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  • solar system design

    i want to design a off grid solar system to run a load of 450W.
    the tech specs of module i want to use is as follows

    Power (W) ------------------------ 270 W
    Max. Power Voltage (V)------------- 36.72
    Max. Power Current (A)------------- 7.45
    VOC Voltage (V)-------------------- 44.49
    SC Current (A)---------------------- 7.95

    Please suggest me
    1.battery size (for 2 days autonomy)
    2.charge controller size ( using Steca germany products)
    3.inverter size ( using studer germany products)
    4.Number of modules in parallel and series
    5.what would be the PV plant size

    i already designed this system but i feel every time i do some minor mistake which make my system oversize........
    So need inputs from experts......
    Thanks

  • #2
    All you have given is the size of the size of the panel. None of your questions can be answered with that info. First thing we need to know is how many watt hours per day, and load demand calculations...

    Second you cannot getaway with only 2-day autonomy, that would wear your batteries out in a very short time.
    MSEE, PE

    Comment


    • #3
      Sunking,

      He gave you the load -- 450 watts. I doubt he means 450 watts continuous, but that's the load he gave.

      450 watts * 24 = 10,800 watt-hours AC per day.
      Derate by 55%, 10.8 KWh / 0.55 = 19.6 KWh DC per day.

      Two days autonomy = 39.2KWh to 50% DoD, 39.2 / 0.50 = 78.4KWh total. That's the battery bank size.

      Assume 48 volt inverters.

      78.4KWh / 48 volts =~ 1,600Ah

      Now for the array, assuming 2.5 hours minimum winter insolation (just 'cuz and he didn't say ...)

      19,200 watt-hours / 2.5 hours = 7,680 watts.

      Panels are 270 watts each, 29 panels needed. They are high Voc panels, so unless he lives somewhere that is always warm, that's 15 strings of 2 to stay below the common 150Voc limit. Don't know about the gear he spec'd and ain't lookin' 'cuz he ain't payin'.

      30 * 270 = 8,100 watts.

      Back calculate the peak charging current at 57.6 volts for start-of-absorb --

      8100 watts / 57.6 volts = 140 amps

      Now for rate of charge --

      1600 / 140 = C / 11.4, which is plenty fine.

      Enjoy!
      Last edited by greenHouse; 12-15-2010, 10:38 AM. Reason: Show more work, make fewer mistakes.
      Julie in Texas

      Comment


      • #4
        Originally posted by greenHouse View Post
        Sunking, He gave you the load -- 450 watts.
        Skipped right over it and missed it.
        MSEE, PE

        Comment


        • #5
          Originally posted by Sunking View Post
          Skipped right over it and missed it.
          You didn't miss much -- I suspect he's wrong, as you can see from the required battery size!
          Julie in Texas

          Comment


          • #6
            Originally posted by vikas maurya View Post
            i want to design a off grid solar system to run a load of 450W.
            Watts without time, well, it was guessed at the load being on always. 24 hours a day.

            Like the old math problem, if 2 trains are travelling toward each other, one at 12 mph and the other at 38 mph, how far apart are they after 10 minutes ?

            we have to guess at the missing part.
            Powerfab top of pole PV mount (2) | 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

            Comment


            • #7
              Originally posted by Mike90250 View Post
              Watts without time, well, it was guessed at the load being on always. 24 hours a day.

              Like the old math problem, if 2 trains are travelling toward each other, one at 12 mph and the other at 38 mph, how far apart are they after 10 minutes ?

              we have to guess at the missing part.
              8 1/3rd miles less than before, unless they were less than 8 1/3rd miles apart?

              Are they on the same track? Are they ghost trains? Do they have to slow down for pedestrians playing on the tracks? And why is one only going 12 mph?
              Julie in Texas

              Comment


              • #8
                Originally posted by Mike90250 View Post
                Watts without time, well, it was guessed at the load being on always. 24 hours a day.

                Like the old math problem, if 2 trains are travelling toward each other, one at 12 mph and the other at 38 mph, how far apart are they after 10 minutes ?

                we have to guess at the missing part.
                ya mike sorry i forget to give time.
                I want to design for 5.5 hours.
                Total load 450 W*5.5 h = 2475 Wh
                so Total energy demand per day is 2475 Wh.
                Now pls go ahead

                And for battery sizing take DOD 70% and battery efficiency 85%
                sunshine availability in India is about 5.5 Hours

                Comment


                • #9
                  Originally posted by vikas maurya View Post
                  ya mike sorry i forget to give time.
                  I want to design for 5.5 hours.
                  Total load 450 W*5.5 h = 2475 Wh
                  so Total energy demand per day is 2475 Wh.
                  Now pls go ahead

                  And for battery sizing take DOD 70% and battery efficiency 85%
                  sunshine availability in India is about 5.5 Hours
                  Where can I find me an 85% efficient battery?!? I want one!
                  Julie in Texas

                  Comment


                  • #10
                    Originally posted by vikas maurya View Post
                    ya mike sorry i forget to give time.
                    I want to design for 5.5 hours.
                    Total load 450 W*5.5 h = 2475 Wh
                    so Total energy demand per day is 2475 Wh.
                    Now pls go ahead

                    And for battery sizing take DOD 70% and battery efficiency 85%
                    sunshine availability in India is about 5.5 Hours
                    In post #3, Julie did all the math for you, you can just insert your new #'s into her calculations, and see what comes out.
                    or just divide the results by 4.3, and you would be close.
                    Powerfab top of pole PV mount (2) | 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

                    Comment


                    • #11
                      Try this battery, Look up in your broswer:

                      HUP Solar-One Battery

                      Comment


                      • #12
                        Originally posted by DeltaFox 25 View Post
                        Try this battery, Look up in your broswer:

                        HUP Solar-One Battery
                        Find me where it says the full-cycle efficiency, as measured in watt-hours, is 85% or better and I'll be impressed.

                        Here's the problem -- something called "activation energy". The activation energy is additional energy needed for a reaction to occur at all. You see it when charging a battery as the easy increase in voltage that doesn't result in any charging current.

                        When you take volts * amps, you get watts. And the "volts in" includes that activation energy, while the "volts out" doesn't. Added to that, the typical amp-hour efficiency is less than 95%, so you are losing both amp-hours and volts on a full-cycle. Finally, a full-cycle charge requires considerable amounts of energy to complete the last 10 to 15 percent of the charge (depending on current at the time the battery reaches the "Absorb" voltage). This charge results in both gassing and heat. All of the heat is waste, as is all of the energy that goes to gassing.
                        Julie in Texas

                        Comment


                        • #13
                          Originally posted by vikas maurya View Post
                          And for battery sizing take DOD 70% and battery efficiency 85% sunshine availability in India is about 5.5 Hours
                          Two problems here. Unless this is a AGM battery, efficiency is around 80% at best. You say 5.5 hours? Is that average, peak summer, or low winter. In a battery system you have to design for worse case or the shortest Sun Hour day in the year. Very few places receive 5.5 hour in winter or rainy seasons.
                          MSEE, PE

                          Comment


                          • #14
                            Originally posted by vikas maurya View Post
                            I want to design for 5.5 hours.
                            Total load 450 W*5.5 h = 2475 Wh
                            so Total energy demand per day is 2475 Wh.
                            Now pls go ahead

                            And for battery sizing take DOD 70% and battery efficiency 85%
                            sunshine availability in India is about 5.5 Hours
                            Solar panel Wattage = [2475 x 1.5] / 5.5 hours = 675 watts if using MPPT controller and you actually get 5.5 Sun Hours which I doubt in winter.

                            No way should even consider taking a battery to 70% DOD, that is suicide, your battery would be worthless in a few short months, and just 1 cloudy day and you go dark. Try 20% DOD or 5 day reserve which is equivalent to 2.5 days real reserve to 50% DOD. You did not say what voltage you wanted the battery, or was it 48 volts?

                            At 48 volts the battery AH capacity needed is [2475 wh x 5 days] / 48 volts = 257 AH @ 48 volts.

                            Charge controller size assuming MPPT is Panel wattage / nominal battery voltage = 675 watts / 48 volts = 14 amps minimum.
                            MSEE, PE

                            Comment


                            • #15
                              SunKing,

                              You're taking DoD as a per-day figure. It's 70% DoD after 2 days no-sun (see 2 days autonomy up above), or 35% DoD per day. A bit steep, but not total suicide. Not that he's told us minimum insolation either, or that 3 days is enough total storage to stone-cold-dead.

                              There are technologies that will handle very deep cycles. Many of the larger traction batteries will do 50% DoD on a daily basis and last for years. The HuP Solar-One batteries are rated 2,100 cycles to 80% DoD. 2100 / 365 is several years. It's all about plate chemistry, thickness and the rock well.
                              Julie in Texas

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