Modeling cable losses on AC side for multiple inverters

[Rob]

How to model a situation where several inverters are connected to one distribution board and then share a long AC cable run to the meter?

I have 300m AC cable run and the software is simulation a seperate AC cable for each inverter back to the meter (an AC cable for each inverter back to the meter).

I would like to simulete: several Inverters-short AC cable from each of them (95mm2, 5 m)-DIstrubution board- Long Distance (300 mm2, 300m) - Meter

 

[developer_mh]

Hi Rob,

this is not possible right now in PV*SOL, but we already have this feature request on our list. What I would recommend right now is to model the cable for each inverter as if it was part of the long distance cable. As an approximation this is possble if your inverters roughly have the same power.

As the loss power is calculated with P = I_big² · R_big, and R_big = ρ · l / A_big. So through your big cable with cross section A_big you have the current I_big, which is the sum of the currents of your individual inverters. If we suppose that all inverter currents are the same at any time (which they aren't, but we have to assume it as an approximation), we can say that I_big = n · I_inv, with n being the number of inverters. So the cross section of each individual inverter cable would have to be divided by n in order to obtain approximately the same result.

Example:

You have one big cable of 300m length with 300mm² cable with 4 inverters connected to it. This is roughly the same, in terms of power loss, as 4 individual cables with 75 mm² (300 / 4).

• If each inverter would have an output current of 10A, the power loss in the big cable would be (4 · 10 A)² · R_big = 1600 A² · 0.028 Ωmm²/m · 300 m / 300 mm² = 44.8 W
• If each inverter has its own cable, then it is (10A)² · 0.028 Ωmm²/m · 300 m / 75 mm² = 11.2 W, but you have that four times, so its 4 · 11.2 W = 44.8 W

Hope that helps, kind regards,

Martin