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Introduction to Wind Energy
Updated on:  Monday, September 17, 2007 02:22 PM

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Many years ago David Johnson, P.E.,  helped a local Denver company develop an electronic control circuits for a small wind generator which they were perfecting.   They  were fortunate to be near the  National Wind Technology Center  (NWTC) test site near Boulder, Colorado.   They were able to do much of the testing using NWTC's automated data acquisition system. 

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For a comparison, suppose a photovoltaic solar energy system was spread out on the same 50 mile by 0.5 mile plot of land that my train system above might require.  If the solar system had a 10% fill factor (only 10% of the area would have solar cells), a 10% conversion efficiency, 6 hours of sun per day it would generate 4.2 million kilowatt-hours or about 420 million dollars worth of electricity each day.  Clearly for a given large area, the photovoltaic system produces more power.
If we can imagine a photovoltaic panel as big as the million square foot wind sail, each panel could generate 10 million watts. This figure is much less than a wind sail train at 30 MPH wind.  But, at 18 MPH, the sail would be no better than the PV area.
A 250,000 volt, 3 phase transmission line should be able to carry about 500 million watts with a current of 1000 amps.

1.356 joules = 1 foot-pound (pound of force x one foot distance)

1 joule = 0.7375 foot-pounds.

So, 10,000 pounds of force applied toward a mass moved 250,000 feet would require 3.4 billion joules of energy.
If we wanted a sailtrain to generate 44 million watts of power for a period of 5 hours, the energy would be 44 x 5 = 220 million watt-hours or 800 billion joules. If the distance is 50 miles or 250,000 feet, then the pulling force on the train would be 3.17 million pounds.  Clearly, a big rope would be needed for such a pull.  It would require over 32 one inch diameter ropes, each rated with a tensile strength of 100,000 pounds.  But, to withstand wind gusts, perhaps each rope would need to be rated at 1 million pounds
 

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