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If a host system is 80% efficient, the minimum horsepower rating of the motor should be ? if the hoist is to provide 20,000 ft-lb/s. a) 45.45hp b) 23,530hp c) 36.36hp d) 42.78hp

Respuesta :

Answer:

answer is option A i.e.45.45 hp

Explanation:

Given data:

load =20000 ft lb/s

efficiency = 80%

we know that

1 hp = 550 ft lb/s

minimum horsepower rating can be obtained by using following formula

minimum horse power rating = [tex]\frac{load}{efficiency * 1 horse power} \\[/tex]

                                                = [tex]\frac{20000}{0.8*550} = 45.45 hp[/tex]

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