I have posed the following questions to at least two maunfacturing organizations, only to get non-answers.

ASD's or adjustable speed drives - or UPS's - have inherent energy limiting throughput, since the double conversion results in galvanic isolation between the input and output. What would be the energy throughput limitation, and how would/could this be calcualted?

It will depend on how the inverter reacts to a minimal impedance, usually a maximum output current for a certain amount of time is allowed before it shuts down. Values I've seen are 200% nominal current for 0.3s for a large UPS. However, a UPS is usually equipped with a static switch that will close on an arc event. This means that the fault contribution on the output is the same as that on the input (save for the UPS and any cable impedance). To nitpick: here's a slight contribution of the inverter in the first half-cycle of the static switch closing but this is hard to express in a symmetrical current for the hazard calculation.

Drives don't have static switches so there you'll have to find out the inverter maximum output current to calculate the hazard. Although I'm curious as to why you want to do this, are you ever exposed to live conductors powered by a drive?

Note that IEEE 1584's empirically derived equations have a minimum defined current (I want to say 700A, but check the standard for the exact figure) so for small devices you might have to work with the theoretical model (Lee's equations).

I pursued the same question with EATON about one of their 480v, 225kVA UPS. IEEE 1584 Sec 5.1 states the range of their model is from 700A up through 106kA. Rated current for the UPS is around 413A (using 0.8 PF) and thus is outside the IEEE model. Therefore EATON concluded that we don't have to worry about it.

I still worry about it, but there's no way to compute the incident energy. I'm curious whether you could sustain an arc at the output of a large UPS.

Some UPS's will switch to internal bypass to clear a fault.