Download - Wind energy I. Lesson 4. Wind power

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Page 1: Wind energy I. Lesson 4. Wind power

Wind Energy I

Michael Hölling, WS 2010/2011 slide 1

Wind power

Page 2: Wind energy I. Lesson 4. Wind power

Wind Energy I

slideMichael Hölling, WS 2010/2011 2

Class content

4 Wind power

5 Wind turbines in general 6 Wind - blades

interaction

7 Π-theorem

8 Wind turbine characterization

9 Control strategies

10 Generator

11 Electrics / grid

3 Wind field characterization

2 Wind measurements

Page 3: Wind energy I. Lesson 4. Wind power

Wind Energy I

slideMichael Hölling, WS 2010/2011 3

Motivation

How much power can be extracted from the wind ?

Page 4: Wind energy I. Lesson 4. Wind power

Wind Energy I

slideMichael Hölling, WS 2010/2011 4

Betz limit

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

A1, u1 A2, u2

A3, u3

! · A1 · u1 = ! · A2 · u2 = ! · A3 · u3

! d

dt· m = m = const.

Page 5: Wind energy I. Lesson 4. Wind power

Wind Energy I

slideMichael Hölling, WS 2010/2011 5

Betz limit

Extracted kinetic energy and extracted power

A1, u1 A2, u2

A3, u3

Eext =12m

!u2

1 ! u23

"

Pext =12m

!u2

1 ! u23

" u3

u1optimal for max. Pext ????

Page 6: Wind energy I. Lesson 4. Wind power

Wind Energy I

slideMichael Hölling, WS 2010/2011 6

Substitution of u2 will lead to a function Pext(u1,u3)

Betz limit

T = m (u1 ! u3)Thrust:

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

Pext =12

· ! · A2 · u2 ·!u2

1 ! u23

"

has to be substituted

Equals mechanical powerthat can be extracted: Pmech = Pext =

12m

!u2

1 ! u23

"

! u2 =u1 + u2

2

Page 7: Wind energy I. Lesson 4. Wind power

Wind Energy I

slideMichael Hölling, WS 2010/2011 7

Betz limit

Extractable power as function of u1 and u3

Pext =12

· ! · A2 · u31

! "# $wind power

·%12

·&

1 +u3

u1! u2

3

u21

! u33

u31

'(

! "# $cp

0.0 0.5 1.00.0

0.2

0.4

0.6

u3/u1

cp

cpmax ??

Page 8: Wind energy I. Lesson 4. Wind power

Wind Energy I

slideMichael Hölling, WS 2010/2011 8

Betz limit

Find the maximum cp by taking the first derivative

cp(x) =12

·!1 + x! x2 ! x3

"Substitute u3/u1 with x:

First derivative of cp(x): c!p(x) =

12

·!1! 2x! 3x2

" != 0

For maximum second derivative of possible solution x1/2

must smaller than zero: c!!p(x1/2) =

12

· (!2! 6x) < 0

Solution: x =13! cpmax.(1/3) =

1627" 59%

Page 9: Wind energy I. Lesson 4. Wind power

Wind Energy I

slideMichael Hölling, WS 2010/2011 9

Betz limit

x

u1 2/3 u1

1/3 u1 u3

u2 u(x)

What happens at the rotor plane ?

Page 10: Wind energy I. Lesson 4. Wind power

Wind Energy I

slideMichael Hölling, WS 2010/2011 9

Betz limit

x

u1 2/3 u1

1/3 u1 u3

u2 u(x)

p(x)p(x)p1

p+2

p-2

p3

What happens at the rotor plane ?

Pressure drop at rotor plane

Page 11: Wind energy I. Lesson 4. Wind power

Wind Energy I

slideMichael Hölling, WS 2010/2011 10

Betz limit

What does this results mean for wind energy converter ?

There is an optimal rotational frequency that the blockage

results in , characterized by tip speed ratio λ.u3

u1=

13

Page 12: Wind energy I. Lesson 4. Wind power

Wind Energy I

slideMichael Hölling, WS 2010/2011 11

Betz limit

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

0 10 20 300.0

0.2

0.4

0.6

0.8

1.0

1.2

u [m/s]

P(u

) /

Pr

P(u) / Pr

0 10 20 300.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

u [m/s]

P(u

) /

Pr

P(u) / Pr

cp(u) / cpmax

cp(u

) /

cp

max