The rotor diameter of a wind turbine is 124 ft. The wind speed is 15 miles per h
ID: 2290960 • Letter: T
Question
The rotor diameter of a wind turbine is 124 ft. The wind speed is 15 miles per hour. The rotor efficiency is 0.37. The density of the air is 1.225 kg/m3.+- a. Calculate the amount of wind power available for electric power generation? 4- b. Calculate the actual wind power that could be extracted using the wind turbine? c. If the generator and the rest of the system efficiency is 85%, what would be the actual power produced by the turbine? - d. What would be the power if the wind speed is doubled? For the same wind speed, what would be the power if the fotor diameter is reduced to 64 ft?Explanation / Answer
(a) amount of wind power available for elkectric power generation P = 0.5 *p*A*V^3
p=density of Air = 1.225 Kgm^-3
A= Area = pi*r^2 = pi * (0.5D)^2 = (22/7)*(0.5*37.79)^2 = 1122.06 m^2
V= wind speed = 15 mile/hour = 6.70 meter/sec
putting the values , P = 0.5*1.225*1122.06*6.7^3 = 206702.90 watts
(b) Actual Wind power that could be extracted P1= E*P=76480.07 watts
E= efficiency = 0.37
(c) Actual Turbine Power P2= E1*P1=0.85*76480.07=65008.05 watts
(d) Power is directly proportional to cube of the wind speed
therefore as wind speed doubles power increases by 8times.
As rotor diameter is reduced to 64 ft i.e 64*0.3048=19.50 meter
Power approximately close to one quarter P2.
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