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  • Feenalicious
    Junior Member
    • Jul 2015
    • 17

    #16
    Details of my system:

    Here are the components I have, including specs:

    4 ea Renogy 100W 12V Monocrystalline Solar Panels
    --Max Power: 100W, Max System Voltage: 600V DC (UL), Optimum Operating Voltage (Vmp): 18.9V, Open-Circuit Voltage (Voc): 22.5V, Optimum Operating Current (Imp): 5.29A, Short-Circuit Current (Isc): 5.75A, Maximum Series Fuse Rating: 15A

    4 ea Super Start Marine Batteries, 12 volt, 105ah, 210 reserve capacity

    1 ea Renogy 40A MPPT Solar Panel Charge Regulator
    --Nominal System Voltage: 12V/24V, Auto recognition, Max. PV Input Power: 400W (12V), 800W (24V), Rated Charge Current: 40A, Rated Discharge Current: 20A, Max. Solar Input Voltage: 100VDC, Self-consumption: <10mA (24V), Maximum Battery Voltage: 32V, Charge Circuit Voltage Drop: ≤0.26V, Discharge Circuit Voltage Drop: ≤0.15V

    1 ea Nature Power 38320 2000V Pure Sine Wave Inverter
    --Power (Continuous): 2000W, Power (Peak): 4000W, Voltage: 120Vac, Current: 16.6A, Frequency: 60±1Hz, Waveform: True Sinewave, DC Output: 5V USB

    This solar system is for powering an off-grid RV.
    Last edited by Feenalicious; 07-15-2015, 07:10 AM. Reason: Amp hours corrected

    Comment

    • sensij
      Solar Fanatic
      • Sep 2014
      • 5074

      #17
      Originally posted by Feenalicious
      I see, I'm not utilizing a key benefit of the MPPT controller, the higher voltage I can get by putting the panels in series, right?
      Yes. MPPT will help even if you stick with parallel, because there is still some PWM loss between the typical panel voltage (17 V) and the battery voltage (~14 V). But if your CC input voltage range is large enough to allow series-parallel, or even entirely series, you'll save money on the wiring and fuses, and should have less transmission loss too.

      However, your brief description of the charge controller said it can handle 12 or 24 V input. That actually sounds more like PWM than MPPT. Do you have a model number or anything?
      CS6P-260P/SE3000 - http://tiny.cc/ed5ozx

      Comment

      • Feenalicious
        Junior Member
        • Jul 2015
        • 17

        #18
        Product code for charger

        Originally posted by sensij
        Yes. MPPT will help even if you stick with parallel, because there is still some PWM loss between the typical panel voltage (17 V) and the battery voltage (~14 V). But if your CC input voltage range is large enough to allow series-parallel, or even entirely series, you'll save money on the wiring and fuses, and should have less transmission loss too.

        However, your brief description of the charge controller said it can handle 12 or 24 V input. That actually sounds more like PWM than MPPT. Do you have a model number or anything?
        PRODUCT CODE: CTRL-MPPT40 http://www.renogy-store.com/40-Amp-M...trl-mppt40.htm

        Comment

        • sensij
          Solar Fanatic
          • Sep 2014
          • 5074

          #19
          According to the manual for that CC, you can run 400 W in parallel on a 12 V system at just under 94% efficiency, or in series-parallel at just over 94%. If you go all 4 in series, efficiency would drop to under 92.5%, probably costing more than you're getting in reduced transmission losses.

          So, with this, I would suggest series-parallel on the array, which lets you cut out the fuses you originally asked about.

          If you eventually go to a 24 V battery bank, your CC efficiency could improve to around 97.5%, if you stayed with 400 W in series-parallel.
          CS6P-260P/SE3000 - http://tiny.cc/ed5ozx

          Comment

          • posplayr
            Solar Fanatic
            • Jun 2015
            • 207

            #20
            For OP these huge wires are the reason you would rather have a 24 or 48 volt input to your invertor.

            The power losses go up in proportion to current squared requiring wire diameters to go up in proportion. In other words a 24V system would use 1/4 the wire diameter or be able to go four times are far (not limited to 5ft).

            At 48 volts you would have a factor of 16.

            Comment

            • sensij
              Solar Fanatic
              • Sep 2014
              • 5074

              #21
              Originally posted by thastinger
              <Snip> A 600W inverter is plenty. Whatever you decide, please understand that your 2000W inverter would be pulling 153A through the cables and understand the risks associated with that.
              Originally posted by Feenalicious
              I was originally planning a larger system when there were two people living here and bought the inverter then. Didn't realize these repercussions from keeping the 2000V inverter. I actually have a 650W inverter I can use.
              Originally posted by posplayr
              For OP these huge wires are the reason you would rather have a 24 or 48 volt input to your invertor.

              The power losses go up in proportion to current squared requiring wire diameters to go up in proportion. In other words a 24V system would use 1/4 the wire diameter or be able to go four times are far (not limited to 5ft).

              At 48 volts you would have a factor of 16.
              Hopefully the OP follows through with using a 650 W inverter with the 12 V system, or fuses the 2000 W down to a current that can be handled safely.
              CS6P-260P/SE3000 - http://tiny.cc/ed5ozx

              Comment

              • posplayr
                Solar Fanatic
                • Jun 2015
                • 207

                #22
                Originally posted by sensij
                Hopefully the OP follows through with using a 650 W inverter with the 12 V system, or fuses the 2000 W down to a current that can be handled safely.

                Not sure what happned to my other post, so here is a retry. Manual says needs no more than 5 ft of #2/0 wire

                DC Input Cable:
                • Low resistance wire (copper) is required for all the DC connections between the inverter and
                the battery bank.
                • For 38210, 38310, uses minimum #2 AWG wire with maximum cable length of 5 feet.
                • For 38215, use minimum #1/0 AWG wire with maximum cable length of 5 feet.
                For 38320, use minimum #2/0 AWG wire with a maximum cable length of 5 feet.

                Comment

                • SunEagle
                  Super Moderator
                  • Oct 2012
                  • 15159

                  #23
                  Originally posted by posplayr
                  Not sure what happned to my other post, so here is a retry. Manual says needs no more than 5 ft of #2/0 wire
                  That is the problem with using high wattage inverters on a 12volt system. The potential current draw is very large requiring big wire and fusing.

                  I did a similar fubar when I purchased a 2500/5000w inverter. I used wiring capable of handling 200amp and fused it to 150a before I really understood my dilemma of draining my small battery 200Ah system.

                  I now have a 600 watt pure sine wave inverter on the system so I kept the wire but downsized the fuse to 60amp.

                  Comment

                  • Feenalicious
                    Junior Member
                    • Jul 2015
                    • 17

                    #24
                    design reconsidered

                    Originally posted by thastinger
                    OP, you should really reconsider your system design here. Your inverter is way to big and, assuming your batteries are 12V 100AH, and you discharged them at a C10 rate, a 500W load is all you could run. If you tried to run a much larger load, the inverter will likely shut down due to voltage sag from the battery bank (unless these are AGM batteries). If they are 100AH batteries, you'll have a 400A connection, you won't be able to terminate that to a 12V battery post without increasing the risk of fire/spark. A 600W inverter is plenty. Whatever you decide, please understand that your 2000W inverter would be pulling 153A through the cables and understand the risks associated with that.
                    Much thanks, Thastinger. Yes, of course, I will be using the 650 Watt Inverter. Unfortunately, I started researching this solar project three months ago then shelved it until now. I've obviously forgotten most of what I learned. Can I trouble you to show me your calculations that gave you the 500W load number and the 153A passing through the cables? I need to understand this completely and that will give me a little "jump start" back into it.

                    Comment

                    • Feenalicious
                      Junior Member
                      • Jul 2015
                      • 17

                      #25
                      &quot;Fuses down&quot;

                      Originally posted by sensij
                      Hopefully the OP follows through with using a 650 W inverter with the 12 V system, or fuses the 2000 W down to a current that can be handled safely.
                      Yes, switching to the 650W. Just wondering, what do you mean by "fuses the 2000W down...?"

                      Comment

                      • Feenalicious
                        Junior Member
                        • Jul 2015
                        • 17

                        #26
                        Originally posted by posplayr
                        Not sure what happned to my other post, so here is a retry. Manual says needs no more than 5 ft of #2/0 wire
                        Yes, this much I got right. The inverter location is less than 5 ft from the battery bank and I bought the correct gauge wire. Thanks for checking.

                        Comment

                        • SunEagle
                          Super Moderator
                          • Oct 2012
                          • 15159

                          #27
                          Originally posted by Feenalicious
                          Yes, switching to the 650W. Just wondering, what do you mean by "fuses the 2000W down...?"
                          By "fusing down" the 2000 watt inverter you are limiting the amount of power it can draw because the fuse will open when too much current goes through the wire.

                          So technically you are converting that 2000 watt (2000w / 12volts = 167amps) to a 500 watt (500w / 12volt = 42amps) by putting in a 40amp fuse instead of a 160amp fuse.

                          Blowing fuses is not the best way to run a system (will eventually cause the wire insulation to fail) but it will limit the amount your draw with that 2000w inverter.

                          Comment

                          • Feenalicious
                            Junior Member
                            • Jul 2015
                            • 17

                            #28
                            fusing down

                            Originally posted by SunEagle
                            By "fusing down" the 2000 watt inverter you are limiting the amount of power it can draw because the fuse will open when too much current goes through the wire.

                            So technically you are converting that 2000 watt (2000w / 12volts = 167amps) to a 500 watt (500w / 12volt = 42amps) by putting in a 40amp fuse instead of a 160amp fuse.

                            Blowing fuses is not the best way to run a system (will eventually cause the wire insulation to fail) but it will limit the amount your draw with that 2000w inverter.
                            Lol, gotcha. Thanks.

                            Comment

                            • thastinger
                              Solar Fanatic
                              • Oct 2012
                              • 804

                              #29
                              Originally posted by Feenalicious
                              Much thanks, Thastinger. Yes, of course, I will be using the 650 Watt Inverter. Unfortunately, I started researching this solar project three months ago then shelved it until now. I've obviously forgotten most of what I learned. Can I trouble you to show me your calculations that gave you the 500W load number and the 153A passing through the cables? I need to understand this completely and that will give me a little "jump start" back into it.
                              Sure, you have a 12V 400AH battery bank. Following the guide of C8/C10 max discharge rate we get 400/10=40A which can be safely pulled from the battery bank or 50Amps with the C8 rate. So 40 amps at 12 volts is 480 Watts and 50 Amps at 12V is 600W.
                              The smaller inverter should run more efficiently at those levels than the 2000W fused down.

                              So, if you decided to stick to the no more than 20% depth of discharge on the battery bank, what could you do with that much power? Let's use 2 100W light bulbs to figure it out. You could drain 80 Amps from the bank assuming it was fully charged, 80Ax12V=960W. Let's assume your inverter is 92% efficient, so 960x.92=880W of useable power in the evening after the sun goes down. So you could run those 2 light bulbs for a bit over 4 hours. Expectation management is my point, it is a small system so be careful to manage your loads.
                              1150W, Midnite Classic 200, Cotek PSW, 8 T-605s

                              Comment

                              • sensij
                                Solar Fanatic
                                • Sep 2014
                                • 5074

                                #30
                                I wasn't really suggesting that using the 2000 W inverter with a smaller fuse was a good idea, just that it was better than using it unfused. With the wire sized for the full inverter rating, the insulation should hold up OK, but limiting the power reduces the risk of a problem at the terminations, which are hard to make properly for those who don't do it regularly (and have the right tools). The fuse isn't necessary, you could just pinkie swear never to load the inverter more than 600 W (or whatever), but the smaller inverter will be a better choice for a number of reasons (as others have said).

                                With respect to discharge rate, keep in mind that if you discharge at C/10 (40 amps = 480 W), that is a 10 hour rate, while the battery Ah rating is usually published at the C/20 rate (20 amps = 240 W). Discharging at the 10 hour rate may reduce the effective capacity somewhat. Based on the specs you shared, I think it might be this battery. The reserve capacity is the number of minutes a 25 A load can be maintained (100 A for four in parallel = 1200 W), a C/4 rate. 210 min = 3.5 hour * 25 A = 87.5 Ah per battery. You could figure C/10 capacity would be something less than 105 Ah, but more than 87.5 Ah.

                                Also be aware that it is difficult to wire four batteries in parallel and keep them perfectly balanced. If you must stay in this configuration, my very unqualified recommendation would be to rotate their position in the bank periodically so whatever mismatches exist in the wiring get distributed among all four of them over time, like rotating tires. I'm not going to defend that recommendation if others with more knowledge shoot it down.
                                CS6P-260P/SE3000 - http://tiny.cc/ed5ozx

                                Comment

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