This is the most frequently asked question that we receive and this page describes the most important concept to understand when playing with wind turbines. Understanding this page is the key to understanding just about everything else in regards to wind turbines.

How much power can a wind turbine make?

This depends on 3 main Variables:   the size of the turbine the speed of the wind the efficiency of the turbine and generator

Consider the power available from the wind: the wind power equation.

Example:

For a VAWT with 4' tall wings and a 3' diameter arc, the swept area A=12 ft² = 1.1m²

Wind speed @ 15 mph = 6.7 m/sec

So the Wind Power at 15 mph is:  P=½(1.2)(1.1)(6.7)³ =  198 watts

Consider if the wind speed doubles to 30 mph (13.4 m/s)...notice that the power increases more than 8 times!!

Wind Power at 30 mph = P=½(1.2)(1.1)(13.4)³ =  almost 1600 watts!

Even though the wind has that much power,  it's not possible to extract it all due to losses. A typical wind turbine efficiency (C_t) is about 40%, and axial flux alternator efficiencies (C_a) are about 60% efficient*. These numbers will vary depending on the designs and their specific performance curves. But, it's a good starting point.

So you could expect about 40% of 60% of  the wind power.  Which is (.40)(.60)(1600watts) = 384 watts.   Not bad for a 12 square foot turbine!

The Wind Power equation, then becomes the Turbine Power Equation:

Turbine Power Equation:  P=½ρAV³C_tC_a

There are no standards for reporting wind turbine outputs,  so the information out there can be very confusing and sometimes misleading if not down right false.  Info about a turbine such as:  "1.5kW "  is really useless information with out knowing the details at this rating. Mainly the wind speed and turbine size (swept area).  Does it mean that the turbine will output 1.5kW? or is that the max output rating for safe operation? or perhaps something else...

When considering a wind turbine project, it is best to understand the wind power equation, P=½ρAV³  using your expected wind speeds and turbine size. Then apply the typical efficiencies to account for losses from the turbine (C_t = 40%)  and from the alternator ( C_a = 60%) and you should have an idea of what to expect in watts at that particular wind speed. 

If you are designing a turbine, do the same thing.  Then you can design the electrical system around the expected power range that you are dealing with.

* In regards to the alternator efficiency,  it varies along it's power curve.  In dual rotor alternators, we typically see about 80- 85% efficiency at the low power ranges. Power in the form of heat is dissipated or lost through the resistance of the stator coils as described by P=I²R, where I is the current (amps) and R is the resistance (ohms) of the stator coils.  As the current increases, more power is lost through heat generated in the coils. This is common for all electro-mechanical machines.  The efficiency will continue to diminish as the power increases. The design goal is to keep the electrical resistance R as low as possible by using the largest/thickest wire gage in the coils with the fewest turns (shortest wire length) needed to produce the desired voltage.  Ideally, your coils will fill all of the available space around the stator assuring that you've used the thickest wire possible.

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