According to "E = A * r * H * PR" global formula to estimate energy generated, the solar irradiance value is needed. I have Horizontal irradiance, Tilted irradiance, and Normal irradiance values but which measure of irradiance to use here? Or is it that all of these measures are used to calculate one value that is the total irradiance falling on a solar panel?

  • Does the tilted irradiance data tell you what angle it's tilted at? You'll want irradiance data that matches the angle of the panels you're doing the calculation for. – LShaver Sep 28 at 14:00
  • @LShaver. Yes, I do have the tilt angle of the panels. But is the "tilted irradiance" the only measure required here? Cause I have all of the mentioned measures and it doesn't make sense to use only "tilted irradiance".if it was like that then why is there a need to measure any of the other measures? – yash Oct 1 at 6:59
  • I presume if the tilted irradiance value is for a different angle than what you're looking for, you could use the horizontal and vertical irradiance values to make an approximation. Could also be useful for sun tracking arrays. – LShaver Oct 1 at 13:39
  • @LShaver. The tilted irradiance value that I have is for the angle that I require. That's no problem at all. Problem is how can I properly make the approximation of solar irradiance. – yash Oct 3 at 5:08
  • If you provide more details about the formula you're using and the data you have it may be possible to help answer your question. Please edit your question and provide a link to the data source, the units of the values, references, etc. – LShaver Oct 5 at 15:43

You don't want to do it this way. It gets really messy and you need to take into account your climate, sunny hours per day through the year, time of year and so on.

Instead try looking up "solar potential atlas" For Canada this is a big spreadsheet listing values by month for every city and village in the nation. I'm sure there are equivalents for your country.

For example for me, if I have an array facing south sloped at my latitude, I get 1250 kWh per kW installed capacity. So a 1 kW array generates for me about $100 worth of electricity per year.

Which value to use? They are all going to be equivalent -- with a bit of trigonometry.

Normal in this case I assume means normal to the sunlight path. You will have an additional factor of cos(d) where d is the angle difference between the angle of the sun and the angle of the PV array. This varies by 47 degrees over the course of the year. Rule of thumb: Angle the array facing the equator tilted at an angle equal to your latitude from horizontal. The sun wanders 23.5 degrees north and south of normal to this plane, which makes about a 5% difference.

A bigger change is due to the sun not always being to your south. With stationary panels. the sun is at an angle to the east in the morning and west in the afternoon.

At my latitude (51 degrees) the horizontal incidence fluctuates by a factor of 10 between summer and winter, but for a latitude tilted panel it varies only by about a factor of 3. In summer the early morning and later afternoon sun is lost because it's rising 40 degrees north of east and setting 40 degrees north of west. The panel is self shading.

You would see a much larger difference if you used a tracking array, because it avoids this self shading.

(For me it wouldn't matter. My max demand is in winter. A non tracking array gets about 80% of what a tracking array does. My money is better spent on more cells, and bigger batteries.)

  • Solar atlas's like these globalsolaratlas.info does help to know the PV output as well as GHI, GTI etc at a place. But I'm actually trying to understand which measure irradiance (or how they are all used) to calculate the PV output rather than just knowing the output. – yash Oct 9 at 5:21

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