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  1. #21
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    [QUOTE=TDegenhart;3549824]Not so. If the current flows in a resistive element, there will heat generated.
    Tom[/QUOT

    Inductive element...

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    Quote Originally Posted by jim rozen View Post
    Ah, actually not. Power factor is near to zero.

    Sorry, wrong.

    The current that you measure ALL contributes to heating. I^2 * R still works. There is NOTHING DIFFERENT about the current. Do not be fooled into thinking that some of the current "does not count" somehow. The current is "all good all the time".

    The difference is not in the current, it is in the relation of current and voltage, which is what it takes to have "power". A low power factor means that most of the current occurs at a low voltage, so that the power is also low. A high power factor would mean the current is a sine wave closer to being in phase with the voltage so that the power for the same current is larger.

    With the power factor low, but still non-zero, most of the actual power may be the loss due to current flowing through the circuit resistance. Very little is running the motor, but that's fine because in those conditions there is very little power needed.

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    Take DC resistance of the windings. Multiply by current squared. Whatever happens inductively, that heat is getting dumped into the stator windings. It's wire with amps going through it. Heat happens.

    Except for the increasing current due to increasing load, this is constant. For an unloaded motor, this is a substantial chunk of the losses.

    Rotor losses do increase with load, because with little slip there is little rotor current. I'm not sure about magnetic losses.

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    Quote Originally Posted by JST View Post
    Sorry, wrong.

    The current that you measure ALL contributes to heating. I^2 * R still works. There is NOTHING DIFFERENT about the current. Do not be fooled into thinking that some of the current "does not count" somehow. The current is "all good all the time".

    The difference is not in the current, it is in the relation of current and voltage, which is what it takes to have "power". A low power factor means that most of the current occurs at a low voltage, so that the power is also low. A high power factor would mean the current is a sine wave closer to being in phase with the voltage so that the power for the same current is larger.

    With the power factor low, but still non-zero, most of the actual power may be the loss due to current flowing through the circuit resistance. Very little is running the motor, but that's fine because in those conditions there is very little power needed.
    Thanks for the clear summation there on power factor/current/voltage/watts.

    To the OP, the motors are likely fine. You just need to feed them better power. If you starve a dog, don’t beat it for not hunting.

  6. #25
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    Quote Originally Posted by SomeoneSomewhere View Post
    Take DC resistance of the windings. Multiply by current squared.
    How about a different way: Power is V(sq)/R correct? Say the motor has a ten ohm dc resistance per winding, what's the power there? Use a round 200 volts for the
    voltage across that winding.

    Ooh. Looks like 4 kW per winding eh? That's a LOT of heat. Non-physical answer.

    What's wrong with this - maybe the motor is not a dc resistor, it's an inductor and it's not being driven by dc. Laws of physics still apply - but the numbers are complex.

  7. #26
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    That would be correct for a STALLED ROTOR condition, but an idling motor is not stalled, the induced current in the moving rotor is generating a back EMF that opposes the applied voltage. This is the reason motors that are started from zero have an inrush current, roughly 6x times running for larger motors. The inrush time period is the acceleration time of the rotor.

    Tom

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  9. #27
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    Yes. And that locked rotor current/kVA is what is marked on the motor as the inrush "class" code in kVA per HP.

    When running, the "effective voltage" is line voltage - back EMF, and is what drives current through the motor. That is a variable as shaft output varies, because the motor slows down, and the back EMF varies with rpm according to the motor parameters. The effective voltage is normally no more than maybe 10% of line voltage, and can be a fractional percent at no-load.

    You really do not need to do any more but remember that this stuff happens when interpreting the results of measurements. A good 99% of motor problems can be solved by just knowing the motor basics, and using a clamp-on current meter!

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    The reason why I^2R works and V^2/R doesn't is that the most accurate model of an inductor is generally as elements in series - each element gets the same current, but not necessarily the same voltage.

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    well you guys seem to be having fun.

    I'm going to reassemble the drive and test the motor underload and see if the real/imaginaries changed.

    I think the Overload relay is a solder melting relay. Square D B11.5. I don't know how to adjust it or where to set it, nor have I pulled it from the mill to inspect it. I haven't had the mill long and I haven't had much time to really use it so its been happening since the start. First time it happened, Phase-o-matic rep asked me to pull one of the capacitors. He also suggested I ahd a 50 microfarad run capacitor; which I haven't done yet.
    img_7260.jpg

    Phase converter is before the starter. I don't have a schematic of the circuit to confirm it's wire properly. Tag is till on the 110v coil. So, I'm guessing these components are orignal.
    img_7267.jpg
    img_7255.jpg
    img_7263.jpg

    Please provide a link if there's a specific sticky or thread I should read. I try to search my questions before I post.

    Thank you for the responses.

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    oh, the fun is only just starting. ...long and short of it is ditch the "start assist" unit (static phase converter) and either get or build a rotary converter, or probably cheaper, more efficient and more versatile get a VFD.

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    Quote Originally Posted by cyanidekid View Post
    oh, the fun is only just starting. ...long and short of it is ditch the "start assist" unit (static phase converter) and either get or build a rotary converter, or probably cheaper, more efficient and more versatile get a VFD.

    Or use the one on the 7.5 to start an idler. Depends on the HP range that covers, but a static converter plus a motor makes a rotary.

    Unfortunately, Phase-a Matic seem to have a 4 to 8HP and an 8 to 12 HPmodel, so odds are the OP has the 4 to 8, probably too small to use with the size motor he needs to run the 7.5 HP machine, which would be at least a 10 HP idler..

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    Thats one of tests I want to run; to use the 7.5 as the idler. I do have the 4 to 8 HP. Phase-o-matic rep said the difference between the smaller converter and the 4-8hp was a 400 micorfarad capacitor. So I put a switch inline to pull the cap in for the lathe and out for the mill. Deep cleaning the lathe so I have put 'er underload yet.

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    Quote Originally Posted by PINCK43 View Post
    Thats one of tests I want to run; to use the 7.5 as the idler. I do have the 4 to 8 HP. Phase-o-matic rep said the difference between the smaller converter and the 4-8hp was a 400 micorfarad capacitor. So I put a switch inline to pull the cap in for the lathe and out for the mill. Deep cleaning the lathe so I have put 'er underload yet.
    If a Phase-A-Matic static "converter" (NOT REALLY..) was on fire, I'd not bother to piss down its virtual throat. Kicked one around the porch for a year just out of mean-ness before sending it to the landfill.

    Even so.. the buggers are built out of perishable components. A relay that can age, wear, be abused, fail to act as expected, and capacitors as can age fail, etc. If you have some preverted rationalization to keep one of the fool things, at least give it a crust of inspection and maintenance now and then.

    As to making them into the starter / control for an RPC? Why wudja doo that with an old and dodgy one?

    They were built to a budget, and it was anything BUT a "generous" budget.

    Better to start fresh. Build a "proper" RPC.

    Then yah KNOW what yah have. And know that all in it was bought right, and of decent quality.

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    Quote Originally Posted by TDegenhart View Post
    That would be correct for a STALLED ROTOR condition,
    Tom
    Which is what you get if you figure the entire thing can be figured out using dc analysis. Just saying....

    It's not a resistor.
    The dc resistance does not matter.
    A clamp on meter does not tell you how hot the motor will get.
    The meter reads just about the same value as the motor is loaded up, the only thing that changes is the power factor.

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    Quote Originally Posted by thermite View Post
    Better to start fresh. Build a "proper" RPC.
    +1

    Or, as mentioned, if it's a one-off, a VFD makes a neat intallation.

    The mental cost of setting up a phase a matic for this is high enough to pay
    for at *least* one good VFD.

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    Quote Originally Posted by jim rozen View Post
    +1

    Or, as mentioned, if it's a one-off, a VFD makes a neat intallation.

    The mental cost of setting up a phase a matic for this is high enough to pay
    for at *least* one ^^^ pallet of ^^^ good VFD.
    Fixed that for yah.


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    Quote Originally Posted by jim rozen View Post
    Which is what you get if you figure the entire thing can be figured out using dc analysis. Just saying....

    It's not a resistor.
    The dc resistance does not matter.
    A clamp on meter does not tell you how hot the motor will get.
    The meter reads just about the same value as the motor is loaded up, the only thing that changes is the power factor.

    Heh.... I think you are running out of arguments.

    Using an ammeter is not the same as thinking in purely DC terms. Especially if it is an AC ammeter

    A voltmeter will not tell you how hot a motor gets either. But an ammeter will allow you to determine if your motor is overloaded, or at full power, running at a low load/unloaded, etc. For a single phase motor, it will tell you if the start switch is working. For a 3 phase motor, it will tell you if the three phases are balanced.

    A motor running unloaded draws maybe 40% of full load amps ("FLA"). A motor at full load draws "FLA", and it is a poor ammeter indeed that cannot distinguish those two values.

    Would a power meter be good? Sure.

    Can a voltmeter be helpful? Sure, you will need that information also. But it will not tell you as much about the motor.

    Just give it up..... motors run on current, and thus a clamp-on ammeter is the best tool to have, if you can only have one meter.

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    Quote Originally Posted by JST View Post
    Heh....
    Using an ammeter is not the same as thinking in purely DC terms. ....
    Sorry, I was replying to somebody somehwere when he said:

    "Take DC resistance of the windings. Multiply by current squared. Whatever happens inductively, that heat is getting dumped into the stator windings. It's wire with amps going through it. Heat happens...."

    I am actually aware that rotary converters run on ac and the currents and voltages are hardly ever in phase with those, no matter what an amp-clamp meter reading
    might be indavertently be interpreted as, erroniously.

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    Quote Originally Posted by JST View Post
    ... a clamp-on ammeter is the best tool to have, if you can only have one meter.
    Truer than average.



    I'm sure you are aware that near-as-dammit all recent ones also have conventional leads and are full-service general-pupose "VOM ++" meters in ADDITION to having an Amp-clamp.

    Even so.. most work yah want two or even three meters to-hand. Decent ones are cheap enough, and it saves time yah can record more than one parameter at a setup.

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    Quote Originally Posted by thermite View Post
    Truer than average.



    I'm sure you are aware that near-as-dammit all recent ones also have conventional leads and are full-service general-pupose "VOM ++" meters in ADDITION to having an Amp-clamp.

    ......
    Even my old "Amprobe" meters can be switched to be voltmeters. Having the functions as two separate meters definitely has the advantage that you can correlate voltage and current readings.

    I will give Jim R the point that you can read "VA" that way, but it takes a power meter to read the actual power in anything other than a heater or similar resistive circuit.


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