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Three-Phase systems

  1. Apr 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Hello, I am given the following problem.
    A balanced three phase 4 wire supply has r.m.s. line voltage of 415V and supplies a star-connected load made of 3 impedances:
    Za = 10 - j10; Zb = 10 + j10; Zc = 0 + j8;

    Calculate the current phase and neutral currents, also the total power dissipation. [NOTE: Assume that Va is drawn vertically up, i.e. its phase angle is 90]


    2. Relevant equations
    I've used Vline = √3 x Vphase
    Also, P = Ea x I a x cos(Φa) + Eb x I b x cos(Φb) + Ec x I c x cos(Φc)


    3. The attempt at a solution
    Please find the attached file, as my solution to the problem. As it could be seen the answer is quite different. In fact their answer is just the sum of the 1st two bits of the power, ignoring the last one. I assume they used cos(90) to get 0 and hence lower power. If this is the case, then the Zc should look like 8∠90° .
    But isn't it found from the equation : Zc = √02 + 82 ∠tan-1 (8/0) ?

    http://i1375.photobucket.com/albums/ag441/gl0ck1/StarLoad_zpslnhuotig.jpg
     

    Attached Files:

    Last edited: Apr 23, 2015
  2. jcsd
  3. Apr 23, 2015 #2

    Hesch

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    Response:
    Error: The requested attachment could not be found.
     
  4. Apr 23, 2015 #3

    Hesch

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    You must find the center voltage, En, at the center of the star-load.

    Then:

    Ia = ( Ea - En ) / Za , etc.

    ( I think? I cannot see if the center of the star-load is connected to neutral )
     
    Last edited: Apr 23, 2015
  5. Apr 23, 2015 #4
    Strange, just sent it to one of my friends, he was able to open it. Will post a photobucket link just in a sec.
    I am asked to find the neutral current, which could be found from the sum of all currents, no?
     
  6. Apr 23, 2015 #5

    Hesch

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    I've been able to open it now (error disappeared).
     
  7. Apr 23, 2015 #6

    Hesch

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    You may argue: If neutral is not connected to center of the star-load, then In = 0.
    But I've found an error: Zc = ( 0 + j8 ) = 8 / 90°
     
  8. Apr 23, 2015 #7
    Could you please, explain to me how you found it to be 8 ∠90°? What formula did you use? Isnt't it the same I used at the 1st post of the thread Zc = √02 + 82 ∠tan-1 (8/0) ? and tan-1 (0) = 0.
    I presume it is connected since I've been asked to find it.
     
  9. Apr 23, 2015 #8

    Hesch

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    I'm not using any formula: Draw the vector ( 0 + j8 ) in the complex plane and measure the angle ( or you can see it by intuition ).

    ( Contrary: ( 8 + j0 ) = 8 / 0° )
     
  10. Apr 23, 2015 #9

    gneill

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    The angle you've associated with Zc is incorrect. As a result the angle calculated for ##I_c## is incorrect.
    The angle calculated for ##I_a## is incorrect.
     
  11. Apr 23, 2015 #10
    Yes you are right, it should be 135, but really don't understand the Zc's angle. Should I do it always like that? Drawing a vector, when the real part is zero?
     
  12. Apr 23, 2015 #11

    gneill

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    It certainly can be helpful to sketch a vector for a complex value in order to get an idea of what the angle should be (such as the quadrant it lies in). But you should know that any purely imaginary number must have an angle of either + or - 90 degrees.
     
  13. Apr 23, 2015 #12
    Thank you both for the replies!
     
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