Elevator efficiency
An 800 kg elevator moves vertically upward at a constant speed of 1.50 m s⁻¹. In addition to its weight, a constant resistive force of 1200 N acts downward. The overall efficiency of the motor and transmission system is 78%. Take g = 9.81 m s⁻².
Calculate:
a) the mechanical power delivered to the elevator;
b) the electrical input power required;
c) the energy transferred into non-mechanical forms during 30 s of operation.
Worked answer
Solution
- Step 1
Required upward force:
F = mg + 1200. - Step 2
F = 800 × 9.81 + 1200 = 9048 N. - Step 3
Mechanical output power:
Pout = Fv = 9048 × 1.50 = 13.57 kW. - Step 4
Efficiency:
0.78 = Pout / Pin. - Step 5
Pin = 17.40 kW. - Step 6
Loss power:
17.40 - 13.57 = 3.83 kW. - Step 7
Energy lost in 30 s:
3830 × 30 ≈ 1.15 × 10⁵ J.
Final answer
Mechanical power = 13.6 kW; electrical input power = 17.4 kW; non-mechanical energy transfer ≈ 1.15 × 10⁵ J.
Common mistake
Multiplying the mechanical output power by 78% instead of using efficiency = output / input.
Concepts tested
- power
- efficiency
- resistive forces
- energy transfer
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