PV Gap

Erin Said:

Solar energy constant at 50 degrees north?

We Answered:

The height above the ground isn't really a major factor in the calculation. It's much more important to know the time of the day and the season.

Example: The maximum amount of irradiance would be received at the summer solstice (June 21), when the angle between the sun rays and the vertical would only be (50-23.5)° = 26.5°, therefore the irradiance would there be sin(26.5°) / tan(26.5°) * 1.2 kW/m² = 1.1 kW/m²
That's the maximum value. On the winter solstice (December 21) though, the best angle you get is (50+23.5)° = 73.5°
Therefore, the maximum irradiance at that time would only be sin(73.5°) / tan(73.5°) * 1.2 kW/m² = 0.34 kW/m².

Of course, this applies only to surfaces that are lying flat on the ground. If you would e.g. build a solar array that moves with the sun's direction, you'd be able to get much higher irradiance values.

Janet Said:

Is possible to rotate the meter in reverse direction with help of solar energy utilisation?

We Answered:

It is possible to feed power back to the power grid but it must be done in specific ways. Solar power is DC and much be converted to AC and then synchronized with the grid. There must also be safety devices to assure the safety of people and the equipment.

There may also be a need for metering and other contractual agreements to deal with the commercial aspects of such an arrangement.

Cory Said:

Why does a square meter of ground at the equator receive more solar energy than a square meter at the N. Pole?

We Answered:

(There are a lot of goofy answers below.)

At the equator the ground is perpendicular to the sunlight, the whole square meter is bathed in the light.

At the N. Pole, the surface is almost parallel to the sun's ray and so little light hits it, the ray go past, unobstructed.

Imagine trying to throw a ball through a pane of glass. If you are looking at it straight on the target is pretty big, if you look at it almost edgewise the target is small. Keep it mind that light travels in straight lines (in these circumstances).

Jeffrey Said:

Neglecting energy absorbed or reflected by our atomosphere, the solar energy hitting 1 sq. meter of Earth's

We Answered:

What luck! I just saw in another question a statement that a megaton was equivalent to 4X10^15 joules. If that is correct then:
Diamond area 90 feet square is 752 square meters.
752*1360=1,022,720 joules per second
4X10^15/1,022,720 = 3,911,138,924 seconds or 124 years

That would be in space, it would take longer on the ground. At least twice as long just because of half the time being at night, and more because the Sun is not always directly overhead and clouds and so on.

Jay Said:

How long does it take a baseball diamond (90 ft on a side) to receive 1 megaton of solar energy?

We Answered:

Solar energy is not measured in megatons.

A megaton is the equivalent effect in a nuclear explosion of one million tons of trinutrotoluene exploding simultaneously in a chemical explosion.

Explosive force does not measure efficiently in terms of heat or light energy in radiant form as in the case of insolation per unit area.

Better talk to your teacher. Sounds like some basic instruction in matter, energy, and units of measure is called for.

Good luck.