short circuit calculations

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wirenut1980

wirenut1980

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Plainfield, IN
Hello, I am reviewing electrical calculations as part of a submittal and was wondering if you guys could help me check my math. The submitter identified the available "let-through" short circuit current from a 12,470 V - 480/277 V, 1500 kVA transformer. They are submitting 600 MCM (4 conductors/phase), the length of the run is 50 feet, and they are using a C value of 25,000 per phase. The formula I see in an excel spreadsheet calculator from this site using the point to point method is as follows:

f=(1.732*L*I) / (N*C*E)
M=1 / (1+f)
Available short circuit=I*M

where:
f='f' factor
L=length of conductors
I=available short circuit current from transformer
N=number of conductors per phase
C='C' factor
E=line-neutral, or line to line voltage, which in this case would be 277 V, or 480V

This gives me f=0.0921, M=0.8624, and the available line to line SC current=49,379 Amps. The excel spreadsheet fault current calculator comes up with an answer of f=0.028 for the line to line SC current ?f? factor. My question is, where is this difference coming from between my calculated answer (0.0921) and the one calculated by the spreadsheet (0.028)?

The submitter came up with 51,905 Amps available at the Main Distribution Panel (the next point downstream of the transformer). The formula they used was as follows:

f=(1.73*L*I) / (C*E)

They used E=480 V to get the line to line SC current, C=25,000, and they did take into account the 4 conductors per phase. Their answers were f=0.103189, M=0.906463, and 51,905 Amps available SC current. I understand how they got their answer, but what does the ?C? value represent? I guess all in all it only amounts to a difference of 3,000 or 4,000 short circuit Amps in the end. Sorry for being long-winded. Thanks!
 
Re: short circuit calculations

From the information you have given you have to make some assumptions.
1. %Z of the transformer 5%, 3.5%
2. Infinite bus of finite bus

For an infinite bus
5%Z = 33.4 ka and for 3.5%Z = 46.3 ka

For a finite bus of 5000 amps at 12.4 kv
5%Z = 26.6 ka and for 3.5% = 34.1ka

I guess all in all it only amounts to a difference of 3,000 or 4,000 short circuit Amps in the end.
Click to expand...
3 of 4 ka can make a big difference.
 
Re: short circuit calculations

Unless things have changed recently, I believe tha ANSI standard minimum impedance for 501 KVA and up is 5.75%. Starting with that, you have only 31,379 amps at the transformer, my calculations show about 1600 amps of attenuation in the service giving about 29,785 amps at the service.
I was never very good at unusual calcuations,c factors and f factors, I always use I=E/Z.
Jim T
 
Re: short circuit calculations

Thanks for the replies. :) I did question where the submitter came up with the available fault current at the transformer secondary. It was just a number that was thrown in there without any calculations to show where they got it from. I was unaware that there were minimum impedances for transformers...interesting! And on second thought, being off by 3-4 kA can be very costly, very true.
 
Re: short circuit calculations

Quicky without doing serious number-crunching, with infinite bus:

rated secondary FLA of 1800A
multiplied by (100 divided by 5.75)
equals 31304A available right at xfmr

If liquid-filled xfmr, I believe the minimum Z may be less, like maybe 3.5%.
Also some may say to add 4X or 6X motor FLA's for motor contribution.

It is conceivable that there could be a reason why the submittal SCA is that high.
Why not just ask.
 
Re: short circuit calculations

kiloamp7 is correct, you ought to ask. However as an engineer with utility experience, I often see fault current calculations with unusually high requirements. Not to say there aren't very high fault currents available some places, but that many designers specify a high fault current incorrectly.
My reference to a minimum of 5.75% was to liquid filled transformers based on my past utility experience. It may have changed.
Jim T
 
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