generators

Discussion in 'Homework Help' started by muni, Sep 16, 2008.

  1. muni

    Thread Starter Active Member

    Jul 29, 2008
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    sir in this attachment please find the question 12 and 13. i'm unable to understand how the answer has been derived. if any one can help me, i'm thanking them in advance
     
  2. vetterick

    Active Member

    Aug 11, 2008
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    Yes, phase shift is indeed the answer, you can only directly add AC voltages that are in phase with each other.

    Question #12 looks like a stepper motor, I don't see any real practical use for it as a generator, but it works to show how much voltage is lost with the configuration.

    Question #13 is a standard 3 phase generator, the generator actually generates 1.73 (square root of 3) times the output voltage, seems a little bit non productive, untill you find the motor connected to it uses 1.73 times less amperage than a single phase one.
     
  3. scubasteve_911

    Senior Member

    Dec 27, 2007
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    12) Is a two-phase generator.. You should know that the phase will be 90 degrees out of phase, so you plot both 70Vrms signals as vectors. One is at 0degrees, the other at 90degrees. Then, you draw them head to tail, then draw a new vector from the origin. Do the trig, it works out to be 90Vrms.

    13) Same idea as 12, except the angle is now smaller.

    Steve
     
  4. muni

    Thread Starter Active Member

    Jul 29, 2008
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  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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  6. scubasteve_911

    Senior Member

    Dec 27, 2007
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    He just rounded up to 90Vrms. If you input RMS into the trig, then you will return RMS.

    Steve
     
  7. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    √9800≈99, not 90. 99 is the answer in his problem sheet.
     
  8. scubasteve_911

    Senior Member

    Dec 27, 2007
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    Am I bad? That was a strange thing, I saw 98.994 and rounded down to 90 :p

    Steve
     
  9. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I'm glad you said that. I thought I was going nuts.:eek:
     
  10. muni

    Thread Starter Active Member

    Jul 29, 2008
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    sir thank you. i too have not noticed answer before asking you reply. i thank ron H sir also for his valuable suggestion.
    now my doubt is in Q12 it was only 2 vectors. but in Q13 there are 3 vectors. how should i do it ? first shall i calculate the third vector with any one of them. and should i go to final calculation?
     
  11. scubasteve_911

    Senior Member

    Dec 27, 2007
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    Question 13 asks "how much voltage would be measured between any two open wires?"

    So, there are 3 pole pairs, thus 360/(3*2) spacing between phases, or 60 degrees. Use trig. to solve for the resultant vector.

    Steve
     
  12. muni

    Thread Starter Active Member

    Jul 29, 2008
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    sir thank you, with this i got the funda of this question. sir i think if there are more no. of pole pairs, then the spacing between the phases will be by the formula 360/(n*2). if any thing wrong, please diect me
    thank you sir
    muni
     
  13. scubasteve_911

    Senior Member

    Dec 27, 2007
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    muni,

    A little advice, try not to rationalize with formulae, try to be able to reason to a formula. This is because, a lot of times, a variable is changed and your handy formula sheet is no longer valid. For example, what if one were to make the windings asymmetrical or have more poles on the rotor?

    Steve
     
  14. DrNick

    Active Member

    Dec 13, 2006
    110
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    Both of these questions are intended to test knowledge of polyphase systems. Both are essentially asking:

    "what is the line voltage of a N phase system, given that the phase voltage."

    question 12 is a N=2 phase system so

    Vline=Vphase*sqrt(2)=70*sqrt(2)=98.99

    question 13 is a N=3 phase Wye so:

    Vline = Vphase*sqrt(3)=70*sqrt(3)=121.24
     
  15. muni

    Thread Starter Active Member

    Jul 29, 2008
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    sir i really thank you. it gave me a very good idea about this.
     
  16. scubasteve_911

    Senior Member

    Dec 27, 2007
    1,202
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    Okay, now I give you the same three phase generators with a single phase missing. Now it is assymetrical and you can no longer use simplistic formulae.

    Always derive from first principals, it will come back to haunt you one day.

    Steve
     
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