Fault current

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jim dungar said:
When was the last time a contractor got a price from only one sales person?
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Sorry, but I do it a lot, especially with Square D products, only have one distributor in the region, so one price is all I am typically getting and I seldom even am interested in other brands on that kind of equipment unless matching up to existing equipment. Not saying others don't have quality equipment, that is just the way it works for me, what is easiest for me to get, and comes from the supply house I get along with the best.
 
I understand why you used the dc resistance, and 1 wire length, to be consersitive, but so I know the facts, I really should be using the ac resistance, and 2 wire lengths if not using a FC calculator?
I also have seen your posts on including the motors in the calc's. thanks

I'm almost afraid to ask, What does or doesn't change when it is 277/480 single phase, and a vfd to run a 3 phase motor?
 
dave121 said:
I understand why you used the dc resistance, and 1 wire length, to be conservative, but so I know the facts, I really should be using the ac resistance, and 2 wire lengths if not using a FC calculator?
I also have seen your posts on including the motors in the calc's. thanks

I'm almost afraid to ask, What does or doesn't change when it is 277/480 single phase, and a vfd to run a 3 phase motor?
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Some of the "conservative" measures taken simply give you a value on the safe side. If you want to eliminate those "conservative" measures your result will be a higher value but will be more accurate. There are other things that can increase the conductor impedance like running them in a ferrous raceway - creates a very basic inductor but does change impedance a little. Available current from the primary side of the utility transformer is not always something that easy to determine either and often times infinite ability of the supply is assumed as it will yield the most conservative value.

Higher voltage just changes the Ohm's law calculations. A specific resistance with a higher applied voltage will have more current through it than at a lower applied voltage. But, a 100 kVA transformer with a voltage rating of 208 will have different secondary winding size and length than a 100 kVA rated at 480 volts, so your transformer impedance and available fault current at the secondary terminals will be different for each of those, and the output conductors for same kVA will be different size because normal operating current to deliver 100KVA will be a little over doubled for the lower voltage system, so you really have a whole different situation for either system and must start from the beginning with the calculations.

I am not certain if motor contribution is passed through VFD's or not. But logic tells me that the energy it would contribute to a fault can not be passed in reverse direction through the rectifier of the drive, so my best guess is it will not contribute to fault current at a line side of the drive location.
 
kwired said:
Some of the "conservative" measures taken simply give you a value on the safe side. If you want to eliminate those "conservative" measures your result will be a higher value but will be more accurate.
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If this comment is related to the calculation that I posted, then elimination of the conservative measures would have resulted in a lower value. I used the DC resistance, knowing that it was lower than the AC impedance. That caused the calculation result to be higher than it actually would have been in the field. But that was OK by me, because I only wanted to show that the fault current at the equipment was lower than 10kA. Once I achieved that goal, it was no necessary to take additional steps to obtain an even lower calculated value. That is the nature of using conservatisms in calulations. Once you have a result you like, you can quit working.

On the other hand, if I needed to include a 10% margin, and therefore needed the calculated result to be below 9kA, I would have used the AC impedance value. That may or may not have been enough, but it would have been a reasonable engineering approach to the problem.
 
charlie b said:
If this comment is related to the calculation that I posted, then elimination of the conservative measures would have resulted in a lower value. I used the DC resistance, knowing that it was lower than the AC impedance. That caused the calculation result to be higher than it actually would have been in the field. But that was OK by me, because I only wanted to show that the fault current at the equipment was lower than 10kA. Once I achieved that goal, it was no necessary to take additional steps to obtain an even lower calculated value. That is the nature of using conservatisms in calulations. Once you have a result you like, you can quit working.

On the other hand, if I needed to include a 10% margin, and therefore needed the calculated result to be below 9kA, I would have used the AC impedance value. That may or may not have been enough, but it would have been a reasonable engineering approach to the problem.
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Kind of what I was trying to say, I think. If you use a conservative means that will have an error on the high side then the goals in most circumstances will be met, and the fault current rating of the equipment will not be exceeded, even though you don't have a calculation that is correct.

Kind of like ordering conductors, you want to allow for more than you need because not having enough is generally going to be a problem, but you may still want to stay within a reasonable margin or cost goes up significantly.
 
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