The strong wind speed of a certain place is 14m / s, the air density is 1.3kg/m ^ 3, and the wind with cross-sectional area of 400m ^ 2 makes the wind turbine rotate, If 20% of its kinetic energy is converted into electric energy, how many watts of power is generated? (the air to the wind blade in T seconds is contained in a cylinder with s as the bottom area and L = VT in length)

The strong wind speed of a certain place is 14m / s, the air density is 1.3kg/m ^ 3, and the wind with cross-sectional area of 400m ^ 2 makes the wind turbine rotate, If 20% of its kinetic energy is converted into electric energy, how many watts of power is generated? (the air to the wind blade in T seconds is contained in a cylinder with s as the bottom area and L = VT in length)

It should be 142.688 kW

Suppose the diameter of the wind wheel is D, the utilization rate of wind energy is Q, and the air density is X. when the wind speed is V, what is the power of the wind turbine
(wind power converts part of the kinetic energy of the wind passing through the wind into electricity)

P = (1 / 2mV ^ 2) / T = (1 / 2xv (volume) V ^ 2) / T
Volume v = x (density) Pai * (D / 2) ^ 2 * l (length)
L (length) / T = V (velocity)
After finishing, you can get it
The final result is p = 1 / 8pai * x * d ^ 2 * V ^ 3

The wind speed of strong wind is 20m / s, the air density is p = 1.3kg/m3, and the effective wind area of a wind turbine is s = 20m2,
The wind speed of a strong wind in a certain place is 20m / s, and the air density is p = 1.3kg/m3. If the kinetic energy of the wind speed passing through the cross-sectional area s = 20m ^ 2 is all converted into electric energy, the expression of power is written, and its size is calculated by using the above-mentioned known quantity (take 2 significant figures)

P＝1/2ρSV^3＝1/2*1.3*20*20^3＝104000W＝104kW

The air density is x, the diameter of the wind wheel is 1m, and the wind speed is 23m / s. what is the power of this wind turbine? Thank you!

Let R be the radius of the wind wheel and u be the wind speed, then there are: a = π R & # 178; time t the volume of air flowing through the wind wheel: v = a × u × t the total kinetic energy of this part of air: e = 1 / 2xvu & # 178; generator power: P = E / T substituting: r = 1 / 2, u = 23, the solution is: P = 1 / 2 π R & # 178; U & # 179; X = 12167 π X / 8, of course, this is not considered