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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.
- Fall '07
- Fluid Mechanics