Problem 3.31 - The mercury in the right leg is displaced...

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Problem 3.31 [Difficulty: 2] Given: A U-tube manometer is connected to the open tank filled with water as shown (manometer fluid is mercury). The tank is sealed and pressurized. D 1 2.5 m = D 2 0.7 m = d 0.2 m = p o 0.5 atm = SG Hg 13.55 = (From Table A.1, App. A) Find: The manometer deflection, l 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 D 1 D 2 d c d Integrating the hydrostatic pressure equation we get: Δ p ρ g Δ h = When the tank is filled with water and pressurized, the mercury in the left leg of the manometer is displaced downward by l/2.
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Unformatted text preview: The mercury in the right leg is displaced upward by the same distance, l/2. Beginning at the free surface of the tank, and accounting for the changes in pressure with elevation: p atm p o + water g D 1 D 2 d + l 2 + + Hg g l p atm = Upon simplification: p o water g D 1 D 2 d + l 2 + + Hg g l = l p o water g D 1 + D 2 d + SG Hg 1 2 = Substituting values into the expression: l 0.5 atm 1.013 10 5 N m 2 atm 1 1000 m 3 kg 1 9.8 s 2 m 2.5 m + 0.7 m 0.2 m + 13.55 1 2 = l 0.549m =...
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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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