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Steve4Physics said:Agreed.No. It is meaningless to refer to the "intensity of the solar panel". The solar panel does not have an intensity. The intensity is the amount of power from the sun per square metre. 5.53W/m² is the intensity *calculated assuming the solar panels are 100% efficient*. So 5.53W/m² is not an accurate value of intensity (because efficiency is not 100%).
But we know the panels are only 20.8% efficient. So intensity must be (roughly 5 times) bigger than 5.53W/m².Yes but round to 3 sig. figs.No. First, 'example' is the wrong word. And we physicists don't use words like 'tremendous' in this context! Just say (for example) "300W/m² is an overestimate ...
'per day' is wrong. You = mean averaged over a day.
13.29055 is too many significant figures.
The most *accurate* values of intensity are the ones calculated allowing for efficiency. So you should be averaging the answers from 1c) and 2c), not from 1b) and 2b).
(204 + 26.6)/2 = 115Wm/s²
So the 300W/m² (assumed to be an annual average), compared to 115W/m², is roughly between two and three times too big.
Thank you for your reply and your continued insight.
2. c) Sorry my mistake, clearly I misused intensity here.
Question 3) "example" was the wording from the original question. Oh right, I will correct my average value also, thank you for remarking on that.
Can I verify that it would be correct to use an angle of 35 degrees when finding the intensity, as I have become a little uncertain?