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: http://zebu.uoregon.edu/1999/ph162/l4.html
Дата изменения: Thu Apr 15 19:06:50 1999 Дата индексирования: Tue Oct 2 00:00:16 2012 Кодировка: Поисковые слова: saturn |
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:
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:
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.
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 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