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Thread: out of ideas

  1. #144
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    Quote Originally Posted by kdocsr05 View Post
    Personally I do not remember ever testing a motor under no load conditions, never had a reason to.
    hey, we have tested these things "a bunch of times"

    Why would I want to bench test a motor under conditions not offered at the site? This serves no purpose except to demonstrate a potential to work under design conditions.
    a good reason would be, you would understand and agree with the techs here that "have" tested these things

    I’m not sure if I agree with the statements that a unloaded signal phase induction motor will burn up after running unloaded for a couple of hours.
    more specifically the motor will get hot, then go off on overload, and yes, if ya ran the motor long enough without a load it will cycle on and off the overload and eventually fail, how long before it fails? could be months, could be a year, "that we have not tested".

    There are conditions in which a single-phase induction unloaded will draw current higher than the nameplate data that I did not see in the link
    so, there ARE conditions which you did NOT see, i guess you just "KNEW" it


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  2. #145
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    Quote Originally Posted by Dowadudda View Post
    i thought this was 3 phase motor
    Op's first post states run caps checked out fine.

  3. #146
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    A point for us to ponder.

    Bench testing a motor under no load or running a motor with too big/too little a load will overheat a motor after some time period according to Fasco.

    They have stated that a running motor takes about 2 hours before it reaches its normal operating temp. So if its been operating two hours under a no load condition and the motor starts to get hotter then since no energy is going into producing mechanical advantage then most of the energy is going into only producing heat, then wouldn't that motors amp draw go up from the resultant heat?

    Heat creates resistance in an electrical circuit. Would it go to or above the rated FLA? Maybe, who knows and who wants to perform this type of test just to see when the results would only prove that an unloaded motors running amps would increase as it gets hotter. We all already know that. But it's possible under these circumstances that a motor without thermal overload or a failed thermal overload could heat up to the point FLA would be exceeded.

    So it's quite possible in my mind and experience that it's true.

    Side note: my redneck cousin didn't like the way his 1956 chevrolet motor ran and demanded a new motor for his new car under warranty. Dealer said no way. So he rigged the motor to run at over 2,000 rpm out of gear and just sitting there. By the morning the motor had frozen up causing the dealer to come get it and put a new motor in it. I know, it's cheezy but that is what he did.

    Why did the motor freeze up under no load. Cause when under a load all the tolerances within the many components of the internal motor become closed since it's doing work. Under no load all of the mechanical parts tolerance are flopping around which eventually led to the destruction of the motor. Interesting huh?
    "The American Republic will endure until the day Congress discovers it can bribe the public with the public's own money.
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  4. #147
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    If you read that fasco fact book, which I have a copy. It clearly says that an air over motor will overheat if it is underpowered. This is due to the fact that it is designed to have so much air blowing over the shell. An over sized condenser fan motor is a good example, oversize it and it burns up. You'll be scratching your head about it unless you realize how it is meant to be cooled, by the condenser fan!

    A fan cooled motor, with a fan on the shaft inside the end bell, is not going to overheat if run unloaded. The thing is cooled by it's own fan and if it pulls 12 amps and is rated at 12 amps the windings will take it forever.

    a motor will overheat if it is overloaded, either mechanically or if the voltage is low. Or it will overheat if it is not being cooled as per design.

    Simply running a motor unloaded will not be the cause of the overheat.

  5. #148
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    Quote Originally Posted by Pneuma View Post
    Simply running a motor unloaded will not be the cause of the overheat.
    you just contradicted the whole paragraph you wrote



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  6. #149
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    Quote Originally Posted by Airmechanical View Post
    you just contradicted the whole paragraph you wrote



    .
    I see how it could seem that way let me try to clear it up.

    If you take an air over motor and run it on a bench it will overheat, beacause it is cooled from air moving over the outside of the motor, air moved by the fan you attach to the drive shaft. But it fails because it's not being cooled, not because it is unloaded.

  7. #150
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    Quote Originally Posted by Pneuma View Post
    If you read that fasco fact book, which I have a copy. It clearly says that an air over motor will overheat if it is underpowered. This is due to the fact that it is designed to have so much air blowing over the shell. An over sized condenser fan motor is a good example, oversize it and it burns up. You'll be scratching your head about it unless you realize how it is meant to be cooled, by the condenser fan!

    A fan cooled motor, with a fan on the shaft inside the end bell, is not going to overheat if run unloaded. The thing is cooled by it's own fan and if it pulls 12 amps and is rated at 12 amps the windings will take it forever.

    a motor will overheat if it is overloaded, either mechanically or if the voltage is low. Or it will overheat if it is not being cooled as per design.

    Simply running a motor unloaded will not be the cause of the overheat.
    I read that book also and my understanding is the motor is designed with the amount of power it needs to pull the designed load. Without the load the extra power is converted into heat rather than expended on the load. I understand it better than I'm explaining it and I don't have the time to go back and copy those sections of the book.
    I lied:
    Load Factor should not be confused with the common
    motor term called Service Factor. Service Factor pertains
    to self-cooled motors, such as the ones designed in accordance
    to NEMA. Service Factor is the percentage over
    nameplate horsepower that a particular motor can be operated
    at while being sufficiently self-cooled. For example, a
    1.3 rating relates to a 30% reserve in horsepower that can
    be drawn on if needed. This is useful when intermittent
    overloads will be encountered
    FACT Replacement by the use of a larger horsepower may
    cause the motor to be underloaded. It could eventually overheat
    and cause overload trips. The actual ampere reading will be
    below the full load amp rating on the nameplate.
    So sayeth FASCO the wise.
    Last edited by tipsrfine; 12-29-2009 at 07:09 PM. Reason: I lied

  8. #151
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    Quote Originally Posted by Pneuma View Post
    I see how it could seem that way let me try to clear it up.
    it's clear in my mind, get it right for yourself

    If you take an air over motor and run it on a bench it will overheat, beacause it is cooled from air moving over the outside of the motor, air moved by the fan you attach to the drive shaft. But it fails because it's not being cooled, not because it is unloaded
    your not making sense, it failed because it was not loaded.
    your changing words around, think of it like this

    1+1=2


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  9. #152
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    Quote Originally Posted by bigtime View Post
    Can this unit pull in return air and/or outside air? Im wondering if the resistance of the return air duct is more than if its pulling outside air. Maybe when it goes to outside air there is less resitance and the fan loads up more.

    If you get another motor, go to a 3hp. Make sure the wiring is adequate for the upgrade.
    Bump.

    Are you sure something is not going on with the oa and return air dampers? The amp draw on the motor could change when the inlet fan pressure changes. If it moves more air in one condition, it will pull more amps than the other. If its set up on the lower inlet condition, and then regularly sees the higher inlet condition, it will run in the service factor. It may cut on/off on the iol.

    Ive never seen a motor messed up due to "power" problems. Seen lots of stuff mess up due to "set-up" problems.

  10. #153
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    Quote Originally Posted by Airmechanical View Post
    your changing words around, think of it like this

    1+1=2


    .
    I have no idea what you mean by that equation and when someone is not clear on what you are saying, then one trick is to change your words around and try to say it a different way.

    Here is what I am trying to say. The lack of a mechanical load does not cause the current to be dispated as heat rather than work, which was put forward by another poster in this thread. Nothing in ohm's law or in electrical theory supports that. At least nothing that I know of, if you know something different please share it.

    Heat is the result of current flowing through the windings, that is a fact that is hard to argue with. 12 amps running through the windings will heat them according to thier resistance and whatever is cooling them. If 12 amps exists when the motor does work and it exists when it doesn't do work, very unlikely also, then the motor will be the same temperature, if the method and means of cooling the motor remains constant.

    If you disagree please share some technical reference. It would not be the first time I learned something new.

  11. #154
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    An AC motor create inductive reactance due to the fact that electric fields are created using AC voltage. Along with that a mechanical function of the motor called "slip" is also factored into the engineering of the motor. These two items must work together to make the motor efficient in producing "work."

    If work is not produced by the motor, then the motor produced heat in place of that work because the motor is not operating under the mechanical conditions it was designed for.

    The best I can do now is to include this link. Under the efficiency formula the process is explained in that what energy is not produced resulting in mechanical advantage, or "work" then the unused energy will produce in heat within the motor.

    And since the motor is not producing more heat than mechanical advantage, that heat will continue to build up of some period of time causing the motor to run at higher than design temperatures.

    http://progress-energy.com/custservi...ry/MISC003.asp

    Ohms law, to my memory, does not take into account inductive resistance products, only inductive and resistive products. I could be wrong about that though because it's been years since I've had to deal with this stuff.
    "The American Republic will endure until the day Congress discovers it can bribe the public with the public's own money.
    - Alexis de Toqueville, 1835

  12. #155
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    I believe that much of what is being said here is accurate even the contradictory statements. The problem is that it depends on what kind of motor we are talking about.
    With a basic split phase induction run motor (where a capacitor is not involved once the motor is up to speed) running at full speed, there are two basic currents: the current to magnetize the windings and the current to do the work. With a low load the motor should run low amps.
    A capacitor run motor (where you keep a capacitor connected to the start winding at full speed) is designed to act like a 2 phase motor at a given load and RPM. At either too much or to little load the efficiency drops off and the wasted electricity is converted to heat. Generally the smaller the MFD of the capacitor, the less this is a factor. I'm no engineer, but I do understand that the relationship of things like: load, back EMF, RPM, inductive reactance and capacitive reactance, are taken into account.
    Last edited by ChuckHVAC; 12-30-2009 at 07:36 AM.
    Anything I say here is only my opinion. Even if you understood what I said.... What I said may not even be what I meant.

  13. #156
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    Back to the OP: I'm still thinking power factor. If you have a good industrial electrican with a PF meter you could confirm or eliminate that issue. Grossly simplified, power factor is the difference between the Watts and the Volt Amps. In some cases volt amps can be twice what the watts are (PF of .5). I assume the pool room has lots of ballasted lighting of some sort. With all the capacitors and inductive loads involved the chances are things not right.
    A quick DIY experiment in this case would be to reverse the power leads to the motor and see if it changes the measured amps.
    Anything I say here is only my opinion. Even if you understood what I said.... What I said may not even be what I meant.

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