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Unformatted text preview: Physics 106  Exam 2 Solutions 1. Convert the flow rate to cubic meters per second, then use power equals flow rate times g times height: 3300 ft 3 s × ( 1 m 3 . 28 ft ) 3 = 93 . 52 m 3 s [93 . 52 m 3 s × 10 3 kg m 3 ] × 9 . 81 m s 2 × 10 m = 9 . 17 × 10 6 J s = 9 . 17 MW 2. r = d 2 = 17 . 5 m The total power in the wind is: π × (17 . 5) 2 × (6 . 1 × 10 4 ) × (20 3 ) = 4 , 700 kW = 4 . 70 MW The total electrical power, accounting for the efficiencies of conversion of wind energy to rotor KE and conversion of rotor KE to electrical power, is: (4 . 70 MW ) × . 59 × . 75 = 2 . 08 MW 3. The total input power into the OTEC device is: (2 × 10 3 kg s ) × ( 10 3 g kg ) × 2 . 5 ◦ C × 1 cal g ◦ C × 4 . 184 J cal = 20 . 92 MW The efficiency of the OTEC heat engine, using the surface and subsurface water temperatures in K, is: η = 1 277 299 = 0 . 0736 Multiplying these two factors, and including the electrical conversion efficiency, gives the power output: 20 . 92 MW × . 0736...
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This note was uploaded on 02/06/2009 for the course PHYSICS 1101 taught by Professor Hasan during the Spring '08 term at Temple.
 Spring '08
 HASAN
 Power

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