I'm curious if you've done the theoretical math on the energy involved in the start up surge for those devices vs the amount of energy in the super caps?
In the past I've wondered about various strategies involving capacitors to deal with current surges and concluded that the total power stored in the caps was a little low to be a real solution for a lot of scenarios.
I'm also no EE, but we can look up the units...
- farad = the SI unit of electrical capacitance, equal to the capacitance of a capacitor in which one coulomb of charge causes a potential difference of one volt.
- coulomb = the SI unit of electric charge, equal to the quantity of electricity conveyed in one second by a current of one ampere.
So I think putting that in terms we're more used to working with, your 24.5v supply spread over 6 series 1.5 farad caps is about 4 volts when they're fully charged (in this scenario). When the caps drop from 4v to 3v (to make the math simple, let's assume the inverter alarms at 18v), you might get 1.5 coulombs (amp-seconds) @ 24.5 volts out of the whole mechanism. 1.5 amp-seconds * 24.5v = 36.75 watt-seconds. (Realistically, I suspect you'd get less because that 1.5 farads is probably the ideal case when starting at 5.5v, but let's just go with that for now.)
I think that means the next question would be, given one of your loads, what is the area under the curve for the start-up wattage in watt-seconds and how does it compare to the 36 watt-seconds in the super caps?