Here is some background information about my project. My solar project is 45kW DC with three SE11400H, and one SE5000H SolarEdge inverters with 400-watt optimizers for 39.2kW on the AC side. I have 120 Mission Solar panels at 375 watts. My racking is an iron ridge ground mount and going to be 5 modules high in one row of approx 160ft.
My question is about the proper wire size for this project. My distance to the meter and service panel is about 400ft. According to my research I have done, the optimizers essentially condition to power by changing the voltage and match the current from the panels to hit the operating voltage of the inverters. The SE11400H is 400 volts, and SE5000H is 380 volts. My thinking is to have 3 separate strings of 12 panels for the SE11400H and 1 string of 12 panels for the SE5000H. Now I'm told to take the wattage of the string which is 12 panels x 375 watts = 4,500 watts and divide that by normal operating voltage of the inverter which is 400 or 380. Doing that by 380, I get 11.84 amps and 400, I get 11.25 amps. I'm rounding both of those numbers up to 12 and using a voltage drop calculator I get a 4.74% drop with 12awg and 2.98% with 10awg.
A. Am I doing the correct calculation for the wire size? B. Do I need to take anything else into consideration besides the cost of wire? C. If I went with 10AWG wire stranded, can I use that for all the wire hookups? For example; wiring from panels to the inverter and the wiring from the inverter to the main service panel?
Sorry for the long possibly confusing post, but any help would be greatly appreciated.
My question is about the proper wire size for this project. My distance to the meter and service panel is about 400ft. According to my research I have done, the optimizers essentially condition to power by changing the voltage and match the current from the panels to hit the operating voltage of the inverters. The SE11400H is 400 volts, and SE5000H is 380 volts. My thinking is to have 3 separate strings of 12 panels for the SE11400H and 1 string of 12 panels for the SE5000H. Now I'm told to take the wattage of the string which is 12 panels x 375 watts = 4,500 watts and divide that by normal operating voltage of the inverter which is 400 or 380. Doing that by 380, I get 11.84 amps and 400, I get 11.25 amps. I'm rounding both of those numbers up to 12 and using a voltage drop calculator I get a 4.74% drop with 12awg and 2.98% with 10awg.
A. Am I doing the correct calculation for the wire size? B. Do I need to take anything else into consideration besides the cost of wire? C. If I went with 10AWG wire stranded, can I use that for all the wire hookups? For example; wiring from panels to the inverter and the wiring from the inverter to the main service panel?
Sorry for the long possibly confusing post, but any help would be greatly appreciated.
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