Wind Turbine Power: Calculating Max Output

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The discussion focuses on calculating the maximum power output of wind turbines using the equation Power = 1/2ρAv^3, where area (A) can be derived from the circular area formula. A key question raised is whether to use the full circular area or just the "doughnut-like" area created by the turbine blades. It is clarified that the total kinetic energy calculation assumes unrealistic conditions, as extracting all energy would leave the wind with zero speed. The conversation also references the Betz Limit, which states that turbines can only achieve a maximum efficiency of 60%. Overall, the efficiency of wind turbines is highlighted as a critical factor in determining their actual power output.
Donald.
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Hello everyone, I've found these forums extremely helpful in the past and I've decided to create an account in order to answer a question that has been going in my mind and I can't seem to find the answer.

The power a wind turbine can generate can be derived from the equation:
Power = \frac{1}{2}\rho A v^3 and the area (A) of the turbine can be calculated using the area of a circle equation: A = \pi r^2.

This equation would give you the theoretical maximum power obtained by the turbine, however all turbine have a center from which the blades are attached. In a 'real life' situation would one use the whole circle or the 'doughnut like' shape area to calculate the maximum power that can be obtained?

The doughnut area could be calculated using the formula:
A = \pi r_{total}^2 - \pi r_{center}^2


Thank you,
Donald.
A = \pi (r_{total}^2 - r_{center}^2)
 
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Your formula is giving the total Kinetic Energy in the volume of air that passes each second. You could never get that much out at one location because that would imply that the wind has to have zero speed afterwards - and how do you get rid of it all?

Any turbine (air / gas / water) can only get a certain amount of the power out (finite efficiency) and I think the efficiency is, in fact, so low that the small adjustments you suggest, although correct, are not really relevant in the overall scheme of things.


This link suggests that you can't do better than 60% in any case. The Betz Limit, apparently. God that man was such a pessimist.
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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