Induction machine torque.

Discussion in 'Homework Help' started by Ledwardz, Jan 3, 2012.

  1. Ledwardz

    Thread Starter Member

    Oct 31, 2010
    37
    2
    Hi, im a little stuck on a past exam paper question, (see the attached file). When looking at the speed torque graph in my notes the maximum torque peak is alway at 0.2 slip and this starting torque is obviously at slip = 1 hence i substituted these values in but can't help feel like i am missing something else. I have hand a look at the equivalent circuit but meh... i can't see how he has derived the expression for torque

    all i can get is 60Protor / 2*pi*Ns
    which was derived from Pmechanical / rotor speed in rad/s
    and Protor = current^2*(R2/s)

    anyways, any help much appreciated, thanks, Lee.
     
  2. Ledwardz

    Thread Starter Member

    Oct 31, 2010
    37
    2
    here is the question
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Your image isn't very clear. In any case the max torque would occur at
     s=\frac{R_2}{X_2}
     
  4. Ledwardz

    Thread Starter Member

    Oct 31, 2010
    37
    2
    can you explain how you derived that? i assume R2 is the effective rotor resistance i.e. the load and X2 the reactance of the rotor winding. but why divide them? how does that give the maximum torque? I assume its something to do with R2/s = X2 ??

    also the question is

    T = 3E2^2R2S / \omegas (R2^2 + SX2^2)

    and i'm asked to derive expressions for maximum torque and starting torque.

    thanks for the help!
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    If you differentiate the given torque equation with respect to the slip 'S' and equate the result to zero you will derive the slip condition for either the maximum (or minimum ?) torque case. It's an application of basic calculus.
     
    Ledwardz likes this.
  6. Ledwardz

    Thread Starter Member

    Oct 31, 2010
    37
    2
    that is very very very very clever. you sir are genius. Thanks alot. :rolleyes:
     
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