Wind Turbines - Betz Law Explained

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How did Betz come up with this?

The work done on the turbine = change in kinetic energy of the wind: W =ΔK. The speed v2 behind the turbine is slower than the speed in front of the turbine v1, and the average speed at the location of the turbine is

$ v_{av} = \dfrac{1}{2}(v_1 + v_2) $

The mass streaming through the turbine, found as above, is

<br />
\begin{eqnarray}<br />
   \dfrac {\Delta m}{\Delta t} &=& \rho A v_{av} \nonumber \\<br />
   &=& \rho A \dfrac{1}{2}(v_1 + v_2) \nonumber<br />
   \end{eqnarray}<br />

while the available wind power due to this is

<br />
\begin{eqnarray}<br />
   P &=& \dfrac{\Delta K}{\Delta t} \nonumber \\<br />
   &=& \dfrac{\dfrac{1}{2} \Delta m(v_1^2 - v_2^2)}{\Delta t} \nonumber \\<br />
   &=& \dfrac{1}{4} \rho A(v_1 + v_2)(v_1^2 - v_2^2) \nonumber<br />
   \end{eqnarray}<br />

and the undisturbed wind power (the power of the wind if it did not pass through the turbine) is

<br />
\begin{eqnarray}<br />
   P_o &=& \dfrac{K}{\Delta t} \nonumber \\<br />
   &=& \dfrac{1}{2} \rho Av_1^3 \nonumber<br />
   \end{eqnarray}<br />

When graphing P/Po, the maximum power output is found at v2/v1 = 0.33 (Fig. 3).

Figure 3.The plot agrees with Betz's conclusions that the maximum power output (of 59.3%) occurs when v2 is 1/3 of v1. To view the spreadsheet used to produce this graph, see Betz_Law_Spreadsheet_Data.xls

Comments

Great teaching aid for my

Great teaching aid for my home schooled high school student in physics. At the present we are in momentum.

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