Wind Energy Conversion System_Lecture.pdf

Wind Energy Conversion Systems
Prof Rukmi Dutta
School of Electrical Engineering and Telecommunications
UNSW Sydney
Australia

2
School of Electrical Engineering and
Telecommunications
#1 in Australia for Electrical Engineering
40 academics, 1700+ students (UG, PG, PhD)
University of New South Wales (UNSW Sydney)
QS World ranking: 19, Times Higher Education ranking:84
60,000 students, 2,900 Academics

3
Course plan
Module 3
Historical development of wind power
Types of Wind turbines
Power in wind
Maximum rotor efficiency
Maximum power point tracking
Types of generators used in WEC
Direct-drive Vs Geared WEC
Power electronic converters for WEC – isolated and grid-connected
Module 4
Offshore and onshore wind farms
Impact of high penetration of WEC on grid
Role of inertia of individual wind turbines
Power quality – fluctuations of active power
Learning goals
After 2 modules, the students will be able to
1. Explain how a wind turbine works
2. Describes types of wind turbines
3. Derive power in the wind from the KE
4. Explain maximum rotor efficiency, losses and
limitations
5. Compute WEC efficiency
6. Compare SCIG, DFIG and PMSG
7. Explain MPPT control
8. Compare geared and direct-drive systems
9. Contrast different WEC systems in terms of PEC
(power electronic converters)
10. Compare onshore and offshore Wind farms
11. Explain the impact of low inertia
12. Explain power quality in WEC

4
Energy Resources
Electrical
Energy
Nuclear
Fossil Fuels
Renewable
Sources
Coal, Oil, Gas
Hydro, Solar, Wind,
Tidal/Wave, Biomass
We need energy resources to convert to electrical energy,

5
A simple generator
In a real generation system, the drive shaft is turned by a turbine, which in turn is rotated by
steam, gas, wind, or water.
volt
f f
d d
e N
dt dt
φ λ
=
− =
−
Faraday’s law of induced EMF:
Where,
N: coil turns
φf: the magnetic flux in Wb
λf : magnetic flux linkage in Wb-t
Wind energy Turbine Mechanical Energy Generator Electrical Energy

6
Wind Energy
Ref: https://www.peninsulacleanenergy.com /429-2/
Kinetic Energy of Wind:
2
1
[Ws]
2
KE mv
=

7
Historical Development
1891, Denmark
30kW
1941, Smidth Aeromotor
50kW
1941, Smith-Putnam
1250kW
1987, MOD-5, Boeing US
3200kW
Darrieus, Canada 1987
4MW
Modern Wind turbine
10- 12MW
(largest 15MW)

9
Quality HAWT VAWT
Design Propeller design with
aerodynamically
optimised blade lift
Simple
Housing of the Gear,
generator and all other
components
Up on the tower Ground level
Yaw mechanism Yes No
Tip speed ratio and
power coefficient
High low
Self-start Yes No
Power or speed
control via blade
pitching
Yes No

10
Mass m of air flowing with a velocity v m/s produces kinetic energy
Mass m can be expressed in terms of volume and density 𝜌𝜌 in kg/m3 ;
volume is a function of swept area A in m2, wind velocity v in m/s, and a
time interval t in s.
2
1
[Ws]
2
KE mv
=
( )
.
m vol Avt
ρ ρ
= × = ×
3
1
[J or Ws]
2
KE A tv
ρ
∴ =
Substituting m in KE above,
3
1
[ ]
2
wind
KE
P Av W
t
ρ
= =
Thus, the wind Power
2
2
D
A π
 
=  
 
Swept area Rotor diameter D ≅ 2×Blade length

11
3
1
[ ]
2
wind
KE
P Av W
t
ρ
= =
( )
2
2
blade length
2
D
A π π
 
≈
 
 
The larger the blade size, the larger the
power Pwind.
( )2 3
1
blade length [W]
2
wind
P v
ρπ
=
Substituting A,

12
Wind power density
( )
2
2
3 3 3
1 1 1
blade length [W]
2 2 2 2
wind
D
P Av v v
ρ ρπ ρπ
 
= = ≈
 
 
3 2
1
/ [W/m ]
2
wind
wind
p P A v
ρ
= =
Wind power per unit area is defined as
the power density of a site,
Since the swept area is a function of a turbine’s blade length, it is often useful to know the wind
power in each square meter of a site to select a turbine.
Monthly variation of wind power density in Silchar, Assam*
*Wind data analysis of Silchar (Assam, India) by Rayleigh's and Weibull methods, R. Gupta and A. Biswas, Journal of Mechanical Engineering Research 2010 Vol. 2
ρ =1.225kg/m3
v = 2.5 m/s

13
Example 1
(a) If the air density of a site is 1.12 kg/m3 and wind speed is 12 m/s, compute the wind power density [W/m2] of the site.
(b) What should be the blade length of a wind turbine to be used in the site of (a) so that 2 MW wind power is available for
the turbine to capture?
Ans: (a) 967.7W/m2 (b) 25.65 m
3 2
1
Power denisty [ / ]
2
v W m
ρ
=
2
3
1
[ ], where
2 2
wind
D
P Av W A
ρ π
 
= =  
 
Try this
The annual average power density at Silchar is 10W/m2. What should the blade length be to produce 1 MW of wind
power? Is it practical?

14
Power coefficient Cp of wind turbine
Turning of the blades and the rotor of a wind turbine capture only part of the available Pwind.
The mechanical power extracted by an ideal turbine:
( ) ( )
3 3 2 2
1 1 2 2 1 2 1 1 2 2 0 0
1 1
,
2 2
turbine
P A v A v m v v m A v A v A v
ρ ρ ρ ρ
= − = − = = =

The force exerted by the air mass on the turbine blades:
( )
1 2
F m v v
= −
The associate power must be Fv0 , which is equal to Pturbine in a loss less system.
( ) ( ) ( )
1 2
2 2
0 turbine 1 2 0 1 2 0
v v
1
Fv P m v v v m v v v
2 2
+
= → − = − → =
( )
1 2
0 0 0
v v
m A v A
2
ρ ρ
+
∴
= =
By Newton’s second law, a reactive force of the turbine blades pushes the air mass m.
(A0 is the swept area of a turbine.)

15
( ) ( )
( ) ( )( )
1 2
2 2 2 2 2 2
1 2 0 1 2 0 1 2 1 2
1 1 1
2 2 2 4
turbine
v v
P m v v A v v A v v v v
ρ ρ
+
 
= − = − = + −
 
 
Substituting air mass m,
( )( )
2 2
2
0 1 2 1 2
2 2
2
3 1 1
0 1
1
1
4 1 1
1 2
2
turbine
p
wind
A v v v v
P v v
C
P v v
A v
ρ
ρ
+ −   
= = = + −
  
  
The ratio of the extractable power by the turbine from the wind power is called power
coefficient Cp
Cp depends on the ratio of the up and downstream air velocities.
The maximum ideal Cp = 0.593 when v2/v1 = 0.333. Albert Betz, a German aerodynamicist, was the
first to calculate the ideal Cp, which is hence
known as Betz’s limit.
Ratio v2/v1

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In a practical wind turbine, the rotor spinning reduces the Pturbine further. Hence, Cp of all
practical turbines is much smaller than Betz’s limit.
The Cp of the practical turbine is the ratio of energy associated with the turbine’s rotational
speed and the wind’s linear speed. This ratio is known as the Tip Speed Ratio (TSR) and can be
determined as below:
Linear velocity of the rotor blade tip
Wind velocity
blade length
tip
wind wind
v r
TSR
v v
r
ω
λ
×
= = =
≈
2
3 3
1 1
[W]
2 2 2
turbine p wind p wind p wind
D
P C P C Av C v
ρ ρπ
 
   
 
= = =
   
 
   
 
TSR is an important parameter of wind turbine design. If the TSR is too small, most of
the wind pass through the open area between the blades and little energy is captured by
the blade. If it is too large, the fast-moving blades appear like a solid wall to the wind.

17
TSR versus Cp
wind
r
TSR
v
ω
λ
×
=
turbine
p
wind
P
C
P
=
Cp is calculated for a certain TSR of a given turbine. The plotting of the Cp vs TSR, allows to
assess the Cp and power capture of a turbine at different wind speed.

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Example 2
(i) A 50kW wind turbine operates with a peak power coefficient Cp when the wind velocity is
11.6 m/s. At this wind speed, the turbine reaches 62% of the Betz limit (0.593). What is the
Cp for this wind turbine?
(ii) The output of a small wind turbine is 50V and 20A at a certain wind speed, and the wind
power Pwind calculated for this speed is 3.9 kW. What is the value of the power coefficient?
(iii) In a sales pitch, a wind turbine manufacturer boasts that their turbine has reached an
efficiency of 75%. Why or why not this claim is believable?
(iv) If the same turbine manufacturer claims their TSR is around 25, based on the Cp Vs TRS
graph of the previous slide, do you think it is a ‘good’ TSR for a turbine? Why does Cp
fall with the higher TSR?

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Key points
• Pwind is proportional to the square of the blade length.
• Wind power density (W/m2) is useful to assess a site’s wind energy capability.
• Not all of Pwind can be captured by the turbine; the ratio of the power captured by the
turbine to the available Pwind is the power coefficient Cp.
• The theoretical limit of Cp (Betz’s limit) is 0.593.
• The ratio of the tip speed of the turbine blades over the speed of the wind is called
the tip speed ratio (TSR).
• Cp Vs TSR graphs are useful to assess turbine performance.
• VAWT has low Cp but can capture more power at low wind speed; HAWT has
achieved the highest Cp so far.

20
Course plan
Module 3
Power in wind
Module 4
Offshore and onshore wind farms
Impact of high penetration of WEC on grid
Role of inertia of individual wind turbines
Power quality – fluctuations of active power
Learning goals
 Explain how a wind turbine works
 Describes types of wind turbines
 Derive power in the wind from the KE
1. Explain maximum rotor efficiency, losses and
limitations
2. Compute WEC efficiency
4. Explain MPPT control
5. Compare geared and direct-drive systems
6. Contrast different WEC systems in terms of PEC
7. Compare onshore and offshore Wind farms
8. Explain the impact of low inertia
9. Explain power quality in WEC

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Review Quiz
Q1. True or false – maximum possible efficiency of a wind turbine is 59.3%.
Q2. True or false – a wind turbine’s optimum power coefficient Cp corresponds to an optimum TSR (tip speed ratio).
Q3. If a remote village has an average wind speed of 4 m/s and a government scheme has directed your company to install wind
turbines for electrification, which of the following will be your recommendation:
(a) Single HAWT with long blades on a tall tower at the centre of the village to meet the total electricity demand of the village
(b) Small VAWT at the rooftop of each household is needed to meet each household’s electricity demand.
(c) Use of alternative technology such as solar instead of wind.

22
Typical datasheet of a wind turbine

23
Turbine Design steps
Select rated output
power at a rated wind
speed
Estimate an optimum
Cp and a rotor
diameter (i.e. blade
length) from the
estimated Cp Vs TSR
plot
Optimise the design
of the rotor and the
blade shape for the
best possible
aerodynamic
performance
Prated
Vwind_rated

25
Basic aerodynamics of a wind turbine
Useful terminologies:
Lift-to-drag ratio (L/D)
Rotor solidity = Total blade area/Rotor swept area (%)
Aspect ratio = (Rotor radius)2 / Area of the rotor blade
Taper = Chord length at the tip/Chord length at the root
Rotor blade geometry

26
Blade design factors that influence Cp
1. Number of blades
2. Aerodynamically optimum shape of the blades
3. Blade twist
4. Blade air foil (lift-to-drag ratio)
5. Blade thickness
Common materials for the modern turbine blades:
Fibre glass /polyester, fibre glass/epoxy, fibre
glass/carbon fibre.

27
Aerodynamics of VAWT
Varity of designs in recent time
• Cp has increased to 0.4 but is still lower than
HAWT.
• Optimum Cp can be achieved at lower TSR,
hence ideally suited for low-speed, high toque
operation.
• High drag and poor L/D ratio.

28
Yaw control
For turbine to capture the wind power, it must be oriented correctly
with respect to the wind velocity. A large deviation (yaw angle)
results in a large loss of the power extraction. The rotor can be
oriented to the wind direction:
1. Using a wind vane – simple and often used in smaller wind
turbines
2. Free yawning of the rotor located at the downstream – rotor must
be positioned such that the point of attack of the aerodynamic
force is behind the axis of rotation.
3. Using motorised yaw drive – exclusively used in large turbines
Wind vane

29
Aerodynamic power control
At high wind speed, the power captured from the turbine can exceed the design ratings. Also,
during a power outage, due to the sudden loss of load on the generator, the turbine speed can
increase too rapidly beyond the design limit. Hence, the turbine must have aerodynamic means
to limit power and rotational speed.
The aerodynamic force can be reduced by means of varying the
angle of attack or reducing the swept area or adjusting rotor
speed (i.e., variable speed control, but effective usually for a
narrow power range).
Hence, the practical methods involve changing the angle of
attack by (i) rotor blade pitch angle control (ii)passive and
active stall control with fixed or variable blade pitch.

30
υ: pitch angle
α: angle of attack

31
Furling and variable speed control
• The process of tuning the rotor out of the wind using a yawing
mechanism is called furling.
• This method, in combination with the variable speed control, is
found to be very effective.
• The output power of the rotor with a wide speed range can be
controlled using the torque and speed control of the electric
generator.
• At very high speed, when such a generator control becomes
ineffective, furling can gradually reduce the power output. This
method does not require complex blade pitch control, even in
large turbines.

32
Effect of rotor wake
In wind farms, the turbines are located close by.
The effect of downstream flow of one turbine can
affect the wake of the turbine behind it.
The effect includes:
1. Reduced flow velocity and hence reduced
energy output in the subsequent wind turbines
2. Turbulence of the front turbine can increase
turbulence loading of the subsequent turbines
increasing their fatigue resulting shorter life
span
3. May affect blade pitch control

33
Inevitable mechanical loads on turbine

34
Power curve of the wind turbine
Optimum power versus wind speed gives
the turbine’s so-called power curve.
At the cut-in velocity, the turbine starts to
deliver power. At rated velocity, the output
power equals the rated value.
The cut-out velocity is the maximum
speed up to which the turbine delivers
rated power. Beyond this speed, blade
pitch control starts.

35
Mechanical Drive train in the Nacelle
Direct-drive
Gearbox

36
Efficiency
Typical values
Direct Drive
out p turbine gear gen wind
P C P
η η η
=
Pwind
Aerodynamic losses
Cp is related to aerodynamic losses, ηturbine is the loss in the various bearings of turbine rotor.

37
Operating characteristic of turbines
The operating conditions influence the power output of the turbine – (i) pitch angle (β) control
and (ii) speed control
At high wind speed, increased pitch angle reduces Cp to limit power to the rated value.
The plotting of Cp versus rotor speeds shows the influence of wind speed on the characteristic—
the peak Cp value can shift to a higher rotational speed with increasing wind speed.
The turbine can yield maximum power and energy if a variable-speed generator operates along
the maximum power point (MPPT) curve.
out p turbine gear gen wind p tot wind
P C P C P
η η η η
= =

38
Example 4
A 40 m-diameter, three-blade wind turbine produces 600kW as the output of the generator when
the wind speed is 14m/s. Air density of the site is 1.225kg/m3.
(i) At what rpm the rotor turns when it operates with a TSR(tip speed ratio) of 4?
(ii)What is the tip speed of the rotor?
(iii)If the generator needs to turn 1800 rpm, what gear ratio is needed to match the rotor speed
to the generator speed?
(iv) What is the efficiency of the full wind turbine system (blades, gear box, generator) under
these conditions?

39
Solution
( )
tip
wind wind
v r
i TSR
v v
ω×
= =
( ) 56 /
tip
ii v r m s
ω
= × =
generator speed 1800
( ) Gear ratio= 67.4
turbine speed 26.7
iii = =
3
2
3
1 40
1.225 14
2 2
( )
1
2112
2
600
0.284 28.4%
2112
o
wind
wind
P
iv efficiency
P
P Av kW
kW
efficiency
kW
η
ρ
η
π
 
× × × ×
 

=
= = =

=
= =
60
2.8 rad/
1
s
4 4
26.7
4 / 2
2
0
v TSR
rpm
r
ω
π
× ×
=
×
= =

40
3 3 3
2 2 2
1 1
, 967.7 W/m
2 2
20 m
1.216
w
w
w
P
P Av v
A
A r
P MW
ρ ρ
π π
= = =
= = ×
∴ =
( ) 291
out p blade gear gen w
P C P kW
η η η
= =
A wind turbine at a site has three rotating blades; each is 20m in length. Wind power density is 967.7W/m2.
Compute the available power from the wind.
If the power coefficient is 0.4, the efficiency of blades and rotor mechanism is 90%, the gear box efficiency is
95% and generator efficiency is 85%, compute the output power.
Solution
Example 3

41
Now try this out
A 90 m-diameter, three-blade wind turbine is rated at 2MW for a wind speed of 12m/s. Air
density of the site is 1.225kg/m3.
(a) What is the total efficiency of the wind conversion system?
Ans: 29.7%,
(b) If the gearbox, generator and rotating blade combined with rotor mechanism each has an
efficiency of 90% , and turbine operates with the maximum power coefficient Cp = 0.48,
will this wind generator system be able to supply a 2MW load at 12m/s? (Note that
generator output must be ≥2MW to supply a load of 2MW.)
out
wind
P
P
η = 3
1
[ ]
2
wind
P Av W
ρ
=
out p turbine gear gen wind
P C P
η η η
=

43
Course plan
Module 3
Power in wind
Module 4
 Offshore and onshore wind farms
 Impact of high penetration of WEC on grid
 Role of inertia of individual wind turbines
 Power quality – fluctuations of active power
Learning goals
 Explain how a wind turbine works
 Describes types of wind turbines
 Derive power in the wind from the KE
 Explain maximum rotor efficiency, losses and
limitations
 Compute WEC efficiency
 Compare geared and direct-drive systems
 Explain MPPT control
 Contrast different WEC systems in terms of PEC
 Compare onshore and offshore Wind farms
 Explain the impact of low inertia
 Explain power quality in WEC

44
Wind energy estimation
If in a site with air density of 1.225kg/m3, winds blow 6m/s for 100 hours, what is the wind energy in a swept
area of 1 m2? 13,230 Wh
If in the same site, the wind blows 3m/s for 50 hrs and 9m/s for another 50 hrs (i.e. average wind speed is
6m/s), what is the wind energy in the same swept area? 23,152Wh.
Using average wind speed for the second case gives a wrong estimation.
Hence, it is customary to use probability density functions to estimate the average velocity.
For example :
3
1 6
[W] (Rayleigh probablity function)
2
wind avg
P A v
ρ
π
 
=  
 
The wind speed is also significantly affected by
the height H and friction it experiences while
moving across. The two expressions are often
used to account for this:
or
0 0 0 0
v H v ln( H / z )
v H v ln( H / z )
α
 
=
 
 
where α: friction coefficient and z: roughness length

45
Example:
750-kW, 48-m wind turbine is mounted on a 50 m tower in an area with 5m/s average wind
at 10 m height. Air density is 1.225kg/m3, and the surface roughness length z = 0.03m.
Assuming Rayleigh statistics probability, estimate the annual energy delivered if the
system’s overall efficiency is 30%.
50 10
0 0
v ln( H / z ) ln( 50 / 0.03 ) ln( 50 / 0.03 )
v v 5 6.39m / s
v ln( H / z ) ln(10 / 0.03 ) ln(10 / 0.03 )
= → = =× =
2
3 3
6
1 6 1 48 6
1.225 6.39 551
2 2 4
551kW, efficiency =0.3
551 8760 / 1.45 10 / ,
wind avg
out
P A v kW
P
Energy h yr kWh yr
ρ π
π π
η η
η
 
 
 
 
= = × × × × =
 
   
 
   
 
= ×
= × × = ×

46
Wind Energy Conversion System

47
Types of generator
Generators
Synchronous
Generator
Electrically Excited
Permanent Magnet
type
Asynchronous
Generator
(Induction
Generator)
Squirrel cage type
Wound rotor type

48
Synchronous Generator
The Stator consists of 3-phase windings and the rotor is either electrically excited
(EESG/WRSG) or permanent magnet type (PMSG).
8-pole PMSG rotor
2-pole SG round rotor 4-pole SG salient rotor
4-pole IPMSG rotor
3-phase stator

49
The wind turbine turns the rotor of the generator and its magnetic field, which induces EMF in the
three stator windings by the Faraday’s law:
f
ph f f
d
ˆ
e N , cos t
dt
φ
φ φ ω
=
− =
( )
( )
a
b
c
e 2E sin t;
e 2E sin t 120 ;
e 2E sin t 240
ω
ω
ω
°
°
=
= −
= −
V (rms), rad/s
w ph f s
ˆ
E K N , 2 f
2
ω
φ ω π
= =
Turns per phase: Nph
Winding factor: Kw

50
Rotating magnetic field
Synchronous Generator
Load
DC
a
b
c
Stator
Rotor
The direction of the rotation and the rotor speed are always synchronous with the rotation of
the rotating stator field i.e. speed of the rotating magnetic field is also the synchronous speed:
rad/s
s
m syn
( f )
2
p
π
ω ω
= =

57
Working of an Induction Generator
Stator : 3-phase with distributed
windings
Cage rotor Wound rotor
Rotating Magnetic Field:
syn
p
ω
ω = mech rad/sec
( ) ( )
r r r sl
e .l e rBl
ω
= × → =
v B
sl syn m
ω ω ω
= −
Induced voltage in the rotor:
Slip speed (relative velocity):
(SCIG) (WRIG)
syn m
sl 1 2
syn syn 1
f f
s
f
ω ω
ω
ω ω
− −
= = =
Slip:
For generation, ωm>ωsync

58
( ) ( )
ˆ ˆ
r r r ag 1 2 r a
l 2
s g 1
e f
rBl E KN s
f N f
s
K E
φ
ω φ
= →
= −= =
The induced voltage of rotor winding is at slip frequency:
The induced voltage of the stator is at supply frequency:
( ) ˆ
1 1 s ag 1
s
e rBl E KN f
φ
ω
= →=
I2
sX2
sE2 R2
(b)
I2 X2
2
R
s
E2
(a)
Rotor circuit
Referred to stator
Stator circuit
R1 X1
V1
I1
Im
Xm
' 2
2
I
I
a
=
E1=aE2
A
A’
2
'
X
'
2
R
s
Per phase Equivalent circuit

59
Mechanical torque and generator power
( )
( )
m
dev
m
Developed Power P
T
rotor Speed ω
=
Mechanical input power (Developed power) = stator
electrical power +rotor electrical power + losses
When ωm>ωsync and slip is negative,
'
'2 2
2
2
R
P = 3I
s
Total developed power:
A negative slip means it is generated active power.
There is no generated reactive power Q.

62
Fixed speed Generator systems (without power electronic converters)
SG directly coupled to the grid:
• Simple
• No reactive power compensation
is required.
• Not commonly used due to
issues with instability and
dynamic loading on the
mechanical drive train.
IG directly coupled to the grid:
• Simple and commonly used with
smaller turbines, especially in
older versions.
• Larger rated slip of smaller IG
allows for reducing dynamic
loading and instability issues.
• High in-rush current; hence may
need a thyristor-based soft starter,
which may inject harmonics to the
grid.
• Reactive power compensation.
Variable slip IG:
• Slip control via variable resistors
connected to the rotor in WRIG
allows for reducing dynamic
loading issues.
• Grid fault ride-through issues
exist
• Reactive power compensation
• Low efficiency.

63
Multi-speed generator system
The availability of the multi-speed generator improves the rotor speed’s adaptation to the wind
speed. For example, one speed for the low-speed wind to supply partial load and the other for
the high-speed wind to supply full load can reduce noise issues and increase energy yield.
Dual generator
• A smaller generator for low speed and a larger one
for high speed may improve efficiency and pf
during the partial load.
• More costly
• Complex gearing and control
Pole-changing generator
• An induction machine with two isolated stator
windings with different numbers of poles allows to
use one single generator with two different speed.
• Generators are more costly.
• Lower efficiency at low speed.

67
• Slip s>0 sub-synchronous operation, s<0 super-synchronous operation.
• At super synchronous operation (i.e. negative slip),
Mechanical input power = stator electrical power +rotor electrical power + losses
Rotor electrical power = slip power = slip × stator power
The slip is around 0.2 to 0.25; hence, the PEC rating needs only 20 to 25% of the stator power
rating.

68
Assessment criteria for the generators
Criteria SG IG Variable speed with PEC
Dynamic response in a fixed frequency grid Bad Good Good
Speed range Bad Bad Good
Controllability Bad Bad Good
Reactive power Good Bad -
Grid perturbations Good Bad Improved
Synchronisation to grid Bad Good Good
Load disconnection Good Bad Good
Efficiency Good Good Good
Cost High Low High
Maintenance and reliability High High Low, reliability improving

70
Doubly Fed Induction Generator (DFIG
Stator winding
Rotor windings
The rotor is fed from the grid via a
bidirectional PEC system while slip power
is recovered for variable speed operation.
It has become the default generator in the
geared system
The DFIG can operate as a motor or
generator in synchronous and over-
synchronous modes.
Other advantages
1. Offers separate active and reactive
power control.
2. Requires a smaller inverter (about 1/3 of
the rated power), hence, lower losses
and cost in the inverter.

71
Doubly Fed Induction Generator
• It has a wound rotor.
Both stator and rotor terminals can be connected to the power supply-
hence the name DFIG.
The slip is around 0.2 to 0.25; hence, the PEC rating needs only 20 to 25%
of the stator power rating.
Slip power =sP

72
Equivalent circuit - DIFIG
SCIG
DFIG

73
Variable speed generator with inverters
Variable-speed operation requires a power electronic converter (PEC) to transform the variable-frequency supply into the
fixed-grid-frequency supply. The PECs are expensive and have additional losses. However, they eliminate the load dynamic
and stability issues of constant speed operation, and the aerodynamic energy yield is much better.
SG and SCIG require full-rated back-back PECs.
DC link decouples the rotor speed from the grid.
Reactive power compensation is necessary even with the
SG.
Acceleration, deceleration and braking are easier,
especially with SG.
WRIG can use a partial power converter when the speed
control is via slip power recovery.
If a diode rectifier is on the generator side, bidirectional
power flow is impossible, and the generator must always
operate in over-synchronous mode with limited speed
variation.
Harmonics injection to the grid.
WRIG

74
Direct-Drive variable speed generators
Direct-drive
Generator
Wind turbulence can cause large stress on the gear wheels and bearings in the gearbox, leading
to early and frequent failure. The offshore turbine faces faster wind speed, hence more
vulnerable. The gearbox also requires the highest maintenance and one of the heaviest element
in the drivetrain. Thus, the elimination of the gearbox can improve reliability and improve
system efficiency by eliminating gear losses.

75
EESG and PMSG for Direct drive (DD)
EESG PMSG
Generator Type MW rpm Wind farm
type
DFIG Gear 9-6 ~1200 onshore
EESG DD 8-7.5 12 onshore
PMSG DD 6 - 10 12 onshore and
offshore
Challenges for the Generators with DD
• Low speed and high pole number means generators
are larger and heavier.
• Manufacturing difficulties due to large size
• Complex cooling systems
Recent trends for large WECs:
Pros and cons PMSG:
1. Rare-earth magnets are costly.
2. Large magnets are difficult to handle
during manufacturing.
3. High efficiency
4. More compact compared to the EESG
(high power density)
5. PMSG also has better dynamics for
control.

76
Determination of geometrical parameters using the classical sizing equation:
The outer diameter of the stator from the sizing equation and allowed rotor shear stress
Length of the stator stack from aspect ratio (le/Dg)
Type and number of the stator /rotor slots - width, height
Diameter of the airgap (or inner diameter of the stator), airgap length
The peak value of the airgap flux density, tooth flux density
Rotor dimensions - pole pairs and frequency
How to Design generators
1. What type of machine - Synchronous, induction?
2. What is the rated power/ speed( frequency)/ voltage?
3. How many poles and slots?
4. What are the design goals – maximum efficiency, minimum torque ripple(to reduce vibration),
compact size, minimum cost,
5. What are the constraints – geometric and manufacturability, must comply with the standard
Analytical/numerical calculation of performances
Design optimisation

77
Design specification – An example
Design a generator, 2.5MW, 690V (line, rms), 50Hz, (maximum slip 0.25, if IG) and maximum
speed 1500 rpm, efficiency >0.95, Y-connected machine.
1500 rpm, 50 Hz > Calculate pole numbers (4 poles)
2.5MW rated power, 0.95 efficiency, 1500 rpm, 0.25 slip > calculate input mechanical torque
Use the sizing equation to find Dg, and from a given aspect ratio (le/Dg) to find le
2
1
2
1
1
2 1 4
1
,
1 4
e
p e i
out out
out g g e
m
p
out i
i
e
n g g
p
P P m
T AB D l
f m
p
T m
T AB D l
K K K
K
K K K
K K
K m
ϕ
ϕ
η
π
ω
η
= =
+
= =
+
=
=
Electrical loading A, magnetic loading Bg, air gap diameter Dg and axial length le
Kp: Power waveform factor, Ki: ratio of peak to rms value,
Ke: EMF coefficient (=2π Kw).
Winding factor 𝐾𝐾𝑤𝑤 = 𝐾𝐾𝑑𝑑𝐾𝐾𝑝𝑝ℎ𝐾𝐾𝑠𝑠𝑠𝑠
Kd: distribution factor, Kph: pitch factor and Ksk: skewing factor
If A is the total electrical loading and Ar is the rotor electrical loading and Kφ is the ratio of Ar/As, As can be expressed as,
As = A- Ar =
𝐴𝐴
1+𝐾𝐾φ
; In a PM machine, Ar = 0 and As = A, Kφ =0
m: number of phases , m1: number of phase in each stator
Kφ: ratio of rotor to stator electrical loading

78
Use a known ratio (1.65 to 1.69 for 4 poles) of Do/Dg for a given pole pair to determine the
outer diameter. Frame size may determine the le.
• Determine stator current from power and voltage assuming unity power factor
• From the current and EMF equations, the number of turns and slot dimensions can be
estimated. This also allows us to determine the current density or electric loading.
• Turns ratio of the rotor and stator winding = 1/slip; find turns ratio
• Find the referred rotor current Is/turns ratio.
• This will allow us to determine the rotor current density and rotor slot design.
• For a PM generator, rotor design involves dimensioning the magnet (and rotor slots if
IPMSG).
• Selection of the magnet material,
• Estimate Bg from the remanence Br of the magnet materials(B-H property) and magnet
(rotor geometry).
2
1
1
,
1 4
out
in g g e
p e i
T m
T D
K K K
K K
AB l
m
Kϕ
η +
=
= =

79
Design specification – an Example
Design a DFIG 2.5MW, 690V (line, rms), 50Hz, maximum slip 0.25 and maximum speed 1500
rpm, efficiency >0.95. Y-connected machine
1500 rpm, 50 Hz > Calculate pole numbers (4 poles)
2.5MW rated power, 0.95 efficiency, 1500 rpm, 0.25 slip > calculate torque
Use sizing equation to find Dg, aspect ratio (le/Dg) to find le:
2
1
1
2 1 4
p e i
out out
out g g e
m
p
K K K
P P m
T AB D l
f K m
p
ϕ
η
π
ω
= = =
+
Electrical loading A, magnetic loading Bg, air gap diameter Dg and axial length le
Kp:Power waveform factor Ke: EMF coefficient (=2π Kw). Winding factor 𝐾𝐾𝑤𝑤 = 𝐾𝐾𝑑𝑑𝐾𝐾𝑝𝑝ℎ𝐾𝐾𝑠𝑠𝑠𝑠
Kd: distribution factor, Kph: pitch factor and Ksk: skewing factor
If A is the total electrical loading and Ar is the rotor electrical loading, and Kφ is the ratio of Ar/As, As can be expressed as,
s = A- Ar =
𝐴𝐴
1+𝐾𝐾φ
Ki: ratio of peak to rms value, m: number of phases , m1: number of phase in each stator

80
• Use a known ratio (1.65 to 1.69 for 4 poles) of Do/Dg for a given pole pair to determine
the outer diameter.
• Determine stator current from power and voltage assuming unity of
• From the current and EMF equations, the number of turns and slot dimensions can be
estimated. This also allows us to determine the current density or electric loading.
• Turns ratio of the rotor and stator winding = 1/slip; find turns ratio
• Find the referred rotor current Is/turns ratio. This will allow us to determine the rotor
current density and slot design.

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