Short circuit calculations for overhead service

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spakkala

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Master Electrician
I am wondering how anyone calculates short circuit calculations where there is an overhead service and no meter main. Mainly looking at 240/120 single phase. For the first portion (transformer to weatherhead), you need to take the 1.5 times the short current at the transformer for the line to neutral calculation. From the weatherhead to the meter, do you also do this? The reason I ask, is that say you calculate 10,000 at the weatherhead from 15,000 at the transformer. If you do the same calculation from the weatherhead on down, you now get more than 10,000 because this is usually a short run. So once you hit your first point where the wire type changes, would you not use the 1.5 multiplier anymore? This also applies from the meter to the first panel. Where I live, meter mains or service disconnects next to the meter are not very common. Most of the time, the main panel inside is the "service equipment". Just looking for other people's thoughts on this.
Thanks,
 
spakkala said:
I am wondering how anyone calculates short circuit calculations where there is an overhead service and no meter main. Mainly looking at 240/120 single phase. For the first portion (transformer to weatherhead), you need to take the 1.5 times the short current at the transformer for the line to neutral calculation. From the weatherhead to the meter, do you also do this? The reason I ask, is that say you calculate 10,000 at the weatherhead from 15,000 at the transformer. If you do the same calculation from the weatherhead on down, you now get more than 10,000 because this is usually a short run. So once you hit your first point where the wire type changes, would you not use the 1.5 multiplier anymore? This also applies from the meter to the first panel. Where I live, meter mains or service disconnects next to the meter are not very common. Most of the time, the main panel inside is the "service equipment". Just looking for other people's thoughts on this.
Thanks,
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You use the 1.5 multiplier to get the L-N fault current at the transformer. You do not use that multiplier again. Once you get the L-N fault, use that figure and work your way down to the service equipment.
 
Thanks! So if the power company gives me a maximum available fault current of the transformer of 20,000, I use 20,000 for the line to line and 30,000 for the line to neutral. Again this is only for 240/120 single phase though. All others would use the 20,000 everywhere. Is this correct?
 
spakkala said:
Thanks! So if the power company gives me a maximum available fault current of the transformer of 20,000, I use 20,000 for the line to line and 30,000 for the line to neutral. Again this is only for 240/120 single phase though. All others would use the 20,000 everywhere. Is this correct?
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If the figure is provided by the POCO, I think you would have to ask them if that is L-L or L-N. Most of the time when we deal with looking at fault currents it is three phase stuff where L-L is higher. The construction of most single phase transformers is such that L-N is higher. You would think they would be giving you worst case, or line to neutral but I would double-check for clarification.
 
I'm not too sure about this power company. They give same line to line as line to ground. Then just a max of that roughly 20,000. I'll assume that is line to line, and figure the line to neutral 1.5 times that to be safe. Appreciate your help!
 
Unless asked for; the fault current value is going to be L-L for single phase service and 3-ph fault 3-phase system. Typically, if you don't ask for the L-N they aren't going to offer it.
 
This is from the letter the POCO sent. I assume they meant max instead of min though for the top one. So this must be stating max line to neutral which would be the larger. In this instance, I suppose I would use 12,738 for line to line, and 20,276 for line to neutral?
1599226375771.png
 
spakkala said:
Thanks! So if the power company gives me a maximum available fault current of the transformer of 20,000, I use 20,000 for the line to line and 30,000 for the line to neutral. Again this is only for 240/120 single phase though. All others would use the 20,000 everywhere. Is this correct?
Click to expand...
20K is what is available at the transformer. In you OP you said 10K at the end of the service drop, why would you bump it back up to 20 to calculate the raceway to meter or main? The available current at the starting point of that segment is 10K.

you might have 30K line to neutral, but again at the source. By the time you get to the end of the service drop resistance of the conductors likely leaves L-N available current at that point something less than the L-L available current, exception to this is for very short runs, but if L-L is dropping by 10K I doubt you have short enough run to be concerned about L-N being high at the end of the run.
 
spakkala said:
I was just giving theoretical numbers at first. Not the actual values;)
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Still, once you figured available current (Source being point A) at point B, you don't start over with point A value to calculate point C you start that calculation with what you had at point B.
 
spakkala said:
This is from the letter the POCO sent. I assume they meant max instead of min though for the top one. So this must be stating max line to neutral which would be the larger. In this instance, I suppose I would use 12,738 for line to line, and 20,276 for line to neutral?
View attachment 2553480
Click to expand...

Those numbers are odd. First, not sure what the max L-G current is (assuming they mean max) vs the values with no qualifier ahead of them. Second, Its a bit odd the L-G and L-L are the same.
 
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