Best way to parallel 3 battery strings? - Paid gig!

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  • niel
    niel Solar Expert Posts: 10,300 ✭✭✭✭
    Re: Best way to parallel 3 battery strings? - Paid gig!
    inetdog wrote: »
    Quite right. For reasons other than just paralleling, you always want to minimize resistance to the point that it is economical to do so.
    And I think that I will just fall back on the summary that figuring out exactly what would be best for balancing in any given situation is probably not worth the effort. :-)

    as far as balancing goes one could do their best to keep things somewhat equal and do the routine eq charges, but there is also the ability to physically move the batteries from their electrical positions just like rotating tires on a car. not perfect, but should suffice.
  • mlodewyks
    mlodewyks Registered Users Posts: 4
    Re: Best way to parallel 3 battery strings? - Paid gig!

    Hi,

    I found this nifty explanation and some samples a while back. I think Method 4 is a nice configuration to spread the load. http://www.smartgauge.co.uk/batt_con.html

    Regards,
    Martin
  • inetdog
    inetdog Solar Expert Posts: 3,123 ✭✭✭✭
    Re: Best way to parallel 3 battery strings? - Paid gig!
    mlodewyks wrote: »
    Hi,

    I found this nifty explanation and some samples a while back. I think Method 4 is a nice configuration to spread the load. http://www.smartgauge.co.uk/batt_con.html

    Regards,
    Martin

    You have found a very good source, which I believe is mentioned in a few of the stickies on the Forum. There are a lot more good pages at smartgauge.com, including ones on Peukert's law and on battery charging and maintenance in general. This was the first site I was directed to when I started up in PV.
    SMA SB 3000, old BP panels.
  • NorthGuy
    NorthGuy Solar Expert Posts: 1,913 ✭✭
    Re: Best way to parallel 3 battery strings? - Paid gig!
    inetdog wrote: »
    There are a lot more good pages at smartgauge.com, including ones on Peukert's law ...

    Some information on this web site may have a merit, but the interpretation of the Peukert's equation is absurd.

    When battery discharges, lead oxide is converted into lead sulphate on positive plates (and other part of electrochemical rection proceeds on negative plates). To convert one molecule of lead dioxide, two electrons must travel from negative plates to positive plates. When battery is charged, there's certain amount of lead oxide on the positive plate. When it is discharged only some small amount of lead dioxide is left. The difference between these two amounts (and nothing else) determines the amount of electrons necessary for the reaction. 1 AH requires 2.247 x 1022 (if I remember correctly) electrons to pass, so it requires 1.123 x 1022 molecules of lead oxide to be converted to lead sulphate. If you have 1.123 x 1022 molecules of lead sulphate, you need to get 1 AH from the battery to convert them. If you have 2.247 x 1022 molecules of lead sulphate, you need to discharge 2 AH and so on. It doesn't really matter how you get there. The amount of AH taken from the battery depends only on the amount of lead oxide converted (ignoring few insignificat factors such as surface charge and self-discharge).

    While discussing the Peukert formula, smartgauge.co.uk seems to imply that other factors, such as internal and external resistance, speed of dischage, or speed of chemical reaction, influence the amount of AH delivered by the battery. If that was true, there would be a mismatch between electrons moved and the amount of lead dioxide converted. But clearly, this is impossible. One molecule of lead oxide requires exactly two electrons for the reaction. Cannot be one, and cannot be three.
  • BB.
    BB. Super Moderators, Administrators Posts: 33,431 admin
    Re: Best way to parallel 3 battery strings? - Paid gig!

    Yes, a battery has lower apparent AH capacity as the output current increases.

    There are chemical, internal resistance, acid/metal matrix interfaces, temperature, etc. that all affect the ability of the battery to deliver and accept current.
    Trojan T-605 Deep Cycle Battery

    1.75-Hr Rate = 131 AH
    5-Hr Rate = 175 AH
    20-Hr Rate = 210 AH
    100-Hr Rate = 232 AH

    Maybe we have some confusion here...

    Yes, Lead Acid Batteries are near 100% efficient when counting electrons... But most of our "appliances" use power / energy... So when we see AH we are really talking about AH*Voltage = Watts or power...

    A battery will start at near 10.5 volts when dead, and as it recharges, it will the voltage will slowly rise to 14.5 volts. And as it discharges, the voltage will fall again.

    So, when we talk about a Lead Acid (flooded cell) being about 80% (really 80-90%) efficient, we are talking about Energy (Power*Time) into/out of the battery.

    In the UPS side of the battery business, they will will actually rate batteries in Watt*Hours and kWH until "dead"... Which is probably more "useful" for us.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • inetdog
    inetdog Solar Expert Posts: 3,123 ✭✭✭✭
    Re: Best way to parallel 3 battery strings? - Paid gig!
    NorthGuy wrote: »
    Some information on this web site may have a merit, but the interpretation of the Peukert's equation is absurd.
    I have to agree with you there. His illustrations of real effects are OK, as well as his emphasis that you have to start with the C20 rate or make other changes. But his insistence that the formula is mathematically and physically exact and related in a fundamental way to the chemistry rather than just being a good approximation over a limited range of current values bothers me a lot. (Especially the notion that for an ideal LA battery it correctly predicts that you could get unlimited (current x time) if you just drew at a low enough rate!)
    SMA SB 3000, old BP panels.
  • NorthGuy
    NorthGuy Solar Expert Posts: 1,913 ✭✭
    Re: Best way to parallel 3 battery strings? - Paid gig!
    BB. wrote: »
    There are chemical, internal resistance, acid/metal matrix interfaces, temperature, etc. that all affect the ability of the battery to deliver and accept current.
    BB. wrote: »
    1.75-Hr Rate = 131 AH
    5-Hr Rate = 175 AH
    20-Hr Rate = 210 AH
    100-Hr Rate = 232 AH

    Yes. That's of course is true.

    If battery has 232 AH stored in it (based on 100-Hr rate), and you discharge it at 1.75-Hr rate, the voltage drops below 1.75V at the end of 1.75 Hr and 131AH will be removed from the battery. The smartgauge site seems to imply that (232-131) = 101AH got lost. That's impossible. They couldn't go anywhere. The corresponding amount of lead oxide is still in the battery. If you remove the big load, the voltage will go up, and you can still get 100*101/232 = 43.5 hours of discharge at the current corresponding to 100-Hr rate.

    All these factors (temperature including) do not remove charge from the battery as smartgauge implies, but rather decrease the voltage.