IEEE 1584
IEEE 1584
(OP)
In calculating the arc flash hazard on a bus with 2 sources, does one calculate the incident energy contribution from each source seperately and then add them to get the total energy?
For example:
Source A: 3.5 cal/cm2 @ 18" working distance and arc flash boundary is 32"
Source B: 3.9 cal/cm2 @ 18" working distance and AFB is 40"
Would the total arc flash hazard be 7.4 cal/cm2 @ 18"?
What would the arc flash boundary then be? 40" or 72" ?
I could not find anything in IEEE 1584 on this but I believe it would add. Thank You for a sanity check.
For example:
Source A: 3.5 cal/cm2 @ 18" working distance and arc flash boundary is 32"
Source B: 3.9 cal/cm2 @ 18" working distance and AFB is 40"
Would the total arc flash hazard be 7.4 cal/cm2 @ 18"?
What would the arc flash boundary then be? 40" or 72" ?
I could not find anything in IEEE 1584 on this but I believe it would add. Thank You for a sanity check.






RE: IEEE 1584
RE: IEEE 1584
That is essentially what I did. I found the fault current coming from each source, thru the sources incoming breaker at that bus, figured the arcing current from that bolted current and used that to find the tripping time. This gave me the time to use to calculate the incident energy available from each source.
So, do I add them together or take the worst case energy and use that?
What would be the arc flash boundary?
RE: IEEE 1584
RE: IEEE 1584
From a practical standpoint, you can simply compute the arc fault based on the maximum total fault current and use the time for the slowest protective device (based on the fault current it actually sees, of course). The computed arc energy will be too high, but the result will be conservative and errs on the side of safety which is always hard to argue with.
RE: IEEE 1584