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Old 04-04-10, 09:30 AM   #5
OffGridKindaGuy
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Quote:
Originally Posted by msirach View Post
Charger is 29 Watts, lamp is 43 Watts, pump is 20watts(I think).
Are these 120 v.a.c. loads ran via the inverter from your battery?

29+43+20= 92 watts a.c. (Demand from the inverter)

For every 1 amp demand from an inverter requires 10 amps from the battery @ 12 volts d.c. (Not including inverter losses)

92 watts @ 120 v.a.c. = 0.76 amps

92 watts @ 12 v.d.c. = 7.66 amps

Total aHr. reserve of your battery.. 1956 aHr.

Divide by 4 = *489 aHr. (25% of the total for 25% discharge level) *Never discharge your battery any lower than 20-25% of the total capacity. (11.8 volts)

With your given load, (7.66 amps) you should be able to operate things for 63.8 hrs. with fully charged, new condition batteries. (Don't forget ya gotta replace this used energy )

The closer you keep the inverter to the battery, the smaller gauge wire will be needed. With less than a 10 amp load, you could get by with 12 ga. wire within 10 ft. of the battery BUT if you plan to operate other things with the inverter, you'd better plan for heaver wire.

What's the "Grand Plan"? Are these the only items you plan to operate with this set up?
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