Solar Energy

Basics of Solar Energy

The Sun --> Always there; lots of Energy

What Makes the Sun Shine? Nuclear Fusion; something we may learn how to do later on the Earth and thus solve our Energy Problem.

How many photons (energy) reach the surface of the Earth on Average?

The energy balance in the atmosphere is shown here:

The main components in this diagram are the following:

What Happens to the 69% of the incoming radiation that doesn't get reflected back:

Cliff Notes Summary

How much energy from the sun reaches the surface of the Earth on Average?

Note that we measure energy in units of Watt-hours. A watt is not a unit of energy, it is a measure of power.


1 Kilowatt Hour = 1KWH = 1000 watts used in one hour = 10 100 watt light bulbs left on for an hour

Incident Solar Energy on the ground:

So over this 8 hour day one receives:

But to go from energy received to energy generated requires conversion of solar energy into other forms (heat, electricity) at some reduced level of efficiency.

We will talk more about PV cells in detail later. For now the only point to retain is that they are quite low in efficieny!

Collection of Solar Energy

Amount of captured solar energy depends critically on orientation of collector with respect to the angle of the Sun.

A typical household Winter energy use is around 3000 KWHs per month or roughly 100 KWH per day.

Assume our roof top area is 100 square meters (about 1100 square feet).

In the winter on a sunny day at this latitude (40o) the roof will receive about 6 hours of illumination.

So energy generated over this 6 hour period is:

300 watts per square meter x 100 square meters x 6 hours

= 180 KWH (per day) more than you need.

But remember the efficiency problem:

At best, this represents 1/3 of the typical daily Winter energy usage and it assumes the sun shines on the rooftop for 6 hours that day.

With sensible energy conservation and insulation and south facing windows, its possible to lower your daily use of energy by about a factor of 2. In this case, if solar shingles become 20% efficient, then they can provide 50-75 % of your energy needs

Another example calculation for Solar Energy which shows that relative inefficiency can be compensated for with collecting area.

A site in Eastern Oregon receives 600 watts per square meter of solar radiation in July. Asuume that the solar panels are 10% efficient and that the are illuminated for 8 hours.

How many square meters would be required to generate 5000 KWH of electricity?

each square meter gives you 600 x.1 = 60 watts

in 8 hours you would gt 8x60 = 480 watt-hours or about .5 KWH per square meter

you want 5000 KWH

you therefore need 5000/.5 = 10,000 square meters of collecting area

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