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Help with voltage losses in cable question

  1. Jun 14, 2014 #1
    the 400kV circuit is then stepped down onto a 132kV circuit (Ratio 3:1) and is used to feed a 50km, 2 wire (Ostrich) cable circuit, calculate voltage loss in the cable run.

    and this a table that went with the question http://gyazo.com/a47c72d45a149cffa9d781ace8a2008d

    and the first question was as follows : If the output from the generator is 500 MVA and it is feeding onto a 200km long, 4 wire (Squab) cable circuit via a step up transformer (Ratio 1:16) calculate the voltage loss in the cable run. I have calculated this question and got a voltage drop value of 441.4 volts.


    Just wondering if anyone could give me a hand on where to start with the second question?

    Thanks.
     
  2. jcsd
  3. Jun 14, 2014 #2

    mfb

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    Staff: Mentor

    Edit: Ah, I think now I understand it.
    Would have been easier with the original questions, in the right order.

    Question 2 is the same as question 1, just with different numbers. Where do you run into problems?
     
  4. Jun 15, 2014 #3

    This is my answer, would you say that seems right to you?

    Cable length = 50 km

    Cable resistance = 0.227 ohms/km (single ostrich)

    Ostrich resistance = 50 x 0.227 = 11.35ohms

    2 cable parallel resistance = 11.35 || 11.35 = 5.68ohms

    Max current per ostrich = 492A


    I = 500x106/400x103 = 1250 Amps

    Step up transformer ratio = 3:1
    Line voltage = 400x103 / 3 = 132kV (O/P on transformer)
    Line current = 1250 × 3 = 3750A
    Voltage drop per run = I×R = 3750 ×11.35 = 43kV


    thanks
     
  5. Jun 15, 2014 #4

    mfb

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    Formatting looks broken in the calculation of I. And you can use the final voltage there directly. Units are missing. Apart from that, it looks right, and the loss is significant...
     
  6. Jun 15, 2014 #5

    Ah yeah the third numbers are powers. Ok thanks
     
  7. Jun 15, 2014 #6
    The simplified lumped circuit of a transmission line will be:
    In a symmetrical and steady state load, neglecting Xc [nevertheless Xc could be elevated] the voltage drop [approximate] will be DV=sqrt(3)*I*[R*cos(fi)+X*sin(fi)].
    X = k log (GMD / GMR)
    GMD=(DAB*DBC*DCA)^(1/3) where: DAB[for instance]=distance between phase A and B [typical].
    GMR [according to manufacturer catalogue]
     

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