First, sounds like an equipment manfuacturer. The end user has to do arc flash, sorry to say.

Anitech Systems wrote:

Providing the above information, can anyone direct us to any reference information (or even software products)

that would allow us to make any kind of meaningful e.g. NFPA 70E 130.5 / Annex D Arc Flash Hazard Analysis Calculations?

IEEE 1584 equation is based on exactly ONE test where they managed to achieve a stable arc at 208 V and that is used as the "assumption" for the entire range of 50-300 V data. In other words, IEEE 1584 is invalid. The reason that there is so little data is simply because arcs are very unstable at low voltages (usually need around 150 V or so to even initiate an arc so there is a good chance it will not restrike and may do so for several cycles). In addition at this point there is no single phase test data used within the IEEE 1584 test data set. So using that methodology is at best questionable and the standard itself says so.

At this point I recommend using NESC-2012 table which is in turn based on actual test data. Incident energy = 4 cal/cm^2.

As no 480 V data...all information downstream of the transformer is invalid because you cannot determine the available fault current. Worse still if you assume a very high fault current this corresponds to fast trip times which also results in incident energy since trip time is short. However if the available fault current is very large then trip times are very long which results in a very high incident energy. Only the actual field results are valid. You just have to stonewall the customers and trust me, this comes up over and over again. Even in my case where I'm the plant electrical engineer and the project engineers think that it should just be a specification on a piece of equipment when I have no idea what they are actually buying ahead of time.

As to figuring out what to do with UPS's or VFD's, on the front end where everything is still copper, you can safely use the available fault current as it goes through the disconnect, terminals, cable, etc. If it's a line interactive or "offline" type where essentially the UPS passes line current through directly to the load then the fault current also passes straight through so this is relatively easy to model. If on the other hand it is a double-conversion type or a drive then there is no direct connection. You can usually safely assume that the available fault current is about twice the drive's maximum current capability since the semiconductors are rated for RMS current before they rupture in very short order (usually 1 or 2 cycles at most), if the drive even lets it get that far. Drive protection will kick in well before then. The only other factor to consider are the capacitors (DC link) and/or batteries. Batteries can be treated separately as a DC system and you need to know series resistance among other factors. Capacitors can be looked at as the amount of stored energy. Convert joules to calories. Then take the working distance and calculate the area of a sphere with the working distance as a radius in square cenitmeters. Divide by 2 to assume a hemisphere (assumes capacitor is in an enclosure). Then your final answer = calories available / area of hemisphere which will be in cal/cm^2. This is ultraconservative again because only a portion of the heat energy actually gets radiated (most theoretical guesses estimate about 25%) but the best you will be able to do.