Battery question

Ronin2004
Ronin2004 Registered Users Posts: 8 ✭
in Solar Beginners Corner #1
Ok so we have decided on having about a 1 kw continous sytem design. using a 24 volt battery bank
I have calculated that that 1000 w*24 hours *1/24 volts*1/.85*1/.5= 5647 AH so to find how many batterys i need i devide that number by the @20 hr amp hour rate of my battery

this is the battery the group wants to go with
Smart Battery™ SB100
12V 100AH Lithium Ion Battery
Size: Group 31
Nominal Voltage: 12.8V
Amp Hours: 100
1000 Cranking Amps
300 min @ 20A
120 min @ 50A
Max Continuous Discharge 100A

after doing the peukerts equation i find that value to be only 2.73 ams
20=(50*(50/2)^(1.05-1))/x^1.05
does that mean it would take 2068 batteries for this system
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Comments

  • NorthGuy
    NorthGuy Solar Expert Posts: 1,913 ✭✭
    #2
    Re: Battery question

    If you want to output 1kW for 24 hours, you need to store 24kWh /0.85 = 28kWh

    12V 100AH battery discharged to 50% will give 12*100*0.5 = 0.6kW

    So, you need 28/0.6 = 47 batteries. 48 is probably a better number.
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  • Cariboocoot
    Cariboocoot Banned Posts: 17,615 ✭✭✭
    #3
    Re: Battery question

    Why you should always do math in order.

    1kW * 24 hours = 24 kW hours.
    Conversion efficiency of inverter 90% = 26.7 kW hours
    Inverter consumption in 24 hours (assuming a reasonably good inverter) = 576 Watt hours
    Needed stored energy: 27,276 Watt hours. Round up for losses.
    28 kW hours @ 24 Volts = 1167 Amp hours
    Battery bank @ 50% DOD = 2334 Amp hours. Round up to nearest available size. Calculate recharging capacity from that.
    Ridiculous to try and get that much capacity from 100 Amp hour batteries. It will be a mare's nest of wiring.

    But hey; feel free to do any way you like.

    Batteries that give specs on cranking Amps and reserve capacity are probably not suitable for RE applications.
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  • mike95490
    mike95490 Solar Expert Posts: 9,583 ✭✭✭✭✭
    #4
    Re: Battery question
    Ronin2004 wrote: »
    ......

    after doing the peukerts equation i find that value to be only 2.73 ams
    20=(50*(50/2)^(1.05-1))/x^1.05
    does that mean it would take 2068 batteries for this system

    Peukerts is designed for Lead Acid batteries. Applying it to LiOn will give vastly different results. Same idea, but the
    ratios are different.
    Powerfab top of pole PV mount | Listeroid 6/1 w/st5 gen head | XW6048 inverter/chgr | Iota 48V/15A charger | Morningstar 60A MPPT | 48V, 800A NiFe Battery (in series)| 15, Evergreen 205w "12V" PV array on pole | Midnight ePanel | Grundfos 10 SO5-9 with 3 wire Franklin Electric motor (1/2hp 240V 1ph ) on a timer for 3 hr noontime run - Runs off PV ||
    || Midnight Classic 200 | 10, Evergreen 200w in a 160VOC array ||
    || VEC1093 12V Charger | Maha C401 aa/aaa Charger | SureSine | Sunsaver MPPT 15A

    solar: http://tinyurl.com/LMR-Solar
    gen: http://tinyurl.com/LMR-Lister ,

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  • NorthGuy
    NorthGuy Solar Expert Posts: 1,913 ✭✭
    #5
    Re: Battery question
    Cariboocoot wrote: »
    Battery bank @ 50% DOD = 2334 Amp hours. Round up to nearest available size.

    2400.
    24 series of 100AH batteries.
    2 batteries per series.
    24 x 2 = 48

    Same result as mine :D

    24 series probably not going to work anyway
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  • Cariboocoot
    Cariboocoot Banned Posts: 17,615 ✭✭✭
    #6
    Re: Battery question
    NorthGuy wrote: »
    2400.
    24 series of 100AH batteries.
    2 batteries per series.
    24 x 2 = 48

    Same result as mine :D

    24 series probably not going to work anyway

    Uh, two 12 Volt 100 Amp hours in series to get 24 Volta & 100 Amp hours.
    Twenty-four such strings in parallel to bring the capacity up to 2400 Amp hours.

    It's that last one that's ludicrous; there is no way those batteries are going to share current equally no matter what.
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