208 3PH question

Like a kid who can't answer an essay questi "The simple answer is No; it is not practical to measure current in the individual phases. This pulsating current cannot easily be measured. I was wrong to try to show that the current was 1/3 of rated load current. It can be seen, though, that Power in each phase must be equal to 1/3 total Power if the phases are balanced. Simple logic supports this."

It's over, Bob. As in past technical arguments that you have attempted here, you just keep digging the hole deeper.

From one:

That's nice, I have plenty of textbooks, it's just that measuring instantaneous current has nothing to do with the thread. Bob is floundering and keeps trying to introduce new material to cover up his previous misstatements.

What's that Lassie? You say that Robert Swinney fell down the old rec.crafts.metalworking mine and will die if we don't mount a rescue by Tue, 30 Dec 2008 17:57:26 -0600:

That's OK. I'm reading every post and learning a few things along the way.

Robert's trying to overanalyze it - You can take a clamp-amp and measure each phase (in case it is imbalanced somehow) and average it out, and know what it costs roughly to run.

The wattmeter at the service is going to be reasonably accurate, or the State Weights and Measures Commissioner will have words with the power company - If the power company is using that meter to sell you electricity they have a duty to be reasonably accurate - I'd say in the +/- 1% to 2% range

Your average Clamp Amp is reading RMS, and is plenty close enough to agree with what the power meter is seeing - unless you are running a variable speed drive that distorts the heck out of the waveform.

Where it gets tricky is in measuring the start surges and quantifying it's effects on the power bill, versus the lower but still real costs of keeping a motor spinning unloaded. Oh, and the cost of a new motor every year if you are short-cycling them.

There are more sophisticated solutions to reducing the start surges in a shop, like electronic soft starters or a Delta-Wye starter on all the large equipment, but the solution starts getting more expensive than the problem....

And with a soft start solution you need the ability to positively control the load till the motor is up to speed - often used on elevator pumps where they can wait to open the hydraulic valves till after the jack valve gets a 'full speed' signal from the motor starter.

And there are other things to consider - quantifying that start surge's effect, combined with all the other motors and welders and such in the shop, and their effect on the instantaneous demand reading on the meter and the Demand Charge adder on the bill.

Some shops with lots of huge motors can have their power bill doubled because of the demand charges, because the utility metering tells the engineers they have to yank out the 100KVA transformer and place a 200KVA - your regular running loads might be under 100KVA, but all the start surges will stress and take out the transformer if they don't use an oversized one.

And when that pole pig in the parking lot (usually oil filled) blows up, it blows up reel guud. (Get out the marshmallows!)

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What's that Lassie? You say that Bruce L. Bergman fell down the old rec.crafts.metalworking mine and will die if we don't mount a rescue by Wed, 31 Dec 2008 18:29:35 -0800:

But what about the power factor? Or does it not matter that much? I only read one leg, and I was surprised to see that when the motor was loaded it was drawing 15A. And 10A when un-loaded.

Five amps seem low for the work that it's doing.

It's not doing 25 HP worth of work, that's for sure. The power factor in this case will just mean it costs you less to run. I would use a PF from a chart at a low loading, for example .75 (for a 75% PF) * the VA equals watts. So 15 amps * 360 * .75 = watts, then you have to calculate hours used and also your demand charge (KW) if the compressor comes on during a peak demand window.

What's that Lassie? You say that ATP* fell down the old rec.crafts.metalworking mine and will die if we don't mount a rescue by Sat, 3 Jan 2009 21:22:49 -0500:

I'd have to agree, not doing 25 HP. The owner says that the compressor(3) cost $10/ hour to run. But I know that he just pulled that figure out of his ass. So I'm trying to figure a rough estimate on what it costs to run one of them.

I was very surprised to see the unloaded current so high. I'm going to check the other two legs on monday. I'll check the other units too, just to see if they are close. I had checked on one that I had just turned on, so maybe it wasn't fully warmed up yet.

As far as the demand charge, wouldn't even be 1% of our total load. We have 21 big CNC machines. And in the summer, 10 big AC units on the roof. We have our own transformer, big as a small car, on a pad outside of the power room.

That's a pretty big motor, I would expect it to pull 10 amps just running. Full load amps is close to 60.

A drop in the bucket compared to all that, but could still be a significant yearly cost.

What's that Lassie? You say that ATP* fell down the old rec.crafts.metalworking mine and will die if we don't mount a rescue by Mon, 5 Jan 2009 20:38:00 -0500:

I would too. But when the valve cycled, the load went to only 15A.

I agree. But I'm trying to get to an hourly cost. So when the owner complains about the cost of having the air on for a few hours over the weekend, I can counter with a realistic figure. And be able to 'show my work' as well. He seems to think that since the motor runs continuously, that 25hp of power is being used. So I want to have the information about the loaded and unloaded current draw, and the duty cycle when I'm using it. That way I can figure an hourly cost to run.

Thanks for all the help.

I haven't followed all of this thread so it may not be pertinent but you can not tell much of anything about energy usage of loaded and unloaded motors from the line currents and voltages alone. The power factor varies greatly with load; I have seen motors with pretty small differences between idling and full load currents. You need a clamp-on wattmeter or something similar to get meaningful measurements.

Don Young

What's that Lassie? You say that Don Young fell down the old rec.crafts.metalworking mine and will die if we don't mount a rescue by Wed, 7 Jan 2009 20:55:36 -0600:

Damn. That's what I was afraid of.

As Don, I haven't read all this thread but as Don mentioned, power factor changes depending on load.

Motors are inductive and the voltage and current waveforms are not in phase, they are most out of phase when idling. That means P is not equal to I x E.

Outside of startup, amperage is fairly constant in an ac electric motor be it idling or loaded.

Thus the comments on using a watt meter. That checks to see what P is and deals with I x E being out of phase with each other.

Adding to the mix is your utility rate. Are you billed in KWH or KVA? If you have 3 phase service, there is a good chance you are billed for KVA. Home owners tend to be KWH.

Why KVA? Well an idling motor, doing no real work, draws current. Even though the current and voltage wave forms are out of phase and thus not using a lot of power, the heating effects of the ampere component is very real. The utilities see this in heating of wires and their transformers, to them this is something that needs to be priced since it does cost them money.

To add to this mess, an idling motor can have capacitors attached to normalize the load so the power factor is closer to unity. Some factories have banks that are switched in and out via monitoring systems if their power factor is under .80 or so. I assume inductors could be used for highly capacitive loads. The reason for this is industrial users are often charged for excessive power factor.

Wes

You can at least calculate the *worst-case* operating cost, which assumes unity power factor. In any real situation, the cost will be less (usually significantly less) than that calculated from clamp-on ammeters. It *can't* cost more than your calculations show -- only less.

Enjoy, DoN.

What's that Lassie? You say that DoN. Nichols fell down the old rec.crafts.metalworking mine and will die if we don't mount a rescue by 10 Jan 2009 05:36:28 GMT:

Thanks, that's good to know.

Do you have access to a good meter like a Fluke 87 and a clamp-on attachment? If you can leave it running with the clamp-on attached you could get max, min and average readings for the whole day.

What's that Lassie? You say that ATP* fell down the old rec.crafts.metalworking mine and will die if we don't mount a rescue by Sun, 11 Jan 2009 00:22:43 -0500:

Nope. Just an old clamp-on that I picked up at a yard sale once. It's a voltmeter too, but it needs funny leads that I don't have. (they screw onto recessed studs)

I know of no 3-phase Kil-o-Watts, so do this. Find a surplus 3-phase power meter, and mount a meter box on a board. Put it in the motor supply leg, note the time/date/meter reading, and let it got for 5 hours or a day.

That's the best way of finding the actual usage.

Your utility might even sell you an old meter, or try eBay. One gotcha is finding a direct-reading one. Bigger ones will depend on external voltage and current transformers.

What's that Lassie? You say that David Lesher fell down the old rec.crafts.metalworking mine and will die if we don't mount a rescue by Tue, 27 Jan 2009 05:55:46 +0000 (UTC):

Thanks, but that's a bit farther than I want to go.

I think I will just assume full nameplate amps when loaded and multiply by the duty cycle.

BTW, I had been wrong on the voltage all along. It's being fed from one of the 460V panels.