Problem 3.49 - ft 3 32.2 ft s 2 0.5 ft lbf s 2 slug ft ft...

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Problem 3.49 [Difficulty: 2] Given: Geometry of chamber system Find: Pressure at various locations Assumptions: (1) Water and Meriam Blue are static and incompressible (2) Pressure gradients across air cavities are negligible Solution: Basic equation dp dy ρ g = or, for constant ρ Δ p ρ g Δ h = where Δ h is height difference For point A p A p atm ρ H2O g h 1 + = or in gage pressure p A ρ H2O g h 1 = Here we have h 1 8in = h 1 0.667 ft = p A 1.94 slug ft 3 32.2 × ft s 2 0.667 × ft lbf s 2 slugft × ft 12 in 2 × = p A 0.289 psi = (gage) For the first air cavity p air1 p A SG MB ρ H2O g h 2 = where h 2 4in = h 2 0.333 ft = From Table A.1 SG MB 1.75 = p air1 0.289 lbf in 2 1.75 1.94 × slug ft 3 32.2 × ft s 2 0.333 × ft lbf s 2 slug ft × ft 12 in 2 × = p air1 0.036 psi = (gage) Note that p = constant throughout the air pocket For point B p B p air1 SG Hg ρ H2O g h 3 + = where h 3 6in = h 3 0.5 ft = p B 0.036 lbf in 2 1.75 1.94 × slug
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Unformatted text preview: ft 3 32.2 ft s 2 0.5 ft lbf s 2 slug ft ft 12 in 2 + = p B 0.416 psi = (gage) For point C p C p air2 SG Hg H2O g h 4 + = where h 4 10 in = h 4 0.833 ft = p C 0.416 lbf in 2 1.75 1.94 slug ft 3 32.2 ft s 2 0.833 ft lbf s 2 slug ft ft 12 in 2 + = p C 1.048 psi = (gage) For the second air cavity p air2 p C SG Hg H2O h 5 = where h 5 6 in = h 5 0.5 ft = p air2 1.048 lbf in 2 1.75 1.94 slug ft 3 32.2 ft s 2 0.5 ft lbf s 2 slug ft ft 12 in 2 = p air2 0.668 psi = (gage)...
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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