The base formula to use is Watts = Volts * Amps
Thus you are using 563.5W.
That means, at 12V, you will be using 47amps. 8 Hours of that would use 376Ah of energy. Assuming your batteries are 90% full (charging to 100% is far too expensive in terms of efficiency) and that you never want to go below 50% charge, your absolute minimum battery
bank size would need to be 980Ah. Considering the voltage goes down as the battery nears 50% plus the standard conversion losses in going between AC and DC you'd need to add 10% - so now you will need about 1100Ah battery capacity. But it get worse - Peukert's Law. The capacity of batteries is given in 24 hour discharge times, if you draw energy out of the battery at a faster rate, the actual capacity of the battery goes down; so add another 20% on top.
You are now at a 1300Ah battery bank (call it 12 Group 34 's).
If you need to replenish 600Watt for 8Hrs of energy via solar
: at 8 hours per day of usable sunlight (I'm being generous and assuming the boat is on Lake Pleasant in Arizona) that would be around 600W per hour, at 12V that would mean the solar cells would need to produce 50Amps, adding 10% losses during charging
for Dr. Peukert plus other losses and the occasional cloud we arrive at 55amps.
I just checked and see that the average for top-end solar panels
seems to be 11.5 watts per square foot, so that would mean you'd need 52 square feet of solar panels
. But this is an absolutely best-case scenario - lots of sunshine, no boom or mast
to cast shadows, new batteries, an efficient top-end inverter
, big cables
, an efficient AC unit, etc.
Just as hogfighter has already stated, running AC off batteries is a difficult and expensive (and heavy) task. A genset or little generator
is the only real option here. On a side note, I've got 800Ah (at 24V, so 1600Ah at 12V) on board and a top-end Victron inverter
and wouldn't dream of running the AC. I'm in the tradewinds and the wind generator
just keeps up with the fridge power requirements.