High Efficiency Motors

The full load current should decrease slightly compared to a "normally" efficient motor; but the starting current might increase (again slightly).

The starting current increases significantly.

In my experience so far, the Full load current increases by about 5-10% and the pf is lower than before. Higher efficiency and lower costs so long as the pf is handled correctly. Not sure re starting as I use with VFD so not relevant.

I'm confused. Shouldn't a high efficiency motor use less energy? How is this consistent with an increase in current? (I assume at the same voltage)

The energy consumed IS lower, as the current is more but at a reduced power factor.

So you have a larger reactive current component but smaller active component than before.

Is this enough?

I thought the power factor of a motor was just a measure of how inaccurate a watt meter would be at measuring it due to the phase shift between the voltage and current produced by that motor...

I remember seeing advertisements in magazines years ago for plans to make "energy misers" using capacitors to phase shift the power of a refrigerator's compressor motor so it wouldn't register as much on the watt meter on the side of your house.

If you measure the RMS voltage and the RMS current independently you should be able to calculate an accurate wattage.

So if the power factor is reduced it means a watt meter will SHOW more power being used because the phases aren't shifted as much and therefore the watt meter is being more accurate but actually the current will decrease because less energy goes to heating up the motor while the physical load remains the same.

Or are you saying that a current meter is also fooled by the phase shift even though it doesn't have a voltage coil like a watt meter does?

Put simply, normally you discuss apparent power and active power. The apparent power is a product of the voltage and current but the active power (watts) is the current that coincides and is in phase with the voltage. The difference in angle between the voltage and current, Φ, will allow you to calculate the active power from the RMS voltage and current, as you describe.

So, if you take a 3 phase motor the active power (watts) P will then be = √3 x (V x I) x cosΦ. The Output power from the motor (mechanical) will be this electrical power divided by efficiency.

If you play with these you will see what I mean. Basically, I hope you will see that you cannot use measured RMS values of voltage and current to calculate power unless you know the power factor.

I hope this helps.

BTW, can someone explain my ''off topic'' vote, I'm disappointed and don't understand in the context of this discussion why I received this.

Oooops! - The Output Power from the motor (mechanical) will be this electrical power Multiplied by efficiency. Sorry about that!

Hi,

The air gap in between stator and rotor will be lesser than normal efficient motors. Moreover the EF motors are classified as EFF1, EFF2.

in my experience EFF motors are not efficient in all type of applications and before going for the EFf motor, you have to conduct the load study and consumption precisely with your existing motor and EFF motor.