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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]

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:

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

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Dear Nick,

thanks for your interesting question. Shading losses in PV*SOL are considered in several different ways, so perhaps it is best to go through that with an example.

Let's build a roof top pv plant with a lot of shading:

So, in this project we have a horizon as far shading object and a big wall as near shading object. When I simulate this project, I get the following energy balance:

So there are three items where shading losses are taken into account:

1. First, we have the shading of the diffuse irradiance, which is caused by the horizon line. These shading losses are the same for all modules and can be found in the first block of the energy balance which refers to irradiance calculations
2. Then we have the module specific partial shading, which is caused by the big wall. These shading losses are different for each module since the wall is standing close to the pv plant. So in some situations one module will be shaded, while another module still receives full sunlight. These losses also take into account that each module kind of "sees its own horizon", so the additional diffuse shading losses caused by the wall will be different for each module.
3. This is an electrical loss, caused by the imhomogeneities of the shading of the modules. In simple terms, if you have a string of modules connected in series, the one with the lowest power output will determine the output of the others, and thus causing extra losses. You can find more detailed explanation here in our online help

I hope this helps to understand the effects of shading more deeply. If you have any further questions, please don't hesitate to ask.

Martin

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Hi Martin,

Thanks for your detailed explanation, I will go through it and let you know if additional questions arise.

Best regards,

Nick

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Hi Martin,

I came up with a couple of quick questions:

1. For the shading of the diffuse irradiance caused by the horizon line, is there a reason why this would be present if there are no obstructions in the horizon?  For instance, if you have a PV array in a flat field with nothing in the horizon, would you expect this quantity to be non-zero?  I'm observing small losses for this in such a case.

2. For the module specific partial shading, is there a way to separate the beam portion from the diffuse portion in the output results?

Additionally, is the beam shading calculation linear for this category since the mismatch is accounted for later on?

Thanks again for you in-depth explanation.

Best,

Nick

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Hi Nick,

thanks a lot for your questions. If there are no objects in the horizon line (all values of the horizon line are zero) and the modules have no inclination, the diffuse shading losses must be zero. However, if the modules are mounted with inclination, they don't see the whole hemisphere anymore, which leads to a reduction of the diffuse irradiance that reaches the modules.

Regarding the second question: Right now it is not possible to separate the direct and the diffuse fraction of the module specific shading. And yes, the shading is applied linearily, since the electrical losses are calculated afterwards depending on the resulting i-v characteristics of each module.

I hope that helps. If you have further questions, don't hesitate to ask.

Kind regards,

Martin

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