ConvergenceTraining wrote:

I guess it's hard for me, a beginner, to understand that the potential incident energy could be so low that the AFB would be smaller than the LAB. But your explanations help a lot.

I should point out that this is very introductory training, and that we are getting some review/contribution from some experienced electrical safety guys. This was just for my own curiosity, to get an answer quicker rather than later.

I have plenty of equipment, even at 4160 V, where the arc flash rating even at 18" or 24" is significantly less than 1.2 cal/cm^2. So my AFB is less than 18", sometimes only 3-6 inches.

This happens frequently near the "end points" of the system. On a relatively small fuse protected motor (say around 2-3 HP) the wire size may be only 12 to 14 gauge. The voltage is still 480 V, so the same LAB applies. But the available fault current with all the impendance of the wiring involved and especially if it is fed from a small (say 10 kVA) transformer is so small, and with only a 1/4 cycle trip time if there is a dead short which trips the fuses, my arc flash boundary and incident energy at working distance is so small that the AFB almost disappears.

Although this is not a hard and fast rule, remember that the AFB (and incident energy) are based on ENERGY. As we are all taught early on, energy is TIME times POWER. Power is of course volts times amps times power factor and perhaps a constant. Even if voltage is large, if time and current are small, the energy is also small. Where this breaks down is that the impact on incident energy is not quite linear...it is not time X volts X amps. There are curves that make the extremes not quite so smooth. But as a rough estimate, doubling or halving any of the 3 values will roughly double or halve your incident energy.

Now all the energy gets released at one assumed point or at realistic distances, it might as well be considered at a point. As it travels outwards, it does so in a sphere. So the area on the sphere is what matters, and the area of a sphere is pi*4/3*(diameter/2)*(diameter/2). So doubling the distance cuts your incident energy roughly to 1/4. Again there is a little fudge factor in here such as the difference between arcing in an MCC (which is really a hemisphere) and arcing in open air.

I use these grossly simplified examples with my own electricians to help them get used to the idea of how and why the values are what they are. They might get dizzy looking at logarithms and exponentials, but simple examples of doubling and halving, and from a very layman's point of view, help the math make a little more sense.