Transcript Slides on Steady-state basics

Overview of Wind Generators
EE 559
2016
Iowa State University
James D. McCalley
Electric Generators
Type 1
Conventional Induction
Generator (fixed speed)
Types 1-3 are induction
Type 2
generators and must have
Wound-rotor Induction
gearboxes to change “slow”
Generator w/variable rotor
rotor rotational speed (5-20rpm)
resistance
to “fast” generator rotational
speed (750-3600rpm) needed
Type 3
by the 60 Hz grid freq seen in
Doubly-Fed Induction
stator winding.
Plant
Feeders
generator
PF control
capacitor s
Pla nt
Fee ders
gene rator
Slip power
as heat loss
Plant
Feeders
generator
Generator (variable speed)
Type 4 maintains generator at
“slow” rotor rotational speed
producing low freq in stator
winding which is transformed
by converter to grid. Also called
“direct-drive.” Usually uses a
PM synchronous generator.
PF control
capacitor s
ac
to
dc
ac
to
dc
dc
to
ac
partial power
Type 4
Full-converter interface
Plant
Feeders
generator
ac
to
dc
dc
to
ac
full power
Two kinds of AC generators
Note: DC gens can be used
(and sometimes are for small
wind turbines), but are not used
for utility-scale wind turbines
because their brush/commutator
configuration is highmaintenance.
Permanent
magnet
synchronous
machine
For type 4 Steam &
Gas
Turbines
Hydrogens
For
Types
2, 3
For Type 1
R. Gasch and J. Twele, “Wind Power Plants:
Fundamentals, Design, Construction & Operation,”
Chapter 11, Springer-Verlag, Berlin, 2012.
Two kinds of AC generators for wind
Synchronous gen, type 1:
*stator freq=k×rotor speed.
Connects to
grid via converter
DC, supplied
by external
circuit or rotor
is permanent
magnet
Full converter
interface
(variable speed)
Induction gen, types 1-3:
stator freq fixed by rotor
speed and rotor freq.
Stator winding (AC)
Rotor winding
Electric control
Connects directly
to grid (60Hz)
AC, induced via
SLIP between
rotor and stator
mag field
None (fixed speed)
or
DFIG (variable
speed)
 fs 
p
n is rpm,
ns s 4
p is pole pairs.
60
Synchronous Machine Structure
ROTOR (field
Winding or
perm magnet)
One pole pair
(p=1)
+
N
120°
STATOR
(armature
winding)
+
DC
Voltage
S
The negative terminal
+
for each phase is
180 degrees from
the corresponding
positive terminal.
Air Gap
Current is supplied to rotor
windings externally, or the rotor can
carry a permanent magnet. The
two magnetic fields (one from
stator and one from rotor) interact
to produce torque.
Induction Machine Structure
ROTOR
(Squirrel cage)
+
120°
STATOR
(armature
winding)
+
The negative terminal
for each phase is
180 degrees from
the corresponding
positive terminal.
+
Air Gap
Voltage is induced within rotor
windings by relative motion (SLIP)
between those windings and
magnetic field produced by stator
(Faraday’s Law). Then two magnetic
fields (one from stator and one from
rotor) interact to produce torque.
Some fundamental machine concepts:
a. Rotating magnetic field
b. Faraday’s law
c. Torque production
Basic Concepts
rotor
Balanced voltages applied to stator windings
provides a rotating magnetic field of speed ns.
60 f s
ns 
p
(fs: 60 Hz,
p: # of pole
pairs)
p
ns
1
3600
2
1800
3
1200
4
900
5
720
6
600
7 514.29
8
450
9
400
Let’s look at that rotating magnetic field from the stator; the
principles apply equally well to the induction machine and to
the synchronous machine.
8
Rotating magnetic field
• There are 3 stator windings, separated in space by 120°,
with each carrying AC, separated in time by ω0t=120°.
•
•
•
For cnvntnl sync gen, windings directly connected to 60 Hz grid.
For types 1-3 WT (ind gen), windings directly connected to 60Hz grid.
For type 4 WT (PMSG), windings connected to grid via converter.
• Each of these three currents creates a magnetic field in
the air gap of the machine. Let’s look at only the a-phase:
 cos0 t  I a cos 
Ba ( , t )  Bmax
•
•
•
•
Ba, in webers/m2, is flux density from the a-phase current.
/_Ia is there because it is the current that produces the field.
α is the spatial angle along the air gap.
For any time t, α=0,180 are spatial maxima (absolute value
of flux is maximum at these points).
Rotating magnetic field: fixed in space
 cos0 t  I a cos 
Ba ( , t )  Bmax
Let’s fix α=0 and see what happens at 4 sequential times:
1. ω0t1,
such that ω0t1+∟Ia is just less than π/2
2. ω0t2= ω0t1+90,
2
1
3. ω0t3= ω0t1+180,
4. ω0t4= ω0t1+270
×
α=0°
●
×
●
3
×
4
●
×
Positive rotation assumed in the counterclockwise direction.
●
Rotating magnetic field: fixed in time
Fix t=t2: observe field densities
at α=0, α=45, α=90, α=135,
α=180, α=225, α=270, α=315.
Fix t=t3: observe field densities
at α=0, α=45, α=90, α=135,
α=180, α=225, α=270, α=315.
α=0°
Radially outward is
positive; radially
inward is negative.
α=90°
●
×
?
×
α=270°
α=180
°
• The magnetic field is sinusoidally distributed around the airgap.
• The spatial maxima are always at the same location.
The above two attributes are consistent with the equation, when
the first sinusoid is fixed.
B ( , t )  B cos t  I  cos 
a
a
max
0

Fixed
Rotating magnetic field
• Now let’s consider the magnetic field from all three
windings simultaneously.
 cos0 t  I a cos 
Ba ( , t )  Bmax
2  
2 

 cos 0t  I a 
Bb ( , t )  Bmax
cos


 

3

2

 cos 0t  I a 
Bc ( , t )  Bmax
3

3
 
2
 
 cos  
3
 




(1)
(2)
(3)
• Add them up, then perform trig manipulation to obtain:

3Bmax
(4)
Babc ( , t ) 
cos0t  I a   
2
Notice that each of (1), (2), and (3) has a spatial maximum which is fixed in space, as
dictated by the second term of each the 3 expressions. But the spatial maximum of
(4) rotates. This is a characteristic of a rotating magnetic field. One can observe this
using the following:
http://educypedia.karadimov.info/library/rotating_field.swf
www.youtube.com/watch?v=eQk0OznWTjM
The shape of the individual winding fields throughout the airgap are spatially fixed, but
their amplitudes pulsate up and down. In contrast, the amplitude of the composite is
fixed in time, but it rotates in space. What you will see in the video are just the
variation of the maximum field point.
Induction Machine – Faraday’s Law
We have established that the existence of three phase
currents in the three stator windings creates a rotating
magnetic field in the air gap. When the stator windings are
initially energized, the rotor is stationary, and so the rotor
windings see this rotating magnetic field. What happens?
The rotor windings are seeing a time-varying magnetic field,
and by Faraday’s law, a voltage will be established in them.
This voltage causes currents to flow, which in turn creates a
rotating magnetic field in the air gap. A key point is that
the rotor must rotate at a different speed than the
rotating magnetic field of the stator, otherwise there is
no relative motion between the rotor windings and the
magnetic field the rotor windings seeSLIP.
13
Induction Machine - Slip
s is “slip” which quantifies how much the rotor “slips” behind the
stator-produced rotating magnetic field. More precisely, slip is
• the amount (in electrical radians/sec) by which
•
•
the rotational speed of the stator rotating magnetic field exceeds
the rotational speed of the rotor
• normalized by rotational speed of stator rotating magnetic field
Can slip be 0 in the steady-state?
No, because if it were, there would be no induced Vno I flow
in rotor windings.
Now, denote rotor velocity in mechanical rad/sec as Ωm, and
p=number of pole pairs on the rotor, then:
 s  m
slip  s 
;
s  2 f s rad/sec=p s ;
f s  60 Hz; m  p m
s
Notes:
p
p
s   s  m   m
1. Some treatments define p=number of poles, in which case
2
2
2. The number of poles on rotor must be the same as the number of poles on stator.
14
Induction Machine - Speeds
We may also write this in terms of mechanical revolutions
per minute (rpm), denoted by n, according to:
 s  m
s
s
Transform Ω to n:
rad  rev 60 sec
60

nm    m



m
sec
2

 2rad min
60
 rad  rev 60 sec
ns    s



s
sec
2

 2rad min
Or, since Ωm=ωm/p, Ωs=ωs/p, transform ω to n:
  rad  rev 60 sec m 60

nm   m

p 2
 p sec  2rad min
  rad  rev 60 sec s 60

ns   s

p 2
 p sec  2rad min
Or, since ω=2πf, transform fs to ns:
 2f s rad  rev 60 sec 2f s 60 60 f s

ns  


p 2
p
 p sec  2rad min
(If p=number of poles, then ns 
120 f s
p
)
15
Induction Machine - Speeds
Summary:
 s  m
slip  s 
;
s
s  2 f s rad/sec=p s ;
s
60
nm   m
2
nm 
m 60
p 2
60
ns   s
2
ns 
f s  60 Hz; m  p m
p s  p m  s   m

p s
s
(2 / 60) ns  (2 / 60) nm ns  nm
s

(2 / 60) ns
ns
s 60
p 2
And finally:
 s  m
s
s
m  s 1  s 
16
Induction Machine - Example
Example: Consider a four-pole 60Hz induction machine.
Compute slip, Ωm, and ωm for the following values of rpm
rotor speed: 0, 500, 1000, 1750, 1800, 1850, 2600, 3600.
Solution: To make these computations, we first compute
synchronous speed:
n (rpm)
s
Ω (rpm) ω (rad/s)
60 f s 60(60)
ns 

 1800rpm or
0
1
0
0
m
p
m
m
2
500
0.722
52.36
104.7
1000
0.444
104.7
209.4
Then we can use the following:
1750
0.028
183.3
366.6
n  nm
s s
ns
60
2
nm   m
 m 
nm
2
60
m  p m
1800
0
188.5
377.0
1850
-0.028
193.7
387.4
2600
-0.444
272.3
544.6
3600
-1
377.0
754.0
s  2f s  2 (60)  377rad / sec
17
Torque production
To produce constant torque, two rotating magnetic fields must
have the same rotational velocity. In the case of generator
operation, we can think of bar A (the rotor field) as pushing bar
B (the armature field), as in Fig. a (N pole repels N pole). In the
case of motor operation, we can think of bar B (the armature
field) as pulling bar A (the rotor field), as in Fig. b (N pole
attracts S pole).
S
S
S
Bar B
Bar A
N
N
Fig a: generator operation
N
Bar B
Bar A
S
N
Fig b: Motor operation
Note: Analytic treatment of torque production is done using the coupling field approach.
We will treat this a little later.
Induction Machine –
Freq of Rotor Currents
Fact 1: The rotor must have speed (ωm) which differs from
speed of rotating magnetic field (ωs) from stator.
Otherwise, no voltage is induced in rotor windings.
Fact 2: Constant torque only exists if the two rotating
magnetic fields (one from stator currents and one from
rotor currents) are at the same speed.
How can Fact 1 and Fact 2 both be true?
Speed of rotating magnetic field from rotor=
rotor speed + {rotating magnetic field speed relative to rotor}
s  m  r
r  s  m
Frequency
(rad/sec) of
rotor currents.
 s  m r
s

s
s
Rotor speed
Speed of RMF from
rotor, relative to rotor
Speed of RMF
from stator
 r  ss  f r  sf s
19
Induction Machine –
Clarification
ωr is the frequency (rad/sec) of the electrical quantities in the rotor winding.
But your added statement may need some further clarification, “due to rotor being in rotating stator magnetic field.”
ωr occurs due to the motion (speed) of the magnetic field and the motion (speed) of the rotor: ωr= ωs-ωm.
From another view, it is the speed of the rotating magnetic field from the rotor referenced to the rotor.
ωm would be the speed of the rotating magnetic field from the rotor if the rotor currents were DC.
It is the difference between
• the speed of the rotating magnetic field from the stator and
• the speed of the rotating magnetic field from the rotor referenced to the rotor
ωm= ωs-ωr
The little omegas (ωr, ωs, ωm) are related to their corresponding big omegas (Ωr, Ωs, Ωm) by ω=pΩ where p is the number of
pole pairs. When p=1, they are equal. When p=2, the little omegas double the big omegas, etc. This shows that the little
omega quantities reflect electrical “rotation” that is easily visualized as the sinusoid associated with the corresponding
electrical quantities (currents and voltages); it is also visualized as the rotating B-field. If there are 2 poles, one full
mechanical rotation takes the corresponding electrical quantities through one full electrical rotation. If there are 4 poles,
one full mechanical rotation takes the corresponding electrical quantities through two full electrical rotations, etc.
20
Transformer Per-phase Equivalent Circuit
An impedance
…An induction machine may be thought of
as a transformer with a rotating secondary.
21
Induction Machine Per-phase
Equivalent Circuit
I’2
Notation issue:
In these slides, we
are using the prime
notation to indicate
the quantity is
referred from rotor
to stator (in the
DFIG notes, we use
primed notation to
indicate rotor
quantity is referred
to rotor).
For transformer, this
is an impedance
The main difference between the transformer equivalent cct
and the induction machine equivalent cct is the loading: the
transformer load is an actual impedance whereas the induction
machine load is a variable resistance that depends on “s”.
22
Induction Machine Per-phase
Equivalent Circuit
I’2
What does the induction mach load represent?
It represents the power provided to or from the shaft.
Thus the induction mach load is a purely “real” load since
energy transfer by mechanical means must be MW only
(this is not the case for the transformer).
23
Induction Machine Per-phase
Equivalent Circuit
Denote load
resistance
as:
I’2
Req=
1 s 
Req  R2 

 s 
R2’ is the rotor resistance referred to the stator (as indicated by the
prime – see slide 20). The variable “s” is the slip, as defined earlier.
“Referred to”: ohmic value of impedances differ in numerical value, according to
Zprim/stat=(Nprim/stat/Nsec/rot)2 Zsec/rot
based on whether they are represented on the primary/stator side or on the secondary/rotor side of
the transformer/machine, where N is the number of turns on the indicated side of the device.
24
Induction Machine - Power
What about real power?
I’2
Req=
P1
PD
P1  3 | V1 || I1 | cos 
R'2 (1  s)
PD  3 | I '2 | Req  3 | I '2 |
s
2
2
25
Induction Machine Per-phase
Equivalent Circuit
Power crossing air gap? Includes mechanical load & rotor winding losses
I’2
PG
Req=
P1
PD
R'2 (1  s )
]
s
R '  R '2 s
2 R' s
2 R'
 3 I '2 [ 2  2
]  3 I '2 [ 2 ]
s
s
s
PG  3 I '2 [ R'2  Req ]  3 I '2 [ R'2 
2
2
26
Induction Machine Per-phase
Equivalent Circuit
I’2
PG
Req=
P1
PD
Compare PG to PD:
Conclusion:
PG  3 I '2 [
2
R '2
]
s
PD  3 | I '2 |2
R'2 (1  s )
s
PD  PG (1  s)
27
Induction Machine - Torque
Torque expression:
Mechanical
radians/sec
PD
PD
pPD
TD 


 m m / p m
R'2 (1  s )
PD  3 | I '2 |
s
But recall that:
Electrical
radians/sec
2
 m   s (1  s)
Substitution yields:
TD  3 p | I '2 |2
R'2 (1  s )
R'
 3 p | I ' 2 |2 2
s (1  s) s
s s
Assuming we know the applied voltage and speed (or slip), then computing
torque requires getting I’2. We can do this using circuit analysis.
28
Induction Machine – Obtaining I’2
I’2
Recall: When 2 impedances are in
parallel, we obtain their equivalent using:
Z eq  Z1 // Z 2 
Z1 Z 2
Z1  Z 2
Req
One method: Given slip, compute load resistance, Req. Then Ohm’s
Law provides current as applied voltage over equivalent impedance
V1
I1 
R1  jX1  RC // jX m  // R'2  jX 2  Req 
Then, by current division, we obtain:
RC // jX m
I 2 
 I1
RC // jX m   R'2  jX 2  Req 
Now, for various values of slip, from s=-1 to s=+1,
compute I’2 and subsequently T. Then plot T vs. s.
29
Induction Machine – Obtaining I’2
I’2
Example: Consider a 6-pole induction generator with
line-line voltage of 220 volts, and the below data..
R1=0.294 Ω
X’2=0.209 Ω
X1=0.503 Ω
R’2=0.061 Ω
RC=1000 Ω
Xm=13.25 Ω
Compute the torque for f=60 Hz, for Ωm=130 rad/sec.
We need to compute:
Req
I1 
V1
R1  jX 1  RC // jX m  // R'2  jX 2  Req 
Therefore…
Compute the line-to-neutral voltage, V1=220/sqrt(3)=127.017volts.
Compute Ωs=2(pi)f/p=2(pi)(60)/3=125.664rad/sec.
Compute slip: s=(Ωs-Ωm)/Ωs=(125.664-130)/125.664=-0.0345
Compute Req=R2’(1-s)/s=0.061(1- -0.0345)/(-0.0345)=-1.8289
Now obtain:
Rc // jX m 
( Rc )( jX m ) 1000( j13.25)

 0.1755  j13.248
RC  jX m  1000  j13.25
Now obtain:
R'2  jX 2  Req  0.061  j0.209  1.8289  1.7679  j0.209
Compute:
RC // jX m  // R'2  jX 2  Req   (.1755  j13.248)(1.7679  j.209)  1.687  .4284
.1755  j13.248  1.7679  j.209
30
Induction Machine – Obtaining I’2
I’2
Example: Consider a 6-pole induction generator with
line-line voltage of 220 volts, and the below data..
R1=0.294 Ω
X’2=0.209 Ω
X1=0.503 Ω
R’2=0.061 Ω
RC=1000 Ω
Xm=13.25 Ω
Compute the torque for f=60 Hz, for Ωm=130 rad/sec.
From previous slide:
R '2  jX 2  Req  1.7679  j 0.209
Req
So:
I1 
Rc / / jX m  0.1755  j13.248
 RC / / jX m  / / R '2  jX 2  Req  1.687  .4284


V1
127.017

 63.1625  j 42.069
R1  jX 1  RC // jX m  // R'2  jX 2  Req  .294  j.503  1.687  j.4248
Then, by current division, we obtain:
I 2 
RC // jX m
.1755  j13.248
 I1 
 (63.1625  j 42.069)  66.6  j 32.711
RC // jX m   R'2  jX 2  Req 
.1755  j13.248  1.7679  j.209
I 2  66.62  32.7112  74.12
TD  3 p | I '2 |2
R' 2
0.061
 3(3)(74.12)2
 231.9 Ntn-meters
s s
376.99( .0345)
31
Equivalent circuit by Thevenin
I’2
Req
An alternative way to obtain I’2 is by use of
Thevenin, looking into the terminals as shown.
Equivalent circuit by Thevenin
Za=R1+jX1
By voltage division:
V1
Zb=
Rc//jXm
Vth  V1
Vth
Zb
Z a  Zb
Note that V1 is the line-to-neutral
voltage, given by V1=VLL/sqrt(3).
Za=R1+jX1
Zb=
Rc//jXm
The two impedances are
in parallel:
Z th  Z a // Z b 
Z a Zb
Z a  Zb
DFIG equivalent circuit Thevenin
R’2
Zth
jX’2
I’2
R’2 (1-s)/s
Vth
I 2 
Vth
Z th  R'2  jX 2  Req 
R' 2
TD  3 p | I '2 |
s s
2
Induction Machine – Plotting T-s characteristic
Example: Consider a 6-pole induction generator with line-line voltage of 220 volts,
and the below data. Plot the torque-speed characteristic for f=60 Hz.
R1=0.294 Ω
X’2=0.209 Ω
X1=0.503 Ω
R’2=0.061 Ω
RC=1000 Ω
Xm=13.25 Ω
Solution: The below calculations were carried out for this machine for a range of slips.
Vth  V1
Zb
Z a  Zb
Z th  Z a // Z b 
Z a Zb
Z a  Zb
Vth
I 2 
Z th  R'2  jX 2  Req 
R '2
TD  3 p | I '2 |
s s
2
35
Type 1 Induction Generator – TD vs. s
But it has a wide
torque range, dictated
by the wind. And so
wind gusts are
passed on directly to
the grid.
Motor
Region of operation
It is referred to as
“fixed speed”
because of the
narrow band of speed
operation shown
here.
36
Generator
Torque
Range
Slip
1
0
Subsynchronous operation
-1
Supersynchronous operation
Type 1 Induction Generator – TD vs. s
Torque-speed
characteristics associated
with wind turbines are not
simply a function of a
mechanical load but also
reflect the aerodynamic
features of the wind-blade
interaction, and as a result,
appear as shown in the
figure, where the numbers
4-10 correspond to wind
speed (m/sec). We
observe that as the wind
speed increases, the
electromagnetic torque
increases significantly, but
the speed of the machine
changes very little. For this
reason, this kind of wind
turbine configuration, using
an induction machine with
a squirrel cage rotor, is
referred
37 to as a fixedspeed machine.
Slip
1
Subsynchronous operation
0
-1
Supersynchronous operation
Type 1 Induction Generator – TD, PD, I1, I’2
It is also interesting to look at the real power consumed
and the currents, given by the below expressions.
V1
I1 
R1  jX1  RC // jX m  // R'2  jX 2  Req 
I 2 
RC // jX m
 I1
RC // jX m   R'2  jX 2  Req 
R '2
TD  3 p | I '2 |
s s
2
PD  3 | I '2 |2
R'2 (1  s )
s
Type 1 Induction Generator
In the top plot, P is what
we called PD on the
previous slide (power
developed) since, as the
bottom figure indicates, it
goes to 0 when I’2 goes
to 0. Also, T is what we
called TD.
In this plot, I2 is
what we called I’2
Observe that when TD,
PD, and I2 are 0, I1 is not
zero and has a value of
Im, which is the
magnetizing current. One
can understand this by
considering the
equivalent circuit, see
next slide.
Type 1 Induction Generator
I’2
Req
Note from our example data (slide 30) that RC=1000 Ω and Xm=13.25 Ω. This is quite
typical in that RC>>XM. Thus, for analysis of currents, we can neglect RC. Therefore, the
current flowing into the shunt leg is just the current flowing into XM, which is the
magnetizing current, IM.
Now consider the operating condition s=0, and considering s=(ωs-ωm)/ωs, we have that
ωs=ωm, i.e., the rotor is at synchronous speed. This means that Req=(1-s)/s=∞, i.e., the
rotor side of our model is open circuited. In this case, under the condition that RC>>XM,
we have that I1=IM, as indicated by the previous diagram.
The operating condition s=0 in an induction machine corresponds to the “open-circuit”
(infinite load impedance) condition for a transformer.
Type 1 Induction Generator
Let’s also look at Q1 and QG.
I’2
QG
Q1
I1 
I 2 
V1
R1  jX1  RC // jX m  // R'2  jX 2  Req 
RC // jX m
I
RC // jX m   R'2  jX 2  Req  1
Q1  3 | V1 || I1 | sin 
QG  3 | I '2 |2 X 2
Type 1 Induction Generator
Positive is
absorbing
Positive is
absorbing
This is PD. Observe
that it is 0 for s=0
This is Q1. Observe that
it is just a little greater
than 0 for s=0. This is
due to the magnetizing
current IM flowing
through X1 and XM.
Sources:
1. “Chapter 6: Electrical Generator Machines
in Wind-Energy Systems.”
2. T. Burton, et. al, “Wind Energy Handbook,”
2nd edition, Wiley, pp 439-440.
Type 1 Induction Generator – Q1, PD
We can also compute Q1 and PD for the same
value of s and then plot Q1 vs. PD.
I1 
V1
R1  jX1  RC // jX m  // R'2  jX 2  Req 
I 2 
RC // jX m
I
RC // jX m   R'2  jX 2  Req  1
Q1  3 | V1 || I1 | sin 
PD  3 | I '2 |2
R'2 (1  s )
s
Type 1 Induction Generator - Q1, PD
Sources:
1. “Chapter 6: Electrical Generator Machines
in Wind-Energy Systems.”
2. T. Burton, et. al, “Wind Energy Handbook,”
2nd edition, Wiley, pp 439-440.
Positive is
absorbing
So Type 1 machine always absorbs reactive power from the network
independent of whether it is gen or motor.
Windfarms withType 1 machines must have reactive compensation at the
substation. Generally, it is supplied via capacitor banks.
Type 1 Induction Generator – Voltage dips
When an induction generator experiences a voltage sag
on the network, the generator speed increases. If it is
not disconnected, it can accelerate to an unstable
condition. This is illustrated on the next slide, where,
following a voltage drop, the machine moves from point
X to point Y, where the machine speeds up because the
mechanical torque (a constant) is higher than the
electrical torque (which has decreased as indicated by
the difference in torque values between point X and Y). If
the acceleration moves beyond the pull-out torque, then
further speed increases cause further decrease in
electrical torque, resulting in more acceleration, an
unstable condition.
Source: O. Anaya-Lara, et. al, “Wind Energy Generation: modeling and control, Wiley, 2009.
Type 1 Induction Generator – Voltage dips
Mechanical torque
stays relatively
constant. At point X,
Tmech=Te. At point Y,
Tmech>Te and so
machine accelerates. If
Te exceeds “pull-out”
torque (the hump), then
further acceleration
(e.g., point Z) will
decrease Te and further
accelerate the machine
which decreases Te
which…. UNSTABLE!
But this behavior (tripping) will violate voltageride-through requirements (see next slide).
Source: O. Anaya-Lara, et. al, “Wind Energy Generation: modeling and control, Wiley, 2009.
Type 1 Induction Generator – Voltage dips
To avoid violation of low-voltage ride-through requirements, windfarms using
fixed-speed induction generators (type 1) may need to deploy dynamic var
capabilities (e.g., SVC or STATCOM) at the substation.
Type 2 Induction Generator – speed cntrl
Why do we want to provide speed control? Because it
enables us to get closer to the maximum power point of
the turbine, as indicated by the below diagrams.
Type 2 Induction Generator – with R control
Some control may be achieved via use of a resistor in
series with the rotor, increasing R’2. This means that the
effective rotor resistance becomes:
R'2,ext  R'2  Rext
This means that we must substitute R’2,ext for R’2 in the
Req expression.
Req ,ext 
R2,ext (1  s)
s
Then the total rotor circuit resistance can be written as:
R '2  Rext 1  s  sR '2  sRext  R '2  Rext  sR '2  sRext

R'  R
 R'  R 

2, ext

eq , ext
R '2  Rext
s
2
ext
s
s
Type 2 Induction Generator – with R control
The circuit model is shown below. Note that we have
combined the equivalent resistance with the rotor
resistance. If you want to compute PD, then you need to
use Req,ext, as shown on the next slide.
I’2
R’2/s
Rext/s
Type 2 Induction Generator – with R control
R’2,ext
I’2
Req,ext
=R’2,ext(1-s)/s
Type 2 Induction Generator with R control
One can see the effect of inserting resistance in series
with the rotor circuit by inspecting the torque equation:
TD  3 p | I '2 |
2
R'2,ext
s s
Vth
I 2 
Z th  R'2,ext  jX 2  Req ,ext 
Req=R’2,ext (1-s)/s
2
TD  3 p | I '2 |
2
R'2,ext
s s
 3p
Vth
R'2,ext (1  s) 

Z th   R'2,ext  jX 2 

s


R'2,ext
s s
The effect this has on the torque-slip (torque-speed)
curves can be observed on the next slide.
52
Type 2 Induction Generator with R control
Observe that for constant
torque, the curves shift to the
left as rotor resistance is
increased, decreasing speed.
Source: O. Anaya-Lara, et. al, “Wind Energy Generation: modeling and control, Wiley, 2009.
Type 2 Induction Generator with R control
One way of controlling the rotor resistance (and therefore
the slip and speed of the generator) is to use a wound
rotor connected to external variable resistors through
brushes & slip-rings. The rotor resistance can be adjusted
by means of converter circuits, as illustrated below.
Aside: Soft starter is a power electronics device used to
build up the magnetic flux slowly and so minimize transient
currents during energization. See Anaya-Lara, pp. 21-22.
Source: O. Anaya-Lara, et. al, “Wind Energy Generation: modeling and control, Wiley, 2009.
Type 2 Induction Generator with R control
Slip rings can be avoided by mounting the variable
resistors and control circuitry on the generator rotor.
However, then one has a heat dissipation problem in a
tightly confined space. Therefore an advantage of
mounting the resistors externally is that it is easier to
dissipate the extra heat which is generators and may be
otherwise a limiting factor at large sizes.
Source: O. Anaya-Lara, et. al, “Wind Energy Generation: modeling and control, Wiley, 2009.
Type 2 Induction Generator with R control
Slip rings can be avoided by mounting the variable
resistors and control circuitry on the generator rotor.
However, then one has a heat dissipation problem in a
tightly confined space. Therefore an advantage of
mounting the resistors externally is that it is easier to
dissipate the extra heat which is generators and may be
otherwise a limiting factor at large sizes.
Source: O. Anaya-Lara, et. al, “Wind Energy Generation: modeling and control, Wiley, 2009.
Type 2 Induction Generator w/ pole changing
Another way to provide some (limited) speed control is to
change the number of poles. This changes the
synchronous speed, according to: ns  60 f s
p
This means the velocity of the rotating magnetic field from
the stator currents changes.
This means the frequency of the voltages and currents
in the rotor winding change.
This means the rotor velocity changes.
Pole changing may be done instead of R-control or with R-control.
Source: O. Anaya-Lara, et. al, “Wind Energy Generation: modeling and control, Wiley, 2009.
Type 2 Induction Generator w/ pole changing
There are three ways to implement pole-changing:
1. Connect two different generators, each with p, to the
gearbox output shaft (expensive!). In this case, the
rating of the generator for low-speed operation would
normally be much less than the turbine rating.
2. Pole-amplitude modulation: This requires a generator
with two sets of winding that can be connected together
in different ways.
3. Provide two independent sets of windings on the same
stator. Generators of this type are available which can
be switched between four and six pole operation, giving
a speed ratio of 1.5.
Source: O. Anaya-Lara, et. al, “Wind Energy Generation: modeling and control, Wiley, 2009.