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Motor Failure During Brownouts

  1. May 9, 2012 #1
    I used to think this question would be a simple question, but I haven't been able to get a good explanation from anyone (including two electrical engineers). So, here goes

    During a brownout (periods where a building does not receive it's normal level of voltage) motors can potentially overheat to the point of failure. As I have been lead to understand, it is because motors are designed to operate at a certain HP (745 watts). So, when the voltage drops, the current will have to increase in order to maintain this HP. However, the windings of the motor may not be able to handle this current, and the excessive heat can pose a real danger to the integrity of the motor. The part I can't understand is - what component on the motor is drawing this extra current. It's my understanding that a motor is a dumb piece of equipment that operates at a given HP when it's on. I just don't understand what part of the motor is drawing this additional current, and how it knows to draw this additional current.
  2. jcsd
  3. May 10, 2012 #2


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    Windings IS the component that can over heat due to excess current when the voltage is below specification (see bold text above by me), which can result in this kind of failure.


    Welcome to Physics Forums
  4. May 10, 2012 #3
    Thanks for getting back to me

    I totally understand that the windings are what will overheat, and I understand they will overheat because they are drawing extra current due to the drop in voltage. What I don't understand is how they draw this additional current. Motors don't typically have sensors that can detect a loss of voltage and then tell it to draw more current so it can continue to function at its rated HP. Without these sensors I just don't get how a motor can "know" that it's not getting full voltage and "knows" that it needs to draw more current (even though doing so will damage it). That is where I'm getting hung up.
  5. May 10, 2012 #4


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    On most AC motors the rotational "synchronous" speed is locked (The EMF and back EMF close to balance "small slip") to the power frequency. To maintain that speed with a load requires a level of power that is obtained by drawing current from the supply voltage on the motor windings. As the needed current increases with voltage drops the wires resistance causes heat and restricts the needed power to the point the rotational speed (and the motor cooling fan) slips below the lock frequency causing large reactive currents (back EMF unbalanced "much larger slip")in the internal wiring that saturate the cores creating harmonic currents with even more heating until the whole thing smokes.
    Last edited: May 10, 2012
  6. May 10, 2012 #5

    jim hardy

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    you ought to familiarize yourself with "Speed-Torque Curve" for induction motors.
    See Microchip's induction motor tutorial, it's their AN887.

    When you do that you'll understand the basic point which is this:

    Power an induction motor can deliver is in proportion to the square of applied voltage.
    An induction motor can withstand a brief overload to perhaps 175-200% of its rated power.

    So a 1hp motor at 70% voltage becomes a 1/2 hp motor.
    If it is connected to a load that requires more than 1/2 hp, it will be overloaded and burn up.
    If it is connected instead to a load that requires less than 1/2 hp it will continue to run just fine.

    old jim
    Last edited: May 10, 2012
  7. May 10, 2012 #6
    The poles of the stator are pulling the poles of the rotor. Because of the Universal law tending everything to equilibrium, the both poles tend to be lined up against each other as fine, as possible (and as designed). During the "brownout" the reduced voltage causes reduced current, reduced current causes lessened magnetic field, the lessened magnetic field makes the pulling force weaker, the distance between the poles increases. Now the stator pole deprived of its counterpole in the rotor behaves less like an electromagnet, but more like a wound copper conductor. So, at the same time as the magnetic current decreases, the resistive current is stepping in. The resistivity of the copper is low, current is high. This high current makes the windings get heated. The motor "burns."
  8. May 10, 2012 #7


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    OP Read this post:

    It does't draw more current because the voltage was low, to compensate. as you said a motor is a dumb piece of equipment that only draws current based on applied voltage and resistance.

    BUT the thing is, as the voltage is decreased, it draws less current as well, but since the motor is now spinning a lot slower, there is less Counter-emf developed that bucks the source voltage, and the "limited" amount of current draw is still more now than it was at higher voltage. you see what I mean?

    Motors draw ONE value of current ALWAYS (for a fixed temperature). The time when a motor draws max current, is when it is not rotating. As it starts to rotate, it develops counter emf, which inhibits the draw of current. That is why during your brownout, there was a low voltage applied, which meant low current was drawn initially, but with this applied power the motor could not turn much, there was little counter emf developed and therefore the motor draws more current than if it ran.

    I really wish I could rid of that saying, tha motors "compensate" by drawing more/less current opposite the voltage, it si simply not so. Only motor's with drives are smart enough for that. As for the rest, its counter EMF.

    The same is true for ALL motors ac or dc, and for all applications. That is why motors burn our on undervoltage, and why they burn out when stalled. if they are not moving, they are not developing counter emf, and they are not limiting current draw. which means current is drawn in excess to what the windings can handle.

    For more info on counter emf read on....

    You know that a motor can turn if conditions for motoring are met, that is a current carrying conductor in a magnetic field experiences a torque, due to the reaction of two magnetic fields. But now that the motor IS turning, you have conditions for generation met as well! Relative motion between a conductor and a magnetic field, that induces a voltage BACK into the armature (in a dc motor, or stator in an induction motor) This voltage (due to lenz' law) is OPPOSITE POLARITY of the source voltage, therefore limiting the total motor voltage. As you can see, the amount of voltage induced is given by E = N*dphi/dt, so the faster the motor spins, the more counter emf is induced, and the less current is drawn. THis accounts for why motors APPEAR to draw more current under load. Because under load, they are not spinning as fast, and therefore are not developing as much counter emf since the conductors are now cut less frequently.

    Hope this helped if you take the time to read it I think it would!
  9. May 10, 2012 #8


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    PS heres another good experiment that will illustrate what I am saying:

    Take an old small DC motor from a car blower, connect it to a small 12 volt source and connect an ammeter in series.

    Start the motor and record the current.

    Now, Hold the motor shaft and record the current.

    The values are the same. A motor draws ONE current for a speed. because the counter emf at that speed is the same.

    I would not be surprised if you also had another question.. that is... why does a motor limit it's current draw as it speeds up? or why does it draw more current under load? the answer is because counter-emf is changing with speed, and is effecting total motor voltage.

    A good question to ask your engineers, is why does the motor not draw hundreds of thousands of amps when there are millivolts in difference across the motor? :) That tends to be a good one, they will say WELL NOW COME ON....... but it is what their explanation would imply?

    So yes, the answer is during a brownout, the voltage applied is low, therefore the motor does not speed up enough to develop counter emf, and can burn out due to excess current draw. (the current that burns out the motor, is still less than normal start-up current, it just cannot handle excess current for EXTENDED PERIODS OF TIME, that is another thing to remember, its not the current value, its the time the current is experienced by the motor, because really the current value in amps which destroyed the windings is still less than normal locked rotor (start up) current in amps)

    (motors can burn out due to overloading as well, if the load caused the motor to run at the same speed as your brownout did, you could expect the EXACT same result)
    Last edited: May 10, 2012
  10. May 10, 2012 #9

    That's some fine work FOIWATER, you'll make sergeant for this.
  11. May 11, 2012 #10


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    motors are very interesting!
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