Problem 3.39 - ⋅ SG oil ρ ⋅ g ⋅ h 4 ⋅ =(2 The two...

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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 ρ ⋅ 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 ρ ⋅ g ⋅ x ⋅ SG oil ρ ⋅ 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 − Δ V π d 2 ⋅ 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = 4 π 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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