Effective irradiance to DC power

[Marius Schmiedt]

Hello all,

I'd like to determine the AC power the PV System provides to the grid from the total incident irradiance. So I basically determine poa_global from ghi and derive DC and AC power respectively from poa_global.

For the poa_global to DC power step I use the "pvwatts_dc" model.

But I'm confused here with the function because the "pvwatts_dc" reads the irradiance (W/m^2) and returns the power (W) without taking any area (m^2) into account. This seems wrong to me.

Is there another or preferred way to determine the DC power from the poa_global instead? Actually I first thought I could simply estimate the power with an approximation of the efficiency of 20-30% and by considering the array area.

Did anyone did something similar before?

Thanks, 

Marius

[cwh...@sandia.gov]

The PVWatts conversion is

power_dc (W) = poa_global (W/m2) / 1000 (W/m2) pdc0 (W / (W/m2)) * temperature adjustment.

So the area is implicitly accounted for in the pdc0 parameter, which is the total DC power of the modules at standard test conditions.

Module ratings work the same way - they are not given as W/m2 of module area, but rather as W.

Cheers,

Cliff