{"id":2536,"date":"2019-09-20T10:24:18","date_gmt":"2019-09-20T17:24:18","guid":{"rendered":"https:\/\/c21-wp.phas.ubc.ca\/?post_type=article&p=2536"},"modified":"2019-10-22T12:56:12","modified_gmt":"2019-10-22T19:56:12","slug":"wind-power","status":"publish","type":"article","link":"https:\/\/c21.phas.ubc.ca\/article\/wind-power\/","title":{"rendered":"Wind Power"},"content":{"rendered":"
Wind energy is the kinetic energy of the air flow; its ultimate source is the solar energy, which drives convection in the Earth’s atmosphere. Wind power has been harnessed for possibly two millenia[note]Dodge, Darrel. Part 1 – Early History Through 1875: Wind Power’s Beginnings <\/i> (online). Illustrated History of Wind Power Development. http:\/\/www.telosnet.com\/wind\/early.html <\/a>[2019-09-20]. [\/note], but only now is it starting to become a significant part of the total power needs of industrialized nations[note] The Guardian windpower page, https:\/\/www.theguardian.com\/environment\/windpower<\/a> [2019-09-20].[\/note].<\/p>\n To calculate the rate at which kinetic energy, $KE = \\frac12mv^2$, can be extracted from an air stream, we will need the rate at which air passes through the rotor of the wind turbine. The rotor is made up of the blades, and the total swept area $A$ of these blades can be approximated by a circle whose radius is the blade length $r$, i.e. $A = \\pi r^2$. Now we can calculate the mass $m$ of air, density $\\rho$, velocity $v$, passing through this circle in time \u0394t<\/i> (Fig.1).<\/p>\n Fig.1. Wolfe Island wind farm (author photo).<\/p><\/div>\n From this, we can see that the mass is (Eqn.1):<\/p>\n $$\\begin{eqnarray} Note that the area $A$ is the area swept by the blades, not the area of the blades. This is because the blade moves much faster than the air and so each particle of air is affected by the blade. Therefore, the kinetic energy $KE$ of this mass of air is as follows (Eqn.2):<\/p>\n $$\\begin{eqnarray} Thus the power $P$ of the wind passing through this circle is (Eqn.3):<\/p>\n $$\\begin{eqnarray} The power that can be extracted from this air stream is less than $P$ because the air cannot stop dead on the other side the blades. Some kinetic energy has to remain in the air stream so it can continue to flow. The theoretical “Betz” limit is about 60%[note]Betz’ Law http:\/\/en.wikipedia.org\/wiki\/Betz’_law<\/a> [2019-09-20]. [\/note]. The limit is reached when the wind velocity downstream of the turbine is 1\/3 of the incident velocity. To understand this result, you have to recognize that the area of the incoming air stream is less than that of the outgoing, which is a simple result of mass conservation.<\/p>\n In addition, there are inevitable losses due to turbulence and inefficiencies in converting energy in the blade rotation to useful electrical energy.<\/p>\n The $v^3$ term emphasizes the need to have a high wind speed in order to capture a useful amount of power. Doubling the wind speed increases the power eight-fold.<\/p>\n
\nm = \\rho A v \\Delta t \\tag{1}
\n\\end{eqnarray}$$<\/p>\n
\nKE &=& \\dfrac{1}{2} m v^2 \\tag{2} \\\\
\n&=& \\dfrac{1}{2} \\rho A t v^3 \\nonumber
\n\\end{eqnarray}$$<\/p>\n
\nP &=& \\dfrac {\\dfrac{1}{2} \\rho A t v^3}{t} \\tag{3} \\\\
\n&=& \\dfrac{1}{2} \\rho A v^3 \\nonumber
\n\\end{eqnarray}$$<\/p>\n