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:

• Short wavelength (optical wavelengths) radiation from the Sun reaches the top of the atmosphere.

• Clouds reflect 17% back into space. If the earth gets more cloudy, as some climate models predict, more radiation will be reflected back and less will reach the surface

• 8% is scattered backwards by air molecules:

• 6% is actually directly reflected off the surface back into space

• So the total reflectivity of the earth is 31%. This is technically known as an Albedo . Note that during Ice Ages, the Albedo of the earth increases as more of its surface is reflective. This, of course, exacerbates the problem.

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

• 19% gets absorbed directly by dust, ozone and water vapor in the upper atmosphere. This region is called the stratosphere and its heated by this absorbed radiation. Loss of stratospheric ozone is causing the stratosphere to cool with time.

• 4% gets absorbed by clouds located in the troposphere. This is the lower part of the earth's atmosphere where weather happens.

• The remaining 47% of the sunlight that is incident on top of the earth's atmosphere reaches the surface. This is not a real significant energy loss.

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.

ENERGY = POWER x TIME

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:

• Average over the entire earth = 164 Watts per square meter over a 24 hour day

• 8 hour summer day, 40 degree latitude 600 Watts per sq. meter

So over this 8 hour day one receives:

• 8 hours x 600 watts per sq. m = 4800 watt-hours per sq. m which equals 4.8 kilowatt hours per sq. m

• This is equivalent to 0.13 gallons of gasoline

• For 1000 square feet of horizontal area (typical roof area) this is equivalent to 12 gallons of gas or about 450 KWH

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.

Efficiency of PV Cells Demo

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.

• Under optimum conditions, one can achieve fluxes as high as 1000 Watts per sq. meter

• In the Winter, for a location at 40 degrees latitude, the sun is lower in the sky and the average flux received is about 300 Watts per sq. meter

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 (40^o) 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:

• 5% efficiency 9 KWH per day
• 10% efficiency 18 KWH per day
• 20% efficiency 36 KWH per day

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 or 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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