In reply to Neil:
> Bit of a hijack here but I've built a cabin on wheels which I power using solar panels and leisure battery, though I'm clueless really as to electrical stuff. I'm interested in the calculations you use there as I'm finding that the battery I have in the hut doesn't last as long as I thought it should.
Look up Peukert effect. Leisure battery ratings assume about 0.1C discarge rate, if you draw more than that the capacity diminishes. Your solar panels (unless of course you have them metered and you're counting charge in) probably aren't delivering anything like the power on their rating plate for a variety of reasons.
> 100W solar panel charges a 115Ah leisure battery via a charge controller. This then kicks out 12V to an 240V invertor which is connected to an RCB then into the circuit in the hut (twin sockets and a couple of 6W LED lamps). The charge controller will charge the battery up to ~14V and let it drop to ~10V to protect against it discharging to deeply. What calculations should I be doing to find the amount of time I can have the lights on for (and maybe charging a phone too)?
2x6W =12W, at nominally 12V they therefore draw 1A (from P=IV)
Your 115AH battery can deliver 115A for 1H (except it can't, Peukert again)
Or: 1A for 115H
Your inverter is probably 95% efficient (I'm assuming they're 'mains' LEDs
Your LED power supply is probably 95% efficient
Your battery shouldn't routinely deliver more than say 70% of capacity to preserve its performance
From this we get 115H x 0.95 x 0.95* 0.70 = 72H of lighting between charges
Your phone assuming it's HUGE very roughly has a 3V, 4AH battery in it, assuming 100% conversion efficiency it will use 4 x (3/12) = 1AH of your leisure battery's capacity. Even at a pessimistic end to end conversion efficiency of 50% it uses essentially nothing.
HTH,
jk
Post edited at 09:13