ndiorio Posted June 10, 2016 Report Share Posted June 10, 2016 Hello, I'm trying to run a basic simulation with one module tilted 30% facing south (azimuth 180 degrees), with a wall partially obstructing the view to the south. I've setup the scene, and when I run it there are outputs: Area South: Irradiance onto tilted surface [kWh/m2] Shading [kWh/m2] I'm finding that the shading losses compared to the irradiance (which I assume is the plane-of-array quantity) are much lower than I would expect. For a full year, I'm seeing: Shading - 25 kWh/m2 Area South: Irradiance onto tilted surface 1921 kWh/m2. Which is a 1.3% annual loss. However, when I compare the PV energy (DC) to a case without the shading obstruction I get: PV energy (DC) - with shading: 229.83 kWh PV energy (DC) - without shading: 348.31 kWh Which is a 34% annual loss. I realize converting from irradiance to DC power has other losses, but for the same system, same configuration with only a shading difference, I would ascribe most of this difference to shading. I would like to compute how much of the irradiance on the PV module is lost due to shading. Can you please explain what the shading outputs indicate and if there is a way to get more detailed outputs? I'd like to understand why I'm seeing so little shading loss reported compared to the hit in PV energy. Thanks, Nick Quote Link to comment Share on other sites More sharing options...
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