From weatherstation data to solar irradiation on specific surface

[BM99]

So, at home I have a weatherstation which measures solar irradiation in the east, south and west in lux. I can read in these measured sensordata. From these measurements, I would like to calculate the solar irradiance on my windows. The solar energy formulas could help me with that. The work of Soteris A. Kalogirou explains these phenomens in Chapter two (see http://1.droppdf.com/files/AvIdq/solar-energy-engineering-processes-and-systems-2nd-ed-2014-.pdf).

So for example, I have a window which (which has a slope of 90°, vertical) is oriented 20° from the south (clockwise is positive). I know what the incidence angle is and I know that I should use the formule for 'Total radiation on tilted surfaces' which is explained on page 100-101 in the pdf. Starting from the total irradiation in the east, south and west known from my weatherstation, does someone know how to use these measurements to calculate the total irradiance on my window in W/m²?

I hope this explanation is clear enough!

Thank you in advance.

 

[Andy Resnick]

Converting from lux to illuminance is not obvious here due to the large amount of (spectrally) non-visible sunlight. Fortunately, sunlight is important enough that conversion factors have been developed: the solar constant is, on the Earth's surface, approximately 1 kW/m^2 and also 128 klx. That conversion factor definitely won't work if the sky is overcast and is also less accurate as the sun moves off zenith.

For fun, you may be able to pseudo-calibrate your weatherstation by sending 'direct sunlight' into one of the ports and then use it on clear days; I'd be curious what your results are. Check the user guide- don't damage the optical detector by doing this.

For example- use a mirror to send noontime sunlight into a port, and the weatherstation tells you that it's receiving (say) 100 klx so you have a 'weatherstation calibration factor' of 1.28. Then, the next day (or whenever), your station pointed in the direction of your window gives a reading of 50 klx, which means your window actually receives 64 klx = 0.5 kW/m^2.

good luck!

 

[BM99]

Converting from lux to illuminance is not obvious here due to the large amount of (spectrally) non-visible sunlight. Fortunately, sunlight is important enough that conversion factors have been developed: the solar constant is, on the Earth's surface, approximately 1 kW/m^2 and also 128 klx. That conversion factor definitely won't work if the sky is overcast and is also less accurate as the sun moves off zenith.

For fun, you may be able to pseudo-calibrate your weatherstation by sending 'direct sunlight' into one of the ports and then use it on clear days; I'd be curious what your results are. Check the user guide- don't damage the optical detector by doing this.

For example- use a mirror to send noontime sunlight into a port, and the weatherstation tells you that it's receiving (say) 100 klx so you have a 'weatherstation calibration factor' of 1.28. Then, the next day (or whenever), your station pointed in the direction of your window gives a reading of 50 klx, which means your window actually receives 64 klx = 0.5 kW/m^2.

good luck!

Thanks for your interesting answer Andy!

I will try that as a validation of my estimation. Somehow, I'm convinced that with the measured total radiation in the three orientations, I could convert these to a total radiation on a window in any orientation. Well maybe not the total radiation, but rather the beam radiation as I suppose the diffuse and ground-reflected solar radiation to be zero. The following formulas should do it i think considering I know every concerned angle.