{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Problem 4.112 - p atm ρ V 2 2 p max ρ = p 1 2 ρ ⋅ V 2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 =
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: p atm ρ V 2 2 + p max ρ = p 1 2 ρ ⋅ V 2 ⋅ = p 98.1 kPa ⋅ = (gage) For Dr. Pritchard blocking the jet, from x momentum applied to the CV R x u 1 ρ − u 1 ⋅ A 1 ⋅ ( ) ⋅ = ρ − V 2 ⋅ A ⋅ = Hence F ρ V 2 ⋅ π 4 ⋅ D 2 ⋅ = F 15.4 N = Repeating for Dr. Fox h 15 m ⋅ = V 2 g ⋅ h ⋅ = V 17.2 m s = Q π 4 D 2 ⋅ V ⋅ = Q 80.8 L min ⋅ = p 1 2 ρ ⋅ V 2 ⋅ = p 147.1 kPa ⋅ = (gage) F ρ V 2 ⋅ π 4 ⋅ D 2 ⋅ = F 23.1 N =...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online