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CEE 265 (11)
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# November26th.CEE265.docx

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Department
Civil And Environmental Engineering
Course
CEE 265
Professor
Phillip Savage
Semester
Fall

Description
CEE 265 November 26, 2013 SOLAR ENERGY RESOURCES Solar Photovoltaic Electricity  How does it work? o Need a material which is a semiconductor o Top band of a semiconductor is unoccupied o An electron has to be promoted from filled band up into conduction band  Need precise amount of energy  Has to do with discrete nature of energy levels  For silicon (the first material used as a photovoltaic material) the band energy is 1.12 eV o Photon from the sun can provide that energy o Then an electron can jump up to the conduction band o But only ½ of sun’s energy can be collected with a solar cell  We have losses due to absorbance, reflectance back into space, etc  What’s left has a jagged profile o How much of this energy is useful?  To get photon with 1.12 eV need photon with wavelength 1.11 microns  Anything less will not be able to move the electron to the top band o Principles of semiconductors limits the useable energy of sun to about ½  Silicon semiconductors are desirable because they are nonreactive, very easy to process, and very cheap (we plenty of it) o But we could improve the performance of other photovoltaics o Silicon has 4 valance electrons and it wants to get to 8 o So if we look at group three elements they have 3 valence electrons  Or group five has 5 valence electrons  Doping o If we take silicon and remove one of the atoms of replace it with a group 5 element the nucleus will have an extra proton  gives it a net positive charge  An extra electron will be provided as well  That electron can move around o Of if we take silicon and replace an atom with group 3 element we get one fewer proton  gives it a net negative charge  There is an electron “hole” now  And that hole becomes mobile now o When we dope semiconductors together they form a depletion region  That means the mobile electrons on the negative conductor will go over and fill the holes in the positive conduction  Forms a region that is very thin with a very strong directional electric field  Prevent re-absorption of electrons back into silicon o If the photons strike atoms in the junction region, those electrons will be immediately swept off to the current conductor  Then you get electricity to run a load  The depletion region pushes the electrons in the direction of the conductor  Junction creates a voltage basically o You can also use other materials that do not involve doping silicon  Photovoltaic Efficiency o Actual PV cells fall short of the 50% limit  Band gap voltage is only 50-75% of full value  Some of the holes and electrons recombine  Some photons are not absorbed (reflected instead)  Power is dissipated from internal cell resistance o How do we build up larger power?  Make PV modules with 36 cells  Put them in series and in parallel to increase current/voltage  Makes solar arrays o The power is current times voltage  Take the voltage in volts and the current amperes we get power in watts  Efficiencies are usually fairly low due to losses discussed above  But from a life cycle standpoint we don’t have too many expenses In-Class Problem 22.1: direct conversion efficiency  Calculate the conversion efficiency of a 10,000 W diesel generator o Consumes 1 gallon of diesel fuel per hour (138,000 BTU/gal) o Conversion: 1 Whr = 3.412 BTU2  How does that compare to a 100 m silicon cell that produces 10,000 W o Assume sun delivers 1,000 W/m 2  Compare the energy out over the energy in o Eout/Ein = 10,000W/(1,000W/m )*(100 m ) = 10% efficiency for PV o Eout/Ein =10,000W/(138,000 BTU/gallon)*(1 gallon/hr)*(1 Whr/3.412 BTU) = 25% efficiency for the diesel generator  We would also want to consider things like pollution from the diesel generator, emissions of carcinogens, the economics of each option, material costs, etc Land Use Requirements for Renewable Resources  Wind power has a large land footprint o We see the same thing for solar power o You need a lot more land area to gather up and concentrate the sun’s energy o We would need a land area the size of Wyoming to replace 1 cubic mile of oil  We may be putting PV cells in low land grade areas, but still a large footprint  1 cubic mile worth of energy would take… o 52 nuclear plants o 104 coal plants o 32,850 wind turbines o 91,250,000 solar panels  So obviously there is a tradeoff when we switch to renewable resources In-Class Problem 22.2: energy payback time and energy return on investme
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