IEEE 1584 1

[wbd]

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.

 

[davidbeach]

Find the fault current with both sources available and make the calculations based on that fault current.

 

[wbd]

Hi davidbeach,

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?

 

[davidbeach]

Total energy while both sources are feeding the fault. If one source is removed before the other, add in the energy from that source while it is the only source, but the boundary wouldn't change.

 

[dpc]

I think the flash boundary should be a function of the total energy released. This is reflected in the IEEE 1584 empirical equations. But you can't simply add the distances together. You can compute the energy released when all sources are connected, then compute the energy for each time segment as the various sources are tripped off. These energies can be added together and the flash boundary can be determined from the IEEE 1584 equations, I believe.

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.

 

[rbulsara]

You may want to check arc fault energy with each source as well as both sources together. Lower fault current with single source could result in higher AF energy as the OCPD opening time will be greater.