Testing of rotor blades of wind turbines Arno van Wingerde ...

Apart from marketing reasons where there is a certain advantage associated with
having the largest/fastest/best … there are also some technical and economical reasons
for upscaling, especially for offshore wind turbines, where certain aspects, such
as installation, maintenance, cables (offshore) and the foundation costs, do not fully
scale to the diameter of the turbine. For deep water offshore, these aspects account
for 40%-60% of the total cost of the wind turbine.
However, there exists a major hurdle called the square-cube law which all but prevents
economical upscaling of the wind turbine. When the blade is simply scaled up,
the size of the swept area, the disk of air captured by the wind turbine, and thereby
the potential energy production, increases with the square of the rotor diameter.
However, just scaling up the 3 dimensions of the blade will result in a weight that increases
with the third power of the structure.
Influence of the dead weight
The cubically increasing dead weight of the blade with the rotor diameter is a major
problem for the industry but, for large blades, the weight of the blade itself becomes a
major load component. Thus extra material is needed to carry the weight which in
turn further increases the weight of the blade.
An example may serve to get a feeling for effect of the dead weight of the blades with
increasing lengths. Consider the maximum length of a rod with a diameter D and a
length l, which is clamped at one side and only loaded by its dead weight, as shown
in Figure 2Error! Reference source not found.. Furthermore the rod is made of
glass-epoxy, the predominant material used for blades of wind turbines, with the specific
density ρ of glass-epoxy of about 2.3, or 23000 N/m 3 and a strength of 800
L
l
D
MPa= 800·10 6 N/m 2 .
Figure 2 Dead weight problem
Notation used:
D: the diameter of the rod
L: the length of the rod
F: the dead weight of the rod (which can be considered to act halfway the rod)
M: the bending moment at the clamped end of the rod, due to the dead weight
W: the elastic moment of the rod
σ max : the maximum bending stress in the rod
The dead weight of the rod, also the load on the rod is: F=L·¼·π·D 2·ρ
(1)
The resulting bending moment M at the clamped end is: M=½·F·L
(2)
The maximum stress is the rod, σ max = M/W
(3)