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Earth fault in delta star (grounded) transformer

  1. Dec 9, 2016 #1
    Hi,

    I am studying transmission lines faults. There was a line that said
    In case of delta star transformer in transmission line, an earth fault on star (grounded) side is seen as a line to line fault on delta side.
    There was no explanation. I tried to google solutions but couldn't find any explanation.
    Here's the diagram
    upload_2016-12-9_12-2-59.png
    The cross is the L-G fault. Why is it seen as a line to line fault on delta side?

    Thanks in advance...
     
  2. jcsd
  3. Dec 9, 2016 #2

    jim hardy

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    Think simple.
    This is one of those rare occasions where an actual physical picture explains it better than a schematic drawing.
    Look at a real transformer.
    3phcore.jpg

    Surely you see from your schematic that the L-N fault shorts out one secondary winding .
    Think in small steps now.

    Which primary winding shares the same leg of transformer's iron core as the secondary winding feeding the L-N fault?
    Does the shorted secondary winding depress flux in that leg?
    Does that depressed flux suppress counter-emf in the primary winding on that leg?
    Is that leg's primary winding connected phase to phase?
    Does that absence of counter-emf result in high primary current in that leg, phase to phase?

    hope that helps.
     
    Last edited: Dec 9, 2016
  4. Dec 9, 2016 #3
    Thanks a lot for the reply and also the stepped response. Now I can discuss the points.
    upload_2016-12-9_23-4-44.png
    Which primary winding shares the same leg of transformer's iron core as the secondary winding feeding the L-N fault? Suppose fault occurs on Y-leg on star side.
    Does the shorted secondary winding depress flux in that leg? So Iy will be maximum and Ir and Ib will be zero. That is entire current would go through y phase. Right?
    Does that depressed flux suppress counter-emf in the primary winding on that leg? Excess Iy on secondary side causes excess Iy on primary side.
    Is that leg's primary winding connected phase to phase? And this Y winding is connected phase to phase.
    Does that absence of counter-emf result in high primary current in that leg, phase to phase? So Iy will be high in Y phase of primary winding.

    But I'm not sure what will effect be on R and B phase of secondary side. Would they go to zero?
     

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  5. Dec 9, 2016 #4

    jim hardy

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    Ahhh R and B are colors ! Had me confused.

    Well, if there's zero volts between Y1 and Y2 on primary side then R and B on secondary would try to become equal and opposite, wouldn't they ?
    Does your textbook show phasor triangle distorted under a fault?
     
    Last edited: Dec 9, 2016
  6. Dec 11, 2016 #5
    The text book doesn't have the phasor triangle or any description. Just the statement I put in the starting thread.

    I didn't understand this:
    Well, if there's zero volts between Y1 and Y2 on primary side then R and B on secondary would try to become equal and opposite, wouldn't they ?
    Why is there zero volts between Y1 and Y2 on primary? The secondary current at y1 and y2 on secondary is fault so it rises very high. So is it like a transformer on short circuit test? But there would be some voltage because secondary current is very high, so secondary flux is high so e = d(flux)/dt is very high, so there must be some voltage, right?
     
  7. Dec 11, 2016 #6

    jim hardy

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    that's right.

    Just enough voltage to overcome leakage reactance, winding resistance, and wires between transformer and fault.

    that's not so. Current makes mmf , and in a transformer secondary and primary mmf's cancel. Flux in a short circuited transformer is low. Remember how a current transformer works ?
     
  8. Dec 12, 2016 #7
    Okay so i understood why voltage at y1, y2 is zero. Current transformer has high number of turns at secondary and CT is always loaded else you have a very high voltage at secondary.

    Reason for Voltage R and B to be equal and opposite: V(y1 - y2) is same as V(r2 - b1) so Vr2 = Vb1. Thus Vr1r2 = -(Vb1b2). i guess...
     
  9. Jan 11, 2017 #8
    I think it makes more sense if you look at the original diagram that you posted with a delta wye transformer. There are 3 primary windings on the primary and 3 on the secondary. You'll notice that each secondary winding is drawn parallel to one of the primary windings. These parallel lines indicate that this pair of windings is on the same leg of the transformer.

    Take the vertical lines for example. On the secondary, this corresponds to a phase to neutral voltage. Its corresponding primary winding is connected from phase to phase. So, if I fault the secondary winding (ground fault on the star side), the primary winding will be supplying phase to phase fault current.

    Does this help?
     
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