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Wind Energy I




                                Wind power


Michael Hölling, WS 2010/2011       slide 1
Wind Energy I                              Class content
                                                            5 Wind turbines in
                                                                                       6 Wind - blades
                                                              general
                                2 Wind measurements                                      interaction
                                                                                  7 Π-theorem

                                                                         8 Wind turbine
                                                                           characterization
                                  3 Wind field                                    9 Control strategies
                                    characterization
                                                                     10 Generator
    4 Wind power


                                                                     11 Electrics / grid




Michael Hölling, WS 2010/2011                     slide 2
Wind Energy I                   Motivation

            How much power can be extracted from the wind ?




Michael Hölling, WS 2010/2011     slide 3
Wind Energy I                              Betz limit

  Approach: free-stream air flow and conservation of mass flow




            A1, u1
                                  A2, u2
                                                             A3, u3

       ρ · A1 · u 1 =            ρ · A2 · u 2 =         ρ · A3 · u 3

                                  d
                                ⇒    · m = m = const.
                                           ˙
                                  dt

Michael Hölling, WS 2010/2011                slide 4
Wind Energy I                            Betz limit

                 Extracted kinetic energy and extracted power




            A1, u1
                                A2, u2
                                                                   A3, u3
                      1
           Eext      = m u1 − u3
                          2    2
                      2
                      1                                       u3
            Pext     = m u2 − u2
                        ˙ 1                           optimal       for max. Pext ????
                      2        3                              u1

Michael Hölling, WS 2010/2011              slide 5
Wind Energy I                           Betz limit

             Substitution of u2 will lead to a function Pext(u1,u3)
             1
   Pext     = · ρ · A2 · u 2 · u 1 − u 3
                                 2     2
             2
                                has to be substituted

  Thrust: T = m (u1 − u3 )
              ˙

  Corresponding power: Pthrust = m (u1 − u3 ) · u2
                                 ˙

  Equals mechanical power              1
  that can be extracted: Pmech = Pext = m u2 − u2
                                         ˙ 1    3
                                                        2
                                         u1 + u2
                                  ⇒ u2 =
                                            2
Michael Hölling, WS 2010/2011              slide 6
Wind Energy I                                 Betz limit

                     Extractable power as function of u1 and u3
                          1                 1      u3  u2 u3
                Pext     = · ρ · A2 · u 1 ·
                                        3
                                              · 1+      3  3
                                                      − 2− 3
                          2                 2      u1  u1 u1
                                 wind power                    cp


                                 0.6                                cp
                                                          max ??
                                 0.4
                            cp




                                 0.2


                                 0.0
                                   0.0            0.5               1.0
                                                  u3/u1

Michael Hölling, WS 2010/2011                   slide 7
Wind Energy I                   Betz limit

       Find the maximum cp by taking the first derivative
                                 1
Substitute u3/u1 with x: cp (x) = · 1 + x − x − x
                                              2     3
                                 2
                                   1               !
First derivative of cp(x): cp (x) = · 1 − 2x − 3x = 0
                                                 2
                                   2

For maximum second derivative of possible solution x1/2
                                    1
must smaller than zero: cp (x1/2 ) = · (−2 − 6x) < 0
                                    2

                           1                 16
              Solution: x = ⇒ cpmax. (1/3) =    ≈ 59%
                           3                 27

Michael Hölling, WS 2010/2011     slide 8
Wind Energy I                               Betz limit

                            What happens at the rotor plane ?


u1
u2                                 2/3 u1

u3                                                       u(x)   1/3 u1



                                                                    x




Michael Hölling, WS 2010/2011                 slide 9
Wind Energy I                                  Betz limit

                            What happens at the rotor plane ?

                                p-2
u1
u2                                    2/3 u1

u3                                                          u(x)      1/3 u1
                 p(x)                             p(x)
p1                                                                             p3

                                                                          x


                                                   Pressure drop at rotor plane
                                      p+2


Michael Hölling, WS 2010/2011                    slide 9
Wind Energy I                   Betz limit

       What does this results mean for wind energy converter ?




 There is an optimal rotational frequency that the blockage
            u3    1
 results in    =     , characterized by tip speed ratio λ.
            u1    3

Michael Hölling, WS 2010/2011     slide 10
Wind Energy I                                   Betz limit

              What is the value for cp for a WEC over wind speed ?

            1.2                                               1.2
                                                                                    P(u) / Pr
            1.0                                               1.0                   cp(u) / cpmax




                                                                    cp(u) / cpmax
P(u) / Pr




            0.8                                               0.8

            0.6                                               0.6

            0.4                                               0.4

            0.2                                               0.2

            0.0                                               0.0
               0                 10             20           30
                                      u [m/s]


 Michael Hölling, WS 2010/2011                    slide 11

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Wind energy I. Lesson 4. Wind power

  • 1. Wind Energy I Wind power Michael Hölling, WS 2010/2011 slide 1
  • 2. Wind Energy I Class content 5 Wind turbines in 6 Wind - blades general 2 Wind measurements interaction 7 Π-theorem 8 Wind turbine characterization 3 Wind field 9 Control strategies characterization 10 Generator 4 Wind power 11 Electrics / grid Michael Hölling, WS 2010/2011 slide 2
  • 3. Wind Energy I Motivation How much power can be extracted from the wind ? Michael Hölling, WS 2010/2011 slide 3
  • 4. Wind Energy I Betz limit Approach: free-stream air flow and conservation of mass flow A1, u1 A2, u2 A3, u3 ρ · A1 · u 1 = ρ · A2 · u 2 = ρ · A3 · u 3 d ⇒ · m = m = const. ˙ dt Michael Hölling, WS 2010/2011 slide 4
  • 5. Wind Energy I Betz limit Extracted kinetic energy and extracted power A1, u1 A2, u2 A3, u3 1 Eext = m u1 − u3 2 2 2 1 u3 Pext = m u2 − u2 ˙ 1 optimal for max. Pext ???? 2 3 u1 Michael Hölling, WS 2010/2011 slide 5
  • 6. Wind Energy I Betz limit Substitution of u2 will lead to a function Pext(u1,u3) 1 Pext = · ρ · A2 · u 2 · u 1 − u 3 2 2 2 has to be substituted Thrust: T = m (u1 − u3 ) ˙ Corresponding power: Pthrust = m (u1 − u3 ) · u2 ˙ Equals mechanical power 1 that can be extracted: Pmech = Pext = m u2 − u2 ˙ 1 3 2 u1 + u2 ⇒ u2 = 2 Michael Hölling, WS 2010/2011 slide 6
  • 7. Wind Energy I Betz limit Extractable power as function of u1 and u3 1 1 u3 u2 u3 Pext = · ρ · A2 · u 1 · 3 · 1+ 3 3 − 2− 3 2 2 u1 u1 u1 wind power cp 0.6 cp max ?? 0.4 cp 0.2 0.0 0.0 0.5 1.0 u3/u1 Michael Hölling, WS 2010/2011 slide 7
  • 8. Wind Energy I Betz limit Find the maximum cp by taking the first derivative 1 Substitute u3/u1 with x: cp (x) = · 1 + x − x − x 2 3 2 1 ! First derivative of cp(x): cp (x) = · 1 − 2x − 3x = 0 2 2 For maximum second derivative of possible solution x1/2 1 must smaller than zero: cp (x1/2 ) = · (−2 − 6x) < 0 2 1 16 Solution: x = ⇒ cpmax. (1/3) = ≈ 59% 3 27 Michael Hölling, WS 2010/2011 slide 8
  • 9. Wind Energy I Betz limit What happens at the rotor plane ? u1 u2 2/3 u1 u3 u(x) 1/3 u1 x Michael Hölling, WS 2010/2011 slide 9
  • 10. Wind Energy I Betz limit What happens at the rotor plane ? p-2 u1 u2 2/3 u1 u3 u(x) 1/3 u1 p(x) p(x) p1 p3 x Pressure drop at rotor plane p+2 Michael Hölling, WS 2010/2011 slide 9
  • 11. Wind Energy I Betz limit What does this results mean for wind energy converter ? There is an optimal rotational frequency that the blockage u3 1 results in = , characterized by tip speed ratio λ. u1 3 Michael Hölling, WS 2010/2011 slide 10
  • 12. Wind Energy I Betz limit What is the value for cp for a WEC over wind speed ? 1.2 1.2 P(u) / Pr 1.0 1.0 cp(u) / cpmax cp(u) / cpmax P(u) / Pr 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 0.0 0.0 0 10 20 30 u [m/s] Michael Hölling, WS 2010/2011 slide 11