Calculating the maximum watt output for a 3 phase 480v
Is there a formula or way to calculate the maximum watts output from a PV System if I know the service is 3 Phase 480v and the max AWG size is 350 (using 310amp copper feeder)?
Assuming the the question can be restated "What is the largest PV system inverter that can be connected...."
For a 3-phase system:
Power = Volts x Amps x PF x 1.73
PF = power Factor, but this is 1.00 or .99 for PV inverters at full power
You can only load the cable to 80% or less depending on temperature. Assuming high temperature is 96-104°F you can use the above.
Looks like 206KW out of the inverter. The PV array can be larger.
The 2011 NEC 705.12(A) for supply side interconnections limits the inverter overcurrent protection to the rating of the service from the utility. As an example, if the service is 400A then your inverter AC overcurrent protection is limited to 400A. This will limit the inverter output to 400/1.25 = 320A and the inverter power rating at 480V 3ph to 266kW. If you have a load side interconnection then your output will be limited by the existing service entrance overcurrent protection, which in this example would also be 400A.
If you are still on the 2008 NEC I would still go by these limits.