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| − | <h2> Wind Power </h2>
| + | == Wind Power == |
| − | <p>The power <i>P </i>of a wind-stream, crossing an area <i>A </i>with velocity <i>v </i>is given by
| + | |
| − | </p><p> <img _fckfakelement="true" _fck_mw_math="P=\frac{1}{2}\rho A v^3" src="/images/math/a/f/f/aff47ae2d0c794edc79c72dc23327697.png" /><br />
| + | The power ''P ''of a wind-stream, crossing an area ''A ''with velocity ''v ''is given by |
| − | </p><p>It varies proportional to air density <span class="texhtml">ρ</span>, to the crossed area <i>A </i>and to the cube of wind velocity <i>v</i>. 
| + | |
| − | </p><p>The Power <i>P </i>is the kinetic energy
| + | <math>P=\frac{1}{2}\rho A v^3</math><br> |
| − | </p><p><img _fckfakelement="true" _fck_mw_math="E=\frac{1}{2}mv^2" src="/images/math/d/0/9/d09bd4120bbded4606433d6eb4539e7c.png" /> | + | |
| − | </p><p>of the air-mass <i>m </i>crossing the area <i>A </i>during a time interval <br />
| + | It varies proportional to air density <math>\rho</math>, to the crossed area ''A ''and to the cube of wind velocity ''v''. |
| − | </p><p><img _fckfakelement="true" _fck_mw_math="\dot{m}=A \rho \frac{dx}{dt}=A\rho v" src="/images/math/b/3/e/b3ef61411620083fa8c4d12f2df4d414.png" />. | + | |
| − | </p><p>Because power is energy per time unit, combining the two equations leads back to the primary mentioned basic relationship of wind energy utilisation
| + | The Power ''P ''is the kinetic energy |
| − | </p><p><img _fckfakelement="true" _fck_mw_math="P=\dot{E}=\frac{1}{2}\dot{m}v^2=\frac{1}{2}\rho A v^3" src="/images/math/e/d/d/eddae381c857bc2114fd643b32111bf9.png" /> | + | |
| − | </p><p>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 (<span class="texhtml"><i>v</i><sub>1</sub></span>) and behind the rotor area (<span class="texhtml"><i>v</i><sub>2</sub></span>) is <span class="texhtml"><i>v</i><sub>1</sub> / <i>v</i><sub>2</sub> = 1 / 3</span>. The maximum power extracted is then given by<br />
| + | <math>E=\frac{1}{2}mv^2</math> |
| − | </p><p><img _fckfakelement="true" _fck_mw_math="P_{Betz}=\frac{1}{2} \rho A v^3 c_{P.Betz}" src="/images/math/0/8/3/08377cd8c23f47f4ce336b72d8422baf.png" /> | + | |
| − | </p><p>where <span class="texhtml"><i>c</i><sub><i>p</i>.<i>B</i><i>e</i><i>t</i><i>z</i></sub> = 0,59</span> 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 <span class="texhtml"><i>c</i><sub><i>p</i>.<i>B</i><i>e</i><i>t</i><i>z</i></sub> = 0,5</span>.
| + | of the air-mass ''m ''crossing the area ''A ''during a time interval <br> |
| − | </p> | + | |
| − | <h2> Unit abbreviations </h2> | + | <math>\dot{m}=A \rho \frac{dx}{dt}=A\rho v</math>. |
| − | <table width="399" cellspacing="1" cellpadding="1" border="0" align="left" style=""> | + | |
| | + | Because power is energy per time unit, combining the two equations leads back to the primary mentioned basic relationship of wind energy utilisation |
| | + | |
| | + | <math>P=\dot{E}=\frac{1}{2}\dot{m}v^2=\frac{1}{2}\rho A v^3</math> |
| | + | |
| | + | 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> |
| | + | |
| | + | <math>P_{Betz}=\frac{1}{2} \rho A v^3 c_{P.Betz}</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>. |
| | + | |
| | + | == Unit abbreviations == |
| | + | |
| | + | {| width="399" cellspacing="1" cellpadding="1" border="0" align="left" style="" |
| | + | |- |
| | + | | m = metre = 3.28 ft.<br> |
| | + | | HP = horsepower<br> |
| | + | |- |
| | + | | s = second<br> |
| | + | | J = Joule<br> |
| | + | |- |
| | + | | h = hour<br> |
| | + | | cal = calorie<br> |
| | + | |- |
| | + | | N = Newton<br> |
| | + | | toe = tonnes of oil equivalent<br> |
| | + | |- |
| | + | | W = Watt<br> |
| | + | | Hz = Hertz (cycles per second)<br> |
| | + | |} |
| | + | |
| | + | <br> |
| | + | |
| | + | <br> |
| | + | |
| | + | <br> |
| | + | |
| | + | <br> |
| | + | |
| | + | <math>10^{-12}</math> = p pico = 1/1000,000,000,000 |
| | + | |
| | + | <math>10^{-9}</math> = n nano = 1/1000,000,000 |
| | + | |
| | + | <math>10^{-6}</math> = µ micro = 1/1000,000 |
| | + | |
| | + | <math>10^{-3}</math> = m milli = 1/1000 |
| | + | |
| | + | <math>10^{3}</math> = k kilo = 1,000 = thousands |
| | + | |
| | + | <math>10^{6}</math> = M mega = 1,000,000 = millions |
| | + | |
| | + | <math>10^{9}</math> = G giga = 1,000,000,000 |
| | + | |
| | + | <math>10^{12}</math> = T tera = 1,000,000,000,000 |
| | + | |
| | + | <math>10^{15}</math> = P peta = 1,000,000,000,000,000 |
| | + | |
| | + | |
| | + | |
| | + | [[Portal:Wind]] |
| | + | |
| | | | |
| − | <tr>
| |
| − | <td> m = metre = 3.28 ft.<br />
| |
| − | </td>
| |
| − | <td> HP = horsepower<br />
| |
| − | </td></tr>
| |
| − | <tr>
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| − | <td> s = second<br />
| |
| − | </td>
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| − | <td> J = Joule<br />
| |
| − | </td></tr>
| |
| − | <tr>
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| − | <td> h = hour<br />
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| − | </td>
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| − | <td> cal = calorie<br />
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| − | </td></tr>
| |
| − | <tr>
| |
| − | <td> N = Newton<br />
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| − | </td>
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| − | <td> toe = tonnes of oil equivalent<br />
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| − | </td></tr>
| |
| − | <tr>
| |
| − | <td> W = Watt<br />
| |
| − | </td>
| |
| − | <td> Hz = Hertz (cycles per second)<br />
| |
| − | </td></tr></table>
| |
| − | <p><br />
| |
| − | </p><p><br />
| |
| − | </p><p><br />
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| − | </p><p><br />
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| − | </p><p><span class="texhtml">10<sup> − 12</sup></span> = p pico = 1/1000,000,000,000
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| − | </p><p><span class="texhtml">10<sup> − 9</sup></span> = n nano = 1/1000,000,000
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| − | </p><p><span class="texhtml">10<sup> − 6</sup></span> = µ micro = 1/1000,000
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| − | </p><p><span class="texhtml">10<sup> − 3</sup></span> = m milli = 1/1000
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| − | </p><p><span class="texhtml">10<sup>3</sup></span> = k kilo = 1,000 = thousands
| |
| − | </p><p><span class="texhtml">10<sup>6</sup></span> = M mega = 1,000,000 = millions
| |
| − | </p><p><span class="texhtml">10<sup>9</sup></span> = G giga = 1,000,000,000
| |
| − | </p><p><span class="texhtml">10<sup>12</sup></span> = T tera = 1,000,000,000,000
| |
| − | </p><p><span class="texhtml">10<sup>15</sup></span> = P peta = 1,000,000,000,000,000
| |
| − | </p><p><br />
| |
| − | </p><p><a _fcknotitle="true" href="Portal:Wind">Portal:Wind</a>
| |
| − | </p>
| |
| | | | |
| | [[Category:Wind]] | | [[Category:Wind]] |
Revision as of 10:09, 16 May 2012
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.
|
HP = horsepower
|
s = second
|
J = Joule
|
h = hour
|
cal = calorie
|
N = Newton
|
toe = tonnes of oil equivalent
|
W = Watt
|
Hz = Hertz (cycles per second)
|
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
Portal:Wind
