battery charge rates

sawmill
sawmill Solar Expert Posts: 93 ✭✭✭
in General Solar Power Topics #1
I was trying to compute the proper charge rate (10-13%)for a friends battery bank and solar system and have encountered some Senior moments in my computations. The batteries are 2 volt(flooded) Globe Union, data plate states-TCC-840 A.H at the 8 hour rate. Bank will be 12 batteries in series for a 24 volt system.

I don't have any information on proposed loads. I am just trying to determine necessary solar to bring this bank back to close to a full charge from a 50% discharge.
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  • niel
    niel Solar Expert Posts: 10,300 ✭✭✭✭
    #2
    Re: battery charge rates

    actually the full range is really 5-13%. 5% represents the battery industry standard of 20hrs and 10% is also another battery industry standard that usually isn't shown for leadacid batteries so much as other types and represents the 10hr rate. 13% is the top end for charging general lead acid batteries that may be basic or generic and unless the battery manufacturer says you can go higher we don't recommend that you do.

    now to your charge rates for your 12 in series 2v batteries at 840ah. that will be 24v at 840ah.
    5%=42amps
    10%=84amps
    13%=109.2amps

    many times those rates are unreachable and we say that sometimes without any other loads being present that a slower rate of 3% could be used and that would be 25.2amps. technically that would be the 33 and 1/3hr rate. in real life these rate times aren't true as the absorb stage will take a bit longer as batteries generally don't charge to 100% via the bulk stage, which is about 80% of the charge process, but taper the rate of charge at the top end thus taking longer time.
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  • sawmill
    sawmill Solar Expert Posts: 93 ✭✭✭
    #3
    Re: battery charge rates

    Thanks Neil for your informative reply. It's always a pleasure reading this forum and the great advice offered.
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  • crewzer
    crewzer Registered Users, Solar Expert Posts: 1,832 ✭✭✭✭
    #4
    Re: battery charge rates

    Sawmill,

    The coulombic efficiency (Ah out/ Ah in) of flooded-cell batteries is ~90%. Accordingly, you’ll need to supply ~467 Ah to replace the 420 Ah removed.

    The absorption stage of the battery charging process is current limited. Accordingly, not many Ah will be delivered during these two hours. If the batteries are 95% full when the charger switches from bulk- to absorb mode, then the charge will deliver the final ~47 Ah during this period.

    If your location and PV array averages the equivalent of five hours a day of “full” Sun, and the equivalent of one full Sun hour is lost to the two hours required for the absorption stage, you’ll then need to be able to deliver ~467 Ah – ~47 Ah = ~420 Ah during this other four “full” hours.

    Accordingly, you’ll need an array and charge controller that can deliver ~420 Ah / 4 hours = 105 A. 105 A is 12.5% of 840 Ah, so that should be OK.

    Assuming an average charge voltage of 27.5 V, that would be 2, 888 W. Assuming array efficiency of 85%, wiring efficiency of 97%, and charge controller efficiency of 95%, you’d need an array rated at (2,888 W / (85% x 97% x 95%) = 3.7 kW.

    You'll need to check season insolation specs for your location, and then perhaps re-crank the numbers.

    HTH,
    Jim / crewzer
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