Electric motor losses, efficiency and IE classes

Efficiency and the energy flow

A motor converts electrical input power into mechanical output power. Nothing is converted perfectly; the rest turns into heat that warms the motor and the room.

η = P_out / P_in = (P_in − P_losses) / P_in

where P_out = T × ω is the power at the shaft. For a three-phase AC motor the input power is

P_in = √3 × V_L × I_L × cos φ

with V_L the line voltage, I_L the line current and cos φ the power factor. For a DC motor it is simply P_in = V × I.

Efficiency is not constant. It is low at light load, peaks at around 60% to 100% of rated load, and drops again when the motor is overloaded. Small motors are less efficient than large ones: a very small motor may be well below 70%, while large machines can exceed 95%.

Copper losses (I² × R)

Current flowing through the resistance of the windings produces heat.

P_cu = I² × R

They are called copper losses (or Joule or resistive losses) even in aluminium rotor bars. In a three-phase induction motor there are two parts: stator copper loss, 3 × I² × R_s, and rotor copper loss, which equals slip times the air-gap power, s × P_gap. Rotor copper loss is the price of slip: the larger the slip, the larger the share of air-gap power lost in the rotor.

Copper loss grows with the square of the current, so doubling the current gives four times the heat. This is the dominant loss at high load, and the reason overloading quickly overheats a motor. It also increases with temperature, because copper resistance rises about 0.4% per °C.

Iron (core) losses

The alternating magnetic flux in the stator core causes two losses in the steel.

• Hysteresis loss: magnetizing and demagnetizing the steel each cycle consumes energy proportional to the area of the B–H loop and to the frequency.
• Eddy-current loss: the changing flux induces circulating currents in the steel itself. They are reduced by building the core from thin insulated laminations, typically 0.35 mm to 0.5 mm thick.

Iron loss depends mainly on the voltage and frequency, because they set the flux, and hardly on the load. It is therefore roughly constant from no load to full load, and it is the main reason lightly loaded motors are inefficient. Higher grades of electrical steel, thinner laminations and a lower flux density all reduce it.

Friction, windage and stray losses

• Friction loss: bearings and, in brushed motors, the brushes rubbing on the commutator or slip rings.
• Windage loss: air drag on the rotor and the power consumed by the cooling fan on the shaft.
• Stray load losses: everything that is hard to model, such as harmonic fluxes, leakage flux in the end region and eddy currents in the conductors that grow with load. Standards either assume a fixed fraction or determine them by test.

Friction and windage depend on speed and are almost constant for a fixed-speed motor. Stray losses rise with load, usually about the square of the current. Together all three are typically a few percent of the rated input in medium-sized motors.

Worked example: a 7.5 kW four-pole motor

A 400 V, 50 Hz, four-pole induction motor delivers 7.5 kW at the shaft at full load. Suppose its losses are:

• Stator copper: 340 W
• Rotor copper: 190 W
• Iron: 160 W
• Friction and windage: 70 W
• Stray load: 40 W

1. Total losses: 340 + 190 + 160 + 70 + 40 = 800 W. 2. Input power: P_in = 7500 + 800 = 8300 W. 3. Efficiency: η = 7500 / 8300 ≈ 0.904, or 90.4%. 4. With a power factor of 0.82, the apparent power is S = 8300 / 0.82 ≈ 10 122 VA. 5. The line current is I = S / (√3 × 400) = 10 122 / 692.8 ≈ 14.6 A.

The 800 W of heat must be removed by the cooling, which is why the cooling fan and the frame fins matter. At half load the output is 3.75 kW; copper losses fall to roughly a quarter, but iron and friction losses stay the same. With these numbers the losses drop to about 372 W and the efficiency stays near its peak at 3750 / 4122 ≈ 91%, while the power factor falls much further.

IE efficiency classes and power factor

The standard IEC 60034-30-1 defines international efficiency classes for line-fed three-phase motors, roughly from 0.12 kW to 1000 kW, for 2, 4, 6 and 8 poles at 50 and 60 Hz.

• IE1: standard efficiency
• IE2: high efficiency
• IE3: premium efficiency
• IE4: super premium efficiency

An IE5 ultra premium class is also in use or under discussion in newer standards work, and converter-fed motors have their own standard, IEC 60034-30-2. The exact limit depends on power, pole count and frequency, so look up the table for your motor. Many regions, including the EU, have phased out IE1 motors for general use and require IE3 or better over much of the power range. Moving up one class typically removes roughly 15% to 20% of the losses, which pays back quickly because motors consume a large share of all industrial electricity.

Power factor is a separate quantity. It is cos φ, the ratio between active and apparent power, and it only exists for AC motors. An induction motor needs reactive current to magnetize its core, so cos φ is typically 0.8 to 0.9 at full load and below 0.3 at no load. A low power factor does not waste energy in the motor itself, but it increases the line current, the cable losses and the demand charge, which is why it is often corrected with capacitors.

Try it in the Curvo Studio simulators

Open the three-phase induction motor simulator. The readouts include Efficiency, Power factor, Output power, Motor torque and Estimated current, and the Performance vs. load chart plots current, slip, efficiency and power factor against load.

• Under Load type choose Constant torque (the Mechanical load slider is disabled for No load), then move the Mechanical load slider from 0% to 100%. With the default 4 kW motor the Power factor climbs from about 0.1 at no load to about 0.8 at rated load, and Efficiency rises from about 79% at 25% load to roughly 86% between 50% and 100% load. This is a lesson model, so its figures are not an IE-class value.
• Pick a small and a large motor from the Nameplate template list and compare their readouts at the same load percentage.
• Lower the Applied line voltage and see how the Estimated current and Efficiency change at the same load.

The DC generator simulator has Shaft input power, Electrical output, Total losses and Efficiency readouts, so you can check the energy balance P_in = P_out + P_losses directly. The brushed DC motor simulator shows the Armature current, from which you can calculate I² × R.

Try it in the simulator