Electric motor efficiency is basically kilowatts in versus kilowatts out. With standards like IEEE 112 (method B) and CSA 390 in addition to measuring the incoming data with instrumentation and output data with a dynamometer, there are steps that have to be taken regarding separating out the losses and running the motor in specific conditions. These include a pre-test heat run so that the resistances are correct at full load and ensuring that you are operating at correct voltages with specific limits on phase balance. While the 2017 edition of IEEE 112 allows for reduced voltage loading in order to effectively increase the power range of a dynamometer, our testing of that method found that test results read a little higher than under full voltage.
But what if you are in a plant with an electric motor that you want to know if multiple repairs has had an impact? Or you wish to understand what happens if you want to compare two motors to verify that the new one is more or less efficient than the old one? As it turns out there are multiple methods one of which is similar to the original ORMEL96 (Oak Ridge National Labs method) which was introduced in 1996. One way is to use the EMPATH system which follows the methodology (https://empathcms.com) or to follow the methodology that follows. We know based on work performed by the US Department of Energy and the Washington State Energy Office in the 1990s that the accuracy of the following approach is relatively accurate (within 1-5%) depending on how it is performed. This can be compared to the efficiencies produced by MEASUR, a US DOE software available here: MEASUR | Department of Energy.
In order to perform the work without the use of the motor manufacturer’s efficiency curves we will need to know the load torque. For this case we will use a 100 horsepower electric motor with a nameplate speed of 1780 RPM, running speed of 1780.32 RPM, nameplate voltage and current of 460 volts and 99 amps with a running average voltage of 474 and current of 98.7 amps. The data is measured by an EMPATH system, so the output torque is determined to be 270.8 ft.lbs. Input power is measured at 76.4 kW, which is slightly overloaded, and there is an impedance unbalance of 3.03%:
First, we need to convert the output torque from pound-feet (lb-ft) to Newton-meters (Nm) and determine the angular speed in radians per second (rad/s):
- Convert output torque from lb-ft to Nm:
1 lb-ft = 1.35582 Nm
Output torque = 270.8 lb-ft × 1.35582 = 366.9899 Nm - Convert angular speed from RPM to rad/s:
1780.32 RPM × (2 * π) / 60 = 186.5427 rad/s
Now, we can calculate the output power (P_out):
P_out = Torque × Angular speed
P_out = 366.9899 Nm × 186.5427 rad/s ≈ 68,498.8 W (68.4988 kW)
Now, let’s correct for impedance unbalance (assuming it’s mainly due to current unbalance):
Impedance unbalance factor = 1 + 2 * (I_unb / 100)^2
Impedance unbalance factor = 1 + 2 * (3.03 / 100)^2 ≈ 1.00183
We will use this factor to correct the input power (P_in) for impedance unbalance:
Corrected P_in = P_in × Impedance unbalance factor
Corrected P_in = 71.5 kW × 1.00183 ≈ 71.6308 kW
Finally, we can calculate the efficiency (η):
η = (P_out / Corrected P_in) × 100
η = (68.4988 kW / 71.6308 kW) × 100 ≈ 95.61%
So, the efficiency of the motor is approximately 95.61% using an output torque of 270.8 lb-ft, which is higher than the EMPATH and ORMEL96 method but within tolerance.
