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Transformer available fault current

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
A client of mine has contracted the services of an engineering firm to perform an arc flash study for one of their projects. There is a work sheet the firm has asked that my client fill out, and one of the things they are asking for is the available fault current (AFC) from the three phase utility transformer. They are asking for both phase to ground and phase to phase numbers, and that puzzles me. The calculation I am familiar with for calculating AFC does not differentiate, although it specifies line to line voltage; does that mean that it is phase to phase AFC, and for line to neutral/ground AFC I should use phase to neutral/ground voltage? That would mean that the infinite bus phase to phase AFC would be larger than the phase to neutral/ground AFC by a factor of sqrt3.
 

Melak 22

Member
Location
Michigan
Occupation
Electrical Engineer
For infinite bus/bolted fault calculations, you'd be correct. But a lot of times the study consultants are looking for primary AFC and the utility will perform these calculations on the software where they have their system modeled. In this case the factors you mention would change depending on the positive and zero sequence impedances within the network.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
For infinite bus/bolted fault calculations, you'd be correct. But a lot of times the study consultants are looking for primary AFC and the utility will perform these calculations on the software where they have their system modeled. In this case the factors you mention would change depending on the positive and zero sequence impedances within the network.
Correct that phase to phase AFC = sqrt3 times phase to neutral/ground AFC for three phase utility transformers using the infinite bus model for the primary?
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
Most software packages have the ability to calculate L-G fault currents as well as L-L-L ones, but do not confuse this with single phase vs three phase short circuit analysis. So, the study engineer asks for stuff that may not be necessary if all you want it to perform arc flash calculations using the IEEE1584 formulas. They will usually proceed with the 3-phase value if that is all you can get.

There are a few outlier situations, such as short distances between the transformer and the equipment, where the L-G fault value is higher than the L-L-L value and so needs to be considered when evaluating equipment AIC and SCCR.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
I do not know but am curious. My first thought is that it is not as simple as L-G AFC X 1.732 = L-L-L AFC 🤔
The equation I have been using for calculating transformer available fault current AFC assuming infinite bus on the primary:

AFC in kA = (kVA)(100) / (sqrt3)(V line to line)(%Z)

My question is/was twofold; 1) is this line to line AFC, and 2) will substituting V line to ground/neutral for V line to line give me line to ground/neutral AFC? They would, of course, be different by a factor of sqrt3
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
The equation I have been using for calculating transformer available fault current AFC assuming infinite bus on the primary:

AFC in kA = (kVA)(100) / (sqrt3)(V line to line)(%Z)

My question is/was twofold; 1) is this line to line AFC, and 2) will substituting V line to ground/neutral for V line to line give me line to ground/neutral AFC? They would, of course, be different by a factor of sqrt3
Yes, it is a reasonable formula for 3-phase L-L maximum fault current for equipment evaluation. However it should not be used for arc flash calculations.

No, you cannot simply divide the L-L-L fault currents by sqrt3 to get the L-G fault current. This has somewhat to do with the different impedances due to the number of conductors in the fault circuit.
 
Also, when transformer impedance is calculated, I believe all three phases are bolted. So the AFC calculated using that impedance figure would be a L-L-L fault. Wouldn't a L-L fault be different and a L-G fault be different again?
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
I would expect L-L, L-L-L, and L-N fault current to be similar in the simplest approximation.

If we assume that the only thing limiting current is the transformer impedance then we are really asking 'What is the % impedance of the transformer across any pair of terminals?'

If we have a bank transformer made up of all the same component transformers, then I'd expect the % impedance (and thus the fault current) of the individual transformers and the entire bank to be the same.

If you have multiple coils on a common core then this would change since the whole transformer would feed partial faults. But I was just giving my initial guess :)

Jon
 

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
AFC in kA = (MVA)(100) / [(sqrt3)(kV line to line)(%Z)]
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
The reason they are asking for the L-G available fault current is that SKM won't let you leave that field blank. Never mind what SKM does with it.
I have a phone call scheduled for later today to talk to the engineer who will be doing the arc flash study. I am certain that all will be crystal clear to me after that call. :D
 

David Castor

Senior Member
Location
Washington, USA
Occupation
Electrical Engineer
If calculating AF per IEEE 1584, the L-G fault current isn't a factor since the short circuit calculation is based on three-phase faults. But I always ask for it from the utility when doing a study. They can provide it and it could be useful for other purposes.
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
If calculating AF per IEEE 1584, the L-G fault current isn't a factor since the short circuit calculation is based on three-phase faults. But I always ask for it from the utility when doing a study. They can provide it and it could be useful for other purposes.
I think SKM uses the SLG value for the device evaluation.
 

David Castor

Senior Member
Location
Washington, USA
Occupation
Electrical Engineer
think SKM uses the SLG value for the device evaluation.
Not sure what you mean by device evaluation, but the SLG fault current doesn't have any impact on the calculated incident energy or arc-flash boundary when using the IEEE-1584 equations.

I think SKM requires some value for the SLG current for sources to do any SC calculation, but it won't change the incident energy.
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
Not sure what you mean by device evaluation, but the SLG fault current doesn't have any impact on the calculated incident energy or arc-flash boundary when using the IEEE-1584 equations.

I think SKM requires some value for the SLG current for sources to do any SC calculation, but it won't change the incident energy.
Device evaluation is when SKM compares the protective device AIC ratings against the calculated short circuit currents. While not part of the IEEE incident energy calculation it is often used as part of NFPA70E compliance.
 

steez

Member
Location
TX
Occupation
Electrical PE
There are a couple issues using infinite bus.

1. You may cause equipment to fail the short circuit analysis that otherwise would not. The SC values are artificially high in infinite bus assumptions.
2. Its not always a worst case for arc flash. Arc flash energy ***can*** sometimes go down with higher fault current as this causes devices to operate faster. A good example with this is when you evaluate emergency systems. Sometimes the generator power scenario is the worst case because its a weaker source than the utility, produces less fault current, and therefore causes devices to operate slower.

Request the data from the utility, they will either give you the source strength at the primary voltage of the XFMR or secondary. If the latter your model would just start with a utility source with no utility XFMR. (this will is not ideal as sometimes including the utility XFMR helps. The primary fusing can lower the AF energy on the secondary.)

If you cant get utility run an infinite bus along with "weaker" source scenarios as a sensitivity analysis. Run the AF analysis as a worst case of those.
 

wwhitney

Senior Member
Location
Berkeley, CA
Occupation
Retired
Arc flash energy ***can*** sometimes go down with higher fault current as this causes devices to operate faster.
Can you briefly mention the math behind that? A statement like (made up, probably wrong) "in some regions of possible fault current, the opening time will vary as 1/I^2, while the power delivered during that time will vary as I, so the energy released varies as 1/I."

Thanks,
Wayne
 
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