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I have an RV with two 100Ah batteries that provide 200Ah of 12V power in total. I would like to have a classic air conditioner (I've looked at swamp ones but they won't work for my needs) and I'd like to be able to run it for about a maximum of 8 hours.

How do I work out if this is possible, given that there also needs to be a 12V-120V inverter? I've found AC units as low as 465W (although I don't know what their max pull is).

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  • \$\begingroup\$ Do you know how power and energy are calculated? Can you multiply by efficiency? Really. Simple arithmetics. \$\endgroup\$ – Eugene Sh. Aug 5 '16 at 1:47
  • \$\begingroup\$ @EugeneSh. The main thing I didn't know was how to incorporate was the inverter. Also, I was looking for suggestions for AC units. \$\endgroup\$ – averageradical Aug 5 '16 at 1:56
  • \$\begingroup\$ The inverter is the efficiency part. You won't get recommendations for products here, as it is off-topic on this site. \$\endgroup\$ – Eugene Sh. Aug 5 '16 at 1:56
  • \$\begingroup\$ @EugeneSh. I didn't know that incorporating an inverter was simply a matter of multiplying a percentage and I didn't know what that percentage was. \$\endgroup\$ – averageradical Aug 5 '16 at 1:57
  • \$\begingroup\$ Remember you can switch the AC on and off : if you can only supply 250W for 12h, you can run a 500W AC for 30 minutes every hour, on a timer (assuming your inverter can supply 500W). \$\endgroup\$ – Brian Drummond Aug 5 '16 at 11:34
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Theoretically, amp-hours times volts gives you watt-hours. Then watt-hours divided by hours gives you watts.

In this case, 12V x 200Ah / 8h = 300W.

The inverter will use some energy, so multiply that again by the efficiency of the inverter. Say it's 85%, then you can run a (300W x 0.85) = 255W air conditioner for 8 hours. Theoretically.

If you can't find an air conditioner with a low enough wattage, you might be able to run a more powerful one at a low power setting, or only run it for part of the time (for example, a 380W air conditioner on for 10 minutes and off for 5 minutes gives a 254W average).

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    \$\begingroup\$ I think starting up an AC will consume much above the rated wattage, so I am not sure the second approach will actually save anything. \$\endgroup\$ – Eugene Sh. Aug 5 '16 at 1:53
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    \$\begingroup\$ The compressor and fan together will draw 3x the normal run current when starting up. Plus the inductive loads and start capacitor can play havoc with some DC-AC converters. Most put out a choppy sine wave at best. Be sure the DC-AC converter is rated for motors. \$\endgroup\$ – Sparky256 Aug 5 '16 at 1:59
  • \$\begingroup\$ @Sparky256 Thanks, is an inverter rated for motors different from a pure sine wave inverter? \$\endgroup\$ – averageradical Aug 5 '16 at 2:01
  • \$\begingroup\$ Agreed, seen AC start up estimates that are roughly double. Also, 200Ah batteries doesn't necessarily mean you can pull 200 amps for 1 hour. Actually it is likly the less you pull, the closer you will get to this capacity. For example 1 amp for 200 hours may be more realistic than 200amps for 1 hour. \$\endgroup\$ – st2000 Aug 5 '16 at 2:03
  • \$\begingroup\$ Funny you should ask. Pure sine-wave inverters are recommended for motor/inductive loads. A motor expects a clean sine-wave, and so does the start capacitor. This way no odd-shaped waveforms muck up the efficiency or cause undue back-EMF into the inverter. \$\endgroup\$ – Sparky256 Aug 5 '16 at 2:06

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