Turbines May4_2009_1.pdf

8/11/2019 Turbines May4_2009_1.pdf

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Wind turbines Sren Gundtoft 4

the mass flow equals v. Momentum equals mass times velocity, with the unit N. Pressure equalsforce per surface, then the differential pressure can be calculated as

( )31 vvvp = [Pa] (2.5)

Now (2.4) and (2.5) give

( )312

1vvv += [m/s] (2.6)

This indicates that the speed of air in the rotor plane equals the mean value of the speed upstreamand down stream of the rotor.

Power production: The power of the turbine equals the change in kinetic energy in the air

( )AvvvP 23212

1= [W] (2.7)

HereAis the surface area swept by the rotor.

The axial force (thrust) on the rotor can be calculated as

ApT = [N] (2.8)

We now define the axial interference factor asuch that

( ) 11 vav = [m/s] (2.9)

Using (2.6) and (2.9) we get v3= (1 2a) v1and (2.7) and (2.8) can be written as

( ) AvaaP 312

12 = [W] (2.10)

( ) AvaaT 2112 = [N] (2.11)

We now define two coefficients, one of the power production and one of the axial forces as

( )2P 14 aaC = [-] (2.12)

( )aaC = 14T [-] (2.13)

Then (2.10) and (2.11) can be written as

P

3

12

1CAvP = [W] (2.14)


T

2

1 CAvT = [N] (2.15)

In figure 2.2, curves for CPand CTare shown.

0,0

0,2

0,4

0,6

0,8

1,0

0 0,1 0,2 0,3 0,4 0,5

a [-]

C_

P

&

C_

T

[-]

C_T

C_P

=16/27

Figure 2.2: Coefficient of power CPand coefficient of axial force CTfor an idealized wind turbine.

As shown, CPhas an optimum at about 0,593 (exactly 16/27) at an axial interference factor of 0,333(exactly 1/3). According to Betz we have

27

16with BetzP,

3

1Betzp,Betz == CAvCP [W] (2.16)

Example 2.1Let us compare the axial force on rotor to the drag force on a flat plate? If a= 1/3 the CT= 8/9 0,89. Wind passing a flat plate with the areaAwould give a drag on the plate of

AvCF2

1DD2

1= [N] (2.17)

where CD1,1 i.e. the axial force on at rotor at maximal power is about 0,89/1,1 = 0,80 = 80%of the force on a flat plate of the same area as the rotor!

3. Rotor design

3.1. Air foil theory an introduction

Figure 3.1 shows a typical wing section of the blade.

The air hits the blade in an angle Awhich is called the angle of attack. The reference line forthe angle on the blade is most often the chord line see more in Chap. 4 for blade data. The forceon the bladeFcan be divided into two components the lift forceFLand the drag forceFDand thelift force is per definition perpendicular to the wind direction.


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FD

w

LF

F

Chord line

Figure 3.1: Definition of angle of attack

The lift force can be calculated as( )cbwCF 2LL

2

1= (3.1)

where CLis the coefficient of lift, is the density of air, wthe relative wind speed, bthe width ofthe blade section and cthe length of the chord line.

Similar for the drag force

( )cbwCF 2DD2

1= (3.2)

The coefficient of lift and drag both depend of the angle of attack, see figure 3.2.

For angles of attack higher than typically 15-20 the air is no longer attached to the blade, aphenomenon called stall.

The ratio CL/CDis called the glide ratio, i.e. GR = CL/CD. Normally we are interested in at highglide ratio for wind turbines as well as for air planes. Values up to 100 or higher is not uncommonand the angles of attack giving maximum are typical in the range 5 10.

NACA 23012

0,0

0,2

0,4

0,6

0,8

1,0

1,2

1,41,6

1,8

0 15 30 45 60 75 90

alpha []

C_

L

&

C_

D

[-]

0

20

40

60

80

100

120

140160

180

GR

[-]

C_L

C_D

GR

NACA 23012

0,0

0,2

0,4

0,6

0,8

1,0

1,2

1,4

1,6

1,8

0 2 4 6 8 10 12 14 16 18 20

alpha []

C_

L&C_

D[

-]

0

20

40

60

80

100

120

140

160

180

GR[

-]

C_L

C_D

GR

Figure 3.2: Coefficient of lift and drag as a function of the angle of attack (left: 0


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11

tip

v

R

v

vX

== [-] (3.6)

Combining these equations we get

( )R

Xrr

2

3arctan= [rad] (3.7)

or

( )Xr

Rr

3

2arctan= [rad] (3.8)

and then the pitch angle

( ) DBetz 32arctan =

XrRr [rad] (3.9)

where Dis the angle of attack, used for the design of the blade. Most often the angle is chosen tobe close to the angle, that gives maximum glide ration, see figure 3.2 that means in the range from 5to 10, but near the tip of the blade the angle is sometimes reduced.

Chord length, c(r):

If we look at one blade element in the distance rfrom the rotor axis with the thickness drthe lift

force is, see formula (3.1) and (3.2)

L2

L d2

1d CrcwF = [N] (3.10)

and the drag force

D2

D d2

1d CrcwF = [N] (3.11)

rR

dr

Figure 3.4: Blade section


Rotor axis

Rotorplane D

dF

dFLL

dU

dU

D D

dF

LdFdFL

D

dF

dTLdTD

Torque Thrustx

y12.07.2008/SGt

Figure 3.5. Forces on the blade element decomposed on the rotor plane, dU (torque), and in the

rotor axis, dT (thrust)

For the rotor plane (torque) we have

x2 d

2

1d CrcwU = [N] (3.12)

with

( ) ( ) cossin DLx CCC = [-] (3.13)

For the rotor axis (thrust) we have

y

2

d2

1

d CrcwT = [N] (3.14)with

( ) ( ) sincos DLy CCC += [-] (3.15)

Now, in the design situation, we have CL>> CD, then (3.12) and (3.13) becomes

( ) cosd2

1d L

2 CrcwU= [N] (3.16)

and then the power produced

rUP dd = [W] (3.17)

If we haveBblades, (3.16) including (3.17) gives

( ) rCrcwBP cosd2

1d L

2= [W] (3.18)

According to Betz, the blade element would also give


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( )rrvP d22

1

27

16d 31 = [W] (3.19)

Using v1= 3/2 wcos() and u= wsin(), then (3.18) and (3.19) gives

( )

9

4

1

9

162

2DL,

Betz

+

=

R

rXX

CB

Rrc

[m] (3.20)

where CL,Dis the coefficient of lift at the chosen design angle of attack, A,D.

Example 3.1

What will be shown later is that a tip speed ration of aboutX= 7 is optimal (see fig. 6.2). Further

more 3 blades seem to be state of the art. Figure 3.6 and 3.7 shows the results of formula (3.20)concerning the chord length i.e. according to Betz.

Figure 3.6. Chord length as function of radius for X = 7 and for different numbers of blades

Figure 3.7. Chord length as function of radius for three blades B = 3 and for different tip speed

ratios


3.3. Pitch angle, , and chord length, c, after Schmitz

Schmitz has developed a little more detailed and sophisticated model of the flow in the rotor plane.

The torqueMin the rotor shaft can only be established because of the rotation of the wake, cf.Appendix A which is a result of the conservation law for angular momentum

v

vu

u

[m/s]

Figure 3.8. Down stream rotation of the wake The wake rotates in the opposite direction to the

rotor

The power can be calculated as

MP= [W] (3.21)

whereMis the torque in the rotor shaft and is the angular speed. According to the conservationrule of angular momentum, the torque in the rotor shaft can only be established because of a swirlinduced in the slipstream in the flow down stream of the rotor. As for the axial speed vit can beshown theoretically that the change in the tangential speed in the rotor plane is half of the totalchange, i.e. we have in the rotor plane

uru += [m/s] (3.22)

or


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( )'1 aru += [m/s] (3.23)

which defines the tangential interference factor a

As mentioned previously index 1 is used for the upstream situation, index 2 and 3 for rotor planeand downstream respectively. In the following index 2 is some times omitted for simplicity.

Now look at the flow in the rotor plane, see figure 3.9. What is important here is the relation

www rrr

+= 1 [m/s] (3.24)

The change in w1is because of the air foil effect. If we assume that the drag is very low (comparedto lift, i.e. CD CD0) then the wvector is parallel to the lift force vector dFL(because

of the conservation law of momentum) and we per definition of the direction of lift force alsohave that the wvector is perpendicular to w see figure 3.9-b4). Based on these considerations wehave the following geometrical relations

( ) = 11 cosww [m/s] (3.25)

and from figure 3.9. b2)

( )sinwv= [m/s] (3.26)

Combining (3.25) and (3.26) we get

( ) ( ) sincos 11 = wv [m/s] (3.27)

From figure 3.9 we further have

( ) = 11 sin2ww [m/s] (3.28)


w 1

1

v

1=u

1

r

1

v

r

w

w

w

r

v

u

uw

v

v

r

w

u

1

w

1

w

v3

r

w3

u

Rotorplane

fardownstream

upstr

eam

a)

b1)

b4)

c1)

b2)

b3)

Rotor plane

Rotor plane

Rotor plane

Rotor plane

12.07.2008/SGt

1

r

ww

v

u

3

3

w

c2)

Figure 3.9. Speed in the rotor plane a) far upstream; b) in the rotor plane and; c) far down stream

Now, let us look at the power! From the conservation of momentum we have

qwF dd L = [N] (3.29)

where dqis the mass flow through the ring element in the radius rwith the width dr, i.e.


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vrrq d2d = [kg/s] (3.30)

Power equals torque multiplied by angular velocity and (neglecting drag) then

( )( )

( ){ }( ) ( ) ( )[ ] ( )

( )[ ] ( )12

121

2

1111

L

sin2sind2

sinsincosd2sin2

sind

sind

dd

=

=

=

=

=

wrr

rwrrw

rqw

rF

MP

[kg/s] (3.31)

In the bottom transaction above we have used the relation sin(x) cos(x) = sin(2x).

We have now a relation for the power of the ring element as a function of the angle but we do notknow this angle? The trick is now to solve the equation d(dP)/d= 0 to find the angle that givesmaximum power. Doing this for (3.31) we get

( ) ( ) ( )[ ] ( )[ ]( )

( ) ( )[ ] ( )[ ]{ }( ) ( ){ }

32sinsin2d2

sin2coscos2sinsin2d2

cossin2sin2sin2cos2d2d

dd

121

2

1121

2

12

121

2

=

=

+=

wrr

wrr

wrrP

[W/] (3.32)

From d(dP)/d= 0, it follows

1max3

2 = [rad] (3.33)

or

rX

R

r

varctan

3

2arctan

3

2 1max ==

[rad] (3.34)

and the for pitch angle

( ) DSchmitz arctan3

2 =

Xr

Rr [rad] (3.35)

Example 3.2

Lets compare Betz and Schmitz formulas for the design of the optimal pitch angle. AssumingX =7;B= 3; D=7,0; CL= 0,88 one gets


Optimal pitch angleX = 5; B = 3; alfa_D = 7,0; C_L = 0,88

-10

0

10

20

30

40

50

60

70

80

0,0 0,2 0,4 0,6 0,8 1,0r/R [-]

beta[]

beta(Betz)

beta(Schmitz)

Figure 3.10: Optimal pitch angel

Note, that only for small r/Rthe two theories differ. And here the power produced is small becauseof the relatively small swept area. At the tip (r/R= 1) the optimal angle is approx. 0,5 for both.

Using the result of (3.27), (3.28) and (3.33) in (3.29) we get

( ) ( ) ( )( )

=

=

=

=

3cos

3sind22

3

2sin

3cos

3sind22

sincosd2sin2

dd

12122

1

1112

1

1111

L

rrw

rrw

wrrw

qwF

[N] (3.36)

where we again use sin(2x) = 2 sin(x)cos(x).

From the air foil theory we have

=

=

3cosd

dd

1L

2

1

L2

L

CrcBw

CrcBwF

[7] (3.37)

where we have used (3.25) and = 2/31.

Combining (3.37) and (3.36) we get

( )

=

3sin

161 12

L

Schmitz

C

r

Brc [m] (3.38)


9/22


or

( )

=

rX

R

C

r

Brc arctan

3

1sin

161 2

L

Schmitz

[m] (3.39)

Example 3.3

Lets again compare Betz and Schmitz formulas for the design of the optimal pitch angle.AssumingX = 7;B= 3; D=7,0; CL= 0,88 one gets

Optimal chord ratioX = 5; B = 3; alfa_D = 7,0; C_L = 0,88

0,0

0,1

0,2

0,3

0,4

0,50,6

0,7

0,0 0,2 0,4 0,6 0,8 1,0

r/R [-]

c/R[

-]

c/R(Betz)

c/R(Schmitz)

Figure 3.11: Optimal chord length

Note, near the tip there are no difference between Betz and Schmitz theory.

4. Characteristics of rotor blades

Wing profiles are often tested in wind tunnels. Results are curves for coefficient of lift and drag and

moment. Data for a lot of profiles can be found in Theory of Wing Sections, Ira H. Abbott and A.E. Doenhoff, ref./3/.

Figure 4.1 shows data for the profile NACA 23012.

Lift, drag and torque (per meter blade width) are defined by the equations

L2*

L CcwF = [N] (4.1)

D2*

D CcwF = [N] (4.2)


M22*

M CcwQ = [Nm] (4.3)

The density of air is at a nominal state, defined as 1 bar and 11C, 1,225 kg/m3.

The curves in figure 4.1 are given at different Reynoldss number, defined as

/Re

wc= [-] (4.4)

For PC-calculation it is convenient to have the curves as functions. For the NACA 23012 profileone can use the following approximation: CD,L= k0+ k1+ k2

2+ k33 + k4

4, with the followingconstants

NACA 23012

CL CDk0k1k2k3k4

1,0318e-11,0516e-11,0483e-37,3487e-6

6,5827e-6

6,0387e-33,6282e-45,4269e-56,5341e-6

2,8045e-7Table 4.1: Polynomial constants for 0 < < 16

As shown in figure 4.2, the data are given in the range of < 20. For wind turbines it is necessaryto know the data for the range up to 90. In the range from st


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CDmaxcan be set at 1. For the NACA 23012 profile, the angle of stall is a little uncertain, but couldin practice be set at 16. Figure 3.2 show the result of the formulas above.

Figure 4.1 show some typical data for an air foil.

Location ofmax. camber

Chord

Max camber

Mean camber lineUppersurface

Lowersurface

Location ofmax. thickness

Max thickness

Leadingedge

Trailingedge

Chord line

Leadingedgeradius

Figure 4.1: Definition of typical air foil data

The chord lineis a straight line connecting the leading and the trailing edges of the air foil. The mean camber lineis a line drawn halfway between the upper and the lower surfaces. The

chord line connects the ends of the mean camber lines. The frontal surface of the airfoil is defined by the shape of a circle with theleading edge radius

(L.E. radius). The center of the circle is defined by the leading edge radius and a line with a givenslope of the

leading edge radiusrelative to the chord.

Data for the NACA 23012 profile is given by the table (upper left corner) on figure 4.2.


Figure 4.2: Data for NACA 23012 (Ref./3/)


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5. The blade element momentum (BEM) theory

In the blade element momentum (BEM) method the flow area swept by the rotor is divided into anumber of concentric ring elements. The rings are considered separately under the assumption thatthere is no radial interference between the flows in one ring to the two neighbouring rings.

Figure 3.3 shows the profile and the wind speeds in one ring. The angle of attack is given by

= [rad] (5.1)

From figure 3.3 we get

( ) rv

a

a1

'11tan +

= [-] (5.2)

If the number of blades isB,we can calculate the axial force dTand the torque dUon a ring elementwith the radius rand the width drand the torque as

rCBcwT dd y2= [N] (5.3)

rrCBcwU dd x2= [Nm] (5.4)

where Cyand Cxare given by (3.15) and (3.13)

If we now use the laws of momentum and angular momentum, we get

( ) rvvvrT d2d 312 = [N] (5.5)

ruvrU d2d 322= [Nm] (5.6)

In (5.6) we are using u3for the tangential speed far behind the rotor plane, even though there issome tangential rotation of the wind. This can be shown to be an allowable approximation, becausethe rotation of the wind normally is small.

Combining (5.3) and (5.5) - and - (5.4) and (5.6) we get

( )=

2y

sinr81

CBc

a

a [-] (5.7)

( ) ( )=

+ cossinr81'

' x

CBc

a

a [-] (5.8)


Here we have used

( )( )sin

11 avw

= [m/s] (5.9)

or

( )( )

cos

'1 arw

+= [m/s] (5.10)

If we now define the solid ratio as

r

Bc

2= [-] (5.11)

and solve the equation (5.7) and (5.9) we get

( )1

sin4

1

y

2

+

=

C

a

[-] (5.12)

and

( ) ( )1

cossin41

'

x

=

C

a

[-] (5.13)

For rotors with few blades it can be shown that a better approximation of aand ais

( )1

sin4

1

y

2

+

=

C

Fa

[-] (5.14)

and

( ) ( ) 1cossin41

'

x

=C

Fa

[-] (5.15)

where

( )

=

sin2exparccos

2

r

rRBF [N] (5.16)


12/22


This simple momentum theory breaks down when abecomes greater than ac= 0,2. In that case wereplace (5.14) by

( ) ( )( ) ( )

+++= 14221212 2c2

cc aKaKaKa [-] (5.17)

where

( )

y

2sin4

C

FK

= [-] (5.18)

Calculation procedure

We can now calculate the axial force and power of one ring element of the rotor by making the

following iteration:

For every radius r(4 to 8 elements are OK), go through step-1 to step-8

Step-1: Start

Step-2: aand aare set at some guessed values. a = a= 0 is a good first time guess.

Step-3: is calculated from (5.2)

Step-4: From the blade profile data sheet (or the polynomial approximation) we find CLand CD

Step-5: Cxand Cyare calculated by (3.13) and (3.15)

Step-6: aand aare calculated by (5.14) and (5.15). Or if a> 0,2 then ais calculated from (5.17).

Step-7: If aand aas found under step-5 differ more than 1% from the last/initial guess, continue atstep-2, using the new aand a.

Step-8: Stop

When the iterative process is ended for all blade elements, then the axial force and tangential force(per meter of blade) for any radius can be calculated as

( ) x2* CcwrU = [N] (5.19)

( ) y2* CcwrT = [N] (5.20)

and then the total axial force and power as

( ) rrTBTR

d0*

= [N] (5.21)


( ) rrUrBPR

d0

*

= [N] (5.22)

6. Efficiency of the wind turbine

6.1. Rotor

Betz has shown that the maximum power available in the wind is given by (2.16). Let us define thispower as

AvP3

1max

2

1

27

16= [W] (6.1)

where we have used Cp= Cp,Betz=16/27.

In (6.1)Ais the swept area of the rotor, and in the following we define this area asA = /4D2i.e.we do not take into account, that some part of the hub area is not producing any power!

We can now define the rotor efficiency as

max

rotorrotor

P

P= [-] (6.2)

whereProtoris the power in the rotor shaft.

The rotor efficiency can be calculated on the basis of a BEM-calculation of the power production ina real turbine see the example in Chapter 7.

Another model will be presented here:

The rotor efficiency is divided into three parts

profiletipwakerotor = [-] (6.3)

where wake indicates the loss because of rotation of the wake, tip the tip loss and profile theprofile losses.

Wake loss:

The wake loss can be calculated on the basis of Schmitz theory. Integrating (3.31) over the wholeblade area and using (3.8) and (3.33) gives.


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( )

= R

rd

sin

3

2sin

44

1

2

13

21

0

3

12

Schmitz

R

rXvDP [W] (6.4)

This can be solved numerically, see an example in Appendix D. Based on this we can define

Av

PC

31

SchmitzSchmitzp,

2

1

= [-] (6.5)

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0 2 4 6 8 10

X [-]

Cp[-] Cp(Betz)

Cp(Schmitz)

Figure 6.1. Coef. of power according to Betz and Schmitz

The difference between Betz and Schmitz is, that Schmitz takes the swirl loss into account andtherefore we can define swirl loss or the wake loss as

Betzp,

Schmitzp,

wakeC

C= [-] (6.6)

Tip loss:

In operation there will be a high negative (compared to ambient) pressure above the blade and a(little) positive pressure under the blade. Near the tip of the blade, this pressure difference willinduce a by pass flow from the high pressure side to low pressure side over the tip end of theblade thus reducing the differential pressure and then the power production!

The model of Betz see ref. /4/, page 153-155 results in a tip efficiency of2

2tip

9/4

92,01

+=

XB [-] (6.7)


Profile loss:

From the power calculation after (3.12) and (3.13) we can see, that the power is proportional to Cx.For an ideal profile, i.e. with no drag, the power would the be higher, from which we can define theprofile efficiency to

( ) ( ) ( )

( ) ( )

tan1

cos

sincos

L

D

L

DLprofile

C

C

C

CCr =

= [-] (6.8)

Using (3.7) we get

( )GRR

Xrr

2

31profile = [-] (6.8)

Assuming the angle of attack to be the same over the entire blade length the glide ratio is constanttoo and then (6.8) can be integrated over the blade length to give

GR

X= 1profile [-] (6.9)

Example 6.1

Assuming the glide ration to be GR = 100 and the blade number to B = 3 then the rotor efficiencycan be calculated as function of the tip speed ratio, see figure 6.2.

Rotor efficiencyBased on: GR = 100; B = 3

0,00,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

1,0

0 2 4 6 8 10

X [-]

eta_

rotor[-]

profile

w ake

tip

rotor

Figure 6.2. Rotor efficiency

Most modern wind turbines have tip speed ration at nominal wind speed and power aroundx= 7,and from the curve it is obvious, that this is close to optimal!


14/22


Example 6.2

Most modern wind turbines have glide ratios around 100 and three blades. Figure 6.3 shows therotor efficiency for 2,3 and 4 blades and with the glide ratio as parameter.

Figure 6.3. Rotor efficiency

ForX= 7 and for a glide ratio GR= 100 it can bee seen, that the number of blades have the

following influence on the rotor efficiency2 blades: 79,5%3 blades: 83,3%4 blades: 85,1%

3 and 4 blades are more efficient than 2 blades, but also more expensive. When a 3 blade rotor inspite of that has become a de facto standard it is due to a more dynamical stable rotor.

6.2. Gear box, generator and converter

Most wind turbines have the following main parts, a rotor, a gear box a generator and an electric

converter, see figure 6.4. Each of these components has losses.


Pmax rotorP LSP genP

gridP

Gearbox

Generator

Converter

Figure 6.4: Main components in a wind turbine

The total efficiency of such a turbine can the be defined as

convgengearboxrotor

max

grid

total ==P

P [-] (6.10)

where

gen

grid

conv

HS

gen

gen

rotor

HSgearbox

max

rotorrotor

P

P

P

P

P

P

P

P

=

=

=

=

[-] (6.11)

where the indices stand for LS = low speed (shaft); gen = generator; conv = frequency

converter and grid = grid net.

Typical values for the efficiencies are at nominal powerGearbox: 0,95-0,98Generator: 0,95-0,97Converter: 0,96-0,98

At part load, the lower values can be expected.

Cooling:

The cooling of the components can be calculated as power input minus power output. As anexample for the gear box: gearbox=Protor -PLS.


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0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30

Wind speed [m/s]

Power[kW

]

0

10

20

30

40

5060

70

80

90

100

0 2 4 6 8 10

Tip speed ratio [-]

Efficiency

[%]

Figure 7.3: Power as function of wind speed (left) and efficiency as function of tip speed ratio

(right)

8. Distribution o f wind and annual energy production

Weibull distribution

The wind is distributed close to the Weibull distribution curve. For practical purposes one can

calculate the probability for the wind being in the interval vi< v< vi+1

( )

=


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Annu al product ion

0

10.000

20.000

30.000

40.00050.000

60.000

70.000

0,5

2,5

4,5

6,5

8,5

10,5

12,5

14,5

16,5

18,5

20,5

22,5

24,5

v_m [m/s ]

E[kWh]

Figure 8.3: Annual distribution

What would be an optimal maximum power,Pmax? From figure 8.1 we can see that wind speedabove 15 20 m/s are very rare. On the contrary, the power production of af wind turbine rises witha power of 3.

Figure 8.4 shows a calculation of the annual production as a function of the maximum power,Pmax.

Annual energy product ion

0

100

200

300

400

500

600

700

0 200 400 600 800 1000

P_N [kW]

E_

ann[MWh]

0,0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

CF[-]

E_ann

CF

Figure 8.4: Annual energy production and capacity factor as function of nominal power

The capacity factor is definded as CF=Eann/ (PN8766h).

Conclusion:To answer the question we must know the price of the turbine, including tower and foundations, butmore than about 200-300 kW does not seems reasonable.


9. Symbols

a - Axial interference factora - Tangential interference factor

A m/s Wind speed, distribution curveA m2 Area, swept area of the rotorA1 - ConstantA2 - ConstantB1 - ConstantB2 - ConstantB - Number of bladesCD - Coefficient of dragCD,max - Coefficient of drag, max value, at = 90CDst - Coefficient of drag, where stall beginsCL - Coefficient of liftCLst - Coefficient of lift, where stall beginsCP - Power production factorCF - Axial force factorCy - Coefficient of axial forcesCx - Coefficient of tangential forcesc m Chord lengthEann Wh Annually produced energyF - Calculation valueFL N/m Lift force (per length of blade)FD N/m Drag force (per length of blade)

k - ConstantK - FactorM Nm Torquen 1/s Rotational speed of rotorp Pa Pressureptot Pa Total pressure (Bernouillis equation)P W PowerPN W Power, nominal windPmax W Max power of a given turbineQM

* N/m Torque per length of blader m Radius to annular blade section (BEM theory)

Re - Reynolds numberT N Axial force (thrust) on the rotorT* N Axial force per width of the bladesU N Tangential force on the rotorU* N Tangential force per width of the bladesu2= u m/s Tangential wind speed in the rotor planev2= v m/s Axial wind speed in the rotor planev1 m/s Wind speed, upstream the rotorv3 m/s Wind speed, down-stream the rotorvTIP m/s Tip speed of rotor bladew m/s Relative wind speed

X - Tip speed ratio


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x - Local speed ratio

A Angle of attack, relative wind in relation to blade chordst Angle of attack, where stall begins

Pitch angle of the blade to rotor plane Relative wind to rotor axis Efficiency Angle of relative wind to rotor plane s-1 Angular velocity

kg/(m s) Dynamic viscosity

p Pa Differential pressure, over the rotor

w m/s Change of relative wind speed

u m/s Change of tangential wind speed

v m/s Change of wind speed

kg/m3 Density of air (here 1,225 kg/m3)

10. Literature

/1/ Andersen, P. S. et alBasismateriale for beregning af propelvindmllerForsgsanlg Ris, Ris-M-2153, Februar 1979

/2/ Guidelines for design of wind turbinesWind Energy Department, Ris, 2002, 2ndeditionISBN 87-550-2870-5

/3/ Abbott, I. H., Doenhoff, A. E.Theory of wing sectionsDover Publications, Inc., New York, 1959

/4/ Gasch, R; Twele, J.Wind power plants - Fundamentals, Design, Construction and OperationJames and James, October 2005


App. A: Conservation of momentum and angular momentum

Momentum

Momentum of a particle in a given direction is defined as

um= (A1)

where mis mass and uis speed of the particle

According to the Newtons 2ndlaw we have

t

pF

d

d= (A2)

whereFis the force acting on the particle

If the mass is constant, we have (Newtons 2ndlaw)

amt

umF ==

d

d (A3)

where ais the acceleration of the particle

If we have a flow of particles with the mass flow qmwe can calculate the force to change thevelocity for u1to u2as

( )12 uuqF m = (A4)

Force equals differential pressure, p, times area,A, i.e. (A4) can be written as

( )

A

uuqp m 12

= (A5)

Example

For a wind turbine we have a wind speed up-stream the turbine of u1= 8 m/s and a wind speeddown stream of u2= 2,28 m/s. In the rotor plane the wind speed is just the mean value of these tovalues, i.e. u= 5,14 m/s. The blade length isR= 25 m. Find the axial force on the rotor and thedifferential pressure over the rotor.

First we calculate the mass flow as

kg/s12369225,114,52522 ==== uRqqVm


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Using (A4) we getF= 12369 ( 2,28 8,0 ) = -70,7 kN. The negative sign tells us that the force is inthe opposite direction to the flow. The differential pressure is calculated by (5) giving p= 36 Pa.

Angular momentum:

r

m

u t

Figure A1: Rotating mass

Figure A1 shows a particle of mass mrotating a radius rwith a tangential velocity of ut. Theangular momentum,L, is given by

IrmrurmurmL ==== 2t2t (A6)

whereIis the moment of inertia and is the angular velocity.

The torque of the particle is given by

It

It

LM ===

d

d

d

d (A7)

where is the angular acceleration

Now consider a particle moving in a curved path, so that in time tit moves from a position at whichit has an angular velocity 1at radius r1to a position in which the corresponding values are 2andr2. To make this change we must first apply a torque,M1, to reduce the particles original angularmomentum to zero, and then apply a torque,M2, in the opposite direction to produce the angularmomentum required in the second position, i.e.

trmM 1211

= (A8)


and

trmM 2222

= (A9)

The torque to produce the change of angular momentum can then be calculated as

( )21122212 rrt

mMMM == (A10)

This formula applies equally to a stream of fluid moving in a curved path, since m/tis the massflowing per unit of time, qm. Thus the torque which must be acting on a fluid will be

( )211222 rrqM m = (A11)

or

1t12t2 ruruqM m = (A12)

Example

Figure 2 shows a wind turbine with 2 blades. The blade length isR= 25 m and the rotational speedis n= 25 rpm which gives an angular velocity of = 2,62 s-1.

R

r

dr

uu

u2t

1u2

Figure A2: Wind turbine with 2 blades

Let us calculate the power for the annular element given by radius r= 17 m and with a thickness ofdr= 10 cm. In a calculation concerning the BEM theory, one can find the axial velocity in the rotorplane at u= 5,14 m/s (a= 0,357) and at tangential velocity of the air after pasing the rotor plane at

u2t= 0,65 m/s (a= 0,0072)

Wi d bi S G d f 38 Wi d bi S G d f 39


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The mass flow through the annular element is

( )( ) ( )( ) kg/s0,68225,114,5171,017d 2222 =+=+== urrrqq Vm

In formula (A12) we have u1t= 0 because there is no rotation of the air before the rotor plane andu2t= 0,65 m/s and r1= r2= r= 17 m. The torque can be calculated at

( ) Nm755000,1765,00,68 ==M

The power can be calculated atP=M = 1,98 kW


App. B: Formulas, spread sheet calculat ions

Formulas in spread sheet, see figure 7.1

14

15

161718

E

=1,5/8

=E14*$E$7

=2/3*ATAN(1/$E$7/E14)*180/PI()=E16-$E$9=1/$E$8*16*PI()*E14/$E$10*(SIN(1/3*ATAN(1/$E$7/E14)))^2

Formulas in spread sheet, see figure 7.2

2223

2425

E

=Design!E14=E22*$E$6

=Design!E17=Design!E18*$E$6

2930313233

E

24,2317401773297

1,29343334743683

=E29+$E$64=MIN(E30*$E$10/(2*PI()*E23);1)=E23*$E$11

37383940

414243

444546

474849

50515253

545556

5758596061

E

0,316464512669469

0,24291710357473

=ATAN((1-E37)/(1+E38)*$E$7/(E23*$E$11))/PI()*180=E39-E31

=IF(E40


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Documents

Turbines May4_2009_1.pdf