*A common question I come across is...*

**How Can I Calculate the Amount of Power Available at a Given Wind Speed?**

Since air has mass and it moves to form wind, it has kinetic energy. You may remember from college that:

kinetic energy (joules) = 0.5 x m x V2

where:m = mass (kg) (1 kg = 2.2 pounds)V = velocity (meters/second) (meter = 3.281 feet = 39.37 inches)

Usually, we're more interested in power than energy. Since energy = power x time and density is a more convenient way to express the mass of flowing air, the kinetic energy equation can be converted into a flow equation as below:

**Power in the area swept by the wind turbine rotor:**

P = 0.5 x rho x A x V3

where:P = power in watts (746 watts = 1 hp) (1,000 watts = 1 kilowatt)rho = air density (about 1.225 kg/m3 at sea level, less higher up)A = rotor swept area, exposed to the wind (m2)V = wind speed in meters/sec (20 mph = 9 m/s) (mph/2.24 = m/s)

This yields the power in a free flowing stream of wind. Of course, it is impossible to extract all the power from the wind because some flow must be maintained through the rotor (otherwise a solid wall would be a 100% efficient wind power extractor). So, we need to include some additional terms to get a practical equation for a wind turbine.

**Wind Turbine Power:**

P = 0.5 x rho x A x Cp x V3 x Ng x Nb

where:P = power in watts (746 watts = 1 hp) (1,000 watts = 1 kilowatt)rho = air density (about 1.225 kg/m3 at sea level, less higher up)A = rotor swept area, exposed to the wind (m2)Cp = Coefficient of performance (.59 {Betz limit} is the maximum thoretically possible, .35 for a good design) V = wind speed in meters/sec (20 mph = 9 m/s)Ng = generator efficiency (50% for car alternator, 80% or possibly more for a permanent magnet generator or grid-connected induction generator)Nb = gearbox/bearings efficiency (depends, could be as high as 95% if good)

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