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 (h ₁ - h ₀ )                     (1)

where

W = work done (J, ft lb _f )

m = mass of elevator and passengers (kg, lb _m )

a _g = acceleration of gravity (9.81 m/s ² , 32.17 ft/s ² )

h ₁ = final elevation (m, ft)

h ₀ = 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 lb _f )

F _c = force acting on the elevator at constant speed (N, lb _f )

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 (h ₁ - h ₀ )              (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 (v ₁ - v ₀ ) / t _a (5)

where

F _a = acceleration force (N, lb _f )

v ₁ = final velocity (m/s, ft/s)

v ₀ = 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 (h ₁ - h ₀ ) / t (6)

where

P = power (W, ft lb _f )

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/s ² )

= 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/s ² ) ((15 m) - (0 m)) / (20 s)

= 14865 W

= 14.9 kW