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Energy Storage 2nd Edition
3.1 Availability of Solar Energy
Indicative to this explanation is a comparison between the solar energy
that is available from photovoltaic conversion mechanisms and the solar
energy that is available via plants’ conversion into combustible fuels, such
as methanol.
The total available energy via combustion from 1 gallon of gasoline in
an internal combustion engine is about 36kWh. The conversion efficiency
of an automobile engine to mechanical output is at most about 20%. Thus,
if we had very efficient electric motors instead, we would need to have
at least 1/5 of the 36kWh × 500 million gallons per day. That number is
3,500 million kilowatt-hours of energy from sunlight per day.
Again, if the conversion efficiency were 100%, dividing the numbers
would reveal that we need about 5,000 square miles of area to provide the
equivalent of gasoline. Unfortunately, we must multiply that figure by 10 to
account for conversion losses, giving us a total of about 50,000 square miles.
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However, the following problem remains: How does one use solar energy
to operate cars without storage?
The practicality and cost of solar energy for consumer use to power
industry and residential homes has become a widespread concern that
stems from questioning energy sources and their limits. A simple estimate
of available solar energy per unit land area, especially if it is to be collected
and transformed directly into electric power, reveals that immense land
areas are required to provide even modest amounts of energy for commercial use as illustrated below.
The following calculations present a rather dismal prospect for widespread use of solar energy as a substitute for the more “conventional”
sources presently in use. Regardless, the problem of providing energy to
meet the increasing demands of the future is very serious. Other than oil
and coal, there are no realistic answers. Nuclear, along with cheap and
reliable energy storage (such as batteries), could be an answer to welldesigned, plug-in hybrid vehicles, giving a range of 200 or more miles on
a single charge.
As an example, consider a middle region of the North American
Continent and use simple approximations. At high noon (normal incidence) on a cloudless day at the Equator, incident solar power surface density is in the range of 0.12 watts per square cm. Since sunlight is available
only half the day (about 10 hours of useful daylight), the energy density is
reduced to approximately 1.2 watt-hours per square cm per day.
And, due to Earth’s rotation on its axis, the sun’s radiation makes a
changing angle to the Earth’s surface perpendicular. Therefore, we can simply approximate that by another 50% factor. Hence, the energy density over
a 24-hour period is further reduced to 0.6 watt-hours per square cm.
The above energy per unit area per day becomes 0.6Wh/cm2 = 3.6 Wh/
2
in , or 3.6 Wh/in2 = 500 Wh/ft2.
Since there are about 5,300square feet per square mile, there are
5300×5300×500 = 28 million × 500WH from 1 square mile, or 15,000 megawatt hours per square mile in any one clear day at the equator.
At latitudes of the mid-United States region, we might need to divide
that number by two, making the total sunlight energy equal to about 7,000
megawatt-hours per day per square mile of area.
Now let’s take a brief look at what the order of magnitude of the needs
are for domestic energy per day or per year in the United States. According
to the website Globilis, the amount of energy produced and consumed
per year in 2002 was 4 million kilowatt-hours. That number reduces to
10,000 million kWh per day consumption.
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Energy Storage 2nd Edition
So, if we want to produce (assuming 100% conversion efficiency of electromagnetic energy from the Sun) that amount of energy, we would need
an area of about 1400 square miles. Present-day efficiencies of solar photovoltaic cells can be in the order of 10% to 20%. Using these lower efficiency
figures, we need 7,000 to 14,000 square miles of accessible area to generate
the electrical energy presently being used by the United States. Imagine
the maintenance and access roads required for such a solar collector field.
Simply the problem of keeping the surfaces of semi-conductors and collectors clean and repaired would be monumental.
Now, consider the gasoline situation. The United States used 21 million
barrels of oil per day in 2007. The yield at the refineries was about 20 gallons of gasoline per barrel of oil, which means the nation actually used over
400 million gallons of gasoline per day – an astronomical figure.
The problem remains unsolved. In addition, we must also have available
storage to make the energy portable and useable whatever peak demands
arise.
Now, consider corn as a source of methanol. The data from the US
Department of Agriculture shows that the average yield of corn per growing period of at least six months is about 150 bushels per acre. A bushel
of corn will yield 2.5 to 3 gallons of ethanol. Ethanol has less than half
the energy content per gallon than gasoline. Thus, we would obtain about
200 gallons of gasoline equivalent per acre of cornfield.
Returning to the gasoline consumption rate above, if the United States
consumes about 400 million gallons per day multiplied by 300 days per
year, then the country consumes about 12,000 million gallons per year.
After dividing the numbers for corn yield, close to 100 million acres of
farmland would be required.
A square mile is equal to 700 acres. After dividing again, it appears that
about 150,000 square miles of farmland is needed under ideal circumstances – no provision is made here for roads, buildings, fertilizer, machinery, etc., – which is not very encouraging.