Problem 3.28 - L ⋅ p atm = Upon simplification x L Δ p...

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Problem 3.28 [Difficulty: 2] Given: Reservoir manometer with vertical tubes of knowm diameter. Gage liquid is Meriam red oil D 18 mm = d 6 mm = SG oil 0.827 = (From Table A.1, App. A) Find: The manometer deflection, L when a gage pressure equal to 25 mm of water is applied to the reservoir. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g = (Hydrostatic Pressure - h is positive downwards) ρ SG ρ water = (Definition of Specific Gravity) Assumptions: (1) Static liquid (2) Incompressible liquid Integrating the hydrostatic pressure equation we get: Δ p ρ g Δ h = Beginning at the free surface of the reservoir, and accounting for the changes in pressure with elevation: p atm Δ p + ρ oil g x L + ( ) + p atm
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Unformatted text preview: L + ( ) ⋅ + p atm = Upon simplification: x L + Δ p ρ oil g ⋅ = The gage pressure is defined as: Δ p ρ water g ⋅ Δ h ⋅ = where Δ h 25 mm ⋅ = Combining these two expressions: x L + ρ water g ⋅ h ⋅ ρ oil g ⋅ = Δ h SG oil = x and L are related through the manometer dimensions: π 4 D 2 ⋅ x ⋅ π 4 d 2 ⋅ L ⋅ = x d D ⎛ ⎝ ⎞ ⎠ 2 L = Therefore: L Δ h SG oil 1 d D ⎛ ⎝ ⎞ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ = Substituting values into the expression: L 25 mm ⋅ 0.827 1 6 mm ⋅ 18 mm ⋅ ⎛ ⎝ ⎞ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ = (Note: s L Δ h = which yields s 1.088 = for this manometer.) L 27.2 mm ⋅ =...
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