[Clarify] Solar radiation measurement

Wondering if an engineer could quantify how similar the tempest panel measurement is to what a commercial solar panel would pull in usable electricity. Yes, panel models may have different efficiency, so how are the tempest panels comparable?

I was able to convert w/m2 to kwh but not sure the value of the output as a gauge of solar site viability. Should be close but how close.

Would like to see a comparison between the tempest readout and an installed solar panel at the same location

Welcome to the forum and Happy New Year!

The sensor in the center of the top of the Tempest is what is used to measure Solar Radiation and has been calibrated to show the full Solar Radiation falling on the sensor then converted to W/m^2. How efficient the Tempest solar panels are is irrelevant to the reading. What may be relevant is that the sensor is sitting horizontal, whereas solar panels may be angled more optimally for a given location.

I’m not quite sure how you converted from W/m^2 to kWh because W/m^2 is a power per area unit and kWh is a total energy unit. The W/m^2 is exactly as the units state, watts per square meter of solar radiation. For example, at one point yesterday my Tempest showed that the Solar Radiation was 200 W/m^2. This means that if I had a 100% efficient square solar panel measuring one meter on a side, I could power two 100W lightbulbs fully. In reality, there is no way to get a 100% efficient system, let alone a solar panel.

Solar panel efficiency is typically below 20% so with the SR example above of 200 W/m^2, only 40W of that power would come out of the solar panel. This means that, at that rate, it would take a minimum of 25 hours to accumulate 1000 watt-hours or 1 kWh of energy.

To get the answer you are looking for, you really need to integrate the SR output over time. Mathematically speaking, this would be the area under the SR curve to get a total energy value per square meter over the time used.

The sensor in the center of the top of the Tempest is what is used to measure Solar Radiation and has been calibrated to show the full Solar Radiation falling on the sensor then converted to W/m^2. How efficient the Tempest solar panels are is irrelevant to the reading. What may be relevant is that the sensor is sitting horizontal, whereas solar panels may be angled more optimally for a given location.

Thanks thats exactly the clarification I was looking for. The question: More or less what is the suitability of the Solar Radiation measurement to determine the output of a proposed solar panel at the same location, essentially to pull more utility out of the device.

My first assumption was the solar panels powered the device, not the top sensor. Still a properly sited device, perfectly horizontal, facing the correct direction could define a data standard to base a software calculation, but I agree now, any assessment would require the user to measure and input a roof or panel angle, maybe more.

Searched, found a thread that mentioned dimensional analysis and Wolfram Alpha, took the shortcut and went to Wolfram Alpha, entered the tempest average solar radiation number in W/m^2 for the month, converted to kWh/m^2, logged into my electric company website, checked the monthly average use for the month. Wanted to believe.

1kWh/m^2/year to W/m^2 - Wolfram|Alpha (wolframalpha.com)

I think at my location the average was 34 W/m^2 for December and its possible depending on how that figure is calculated, the unit was installed after Dec 1st and the data was incomplete. The goal was to see if it was possible to equate the daily/monthly/yearly W/m^2 to the monthly electric utility use to determine site suitability for a solar electric system, based on panel size, orientation angle and solar gain/electric conversion efficiency. Therefore Im assuming what you are saying is start by multiplying 34 W/m^2 by 0.20(percent), equaling 6.8 W/m^2 and we just now getting started with some truly thrilling numbers…

Im going to think about it a bit more.

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