Keeping it simple: Steps to determine motor’s actual load

Oversized motors cost more to operate—sometimes a lot more. Fortunately, there’s a simple procedure for determining the actual hp required by a load, without expensive equipment or engineering.
By Chuck Yung October 20, 2014

Figure 1: Percent load versus current. Source: EASAContrary to popular opinion, bigger isn’t always better—especially when it comes to electric motors. Plant maintenance and engineering departments like having a little extra power available “just in case,” so they sometimes specify larger motors than applications require. But oversized motors cost more to operate—sometimes a lot more. Fortunately, there’s a simple procedure for determining the actual hp required by a load, without expensive equipment or engineering. Bear in mind that loads should be determined when the motor is operating at its maximum load. Loads that vary widely are good candidates for variable-frequency drives (VFDs), which offer the added benefit of controlling rate of production.

Estimating actual load

As Figure 1 illustrates, percent load and current are essentially linear, from no-load up to motor’s nameplate current. An easy mistake, however, is to assume that zero load = zero current. That assumption will result in errors in determining hp, with the error inversely proportional to the load (shaded area of Figure 1). The biggest errors occur when considering motors most in need of “right-sizing” [i.e., 50% of full-load amps (FLA) does not = 50% load].

While it’s possible to determine the percent load a motor carries from the graph in Figure 1, the actual load of a motor can be approximated mathematically from good input data:

Operate the motor uncoupled and record the current at no-load (0% load). Don’t take any shortcuts here, because the measured current will be higher if the motor is coupled. To avoid errors, always use the uncoupled current.

Next, document the nameplate current and then measure the current at the motor’s actual load. Since an undersized motor presents other problems, it’s best to measure the current over the operating cycle of the process. If the load varies, record the current during peak load.

Cost of “safety margin”

The additional cost of operating seriously under-utilized motors often includes utility surcharges for poor power factor (more about that later). Cyclical power users may also be subject to demand charges based on the peak power usage. That means a single high-usage episode (e.g., across-the-line starting of a large motor) could trigger a higher kW/hour rate for an entire billing period. Identifying under-utilized motors offers many users an opportunity to reduce operating costs.

Hidden costs of oversized motors

Inrush current, the current a motor draws at the moment of starting is not load-dependent. For a motor of a given size, it is the same regardless of actual load. That means a 100 hp motor starting uncoupled draws the same initial current as it does when starting a 100 hp load. Because starting current is roughly six times nameplate current (depending on the NEMA code letter), it can be significantly higher for an oversized motor than a “right-sized” model.

A motor’s actual inrush current, also known as locked-rotor amps or LRA, can be calculated from the kVA/hp values associated with its NEMA code letter (Table 1):

LRA = CL x hp x 1000/1.732 x Voltage

(For CL, plug in the kVA per hp value from the table.)

Table 1: NEMA code letters for locked-rotor KVA (Reference: NEMA Standards MG1-2011, 10.37.2)

For example, the LRA for a 125 hp motor with the code letter G (5.6 – 6.3 kVA/hp) should fall between 878 and 988 amps:

5.6 x 125 x 1000/1.732 x 460 = 878 amps

6.3 x 125 x 1000/1.732 x 460 = 988 amps

Real life example

A 125 hp motor on a fan application had a nameplate rating of 148 amps but drew 44 amps (slightly less than 1/3 of FLA) when it was run uncoupled. Testing confirmed that, when operating at normal load, the motor drew only 63 amps. The actual hp required was calculated at less than 23 hp:

hp = 125 [1-(148-63/148-44)] = 22.8 hp

Substituting a higher efficiency 25 hp replacement motor dramatically decreased the starting current from 890 amps to 198 amps (at the same code letter G). The “full-load” current also decreased from 63 to 29 amps. Obviously, the plant had been paying for a lot of wasted electricity.

The power factor of the original 125 hp motor was also very low when operating a 22.8 hp load, which may have triggered surcharges from the utility.

Excessively high starting current can increase maintenance costs due to additional wear and tear on motor starters and contacts, but higher inrush current is also a feature of higher efficiency motors.

Note: Before “rightsizing,” be sure the lower hp motor can provide the required starting inertia of the load.

Power factor and efficiency

Power factor (PF) is important, because it can be used to determine efficiency. To calculate power factor, use the following formula:

PF = Input watts / [1.732 x Volts x Amps]

Efficiency can also be calculated if power factor is measured using one of several instruments available to today’s electrician. To calculate efficiency for a three-phase motor:

Efficiency = 746 x hp / [1.732 x Volts x Current x PF]

In the case of the 125 hp motor, an electrician measured a PF of 0.7, so its calculated efficiency when driving the 22.8 hp load was:

Efficiency = 746 x 22.8 / [1.732 x 460 x 63 x .7] = 0.48

The motor was operating at only 48% efficiency. Compare that to an NEMA Premium replacement motor with 95.4% efficiency. According to the Motor Master software available from the U.S. Dept. of Energy, the original 125 hp motor would use more than 457,700 kWH annually if it ran 24/7 (8,760 hours).A 25 hp EPACT replacement would use only 167,200 kWH. At 7 cents per kWH, the first year savings would top $20,000 (U.S.). Where do you want to spend your money?

Chuck Yung is a senior technical support specialist at the Electrical Apparatus Service Association (EASA), St. Louis, Mo. This article originally appeared on EASA.

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