Any one has tried this program for single phase arc flash incident energy calculation? The first input is arc current? How to get that info since IEEE only give arc fault current calculation for three phase bolted fault current only. Or I just missed anything here? Also, if you got the incident energy from the calculation at the input the distance, how to get the arc flash boundary? Any inputs will be greatly appreciated!

Actually, for this program above 15 kV the assumption is I arc = I bolted. As the voltage increases the difference between the arcing current and bolted current converge. So you can simply use the line-ground bolted current from a traditional short circuit study. Hope that helps.

Thanks for the reply, Jim. What if my system is at lower voltage? Say 7200V, Any suggestions?

Same approach. Above 1 kV, Iarc is Ibolted. But below 15 kV, 1584 gives more reliable results.

I also have a few single phase 120/240 panels off the 50kVA single phase transformer. Can I alos treat Iarc= I bolt by using Duke power calculator? When to decide the duration time, should I use bolted fault current or use arc current if they can't be treated as same for incident energy calculation? Much appreciated!

Be cautious assuming arcing current is equal to bolted current in your calculations. For that assumption to be true, arcing resistance would have to be equal zero, hence power released by arc and incident energy would be equal zero. Also, by using bolted current when determining arc duration you risk grossly underestimating arc duration.

Not sure why you are trying to calculate an incident energy from a single phase source. IEEE 1584 is for 3 phase and the majority will default a single phase at these voltages to a value of <1.2 cal/cm^2.

Thanks Barry for the input. However, the IEEE 1584 doesn't provide exemption for a service rated at 240V. My system is rated at 120/240V equipment though.

Please note that although IEEE 1584 guide has been developed primary for three phase arc fault analysis, it is also <deleted link to external commercial website>

IEEE does not provide an "exemption". It's the other way around. IEEE only COVERS 208-15 kV, 50/60 Hz, 700 A-106 kA, 13-152 mm gaps, 3 phsae.

120/240 V single phase equipment is not covered.

Your choices are the following:
1. Use the table in CSA Z462 for 240 V or less. So everything will be either 1.2 or 4 cal/cm^2, task specific.
2. Use the table in IEEE C2-2012 for 240/120 V or less. Although this is "utilities" the equipment specified works in industrial facilities, too. Everything will be 4 cal/cm^2.
3. Use IEEE 1584 for 240 V and ignore its limitations, assuming that 3 phase results are worse than single phsae; thus, conservative result. Not supported by the standard.
4. Search for and use test data that exists such as that published by EPRI. As long as you can match up conditions, this is a viable solution.
5. Use/abuse the 1584 "208 V exception". This is likely to decrease with the next edition and again, it's 3 phase, but it gives you something to go by.

Frankly I've combed the OSHA database of injuries. In the case of 120 V there are burn injuries but every one was within the "working distance" boundary that IEEE 1584 uses such as an IT worker that was severely burned on the hand and hospitalized from plugging/unplugging a 120 V cord into a defective power strip which arced. But I can't find a single case where following IEEE 1584 and 70E would have made a difference (they don't attempt to protect within the working distance). I have found cases of injuries and even a fatality with 240 V equipment so clearly an arc flash injury and/or fatality is possible. Keep in mind these are extreme cases but they do exist. So the IEEE 1584 "rule" or something like it is appropriate in the 240/120 range and 4 cal/cm^2 PPE is probably a necessity for cases beyond such a cutoff. The challenge is defining the cutoff itself.

Much appreciated!