Power available in wind is calculated by formula
Power Available = ½* air density * swept area * wind velocity3
Air density at sea level is 1.23kg/m3
I will consider it as 1kg/m3 at Faisalabad (Pakistan).
Swept area is in m2. It is calculated by formula πr2.
Where r is radius of the rotor. Diameter of the rotor is 2r.
Wind velocity is in m/s.
Therefore, at a wind velocity of 5 m/s and rotor diameter of 1m,
Power available in wind is 196.25 Watts.
If we double the wind speed power increases by a factor of 8. i.e. at 10m/s
Power available in wind is 1570 Watts.
When the diameter of rotor is doubled power increases by factor of 4.
If wind velocity is 5 m/s and diameter is 2m, then the power available is 785 Watts.
But it is not possible to extract all the Power available in wind.
In 1919 Betz calculated the limit to how much a wind turbine can extract from wind.
This limit is called Betz limit.
Betz Limit = 59.26%
It is the maximum amount of power that can be extracted from wind.
An ideal wind machine will work at Betz limit. But in practice there is no ideal machine.
There are power losses because
- No blades are 100% efficient
- No alternator/ generator is 100% efficient.
- Friction losses
- Magnetic drag and resistance losses
Mike Klemen calculated a co-effient of Power= 35%
A good turbine extract about 35% of energy available in wind.
You can take a look at Mike Klemen work here
http://www.ndsu.nodak.edu/ndsu/klemen/Perfect_Turbine.htm
|
Data on power available in wind |
|||||
|
Wind Speed ( m/s ) |
Rotor Diameter ( m ) |
Area ( m2 ) |
Power Available ( Watts) |
Betz limit ( Watts ) |
Good Turbine ( Watts) |
|
1 |
1 |
3.14 |
1.57 |
0.930382 |
0.5495 |
|
5 |
1 |
3.14 |
196.25 |
116.2978 |
68.6875 |
|
10 |
1 |
3.14 |
1570 |
930.382 |
549.5 |
|
1 |
2 |
12.56 |
6.28 |
3.721528 |
2.198 |
|
5 |
2 |
12.56 |
785 |
465.191 |
274.75 |
|
10 |
2 |
12.56 |
6280 |
3721.528 |
2198 |
Speed of wind can be estimated using Beaufort Scale.
