I found this article to be helpful to my understanding of IEEE 1584 calculations. It is nice to see an example based on the utility style equipment that I'm doing calculations for.

[url="http://www.brainfiller.com/documents/PadMountArcFlashJimPhillips.pdf"]http://www.brainfiller.com/documents/PadMountArcFlashJimPhillips.pdf[/url]

However, using IEEE 1584, I cannot duplicate the predicted incident energy (cal/cm2) calculated in Table 3 of this article. All of the required parameters, except the distance exponent, seem to be specified in the article.

I could use some of the intermediate calculations to understand where I'm going wrong. Knowing the distance exponent would also be helpful.

Here are the parameters I used:

Specified in article,
V = .480 kV (480V xfmr secondary)
Ibf = 30.02 kA (bolted three phase fault on xfmr secondary)
D = 457.2 mm (18" working distance)
t = 0.217 sec (13 cycle clearing time)
G = 25.4 mm (1" gap)

Assumed,
x = 1.641 (distance exponent corresponding to 1" gap in Table 4)
K = -0.097 (arc in a box)
K1 = -0.555 (arc in a box)
K2 = -0.113 (grounded)
Cf = 1.5 (voltage <1kV)

My calculated incident energy = 10.53 cal/cm2,
Articles calculated incident energy = 11.87 cal/cm2.

Why the difference?

Thanks for your interest in the article. It was a pretty interesting arc flash test. Since it was a "quasi-open" enclosure, we knew it would be between open air and in a box, we just were not sure where the results would end up.

As far as the difference in the answers, I did not mention a "minor detail" We used an X distance factor of 1.473 which is for switchgear. The reason is we did not have any good information for the new enclosure and decided it was larger than a panel so switchgear seemed like the next logical choice.

The clearing time was 0.214 seconds which was 12.8 cycles. This was the time from the time current curve and then in the lab we round it up to 13 cycles to account for a zero crossing arc extinction.

I hope it helps.

Thanks!

It does help to know the X distance factor, but I still cannot calculate the 11.87 cal/cm. Now I get 10.04 cal/cm2.

I re did the calcuations last week by hand and just re ran them with arc flash S/W.

What arcing current did you get? It should be 16.76 kA. That with all the other info should yield 11.8 cal/cm2

Let me know how it goes. You can send me a PM also and we can sift through the calcs.

Okay, now I see what I'm doing. I forgot my spreadsheet was set up to use the 85% arcing current for <1kV. Now I'm getting a number much closer to yours.

Thanks!

That is good to hear. You were correct in also looking at 85% of the arcing current. as well. IEEE 1584-a 2004 says the calculation should be performed at both100% and 85%. This is done just in case the 85% of arcing current produces a longer clearing time that can lead to greater incident energy. Glad it all worked out.

It seems that the distance factors are to be used with the corresponding gaps in table 4. Why then didn't you stick with the table and change your test set up to have a 32mm bus gap, corresponding to 1.473 distance factor, instead of 25mm?

Really good question. It was simply a matter of just wanting to getting the arc to sustain since the pad mount spades were so far apart. We were not that concerned with the table values in the lab at the time as we were getting a good flash.

All we really wanted to do was compare the incident energy for an open top and side vs. closed top and side configuration. The 1.473 value was used since the enclosure seemed more like switchgear than a panel. I'm sure the actual exponential value was likely different than any table value since we did not have the exact same size enclosure as the IEEE 1584 tests. We did the best with what we had.

All we were attempting to prove (and did prove) was there is quite difference in incident energy depending on the enclosure configuration. - Good question!