how do you calculate electricity generated by windmills?

Answer 1

Power is the product of force and velocity. The pressure produced by the wind of speed v depends on the relative speed between the blade and the wind (vr) and on the density of air (d):

P = (d * vr^2)/2

The force is the product of pressure and area (A) swept by the blades. The power (Q) is the product of the force and the generator velocity (vg). The wind velocity is the sum of the vr and vg.

So the power is Q = P * vg = (d * (v - vg)^2)*vg * A/2

By taking the derivative, you can maximize power when vg = v/3. The maximumpower then is:

Q = (d * (v - vg)^2)*vg/2 = (2 * d * A * v^3)/27 = 0.074 d*A*v^3

The density of air is 1.2 kg/m^3

This will give you the maximum mechanical power from a windmill. Electric generators operate around 80% efficient so multiply the above by 80%. If you put this all together you find that a stiff wind (15 mph) can generate 2 watts per square foot.

All in all, the 2 watts per square foot is much of the reason why you don't see many windmills. A typical commercial power plant produces 1000 megawatts. For a windmill field to do this, you need 500,000,000 square feet of windmills. That means 16,000 windmills, each 200 ft in diameter.

Answer 2

The best crude indication of a wind turbine's energy production capabilities is its rotor diameter--which determines its swept area, also called the capture area. A wind turbine may have an impressive "rated power" of 100 kW, but if its rotor diameter is so small that it can't capture that power until the wind speed reaches 40 mph (18 m/s), the wind turbine won't rack up enough time at high power output to produce a reasonable annual energy output.

Expected energy output per year can be reliably calculated when the wind turbine's capacity factor at a given average annual wind speed is known. The capacity factor is simply the wind turbine's actual energy output for the year divided by the energy output if the machine operated at its rated power output for the entire year. A reasonable capacity factor would be 0.25 to 0.30. A very good capacity factor would be 0.40.

NOTE: Capacity factor is very sensitive to the average wind speed. When using the capacity factor to calculate estimated annual energy output, it is extremely important to know the capacity factor at the average wind speed of the intended site.

Lacking a calculated capacity factor, the machine's power curve can actually provide a crude indication of the annual energy output of any wind turbine. Using the power curve, one can find the predicted power output at the average wind speed at the wind turbine site. By calculating the percentage of the rated power (RP) produced at the average wind speed, one can arrive at a rough capacity factor (RCF) for the wind turbine at that site. And by multiplying the rated power output by the rough capacity factor by the number of hours in a year, (8,760), a very crude annual energy production can be estimated. For example, for a 100 kW turbine producing 20 kW at an average wind speed of 15 mph, the calculation would be:

100 kW (RP) x .20 (RCF) = 20 kW x 8760 hours = 175,200 kWh

See: http://www.awea.org/faq/basicen.html