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Distribution transformer short-circuit test issue

  1. Nov 21, 2014 #1
    I have a question considering the transformer's short-circuit test and the impedance (or short-circuit voltage).

    The transformer's general data is:
    • Rated power: 50kVA
    • Primary voltage: 10000V (10kV)
    • Secondary voltage: 420V (0,42kV)
    • Primary current: 2.89A
    • Secondary current: 68.73A
    • Vector group: Yzn5
    • No-Load Losses: 190W
    • Load-Losses: 1100W

    The two transformers were tested by all types of routine tests:
    1. Voltage turns-ratio test
    2. Windings resistance test
    3. Open-circuit test
    4. Short-circuit test
    5. Dielectric tests
    All tests were clear to me, but the short-circuit test.

    The winding resistance test, at some temperature (13°C) gave us the low-voltage and high-voltage transformer winding resistance.
    The first transformer had RABC = 41.74Ohm's (RABC = (RAB + RAC + RBC) / 3) on the primary windings, and the Rabc = 54.03miliOhm's (Rabc = (Rab + Rac + Rbc) / 3) on the secondary windings.
    The secondary transformer had RABC = 41.15Ohm's (RABC = (RAB + RAC + RBC) / 3) on the primary windings, and the Rabc = 58.58miliOhm's (Rabc = (Rab + Rac + Rbc) / 3) on the secondary windings.
    We used these values to calculate the Copper-Losses for both transformers, by using the formula PCu = 1.5*(RABC*I1N2 + Rabc*I2N2). The I1N and the I2N are the rated currents on the primary and the secondary side.
    The calculated Copper-Losses for the first transformer were 906W.
    The calculated Copper-Losses for the second transformer were 931W.

    The short-circuit test is done by short-circuiting the low-voltage side of the transformer and applying the test voltage on the high-voltage side, and slowly increasing this voltage, until the rated current (or the short-circuit current) is not measured on the high-voltage side.
    We record then this voltage and call it short-circuit voltage.
    We record this current and call it short-circuit current (this should be the rated secondary current).
    We record the losses and call them the load losses.
    On the first transformer we reached the rated secondary current of 68.73A at 396V, and recorded losses of 1060W.
    On the secondary transformer we reached the rated secondary current of 68.73A at 365V, and recorded losses of 800W.

    Here is the issue !!!
    We calculated the Copper-Losses for the first transformer at 906W and we measured the 1060W.
    The difference from the 906W up to the 1060W are the extra losses of the 154W, which we find OK.
    The calculated impedance (or the short-circuit voltage) for the first transformer is 3,96%.
    We calculated the Copper-Losses for the second transformer at 931W and we measured the 800W.
    The difference from the 931W up to the 800W are the extra losses of the -131W, which we find NOT OK.
    The calculated impedance (or the short-circuit voltage) for the second transformer is 3,65%.

    What does it mean? How is this possible?

    The only technical difference between these two transformers is the cross-section of the low-voltage Copper wire.
    The cross-section of the low-voltage Copper wire for the first transformer is 24mm2.
    The cross-section of the low-voltage Copper wire for the second transformer is 21mm2.
    The cross-section of the high-voltage Copper wire is the same for both transformers 0.78mm2.

    So, can this difference in the low-voltage Copper wire cross-section be responsible for the issue with the calculated and the measured data, for the second transformer?

    Thank You for Your answers !!!
     
  2. jcsd
  3. Nov 22, 2014 #2
    So, these manufacturer's data are exactly identical for both transformers? Is the manufacturer the same one or not(if yes, it's strange that cross-section of low voltage winding wires aren't same for both transformers...)?
     
  4. Nov 22, 2014 #3

    NascentOxygen

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    Staff: Mentor

    Hi darko1978. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    What instrument/s did you use to measure the losses for the s/c test?

    You are trying to account for some differences between measured and calculated losses? I'm no expert on transformers, but some things come to mind, that may or may not be applicable.

    Perhaps different coppers were of different purities? Skin effect is not necessarily insignificant.

    At full load current the waveform will be starting to distort, owing to non-linearities in the core material. Does the accuracy of your measuring instruments depend on current and voltage being highly sinusoidal? --- meaning that the presence of harmonics will cause errors in the reading?
     
    Last edited by a moderator: May 7, 2017
  5. Nov 22, 2014 #4
    We are using the digital multimeter with current transformers 10/5A, in this case.
    Maybe we have a wrong measurement?
     
    Last edited by a moderator: May 7, 2017
  6. Nov 22, 2014 #5
    Yes, the manufacturer is the same.
     
  7. Nov 22, 2014 #6
    Very interesting. Transformers are new, nobody changed nothing at them? I wonder what would be motivation for making practically same transformers, with exactly same load and no-load loses, but with different wire cross sections... Maybe for research purposes? Manufacturing two lines with different winding parameters isn't financially good choice.
     
  8. Nov 23, 2014 #7
    The other test result-short-circuit voltage- is in limits required by IEC 60076-5, 1 [4% minimum, 10% tolerance].
    In my opinion, 800 W losses was measured only on two phases [the third current was not connected-it seems to me 1/3 is missing].
     
  9. Nov 23, 2014 #8

    NascentOxygen

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    Staff: Mentor

    You'd be more comfortable with losses being 1200W, rather than 800W? You can see the benefit of repeating each series of measurements, now! It is best to never have to rely on just one data point.

    Can you compare your measurements with those of other students who separately carried out the same investigation?
     
  10. Nov 26, 2014 #9
    Sorry for the delay. In my opinion the no-load losses could be 150W steel losses and 40W copper losses.
    The calculated copper losses using resistance measurement result [for 50 dgr.C-as I think the transformer did not reach the rated temperature-may be 90oC] = 1070W.So total losses have to be 1070+150=1210W
    If Ic [phase C current measurement=0) then calculated copper losses will be 1210*2/3=806 W.
    According to IEC 60076-1 Table 1-pos.a: tolerance of total losses is +10%.
    That means if 1100*1.1=1210W total losses it could be permissible.
     
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