alex Posted May 5, 2015 Share Posted May 5, 2015 I usually run my simulation using the orientation field option with unlimited shed(single rows of one module width). In the orientation field, If I choose unlimited sheds and I choose electrical shading effect for the simulation, what is the default value?If I used the according to module string option, I can choose the fraction for electrical effect. In fact, I need to run a simulation using 60% fraction for electrical effect.However, I would like to compare both options but I need to know what is the default value for the orientation field option. Any response is really appreciated. Link to comment Share on other sites More sharing options...
André Mermoud Posted May 5, 2015 Share Posted May 5, 2015 In your first question, I don't understand which "Default value" you are talking about. The "number of strings" in the width of the row: if you have one only module, it will be one. If you have 2 modules in width, this will be two if the modules belong to 2 different electrical strings, and one if they belong to the same string. The cell's width is 15.6 cm if you have 6" cells, and 12.5 cm if you have 5" cells. I think you did not understand the meaning of the "fraction for electrical effect". When you calculate shading losses, you have the "linear" losses" (the real deficit of irradiancre due to shades) and the "Electrical losses" corresponding to the electrical effect due to string connexions (i.e. the yellow part on the 3D shading animation). The "Fraction for electrical losses" indicates to which extent the Yellow part is taken into account in the losses. For shed arrangement this fraction will always be 100% (i.e when the bottom cells row is shaded, there is no electrical production anymore for this string). Please see our FAQ How to evaluate the the effect of by-pass diodes in shaded arrays ? Link to comment Share on other sites More sharing options...
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