Industrial Motors

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)

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
Where:
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.

Locked-rotor amps. The locked-rotor current for the 50 hp motor is also 56% lower. Other than the reduction in current itself, the reduced starting current is mechanically less stressful on the motor windings, so they should last longer.

Cautions?

Starting torque. The input power measurement data in the above example shows that the 50 hp motor would be acceptable under steady-state, normal operating conditions. It’s equally important that the motor be able to start the load. Variable-torque loads like fans and blowers have low starting torque requirements and load-torque curves similar to those in Load 1 in Figure 1, so with these kinds of applications, there’s little risk to making this type of change. The key is that the motor speed-torque curve must always be above the load-torque curve for acceleration to occur.
Inertia of the load. The inertia (Wk2) of the load is another concern and evaluation criterion. The angular acceleration of the machine is propFortional to the torque and inversely proportional to the inertia. If the inertia is very large, the acceleration time can become long and cause excessive heating in the motor.

If a more-constant torque load curve such as Load 2 in Figure 1 were present, the 50 hp motor in the example would fail to accelerate, because its speed-torque curve would intersect the load-torque curve. However, use of a variable-frequency drive (VFD) in this case would be helpful, as shown by the VFD series of curves in Figure 1. The VFD would allow for a constant magnetic field and thus constant torque during acceleration.

Special applications. Be careful when right-sizing special applications, such as those with intermittent duty or widely varying loads. If the load exceeds the breakdown torque of the motor at any point, a stall will occur. In applications with a wide variation in load that repeats in a given cycle, additional calculations should be performed based on the loading profile and duration at each load.