SolHW4ME2320 - watts P s lbft s ft ft lbm slug ft lbm V VA...

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HOMEWORK No.4 1/2 2 – 48 GIVEN: p = 40 psia; V = 1.2 ft 3 /s FIND: Power required (P) SOLUTION: hp P s lbft in lb s ft p V p VA p m e m P mech 57 . 12 / 6912 ) / 40 )( / 2 . 1 ( 2 3 = = = = = = = ρ 2 – 55 GIVEN: A = 3 ft X 3 ft V = 22 ft/s; ρ = 0.075 lbm/ft 3 FIND: Minimum Power Consumption (P) SOLUTION:
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Unformatted text preview: watts P s lbft s ft ft lbm slug ft lbm V VA V m e m P mech 3 . 151 / 51 . 111 2 ) / 22 ( ) 9 )( 2 . 32 / 1 )( / 075 . ( 2 2 3 2 3 2 2 = = = = = = = & & 2 54 2/2...
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This note was uploaded on 04/17/2008 for the course ME 232 taught by Professor Monefort during the Spring '08 term at Western Michigan.

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SolHW4ME2320 - watts P s lbft s ft ft lbm slug ft lbm V VA...

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