Problem 3.39 - ( ) SG oil g h 4 = (2) The two unknowns here...

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Problem 3.39 [Difficulty: 2] h 2 h 1 h 3 h 4 x Oil Air Hg Given: Data on manometer before and after an "accident" Find: Change in mercury level Assumptions: (1) Liquids are incompressible and static (2) Pressure change across air in bubble is negligible (3) Any curvature of air bubble surface can be neglected in volume calculations Solution: Basic equation dp dy ρ g = or, for constant ρ Δ p ρ g Δ h = where Δ h is height difference For the initial state, working from right to left p atm p atm SG Hg ρ g h 3 + SG oil ρ g h 1 h 2 + () = SG Hg ρ g h 3 SG oil ρ g h 1 h 2 + () = (1) Note that the air pocket has no effect! For the final state, working from right to left p atm p atm SG Hg ρ g h 3 x () + SG oil ρ g h 4 = SG Hg ρ g h 3 x
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Unformatted text preview: ( ) SG oil g h 4 = (2) The two unknowns here are the mercury levels before and after (i.e., h 3 and x) Combining Eqs. 1 and 2 SG Hg g x SG oil g h 1 h 2 + h 4 ( ) = x SG oil SG Hg h 1 h 2 + h 4 ( ) = (3) From Table A.1 SG Hg 13.55 = The term h 1 h 2 + h 4 is the difference between the total height of oil before and after the accident h 1 h 2 + h 4 V d 2 4 = 4 1 0.5 in 2 0.2 in 3 = 1.019 in = Then from Eq. 3 x 1.4 13.55 1.019 in = x 0.1053 in =...
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