What percentage of light hits earth’s surface directly? 3 percent 5 percent 31 percent 47 percent

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.

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
  
  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 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
  
  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 sources

  

  There 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

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.

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

A typical household Winter energy use is around 2000-3000 KWHs per month or roughly 70-100 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 (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%).

Photons into Electrons: PhotoVoltaic Devices

Charge Generation

Photoelectric Effect

When photons strike a metal, their energy is used to liberate loosely bound electrons and therefore induce a current.

Efficiency of this process depends upon the material

This is the principle behind Digital Cameras (all college graduates should know how a digital camera works -- its a literacy test).

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.

Schematic structure of energy bands in Silicon:

Hence, if a silicon atom receives at least 1.11 Electron Volts from some source, a valence electron will move to the conduction band. Once an electron is in the conduction band, the material can carry a current and the material is now a conductor.

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.