{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Problem 3.49

# Problem 3.49 - ft 3 ⋅ 32.2 × ft s 2 ⋅ 0.5 × ft ⋅...

This preview shows page 1. Sign up to view the full content.

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 8 in = 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 4 in = 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 6 in = h 3 0.5 ft = p B 0.036 lbf in 2 1.75 1.94 × slug ft 3 32.2
This is the end of the preview. Sign up to access the rest of the document.

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 ρ 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 ρ 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)...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online