Interested to know how to calculate how much energy photovoltaic (PV) panels would produce given x amount of solar irradiation per day - amount of direct sunlight and how much that might be effected by y amount of cloud cover. Also does temperature or season of the year have a significant effect on the amount of irradiation or direct sunlight used by solar panels to generate electricity?

Lots of solar articles tell you you need Sun to generate solar power but not how much irradiation from the sun is required to get a solar panel system of size k to full capacity for a day.

Could you please share resources or a calculation for converting the amount of solar irradiation or direct sunlight required to power a solar system of say 10KW.

2 Answers 2


Solar Irradiance

The amount of energy striking the earth from the sun is about 1,370W/m2 (watts per square meter), as measured at the top of the atmosphere. This is the solar irradiance. The value at the earth's surface varies around the globe, but the maximum measured at sea level on a clear day is around 1,000W/m2. The loss is due to the fact that some of the sunlight's energy is absorbed by the atmosphere on the way down.

Panel Efficiency

When this sunlight strikes a solar panel, about 10-20% of the energy is converted to electricity. So a good (20% efficient) 10kW array would measure 50 m2, or about 7m by 7m.

Panel Location and Orientation

Theoretically, the maximum output you can get from a solar panel will be for a panel lying flat at the equator under a clear sky when the sun is at its zenith, such that sunlight strikes the panel at a 90° angle. At this moment, a 10kW solar array will produce 10kW of power*. (This takes into account panel efficiency, conduction losses, charger efficiency, etc).

From this ideal, three factors reduce the power output of a panel (in order of importance, assuming a brand new, clean panel):

  1. Cloud cover, smoke, haze (which you can't really control, other than putting panels in the desert)
  2. The angle of the sun's rays relative to the panel moving away from 90° (which you can control by keeping the panel tilted toward the sun)
  3. The thickness of the atmosphere through which the sun's rays must pass (which you can't control, as sunlight passes through more of the atmosphere at sunrise and sunset)

Capacity Factor

For any given time period, we can define the capacity factor (cf) of a particular solar panel or array. This is the amount of energy output given vs the maximum possible output (the same as the panel's rating):

cf = Eactual/Etheoretical

So the panel on the equator with the sun at it's zenith would have cf=1 in that moment.

Note that the panel rating, given in watts (W) or kilowatts (kW) is a unit of power. Capacity factor is defined in terms of energy: watt-hours (Wh) or kilowatt-hours (kWh), which is power over time:

E = P × t

So if the sun at its zenith paused for one hour in the scenario described above, the panel would produce exactly 10 kWh. In reality, it will be slightly less than this, since the angle will change during that hour.


There are lots of different ways to calculate the capacity factor for a given panel installation, and it can be calculated over lots of different time periods -- the usual case is to calculate an average cf over the course of a year, for calculating actual energy output (in kWh) and payback period.

One excellent tool for simulating capacity factor of a particular installation is PVWatts, from the US National Renewable Energy Lab (NREL). With this tool you select the type of panel, orientation, angle, type of tracking system, and a few other (optional) characteristics. Capacity factor is included in the output with a lot of other valuable information. Weather data from around the world is included in the calculation.

*This will vary with altitude, and with location for panels which are still optimally tilted/oriented, since the thickness of the atmosphere does matter. I haven't looked into research on this personally, but I imagine the variation would be small and/or hard to measure accurately.

One other note -- I recently read an interesting paper modeling ideal azimuth (orientation with respect to compass directions) and tilt to get the highest annual capacity factor. In the author's model of the US, local weather played a big factor, and southward facing was not always found to be optimal.


Additional information to @LShaver's answer:

Google for a solar power potential atlas or map.

These produce tables or maps that estimate the amount of power that can be generated over the course of a year for a 1 kW panel given a specific location. Some will give month by month breakdowns too, or give the relative values for different orientations of the panel.

They are based in part on latitude which gives you seasonal day length and solar elevation, but also in part on the climate record for that location, which comes from your local weather station's observations.

Typical Numbers:
Where I live in Alberta a 1 kW panel oriented to be at right angles to the sun on equinoxes, will generate about 1250 to 1350 kWh/year.

The same array in Germany generates about 800 kWh, while one in the SW deserts in the U.S. will generate 1500-1600 kWh.

The capacity factor of solar power bites. Consider: There are not quite 9000 hours in a year. If you get 1300 kWh/year for a 1 KW array, you have a capacity factor of about 15%

The flip side: No moving parts. Better predictability. Little maintainence. (Wash the bird crap off...)

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