How does the sun angle influence the amount of solar energy received at a given location on earth?

Answer 1

It is a geometry problem. If the Sun is at the zenith, or directly overhead and shining directly down on the ground, you get maximum illumination. Imagine a solar panel laying flat on the ground. It is maybe a square panel one meter by one meter. It gets one square meter of sunlight. Hold it horizontally above the ground and it makes a 1 meter square shadow. That shadow is the sunlight that hits the panel instead of the ground. Now imagine another panel standing on edge. All the sunlight slips past it to hit the ground. It makes no shadow. There is no sunlight absorbed by the panel, it all just slips by and hits the ground. Now hold the panel at some angle, More that zero light will hit it, but less than 1 square meter of sunlight will. Maybe it makes a shadow 1 meter by half a meter, so it is catching only half as much light as it could if you held it the right way.

Now the Sun is usually not directly overhead. Even at noon, the Sun cannot be directly overhead unless you are in the tropics. You can see this with a globe and light source. Especially at the poles the Sun never gets anywhere near directly overhead, and in winter it never even rises. There is 6 months of night at the pole.

Answer 2

Mathematically, its represented because of the fact the photograph voltaic consistent * cos phi, the place phi is the degree of selection. The photograph voltaic consistent is a pair of million,367 W/m^2, so in case you needed to calculate how a lot insolation there is at a particular selection, you will use (a million,367 W/m^2)*cos(ranges of selection). instruments are in W/m^2. A watt is a million J/s. regularly, the closer you're to the equator the greater photograph voltaic insolation.