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4 - 6V in series = 24V

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  • #16
    Hi,

    Yes I have bounced about a bit, call it exuberance. Reality is settling in though.

    What I want is to be ready for an extended outage and to use batteries as part of our solution to this inconvience.

    We've had a couple of short LOCAL interruptions already, the longest lasting 3 days. It was a good learning experience but to rely on a generator when all you need, at the time, is a few lights seemed wasteful. Generator did fine with the big things but to run a few lights after it got dark not so much.

    So I guess what I'm looking for is where can DC be a more efficient tool and then how can I recharge the batteries if the grid is down for an extended time.

    I'll get one of those usage meters (Kill a Watt ???) to see what our big ticket items draw if for no other reason than it'd be good to know anyway.

    Thanks, mj52

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    • #17
      Yes, its called a Kill-a-watt meter. Plug it into the wall outlet, plug the load into it and let it run for 24 hours then come back and push the KwH button on the meter to get your 24 hour draw.
      You'll need to figure out your loads before you can decide on an inverter and batteries etc.
      1150W, Midnite Classic 200, Cotek PSW, 8 T-605s

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      • #18
        Another question please,

        Trying to get my head around this.

        A 12V computer fan uses 9W (12V x 0.75A= 9W)

        my battery has a 100hr discharge rate: 250 Ah

        OR even better use a hypothetical example to more easily explain the math involved.

        I have a source and a load and I can draw this for that long till I reach a point where the source has been drained to here and any further work done may case premature aging of the source.

        I like to understand the math part of things, this may help me to work through my own questions or at least explain where my line of thinking comes from as i drive off the cliff.

        Thanks, mj52

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        • #19
          The math is easy, just simple 5th grade math.

          AH = Amps x Hours
          Amps = AH/H
          Hours = AH/A

          Having said that it only works at the battery specified hour rate which is usually the 20 hour rate. So for say a 100 AH battery the formula is accurate only at 5 amps or 100 AH / 20 H = 5 amps. If you discharge at some other rate you have to apply a Correction Factor to compensate for Peukert Law. If you discharge at a faster rate than 20 hours the CF is less than 1, and if slower than 20 hours is greater than one. Take this battery for example. At 20 hours is a 220 AH battery. If you discharge at only 2.93 amps is a 293 AH battery. If you discharge at the 1 hour rate (79 amps) is only a 79 AH battery.
          MSEE, PE

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