Loaded question: How much do you know about motor load?

Careful evaluation of motor load requirements could yield long-term savings.

By Mike Howell, Electrical Apparatus Service Association (EASA)

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Right-sizing of three-phase induction motors for different applications – and striking a balance between reliability and efficiency – isn’t always easy, but it can be cost-effective. Before the days of comprehensive predictive and preventive maintenance programs, the conventional approach to reliability was conservatism, both in design and in application. That’s to say that on a 25 hp application, you’d find a very conservatively designed 60 hp motor that was really 75 hp “under the hood.” And yes, the motor would last a very long time, but it would have an inflated (and often ignored) operational cost. Many of these robustly engineered applications are still out there, and it can be well worth the effort to identify and correct them.

New vs. existing applications

With new applications, load speed-torque characteristics, load inertia and duty cycle are all essential considerations for selecting the right motor design and output rating. Invest time and resources up front in these cases, taking into account expert advice from the motor supplier’s application specialists and/or the driven equipment manufacturer, and you’ll likely get a good startup and a reliable installation.

But what about existing applications, especially the ones that have been around forever? Sometimes it’s easy to identify low-hanging fruit that can yield significant energy and long-term maintenance savings without jeopardizing operational reliability. The best time to do so usually comes during maintenance shutdowns or when it’s time to repair or replace a motor. To begin the process, you need to know the actual load of the application.

Determining the actual load

With installed machines, you have several options for determining the actual load. The easiest of these methods relies on input power measurements. Collecting this data has become much more practical in recent years because basic power quality meters are now more affordable.

The power rating on a motor’s nameplate represents its output power – i.e., the power available at the shaft. Because this is difficult to measure in the field, a reasonable approach for estimation is to use published or approximate efficiency information as the basis for comparing the rated input power with input power measurements. Most motor nameplates list full-load efficiency. If this isn’t available, Table 1 provides conservative efficiency estimates for most machines.

If the machine is rated in horsepower (hp), first multiply the rated hp by 0.746 to convert it to kilowatts (kW). Then calculate or estimate the input power at rated load as follows:

Pin = Pout / eff
Pin = rated input power, kW
Pout = nameplate output power, kW
eff = efficiency at rated load, per unit

Example: For a 100 hp, 4-pole induction motor (NEMA design B, code letter G) with a nameplate efficiency of 0.95 (95%):

Convert the rated output power to kW
100 hp x 0.746 kW / hp = 75 kW

Calculate the rated input power
75 kW / 0.95 = 78.9 kW

Determine typical loading by measuring the average input power at peak load.

In this process, always involve someone who is familiar with the application operation; it’s important to obtain representative sample measurements at the maximum load the motor sees during normal operation.

For this example, let’s say power measurements for the maximum load during normal operation averaged 39.3 kW. Expected efficiency is typically fairly flat (±5%) at 40% or better load. Assuming this condition, the input power can be estimated as 39.3 kW x 0.95 = 37.3 kW or 50 hp. Therefore, this 100 hp machine is roughly 39.3 / 78.9 x 100% = 50% loaded.

What happens if we replace the 100 hp motor with a 50 hp motor? Table 2 provides some performance data from a typical manufacturer for NEMA design B, code letter G, 50 hp and 100 hp open drip-proof machines.

Worth noting

Line amps. The first notable difference between the two machines is a 20% reduction in line amps at 50 hp, thanks to a 24% increase in power factor. Because electric utilities often add a surcharge for low power factor, improving it could reduce energy costs. Additionally, lower current means less wear and lower temperatures on starters, feeder cables, and other motor system components – i.e., improved reliability and decreased maintenance and repair costs.

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  • Calculating required motor horsepower is difficult: partially, because there are so many variables associated with knowing how much horsepower the load will require, and partially, because most engineers don't like to fail. Our motto seems to be, "more is always better", but as electrical energy costs continue to escalate, we will have to be a little more careful when sizing motors. In the example shown, the biggest impact is to the power factor. This is something that many people either ignore, or don't understand, but the good news is that in conjunction with a VFD, the power factor effects are somewhat mitigated, due to the fact that the DC bridge decouples the inverter from the AC line. This is somewhat off the subject, but worth mentioning, anyway: VFD costs have dropped to the point where one needs to consider the real, "economy" of applying across-the-line control. It is reaching the point where you're dollars ahead by applying a VFD to the solution, because you can reduce shock, trim output velocity to optimize the process, and have reversing capacity at your fingertips at no additional cost. In any event, we need to become much more energy-conscious.


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