Faults with Transformers

[hello21]

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?

 

[Strathead]

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!

 

[charlie b]

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.

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.

 

[charlie b]

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.

 

[Strathead]

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.

 

[gar]

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.

.

 

[hello21]

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)?

 

[Ingenieur]

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

 

[hello21]

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?

 

[Ingenieur]

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

 

[hello21]

not the accepted way, in 30+ yrs can't remember ever seeing it done

line reactors are used, cheaper and can do better

Right I agree there may be a cheaper way, but this may be just one of the pros used to install it. I'm not saying to do it financially, but just asking about the theory and calculations.

 

[Ingenieur]

Right I agree there may be a cheaper way, but this may be just one of the pros used to install it. I'm not saying to do it financially, but just asking about the theory and calculations.

it isn't a reason to install it, it's a consequence, just like a long wire run
reactors are cheaper and more effective
i limiting fuses
proper protection sizing
many better methods


I would not focus on this
you need basics based on your question as to whether sec faults affect the prim

but yes, ANY Z in the fault path reduces its magnitude

 

[gar]

hello21:

Your post #1 seemed to imply that you did not understand that any secondary current was reflected to the primary.

In post #7 you now introduce the question of whether an isolation transformer, as compared to no transformer, limits fault current. Others answered this for you. At this point I do not know whether you understand that secondary current reflects to the primary.

I question whether you have a good background in electrical circuit theory. You need to study basic electrical circuit theory, and then try to do an analysis by yourself. If you did that I don't think you would have needed to start this thread. Start from basics.

In post #9 you seem to understand that a transformer vs no transformer can reduce fault current. This results from the leakage inductance of the transformer providing series impedance in the circuit vs no transformer. The greater the transformer leakage inductance the greater is the series impedance and thus lower fault current.

The following is a useful discussion on a transformer equivalent circuit, was simply the first one Google brought up.
https://www.electricaleasy.com/2014/04/equivalent-circuit-of-transformer.html

Don't rely on hearsay information. Possibly use it to go study the subject. It may be valid, or totally incorrect.

Just because you can plug values in an equation and get some answer does not mean the result is of any value.

.

 

[hello21]

Sorry for asking.

 

[gar]

180712-0644 EDT

hello21:

You should not be sorry for asking.

What I am seeing is that you are picking up smatterings of information from various sources, quite possibly from persons without a good understanding of basics, and that you have not been taught, or learned on your own, basic electrical circuit theory.

Two hundred years ago little was understood about electricity. Gradually thru the 1800s various experimenters and theoreticians developed concepts on the basic operation of electrical devices, and circuits.

Ohm studied the relationship of voltage, current, and conductivity. The results later became known as Ohm's law. But Ohm did not study power, and therefore the power law is not part of Ohm's law. Joule is the person that studied power and its relation to electricity. Ohm invented a way to produce a continuously variable voltage for his experiments. This was long before we had any electronic or mechanical means to do this.

In developing an efficient DC generator (dynamo) Edison discovered magnetic saturation of an iron core.

Edison while working on the development of his incandescent light bulb discovered electron flow, but did not know it. Circa 1880. See http://ethw.org/Edison_Effect . The electron was not discovered until 1897 by J. J. Thomson. See https://www.nobelprize.org/educational/physics/vacuum/experiment-1.html .

The early study of transformers is discussed at https://en.wikipedia.org/wiki/Transformer .

I don't know how your original post could have been presented in some different way that would have more clearly defined your question. Often times on any problem the major problem is to figure out what are the correct questions to ask.

Don't hesitate to come back and ask more questions.

.

.

 

[Ingenieur]

Sorry for asking.

no need
but do not get upset when people disagree with your assertions, this place has literally 1000's of years of experience...for free
learn from it

research 'current limiting reactors' to see their advantages over an iso xfmr

 

[mbrooke]

Sorry for asking.

Don't be sorry. You are actually right. Transformers are certainly used to limit fault current- and in terms of blocking zero sequence current (delta-wye), they are the best option. Even a transformer with a low impedance will limit the fault current if its kva is small relative to the source supplying it.


FWIW in data centers there has been talk of going to 240/416Y instead of stepping 480 down to 120/208 at various locations, but the issue of increased fault current has made it a challenge.

 

[Ingenieur]

Don't be sorry. You are actually right. Transformers are certainly used to limit fault current- and in terms of blocking zero sequence current (delta-wye), they are the best option. Even a transformer with a low impedance will limit the fault current if its kva is small relative to the source supplying it.


FWIW in data centers there has been talk of going to 240/416Y instead of stepping 480 down to 120/208 at various locations, but the issue of increased fault current has made it a challenge.

transformers are not used to limit fault current
transformers are used to transform voltage/current levels, or for electrical isolation
ANY Z will limit fault current

on a delta:wye it doesn't 'block' 0 seq I, a gf can't exist in an ungrounded delta
but its phase current will rise on the grounded wye leg, although it will remain pos seq balanced (ie, a current will circulate in the delta)

 

[Ingenieur]

starting on 174 http://www.zmuda.ece.ufl.edu/Fall_2013_Power_Systems/6-Symmetrical_Components.pdf
good explanation

page 193 shows yg:d

a ground fault on the y will cause a current to circulate in the delta, in essence, that is the 'ground fault current'

 

[mbrooke]

transformers are not used to limit fault current
transformers are used to transform voltage/current levels, or for electrical isolation
ANY Z will limit fault current

And a transformer is Z- which means it can limit fault current. A transformer is one more thing on the list that can do so.

on a delta:wye it doesn't 'block' 0 seq I, a gf can't exist in an ungrounded delta
but its phase current will rise on the grounded wye leg, although it will remain pos seq balanced (ie, a current will circulate in the delta)

Explain.


A line to ground fault on the secondary will not generate any ground or neutral current on the primary. As such zero sequence can not pass- draw the equivalent circuit out, there is no path.