In the 1584 method the arc flash energy at a given distance is found by scaling the calculated normal energy based on the clearance time and working distance. The distance exponent (x-factor) varies based on the equipment type and my understanding of the explanation for this was that some equipment focuses the energy, thus it drops off slower with distance and has a lower x-factor to account for this. Jim has a good plot of this on page 136 of his complete guide.

The way the equations work means that the energy for the open air case increases more rapidly than the other equipment types when the distance is less that 610mm (see attached graph). My question is, is this scaling method valid for distances smaller than the normalised 610mm distance? Surely this focussing effect occurs from the source of the arc (or very close to it) and not at the 610mm distance, so the open air energy should always be lower even at closer working distances. Any explanations or comments of the limits of the model will be greatly appreciated.

Nice graph. Your graph is based on each case having the same incident energy at 24 inches (each case intersects at 24 inches). Calculating the incident energy for shorter distances shows that each case has a different value which is expected. (and odd looking)

However if the graph assumes each case has the same incident energy at a closer distance like 18 inches (i.e. the intersection is at 18 inches instead of 24 inches) the graph makes a bit more sense when moving to further distances. I attached a graph that I use in training to help explain this.

IEEE 1584 shows distance data down to:
18 inches/457mm for LV switchgear (X=1.473).
12 inches/305 mm for LV panels (X=1.641)
24 inches/610 mm for MV switchgear (X=0.973)

Hope it helps!

Thanks Jim. My question now is how did you produce that graph? The normalised incident energy from the IEEE1584 equations is given at a working distance of 24 inches. This is then rescaled to whatever working distance you require but that rescaling requires you to know the x-factor. From my graph it can be seen that different x-factors result in different energies at a given distance (presuming you have moved away from the 24 inch position). It seems to me like you would need a new equation to give you the normalised energy at the 18inch distance before you could produce your graph. What have I missed?

I select an energy (somewhat arbitrary) of around 25 cal/cm2. Then I plotted the effect of moving away from the arc using the different X distance factors (enclosures). I believe what you did is assumed the Ei at 24 inches and went both direction. I'm sure you would have a similar result if you took my value at 18 inches and calculated closer. What I was trying to show was how the Ei drops differently with distance. You don't need a new equation for 18 inches, you just take the difference in distance (24/18) and raise to the X. Similarly if you wanted to know what it is at 36 inches, it would be (24/36)^X. Great discussion!

When D is less than 24 inches, scaling for distance using (24/D)^x results in the energy for the open air case rising quicker with distance than for say, the switchgear case. This is what I was trying to show in my previous graph. I would like to understand whether that is what happens in reality because surely the case where the energy is focused (i.e. switchgear) should always have a higher energy level, even at the smaller distances?

The problem with the graph is it is based on the energy being the same at a specific distance, 24 inches in this case. If the Ei is the same for each case at 24 inches, then moving closer to the source does show the incident energy rising more quickly for the case of an arc flash in air in open air. BUT, what is not discussed is if the energy is the same at 24 inches for each type of equipment (X distance factor), then the actual arc energy for each case (short circuit current, clearing time etc.) it is NOT the same. i.e. the energy at the arc for each equipment type (X distance factor) would be different and the energy at the arc for an arc in air would be significantly higher than the other cases.

Here is probably an oversimplified illustration but I'll use it anyway.

3 cars begin at the same starting line and travel at 50, 75 and 100 mph in the same direction for the same amount of time. In one hour, the distance traveled by the three cars is 50, 75 and 100 miles for each. They all have a totally different final distances traveled because they all began at the same place but had different rates of speed.

Continuing with the same analogy, if we now say they all ended up at the same place (which they could not if they began at the same starting line using different speeds), and have them head towards the starting line, it changes everything. The 100 mph car is moving faster (getting there first). If they each travel for one hour, they will not end up at the same starting line because they are all traveling at different speeds.

This parallels the case with the X distance factor. X would be like the speed in the analogy above. The premise of the case is we assume if they all have the exact same energy at the arc itself (starting line), the energy at a fixed distance away (finish line) will be different depending on the equipment and X distance factor (X). For each case having the same arc energy, the Ei would be much lower at a give point (24 inches) for air than for switchgear or a panel.

Now if we say Ei is the same for all of the enclosures at 24 inches (finish line) and that the energy increases differently as you move closer, this is correct. Further if you state that the energy for the arc in air increases faster (different speed), this is also correct. Where the logic flaw lies, is for this case, the actual energy at the arc for each case (starting line) could not be the same if all have different X distance factors. .

If the energies are the same at 24 inches, with different X values, the energy at the arc would have to be completely different for each case in order for them all to be the same at 24 inches. So saying the energy for an arc in open air increases faster with closer distance is correct but it also means the actual energy in the arc would be sky high compared to the other cases. It is not increasing quickly because it is in air, it is increasing quickly because the energy in the arc is extremely high for this case. i.e. higher short circuit current, clearing time etc. It is impossible for the different cases to have the same energy at the arc AND the same energy some distance away. This is where the logic fell apart.

Hope it helps clarify things a bit better.

Ahh...it all makes sense. Thanks very much for your help with this Jim.