Group Assignment 3

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Introduction to Wind Energy

Group Assignment 3

Dynamics

Group number 12

Editors

Group members Student number

Sergio Torres 4116127

Joseph Vitolla 4118308

Other group member

Konstantinos Gorgogetas 4119096

Ana Maria Núñez 4123093

Aymeric Buatois 4125738

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Contents

1. Motion of the transmission ................................................................................................. 3

2. Equilibrium point ................................................................................................................ 6

3. Eigenfrequency of the transmission system ....................................................................... 9

4. Torsion angle transmission ............................................................................................... 12

Table of Figures Figure 1. Model Transmission ................................................................................................... 3 Figure 2. System equivalent to geared system. .......................................................................... 3 Figure 3. Campbell diagram .................................................................................................... 11

Figure 4. Torsion angle of transmission in response to a wind gust ........................................ 12 Figure 5. Response of the torsion angle of transmission to a severe wind gust ...................... 12

Figure 6. Detail of the response of the torsion angle transmission to a severe wind gust ....... 13 Figure 7. Comparison of the response of the torsion angle and rotor flap angle to a severe

wind gust .................................................................................................................................. 14 Figure 8. Response of the torsion angle to a sinus gust ........................................................... 15

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1. Motion of the transmission

In the following question, we are using the notation:

J Inertia Kg.m2

M Momentum / Torque N.m

α Angular acceleration. Rad/s2

Ω Rotation speed Rad/s

θ Angular position Rad

ν Transmission ratio. -

r Radius M

m Mass Kg

k Stiffness

ς Damping

a) Derive the equation of motion of the transmission. Usually the combination of

slow shaft / transmission / fast shaft is replaced by an equivalent system of just one

shaft: the slow shaft (as shown during the lecture). Argue why the generator moment

should be multiplied by the transmission ratio and the inertia of the generator by the

transmission ratio squared.

Figure 1. Model Transmission

Usually the combination of slow shaft / transmission / fast shaft is replaced by an equivalent

system of just one shaft: the slow one as is shown in the Figure 2.

I1 I21M , kt2M

1 2

Figure 2. System equivalent to geared system.

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To demonstrate why the generator moment should be multiplied by the transmission ratio, the

moment definition is used as follows:

FrM Equation 1: Momentum definition

g

gfast

fast

slow

gslow

gfastg

MM

FrM

r

r

FrM

FrM

*

)(

2

2

2

It is possible to refer shaft stiffness (k) and inertias (J) to equivalent values on a single shaft

(It is assumed that the shafts themselves have no inertia). This is done by multiplying all

stiffness and inertias of the geared shaft by ν2 where ν is the speed ratio between the two

shafts. It is also shown as follows, using the concept of inertia:

2*rmJ

Equation 2: Concept of inertia

To demonstrate why the inertia of the generator must be multiplied by transmission ratio

squared, the concept of inertia is used:

g

fastg

fast

slow

slowg

fastgg

JJ

rmJ

r

r

rmJ

rmJ

2

2

2

2

2

2

2

*

*

*

Derivation of the equation of motion:

Gears are frequently used to transfer power from one shaft to another, while maintaining a

fixed ratio between the speeds of the shafts. While the input power in an ideal gear train

remains equal to the output power, the torques and speed vary in inverse proportion to each

other. So:

gr

gr

grrt JJ

JJ

JJJJJ

2

2

2

2

11111

Taking into account the mass / spring / damper system, where:

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g

gr

rr

gr

g

gr

gr

gg

r

rr

rr

r

tt

MJJ

JM

JJ

Jk

JJ

JJ

JJandMM

MJJ

JM

JJ

Jk

JJ

JJ

MkJ

**

**

22

2

2

2

2

22

2

22

2

2

2

b) Convert the 2nd

order differential equation to 2 1st order differential equations.

The method used to convert the 2nd

order differential equations in 2of 1st order was the

separation of variables.1

g

gr

rr

gr

g

gr

grM

JJ

JM

JJ

Jkyy

JJ

JJii

yi

***

)

)

22

2

2

2

c) Compare your results with the listing of the MATLAB file dynmod.m.

In the MATLAB file can be found the follow equations:

deps=epsd; ddeps=1/Jtot*(Jtot/Jr*Mr+Jtot/(nu^2*Jg)*nu*Mg-dr*epsd-kr*eps);

Where:

epsd: torsion angular velocity transmission (rad/s)

deps: torsion angular velocity transmission (rad/s)

ddeps: torsion angular acceleration transmission (rad/s2)

Jtot: Total inertia transmission (kg*m2)

nu: transmission ratio

Jg: Inertia generator (kg*m2)

Mg: generator torque

kr: stiffness transmission (Nm/rad)

eps: torsion angle transmission (rad)

So, we can see that deps=epsd is equal to y , as was shown in b. Furthermore for the

second:

1 http://www.sc.ehu.es/sbweb/fisica/cursoJava/numerico/eDiferenciales/rungeKutta1/rungeKutta11.htm#Sistema

de ecuaciones diferenciales de segundo orden

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kyMJ

JM

J

J

Jy

MJ

JM

J

JkyyJ

MJJ

JM

JJ

Jkyy

JJ

JJ

g

g

tr

r

t

t

g

g

tr

r

tt

g

gr

rr

gr

g

gr

gr

***1

***

***

2

2

22

2

2

2

So, can be concluded that the same equation was obtained previously.

2. Equilibrium point

a) Indicate how an equilibrium point can be determined; by which equilibrium of

moments is the equilibrium point determined?. What is the explicit expression?

The equilibrium point can be reached when the differential equation for dynamics in wind

turbine are zero. It means that there is not any change with the time.

So, flap angular velocity, tower top velocity and torsion angular velocity are equal to zero.

As is known, the rotor torque creates the generator torque, so, in equilibrium Mg=Mr.

In order to obtain the rotor torque, the blade element method is applied, taking the following

equations:

drrcCosCdSinClVdM

drCosCdSinClcVdF

drcVCddrcVCldF

CosdDSindLdF

drcVCddDAQCdD

drcVCldLAQClL

CosDSinLF

r ****2

1

****2

1

****2

1****

2

1

**

****2

1**

****2

1**

**

2

2

22

2

2

To obtain the generator torque, the following equations are used:

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V

RPM

MM

shaftslowtofering

R

V

PM

PM

MPP

shaft

g

gg

rr

rg

r

shaft

g

rg

g

shaft

g

gggshaft

*

*

*

:Re

*

*

*

*

2

2

Replacing in Mg=Mr at the equilibrium point:

V

RPdrrcCosCdSinClV

shaft

*

*****

2

1 2

b) Compare your answer with the listings of equi.m and equi_fun.m.

In equi.m for the equilibrium state is stated:

betad=0; xd=0;

It is the same as was stated in a, where xand .

Furthermore, equi.m has two conditions to determinate the equilibrium point. The first is in

partial load operation (V≤Vn) and full load (V>Vn).

For V≤Vn, the equilibrium point is determinate when there is equilibrium between rotor

torque and generator torque, as was said before, Mr=Mg.

In order to obtain the steady condition it is used the fun_equi of MATLAB through the

determination of the difference between aerodynamic rotor torque and generator torque.

To find the stationary generator angular velocity the MATLAB file uses: omg=nu*omr

That is the same as the one used in the present report.

rg *

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For the generator torque the MATLAB file gen.m is used, at nominal conditions:

omgn=nu*lambdan*Vn/R;

As previously mentioned, and stated in the report:

R

V

R

V

g

rr

rg

**

*

*

In MATLAB to determinate the generator torque, the mechanical generator power is

considered, as follows:

Psh=Pn/eta Mg=Psh/omg;

The same was also expressed in this document:

V

RPM

MM

shaftslowtofering

R

V

PM

shaft

g

gg

rr

rg

r

shaft

g

*

*

*

:Re

*

*

*

2

2

To prove the rotor torque, in MATLAB there is the file bem.m, which also uses the file

aero2.m with the inductor factor determined. So, the rotor torque is:

dMr=Nb*ri.*(kp.*dL.*sin(phi)-dD.*cos(phi));

Where:2

Nb: number of blades

ri: radial position blade elements

kp: power loss factor

dL: lift force blade element

dD: drag force blade element

For dL and dD MATLAB has:

2 Taken from aero.m in MATLAB

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dL=Cl.*0.5*rho.*W.^2.*ci.*dr; dD=Cd.*0.5*rho.*W.^2.*ci.*dr;

Where:

rho: air density

W: resultant velocity

ci: chord blade element

dr: length blade element

Finally, in the present document for the rotor torque was obtained:

drrcCosCdSinClVdM r ****2

1 2

Which is the same in the MATLAB files.

c) Determine the equilibrium point for a wind speed of 8 m/s (use equi.m).

For this velocity through MATLAB file was obtained:

- Stationary flap angle β=0.0159 (rad)

- Stationary tower top displacement: x=0.2642 (m)

- Stationary rotor angular velocity: Ωr=1.3932 (rad/s)

- Stationary torsion angle transmission: ε=0.0040 (rad)

- Stationary generator angular velocity: ωg=136.529 (rad/s)

- Stationary induction factor: a=0.3259

- Blade pitch angle = - 1.5 (°)

- Stationary axial force= 2.2362*105 (N)

- Stationary aerodynamic flap moment = 2.1758*106 (N-m)

- Stationary aerodynamic rotor torque = 7.1893*105 (N-m)

3. Eigenfrequency of the transmission system

a) Determine the eigenfrequency of the transmission. Does this eigenfrequency depend

on the rotational speed?

The eigenfrequency of the transmission system depends only on the stiffness of the elements

and the masses involved. It is independent from the rotational speed.

From the equation of motion, the eigenfrequency can be obtained as:

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Where

For Vestas V90 turbine parameters are given:

Transmission ratio =98

Stiffness transmission kt= 1.8 * 108 [Nm/rad]

Inertia generator Jg= 60 [kg*m2] → J2= 2

* Jg = 982 * 60 = 576240 [kg*m

2]

Inertia rotor J1= 3*Jblade = 3* 3.9*106 = 11.7*10

6 [kg*m

2]

Then, is calculated as:

b) Sketch (i.e. calculate just a few points) this relation between eigenfrequency of

the transmission and rotational speed in a Campbell diagram. Give excitations and

eigenfrequencies a different line type (or color).

We select the nominal wind speed to find the rotor speed:

Vn=12 m/s λ=7.8 R=45 m

Then,

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The harmonics plotted on the following Campbell diagram are:

1-P:

3-P:

6-P:

Figure 3. Campbell diagram

The harmonics are plotted with 20 % of margin for each P-frequency. As it can be seen from

the Campbell diagram, there is not intersection point between the eigenfrequency value and

the curves for 1-P, 3-P or 6-P, meaning that there won’t be undesired resonance in the

operation of the turbine for the transmission system.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.350

0.5

1

1.5

2

2.5

3Campbell diagram

Rotational speed (Hz)

Chara

cte

ristic f

requency (

Hz)

6-P

3-P

1-P

Eigenfrequency

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4. Torsion angle transmission

a) Determine the response of the torsion angle transmission to a smooth wind

gust, by means of gust1.m, for a mean wind speed of 8 m/s.

Figure 4. Torsion angle of transmission in response to a wind gust

a) Determine the response of the torsion angle transmission to a ‘severe’ wind gust, by

means of step; the mean wind speed is 8 m/s.

Figure 5. Response of the torsion angle of transmission to a severe wind gust

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b) Estimate from the previous result (severe gust) the frequency of the variations of the

torsion angle transmission, during the first seconds of the response. Does it coincide

with the frequency determined at 3); explain.

Figure 6. Detail of the response of the torsion angle transmission to a severe wind gust

From the graph, we can deduct the frequency: Hz2.842

13

tt

f

Although very similar to the eigenfrequency calculated in section 3, the values for this

frequency is different meaning that there is no resonance in the system. However a very

simple model is used for the calculations and many assumptions are taken. For further

analysis it would be necessary to use a more complex model and more powerful simulation

tools in order to get a more accurate result.

t1

t2

t3

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c) Compare qualitatively the response (on the severe gust) of the torsion angle with the

response of the rotor flap angle (first element of the state vector x).

Figure 7. Comparison of the response of the torsion angle and rotor flap angle to a severe wind gust

The flap response is more damped because of the reaction of the system. When the wind

increases, the system will pitch the blade to adjust to the new wind speed, resulting in a

quicker stabilization of the flap. Furthermore, materials employed for the fabrication of the

transmission system and the rotor system are different. In the former case, materials are more

rigid (larger stiffness coefficients) which result in a more damped response, if they are

compare to the blade system for which more flexible materials are used.

d) Determine the response of the torsion angle transmission to a sine wind gust by

means of gust2.m. In reality a sine wind is of course not possible; it represents the 1P

(once per revolution) variations of the wind speed at the location of a blade section

due to ‘rotational sampling’, tower passage, wind shear and yawed flow. In fact, you

studied the varying wind speed due to wind shear in the 3rd

individual assignment,

question 4. Please take care to use the correct rotational speed which corresponds to

the mean wind speed of 8 m/s.

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Figure 8. Response of the torsion angle to a sinus gust

e) After some period (say 60 s) the transient oscillations are damped out and the

response has also a sine form. Estimate again the frequency and explain your answer.

The frequency can be calculated from the plot on section e). It can be seen that the period

from the signal is T=4.5 s. Frequency can be calculated as

There is a stabilization period after which the response remains damped with the same

frequency of the wind gust. There is a phase shift that can be explained as a delay in response

caused by the inertia of the system.

Reference:

- Manwell, J. et al., “Wind Energy Explained, Theory, Design and Application”, USA,

2002.