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lock roto & voltage drop

Discussion in 'Electronic Basics' started by D, Jul 2, 2004.

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  1. D

    D Guest

    Hello I have a question on induction scenarios.
    I read that an electric motor will pull more amps
    at start up (lock rotor state). I have not as yet fully
    visualized why more current flows thru the windings
    when they are not spinning and/or spinning as fast.
    But this brings another curious thought to mind. Does
    the same theory apply to say like ceiling fans? I.e. when
    the fan is on a rheostat or different speed switches, doesn't
    this drop the voltage so the fan rotates slower and does
    this cause more current to flow thru the windings also??
    And one other question, if the correct wire size for the
    amp rating on a motor is close to maxed out
    (say like #12 copper and the pump pulls 16.5amps normal
    running speed) but I need to run more than 100 ft ( say like
    140 ft ) of wire to get to the motor, do I need to use the
    next higher (#10 copper) wire to keep the voltage from
    dropping too low over the length of the conductor??
     
  2. Rich Grise

    Rich Grise Guest

    google "back emf".
    The slower rotation causes lower back emf, yes, but there's
    a lower supply voltage, which predominates.
    I'd use the next higher size for _any_ run, and go up to
    #8 for 100'; and I'd still keep an eye on the motor (or
    thermometer) for a few days to feel sure that it's working
    OK.
     
  3. Also the fan design and windings are different and will withstand much more
    abuse.
     
  4. In general, the motor draws the least current when the rotor spins in
    lock step with the AC frequency, and the current rises smoothly as the
    rotor slip increases. This implies that at maximum possible slip
    (locked rotor,) the current is highest.
    In induction motor can be visualized as a transformer that has half of
    the core rotating and with a short circuited secondary. In an
    ordinary transformer, the lowest current draw occurs when the
    secondary is open circuited, and hits maximum draw if the secondary is
    shorted. The induction motor does not draw high current into its
    shorted secondary, if the core and secondary can flip over each half
    cycle.
    If the fan was slowed by an increased torque load, then, yes, the
    current would increase as it slowed. But if the slowing is caused by
    a lowered applied voltage (and the torque load actually falls as the
    fan slows) then the current may drop as the fan slows. The details
    involve the design of the rotor and how resistive the rotor shorting
    bars are, and how deeply they are buried in the rotor iron.
    This is a good idea if the motor is starting under a significant
    torque load. The temporary voltage sag may not allow enough locked
    rotor current to get it moving so that the speed can rise quickly
    enough to bring the current down to rated. In that case, the fuse is
    likely to blow, to protect the wire.
     
  5. Don Kelly

    Don Kelly Guest

    -
    You are unlikely to find references on "back emf" (a much misused term) of
    much help when considering induction motors.
    John Popelish has it correct -minimum input impedance at standstill
    (assuming the motor is not rotating in the reverse direction at the time it
    is energised. )
    --------------------------
    ------
    John has this one right as well.
    --
    Don Kelly

    remove the urine to answer

    -----------
     
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