What Ungrounded System's Ground Fault Current t?

[Jagger71]

I am trying to calculate the ground fault current (do to the system capacitance) of a single phase to ground fault through a finite resistance ground fault connection on an ungrounded delta system.

Does anyone know the method? I believe I have the formula's for a simple solid ground fault but not if the ground fault has a finite resistance (say 600 ohms).

Thanks for any help.

 

[RogO]

In the case of 1 phase going to "ground" in an ungrounded system current would not flow, you would not trip a breaker or blow a fuse.
If 1 phase gets unintentionally grounded what you would have then in effect is a grounded system. The only time that you would have real current flow is when another phase connects to ground then you would have a phase to phase fault minus the resistance of the conductors and resistance between the fault points.

I hope this is what you had in mind.

 

[Jagger71]

There is a small flow of current depending on the the system capacitance and voltage (sometimes refered to as the system charging current). (For 13 kV the charging current can be in the 5A range)

For a solid ground fault the current is in the order of twice the charging current but I am unsure how to calculate when the ground fault has a finite resistance.

 

[RogO]

I was on the wrong track maybe. Are you formulating the ground fault settings on your relays or are you making short circuits calulations?

 

[mpparent]

My guess is:

1/(z1+z2+z0+3z), where:

z1 - pos. sequence impedance
z2 - neg. " "
z0 - zero " "
z - finite resistance


I'm guessing it would be the same as a grounded single line to ground fault through an impedance since you are technically "grounded" through your resistance. Remember though, this is a hunch, and hopefully some of the experts on the forum will chime in.

Mike

 

[DanDel]

As mentioned already, there will be virtually no ground fault current flow through a 600 ohm resistance (Ohm's Law).
To calculate the capacitive current flow in an AC circuit, you must first determine the total impedance, based on the resistance and capacitance of the circuit. Finding the capacitance will be difficult if not impossible, and in the end you will find that, as expected, the current, when compared to a true 'fault', is virtually negligible.

Using Ohm's Law, the equation is: I = E/(R - jXc)
Where E is the voltage to ground, R is the resistance, and Xc is the capacitive reactance (j is the square root of -1). To get Xc, use Xc = 1/(2*pi*f*C), where f is the system frequency, and C is the capacitance in Coulombs.

 

[hbendillo]

You should be able to look up capacitance values for the different elements of your system by using tables in the NEC or other sources. I.E. you get capacitance per linear foot for a given conductor type. From that do a short ciruit calculation and thus your results.

 

[Jagger71]

OK, thanks all for your responses. I believe the correct way to analyse this situation is as an unbalanced three phase star connected load (without ground connection). The two balanced phases have an impedance due to the system's capactiance to ground. The third phase has an impedance determined by the fault resistance (in parralel with the system capacitance).

The resulting current is 3V/(2Zf+3Zs+Zc)

where Zf is the fault impedance, Zs is the balanced systems positive sequence impedance, and Zc is the impedance due to the system capacitance to ground.

Note for balanced load Zf=Zc=Zl (Zl=Z load) and I=V/(Zs+Zl)
and for Zf=Zc=0 I=V/Zs which is just the three phase short circuit as expected.


Any comments?

 

[DanDel]

I don't think you will find your answer with sequence impedance modeling.
Remember, were talking very small current levels when compared to real fault currents.
Your Zf contains R and +jX, and your Zc contains -jX, but you can't combine them correctly with sequence impedances.
If you're looking for steady-state current for a single-phase AC circuit with R and jX components, you should use the equations from my previous post.
If you are looking for anything different, you will probably need to use time-domain analysis or transforms.
Why exactly do you need this value?

 

[davidbeach]

DanDel,

I think I need to take issue with your statement "I don't think you will find your answer with sequence impedance modeling." Meaningful fault analysis of polyphase AC power systems is best acheived through symmetrical component analysis. Your equation "I = E/(R - jXc)" can be used to find the 3 phase fault currents but is not particularly helpful in finding ground fault currents without first pulling the system apart and finding the appropriate voltages and impedances. Remember the fault will cause a zero- and negative-sequence voltage rise while there will be a positive-sequence voltage drop at the fault.

It may well be difficult to arrive at the sequence networks, but if an accurate representation can be derived, the fault calculations will be relatively easy. System charging capacitance is one of the most difficult values to derive with sufficient accuracy to be useful in a calculation of this type and minor variations will have a significant impact on the results.

Unless the system in question has considerable capacitance or the question deals with extremely sensitive fault current measurements, the answer to the original question is that the fault current is zero [±] some small rounding error because the data is not good for more precission than that.

 

[DanDel]

davidbeach, the equation "I = E/(R - jXc)" is certainly not the easiest method to find the 3 phase fault currents, but it was given because Jagger71 asked for "ground fault current (do to the system capacitance) of a single phase to ground fault through a finite resistance ground fault connection".
I think everybody in this thread, including yourself, agrees that the current he is looking for is really not considered 'fault current', since it is quite possibly in the order of milliamps.
This is why I am curious why this value is needed by him.
I am equally curious why you suggest that using 3-phase sequence calculations for a single-phase fault makes more sense that single-phase analysis.
And I don't believe single-phase current flow of this magnitude will "cause a zero- and negative-sequence voltage rise while there will be a positive-sequence voltage drop at the fault"

 

[davidbeach]

DanDel,

Single phase analysis of three phase power systems only works when the three phases are balanced, which is not the case during a single phase to ground condition. Once you introduce imbalances, the calculations are much easier if one uses symmetrical components. It is the symmetrical commponents that will eventually allow a value of 'Z' to be derived that can be plugged into you formula.

As for the voltage rise

J. Lewis Blackburn, Protective Relaying, 2nd Ed. pg. 110 ...postitive-sequence voltage is always maximum at the generators and minimum at the fault. The negative- and zero-sequence voltages are always maximum at the fault and minimum at the generator or grounded neutral.

While in the case presented there will be little current flowing to the 'fault' there will still be the voltage drop/rise happening. As the phase voltage of the faulted phase is pulled toward ground, the positive-sequence voltage goes down while the other two go up. The sum of the three sequence voltages will equal the prefault voltage.

 

[DanDel]

davidbeach, was not the initial question "I am trying to calculate the ground fault current (do to the system capacitance) of a single phase to ground fault through a finite resistance ground fault connection on an ungrounded delta system."?
If the initial question were "How do I use sequence impedances to calculate the ground fault current (do to the system capacitance) of a single phase to ground fault through a finite resistance ground fault connection on an ungrounded delta system.", then maybe I could see your insistence on using them.
Sequence impedances and symmetrical components are very useful in calculating what happens in a three-phase system during three-phase and single-phase faults. The issue here is that this is not a fault, but a circuit analysis question(whether you or Jagger71 realize it). How close do you think the phase voltage will be pulled toward ground in this case where maybe 1 or 2 amps of zero-sequence current will flow (if that)?
I still would like to know why this calculation is needed.

 

[davidbeach]

DanDel,

In the case of an ungrounded system, with one phase to ground fault, the voltage of the faulted phase will be pulled all the way down to ground and the other two phases will have a phase to ground voltage equal to their phase to phase voltages. Voltage shift does not require current flow. In the example just cited, the three sequence voltages will all be 1/3 the prefault voltage magnitude where as the prefault sequence voltages will have been 1 per unit positive sequence and zero per unit for the other two.

If you can accurately determine the correct impedance to plug into your formula without the use of symmetrical components analysis, you are far better at circuit analysis than many of the great authors of the field (of whom I dare not compare myself). I have yet to see a circuit analysis text tackle a single line to ground fault without symmetrical components. A fully developed analysis that does not use symmetrical components but arrives at the same fault current value as a symmetrical component analysis would be most welcome and could be your claim to fame.

In the problem at hand, even modified sufficiently that a meaningful result can be achieved, what value do you propose for your R-jX_C? Do you use prefault or post fault X_C values (voltage changes, therefore capacitive reactance changes)? Where did the +jX_L disappear to? If it were a single phase system it would make the calculations much easier, but circuit analysis of three phase systems under unbalanced conditions does not lend itself to such simplifications.

Now, I've said what I have to say, I'll let you have the last word, and that, hopefully, will be the end of this, since it doesn't look like either of us will convince the other.

 

[DanDel]

davidbeach, thank you for your generous offer of the last word, but I think if you read my other responses to the initial question carefully, you will find the answers to the questions you are asking in your last post.

 

[whinton]

I have worked with ungrounded power systems for over 30 years and have only one positive way to measure ground current and I use this EVERY time we go looking for the cause of our grounding problem. We start at the substation and measure the current in the ground wires in our secondary cables fed from the substation circuit breakers. The ground wires are carrying current that is not returning in the phase conductors. We use a Fluke 123 scope-meter as it has a very low range but uses a large current clamp. Our overvoltage problems stopped when we installed Phaseback. The link is:

http://www.applied-energy.us

I hope this helps
Bill