Problem 3.118 - Problem *3.118 [Difficulty: 2] Given:...

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Problem *3.118 [Difficulty: 2] Given: Spinning U-tube sealed at one end Find: Pressure at A; water loss due to leak Assumption: Water is incompressible; centripetal acceleration is constant Solution: Basic equation From the analysis of Example Problem 3.10, solving the basic equation, the pressure p at any point ( r , z ) in a continuous rotating fluid is given by pp 0 ρω 2 2 r 2 r 0 2 + ρ g zz 0 () = (1) where p 0 is a reference pressure at point ( r 0 , z 0 ) In this case A = p 0 p D = A = z D = z 0 = H = r0 = r 0 r D = L = The speed of rotation is ω 300 rpm = ω 31.4 rad s = The pressure at D is p D 0 kPa = (gage) Hence p A 2 2 L 2 ρ g 0 = 2 L 2 2 = 1 2 1.94 × slug ft 3 31.4 rad s 2 × 3in 2 × 1f t 12 in 4 × lbf s 2 slug ft × = p A 0.42 psi = (gage) When the leak appears,the water level at
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