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| The concept of the theoretical efficiency |
Dr. Albert Betz’ theory about windmills from 1926 Nomenclature
![]() Using the law about continuity, we get: The ratio between the velocities is: A geometrical relation gives: The difference between the powers before and after the mill is equal to the power that enters the mill. The kinetic energy in one cubic meter of air, could be found in this way: The dimension is joule per cubic metre. Now! one J/m3 is equal to the pressure measured in pascal Pa. Before the mill we have the area A0 and after the mill A3. The three areas are increasing and the three velocities are decreasing. In total, we have six unknowns by using two mean values we finally have just two unknowns. ![]() Overall, we have for the power: For the mill this would be: ![]() The mill receives this power. In front of the mill, we have the power: ![]() The ratio between these two powers is the efficiency.
Finally, we get the beautiful result:
![]() Compare this with the formula by Carnot. ![]() (See: University Physics under “The Carnots Cycle”). The law of continuity should bed used on the space in front of and behind the mill. By means of:
These expressions are substituted into the power law given above:
or: ![]() We will try to optimise the power equation. Then we have to differentiate the equation and solve a second order equation. ![]() There is a misprint in the last equation…find this misprint There are two solutions to the equation: ![]() This means that the mill will get its maximum power if it is constructed so, that the ratio between the two velocities is equal to one to three. The maximum efficiency will be: ![]() Multiplying with 3/3, we get the ratio to 24/27. This rewriting will be obvious if we look into the next chapter. The result do we find in many books but not often, the calculation is given. We will now try to find some other efficiencies but in relation to the three areas. This ratio has the dimension watt per square meters.. From here, we get the ratios: ![]() ![]() Using the ratio 1/3 which gives the maximum power, we get: , , ![]() |