The average wind speed in an area is 6 m s, and the density of the air is known to be 1 2 kg m3

6 seconds per meter multiplied by 1 second, multiplied by 1,200 square meters, multiplied by kilograms per cubic meter, should be the mass of air per second.

One kilogram can produce joules of electricity, so what about so many kilograms per second now? It's very simple, right? Multiply the mass by the following.

In time t, the distance traveled by the wind l=vt, the volume of the wind v=ls=vts, the mass of the wind m= v= vts, and the kinetic energy of the wind ek = mv = stv

e electricity = ekp = e electricity t = sv

p= w

It is easy to know that its power generation in time t is.

q=△ek=(ρsvt)*v^2/2

So its power is:

p=q/t=ρsv^3/2=

First of all, Sakura calculates the kinetic energy.

Set the time for one second for the Ant Bush.

Mass scramble m=

1/2 *m*5*5*

p=w/t=

Hope it can help you

The kinetic energy of the air flow acting on the windmill in time t m= r 2 v·t The kinetic energy of these air currents is 1 2mv 2, and the converted electrical energy e= 1 2mv 2 10

So the power of the windmill to drive the generator is p=e t= r 2 v 3 10

Substituting the data yields p= kw

Work done by traction: w=fl=1 2mv2 (l is displacement) and then by the power formula: p=fv and mass:

m=pv (p is density) These three formulas are sufficient. Note: v=sl (volume equal to base area multiplied by length).

It's a bit detailed, do you understand??

To give you a rough calculation: fan power = 1 2 * cp p * s * v 3Among them, cp is the utilization rate of wind energy, and now the better wind energy utilization rate of the wind turbine is between, then the fan power = , the fan power is still lost, and your power generation efficiency is 80%, then.

First of all, it should be noted that after the fan exceeds the rated wind speed (generally 7 8 meters per second), certain measures will be taken in order to stabilize the power of the generator. So it's not that the higher the wind speed, the more power the generator will be.

pt=[(1 2) v1*t*s*v1 2-(1 2) v2*t*s*v2 2]*80%, that is, the kinetic energy of the mass in the panic period per unit time after passing through the wind turbine is reduced by 80%, where v1=20m s, v2=12m s

Average the situation in 1 second.

The pressure of a 10-level wind on an object with an area of f = air guess density) * upper limit of wind speed of class 10) =

Air volume = wind speed * cross-sectional area, taking the diameter of 600mm and the wind speed of 12 meters per second as an example, the air volume = 12 * 3600*

Suppose you take a piece of area s perpendicular to the wind, because the wind speed is v, then in time t, the volume of air blowing on the area s v=svt, according to the density knowledge, m= svt, the wind encounters an obstacle velocity is reduced to 0, and the pressure of the wind is f, according to the momentum theorem:

The density of air is about kilograms per cubic meter, and the resulting pressure is about 300 Pa.