Ac cable drop calculation

[jjavier]

Hi

In PV systems, you must calculate the output current of the inverter *1.25, to select the correct wire. But when you are going to calculate voltage drops, Do we need to multiply the power of the inverter * 1.25 to do this calculation?

 

[GoldDigger]

I would not.
The Code does not get specific about either voltage drop requirements or calculations for most purposes (with fire pumps a notable exception.)

 

[ggunn]

Hi

In PV systems, you must calculate the output current of the inverter *1.25, to select the correct wire. But when you are going to calculate voltage drops, Do we need to multiply the power of the inverter * 1.25 to do this calculation?

No. To calculate Vd you use nominal current values, i.e., inverter rated output on the AC side and Vmp with no multiplier for excess insolation on the DC side.

 

[Carultch]

Hi

In PV systems, you must calculate the output current of the inverter *1.25, to select the correct wire. But when you are going to calculate voltage drops, Do we need to multiply the power of the inverter * 1.25 to do this calculation?

Voltage drop calculations are based off the current value directly on the datasheet for the AC side, and the nominal voltage. No 1.25 safety factor. The 1.25 safety factor is to account for the heat accumulation of continuous loads, when using a standard rated OCPD or other equipment that is not specifically rated as continuous duty, and is for ampacity requirements.

On the DC side, it is based off Vmp and Imp at standard test conditions.

On a power optimizer's system, it is based off nominal operating string voltage and the current that corresponds to the array's STC power at that voltage.

 

[jaggedben]

...On a power optimizer's system, it is based off nominal operating string voltage and the current that corresponds to the array's STC power at that voltage.

...or the max optimizer output current, whichever is less.

 

[310 BLAZE IT]

Hi

In PV systems, you must calculate the output current of the inverter *1.25, to select the correct wire. But when you are going to calculate voltage drops, Do we need to multiply the power of the inverter * 1.25 to do this calculation?

1.25 is not always required too see the code section you can just correct and adjust it for temperature and use instead. Also you can use a 100 percent rated breaker.

Sent from my SM-N920V using Tapatalk

 

[ggunn]

1.25 is not always required too see the code section you can just correct and adjust it for temperature and use instead. Also you can use a 100 percent rated breaker.

Sent from my SM-N920V using Tapatalk

Actually, I never multiply the inverter current by 1.25 for anything but the rating of the OCPD (I never use 100% rated breakers).

The rated output inverter current is Imax; The conditions of use derated 90 degree ampacity and the continuous use derated 75 degree ampacity must both be greater than Imax. This is true for both AC and DC circuits, except that the DC Imax is 1.25 times Isc, adjusted for the possibility of more current output than Isc if irradiance exceeds 1000W^m2.

I realize that the continuous use derating of the 75 degree ampacity is a multiply by 0.8 which is the reciprocal of 1.25, but to me establishing Imax and then treating AC and DC the same by derating conductor ampacity is a lot less confusing than trying to remember when to multiply by 1.25 and if once or twice.