The Enercon E-125 wind turbine has a nameplate capacity of 7.5MW. Its rotor diam

Question

The Enercon E-125 wind turbine has a nameplate capacity of 7.5MW. Its rotor diameter is 126m. Its ideal tip speed ratio is approximately 5. Assume an Enercon turbine reaches its nameplate output in 11m/s winds, and is at its ideal tip speed ratio.
What is the rpm (revolutions per minute) of the wind rotor?
How fast is the tip of the blade moving?
What is the efficiency of wind energy-to-electrical energy conversion at this condition?
If the Enercon turbine (a direct drive, gearbox-free design) has a 60 pole electric rotor/generator, what is the output AC frequency under the above conditions?

Explanation / Answer

Given Wind Turbine

P = 7.5 MW

Rotor diameter d = 126 m

Ideal tip speed ratio = 5

Wind speed at which turbine reaches to maximum power generation

a) The rpm of the wind rotor

we know tip speed ratio = tip speed of turbine blades/speed of wind

5 = tip speed of blade / 11,

CIrcumference of wind turbine is pi*d = pi*126 = 395.84 m

Tip speed is 55 m/s , time taken to one revolution is 7.197 seconds . No.of revolutions in 60 seconds is 8.33 rpm

N = 8.33 rpm

b)

Tip of the blade moves with speed of 55 m/s

tip speed ratio = speed of tip of blade/speed of wind.

c)

Theoritical power output is

P = (Density of air = 1.225 kg/m2)*(Velocity of air = 11)3*(Area swept by wind blades=pi*d2/4)*0.5

P = (0.5*1.225*113*pi*1262)/4

= 10165181.1 W = 10.165 MW

Efficiency = (Actual power output/Theoritical power output) = (7.5/10.165) *100 = 73.78 %

d)

Given poles of generator is Poles = 60

We know frequency f = N*Poles/(120) = 8.33*60/120 = 4 Hz

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