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24. An elevator weighing 500 kg is to be lifted up at aconstant velocity of 0.4 m s. What should be theminimum horse power of the motor to be used ? TakeAns. 288 hpl |
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Answer» GivenForce = m g = 500 kg x 9.8 m/s²=4900 NVelocity = 0.4 m/sNow, power=Force x velocity=4900 x 0.4=1960 WNow 1 W = 0.00134 hpSo, 1960 W=1960 x 0.00134=2.6 hp Given -mass of lift = 500 kgg = 10 m/s2velocity (v) = 0.4 m/s displacement D in 1 sec = 0.4m to find - HP of motor used solution -assuming that the motor is an ideal one-work done = force x D= m x g x D= 500 x 10 x 0.4 = 2000 Nm. time taken = 1 sec Power = work done / time= 2000 / 1 = 2000 W = 2kW = 2.68 HP [ 1 watt =0.00134 HP] hence required hp of motor = 2.68 HP (ANS) |
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