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Page 1 of 15
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
Page 2 of 15
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
Page 3 of 15
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.
Page 4 of 15
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:
Page 5 of 15
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
Page 6 of 15
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:
Page 7 of 15
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 *
Page 8 of 15
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
Page 9 of 15
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:
Page 10 of 15
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,
Page 11 of 15
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
Page 12 of 15
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
Page 13 of 15
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
Page 14 of 15
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.
Page 15 of 15
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.