Oversized Motor - Poor Power Factor

[SKahle89]

All,

First let me preface this post by stating that I am in no way an electrical engineer. My experience with motors and electricity is limited; so please respond in a way that typical engineer could understand.

I have a very oversized motor driving an reactor agitator. The agigitators were changed back in the 1970's, but the project manager chose not to replace the motors. The new agitators require a 75HP motor; the existing motors are 150HP. I grabbed my I&E tech and hooked up a portable power meter with very disturbing results:

HIGH-SPEED
POWER
kW 17
kVAR 17
kVA 71
PF 0.216

VOLTAGE
V AN 283
V BN 283
V CN 283
V AB 485
V BC 485
V CA 485

AMPERAGE
A N 162
A A 83
A B 84
A C 85

Notice the power factor of ~0.20. As everyone knows, energy reduction is a hot topic in industry right now. My question is, if I replace these motors with more appropriately sized ones, am I going to realize any savings due to decreased reactive load? Obviously there is some savings to be had by upgrading to a new high efficiency motor, but I am asking specifically about reactive load/power factor improvement. Thanks!

 

[dpc]

There will be some improvement, but it is a secondary effect. Reactive "power" does not directly use any "real" energy. But it does require additional current and this additional current creates real power losses (due to resistance) in the feeder cables, transformers, and in the motor. By improving power factor, the current is reduced (for the same amount of real power to the motor), so losses are reduced.

But direct energy savings by improving power factor is not as great as often expected.

"The more the universe seems comprehensible, the more it also seems pointless." -- Steven Weinberg

 

[retiredbum]

With all due respect, some of the numbers appear to be incorrect. Your measured line current and the voltage seems to confirm the apparent power is indeed about 70 KVA. That, in itself, implies that either the real power of 17 Kw and/or the reactive power of 17 Kvar is wrong. It simply does not balance. Kw*2 + Kvar*2 = Kva*2 Am I missing something here?

Also 70 Kva is in the ball park for the load on a 75 HP motor.

Check your files for information on the 150HP motors or, if possible, call the manufacturer. There are usually graphs which will allow you to estimate the motor efficiency.

 

[racobb]

With all due respect again, none of your numbers make any sense! 17 kW and 17 kVAR is not 71 kVA. A balanced 485 volt system should give you 280 to ground voltage, although that one is pretty close. Your kW and kVAR numbers indicate a 70% PF....not 20%. With such a well balanced load, what's up with the 162 amps neutral current. The motor should not even need a neutral! Where are you taking these measurements? You really need to take another look at your measurements.

Don't know what you are using for the data, but if it is giving you the 20% PF you should be smelling a rat!

Having said all of that, my experience with motors tells me that to get a 20% PF, the motor will have to be idleing on line with absolutely no load. Have seen it on sewing machines in textile factories back when we used to have them in the southeast US.

Alan

Democracy is two wolves and a sheep deciding what to have for dinner. Liberty is a well armed sheep!
Ben Franklin

 

[racobb]

After thinking about your numbers, I wanted to ask again just how are you getting the data. If you are hooking up a portable recorder or meter, you need to check the connections well. Haven't even tried to do the math at this point, but is a thought.

Alan

Democracy is two wolves and a sheep deciding what to have for dinner. Liberty is a well armed sheep!
Ben Franklin

 

[waross]

The fairly well balanced phase amperages and the supposed neutral amperage of almost double is an indication that you should stop before you hurt yourself.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[davidbeach]

Very well stated, Bill.

 

[jraef]

Aside from the above math inconsistencies, here is the bottom line. What dpc said is correct. Poor pf is only a cost issue if the utility penalizes you for having poor pf, or if you are generating your own power. There would be an added percentage of I[sup]2[/sup]R losses in the conductors feeding the motors, but in general those are fairly low because most likely your conductors are still sized for the 150HP motor load, so the losses are low to start with.

What is happening however is that the motors are not as efficient as they could be here, because the magnetizing current is higher due to the larger frame, therefore the percentage of useful work compared to total power consumed is lower. You would benefit greatly from replacing those motors with energy efficient 75HP versions, and I'd venture to say that many utilities here in the US would provide some kind of financial incentive to do so. it's definitely worth looking in to.

But just correcting the pf is not likely to be worth much in actual energy savings.


"If I had eight hours to chop down a tree, I'd spend six sharpening my axe." -- Abraham Lincoln
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[racobb]

Correct, but I find myself explaining to people all of the time that PF is a number without dimension. What utility cares about a 70% PF if it is 1 kw and 1 kvar. But let it become 10000 kw and 10000 kvar and we take notice. Unless this is the only motor on the service I expect the utility will never see it. And I agree, correcting the pf is not the issue.

Alan

Democracy is two wolves and a sheep deciding what to have for dinner. Liberty is a well armed sheep!
Ben Franklin

 

[retiredbum]

I also agree. His first step is to hire someone who knows what he is doing.

He should also contact his utility. Without the proper metering and a corresponding rate structure, improving his pf may have little or no effect on his bill. As Alan points out, the meter may never see it.

If his interest is with "energy reduction", he should see a professional who can evaluate and prioritize improvements to his entire facility.

DC

 

[jghrist]

The high neutral current and the fact that kVA is much larger than sqrt(kW² + kvar²) indicates a high third (or 6th, 9th, 12th etc triplen) harmonic current. I wouldn't expect this from a motor unless there are high voltage harmonics. I think you should get someone to measure the harmonics and if they are indeed high, figure out why.

As others have said, low power factor is usually not a problem with a low load. It depends on your utility rate structure. The conductors are also probably oversized for the new agitators, so I²R losses are probably low even with the low power factor.

 

[SKahle89]

All,

Thanks for you responses. The data reported is what I was provided with; I recognize that it doesn't make much sense. Attached is the motor curve for the original 150HP motor. You'll notice it says 150/37.5; that is because this is a two speed motor, but I am only concerned with the high speed.

Scratch the power readings; assuming the amperage readings are correct at 85amps/phase; I think I am looking at a motor output of 57%, a power factor of about 0.64, and an efficiency of 91% based on the motor curves. What is the proper way to calculate my REAL power consumption?


KW = %outHP * MaxHP * Conversion / Efficiency
70.1 = 0.57 * 150 * 0.746 / 0.91

 

[jghrist]

I think your motor output is 37% not 57%.

 

[SKahle89]

jghrist,

Yeah you are right; don't know where that came from.

Other than the typo; is the method correct if I can accurately measure the amperes per phase?

 

[LionelHutz]

Looks OK. I get about 45kW.

It seems you need a 60hp motor.

 

[burnt2x]

SKahle89,
I am a bit troubled by the nature of the motor duty. What is the agitated substance? If the substance is homogenous, motor could be overzized, but then if the substance is like a slurry, you could expect "sanding" problems which will require designing motor drives to overcome high starting torques to come up to speed; hence a larger or slower motor.
Just asking.