If an object is set up perpendicularly to the wind, the wind exerts a force FD on the object. The wind speed v, the effective object area A and the drag coefficient cD, which depends on the object shape, define the drag force:
Figure 5.5 shows drag coefficients for various shapes. With PD = FD • v, the power to counteract the force becomes:
If an object moves with speed u by the influence of the wind in the same direction as the wind, the drag force is:
and the power used is:
Cone with hemisphere c0= 0,16...0.2
Hemisphere with cone cD= 0,07...0.09
Sphere i'i>= 0.3...0.4 Streamlined form i'i>= 0.3...0.4 Streamlined form rD- 0.055
Source: Hering et al, 1992
The following example calculates approximately the used power of a cup anemometer that is used for the measurement of the wind speed v. It consists of two open hemispherical cups that rotate around a common axis. The wind impacts the front of the first cup and the back of the second cup (Figure 5.6).
The resulting force F consists of a driving and a decelerating component (Gasch and Twele, 2002):
The used power is:
The ratio of the circumferential speed u to the wind speed v is called the tip speed ratio A:
The tip speed ratio of drag devices is always smaller than one. Using the tip speed ratio, the power is:
Hence, the power coefficient of the cup anemometer becomes:
The maximum value of the power coefficient of the cup anemometer is about 0.073. This is much below the ideal Betz power coefficient of 0.593. The cup anemometer reaches its maximum power coefficient at a tip speed ratio of about 0.16, when the wind speed v is about six times higher than the circumferential speed u.
The optimal power coefficient cPoptD of a drag device can be calculated using cP -
Using u/v = 1/3 as well as the maximum drag coefficient of cDmax = 1.3, this gives:
This value is also much below the ideal value of 0.593. Therefore, modern wind turbines are lift devices, rather than drag devices, and these achieve much better power coefficients.
Was this article helpful?