Electric motors consume roughly 70% of all industrial electricity. A motor that operates at 91% efficiency instead of 95% does not sound like much of a difference, but on a 100 HP motor running 8,000 hours per year at $0.08/kWh, that 4% gap costs $1,900 per year in wasted electricity. Over a 20-year motor life, the energy cost dwarfs the purchase price by a factor of 10 to 50. Motor efficiency is not an abstract specification; it is an operating cost you pay every hour the motor runs.
The efficiency number on the motor nameplate is measured at full rated load. But most motors in industrial plants run at 50% to 80% of their rated load because engineers (correctly) size motors with a service margin. The problem is that motor efficiency drops at light loads, and the power factor drops even more. An oversized motor running at 40% load has lower efficiency, worse power factor, and draws more reactive current than a right-sized motor at 75% load doing the same work. This guide covers how motor efficiency behaves across the load range, when right-sizing or upgrading makes financial sense, and what the NEMA Premium efficiency standard actually guarantees.
How Motor Efficiency Varies with Load
A motor's losses come from two sources: fixed losses (core losses from magnetizing the iron, friction and windage) that are roughly constant regardless of load, and variable losses (I²R copper losses in the stator and rotor) that increase with the square of the current. At no load, the motor draws magnetizing current and produces core losses plus friction, but no useful work. Efficiency is zero. As load increases, the useful output grows faster than the additional copper losses, and efficiency rises.
Most motors reach peak efficiency between 75% and 100% of rated load. A typical NEMA Premium 50 HP motor might have 95.4% efficiency at full load, 95.6% at 75% load (the peak), and 94.1% at 50% load. The efficiency curve is relatively flat from 50% to 100% load. Below 50% load, efficiency drops more steeply: at 25% load, the same motor might operate at 89% to 91% efficiency.
Power factor tells the other half of the story. At full load, a typical motor operates at 0.85 to 0.90 power factor. At 50% load, power factor drops to 0.70 to 0.80. At 25% load, it can fall below 0.50. Low power factor means the motor draws more current than necessary for the actual work, increasing I²R losses in the distribution system (cables, transformers, switchgear) and potentially triggering utility power factor penalties.
The practical takeaway: a motor running between 50% and 100% load is operating efficiently. Below 50% load, efficiency and power factor degrade significantly. Below 25% load, the motor is wasting a substantial fraction of its input energy as heat, and replacement with a smaller motor should be evaluated.
How to Measure Motor Loading in the Field
Motor loading is the ratio of actual shaft output to rated output. The most practical field method is the input power method: measure the three-phase input power with a power analyzer or clamp-on power meter, then calculate: Load % = (P_input × η_rated) / P_rated × 100, where P_input is measured input power in kW, η_rated is the nameplate efficiency (as a decimal), and P_rated is the rated output power in kW (HP × 0.746).
The current method is simpler but less accurate: Load % ≈ (I_measured / I_nameplate) × 100. This approximation works reasonably well above 50% load but underestimates loading at light loads because the no-load (magnetizing) current is a larger fraction of total current. A 50 HP motor with a nameplate FLA of 60A might draw 25A at no load. At 50% mechanical load, it draws about 42A, not 30A. The current method would estimate 42/60 = 70% load, overstating the actual 50% mechanical load by 20 percentage points.
For accurate loading determination, use a three-phase power analyzer that measures true RMS power (watts), not just current. Record measurements over a representative operating period, ideally 24 hours or more, to capture the full load cycle. Many motors run at different loads depending on process conditions, time of day, or batch operations. A single snapshot reading can be misleading.
Slip speed is another field method for squirrel cage motors: Load % ≈ (n_sync - n_actual) / (n_sync - n_rated) × 100, where n_sync is synchronous speed, n_actual is measured speed (from a strobe or tachometer), and n_rated is nameplate speed. This method is accurate to about 5% for standard NEMA Design B motors. It requires measuring the actual shaft speed, which is easy with a laser tachometer but impractical on enclosed or inaccessible motors.
When to Right-Size: The Economics of Motor Replacement
Replacing a working motor with a smaller, more efficient one only makes sense when the energy savings pay back the investment in a reasonable time. The annual energy savings from right-sizing are: Savings = P_load × hours × cost × (1/η_old - 1/η_new), where P_load is the actual shaft load in kW, hours is annual operating hours, cost is the electricity rate in $/kWh, and η is efficiency at the actual load point.
Example: A 100 HP motor (nameplate efficiency 93.0%) running at 40% load (40 HP actual) for 6,000 hours/year at $0.09/kWh. At 40% load, this motor operates at roughly 90.5% efficiency. Replacing it with a 50 HP NEMA Premium motor (nameplate efficiency 95.4%) that runs at 80% load, where its efficiency is approximately 95.6%. The annual savings: 29.84 kW × 6,000 × $0.09 × (1/0.905 - 1/0.956) = 29.84 × 6,000 × 0.09 × (1.105 - 1.046) = 29.84 × 6,000 × 0.09 × 0.059 = $950/year.
A 50 HP NEMA Premium motor costs roughly $3,000 to $5,000 installed. At $950/year savings, the simple payback is 3 to 5 years. That is a reasonable investment, especially if the motor has many years of remaining service. If the motor is near the end of its life and will need replacement anyway, the incremental cost of choosing the right size and efficiency level makes the payback even shorter.
The decision threshold: if a motor consistently operates below 40% load, right-sizing is almost always justified. Between 40% and 60% load, evaluate the economics. Above 60% load, the efficiency gain from right-sizing is usually too small to justify the cost unless you are also upgrading from a standard efficiency to a premium efficiency motor.
VFDs and Motor Efficiency: The Hidden Losses
Variable frequency drives (VFDs) save enormous energy on variable-torque loads (fans, pumps, compressors) by reducing motor speed to match the actual demand. The affinity laws show that reducing speed by 20% reduces power consumption by about 49%. But VFDs are not free from an efficiency standpoint. They introduce additional losses that partially offset the speed-reduction savings.
VFD losses fall into three categories:
- Drive losses: The VFD itself consumes 2% to 4% of the throughput power in switching losses, conduction losses, and cooling fan power. Modern drives are more efficient than older models, but this loss is always present.
- Motor harmonic losses: The PWM (pulse-width modulated) output of a VFD is not a pure sine wave. The harmonic content causes additional core losses and copper losses in the motor, reducing motor efficiency by 1% to 3% compared to operation on utility power at the same speed.
- Cable losses: The high-frequency switching of a VFD creates common-mode currents that cause additional heating in motor cables, especially on long cable runs. Proper shielded cable and correct grounding minimize this effect.
The net result: a VFD-driven motor system is typically 4% to 7% less efficient than the same motor on utility power at the same operating point. This efficiency tax is overwhelmed by the speed-reduction savings on variable loads (where a 20% speed reduction saves ~49% power), but it means VFDs do not save energy on constant-speed, constant-load applications. Putting a VFD on a motor that always runs at full speed and full load just wastes 4% to 7% of the energy. Use VFDs where the load genuinely varies. Use starters (across-the-line or soft start) where it does not.
When specifying motors for VFD duty, select "inverter duty" or "inverter ready" motors. These are built with enhanced insulation systems (often Class H or better) to withstand the voltage spikes from PWM switching, and their windings and laminations are optimized to minimize harmonic losses. Running a standard-duty motor on a VFD shortens its insulation life and reduces efficiency more than running an inverter-duty motor.