Chuck Yung
EASA Technical Support Specialist
Contrary to popular opinion, bigger is not always better. A case in point is the electric motor, the workhorse of industry. There is a natural tendency to want a little extra power, “just in case.”
That’s why auto makers still sell cars with 300 hp engines, even though the speed limit may be under 70 miles per hour. But, just like those gas-guzzlers, operating an oversized electric motor may cost additional money; sometimes, a lot of money. Here is 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. A load that varies widely is a good candidate for a variable frequency drive (VFD).
Estimating actual load
As the graph (Figure 1) illustrates, percent load and current are essentially linear from no-load up to nameplate current of the motor. An easy mistake is to assume that zero load = zero current. That assumption results in errors in determining hp, with the error (shaded area, Figure 1) inversely proportional to the No Load load. The biggest error occurs when considering those motors most in need of “rightsizing” (i.e., 50% of full load Figure 1 amps (FLA) does not = 50% load).
While it is possible to use the graph to determine the percent load a motor carries, the same information can be found mathematically. With good input data, the actual load of a motor can be approximated by the following formula:
hprequired = hpnameplate [1-(FLA-actual amps/FLA-no load amps)]
Where:
FLA is Full Load Amps
hp is Horsepower
Operate the motor uncoupled and record the current at no-load (0%). Don’t take any shortcuts here; if the motor is coupled, the “no-load” current will be higher than the current uncoupled. To avoid errors, the uncoupled current must be used. Document the nameplate current, and the current at the motor’s actual load. Since an undersized motor presents other problems, the safest method is to record the current over the operating cycle of the process. If the load varies, record the current during peak load.
Cost of “safety margin”
To begin with, the utility may impose a charge for poor power factor (more about that later) when a motor is seriously under-utilized. In addition, cyclical power users may be subject to a demand charge based on the peak power usage. The way that can work is that one episode of high usage raises the kW-hour cost of electricity for the entire billing period. That means severe penalties for across-the-line starting of large motors. Identifying under-utilized motors offers many users an opportunity to reduce those poor power factor charges.
Hidden costs of oversized motors
Inrush current — the current a motor draws at the moment of starting — is not load-dependent. Inrush current for a given motor 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. Since starting current is roughly 6 times (depending on the NEMA code letter) the nameplate current, that represents a lot of energy. Actual inrush current (also known as Locked Rotor Amps or LRA) can be determined from the motor nameplate. NEMA prescribes a code letter (CL) be used to indicate the LRA of a motor.
From Figure 2, plug in the upper value as CL, into the following formula:
LRA = CL x hp x 1000/1.732 x Voltage
For a 125 hp motor with code letter G (5.6 – 6.3), the LRA will fall between 878 and 988 amps.
[5.6 x 125 x 1000/1.732 x 460 = 878]; [6.3 x 125 x 1000/1.732 x 460 = 988].
Compared to the LRA for a replacement 25 hp – 198 amps (at the same code letter G) – the difference in in-rush current is dramatic. Wear and tear on motor starters, replacement costs for contacts, etc. unnecessarily drive up the maintenance costs considerably.
Real life example
A 125 hp motor was driving a fan to provide make-up air to a plant. The nameplate current rating was 148 amps. Operating uncoupled the motor drew 44 amps, slightly less than 1/3 of FLA. When subjected to its normal load, the motor drew 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 25 hp replacement motor decreased the starting current from 890 amps to only 198 amps. “Full load” current decreased from 63 amps to only 29 amps. The plant was paying for a lot of wasted electricity. In addition, the power factor of the 125 hp motor operating a 22.8 hp load was very low. (Before “rightsizing,” be sure that the lower hp motor can handle the 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 3-phase motor:
Efficiency = 746 x hp / [1.732 x Volts x Current x PF]
In the 125 hp example, an electrician measured a PF of 0.7, so the calculated efficiency of the 125 hp motor driving the 22.8 hp load was: Efficiency = 746 x 22.8 / [1.732 x 460 x 63 x .7] = .48
The motor was operating at only 48% efficiency. Compare that to a replacement EPACT motor with 94.5% efficiency. According to the MotorMaster software available from the U.S. Department Of Energy, if the motor-operated 8,760 hours per year (that’s 24/7), the original motor used over 457,700 kWh per year compared to an EPACT replacement 25 hp which would only use 167,200 kWh per year. At 7 cents per kW-hour, the first year savings would be over $20,000 U.S. Where do you want to spend your money?
AVAILABLE IN SPANISH
Print