The Sun Always there; lots of Energy
How many photons (energy) reach the surface of the Earth on Average?
The energy balance in the atmosphere is qualitatively shown here:
Averaged over the entire earth = 164 Watts per square meter over a 24 hour day So the entire planet receives about 85,000 Terrawatts of Power our current worldwide power use is about 14 TW (of which 3.5 are required for the US alone). Since the Sun gives us 84,000 and we only need 15 is this a solution?
In principle we would need 14 Terrawatts (= 14 x 1012 watts)
divided by 164 watts per square meter = 1011 square meters of
collecting area.
But since our efficiency will only be about 10% (see more below), then we
need 1012 square meters which is equal to one million square
kilometers.
Now suppose we could construct an equatorial grid, with one station per
time zone. Then we would need 24 installations each of size 200 x 200
km. This is formidable.
and remember, little of the world current runs on
renewable energy sources
There is a large amount of infrastructure (e.g. cost) required to convert from potential to deliverable energy.
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.
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 the incident solar energy 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. We have to build a device to convert incident solar power into deliverable power. For now, again we don't care what the device is, only that is efficiency lies somewhere in the range below:
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
These considerations suggestd that Solar PV rooftops (e.g. solar
shingles) are likely a generically good idea.
What about the prospect of "solar farms" and remote power delivery?
The relative inefficiency can be compensated for with collecting area.
Examples:
A site in Eastern Oregon receives 1200 watts per square
meter of solar radiation in July. Asuume that the solar
panels are 10% efficient and that the are illuminated
for 10 hours.
How many square meters would be required power Eugene
at 300 Megawatts?
each square meter gives you 1200 x.1 = 120 watts
Remember, we want 300,000,000 Watts, not just
120.
There are one million square meters in one square km.
so, each square km gives your 120 x 1,000,000 = 120 MW.
Therefore 2.5 square kilometers would be enough to Power Eugene
Ten square kilometers would give you about 1200 MW
keep this as a figure of merit reference under optimum conditions.
Now, of course, you wouldn't have continuous coverage, the individual
collectors would have to be spaced out
And so, in practice, the actual land use is about twice as great per MW generated. In addition, 1200 Watts per square meter is July is about 2.5 times larger than the annual average. These considerations then produce the following rubric:
10,000 Megawatts of Average Annual Solar Power (when the sun is shining) requires 500 square kilometers of collecting area. (at 10% efficiency)
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