Fortunately I have been recording my solar panel yield weekly since they were first installed, so I have a lot of data to go on. In order to estimate the improvement in yield I have to work out what we would have got with the old inverters. In a perfect world, with perfect weather data and an ideal installation this would be straightforward. In practice, the best weather data I have managed to find is monthly total sun hours from the MET office for East Anglia [1]. (There seem to be some issues with the sunshine data recorded at the local Cambridge Digital Laboratory.) However 'sunshine hours' isn't quite the same as total solar energy. Also, although our panels face almost due South they suffer from shading in the early morning due to a wall to the East. So I have used statistical techniques for my estimation.
So as not to overwhelm you with details, I will give you the charts and the final results first. The detail is in the appendix.
The first step is to work out how to predict yield using data from before the SolarEdge installation. The appendix shows the detail of the model I have used. In the graph below, I have plotted estimated PV yield by month (solid black circles) and actual yield (open circles). The difference between them (shown as a thin black line) is the error in my model. However they are generally pretty close.
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The red part varies with the season but not with the weather - this allows for the fact that even when it is not sunny the PV panels still get diffuse light and the green part is linked to how much sunshine we have had. The brown bit at the bottom is a constant which is negative - this might be due to our shading problem.
The second step is to use this predictor function to work out what we would have got from the panels with the old inverters. This graph compares the predicted with the actual yield, starting in July last year (it was installed during June). Using one whole year of data the improvement seems to be 5.9%.
Comparison of predict PV yield with actual yield after the SolarEdge equipment was installed -based on one year it seems to be 5.9% better.caption |
(This post is effectively an update from an earlier one in May which I have now deleted. As well as having a whole year of new data I have improved the estimation slightly.)
Appendix - the prediction function.
By educated trial and error I have devised a formula for the yield with the following form:
pv = ksun * sun.hours * angle.adjust + kelev * sin.elevation + k0.
where ksun, kelev and k0 are constants that I have estimated by regression analysis (least squares fit).
elevation is the elevation of the sun at midday in the middle of the month. Calculating this is too complex to describe here. I got it from [2].
sun.hours is the number of hours of sun/day, calculated from the MET data [1] by dividing the monthly total by the number of days in the month.
angle.adjust is an adjustment factor depending on the sun elevation and the angle of our panels. Our panels are at 48 degrees to the horizontal (measured with a mobile phone app which was much easier than a ruler) so
sun.angle.adjust = sin(elevation+48)
sin.elevation is a measure of the power of the sun at the time of year
sin.elevation = sin(elevation)
The angle of the panels affects the bright sunshine component (ksun) but not the indirect sun.
If any of you have historical data for your PV yield then you can try this too. Let me know if you want my code as a reference (it is in R).
[1] UK and Regional Series (MET office)
[2] Elevation Angle (PV Education.org)
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