Power is the rate of energy transfer.
A watt (W) is one joule of energy transferred per second. One kilowatt (kW) is 1,000 W. Power tells you how quickly energy is being transferred at a particular time.
For steady DC, electrical power is P = V × I. A 24 V load taking 0.5 A receives 12 W.
Energy includes time.
20 W × 5 h = 100 Wh = 0.1 kWh. A kilowatt-hour is a quantity of energy, not a rate of use.
For constant power, energy = power × time. If power changes, add the energy from the different periods rather than assuming full power all day.
Input power and useful output differ.
Equipment does not convert all its input into the desired result. Motors also produce heat, for example. Efficiency compares useful output with input under specified conditions.
A motor’s nameplate commonly states rated mechanical output. Actual electrical input depends on the operating point and losses; it is not automatically equal to the marked kW.
A lift’s energy use changes through the day.
Travel, waiting, lighting and control equipment contribute differently. Some installations can also return energy to the supply during generating operation.
A meaningful comparison needs a defined period and operating conditions. A smaller motor rating alone does not establish that a lift will use less energy.
A closer look
For AC loads, multiplying RMS voltage by RMS current gives apparent power in volt-amperes for a single-phase circuit. Real power in watts also depends on power factor. Do not apply the simple DC calculation to every motor circuit.
Check your understanding
A constant 2 kW load runs for half an hour. How much energy does it use?
Reveal the explanation
1 kWh: 2 kW × 0.5 h. Using 2 kW for one hour would use 2 kWh.