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Unformatted text preview: downward by l/2. The oil 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 water g D 1 D 2 d + l 2 + + oil g l p atm = Upon simplification: water g D 1 D 2 d + l 2 + oil g l = D 1 D 2 d + l 2 + SG oil l = l D 1 D 2 d + SG oil 1 2 = l 2.5 m 0.7 m 0.2 m + 1.75 1 2 = l 1.600m =...
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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