Question

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

Answer 1

answer is option A i.e.45.45 hp

Step-by-step 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 =
`(load)/(efficiency * 1 horse power) \\`

=
`(20000)/(0.8*550) = 45.45 hp`