Newbie orientation

john787john787 Registered Users Posts: 3
Hello, I'm looking into setting a Off Grid Solar power system with a 7,000 Watt 24 Volt 240VAC inverter. My intent is to run the system 12 hours. Could you guys help me calculate how many batteries would I need?

Thanks in advance,
John

Comments

  • mcgivormcgivor Solar Expert Posts: 3,854 ✭✭✭✭✭✭
    You need to provide detailed information about  the loads before anyone can provide somewhat accurate feedback. Your geographic location would be needed to calculate the PV needed to change the battery which needs to support the loads.
    1500W, 6× Schutten 250W Poly panels , Schneider MPPT 60 150 CC, Schneider SW 2524 inverter, 400Ah LFP 24V nominal battery with Battery Bodyguard BMS 
    Second system 1890W  3 × 300W No name brand poly, 3×330 Sunsolar Poly panels, Morningstar TS 60 PWM controller, no name 2000W inverter 400Ah LFP 24V nominal battery with Daly BMS, used for water pumping and day time air conditioning.  
    5Kw Yanmar clone single cylinder air cooled diesel generator for rare emergency charging and welding.
  • john787john787 Registered Users Posts: 3
    The system would be for a house with 2 refrigerators and 3 Inverter A/Cs. I would say 6000 watts approximately. I live in Puerto Rico.

    Thanks,
    John
  • PhotowhitPhotowhit Solar Expert Posts: 5,658 ✭✭✭✭✭
    john787 said:
    The system would be for a house with 2 refrigerators and 3 Inverter A/Cs. I would say 6000 watts approximately. I live in Puerto Rico. 
    Really hard to say without a realistic idea of how much power would be used. A system that can run 6-7000 watts shouldn't have to run 6-7000 watts all the time.

    Is your goal just a battery backup? ...or do you want a solar electric system to run your home all the time?

    There are several off grid solar electric use calculators online, here's one;

    https://www.solar-electric.com/solar/calc

    Another way to figure out your use, would be to use a Kill-A-Watt meter and plug each item into it and let it run for a couple days. This will allow you to get a realistic idea of the energy used over time (watt hours).

    https://www.walmart.com/ip/P3-Internatoinal-P4400-Kill-A-Watt-Energy-Monitor/14282370


    Home system 4000 watt (Evergreen) array standing, with 2 Midnite Classic Lites,  Midnite E-panel, Magnum MS4024, Prosine 1800(now backup) and Exeltech 1100(former backup...lol), 660 ah 24v Forklift battery(now 10 years old). Off grid for 20 years (if I include 8 months on a bicycle).
    - Assorted other systems, pieces and to many panels in the closet to not do more projects.
  • BB.BB. Super Moderators, Administrators Posts: 32,027 admin
    Welcome to the forum John,

    Your best place to start is conservation and choosing "solar friendly" loads. A/C and refrigerators are really large loads for a solar power system. To give you the math using our rules of thumbs (gives you a relatively optimum/cost effective/reliable system design). It will not be small/cheap to run multiple large A/C and other loads. Picking "inverter" type A/C systems helps as wall as Inverter type or Linear Compressor based refrigerators.

    To give you the math... Lets assume that your averages loads are 1/2 the inverter capacity or 3,500 Watts. A standard Flooded Cell Lead Acid battery bank design would be for 2 days storage and 50% planned discharge (for longer battery life). Such a bank would be;
    • Total daily load (guessing): 3,500 Watt average load * 12 hours per day = 42,000 WH per day (or 42 kWH per day)
    • 42 kWH per day * 30 days per month = 1,260 kWH per month (if you had a utilty bill to compare: a good amount of energy usage)
    • 3,500 Watt average load * 12 hours per day * 1/0.85 inverter eff * 2 days storage * 1/0.50 max discharge * 1/24 volt battery bank = 8,235 AH @ 24 volt battery bank
    That is a huge battery bank, and for various reasons, going to a 48 volt battery bank would be suggested:
    • 3,500 Watt load * 12 hours per day * 1/0.85 inverter eff * 2 days storage * 1/0.50 max discharge * 1/48 volt battery bank = 4,118 AH @ 48 volt battery bank
    Both battery banks store the same amount of energy, but the wiring and charge controllers will be more "cost effective" with the higher bank voltage (Power = Voltage * Current, 2x voltage means 1/2 current--smaller wires, more Wattage from charge controllers).

    There are two calculators for solar arrays... One is based on on battery bank capacity (larger bank, more charging current/power needed). 5% can work for weekend/sunny weather systems... 10%-13%+ suggested for full time off grid usage:
    • 4,118 AH * 58 volts charging * 1/0.77 panel+controller deratings * 0.05 rate of charge = 15,509 Watt array minimum
    • 4,118 AH * 58 volts charging * 1/0.77 panel+controller deratings * 0.10 rate of charge = 31,019 Watt array nominal
    • 4,118 AH * 58 volts charging * 1/0.77 panel+controller deratings * 0.13 rate of charge = 40,324 Watt array "typical" cost effective maximum
    And sizing the array based on your location (hours of sun per day), your daily loads, and your seasonal loads (summer A/C, etc.). Assuming you are around San Juan, PR. Fixed array, facing south:
    http://www.solarelectricityhandbook.com/solar-irradiance.html

    San Juan
    Average Solar Insolation figures

    Measured in kWh/m2/day onto a solar panel set at a 72° angle from vertical:
    (For best year-round performance)

    JanFebMarAprMayJun
    5.43
     
    5.88
     
    6.29
     
    6.13
     
    6.00
     
    6.49
     
    JulAugSepOctNovDec
    6.38
     
    5.86
     
    5.71
     
    5.69
     
    5.25
     
    5.29
     

    Lets pick 5.29 hours of sun per day (December)... Looks like A/C and dehumidification may be a 365 day a year operation:
    • 42,000 WH per day * 1/0.52 off grid AC system efficiency * 1/5.29 hours of sun per day = 15,268 Watt array "December break even"
    Realistically, you have some choices here... If you "required" the A/C to always run 12 hours per day--Then you should only plan on using ~50% to 65% of "prectied daily output". If you can turn A/C off or least reduce usage during winter, bad weather, and/or less than sunny weather--Perhaps you could get away without an even larger solar array. For example a 65% base load would mean:
    • 15,268 Watt array * 1/0.65 base load "fudge factor" = 23,489 Watt array (65% base load design)
    You have lots of sun in PR... So your 10% rate of charge array of 31,019 Watts is already much larger than even your 65% base load array of 23,019 Watt Array...

    Lead Acid battery manufacturers do recommend a 10% minimum rate of charge...

    The above is a stake in the sand (a very expensive stake)... My assumptions/guesses on energy usage and location are just guesses. Your details can change that... And here are other batteries types (like LiFePO4 or others) that can change the above assumptions.

    And conservation (wall and roof insulation, awnings on windows, etc.), using mini-split vs standard central air (modern mini-splits can be very energy efficient vs ducted central A/C) all can help.

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
  • john787john787 Registered Users Posts: 3
    Thank you Photowhit and Bill for your responses. Sorry for not giving all the correct information, I'm completely out of my element here. The idea indeed is to use this a backup. Power in Puerto Rico is very unstable and we are prone to hurricanes every year so the idea is to have this battery system that could give me power at night (12 hours) if I were to lose power.
  • BB.BB. Super Moderators, Administrators Posts: 32,027 admin
    John, you have told us pretty much everything except how many Watt*Hours or kWH per day you want to use...

    If you are already there, looking at your electricity bill is a good place to start.

    And, you can (generally) take a look at your meter at the times you want to run your loads (8pm to 8am) and see what was recorded on your utility meter.

    And remember, that if you only run things like A/C, refrigerators, etc. 12 hours per day--There is heat that builds up during the day... So when you first turn the stuff on, it may take several hours of running to bring the (fridge/room) temperatures down.

    If you have central air--Looking at Mini-Splits can save you lots of energy--Just cool the bedroom(s) and living room... And let the rest of the place got hot (when power is out).

    Also, if you do not have lots of insulation yet--And/or an older central air system (or older window units), reviewing what you can do for energy conservation will not only let you use a smaller solar power system (and less expensive), but also save you money on your utility bill the rest of the time.

    Other things can also drive up utility bills. Older filament lamps, a big desk top computer, older TV, older refrigerators/freezers, etc. -- They generate heat inside the building--And A/C has to move that heat outside too (you spend extra money running older inefficient appliances, and more money again to move the heat outside).

    A typical North American home uses around 300-500 kWH per month. Add electric stove and some AC, upwards of 1,000 kWH per month. A/C in Texas in a large home, upwards of 3,000 kWH per month.

    3,000 WH per month = 3,000,000 WH per month
    3,000,000 per month / 30 day pwer month = 30,000 WH per day
    A medium sized solar power system is around 3,300 WH per day (typical small home/cabin, using propane/gas for heating, cooking, hot water).

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
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