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I am really new into this. I hope I phrased my question appropiately

Assuming I have 200 Ah / 6 volt battery rating, that would mean it could cover 1.2 kWh load.

And if I need to run my 400 kWh / 220 volt AC house monthly, will I need at least 400 of them?

Or the conversion from 6 volt DC to 220 volt AC require additional calculation?

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  • \$\begingroup\$ If your home averages 550 watts continuous (400 kWh/month), then discounting any inefficiencies in converting 6 VDC into 220 VAC RMS (which there will most certainly be some), you would come out at 333 of them. So 400 in round numbers, sure. (That's without any recharging or generator, etc.) Of course whatever you do will have to handle your peak demands. Are you planning to be without external sources for a month at a stretch? \$\endgroup\$
    – jonk
    Aug 22 at 2:09
  • \$\begingroup\$ Thank you for your input. Would you able to explain how you get those 333 numbers? I see when you mention 550, its wattage per hour in average. Also, since the 6DC-220AC have no significant impact, is it safe to say that if I need to decrease total amount required, I would need battery with higher voltage? (I.e with 12 volt, certainly will only need 200 of them) - As for your question, in this case I assumed using primary cell, replacing with new one if it runs out \$\endgroup\$
    – mightyhorn
    Aug 22 at 2:32
  • \$\begingroup\$ Go to google search and type in "(550 watt)*(1 month) in watt*hour =" and hit ENTER or else SPACE and see what pops up. I get "= 401766.419 watt * hour". That's your figure. Now type in "(400000 watt*hour)/(6 volt*200 amp*hour) =" and again hit ENTER or SPACE. I get "= 333.333333". (I used quotes to highlight what to type, but don't use the quote marks.) And normally in a case like this, I'd consider batteries to be rechargeable ones and I'd want to store energy safely, as diesel or clear kerosene (which is about the safest, high-density energy storage that exists.) \$\endgroup\$
    – jonk
    Aug 22 at 2:38
  • \$\begingroup\$ And no, your wording, "I see when you mention 550, its wattage per hour in average," is poorly written. A watt per hour isn't something people usually talk about. In fact, I can't recall a time. "Per" means "divided by." But a watt-hour is a watt TIMES an hour. So it's different. On the other hand a Joule per second is a Watt! So you can have Joules per hour, if you want, and that's useful. Just not Watts (or wattage) per hour. \$\endgroup\$
    – jonk
    Aug 22 at 2:55
  • \$\begingroup\$ I understand now, thank you! \$\endgroup\$
    – mightyhorn
    Aug 22 at 3:56
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400kWh per month equates to 555W average. A 1200Wh battery would support that load for a little over 2 hours. A reasonably good inverter will provide about 90% efficiency. If you’re contemplating using a number of batteries then it’s typical to put a number in series to provide 12, 24 or 48V which can give somewhat higher efficiency, but still below 100% of course. Also consider the peak loading - if you actually use 5kW for 10% of the time and almost nothing the rest of the time then you’ll need enough capacity, and draining batteries at a high rate results in less output. Back to the calculations- 12 batteries for 24 hours, so 400 would be sufficient for a month give or take a few percent.

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  • \$\begingroup\$ "put a number in series to provide 12, 24 or 48V which can give somewhat higher efficiency" => Can you elaborate more on this "efficiency"? And if it applies to any higher voltage or to only those (12,24, and 48) numbers, since from my standpoint - Would the number of batteries required will stays the same? \$\endgroup\$
    – mightyhorn
    Aug 22 at 2:57
  • \$\begingroup\$ Sure, low voltage (and therefore high current) setups tend to have lower efficiency because resistive losses are very much more pronounced - if your 6V supply drops by 500mV at 100A then you’ll lose 8% of your power, but a 48V battery would only need to supply 12A and would drop 60mV or so with the same cabling. The same applies to a degree in the inverter itself, higher voltages lend themselves to better efficiency. \$\endgroup\$
    – Frog
    Aug 22 at 3:01
  • \$\begingroup\$ But yes the number of batteries would be essentially the same. \$\endgroup\$
    – Frog
    Aug 22 at 3:02
  • \$\begingroup\$ I see. That would mean in this 400 battery setup it's much better to put every 40 batteries (6V) in series, right? So it would get 240V (higher voltage = higher efficiency). Or I understand it wrong? \$\endgroup\$
    – mightyhorn
    Aug 22 at 3:18
  • \$\begingroup\$ It depends on the inverter you intend to use. 24 and 48V are relatively safe to handle and so they are quite common, while 240VDC is very hazardous (more so then AC) and so tends to be used only in larger installations where the benefits are more pronounced. \$\endgroup\$
    – Frog
    Aug 22 at 3:51

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