jacob Posted March 7, 2022 Posted March 7, 2022 Hello, I'm looking for any guidance on how to calculate the percent shading of a mesh. In particular our jurisdiction is requiring a mesh covering on the back of our ground mount arrays to prevent access to wiring. The modules being used are bifacial so unfortunately this mesh will reduce the effectiveness of the backside. I'm trying to figure out how big of an effect that is. We have selected a mesh with the largest openings we feel we can get away with. In terms of total area, 5% is wire and 95% is open air. Can I say that the shading is therefore increase by ~5% when covered by this mesh or is it more complicated then that?thanks,
Michele Oliosi Posted March 10, 2022 Posted March 10, 2022 A disclaimer: my answer is just based on physical intuition, not on a specific study. Because of the mesh thickness it could be a bit worse than that for diffuse light. If you want to be on the safe side you can increase that factor. However the best course of action is probably to make some measurements on a test module to ascertain these losses.
MicheleANE Posted March 29, 2022 Posted March 29, 2022 Hi, You could also make an experiment in PVsyst: Using just a table with the correct size and tilt, place it facing backward as if it was the rear face of the module. Then run 2 simulations, one with the net and one without it and see the difference in shadings. It's still not real but could give a guide for your assumptions. Best regards
jacob Posted May 17, 2022 Author Posted May 17, 2022 That is a really good idea MicheleANE. I gave it a go but got stuck because PVsyst only allows tilts between 0-90. I can't think of a way to work around this. I've been trying to get a rough estimate of the solid area vs open area of the mesh. I think the solid is about 14%. I'm thinking I could plug that value into the backside shading value but I already have 15% to account for the rack. I don't think it would be appropriate to add the two, but also my gut tells me it wouldn't be appropriate to just take the larger of the two either.
dtarin Posted May 18, 2022 Posted May 18, 2022 Try creating a net with the handrail object. Set width of rail to whatever the thickness is, and the number of vertical rails to the spacing of the net, and place in the front of your fixed tilt system oriented normally. Then run a simulation with and without, take the delta, and add to rear shading. I tried running this with a 4x7 table and a large "net" structure with thin object shading set to 5%, but after running for hours it crashed. Perhaps a single module will be more successful.
jacob Posted June 2, 2022 Author Posted June 2, 2022 Thanks dtarin. I gave this some more thought. I already put in a 15% shading loss on the backside due to the racking based on experiments someone did. From what I understand this means the cell with the worst shade (the weak link) is shaded 15%. The other cells may be shaded less but the worst case cell sets the current for the rest of the string. When I add the mesh that cuts what irradiance that was getting to all the cells down by another 14%. So instead of subtracting 14% and 15% (total 29%) I am thinking I need to subtract out 14% from what was remaining after the 15% was taken out (total 26.9%, mesh accounting for 11.9%). In other words Irr - 1x15% - .85x14% I tried out the handrail idea on one table and it worked. Maybe not surprisingly it splits out what I put in. Aka I created a handrail that is 14% solid and I set the thin object factor to 14% and I got a shading loss of ~14%. Therefore I really needed to add the racking in to get a meaningful result. I did that and the result was that the mesh accounted for additional loss for 14.9%. This is obviously more than 14% and certainly more than the 11.9% I was expecting. Not sure what to make of it. I thought I had a good hypothesis going but these results certainly don't support that hypothesis. I can't think of a logical reason why I got the results I did though.
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