Using solar W/m2 to determine solar panel efficiency

I prefer centigrade my self. :crazy_face:


HA! So this was a good exercise at minimum! So, I will wait on some other big solar days and see how the Tempest compares against solar generation.

Keep in mind, the Tempest is made up of consumer grade products. You will never see the same results and the sensor is parallel to the earth where as your panels are not.


@dsfg I use micro inverters from Each coverts DC to AC. For a couple of reasons. 1) i didn’t want a central inverter in my house, as they sometimes emit a very high pitched sound (not from the fans) 2) Some panels receive a bit a shadow during the day. Now each panel runs at its optimum power point independently, thus creating more power overall. 3) easy to expand as I don’t have to buy a bigger main inverter. (last year I bought some extra) 4) its saver as there isn’t a very powerful DC cable running on the roof, that might easily overheat. Also in case of fire, it is saver for the fireman. 5) they are supposed to last 25 years! (much longer than a central inverter). And in case one breaks down, the rest will still function normally.

The main disadvantage is of course that it is more expensive.


Hello. What can you tell from the history of solar radiation looking at last year? Leaning towards a solar panel installation and wondering if the radiation can hint about how many KWh last year could have produced? //Matias

Hi @matias.bjorklund ,
This thread that I moved your post to discusses your question.
You probably realise that W/m2 is measured on a flat surface horizontally which means the measurements from your Tempest if you live near the equator will read a greater amount of radiation than if you lived near the poles for the same amount of radiation from the sun.
edit. You can also use your tempest to help decide what angle to mount your panels. Rather than converting the horizontally recorded W/m2 values from the Tempest to the designed best angle for your panels you can mount your Tempest at the desired angle of your panels end record how much energy it receives. This will be much more accurate because it takes into account the cloudy times of the day at the correct angle to measure the true energy received. Ideally you would have more than one Tempest and compare the results to find the best location and angle for your panels.
cheers Ian :slight_smile:

Hi Ian. Understand that its not compareable from a flat to angled position. What I like to know is how much the total/year is. Looking at the tempest history 2021 I see only what day had max radiation and also the average. Is total avg*365 or is it avg those day the sun been showing?

Hi @matias.bjorklund ,
The average for the year (or for month or week or day) is the average value for that period.
The average is calculated by adding the W/m2 for every minute in that period divided by the number of minutes in that period.
If you wish to know how many kiloWatt hours of sunlight energy were available that year then multiply the average value by how many hours in a year.(not mentioning conversion from the horizontal plane and area which must be factored in, but ignored now for simplifying the average into an energy amount)
My house recieves an annual average of 237W/m2, times 365days times 24hours = 2076kWh per year.
Or 237 x 24 = 5.7kWhr/m2 per day.
Now to convert angles and efficiencies to guess actual electricity generation is more complex but
I have 1.2kW of panels installed and I roughly generate 5kWh per day.
Cheers Ian :slight_smile: