In IEEE 1584 it is required to do two calculaltions of incident energy. One with 100% of arcing current and one with 85% of arcing current.

Will this rule apply to low AND medium voltage locations or ONLY to low voltage (less than 1,000V) locations?

Thanks.

Recs

There's nothing limiting the voltage range for which this double calculation is to be done. So it's always to be done.

IEEE 1584a-2004 says you need the two calculations for applications with system voltage under 1000 V.

NFPA 70E 2009 doesn't have that limitation.
There's the empirical equation for Ia for V <= 1kV, then for Ia for V > 1kV, then the comment about 100% Ia and 85% Ia.

Why remove that requirement for V > 1kV in IEEE 1584? The reasoning seems sound to do it...

I say you need to do it for both. I've seen the instantaneous trip values on breakers set higher than the actual calculated fault current. I like to set the Int trip to a min of 70% of calculated to make sure they function during a arc fault.

The equations I think you are referring to in the 70E are from the IEEE 1584. In other words, the 1584 rule came first, but the 70E folks just copied it, and left out the mention of rule being for low voltage.

As I understand it, the reasoning behind the 85% rule for low voltage was due to the variability in the arc and the resulting arcing current. The arc was more stable, and the arcing current is nearly equal to bolted fault current for medium voltage.
The 85% rule is meant to account for the variability in the arc and arcing current, not to cover for other other situations.

That's correct.

So, to answer RECS question, IEEE 1584 requires to do 2 calculations for low voltage locations only.

But you could also want to see the effect of a 15% arcing current variation on the incident energy at medium voltage.