Sander Posted January 10, 2019 Report Share Posted January 10, 2019 Hi, I'm currently looking into a simulation with microinverters. Not having used them before I was interested in calculating the cable losses. I expected high losses on the AC side. (from counter(meter) to microinverters) With cable lenghts easily exceeding the 25+ meter mark, the losses are extremely low. The DC losses however seem to be too high. Am I missing something or are the calculations switched? * using the cable tab to caculate the losses in detail Best regards, Sander Quote Link to comment Share on other sites More sharing options...
developer_mh Posted January 10, 2019 Report Share Posted January 10, 2019 Hi Sander, could you provide us with a project file so that we can investigate your test case better? You can either post it here or send it over by private message. Thanks a lot, Martin Quote Link to comment Share on other sites More sharing options...
Sander Posted January 10, 2019 Author Report Share Posted January 10, 2019 Hi, Included you will find my test project. Sander Enphase test.pvprj Quote Link to comment Share on other sites More sharing options...
developer_mh Posted January 10, 2019 Report Share Posted January 10, 2019 Thanks for the project. For me, this looks fine. The nominal output current of the Enphase inverter is max. 1.5 A (350 W / 230 V), usually it will be far less since you only have the 320 Wp from the module when the irradiance is at 1000 W/m². But even if you have 1.5 A output from the inverter, your ohmic power loss is only about 0.4 W (P = I² x R). The resistance of your cable is 0.18 Ω ( ρ*l/A, 0.018 Ω·mm²/m * 25m / 2.5mm²). So the relative losses on the AC side in your worst case are 0.4 W / 350 W = 0.12 %. In the DC side, you have 9.6 A MPP current (under STC, like above) and a cable resistance of 0.018 Ω·mm²/m * 2m / 4mm² = 0.009 Ω, so a power loss of 0.83 W (or relative 0.83/320 = 0.36%). So, when comparing ohmic losses, be sure to always have an eye on the currents. Cheers, Martin Quote Link to comment Share on other sites More sharing options...
Sander Posted January 11, 2019 Author Report Share Posted January 11, 2019 Agreed Martin, but thats just the AC loss for one inverter, with the included 25m cable lenght. (Your formula should however use the double lenght if i'm correct.) In this case we have 8 parallel microinverters. The biggest problem however is that we already have loss on the AC coupling cable between the inverters we will have to take into account, when sizing the AC kabel. (I noticed I should use 0m lenght for the DC kabel, following the instructions in PV*Sol which i will do next time) Quote Link to comment Share on other sites More sharing options...
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