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Problem 3.117

# Problem 3.117 - L 2 2 ⋅ − = Thus the minimum pressure...

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Problem *3.117 [Difficulty: 2] Given: Spinning U-tube sealed at one end Find: Maximum angular speed for no cavitation Assumptions: (1) water is incompressible (2) constant angular velocity Solution: Basic equation In components r p ρ a r = ρ V 2 r = ρ ω 2 r = z p ρ g = Between D and C, r = constant, so z p ρ g = and so p D p C ρ g H = (1) Between B and A, r = constant, so z p ρ g = and so p A p B ρ g H = (2) Between B and C, z = constant, so r p ρ ω 2 r = and so p B p C p 1 d 0 L r ρ ω 2 r d = p C p B ρ ω 2 L 2 2 = (3) Integrating Since p D = p atm , then from Eq 1 p C p atm ρ g H + = From Eq. 3 p B p C ρ ω 2 L 2 2 = so p B p atm ρ g H + ρ ω 2 L 2 2 = From Eq. 2 p A p B ρ g H = so p A p atm ρ ω 2 L 2 2
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Unformatted text preview: L 2 2 ⋅ − = Thus the minimum pressure occurs at point A (not B). Substituting known data to find the pressure at A: p A 14.7 lbf in 2 ⋅ 1.94 slug ft 3 ⋅ 1600 rev min ⋅ 2 π ⋅ rad ⋅ rev × min 60 s ⋅ × ⎛ ⎝ ⎞ ⎠ 2 × 1 2 × 3 in ⋅ ft 12 in ⋅ × ⎛ ⎝ ⎞ ⎠ 2 × lbf s 2 ⋅ slug ft ⋅ × ft 12 in ⋅ ⎛ ⎝ ⎞ ⎠ 2 × − 2.881 lbf in 2 ⋅ = = At 68 o F from steam tables, the vapor pressure of water is p v 0.339 psi ⋅ = which is less than the pressure at A. Therefore, cavitation does not occur.:...
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