hm,
the easiest would be to calculate the power demand.
(just add what's on the labels)
say 1kW fridge, 1kW well, 200Watts light
and double that to get the minimum amount fuel used
example:
fridge 24h a day, well IDK 10?, lights maybe 5?
makes 24000 + 10000 + 1000 (35k) Watthours
roughly double that (efficiency losses overall efficiency of 50% seems plausible to me)
so 70kWh a day
Energy density for propane is 13,777.8 Wh/kg let's say 13.8kWh/kg
means you need roughly 5kg of LPG propane per day (~11 lbs)
tank will depleted in about two days. (bit less)
you can look up the engine details or better generator details if you like,
maybe you're lucky and you get an efficiency rating with it..
then you can change the 50% assumption to a closer to real world value,
but the idea is still the same in the end.
hoping that the enigne itself governs itself to accomodate for low demands
if it doesn't you can almost certainly assume the total engine power (not generator power)
to be used at all times.. likely we talk 5.2kW engine
and that'd be 125 kWh a day (or 9kg of lpg propane a day [19,8 lbs.. the full tank so to speak])
'sid
