fault current using utility data vs. infinite bus method

[PeterOven]

I would appreciate if you could point out any errors in the following reasoning.

2500 kVA transformer, 12.47kV primary, 480Y/277V secondary, 5.75% Z.

Using the infinite primary bus method, I get fault current of 52296A at secondary terminals (bolted 3 phase fault).
(2500 kVA / (480V * 1.732)) / 0.0575 = 52296A

MVA method:
Transformer admittance: 2.5 MVA / 0.0575 = 43.47 MVA
43.47 MVA / (480 * 1.732) = 52288 A (due to rounding errors)


The utility engineer supplied the following data for the 12.47 kV power lines coming into the plant:

3-Phase FAULT CURRENT:	1552 A
Line-Line FAULT CURRENT:	1344 A
1-Phase FAULT CURRENT:	1727 A
System X/R:	4.11
Line to Line Voltage (kV):	12.47

When I do the calculation with the 1552 A value, this is what I get, using the MVA method:
Utility fault current MVA: 12.47 kV * 1.552 kA = 19.35 MVA
Transformer admittance: 2.5 MVA / 0.0575 = 43.47 MVA
Series combination: 1 / ( 1/19.35 + 1/43.47 ) = 13.39 MVA fault at transformer secondary
Convert to amps: 13390000 / (480 * 1.732) = 16106 A fault current at transformer secondary

The reason I am asking is that this seems rather low compared to the infinite bus method, and this is my first job doing the calculations, so I just wanted to put it out there to see if it looked reasonable. Do I need to use the X/R somewhere? This is in a small town, so it is not right along a main transmission line.

thank you

 

[kingpb]

You forgot the sqrt 3 for the utility MVA contribution, should be around 35.5MVA

 

[PeterOven]

You forgot the sqrt 3 for the utility MVA contribution, should be around 35.5MVA

Thank you.

so the corrected utility calcuation is:
Utility fault current MVA: 12.47 kV * 1.552 kA * 1.732 = 33.52 MVA
Transformer admittance: 2.5 MVA / 0.0575 = 43.47 MVA
Series combination: 1 / ( 1/33.52 + 1/43.47 ) = 18.93 MVA fault at transformer secondary
Convert to amps: 18930000 / (480 * 1.732) = 22765 A fault current at transformer secondary

 

[Bugman1400]

Converting to per-unit, at 12.47kV, I calc the source impedance (Zs) to be 2.98pu. For the xfmr, at 12.47kV, I calc its impedance (Zt) to be 2.30pu. So, a three phase fault on the secondary terminals of the xfmr should equal Ipu = 1.0/(2.98 + 2.30) = 0.1894pu.

Ibase at 480V is equal to 120,281. So, Ipu X Ibase = 22,780 Amps

 

[kingpb]

12.47kV x 1552A x 3sqrt = 33.52MVA from system (3-phase)

2500KVA/0.0575 = 43.5MVA (xfmr)

MVA Method says series MVA combines like resistors in parallel:

1/((1/33.52MVA)+(1/43.5MVA)) = 18.93MVA

18.93MVA/480/3sqrt = 22.8KA

The difference in infinite bus and MVA method is that you are using actual maximum possible (Infinite) versus MVA method which is a close approximation to actual.

I modeled it in ETAP (5 min effort) and the result is 22.8KA which is the same as MVA method.

The question is: what does the utility plan for (future) on this line? You need to have them tell you what the maximum fault current planned on the line will be.

Also, if that is a utility transformer, the impedance will not be 5.75%. That is an ANSI standard, but utilities typically put in much lower impedance equipment.

 

[Phil Corso]

Peter, here is an example using my Gi.Fi.E.S. method (pronounced Jiffy’s) for solution, where:
Gi represents "Given";
Fi for "Find";
E for "Equation"; and
S for "Solution!" Following is the solution,

Example 1
Given:
A 3-phase transformer having the following parameters; design or rated capacity:
o 112.5kVA; o kVpri = 0.48; o kVsec = 0.208; o Imp Z=1.5%; infinite source.
Find:
Short-circuit current on secondary, Isc.
Equation(s):
Infinite means no impedance between the transformer primary and the source voltage, then the short-circuit current magnitude is the transformer's rated kVA, divided by its impedance in per-unit,
o kVAsc = kVA / (Z/100), and then,
o Isc = kVAsc / (1.732 * kVsec)
Solution:
o kVAsec = 112.5 / 0.015 = 7,500 kVA.
o Isc = 7,500 / (1.732 * 0.208) = 20,820A = 20.8 kA

Example 2
Given:
Parameters above except that SC fault-current level at xfmr-primary is given in Amperes = 9,500 A
Then, primary SC Duty, kVApri = 1.732 * Isc * kVpri ~ 7,900 kVA
Find:
Short-circuit current on secondary, Isc.
Equation(s):
Because only reactance’s are involved, that is, resistances are ignored, I will use a short-cut procedure called the kVA Value Method.
Solution:
Combine the pri and sec fault kVA’s together as if they were two resistances in parallel, that is,
o kVAequiv = (kVApri * kVAsec) / (kVApri + kVAsec).
Solve equations as required to find equivalent SC-Duty on 208V bus:
o kVAequiv = (7,900*7,500) /(7,900+7,500) = 3,850 kVA.
o Isc = 3,850/ (1.732 * 0.208) = 10.7 kA

Please let me know if additional detail is required.
Regards, Phil Corso

 

[kingpb]

The X/R is used for more formal calculations. An X/R for the system of 4.11 is low which would hurt if you plan on putting in a bunch of big reactance hogging loads. More importantly, it is also needed for determining the 480V breaker rating. Refer to IEEE C37.13 for what this means and for the multiplying factor formula; but basically if the X/R is greater than 6.6, then you have to have a MF that increase the fault current to account for asymmetrical component.

This is very often overlooked when sizing LV equipment, and normally wouldn't cause any consternation due to the typical margins used when buying the equipment, however, I have seen it bite people in the butt on occasion.

 

[PeterOven]

The question is: what does the utility plan for (future) on this line? You need to have them tell you what the maximum fault current planned on the line will be.

Also, if that is a utility transformer, the impedance will not be 5.75%. That is an ANSI standard, but utilities typically put in much lower impedance equipment.

The values I listed in my original post are the existing fault current values. The nearby substation will be replaced within the next year, so the utility engineer also gave me the new values based on the new substation:

3-Phase FAULT CURRENT:	1536	Amps
Line-Line FAULT CURRENT:	1330	Amps
1-Phase FAULT CURRENT:	1834	Amps
System X/R:	5.27
Line to Line Voltage (kV):	12.47

Not much of a change, in fact it is less on the three phase and line-line faults. The X/R did change quite a bit though.

I will have to check on the transformer impedance, I did use the value from a standard table distributed by the utility, not from the nameplate.

 

[ron]

Don't forget to add any motor contribution or other sources.

 

[Iron_Ben]

Actually, 5.75% is the right guess and the number that I would have chosen. You'll plug in the nameplate impedance when it's available but it won't be too far from that. Low impedance transformers, say 1.5 to 2%, are the rule in the smaller sizes, perhaps up to 100 kva single phase pole top, something like that. The ISEE/ANSI standard for three phase padmounts 750 kva and above is 5.75%.

 

[PeterOven]

thanks again for everyone's input

 

[templdl]

We know the breakers are manufactured with a gived kaic. Also one could assume that there is an unlimited fault current at the primary of the trainsformer ,and asume a 200% motor contribution. With that information you can determine the available fault current matching that with the appropriate kaic.
If you find that the available fault current is just high enough to cause you to select a give OCPD then it may pay off to "fine tune" your available fault current calculation such as what the real available POCO fault current is and recalculate yourr ka, etc in order to find out is you can step down to lower kaic rated devices which would equate to a reduced cost.
But to go through the agony of doing the calculation to get the available fault current calculsatios and not having anything beyond simply using the unlimited pri fault current adding for 100% contribution and not gain anything is impractical.
Does that make any sense?

 

[wbdvt]

I think all depends on what you need to find. For design work, an infinite bus is used to determine the AIC rating of breakers/equipment.
For an existing installation, the actual utility fault current is used for incident energy (arc flash) calculations and the X/R is also used for equipment duty. Many times a facility is older and started with smaller electrical equipment and then loads increased, causing utility to install larger transformer (increasing fault current), utility system has increased, all of which leads to equipment that may not be adequately rated.

 

[templdl]

I think all depends on what you need to find. For design work, an infinite bus is used to determine the AIC rating of breakers/equipment.
For an existing installation, the actual utility fault current is used for incident energy (arc flash) calculations and the X/R is also used for equipment duty. Many times a facility is older and started with smaller electrical equipment and then loads increased, causing utility to install larger transformer (increasing fault current), utility system has increased, all of which leads to equipment that may not be adequately rated.

Yes, this is very common. The POCOs are upgrading their distributions which results in a greater fasult current capability.
After evaluation the avsilable fault current issue in a effort to see how much trouble the customer is in one easy option is to add an air core current limiting resactor at the SE. While as a sales and applications engineer for a dry type transformer manufacture custom designing and building them became a populsr solution to this problem.
It seems as though people aren't aware iof ACRs or that they are completely forgot about. You must first first find a good manufacture who has an engineer that has experience in ACR design.

 

[PeterOven]

We know the breakers are manufactured with a gived kaic. Also one could assume that there is an unlimited fault current at the primary of the trainsformer ,and asume a 200% motor contribution. With that information you can determine the available fault current matching that with the appropriate kaic.
If you find that the available fault current is just high enough to cause you to select a give OCPD then it may pay off to "fine tune" your available fault current calculation such as what the real available POCO fault current is and recalculate yourr ka, etc in order to find out is you can step down to lower kaic rated devices which would equate to a reduced cost.
But to go through the agony of doing the calculation to get the available fault current calculsatios and not having anything beyond simply using the unlimited pri fault current adding for 100% contribution and not gain anything is impractical.
Does that make any sense?

This was a case where a boiler was being installed and the control panel was rated for 22kA. Without the utility data, the available fault current was too high to allow the panel to be installed without lengthening the feeders significantly. Motor contribution in this plant actually results in a higher fault current at the main panel in the plant than at the transformer secondary. Using the utility data, I calculated the available fault current at the control panel to be 20kA. Success.

 

[petersonra]

I don't see how you can account for any of the myriad of things that might happen down the road, so IMO your best bet is to base your designs on what is there now. Most POCOs have standards for what they supply that you can use. Often they even tell you what the worst case SCC will be.

If something changes down the road, it is up to someone to deal with it when and if the change is made.

This idea that we are going to design in changes we think might happen seems silly to me, given how few of us are able to accurately predict anything beyond what we might have for lunch.