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3-PhaseInduction Motor slip baffling problem

  1. Jun 29, 2012 #1
    Hello, I was trying to solve a problem about induction motors but was baffled at one point.
    Here's the question.

    Nameplate of a 3 phase induction motor has the following information:

    Power = 3kw, speed = 1430rpm, f= 50hz, PF = 0.85, V= 380v
    Neglect Stator Impedance.

    Q1. Calculate Torque : T= (Motor Power)/(angular velocity of motor) = 20.02 Nm (confirmed from answers)

    Q2. Calculate the air gap power ?

    What I did :
    I know that the air gap power is the same as the rotatory magnetic field power which is at synchronous speed. I also know that this power is all sent to the rotor as (I^2)R/(slip) meaning that :

    P (air gap)= Power of Motor / (1 - slip)

    My problem I don't seem to find a way to calculate the slip because I dont know the number of poles used since:

    slip = ( synchronous speed - motor speed ) / synchronous speed

    where synchronous speed (rev/min) = ( 120 * frequency ) / # of poles

    Please, if someone has an idea, how the air gap power can be calculated I will appreciate that.

    Ans = Air gap power = 3.144KW
  2. jcsd
  3. Jun 30, 2012 #2
    Lets try some educated guessing:

    ( p = # poles, n=synchronous speed, s=slip )

    1. p=2 -> n=3000 rpm -> s= 0.52
    2. p=4 -> n=1500 rpm -> s= 0.047
    3. p=6 -> n= 750 rpm -> s= -0.9

    In what order of magnitude is the slip in an induction machine at rated operation ?

    Based on this reasoning and ruling out all the alternatives, the answer is left alone.
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