WIND ENERGY SYSTEMS - Cd3wd
WIND ENERGY SYSTEMS - Cd3wd WIND ENERGY SYSTEMS - Cd3wd
Chapter 4—Wind Turbine Power 4–34 Figure 20: Aerodynamic torque variation for Sandia 17-m Darrieus at 50.6 r/min. Figure 21: Definition of rotor angle for Sandia 17-m Darrieus. elastic range. Permanent deformation occurs when a material exceeds its elastic range. The shaft can be thought of as a spring with a torsional spring constant k T where k T = T θ The angle θ has to be expressed in radians, of course. A large value of k T (39) represents a Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
Chapter 4—Wind Turbine Power 4–35 Figure 22: Shaft twisted under an applied torque. stiff shaft, while a small value represents a soft or flexible shaft. The torsional spring constant is also given by k T = JG L (40) where J is the polar area moment of inertia, G is the shear modulus, andL is the length of the shaft. The shear modulus is the proportionality constant between a shear stress and the resulting deflection or strain. A typical value for the shear modulus for steel is 83 GPa (83 ×10 9 Pa). A twisted shaft contains potential energy, just like a compressed spring. The amount of this potential energy is given by U = k T θ 2 2 J (41) This potential energy has to be supplied to the shaft during system start-up and will be delivered back to the system during shut-down. Also, when a wind gust strikes the turbine, part of the extra power will go into shaft potential energy rather than instantly appearing in the electrical output. This stored energy will then go from the shaft into the electrical system during a wind lull. We see then that a shaft helps to smooth out the power fluctuations in the wind. Example Assume that the high speed shaft of a wind turbine has a torque of 1140 N·m/rad, an angular velocity of 188.5 rad/s, a diameter of 0.0473 m, and a shear modulus of 83 GPa. The length is 2 m. Find the rotation angle θ and the energy stored in the shaft. From Eq. 37 the moment of inertia is J = π(0.0473/2)4 2 From Eq. 40, the torsional spring constant is =4.914 × 10 −7 m 4 Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
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Chapter 4—Wind Turbine Power 4–35<br />
Figure 22: Shaft twisted under an applied torque.<br />
stiff shaft, while a small value represents a soft or flexible shaft.<br />
The torsional spring constant is also given by<br />
k T = JG L<br />
(40)<br />
where J is the polar area moment of inertia, G is the shear modulus, andL is the length of<br />
the shaft. The shear modulus is the proportionality constant between a shear stress and the<br />
resulting deflection or strain. A typical value for the shear modulus for steel is 83 GPa (83<br />
×10 9 Pa).<br />
A twisted shaft contains potential energy, just like a compressed spring. The amount of<br />
this potential energy is given by<br />
U = k T θ 2<br />
2<br />
J (41)<br />
This potential energy has to be supplied to the shaft during system start-up and will be<br />
delivered back to the system during shut-down. Also, when a wind gust strikes the turbine,<br />
part of the extra power will go into shaft potential energy rather than instantly appearing in<br />
the electrical output. This stored energy will then go from the shaft into the electrical system<br />
during a wind lull. We see then that a shaft helps to smooth out the power fluctuations in<br />
the wind.<br />
Example<br />
Assume that the high speed shaft of a wind turbine has a torque of 1140 N·m/rad, an angular<br />
velocity of 188.5 rad/s, a diameter of 0.0473 m, and a shear modulus of 83 GPa. The length is 2 m.<br />
Find the rotation angle θ and the energy stored in the shaft.<br />
From Eq. 37 the moment of inertia is<br />
J = π(0.0473/2)4<br />
2<br />
From Eq. 40, the torsional spring constant is<br />
=4.914 × 10 −7 m 4<br />
Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001