1 horsepower is 1hp = 3.3x10 4 lb f .ft min = 550 lb f .ft s so the pump output must be W = 3x10 4 ft.lb f s 1hp 550 ft.lb f s = 54.5hp . 2. In the figure below, an open tank with water at height h drains through a rounded orifice into the air.
Solution Assume quasi-steady conditions. Bernoulli’s equation is 1 2g v 2 2 v 1 2 ( ) = z 1 z 2 + p 1 p 2 ρ g , where by 1 a point at the top of the jet is inferred, and by 2 a point lower down by a distance z. Since the pressure is equal to atmospheric pressure everywhere in the jet, we have v 2 2 v 1 2 = 2gz . Since the velocities and areas must be related by v 1 A 1 = v 2 A 2 , we have v 2 = A 1 A 2 v 1 so A 1 2 A 2 2 = 1 + 2gz v 1 2 . For a rounded orifice, to a good approximation v 1 = 2gh , so the area changes according to A 1 2 A 2 2 = 1 + z h . 3. Deen 11.13
To initiate flow, the minimum suction strength needed at point C is ρ gH B . b)

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