I just got my hands on an updated Study done here in our plant. I happened to notice a value that I must not have seen in the previous one. Over 23,000 cal/cm^2... can that be a real value?

I asked a question about maximum incident energy a while back in this form's "question of the week". The results are below:

[url='http://www.arcflashforum.com/threads/1554/'][Incident Energy Survey Question][/url]

Believe it or not it sounds like it is almost correct if:
1) The instantaneous is too high
2) They did not use the 2 second cut off by IEEE
3) The maximum time defaulted to 1000 seconds which is the top of the time current curve.

If all 3 are true, then this is only 0.383 cal/cm2/cycle (60 Hz)

Check the clearing time in the study to see if this is correct. See what happens if the instantaneous is lowered so the tripping time is faster.

As a side note, the latest version of SKM's Powertools for Windows no longer defaults to the IEEE 2 second rule (it did before), instead using 1000s when no protective device trips. Check your study for the trip time, it may hold your answer. Reducing it to 2 seconds will lower the energy, but you have to assess if this a reasonable assumption first (i.e. enough room to maneuver).

The two previous responses are probably what happened. I will add that when we achieve suspicious arc flash results I often check what device is clearing the fault or model a sequence of clearing events to make sure the expected devices are working in the model.

Value is not that high 15.1kA. But the clearing time is 1000 sec. Boundary is listed as 1800 ft. This is downstream in switchgear that is fed via transformer from our 15kV distribution inside the plant.

We have two instances of 19,817 cal/cm2 at our 15 MVA feeders.

How do you change it back to 2 second default in SKM?

I believe that the problem is that the incident energy in cal/cm^2 was initially accepted by industry as the only measure of determining arc flash hazard. In fact, heat flux intensity measured in cal/cm^2 /sec should be factored in when evaluating the hazard. As an example, staying on a beach for 20 minutes makes yourself exposed to approx 1000W/m^2 x 20min x 60 sec = 0.1W/cm^2 x 1200sec = 120J/cm^2, or 29cal/cm^2 incident energy. But it doesn't really mean you have to wear category 4 suit to protect against the sunshine, or does it?

I encourage you to follow the http://www.iaei.org/magazine/2012/07/evaluation-of-onset-to-second-degree-burn-energy-in-arc-flash-hazard-analysis/ link to access the IAEI article examining the issue. In the scope of software solution, you may want to consider new AFA V5.0 software for arc flash hazard analysis and labeling, currently the only software program capable to perform arc flash boundary calculations based on Equation 2 for evaluated onset to second degree burn energy from the article cited above.

In short, one can expose himself to a very high incident energy without much damage when heat flux intensity is low enough (the beach example above). On the other side, even the incident energy less that the industry accepted threshold of 1.2 cal/cm^2 to a second degree burn could cause the burn when delivered in short time interval (case of high heat flux).

arcad: Good stuff. We just had a lengthy discussion about this exact subject today at an IEC meeting in Paris. The IEEE 1584 equations did and still do focus on total incident energy. Fortunately more people like you, me and many others - including many that are part of the Arc Flash Forum family are finally making headway with others that this is not all there is to it. Thanks!

Very interesting. Good read- thanks Arcad.

I just saw a client's arc flash study last week - the calculation was done with a 13 second time delay on the OCPD. The arc flash value was calculated at 550 cal/cm2. The study was recalculated just to get a ballpark with the IEEE 2 second rule... obviously the value dropped significantly (do not have the data in front of me currently). What is the overall thought in this case: The switchgear was in a remote area with only one access door, of course at the far end. The victim in an arc flash would possibly be trapped in the area for the duration of the fault. Should the calculation remain at 13 seconds in this case?

Would the person remain stationary or would he has an ability to escape towards the door? Also, arc blast would alsmost certainly produce pressure high enough to propel the person away from the switchgear limiting, in such a fashion, the individual's exposure to heat flux emanated from the arc, and therefore incident energy exposure. Check [url='http://www.arcadvisor.com/faq/arcblast.html']online pressure calculator[/url]to evaluate pressures produced by arc.

It's usually assumed that the outer limit for the arcing time is no more than 2 seconds. Although this is not a hard and fast rule, it accounts for the likelihood that the arcing material in an arcing field will likely be either burned off or expelled by the force of the blast. In any case, this would extinguish the arc event. Also, IEEE 1584 states "If the time is longer than two seconds, consider how long a person is likely to remain in the location of the arc flash. It is likely that a person exposed to an arc flash will move away quickly if it is physically possible and two seconds is a reasonable maximum time for calculations." [Annex B page 76]

I would personaly hold on to the two (2) second rule for arcing durations exceeding 2 seconds while doing the analysis.