Normally we perform arc flash studies and provide labels for complete electrical systems. In this case, we have an existing building where solar has been added and is interconnected to the building's existing AC panel. I've been asked by the customer to perform an arc flash study for just the solar system, so labels can be applied appropriately to combiner boxes, inverters, panels, etc. However, I told the customer that we cannot do it for a partial system. To perform the study accurately, we need to have the drawings for the whole electrical system. Our customer, who is the installer, says they are only obligated for the solar system and the building is old with no drawings of the existing electrical system. I cannot think of a work-around for this situation to provide accurate arc flash labels for just the solar system. Is there another approach that is allowed by code/OSHA?
Thanks for your help.
You're missing the essential keys such as service type and size and the request is kind of ambiguous.
I will cover it general sense in most conservative way to get the worst possible value.
The source starts at the transformer.
So, if you call the PoCo's contractor service dept and call in with the meter or account #, they should be able to get back to you. Clarify if their value is at the transformer or at service drop including feeder impedance which gets you the "up to" value.
Those transformers are way too big to bench test using real life conditions, so they test at reduced voltage and calculate it.
If the transformer is 4,000v to 200v, 1000A 200kVA, the transformer is setup with a crowbar across the output terminals with a clamp meter around it. The primary side is energized by ramping up from 0 until the clamp meter reads the rated current on 2nd side. If it reaches 1,000A at primary voltage of 200v, that's 200/4,000 = 5% and available fault current at transformer is calculated by taking the reciprocal. In this case, it comes out to 20kA with the assumption that primary source is infinitely stiff. It it's a ring circuit with multiple transformer feeding the ring, it's more complex to calculate.
You work downwards from what they give you based on wire size and distance to the point you want the fault current for. The impedance is a bit higher for steel conduit than plastic. As GoldDigger mentioned, large motors add to fault current.
If I've got a 12v cordless drill battery on the other end of a jumper cable, the "up to" fault current will be small enough that deducting for cable wouldn't make much sense.
If it's a bank of forklift batteries, the fault current at the terminals will be insanely high, so I'd have to derate for the cable to get the value to a realistic value.