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| | == Wind Power == | | == Wind Power == |
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| | The power of a wind-stream is transformed into mechanical energy by a wind turbine through slowing down the moving air-mass which is crossing the rotor area. For a complete extraction of power, the air-mass would have to be stopped completely, leaving no space for the following air-masses. Betz and Lanchester found, that the maximum energy can be extracted from a wind-stream by a wind turbine, if the relation of wind velocities in front of (<math>v_1</math>) and behind the rotor area (<math>v_2</math>) is <math>v_1/v_2=1/3</math>. The maximum power extracted is then given by<br> | | The power of a wind-stream is transformed into mechanical energy by a wind turbine through slowing down the moving air-mass which is crossing the rotor area. For a complete extraction of power, the air-mass would have to be stopped completely, leaving no space for the following air-masses. Betz and Lanchester found, that the maximum energy can be extracted from a wind-stream by a wind turbine, if the relation of wind velocities in front of (<math>v_1</math>) and behind the rotor area (<math>v_2</math>) is <math>v_1/v_2=1/3</math>. The maximum power extracted is then given by<br> |
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| − | <math>P_{Betz}=\frac{1}{2} \rho A v^3 c_{P.Betz}</math> | + | <math>P_{Betz}=\frac{1}{2} \rho A v^3 c_{P.Betz}</math> |
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| − | where <math>c_{p.Betz}=0,59</math> is the power coefficient giving the ratio of the total amount of wind energy which can be extracted theoretically, if no losses occur. Even for this ideal case only 59% of wind energy can be used. In practice power coefficients are smaller: todays wind turbines with good blade profiles reach values of <math>c_{p.Betz}=0,5</math>. | + | where <math>c_{p.Betz}=0,59</math> is the power coefficient giving the ratio of the total amount of wind energy which can be extracted theoretically, if no losses occur. Even for this ideal case only 59% of wind energy can be used. In practice power coefficients are smaller: todays wind turbines with good blade profiles reach values of <math>c_{p.Betz}=0,5</math>. |
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| | == Unit abbreviations == | | == Unit abbreviations == |
Revision as of 13:44, 31 July 2011
Work in Progress - This article is going to be finished until 7.8.2011
Wind Power
The power P of a wind-stream, crossing an area A with velocity v is given by
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P=\frac{1}{2}\rho A v^3
It varies proportional to air density Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \rho
, to the crossed area A and to the cube of wind velocity v.
The Power P is the kinetic energy
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): E=\frac{1}{2}mv^2
of the air-mass m crossing the area A during a time interval
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): \dot{m}=A \rho \frac{dx}{dt}=A\rho v
.
Because power is energy per time unit, combining the two equations leads back to the primary mentioned basic relationship of wind energy utilisation
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P=\dot{E}=\frac{1}{2}\dot{m}v^2=\frac{1}{2}\rho A v^3
The power of a wind-stream is transformed into mechanical energy by a wind turbine through slowing down the moving air-mass which is crossing the rotor area. For a complete extraction of power, the air-mass would have to be stopped completely, leaving no space for the following air-masses. Betz and Lanchester found, that the maximum energy can be extracted from a wind-stream by a wind turbine, if the relation of wind velocities in front of (Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): v_1
) and behind the rotor area (Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): v_2
) is Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): v_1/v_2=1/3
. The maximum power extracted is then given by
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): P_{Betz}=\frac{1}{2} \rho A v^3 c_{P.Betz}
where Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): c_{p.Betz}=0,59
is the power coefficient giving the ratio of the total amount of wind energy which can be extracted theoretically, if no losses occur. Even for this ideal case only 59% of wind energy can be used. In practice power coefficients are smaller: todays wind turbines with good blade profiles reach values of Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): c_{p.Betz}=0,5
.
Unit abbreviations
m = metre = 3.28 ft.
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HP = horsepower
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s = second
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J = Joule
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h = hour
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cal = calorie
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N = Newton
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toe = tonnes of oil equivalent
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W = Watt
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Hz = Hertz (cycles per second)
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Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): 10^{-12}
= p pico = 1/1000,000,000,000
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): 10^{-9}
= n nano = 1/1000,000,000
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): 10^{-6}
= µ micro = 1/1000,000
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): 10^{-3}
= m milli = 1/1000
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): 10^{3}
= k kilo = 1,000 = thousands
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): 10^{6}
= M mega = 1,000,000 = millions
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): 10^{9}
= G giga = 1,000,000,000
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): 10^{12}
= T tera = 1,000,000,000,000
Failed to parse (PNG conversion failed; check for correct installation of latex and dvipng (or dvips + gs + convert)): 10^{15}
= P peta = 1,000,000,000,000,000
