Calculate Neautral Current in Faulted Situation

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AHarb

Member
Location
Atlanta, GA
Hello,

I was reading through some old posts on how to calculate the neutral current on an unbalanced three phase system. Someone stated that the formula to calculate this is:

I = sqrt((Ia?+Ib?+Ic?)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))

We had a fault on our system the other day and the sequence of events showed the fault currents were Ia = 105 A, Ib = 6760 A, and Ic = 1605 A. The Cooper Form 6 relay in our breaker locked out due to the high current lockout feature. The breaker measured 8425 A on the neutral which is higher than what we show the available fault current is at the location of the fault. I decided to use the formula above to calculate the neutral current using the Ia, Ib, and Ic values. I came up with 6046.20 A which is way lower than what our relay shows.

I was just curious if the formula above is true even in a faulted situation. I'm trying to determine if the relay may have been inaccurate and should not have locked out.

Thanks in advance.
 

kwired

Electron manager
Location
NE Nebraska
This is something I have wondered about myself at times. We generally do fault current calculations using the line to line available fault current, though I have always noticed when looking at any chart listing fault current of any transformers that the line to neutral available current is always higher than line to line. It makes some sense though - less source windings involved would be less impedance in the source
 

roger

Moderator
Staff member
Location
Fl
Occupation
Electrician
The formula does not take other factors such as PF into consideration. Basically the formula only works in the class room for evereything being in unity.

Roger
 

Besoeker

Senior Member
Location
UK
Hello,

I was reading through some old posts on how to calculate the neutral current on an unbalanced three phase system. Someone stated that the formula to calculate this is:

I = sqrt((Ia?+Ib?+Ic?)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))

We had a fault on our system the other day and the sequence of events showed the fault currents were Ia = 105 A, Ib = 6760 A, and Ic = 1605 A. The Cooper Form 6 relay in our breaker locked out due to the high current lockout feature. The breaker measured 8425 A on the neutral which is higher than what we show the available fault current is at the location of the fault. I decided to use the formula above to calculate the neutral current using the Ia, Ib, and Ic values. I came up with 6046.20 A which is way lower than what our relay shows.

I was just curious if the formula above is true even in a faulted situation. I'm trying to determine if the relay may have been inaccurate and should not have locked out.

Thanks in advance.
The formula is valid only if the power factor of all three phases is the same.I agree with your 6046A if all phases at at unity factor but the fault current is likely to be at a quite a low power factor.

Make Ib say, 0.1, and In is then greater than 8000A.
 

Besoeker

Senior Member
Location
UK
The formula does not take other factors such as PF into consideration. Basically the formula only works in the class room for evereything being in unity.

Roger
They don't have to be unity, just all the same PF. That keeps equal phase displacement between the currents.
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
Retired Electrical Engineer - Power Systems
Actually that formula is for 'balanced' circuits, it is not for a L-N fault.

Depending on the configuration of the circuit (primarly the distance from the transformer) it is possible for the single L-G fault current to be higher than that of a 3-phase L-L-L fault.

The mathematical solution involves the use of matrix math and Symmetrical components (Positve, Negative, and Zero sequence currents). this is why 'real' fault studies are not for DIY engineers.
 

mull982

Senior Member
If you know the angles associated with each of the faulted phase currents you listed you could treat each fault current as a vector, add them vectorally and the resultant vector will give you the magnitude of the fault current (or neutral current)

Also as was mentioned above the L-G fault current can be higher than the 3-ph bolted fault current close to a transformer or generator because the zero sequence impedance of the transformer or generator is less than the pos/neg sequence impedance. However as you move further from the transformer the zero sequence impedance increases faster than the pos/neg sequence impedance does and therefore the L-G fault current becomes less than the 3-ph bolted fault current at a location further from the transformer.
 
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Besoeker

Senior Member
Location
UK
Actually that formula is for 'balanced' circuits,
It doesn't actually have to be balanced loads. Just the same PF for each to make that formula work.

it is not for a L-N fault.
I totally agree. In fact I don't see merit in teaching it at all - it's applicable to a very specific set of circumstances that generally won't pertain to real life applications and will mostly give a wrong answer which could be far too low. I've commented on this before. I think teaching students a formula that give wrong answers is a bad idea. Better not to teach them it at all.

this is why 'real' fault studies are not for DIY engineers.
Again, I completely agree.
 

AHarb

Member
Location
Atlanta, GA
AHarb...
Can you provide info about:
1) Duration of fault?
2) Extent of damage to enclosure metal, and electrical apparatus?
3) Voltage?
Phil Corso
1) I don't have exactly what happened because I didn't download the oscillographic data from the relay. This would've given me the exact timing of the events that took place. The SOE shows that the control locked out and then tripped 0.114 seconds later so that should be pretty close to how long the fault was present.
2) The fault was caused by a failed lightening arrestor that did not blow in the clear. There was no damage to any other equipment since the control locked out quickly.
3) The normal voltage is 7200 V (line-neutral). During the fault, the sequence of events shows Va = 7500 V, Vb = 2760 V, and Vc = 7260 V.
 

AHarb

Member
Location
Atlanta, GA
AHarb...
What were pre-fault voltages, currents, and angles (if known) for all 4 conductors?
Regards, Phil
Unfortunately, I do not have that data. The only thing I downloaded from the relay after the event occurred was the sequence of events which only records voltage and current when an event happens. I actually went back out to try getting the oscillographic data to get the PF angles and other information, but it was lost because this data is stored in volatile memory and the relay loses the data when rebooted.

I do have the voltages and currents from the last time that a non-fault event occurred. I guess this is what the system would be like in it's normal state. It was only recorded because the non-reclosing switch was activated. These values are a month old, but they are:

Va = 7500 V, Vb = 7380 V, Vc = 7380 V
Ia = 108 A, Ib = 112 A, Ic = 113 A, and In = 12 A

Sorry I don't have more information.
 

Phil Corso

Senior Member
AHarb...
Obviously the unaffected phase Ia =105A in your OP is load-current! The other two phase values were power-follow current thru their respective arresters. Phase c current resulted from its arrester being enveloped by arcing-current products triggered by the phase b arrester failure.
In conclusion, when arcing-current damage is present the the b & c currents have both resistive and magnetic components. Thus, the equation you "questioned" is invalid for the disturbance your facility experienced!
Regards, Phil Corso
 

Smart $

Esteemed Member
Location
Ohio
It's invalid for pretty much all real life applications.
Pretty much all? So you're saying 99.999% of all real life applications have reactive and unequal power factors, and/or non-linear loads? I'd say a majority of applications, even a very high percentage... but not "pretty much" all :blink:
 
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Besoeker

Senior Member
Location
UK
Pretty much all? So you're saying 99.999% of all real life applications have reactive and unequal power factors, and/or non-linear loads? I'd say a majority of applications, even a very high percentage... but not "pretty much" all :blink:
The formula is for calculating the resulting neutral current in a four wire system allowing for unequal currents in the lines.
However, the formula works only for the very specific case of those load currents having exactly matching phase shift and zero harmonic content.
How likely is that for unequally loaded phases?
 

Smart $

Esteemed Member
Location
Ohio
The formula is for calculating the resulting neutral current in a four wire system allowing for unequal currents in the lines.
However, the formula works only for the very specific case of those load currents having exactly matching phase shift and zero harmonic content.
How likely is that for unequally loaded phases?
Ultimately, I do not know an exact probability. What I do know is that I have worked on industrial product-heating subsystems, which had unity power factor and no harmonics. I have also worked on subsystems which had unbalanced but identical reactive loads. I can say from experience that there is a probable factor. I just can't put an exact number on it.
 
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