An explanation of Betz’ Law

**How did Betz come up with this?**

The work done on the turbine = change in kinetic energy of the wind: *W* =*ฮK*. The speed *v*_{2} behind the turbine is slower than the speed in front of the turbine *v*_{1}, 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

\begin{eqnarray}

\dfrac {\Delta m}{\Delta t} &=& \rho A v_{av} \nonumber \\

&=& \rho A \dfrac{1}{2}(v_1 + v_2) \nonumber

\end{eqnarray}

while the available wind power due to this is

\begin{eqnarray}

P &=& \dfrac{\Delta K}{\Delta t} \nonumber \\

&=& \dfrac{\dfrac{1}{2} \Delta m(v_1^2 – v_2^2)}{\Delta t} \nonumber \\

&=& \dfrac{1}{4} \rho A(v_1 + v_2)(v_1^2 – v_2^2) \nonumber

\end{eqnarray}

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

\begin{eqnarray}

P_o &=& \dfrac{K}{\Delta t} \nonumber \\

&=& \dfrac{1}{2} \rho Av_1^3 \nonumber

\end{eqnarray}

When graphing *P*/*P*_{o}, the maximum power output is found at *v*_{2}/*v*_{1 } = 0.33 (Fig. 3).

**Figure 3.**The plot agrees with Betz’s conclusions that the maximum power output (of 59.3%) occurs when *v _{2}* is 1/3 of

*v*. To view the spreadsheet used to produce this graph, see Betz_Law_Spreadsheet_Data.xls.

_{1}