The Sun
How many photons (energy) reach the surface of the Earth on Average?
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.
- 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
this has caused some of use stratospheric cooling as an argument against
the occurence of global warming. The two are not connected
at all.
- 4% gets absorbed by clouds located in the troposphere. This
is the lower part of the earth's atmosphere where weather happens.
This part of the equilibrium cycle is being changed as the troposphere,
particularly at tropical latitudes, is getting cloudier.
- 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 Therefore it makes absolutely no sense to put solar panels in orbit and then "beam" the energy back to the surface.
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 TIME1 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 So the entire planet receives 84 Terrawatts of Power our current worldwide consumption is about 12 Terrawatts so is this a solution?
and remember, little of the world current runs on
renewable energy sourcesThere is a large amount of infrastructure (e.g. cost) required to convert from potential to deliverable energy.
- 8 hour summer day, 40 degree latitude 600 Watts per sq. meter
- 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
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
2000 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
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:
- 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 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
Solar Energy: Collection, Energy Generation and Heat Transfer
How Solar Energy is Used:
- Two Choices:
- Heat Water into Steam
- Turn photons into electrons
Solar Thermal Power:
- Focus sunlight on a bucket of water
This requires about 2000 heliostats Maintenance, initial costs make energy
expensive (25-50 cents per KWH).
- Direct conversion into electricity Photovoltaics; conversion of solar photons into electrons that flow down a semi-conductor. Main problem is low efficiency (about 10%).
Charge Generation
To make use of the photoelectric effect, we need material that is a good conductor of electricity and which can be manufactured in bulk at reasonable cost. This conditions strongly constrain the available choices. For most practical aspects, Silicon is the material of choice.
Silicon:
- is abundant on the earth and readily found in the crust.
It is a direct product of fusion inside stars. It can be easily
recovered from the crust and mass-produced. Computers
are cheap because silicon works well for circuit boards and is
an easily recoverable material from the earth's crust.
The result is a world-wide economy centered around semi-conductor technology.
- Has four outer (valence) electrons to bond silicon atoms together
in a crystal
- Under normal circumstances, there are no free electrons available
in silicon to conduct electricity. All the electrons are used to
bind the atoms in place to form the crystal.
- The conduction band is empty and therefore no current can be carried by the material.
So much energy is 1.11 Electron Volts?
- 1.11 eV corresponds to the energy that a photon of wavelength 1.12 microns has.
- 77% of the energy from the sun is carried in photons with wavelength less than this and therefore can move a valence electron in silicon into the conducting band.