(a) Find the useful power output of an elevator motor that lifts a 2300 kg load
ID: 1362137 • Letter: #
Question
(a) Find the useful power output of an elevator motor that lifts a 2300 kg load a height of 45.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10000 kg--so that only 2300 kg is raised in height, but the full 10000 kg is accelerated.W
(b) What does it cost, if electricity is 9.00¢ per kW·h?
cents (a) Find the useful power output of an elevator motor that lifts a 2300 kg load a height of 45.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10000 kg--so that only 2300 kg is raised in height, but the full 10000 kg is accelerated.
W
(b) What does it cost, if electricity is 9.00¢ per kW·h?
cents
Explanation / Answer
Here ,
a)
useful power output = increase in mechanical energy of load/time
useful power output = ((10000+ 2300) * 9.8 * 45 + 0.5 * (10000+ 2300) * 4^2)/12
useful power output = 5.522 *10^6 W
the useful power output of the motor is 5.522 *10^6 W
b)
running cost = energy * unit charge
running cost = 5.522 *10^6 * (12/3600) * 1*10^-3 * 9 cents
running cost = 165.7 cents
the running cost of elevator is 165.7 cents
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