1 Short Summary of the Mechanics of Wind Turbine Korn Saran-Yasoontorn Department of Civil Engineering University of Texas at Austin 8/7/02

2 Summarized from  Wind energy explained: theory, design and application./ Manwell, J. F. / Chichester / 2002  Wind turbine technology: fundamental concepts of wind turbine engineering. / New York / 1994  Wind energy conversion systems/ Freris, L.L./ Prentice Hall/ 1990  Wind turbine engineering design/ Eggleston, D.M. and Stoddard, F.S./ New York/ 1987  Introduction to wind turbine engineering/ Wortman, A.J./ Butterworth Publishers/ 1983

3 Yaw system Hub Drive train Column Rotor Simple model of Micon 65/13 1D steady wind flow (12m/s)

4 Fundamental Concepts Mass flow rate =  Av Energy per unit volume = 1/2  v 2 Power = rate of change of energy = force * velocity = (  Av) 1/2  v 2 = 1/2  Av 3 Dynamic pressure = force/area = power/vA = 1/2  v 2

5 Actuator Disk Model with no wake rotation U1U1 U2U2 U3U3 U4U4 rotor disk Assumptions: i.Homogeneous, incompressible, steady wind ii.Uniform flow velocity at disk (uniform thrust) iii.Homogenous disk iv.Non-rotating disk stream tube boundary upstream downstream

6 Conservation of Linear Momentum where T is the thrust acting uniformly on the disk (rotor) which can be written as a function of the change of pressure as follow

7 Bernoulli’s Equation (energy conserved) Relate above equations and define the axial induction factor, a as we obtain

8 Power output of the turbine is defined as the thrust times the velocity at the disk. Hence Wind turbine rotor performance is usually characterized by its power and thrust coefficients

9 Notice that  Wind velocity at the rotor plane is always less than the free-stream velocity when power is being absorbed.  This model assumes no wake rotation, i.e. no energy wasted in kinetic energy of a twirling wake.  The geometry of the blades does not involve the calculations.

10  If the axial induction factor of the rotor is founded, one can simply calculate for the thrust and power output.  An ideal turbine generates maximum power. After some manipulations, one can find that the axial induction factor, a, for the ideal turbine is 1/3.  Even with the best rotor design, it is not possible to extract more than about 60 percent of the kinetic energy in the wind

11 Wind Velocity Total Pressure Dynamic Pressure Static Pressure p3p3 p2p2 upstream disk downstream u1u1 u2u2 u4u4 1/2  u 2 p0p0 p0p0

12 Actuator Disk Model with wake rotation U1U1 U2U2 U3U3 U4U4 rotor disk U U(1-2a) U(1-a) dr r The thrust distribution is circumferentially uniform. (infinite number of blades)

13 Conservation of Linear momentum Conservation of Angular Momentum

14 Bernoulli’s Equation (energy conserved) Define the angular induction factor a’ as Hence,

15 Equating the thrust on an annular element derived from the conservation of linear momentum and the Bernoulli’s equation gives where For an ideal turbine that produces maximum power output,

16 In summary Notice that  the geometry of the blades still does not involve the calculations.  if the turbine is assumed to be ideal generating maximum power, one can find a and a’ in each section.  once a and a’ are founded, the total thrust and rotor torque can be determined by integration along the blade spanwise.

17 Blade Element Theory Blade geometry is considered in this part and we may use this to calculate the induction factors that relates the thrust and rotor torque. blade element r dr rotor blade R

18 Lift and Drag Forces u Ωr u rel FLFL FDFD Note that C L and C D vary with cross section (top view)

19 Typical Variation of Aerofoil Coefficients Values of Coefficients angle of attack (degrees) ClCl CdCd flow separation

20 Relative Velocity u(1-a) Ωr(1+a’) u rel FLFL FDFD Wind velocity at the rotor blade is u(1-a) in horizontal direction. Also, the wind rotates with the angular velocity of ω/2 (=Ωa’) while the angular velocity of the rotor is Ω in the opposite direction.

21 u(1-a) Ωr(1+a’) u rel dF L dF D dF N dF T θpθp φ  Blade Geometry

22 From blade geometry, one simply obtains the following relations.

23 Finally, the total normal force on the section and torque due to the tangential force operating at a distance, r, from the center are

24 Since the forces and moments derived from momentum theory (actuator model) and blade element theory must be equal, from momentum theoryblade element theory

25 One can solve for C and α at each section by using this equation and the empirical C vs α curves. Once both parameters are known, a and a’ at the section can be determined from

26 Iterative solution for a and a’ 1. Guess values of a and a’ 2. Calculate φ 3. Calculate angle of attack, α 4. Calculate C l and C d 5. Update a and a’ 6. Check if a > 0.5 (In the case of turbulent wake this analysis may lead to a lack of convergence to a solution)

27 Note that to keep the lift and drag coefficients, and thus the angle of attack, constant over the spanwise of the blade, it is necessary to twist the blade along the length. This however may increase the complexity of their manufacture.

28 Tip loss factor The tip loss factor allows for the velocities and forces not being circumferentially uniform due to the rotor having a finite number of blades. The Prandtl tip loss factor can be express as