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Unformatted text preview: P gin 144.196hp = P gin 107.5 kW 1 hp 0.746 kW = So the engine must provide 107.5 KW at the shaft. There are 746 W/hp  converting to HP yields: P gin 1.075 10 5 watt = P gin P gout = 0.93 = P gout 100 kW = At rated output, the power out of the generator is 100 KW. Since we are given the efficiency, we can solve for the power into the generator: Problem 19 :...
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue UniversityWest Lafayette.
 Spring '12
 Piller

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