Solution_SP3

Solution_SP3 - buoyancy Natural 1 606 1 1000 4 = = = = = =...

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3/8/2008 ME 145 - Solution to Supplemental Problem 5 ( 29 ( 29 ( 29 ( 29 ( 29 kW . MW . . . p v m W MPa p MPa . p MPa . Pa x . . . gH p . v . MPa v MPa p m pump f pump pump drop R f B f f f 7 12 0127 0 0077 0 00165 0 1000 0.0077 0.0523 - 0.06 Needed Pump 06 0 0523 0 10 23 5 250 1 606 15 8 9 rise pressure
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Unformatted text preview: buoyancy Natural 1 606 1 00165 15 15 1000 4 = = = = = = → → = = =-=-= = = = = = = ∆ ρ...
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This homework help was uploaded on 04/09/2008 for the course ME 145 taught by Professor Manno during the Spring '08 term at Tufts.

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