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Thread: Faults with Transformers

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    Faults with Transformers

    Hi,

    My understanding of a typical delta-wye step-down transformer is that the primary and secondary windings are electrically isolated, and you are inducing a current in the secondary windings using an alternating magnetic field. But what happens when there is a fault in the secondary?

    A line to ground fault on the secondary side of the transformer, the fault current should make its way on the equipment ground back to the XO of the transformer then go on the phase conductor and hopefully trip out a breaker (usually the closest breaker to the fault if appropriately coordinated).

    I expect line to line faults to behave similarly, except fault is propagating on phase conductors only.

    I assumed that the fault current would not be induced in the primary, and so regardless of the fault type and magnitude, the primary side of the transformer would not see a fault. Am I way off here?

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    Quote Originally Posted by hello21 View Post
    Hi,

    My understanding of a typical delta-wye step-down transformer is that the primary and secondary windings are electrically isolated, and you are inducing a current in the secondary windings using an alternating magnetic field. But what happens when there is a fault in the secondary?

    A line to ground fault on the secondary side of the transformer, the fault current should make its way on the equipment ground back to the XO of the transformer then go on the phase conductor and hopefully trip out a breaker (usually the closest breaker to the fault if appropriately coordinated).

    I expect line to line faults to behave similarly, except fault is propagating on phase conductors only.

    I assumed that the fault current would not be induced in the primary, and so regardless of the fault type and magnitude, the primary side of the transformer would not see a fault. Am I way off here?
    The way you are looking at it is not bad. The actual flow of electricity is hard to envision in my brain at least. When teaching my second year apprentices I teach them to think of themselves as a single electron that can split and go multiple ways only when required to keep it simple. As such the electron starts at the source and ends at the source.

    In your secondary scenario, the electron starts at the transformer L1 say, it has to get back to XO, L2 or L3 in order for current to flow. (then it returns to L1 across the transformer winding but not important for this) So during normal operation the electron travels from L1 through the breaker, through the load, back through the neutral at the panel and back to XO. When you have a ground fault prior to the load, the electron gets diverted along the ground wire taking the shortest path to the place where the neutral and the ground bond together for that source. Remember its only goal is to get back to L1 the shortest easiest (least resistance) way possible. If it can't get back, no current flow. If the path is too easy (like a short or ground fault) the breaker trips.

    So on the primary side, the ground fault scenario you present the electron would never reach the transformer because the easy path would be back along the ground.

    Hope this helps.

    [edit] Oh, so no you aren't far off. Very good!


    I know what I don't know, and I know where to go to find it!

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    Quote Originally Posted by hello21 View Post
    A line to ground fault on the secondary side of the transformer, the fault current should make its way on the equipment ground back to the XO of the transformer then go on the phase conductor and hopefully trip out a breaker (usually the closest breaker to the fault if appropriately coordinated).
    If current could make its way from the transformer XO to a phase conductor, that itself would constitute a fault. That is not the way this event would be terminated.
    Quote Originally Posted by hello21 View Post
    I assumed that the fault current would not be induced in the primary, and so regardless of the fault type and magnitude, the primary side of the transformer would not see a fault. Am I way off here?
    Sorry, yes. The current on the primary side of a transformer is equal to the current on the secondary side multiplied by the ratio of secondary voltage to primary voltage. For example, on a 480 – 120/208 step down transformer, if the secondary current (all three phases – let’s talk about balanced loading for now) is 100 amps, then the primary current will be 100 x (208/480), or 43.3 amps. During a (again, balanced three phase) secondary fault that has, for example, a fault current of 10,000 amps, the current on the primary side would be 4,330 amps. That should trip the primary side OCPD.

    For a line-to-ground or line-to-line secondary fault, it would be harder to describe the math. But the notion of the primary current becoming elevated would remain.


    Charles E. Beck, P.E., Seattle
    Comments based on 2017 NEC unless otherwise noted.

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    I should clarify that Strathead's reply and mine do not contradict each other. Rather, they are describing different fault situations.

    Downstream of the transformer there has to be an overcurrent device. Most often I see that as the main breaker of the first panel. That panel will have breakers serving various loads. Strathead's scenario has to do with a fault downstream of a branch circuit breaker. My scenario has to do with a fault upstream of the main breaker. But in both cases, there will be an increase in the current seen in the primary side, and that could be enough to trip the feeder breaker to the transformer. Or it might not.
    Charles E. Beck, P.E., Seattle
    Comments based on 2017 NEC unless otherwise noted.

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    Quote Originally Posted by charlie b View Post
    I should clarify that Strathead's reply and mine do not contradict each other. Rather, they are describing different fault situations.

    Downstream of the transformer there has to be an overcurrent device. Most often I see that as the main breaker of the first panel. That panel will have breakers serving various loads. Strathead's scenario has to do with a fault downstream of a branch circuit breaker. My scenario has to do with a fault upstream of the main breaker. But in both cases, there will be an increase in the current seen in the primary side, and that could be enough to trip the feeder breaker to the transformer. Or it might not.
    I assume you are learning. As such we (the collective Mike Holt family) can really help you and confuse you. My answer was intended to get you started. If you draw out any scene and use the "one little electron" game, it should simplify it for you. it gets dicey using the electron when you have a transformer. Whatever helps you to understand is the best way to deal with it. Perhaps think of it as a relay, where the electron on one side passes the baton off to the electron on the other side. But remember that more work by one means more work by the other. Charlie's answer is a more direct answer which I hope you can use the little electron to understand.


    I know what I don't know, and I know where to go to find it!

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    180711-1632 EDT

    I am confused by the original post and some of the responses.

    I would suggest that the original post be simplified to a single single phase transformer. A true 3 phase transformer or 3 single phase transformers to perform the function of a 3 phase transformer just makes it more difficult to understand what happens with some sort of overload on the secondary of a transformer.

    The last paragraph of hello21's post #1 says to me that a fault current on the secondary side of a transformer does not reflect to the primary side.
    I assumed that the fault current would not be induced in the primary, and so regardless of the fault type and magnitude, the primary side of the transformer would not see a fault. Am I way off here?
    Depending upon what was really meant by this paragraph the answer is --- you are way off.

    Consider a tightly coupled transformer, a transformer with a high permeability core coupling primary and secondary (a typical iron core transformer). To a good approximation the current on either side of the transformer relative to the current on the other side is the turns ratio of the primary and secondary windings.

    This ratio is such that the lower voltage side has the higher current. A transformer with a 120 V primary and a 20 V secondary has a turns ratio of 6 to 1. Put any kind of load on the secondary that causes 6 A of secondary load, then the current on the primary side becomes 1 A. Change the secondary load so that secondary current is 12 A, then primary current becomes 2 A.

    A short of any kind or overload on the secondary that produces a change in secondary current will have that change reflected to the primary by the turns ratio.

    How, where, and why a secondary fault occurs has no bearing on what happens on the primary unless it produces a change in secondary current. But as soon as something causes secondary current to change, then primary current changes.

    .

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    So may I ask, other than reducing the available fault current of the system, how does an isolation transformer help with faults? I have read that systems have isolation transformers are installed to help with faults, but I'm not exactly sure how. If you have a 480V, 3P system, and you install an isolation transformer that is 480V Delta/480-277V Wye (or Delta to Delta) how will this help with 3 phase faults (line to line and line to ground)?
    Last edited by hello21; 07-11-18 at 10:03 PM.

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    first
    primary principle of a xfmr
    power in = power out (losses are small and can usually be ignored, especially for this discussion)

    a line to line sec flt draws a large current
    voltage collapses but not alot, it depends on xfmr Z
    if the xfmr drops to 400 v, you have 400 v drop along the conductors
    and huge current and power, they get hot, ie, dissapte power
    if the nml xfmr rating is 480 and 100 A
    and the fault is 400 and 2000
    you see the power has increased 16 times
    it has to come from somewhere...the source via xfmr prim
    so yes, sec faults are reflected to the prim

    an iso xfmr is used to electrically (no conductors, but an air gap) isolate the load noise from the prim (say vfd, mri or something) or prim noise from the sec (sensitive electronics, etc)
    although its Z will somewhat attenuate fault current that is not the primary purpose
    line reactors are used for that

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    Quote Originally Posted by Ingenieur View Post
    first
    primary principle of a xfmr
    power in = power out (losses are small and can usually be ignored, especially for this discussion)

    a line to line sec flt draws a large current
    voltage collapses but not alot, it depends on xfmr Z
    if the xfmr drops to 400 v, you have 400 v drop along the conductors
    and huge current and power, they get hot, ie, dissapte power
    if the nml xfmr rating is 480 and 100 A
    and the fault is 400 and 2000
    you see the power has increased 16 times
    it has to come from somewhere...the source via xfmr prim
    so yes, sec faults are reflected to the prim

    an iso xfmr is used to electrically (no conductors, but an air gap) isolate the load noise from the prim (say vfd, mri or something) or prim noise from the sec (sensitive electronics, etc)
    although its Z will somewhat attenuate fault current that is not the primary purpose
    line reactors are used for that
    But I thought the fault current is limited by the transformer by doing an inifinite bus calc. So if you have a 480V to 480V, 3P, 300 kva transformer with 3% impedance you can only have around 12ka. So if the available fault current was much higher without the transformer then this would be one way to reduce that right?

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    Quote Originally Posted by hello21 View Post
    But I thought the fault current is limited by the transformer by doing an inifinite bus calc. So if you have a 480V to 480V, 3P, 300 kva transformer with 3% impedance you can only have around 12ka. So if the available fault current was much higher without the transformer then this would be one way to reduce that right?
    not the accepted way, in 30+ yrs can't remember ever seeing it done

    line reactors are used, cheaper and can do better

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