Help regarding AC induction motor starting current

Discussion in 'General Electronics Chat' started by Sattamassagana, Apr 9, 2013.

  1. Sattamassagana

    Thread Starter New Member

    May 29, 2011
    3
    0
    Hello,

    I am a bit troubled over calculating the starting current for a motor I want control.
    I have here a 0.15KW 3 phase induction motor. It will be wired in Delta configuration.

    Specs:

    Nominal speed: 880rpm
    Nominal current: 0.72A
    Starting to nominal current ratio: 2.1A
    Nominal torque: 1.63Nm
    efficiency: 50%
    Nominal power factor: cosφ=0.58
    Starting to nominal current ratio: 2.1A
    Starting to nominal torque ratio: 2.2Nm

    On the nameplate I saw that A(Amps) is rated at 1.25A when delta wired @ 230V.

    Also, the motor is fitted with a worm gear unit so that

    Output speed: 9 rpm
    Output torque: 72Nm


    What I' d like to know is


    1) Do the Amps rated on the nameplate refer to the Full load current?



    2) If yes, am I correct with calculating the starting current?


    (Starting to nominal current ratio) * Full Load Current= 2.62A


    3) Does the worm gear unit reduces or increases the duration of the starting current, and is it possible to quantify it?



    Thank you in advance
     
  2. PackratKing

    Well-Known Member

    Jul 13, 2008
    850
    215
    The nameplate on the motor should list " Locked-rotor Amps " or LRA

    Since all 3-phase motors start " with a bang " your start current should be a very short pulse, likely only 3 times load current, that is if you could find a meter short of an oscilloscope, capable of measuring such a short pulse.
    The worm drive should not affect the start parameter.
     
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