Basic electrics calculations.

peterthepainter
peterthepainter Registered Users Posts: 3
I'm trying to understand an inherited Solar system..
I have solar panels on the roof, charging 6 2volt batteries connected in series, rated at 1200 amphours. 230v is then supplied to the house via an inverter.
So, an I correct that this gives me the equivalent of 1 12volt battery of 1200ah?
My basic understanding is that when the batteries are fully charged, I should have a THEORETICAL maximum of 12x1200 = 14400watt hours available. (Yes, I realise that I should never fully discharge the batteries and keeping them above 50% charge of higher is desirable). So does this IN THEORY mean that I could run a 230volt 1KW appliance for 14hrs?

Firstly, is the above true?
Secondly, obviously I'm not going to run my batteries flat. I believe, if I understand the meters correctly, that my usage is approximately 3.5 - 4 KWhrs / day. Is this a reasonable usage to expect such a system to cope with? (I'm away from the house at the moment and don't have details of the solar panels, but I believe that in normal circumstances they will produce the 4KW needed)

Thank you for your help.

Comments

  • mcgivor
    mcgivor Solar Expert Posts: 3,854 ✭✭✭✭✭✭
    Sometimes it's better to work with one rating, let's use Ah as an example, your battery is 1200Ah at 12 V, but in order to stay within 50% depth of discharge, in reality it is 600Ah. So a 1Kw load at 230V draws 4.34 A, the ratio of conversation is approximately 20:1, so multiply 4.34 by 20, this gives 86.8A, then divide 600Ah by 84.8, this would be 6.9 hours to 50% depth of discharge. Bare in mind this is a fast calculation, not talking into account losses in inversion, conductors and so forth, but gives an estimate.
    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.
  • Estragon
    Estragon Registered Users Posts: 4,496 ✭✭✭✭✭
    In addition to losses in battery charging, inverter, etc., some other factors involved in theoretical load run time:
    - battery voltage will drop as it discharges, so current will increase for a given load, increasing losses
    - deep cycle batteries are typically capacity rated at a 20-24 hour rate of discharge. Apparent capacity will be higher if discharged more slowly, and lower if discharged more quickly (as in your 14hr example).
    - not all loads are equal. A 1kw purely resistive load with a power factor near 1.0 will be close to expected run time. Other loads (eg motors) can have much lower power factors, and can require much more power vs "useful work" output.
    Off-grid.  
    Main daytime system ~4kw panels into 2xMNClassic150 370ah 48v bank 2xOutback 3548 inverter 120v + 240v autotransformer
    Night system ~1kw panels into 1xMNClassic150 700ah 12v bank morningstar 300w inverter
  • Photowhit
    Photowhit Solar Expert Posts: 6,002 ✭✭✭✭✭
    To expand on what @Estragon and @mcgivor have said, It is a system, and you need to be able to 'refuel' that top half of the battery. So your array should be properly sized to replace it and promptly. You can check solar insolation for your area to see how much sunlight you get on an average day. You are using 230V so not in the US... Some areas seem to get sun all day. deserts are often like this, but fixed panels, properly angled will only 'see' about 6 hours of sun. Most of the world averages about 4 hours over the year, less in winter, more in summer.

    Energy used during the day, uses less of the solar power generated! Storing power in a lead acid battery requires about 20% more energy than using the energy as it's produced. Yes! you use energy from the system, it can supply energy during the day while it is recharging.

    Your 600 amps at 12 volts, require replenishing. And the battery require some waste, think of it as putting back equal amps but requiring higher voltage to move the amps from the panels to the battery. Also panels rarely make as much energy as they are rated. A 100 watt panel, typically produces about 75 watts. So to replace the 600 amps, in a typical day(4 hours) the calculations would look something like this;

    Need 600 amps replaced
    extra energy means will need 600 x 1.2 = 720 amps
    720 amp hours at 12 volts(nominal), 720x12=8640 watt hours
    8640 watt hours divided by 4 hours per day =2135 watts coming from array
    (2135 watts needed from array) ÷ .75 = @ 2850 watts of array

    Basically 2850 watts of array to use the battery at 50%, understand that off grid solar Must be wasteful! You need to produce more energy than this to maintain proper battery health. Most people try to use the top 20% of their battery saving a reserve for cloudy days. Systems are designed to use the top 50% but usually they have an alternative means of charging, like a generator.

    So typically you would want a solar array capable of providing 10-13% of your battery capacity during full exposure. So a 1200 amphour 12 volt battery bank, would require an array that can produce about 120 - 165 amps. 

    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.
  • peterthepainter
    peterthepainter Registered Users Posts: 3
    Thanks Mcgivor, Estragon and Photowhit. Much to learn and understand here! I'm getting there ..and will no doubt be back asking more questions.
  • mike95490
    mike95490 Solar Expert Posts: 9,583 ✭✭✭✭✭
    edited February 2018 #6
    Sometimes it's easiest to convert everything to Watt Hours
    Sometimes to the units shown on your monitor gear.





    the following is a complex math construct:



     note = Amps is represented by the letter I (current)
    Volts x Amps = Watt

    The picture above shows Volts/Resistance = I (amps)
    I x R = V

    Sometimes V is also shown as E (electromotive force)


    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

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