Fredy Posted February 6, 2014 Share Posted February 6, 2014 When i have more than 1 Inverter Type in 2 sub-fields how do you calculate the Ohmic wiring resistance for each Sub-field.This Solar system does not have main boxes the strings are directly connected to the invertersThis gives me an extremely low ohmic loss if i calculate it with the parallel of all strings like 0.005 mOhmIf I calculate the equivalent resistance of the average of strings per inverter, the result is around 7 mOhm.But if I put 7 mOhm in both sub systems the one with more inverters gives me more loss Fraction (%)1. How do you relate the resistance with the number of inverters?2. Does the global wiring resistance refers to the equivalent resistance of the parallel of strings in just one inverter or does it refers to the equivalent resistance of all the inverters in the system Link to comment Share on other sites More sharing options...
André Mermoud Posted February 7, 2014 Share Posted February 7, 2014 The wiring resistance is defined for each sub-array separately. It is the equivalent resistance of all wires, as seen from the input of all inverters of the sub-array in parallel. You have just to put all the string resistances in parallel, and add the resistances of all connexions from the roof junction boxes to the inverter inputs. This is straightforward for any electrician engineer. The tool "Detailed computation" helps you for this calculation. Now the resistance is around 5.5 mohm/m for 4 mm2, and 8.8 mohm/m for 4 mm2 (listed in PVsyst, button "Wires"). As an example, If you have 40 m of 4mm2 cable for one string, this will represent 220 mOhm/string. I.e. 22 mOhm for 10 strings in parallel.I don't see how you can get 0.005 mOhm except for a multi-100 MWp instalation (this would represent 44'000 strings). Link to comment Share on other sites More sharing options...
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