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.
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):
- Cloud cover, smoke, haze (which you can't really control, other than putting panels in the desert)
- 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)
- 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)
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.