Problem 4.112 - p atm V 2 2 + p max = p 1 2 V 2 = p 98.1...

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Problem *4.112 [Difficulty: 3] CS x c y R x d V, A Given: Water jet shooting upwards; striking surface Find: Flow rate; maximum pressure; Force on hand Solution: Basic equations: Bernoulli and momentum flux in x direction p ρ V 2 2 + gz + constant = Assumptions: 1) Steady flow 2) Incompressible flow 3) Atmospheric pressure throughout 4) Uniform flow Given data h1 0 m = ρ 1000 kg m 3 = D1 c m = Using Bernoulli between the jet exit and its maximum height h p atm ρ V 2 2 + p atm ρ gh + = or V2 g h = V 14.0 m s = Then Q π 4 D 2 V = Q 66.0 L min =
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Unformatted text preview: p atm V 2 2 + p max = p 1 2 V 2 = p 98.1 kPa = (gage) For Dr. Pritchard blocking the jet, from x momentum applied to the CV R x u 1 u 1 A 1 ( ) = V 2 A = Hence F V 2 4 D 2 = F 15.4 N = Repeating for Dr. Fox h 15 m = V 2 g h = V 17.2 m s = Q 4 D 2 V = Q 80.8 L min = p 1 2 V 2 = p 147.1 kPa = (gage) F V 2 4 D 2 = F 23.1 N =...
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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