A VFD does the same thing. The switching power supply charges the UPS DC bus in unity with voltage(although it isn't a pretty sine wave) so you have close to unity power factor. The output power factor of the UPS depends on what is plugged into it. It can accommodate a load with .8 power factor at it's full nameplate power rating. If the supplied load has a lower power factor then .8 the UPS can't be fully loaded. Generators are this way. They rate generators in KW(real) at a given PF.
I would be more inclined to think that the ups could supply 80% of the input, than I would be inclined to think that you could supply it with 111% of what it is rated at.
If you put a scopemeter on the ups input and monitored the current I would bet money you wouldn't think it looks like a resistive load. It would look look pretty nasty, which is why harmonics can come into play.
I'm not sure where you come up with a & b, but assuming you have 100KVA, 80KW and 60KVAR, your power is 80KW. If you assume your .9 efficiency then your 80KW / .9 = 88KW reflected on the line side.
I took the 100 KVA figure on the output where PF is .8.
I tried to back calculate what this would require on the Input
On the input PF approx = .99 or 1.00 so I translated (perhaps mistakenly) the 100 KVA into a 100 KW draw at the input. Am I wrong for doing this?
They are two different things.
An output power factor of 0.8 limits output power to 80 kW.
The input pf depends on the input circuit topology. A 0.98 pf infers an active front end - the bit that converts ac to dc. Unless the manufacturer is being a bit loose on defining pf as just displacement rather than total.