To calculate the size of cable feeding the inverter, we need to know 1) the maximum amount of power being provided by the inverter 2) the efficiency of the inverter at configured values and 3) the voltage we are going to be feeding the inverter with.
The power being supplied by the inverter is equal to the input power derated by the inverter efficiency.
`P_"in"=(P_"out")/(Efficiency)`
We know we plan to provide a maximum of 220 VAC at 100 amps.
That gives us a wattage of 22,000 watts at the output of the inverter
The manufacturer should publish the efficiency of the inverter in its specs sheet. A typical efficiency falls between 85% and 95%
I will use 90% efficiency for this design.
I will also use 48 VDC for the input voltage. There are some inverters that go as high as 384 VDC input voltages, but that comes with another set of issues in the battery array. 48 VDC is a pretty common input voltage.
`P_"in" = (P_"out")/(Efficiency)`
`P_"in" = (22000 W)/.90`
`P_"in" = 24,444.44 W`
`I_"in" = (P_"in")/(V_"in")`
`I_"in" = (24,444,44 W)/(48 VDC)`
`I_"in" = 509.25 A`
The feed wire to the inverter will need to carry over 500 amps. This could be a single wire of 1000 MCM (about 1.32 inches in diameter) or multiple smaller wires. It is not uncommon to use the same wire as was used to feed the loadcenter. In this case you would use as many as would be needed to carry 509 amps or more.
Do not forget that this wire needs to be derated based on the length of the
run. So keep the runs short to save money and increase efficiency.
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