Tuesday, 21 March 2017

How well are my new inverters doing?

We installed our PV panels in 2011 and last year we upgraded the inverter system, installing SolarEdge equipment. We were promised better yields, but are we getting them? In this post I show how I have estimated the improvement, correlating yield with the old system with weather data from the MET office and using this to predict what we would have got recently if we had not upgraded. The prediction function I derived was 98% accurate, which I was very pleased with. Using this I have estimated the improvement to be 4.6% but we don't have a whole year with the new inverters and I think this is probably an underestimate.

The reason I am doing this now is because we are going to discuss solar panel performance at the next Transition Cambridge Energy Group meeting on Thursday.

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. This graph shows an attempt based on the time of year only, ignoring the weather. The solid dots are the predicted yields and the empty dots are the actual yield: the black lines show the error in the prediction - the shorter the line the better the fit. You can see the fit it not bad at all. The prediction accounts for 92% of the variation in yield.
Prediction of PV yield based on time of year only. R-squared=92%
However I can do better. This next graph adds in a factor for sunshine as recorded by the MET office. This prediction accounts for 98% of the variation in yield.
Prediction of PV yield based on time of year and sunshine, R-squared=98%

The weather related bit (green) is about half the yield in summer and this relates to bright sunshine. The yellow part is still important because panels also use diffuse sunlight. Thin cloud is not too much of a problem. Finally, the red part is a fudge factor which I think relates to the shading problem.

I did also wonder if age might also be a factor - if we were starting to see any deterioration in the panels. However, adding age into the model did not improve the prediction. It would be worrying if it did as they are only 6 years old.

In the second step, I have used 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).
Comparison of predict PV yield with actual yield after the SolarEdge equipment was installed - this data is preliminary as we have less than a year of data but so far it seems to be 4.6% better.

We don't have even one whole year yet, and from the look of the graph the improvement is greater in summer than in winter so our estimate so far of 4.6% is probably on the low side. I will redo this calculation in a few months when we have a whole year.

Appendx - the prediction function.

By educated trial and error I have devised a formula for the yield with the following form:

pv = ksun * sun.hours * sun.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.

sun.angle.adjust is an adjustment factor for the sun elevation and the angle of our panels. Our panels are at 35 degrees to the horizontal so
sun.angle.adjust = cos(elevation-35)

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.

The first graph above was based on the simpler function:
pv = kelev * sin.elevation + k0.

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)


  1. Great post (even though I'm a little lost on the maths)!
    A couple of questions (though I intend to make it to the meeting on Thursday)

    Are these new standard inverters (ie. would any new installation come with similar ones), or are they special models?
    How long would it take with a 4.6% increase in yield to finance them?

  2. The new equipment was a new inverter and per-panel optimisers from SolarEdge. I got a very good deal on this and I daren't say here where from. However we do not expect to make a profit - I was as much motivated by the SolarEdge monitoring system which allows us to see how well each individual panel is doing and get real time data from their app :-) 4.6% increase is not enough for the upgrade to pay for itself over the lifetime of the panels. However, as I say above, this is likely to be an under estimate and I am hoping to just about break even.