Three-phase salient-pole synchronous generator

Discussion in 'Homework Help' started by Jess_88, Oct 1, 2012.

  1. Jess_88

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    Apr 29, 2011
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    Hey guys :)

    I'm having some trouble with this question.

    Three-phase salient-pole synchronous generator has the following constants in per unit on the machine kVA rating as a base (all resistances are neglected):

    Xd = 1, Xd' = 0.35, Xd''=0.25, Xq = 0.6
    Ea = 1.8<15°, Vterminal = 1<0°

    - Compute power angle (δ), generator current magnitude (in pu) and phase angle (Ia<θ).

    Now, the power angle is easy (just 15°) but I have no idea what equations I can use to fined the generator current. I'v been reading every text book I can fined and nothing really seems to fit. There just doesn't seem to be much information on this sort of thing.

    What has given me the most promising results is:
    E = Vt + Ia(Xd + Xq)
    Ia = (E - Vt) / (Xq + Xd)
    ... but I have been told this is wrong :(

    I'm guessing I have to do a phasor diagram and use some trig... but I cant see how it could be done with the given information.

    Any help would be great,
    thanks guys :)
     
  2. Kermit2

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    The generator supplies only the current demanded by the motor in this scenario.

    Synchronous motors draw/create current based on the power angle(load). No load - no lag angle - no current
     
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  3. t_n_k

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    You are correct in assuming you must look at the synchronous generator phasor diagram. This is pretty standard - as you would presumably know. Ignoring any resistive terms [as the question states] makes the solution simpler. One would not include the transient or sub-transient reactances in the [steady state] analysis. The answer will "come out" using simple geometrical relationships for the direct and quadrature terms.
     
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  4. Kermit2

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    an example from an old textbook.

    [​IMG]
     
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  5. t_n_k

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    Hello Kermit2,

    If you re-check the OP's post you'll note there's no mention of a motor in this question.

    I believe in this case one simply assumes the generator is being driven [by something] at synchronous speed and the generator output [perhaps to an infinite bus] meets certain criteria based on the power angle [already known] and other stated parameters such as terminal voltage & machine reactances.
     
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  6. Kermit2

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    I believe the vector summations would be the same in either case, but you are right, and the mention of the motor/gen example was probably not the best approach.
     
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  7. t_n_k

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    I have to disagree somewhat - the vector summations wont be the same as those shown in your attachment. With a salient pole synchronous machine one should consider both the direct and quadrature components of the machine reactances. This makes for a more [but not overly so] complicated vector diagram.
     
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  8. Kermit2

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    I was suggesting an approach to the problem, and did not intend to answer the problem for him, just a shove in the right direction. I would be hard pressed to do any of these maths now anyway since my education is almost as old as the text book I used.

    I will defer to your superior knowledge of the subject.
     
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  9. t_n_k

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    You overestimate my knowledge on the subject and probably undervalue your own.
     
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  10. Jess_88

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    Apr 29, 2011
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    Thanks for the input guys :)

    So im pretty sure I have the theta angle worked out from my phasor diagram
    x=tan^(-1)(0.6/1)
    theta = x - 75 = -44
    which looks good.

    Now I'm pretty sure I wouldn't use a phasor diagram for the current magnitude (at lest I think...). I would need a formula incorporating the voltage drop across Xd and Xq and Vt. I'm just unsure which formula to use :(
     
  11. t_n_k

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    Unless you draw the phasor diagram you won't be able to solve the problem. Your answer for θ of -44° isn't what I calculate.
     
  12. Jess_88

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    Apr 29, 2011
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    Here is my phasor diagram.
    Does it look ok?
    [​IMG]

    I'm getting pretty confused with my trig :(
     
  13. Jess_88

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    Apr 29, 2011
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    I noticed something in my notes... but its doesn't exactly make sense to me.
    Is this true? (looking at Xq)
    [​IMG]
     
  14. t_n_k

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    This is how I saw it ....
     
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  15. Jess_88

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    Apr 29, 2011
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    Thanks t_n_k,

    Could you explain a how you got your 75 angle? I think there may be a relationship I'm not aware of.
    I was looking at it from my phasor as, 90 degree difference from Iq to Id. So subtracting 15 from 90 to give -75 from the x axis.
     
  16. t_n_k

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    Yes - since IqXq is orthogonal with Ea it must be true with \delta = 15° that the opposite angle is 75°. Just to be clear, this applies to the right-angled triangle formed by the sides V_{T} \ , \ I_{q}X_{q} and \(E_{a}-I_{d}X_{d}\).
     
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  17. Jess_88

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    Apr 29, 2011
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    Sorry, I think I'm a little slow...
    Did you calculate Id and Iq as well to do this?
     
  18. t_n_k

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    Not to find the 75° - which by the way is essentially superfluous information. It was meant to indicate the right angle triangle lying along the VT and Ea phasors.

    The relevant relationships for finding Id & Iq can be extracted from the formulas shown at the top of page on the phasor diagram of post #14
     
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