Way OT electrical question

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Greetings All, I would like to show the actual dollar savings if we shut down machinery at work during lunchtime. I know basic electric theory, but could use a hand from someone with an industrial background. Basically, no one is shutting down our punch presses during breaks. The motors are 3 phase and either 220 or 440. I could find out the kilowatt rates that we pay. What else do I need and how do I plug the numbers in to come up with a dollar amount for every hour that they let the motor run? I know that I need the amperage from the motor nameplate. I have seen electric questions here bring on a flood of both replies and arguments, I don't want to see that happen. If the thread gets too long, just let it die and we'll all keep to WWing........ Many thanks, Mark
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Given that this is (presumably) an industrial/commercial utility account, the rates may well be more complicated than just $x per kWh used, with adjustments or charges made for peak loads, power factors, moon phases, and your deity of choice only knows what else.
That said, measuring the power consumption of a motor merely by the faceplate rating (especially when said motor is presumably not actually driving the equipment, but just idling at speed) is unlikely to be too accurate. A motor's electrical consumption is, by applying the law of conservation of energy, equal to the sum of the energy it puts out--chiefly mechanical work and heat. Since the mechanical work is rather insignificant in this case (again, assuming the attached machinery isn't doing anything much), it's probably not a vast amount of power being consumed. That's not to say it mightn't be worthwhile to idle them during lunch, of course.
To measure it accurately, you really need to measure the current and power factor. I'd venture to guess the power factor is pretty low in this case.
(N.B. I'm not an electrician or anything related, so this may be misleading or downright wrong. Caveat lector.)
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Andrew Erickson

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True. Considering the power factor, starting all those machines at the same time you can push up the power factor and actually increase the cost as compared to running at no load for 30 minutes.
Frankly, I don't completely understand the power factor thing (it can be corrected with capacitors) but it will probably take an EE to get an accurate assessment of the situation.
The simplistic approach is to take the nameplate x kWh rate and get a raw number. If management does not know any better they may accept your numbers, give you a big bonus and think they are saving a bundles. Realistically, you'd want to check the current draw at a minimum, using an Amprobe. Check that start up though, it may be quite costly.
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The problem you're going to face is that the dataplate amperage is most likely the Full Load Amperage (FLA) of the motor. I'm assuming that the motors are not under load - essentially idling - during the break time. If that's the case, the amperage draw\ will be much less than the FLA. For example, the 5HP motor on my cyclone has a dataplate amperage of 20.8 amps. However under no load, the measured current is less than 6 amps - less than 30% of the FLA.
Depending on accessibility and/or company safety policy, you might be able to measure the actual amperages of the running motors using a clamp ammeter. If not, the motor manufacturer might be willing to provide the no load amperage specs for the motors. I don't know the characteristics of the motors or their load condition during the periods that concern you, but I believe it's a near certainty that simply using the dataplate values will give wildly inaccurate results.
Tom Veatch Wichita, KS USA
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If you can persuade management to cooperate in an experiment, the simplest method is probably to check the meter while all the motors are idling, then shut them off and check it again. By counting the revolutions of the spinning disk in the meter during a fixed time (one minute should be plenty) and using some arithmetic the power savings can be calculated. Sometimes the necessary info is printed on the meter itself, if not the utility co. can tell you what the conversion factors are.
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Oh Mark - you're biting off quite a piece here. There's more to it than the cost of the electricity to keep machines running., There's the cost of start/stop. On many machines, the hardest part of their lives is when they start/stop, The overall maintenance costs may well exceed any savings from shutting them down for a lunch hour.
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I'm kind of torn as to whether the restarting will draw more current than what is saved during the lunches. It only takes app 20 seconds to spool up the motor to a ready state, so I'm guessing that is more than offset by the 35 minutes it would be shut off. And additionally, there are also scrap chutes which are powered by separate motors. And we would be saving on the wear and tear by running at least 2 hours less 5 or 6 days a week. All totaled, the machines would run app 2 1/2 hours less per day (we run 2 shifts), 12 1/2 to 15 hours a week. And that's per press. I can't help but think it's cheaper than running them. But does anyone remember all of the debating during the gas crisis (pick any one in the last 30 years) about whether it was cheaper to idle your car while in line for gas, or shut it off and restart to move up?

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"mark" wrote:

The only way to know for sure it to hook up a recording KWH meter for a couple of weeks.
Test1: Let the press run continuously.
Test2: Stop the press during breaks.
Run each test for a week and compare results.
Lew
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You don't have to have good data in this case. All you have to have is the "conservation" attitude and a loud voice to accuse your (former) employer of despoiling the planet by leaving the lights on and the motors running. Such issues do not require facts, except from those who would defend the status quo.
If they decide to turn everything off, you can compare electric bills. Maintenance is not a player, because you can always argue the failure would have occured regardless.
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"mark" wrote:

The answer to your question lies in the application, namely punch presses.
The motor is connected to a flywheel which stores energy to be delivered to the punch when the clutch is activated.
When the punch is made, the flywheel slows down and delivers the energy to the punch to make the part.
The balance of the cycle is used to bring the flywheel back to design RPM.
As soon as that happens, the motor basically coasts consuming minimum power.
So during your breaks, the press is basically operating at idle or minimum power consumption.
If the press were to be shut off, the flywheel would coast to a stop losing all it's stored energy, then you would have to replace all that energy by bringing the flywheel up to speed before starting operations.
This is a case where keeping the press running during breaks is probably the most cost effective operating procedure.
Lew
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Sorry, I'm not buying your logic.
If the flywheel coasts to a stop during the break, it does so because of frictional losses. By leaving the motor on, the motor has to continuously overcome the losses that are always there. The motor is always doing work. If the flywheel coasts to a complete stop and sits idle for a time, then no more frictional losses are being incurred during that time. However leaving it on, the motor has to continue to do that work. The question becomes whether or not the integral of those losses is more or less than the work required to spin the flywheel back up to speed at the end of the break.
- Owen -
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Owen Lawrence wrote:

This is not a question that you're going to answer by logic. You need to measure the power consumed during the break, measure the power consumed during start up, and compare the two. If it looks like there's a saving, then you need to look at other factors, like the effect of start/stop cycles on TBO.
What you need is a plan for measuring the consumption for each case, and for each machine if they are of different kinds. Find out what such measurements would cost and compare those costs to your best guess as to the potential savings and see if it makes sense to do the tests. If it does then your boss should be quite happy to go ahead with them.
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I'm not buying anyone's logic on this. Power factor is going to come into play tossing out the logic of both of you. The PF will be affected when all the motors are started at the same time, thus affecting the rate for the entire time.
Go take a look at the meter. Some have a pointer on a PF scale. I don't know enough about it to give the definitive answer, but it does affect cost.
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"Owen Lawrence" wrote:

Not a problem.
My response was not an absolute answer, rather was an indication of probable results.
There are too many variables to give a definitive answer with out actual test data which is why my post about using recording KWH meter.
Lew
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Yup. Once I actually witnessed two PhD engineers sparring this way; eventually one of them (a professor doing expensive consulting for the company) finally threw up his hands and said, practically shouting, "Aww don't introduce another variable!"
But it's fun to chat, anyway. I hope the OP finds what he needs.
- Owen -
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Am I missing something here?
MONDAY: 1159hrs - read meter 1200hrs - go to lunch, leave machines on. 1259hrs - return from lunch read meter
TUESDAY: 1159hrs - read meter 1200hrs - go to lunch, turn machines off. 1259hrs - return from lunch, turn machines on, read meter
Compare kwh readings
Lumpy
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Yeah that would be way too simple. Big electricity purchasers also have to factor in the extreme loads in addition to the amount of kilowatts that they consume. You really have to see how the electricity provider interprets your usage to determine if there is a savings or not.
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Lumpy wrote:

Yes, you are missing something. That method does not account for other loads in the building.
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That is how I would do it. If you are worried about other loads in the building, do it everyday for a week. That should factor out the unusual loads (statistically speaking). This would get you pretty close.
Montyhp
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Lumpy wrote:

J. Clarke:

Please explain what "other loads" will be mis-accounted when doing the comparison readings.
Lumpy
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