Elevators - Force and Power
Work done by Lifting the Elevator
The work done by lifting an elevator from one level to an other can be expressed as
W = m a g (h1 - h0 ) (1)
where
W = work done (J, ft lbf )
m = mass of elevator and passengers (kg, lbm )
a g = acceleration of gravity (9.81 m/s2, 32.17 ft/s2)
h1 = final elevation (m, ft)
h0 = initial elevation (m, ft)
The generic equation for work done by a force can be expressed as
W = F c s (2)
where
W = work done by force (J, ft lbf )
F c = force acting on the elevator at constant speed (N, lbf )
s = distance moved by elevator (m, ft)
Forces acting on the Elevator
Since works done in (1) and (2) are equal - the equations can be combined to
F c s = m a g (h1 - h0 ) (3)
Force at constant Speed
Since the difference in elevation and the distance moved by the force are equal - (3) can be modified to express the force required to move the elevator at constant speed to
F c = m a g (4)
Force at start/stop
When the elevator starts or stops - the acceleration or deceleration force in addition to the constant speed force can be expressed as
F a = m (v1 - v0 ) / t a (5)
where
F a = acceleration force (N, lbf )
v1 = final velocity (m/s, ft/s)
v0 = initial velocity (m/s, ft/s)
t a = start or stop (acceleration) time (s)
Power required to move the Elevator
The power required to move the elevator can be calculated as
P = W / t
= m a g (h1 - h0 ) / t (6)
where
P = power (W, ft lbf )
t = time to move the elevator between levels (s)
Example - Force and Power to Lift an Elevator
An elevator with mass 2000 kg including passengers are moved from level 0 m to level 15 m . The force required to move the elevator at constant speed can be calculated as
F c = (2000 kg) (9.81 m/s2)
= 19820 N
The power required to move the elevator between the levels in 20 s can be calculated as
P = (2000 kg) (9.81 m/s2) ((15 m) - (0 m)) / (20 s)
= 14865 W
= 14.9 kW
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• Mechanics
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