Transcript DFIG-Steady state

Double-fed electric machines –
steady state analysis
J. McCalley
Four configurations
We will study
only this one,
the DFIG.
2
Basic concepts
Power
Grid
DFIG
Rotor
DC Link
AC
DC
DC
AC
Rotor is wound: it has 3 windings.
Stator has three windings.
Induction machine looks like a transformer with a rotating secondary (rotor).
In DFIG, we will inject a voltage control signal via that converter.
3
Basic Concepts
rotor
Balanced voltages applied to stator windings provides a
rotating magnetic field of speed
which induces an emf in the rotor windings according to eind
• eind=induced emf in one conductor of rotor
• v=velocity of conductor relative to stator flux rotation
• B=stator magnetic flux density vector
• L=length of conductor in direction of wire
60 f s
ns 
p
 (v  B )  L
(fs: 60 Hz,
p: # of pole
pairs)
4
Basic concepts
 s  m
slip  s 
;
s

ns  nm
;
ns
s  377 rad/sec;
ns 
We can manipulate to get:
m  p m
60 f s
rpm
p
nm  ns (1  s)
Mechanical
rad/sec
 m   s (1  s)
The induced rotor voltages have frequency of :
Substitution into slip expression above yields:
Observe three modes of operation:
r  s  m

s  r  r  ss  f r  sf s
s
ωm< ωs ωr>0s>0Subsynchronous operation
ωm= ωs ωr=0s=0Synchronous operation
ωm>ωs ωr<0s<0Supersynchronous operation
5
Per-phase steady-state model
STATOR VOLTAGE EQUATION:
V s  E s  ( Rs  jXs ) I s
at fs
V s =stator voltage with frequency fs
E s = emf in the stator windings with frequency fs
I s = stator current with frequency fs
Rs =stator resistance
These quantities
are referred to
stator side.
X s =stator leakage reactance
ROTOR VOLTAGE EQUATION:
V r  Ers  ( Rr  jX r ) I r
V r =rotor voltage with frequency fr
Ers =induced emf in the rotor windings with frequency fr
I r =induced rotor current with frequency fs
Rr =rotor resistance
X r =rotor leakage reactance= r Lr
at fr
These quantities
are referred to
rotor side,
indicated by
prime notation.
6
Referring quantities
Application of Faraday’s Law allows the stator back emf and the induced rotor
voltage to be expressed as:
Ks, Kr: stator and rotor winding factors, respectively,
E s  2K s N s f s  m
which combine the pitch and distribution factors.
Ns, Nr: number of turns of stator & rotor, respectively.
Ers  2K r N r f r  m
fs, fr, frequency of stator & rotor quantities, respectively
φm : magnetizing flux
Solve both relations for φm and equate:
Ks Ns fs
Es
Ers
Es
m 




E
Kr Nr fr
2K s N s f s
2K r N r f r
rs
But recall:
f r  sf s 
K N f
K N
Es
 s s s  s s
E rs K r N r sf s K r N r s
Ns
Es

The ratio Ks/Kr is normally very close to 1, therefore
Ers N r s
Es
a
sE s
Ns




E

Define the effective turns ratio: a 
rs

E
s
a
Nr
rs
Es
E

Define the induced rotor voltage referred to the stator side:
rs
a
7
Referring quantities
We just derived that:
E rs 
sE s
a
(*)
At a locked rotor condition (s=1), the device is simply a static transformer,
and we have:
Es
E rs 
a
 E s  a E rs
This tells us it we want to move a voltage from rotor side to stator side, we multiply it
by a=Ns/Nr. We can obtain similar relationships for currents and impedances, and so
we define the rotor quantities referred to the stator according to:
E rs  Ers a
I r  I r / a
jωsLσs
jωrLσr
3
3
Vs
Rs
Lr  Lr a 2
3
3
Is
Rr  Rr a 2
Es
3
3
3
3 Ers
Rr
Ir
Rotor quantities are referred
to the stator-side, indicated
by unprimed quantities.
Vr
This is locked rotor condition
(s=1), therefore ωr=ωs and
Ers=Es
We can account for other
slip conditions using ωr=sωs
and from (*), aE’rs=sEs.
8
Referring quantities
Rs
Is
Ir
3
3
3
3
Es
Vs
Rr
jsωsLσr
jωsLσs
3
3
3
3 Ers=sEs
Vr
Now write the rotor-side voltage equation (referred to stator):
V r  s E s  ( Rr  jss Lr ) I r
Divide by s
Is
Rs
Vr
R
 E s  ( r  j s Lr ) I r
s
s
jωsLσr
jωsLσs
Es
Ir
3
3
3
3
Vs
Rr/s
and we get the following circuit:
3
3
3
3 Es
Vr/s
The voltage on both sides
of the xfmr is the same,
therefore, we may eliminate
the xfmr. .We represent a
magnetizing inductance
jωsLm in its place.
9
Referring quantities
Is
Rs
jωsLσs
jωsLσr
Es
Ir
3
3
3
3
Vs
Rr/s
3
3 jωsLm
Vr/s
10
Power relations
Is
Rs
jωsLσs
jωsLσr
Es
Ir
3
3
3
3
Vs
Rr/s
3
3 jωsLm
Vr/s
We modify the above circuit slightly in order to clearly separate slip-dependent terms
from loss terms:
Rr Rr  sRr  sRr sRr Rr  sRr
Rr (1  s )



 Rr 
s
s
s
s
s
Vr Vr  sVr  sVr sVr Vr  sVr
V (1  s)



 Vr  r
s
s
s
s
s
Change the circuit accordingly….
11
Power relations
Is
Rs
jωsLσs
jωsLσr
3
3
3
3
Vs
Es
Rr
Vr(1-s)/s
Ir
+Rr(1-s)/s
3
3 jωsLm
Vr
It is possible to prove that the mechanical power out of the machine is the power
associated with the slip-dependent terms R2(1-s)/s and Vr(1-s)/s. To do so, use:
Power balance relation:
Pmech  Ps  Pr  Ploss ,s  Ploss ,r
where Ps and Pr are powers entering the machine through the stator & rotor windings,
respectively, and Ploss,s and Ploss,r are the stator and rotor winding losses, respectively.
Expressing the right-hand-terms of the power balance relation in terms of the above
circuit parameters leads one to identify the slip-dependent terms as Pmech.
Knowing that the slip-dependent terms are those responsible for mechanical power,
we may obtain the power expressions from the circuit, as on the next slide.
12
Power relations
Rs
Is
jωsLσs
jωsLσr
3
3
3
3
3
3 jωsLm
Es
Vs

Pmech
Veq=
Vr(1-s)/s I
Rr
r
+Req=
Rr(1-s)/s
Vr

Pmech  3 I r2 Req   3 Re V eq I r



*
Rotor current (Ir) direction is out of positive side
of voltage source; therefore it supplies power to
circuit. But a normal (positive) resistance Req


R
(
1

s
)
1

s




*
 3 I r2  r
  3 Re V r 
 I r  always consumes power. So these two terms
s


  s   should be opposite sign. Defining Pmech>0 (see
below) as motor mode implies Req term should
1 s 
*
2  Rr (1  s ) 
 3 Ir 
  3
 Re V r I r be added and Veq term should be subtracted.

s

 s 


If Pmech>0the machine is delivering power through the shaft: MOTOR!
If Pmech<0the machine is receiving power through the shaft: GEN!
If 0<s<1Req term is positive Veq term is positiveSupplying P to cct
If 0>s>-1Req term is negative Veq term is negativeConsuming P from cct.
13
A first torque expression
Rs
Is
jωsLσs
jωsLσr
3
3
3
3
Vs
3
3 jωsLm
Es

Veq=
Vr(1-s)/s I
Rr
r
+Req=
Rr(1-s)/s
Vr

 R (1  s)   1  s 
*
Pmech  3 I r2  r
  3
 Re V r I r
s

  s 

p
p 2  Rr (1  s) 
p 1 s 
*
Pmech  Temm  Tem m  Tem  Pmech
3
Ir 
3

 Re V r I r
p
m
m 
s
 m  s 
(p: # of pole
p 2  Rrm 
p  m 
  3

 Re V r I *r
pairs)
Tem  3
I r 
m
m  r 
 r 
 
r
;
Recall from slide 5: s 
s 
   (1  s)  1  s  m
and
m
Therefore:
s
s
1  s m  s m


s
 s r  r


3 p I r2 Rr
r
Tem 

3p
r
3 p I r2 Rr
r

*
Re V r I r

3p
r


Vr I r cos  v  i 
14
A second (equivalent) torque expression
Is
Rs
jωsLσs
jωsLσr
3
3
3
3
3
3 jωsLm
Es
Vs
Veq=
Vr(1-s)/s I
Rr
r
+Req=
Rr(1-s)/s
Vr
 
Stator power:
Ps  3 Re V s I s
Stator voltage:
V s  I s Rs  js Ls   I s  I r  js Lm
*
 

Substitute Vs into Ps: Ps  3 Re V s I s  3 Re I s Rs  js Ls   I s  I r  js Lm I s
*

 3 ReR I
*
 3 Re Rs I s I s  js Ls I s I s  js Lm I s I s  js Lm I r I s
2
s s
*
*
*
*
 js Ls I s2  js Lm I s2  js Lm I r I s
*



The middle two terms are purely imaginary, therefore:

*

First term is purely real, only the second term contains real and imaginary, therefore:
P  3R I  3 Rej L I I 
2
s s
*
s
Ps  3 Re Rs I s2  js Lm I r I s
s
s
m
r
15
A second (equivalent) torque expression
Rs
Is
jωsLσs
jωsLσr
3
3
3
3
Es
Vs
Rotor power:
Veq=
Vr(1-s)/s I
Rr
r
+Req=
Rr(1-s)/s
Vr
3
3 jωsLm
 
Pr  3 Re V r I r
*
1 s
1 s


 I r  Rr  Rr
 j s Lr   I s  I r  j s Lm
s
s


Rotor voltage: V r
 R

 I r  r  j s Lr    I s  I r  j s Lm
s
 s

V r  I r Rr  js s Lr   I s  I r  js s Lm
*
Substitute Vr into Pr:
r
s
r
r
r r
r
s r
V
r
V
r



P  3 Re V I  3 Re I R  js L

 3 ReR I
  I
I
 jss Lm I *r 
 3 Re Rr I r I r  jss Lr I r I r  jss Lm I r I r  jss Lm I s I r
2
r r
*
*
*
*
 jss Lr I r2  jss Lm I r2  jss Lm I s I r
*


The middle two terms are purely imaginary, therefore:

*

First term is purely real, only the second term contains real and imaginary, therefore:
P  3R I  3 Rejs L I I 
2
r r
*
r
Pr  3 Re Rr I r2  jss Lm I s I r
r
s
m
s
16
A second (equivalent) torque expression
Now substitute Ps and Pr into the power balance equation:
Pmech  Ps  Pr  Ploss ,s  Ploss ,r

Ps  3Rs I s2  3 Re js Lm I r I s
*



Pr  3Rr I r2  3 Re jss Lm I s I r


*


Pmech  3Rs I s2  3 Re js Lm I r I s  3Rr I r2  3 Re jss Lm I s I r  Ploss ,s  Ploss ,r
*
*
Observe we have loss terms added and subtracted in the above, so they go away.



Pmech  3 Re js Lm I r I s  3 Re jss Lm I s I r
*
*

Now consider what happens when you take the real part of a vector multiplied by j
ja
(or rotated by 90 degrees):
a
Im(a)
Observe that
Re(ja) = - Im(a)
Re(ja)
Therefore:



Pmech  3 Im s Lm I r I s  3 Im ss Lm I s I r
*
*

17
A second (equivalent) torque expression



Pmech  3 Im s Lm I r I s  3 Im ss Lm I s I r
*
*

Let’s consider another vector identity: taking imaginary part of a conjugated vector:
Observe that
Im(a*) = - Im(a)
a
Im(a)
Im(a*)
a*
Therefore:

 
 3 Im L I I  3 Ims L I I 
 3 L ImI I  Ims I I 
 3 L ImI I 1  s 
Pmech  3 Im  s Lm ( I r I s )*  3 Im s s Lm I s I r
s
Recall:
m
*
s
*
r
s
m
s
*
r
s
m
s
*
r
m  s (1  s)
s
s
m
*
r
s
*

*
r
Recall:
Tem  Pmech
p
m
 
Tem  3 pLm Im I s I r
*
 
Therefore: Pmech  3m Lm Im I s I *r
18
Two equivalent torque expressions
Tem 
3 p I r2 Rr
r

3p
r
Vr I r cos  v  i 
Torque expression #1: Need rotor
speed, rotor voltage and rotor current
 
Tem  3 pLm Im I s I r
*
Torque expression #2: Need stator
current and rotor current
A third set of equivalent torque expressions follow….
19
Additional equivalent torque expressions
If we assume the magnetic core of the stator and rotor is linear, then we may express
flux linkage phasors of each winding (stator winding and rotor winding, respectively):
Mutual
inductances
 r  Lm I s  Lr I r
Stator winding  s  Ls I s  Lm I r
Rotor winding
Self inductances
ASIDE: Each self inductance is comprised of mutual and leakage according to:
Ls  Lm  Ls ;
Therefore:
Lr  Lm  Lr
 s  Lm I s  Ls I s  Lm I r
 r  Lm I s  Lm I r  Lr I r
 Lm ( I s  I r )  Ls I s
From stator winding equation:
Is 
 s  Lm I r
;
Ir 
 s  Ls I s
 Lm ( I s  I r )  Lr I r
From rotor winding equation:
;
Ir 
 r  Lm I s
;
Is 
Lr
*
Choose one of these equations and
Tem  3 pLm Im I s I r
Ls
Lm
substitute into torque expression #2….
 
 r  Lr I r
Lm
20
Additional equivalent torque expressions
From stator winding equation:
Is 
From rotor winding equation:
 s  Lm I r
Ir 
Ls
 r  Lm I s
Lr
Substitute into torque expression #2….
 
Tem  3 pLm Im I s I r
*
Using stator winding equation:
Using rotor winding equation:
   Lm I r * 
Tem  3 pLm Im s
Ir
Ls


L
*
*
 3 p m Im  s I r  Lm I r I r
Ls
    L I * 
m s
 
Tem  3 pLm Im I s  r
Lr
 
 
L
*
*
 3 p m Im I s  r  Lm I s I s
Lr

 3p

Lm
*
Im  s I r  Lm I r2
Ls
 
Lm
*
 3p
Im  s I r
Ls


Purely
real


 3p
Lm
*
Im I s  r  Lm I s2
Lr
 3p
Lm
*
Im I s  r
Lr
 


Purely
real
21
Airgap and slip power
On slides 15 and 16, we derived the following relations for the power into the
stator and rotor respectively:

Ps  3R I  3 Re js Lm I r I
2
s s
*
s


Pr  3Rr I r2  3 Re jss Lm I s I r
Subtracting losses from both sides, we obtain:

Ps  3Rs I s2  3 Re js Lm I r I s
*


Pairgap  Ps  3R I  3 Re js Lm I r I


Pr  3Rr I r2  3 Re jss Lm I s I r
This quantity is the power that flows
from the stator terminals to the rotor
(negative for generator operation). In
other words, it is the power across
the airgap. Therefore:
2
s s
*
*
s

*

This quantity is the power that is
transferred from the grid to the rotor
through the converter (negative
when it is into the grid). It is called
the slip power. Therefore:

Pslip  Pr  3Rr I r2  3 Re jss Lm I s I r
*

Bring out front the “s” in the slip power expression and use Re{ja}=-Im(a) (both):

Pairgap  Ps  3R I  3 Im s Lm I r I
2
s s
*
s


Pslip  Pr  3Rr I r2  s3 Im s Lm I s I r
Use Im(a*) = -Im(a) on slip expression:

Pairgap  Ps  3R I  3 Im s Lm I r I
2
s s
*
s


*
Pslip  Pr  3Rr I r2  s3 Im s Lm I s I r
*


The term 3Im{} in the slip power expression is Pairgap. Therefore:
Pslip  sPairgap
22
Airgap and slip power
So we just proved that: Pslip  sPairgap

Pairgap  Ps  3Rs I s2  3 Re js Lm I r I s
*
where


Pslip  Pr  3Rr I r2  3 Re jss Lm I s I r
*

Our power balance relation states:
Therefore:
Pmech  Ps  Pr  Ploss ,s  Ploss ,r  Ps  Ploss ,s  Pr  Ploss ,r

 

Pairgap
Pslip
Pmech  Pairgap  Pslip
Substituting Pslip  sPairgap
we obtain Pmech  Pairgap  sPairgap  1  s Pairgap
m
Pairgap
s
p m
p
p
Tem  Pmech

Pairgap

Pairgap
m  s
m  s
Recall: 1  s 
m
s
 Pmech 
Substituting: Pslip   sPairgap  Pairgap
Tem 
1

Pslip
s
r
s
s
p
Pslip
ss
s  p
T

Pslip
 em
r  s
p
T

Pslip
 em
23
r
Approximate relations between active powers
On slides 15 and 16, we derived the following relations for the power into the
stator and rotor respectively:

Ps  3R I  3 Re js Lm I r I
2
s s
*
s


Pr  3Rr I r2  3 Re jss Lm I s I r
*
If we neglect the stator losses (3RSIs2) and rotor losses (3RrIr2):

Ps  3 Re js Lm I r I s
*


Pr  3 Re jss Lm I s I r
*


Bring out front the “s” in the rotor power expression and use Re{ja}=-Im(a) (both):

Ps  3 Im s Lm I r I s
*


Pr  s3 Im s Lm I s I r
Use Im(a*) = - Im(a) on the rotor power expression

Ps  3 Im s Lm I r I s
*


Pr  s3 Im s Lm I s I r
The term 3Im{} in the rotor power expression is PS. Therefore:
Recall the power balance relation:
Neglecting losses:
Substituting Pr expression: Pmech  Ps  sPs  (1  s) Ps
m
1

s

Recall:
s

*


Pr  sPs
Pmech  Ps  Pr  Ploss ,s  Ploss ,r
Pmech  Ps  Pr
m
P

Ps
 mech
s
p m
p
p
Tem  Pmech

Ps

Ps
m  s  m  s
*
24
Tem 
r
s
s
p
Pr
ss
s  p
T

Pr
 em
r  s
p
T

Pr
 em
r
Active power relations - summary
Both
Exact


  s
P  3R I  3 Rejs L I I 

1 s 
P
 P  3R I  3 Rej L I I 

p
T

P
P  P  3R I  3 Rejs L I I 

Ps  3Rs I s2  3 Re js Lm I r I s
*
r
2
r r
airgap
slip
s
s
2
s s
s
r
m
*
r
2
r r
s
s
Pslip  sPairgap
Pmech  Ps  Pr  Ploss ,s  Ploss ,r
Pmech  Pairgap  Pslip
Pmech  1  sPairgap

Pmech  m Pairgap
s
p
Tem 
Pairgap
s
p
Tem 
Pslip
r
r
m
s
m
*
s
r
m
*
r
s
s
em
mech
m
Approximate


P  3 Rejs L I I 
P
 P  3 Rej L I I 
P  P  3 Rejs L I I 
Ps  3 Re js Lm I r I s
*
r
s
airgap
slip
m
s
s
r
*
r
s
s
r
m
m
s
*
s
*
r
Pr  sPs
Pmech  Ps  Pr
Pmech  (1  s) Ps
m
Ps
s
p
Tem 
Ps
s
p
Tem 
Pr
r
Pmech 
25
Power balance
Pmech  Ps  Pr  Ploss ,s  Ploss ,r  Ps  Ploss ,s  Pr  Ploss ,r

 

Pairgap
Without losses
With losses
Pgrid
Ps
Pairgap
Ploss,s
Pslip
Pmech
Pslip
Pr
Pgrid
Ps
Pairgap
Ploss,r
Pslip
Pr
Pmech
These figures assume proper sign convention
(power flowing to the rotor is positive).
26
Generator modes
Ps 
Mode
s
p
Tem
r  ss
Pr 
 r
Tem
p
Slip and speed
Pmech
Ps
Pr
1. Motor
(Tem>0)
s<0, ωm>ωs
(suprsynchrnsm)
>0 (mch delivers
mech pwr)
>0 (mch receives
power via stator)
>0 (mch receives
power via rotor)
2. Generator
(Tem<0)
s<0, ωm>ωs
(suprsynchrnsm)
<0 (mch receives
mech pwr)
<0 (mch delivers
power via stator)
<0 (mch delivers
power via rotor)
3. Generator
(Tem<0)
s>0, ωm<ωs
(subsynchrnsm)
<0 (mch receives
mech pwr)
<0 (mch delivers
power via stator)
>0 (mch receives
power via rotor)
4. Motor
(Tem>0)
s>0, ωm<ωs
(subsynchrnsm)
>0 (mch delivers
mech pwr)
>0 (mch receives
power via stator)
<0 (mch delivers
power via rotor)
For each mode, we may use the three relations to track the sign Ps, ωr, and Pr from the
signs of Tem and s. For example, for mode 2, Tem<0Ps<0 and Tem<0, s<0 ωr<0Pr<0
Focusing on the generator modes, we observe the standard induction machine
generating mode, supersynchronism, where ωm>ωs (mode 2). We also observe a
subsynchronous mode (mode 3), where ωm<ωs, which is available to the DGIG as a
27
result of the machine receiving power from the grid via the rotor circuit.
Generator modes
Recall the approximate
relation
Pr  sPs
Mode 2
Pm= Pmech
m  s
In fact, DFIGS always
run within about
-0.3<s<0.3.
Mode 3
m  s
These figures show actual flow
direction for generator operation.
They also neglect losses.
Operation must have
|s|<1, so rotor power is
always smaller than
stator power.
Therefore, the rating of
the PE converter circuit
need be only about 30%
of the stator winding
rating.
28
This figure assumes proper
sign convention (power
flowing to the rotor or into
the stator is positive).
Pmech  Ps  Pr
P
Ps  mech
1 s
A question on rating
Pg  Ps  Pr
 sPmech
Pr 
1 s
Pgrid
Ps
Assume an operating condition
such that Pmech=PWTrating. Then
Without losses
Pairgap
Pslip
Pr
Pmech
Pmech  Pg  PWTrating
PWTrating
 sPWTrating
Ps 
Pr 
1 s
1 s
For example, consider Pmech=PWTrating=-2 MW. In supersynchronous mode, with s=-0.3,
2
Ps 
 1.5385 MW.
1  0.3
Therefore stator winding must be rated for 1.5385 MW.
But in the subsynchronous mode, s=+0.3, then
Ps 
2
 2.8571MW
1  0.3
Question: Does this mean that the stator of a 2 MW turbine must be rated for 2.8571?
Answer: No. In subsynchronous mode, the mechanical power from the generator shaft
is lower that that in the supersynchronous mode. If Pmech increases beyond a certain
level, then machine speed increases into the supersynchronous mode. So above
situation never occurs. We can obtain the maximum power in subsynchronous mode as:
Pmech  Ps (1  s)  1.5385(1  0.3)  1.0769MW
29
Question on sign of losses
Pmech  Ps  Pr  Ploss ,s  Ploss ,r
Question: Since stator losses (3RSIs2) and rotor losses (3RrIr2) are always
positive, and since we get sign changes with the numerical values of Pmech,
Ps, and (sometimes) Pr, do the loss terms in the above equation need to
have different signs for motor operation than for generator operation? That
is, do we need to do the following?
Motor operation: Pmech  Ps  Pr  Ploss ,s  Ploss ,r
Generator operation: Pmech  Ps  Pr  Ploss ,s  Ploss ,r
Answer: No. Our original equation applies for both motor & generator operation.
Remember: Pmech is positive for motor operation; Ps, and Pr are positive when
flowing into the device from the grid.
It may help to think about the equation in two different, but equivalent forms.
Motor operation:
Generator operation:
Pmech  Ps  Pr  Ploss , s  Ploss ,r
 
Ps  Pr  Pmech  Ploss , s  Ploss ,r

 
Output
Input
50 = 45 +10 - 3 - 2
Output
Input
- 50 = - 55 + 3 + 2
30
Per-unitization
In general, per-unitization enables inclusion of DFIGs within a system model.
It also facilitates identification of inappropriate data. Finally, a per-unitized
voltage provides the ability to know how far it is from its nominal value
(usually also the “normal” value) without knowing that nominal value.
The procedure is to choose three base quantities and compute other
necessary base quantities. We will choose our base quantities as
• rated rms line-to-neutral stator voltage, Vbase=|Vs|rated (rms volts);
• rated rms stator line current, Ibase=|Is|rated (rms amperes)
• rated stator synchronous frequency, ωbase= ωs,rated (rad/sec))
Then we compute:
• Base impedance:
• Base flux:
Vbase
Z base 
I base
Vbase
base 
base
• Base inductance:
• Base speed:
d 



   Vt 
 Justification : v 
dt t


• Three-phase
base
base base
power base:
S
 3V
I
• Base torque:
Lbase
 m,base
Tbase
base

I base
base

p
Sbase

m,base
31
Per-unitization – stator side
Once all base quantities are obtained, then per-unitization is easy:
• Stator voltage in pu:
• Stator current in pu:
• Stator flux in pu:
• Stator active power in pu:
• Stator reactive power in pu:
Vs
V s , pu 
Vbase
Is
I s , pu 
I base
s
 s , pu 
base
 
*
s s
Re V I
P s, pu  3
Sbase
 
*
s s
Im V I
Q s, pu  3
Sbase
As usual, only the magnitude is transformed (angle remains unchanged).
32
Per-unitization – rotor side
• Rotor voltage in pu:
• Rotor current in pu:
• Rotor flux in pu:
• Rotor active power in pu:
• Rotor reactive power in pu:
Vr
V r , pu 
Vbase
Ir
I r , pu 
I base
r
 r , pu 
base
 
For the rotor side, we
use the same base
quantities as on the
stator side (with
actual quantities
referred to the stator
side).
*
r r
Re V I
P r , pu  3
Sbase
 
*
r r
Im V I
Q r , pu  3
Sbase
As usual, only the magnitude is transformed (angle remains unchanged).
33
Per-unitization – torque, speed, R, L
• Torque in pu:
• Speed in pu:
• Resistances in pu:
• Inductance in pu:
Tem
Tem, pu 
Tbase
m
m, pu 
m,base
Rr
r pu 
Zbase
L
l pu 
Lbase
On the rotor side, we
use the same base
quantities as on the
stator side (with
actual quantities
referred to the stator
side).
Note that the
resistances and
inductances when
expressed in pu are
lower case.
As usual, only the magnitude is transformed (angle remains unchanged).
34
Voltage equations expressed in per unit
From slides 15, 16, we obtain voltage equations for stator and rotor circuits:
V s  I s Rs  js Ls   I s  I r  js Lm
V r  I r Rr  jss Lr   I s  I r  jss Lm
which we rearrange by collecting terms in jωs:
V s  I s Rs  js Ls I s  I s  I r Lm 
V r  I r Rr  jss Lr I r  I s  I r Lm 
From slide 20, we obtain the equations for stator and rotor flux linkages:
 s  Lm ( I s  I r )  Ls I s
 r  Lm ( I s  I r )  Lr I r
(*)
We recognize the flux linkage expressions in the voltage equations. Therefore:
V s  I s Rs  js  s
V r  I r Rr  jss  r
Now we can replace voltages, currents, and flux linkages with the product of their
per-unit value and their base quantity, then the base quantities can be used to perunitize the resistances and frequency to obtain:
V s, pu  I s, pu rs  j s, pu
V r , pu  I r , pu rr  js r , pu
35
Voltage equations expressed in per unit
V s  I s Rs  js  s
V r  I r Rr  jss  r
Replace voltages, currents, flux linkages with the product of their pu value and
their base quantity, then base quantities are used to per-unitize resistances and
frequency to obtain:
V s, pu  I s, pu rs  j s, pu
V r , pu  I r , pu rr  js r , pu
Now consider the flux linkage equations:
 s  Ls I s  Lm I r
 r  Lm I s  Lr I r
Replace currents and flux linkages with the product of their pu value and their
base quantity, then base quantities are used to per-unitize inductances to obtain:
 r , pu  lm I s, pu  lr I r , pu
*
Per-unitize one of the torque equations (#2) Tem  3 p Im  r I r
 s, pu  ls I s, pu  lm I r , pu
Tbase 
 
as follows:
*
*
Sbase 3 Vbase I base

3 p Im r I r 
r
Ir 


*

T



3
Im

3
Im

I


r , pu 
em
,
pu
r
,
pu
m,base m,base / p
3 Vbase I base
V
/

I

 base m,base base 

m,base / p
Per-unitize the power expressions to obtain:
Ps , pu  Vs , pu I s , pu cos( v   i );
Qs , pu  Vs , pu I s , pu sin( v   i )
Pr , pu  Vr , pu I r , pu cos(v  i );
Qr , pu  Vr , pu I r , pu sin(v  i )
36
Homework #3
Homework #3: This homework is due Monday, March 26.
A. Using previous relations provided in these slides, derive the following
torque expressions.
 
 3 p Im , I 
L
 3p
Im ,  
L L
1.
Tem  3 p Im  s , I s
2.
Tem
3.
Tem
*
r
*
r
m
r
*
r
s
(and identify σ)
s
B. Use Q = 3Im{V I*} and the equivalent circuit to derive reactive power
expressions, in terms of Is and Ir for
1. The stator, Qs
2. The rotor, Qr
C. For each DFIG condition below, compute Pairgap and Pslip and draw the power
flows similar to slide 28.
1. Pmech=-1 MW with s=+0.30 (subsynchronous operation).
2. Pmech=-1MW with s=-0.30 (supersynchronous operation).
D. Complete the table on the next slide (the boxed section) by computing the
per-unit values of the indicated five resistances/inductances for the 2 MW
machine.
37
Homework
u (or a)
Rs
Lσs
Lm
R’r
Lσr
Rr
Lσr
Ls
Lr
Vbase
Ibase
Rs
lσs
lm
rr
lσr
38
Phasor diagrams for generator operation
We have developed the following relations:
V s, pu  I s, pu rs  j s, pu
(1) Stator voltage equation
V r , pu  I r , pu rr  js r , pu
 s, pu  Ls I s, pu  Lm I r , pu
 r , pu  lm I s, pu  lr I r , pu
(2) Rotor voltage equation
(3) Stator winding flux equation
(4) Rotor winding flux equation
Draw phasor diagram per below (CCW rotation is pos angle):
Step 1: Draw Vs as reference (0°).
Step 2: For gen, Qs>0, lag; for gen Qs<0, lead. Draw Is phasor.
Step 3: Use (1) to draw the stator flux phasor λs:  s, pu   j(V s, pu  I s, pu rs )
Step 4: Use (3) to draw the rotor current phasor Ir: I r , pu   s, pu / lm  ls I s, pu / lm
Step 5: Use (4) to draw the rotor flux phasor λr:  r , pu  lm I s, pu  lr I r , pu
Step 6: ….
Vs - Isrs
Isrs
Vs
lm Is
Ir=λs/lm – ls Is/lm
Is
λs/lm
– ls Is/lm
λr=lm Is+lr Ir
λs= -j(Vs – Isrs)
lr Ir
39
Phasor diagrams for generator operation
Draw phasor diagram per below (CCW rotation is pos angle):
Step 1: Draw Vs as reference (0°).
Step 2: For gen, Qs>0, lag; for gen, Qs<0, lead. Draw Is phasor.
Step 3: Use (1) to draw the stator flux phasor λs:  s, pu   j(V s, pu  I s, pu rs )
Step 4: Use (3) to draw the rotor current phasor Ir: I r , pu   s, pu / lm  ls I s, pu / lm
Step 5: Use (4) to draw the rotor flux phasor λr:  r , pu  lm I s, pu  lr I r , pu
Step 6: Use (2) to draw the rotor voltage phasor Vr: V r , pu  I r , pu rr  js r , pu
Vr=Irrr+jsλr, s<0
super-syn
lm Is
Vr=Irrr+jsλr, s>0
Irrr
Is
Ir=λs/lm – ls Is/lm
jsλr, s<0
λs/lm
Observe that the angle of Vr is
heavily influenced by the sign of s.
jsλr, s>0, sub-sync
Vs - Isrs
Isrs
Vs
– ls Is/Lm
λr=lm Is+Lr Ir
λs= -j(Vs – Isrs)
lr Ir
40
Question: How to know quadrant of Is?
Consider the circuit below, which is analogous to our stator winding circuit.
At any operating condition, we may
I
characterize the circuit as an impedance
Machine
Z=R+jX=Z/_θ, as indicated. Then we may
express the current according to
Z
V
R  jX 
I  I i 
Z
V
V
V

  
Z Z Z
Observe that current angle is always
negative of impedance angle, θi=-θ
Real pwr Reactive pwr
P>0
motor
R>0
Q>0
absorbing
X>0
Real pwr Reactive pwr
P>0
motor
R>0
Q<0
supplying
X<0
Z
Real pwr Reactive pwr
V
I
Lag
I
P<0
gen
R<0
Q>0
absorbing
X>0
Z
Lead
V
I
Lag
Real pwr Reactive pwr
V
Z
P<0
gen
R<0
Q<0
supplying
X<0
I
V
Z
Lead
41
Example Problem
The 2 MW DFIG given by the data on slide 38 is delivering, from the stator, rated
load (2 MW) at rated voltage with zero stator reactive power in a 50 Hz grid. The
slip is s=-0.25 (super-synchronous). Compute:
(h) Rotor real power
(d) Stator flux
(a) Synchronous speed
(e) Rotor current (i) Rotor reactive power
(j) Total real power generated
(b) Line-to-neutral voltage
(f) Rotor flux
(c) Line current
(g) Rotor voltage (k) Tem
(a) Synchronous speed: s  2f s  2 (50)  314.16 rad/sec
Alternatively, the synchronous speed was given as 1500 rpm, therefore:
1500 rev 2rad min
s 
 157 .08rad / sec  s  ps  2(157.08)  314.16rad / sec
min
rev 60 sec
690
0  398.40 volts
*
*
3
6
 Ps    2 10 
*

  
  1673.4180 amps
P

j
0

3
V
I

I

(c) Line current: s
s
s s
 3V s   3  398.40 
(b) Line-to-neutral voltage:
(d) Stator flux
Vs 
V s  I s Rs  js  s


(V s  I s Rs ) 398.40  (1673.4180)2.6 103
 s 

 1.28  90webers
js
j314.16
42
Example Problem
The 2 MW DFIG given by the data on slide 38 is delivering, from the stator, rated
load (2 MW) at rated voltage with zero stator reactive power in a 50 Hz grid. The
slip is s=-0.25 (super-synchronous). Compute:
(h) Rotor real power
(d) Stator flux
(a) Synchronous speed
(e) Rotor current (i) Rotor reactive power
(j) Total real power generated
(b) Line-to-neutral voltage
(f) Rotor flux
(c) Line current
(g) Rotor voltage (k) Tem
(e) Rotor current
Ir 
 s  Ls I s
Lm
 s  Ls I s  Lm I r
1.28  90  2.587103 (1673.8180)

 1807.4  16.5amps
3
2.5 10
This is the referred rotor current!
We can obtain the actual rotor current from a (or u) =0.34:
I r  a I r  (0.34)1807.4 16.5  614.5 16.5amps This phasor is at the rotor
frequency, of
fr=sfs=-0.25(50)=-12.5 Hz
(f) Rotor flux   L I  L I
r
m
s
r
r
 r  2.5 10 1673.8180  2.587103 1807.4 16.5  1.358  77.4 weber
(g) Rotor voltage V r  I r Rr  jss  r
3
V r  (1807.4 16.5)2.9 103  j(0.25)(314.16)(1.358  77.4)  102.2 165.9volts
V r 102 .2  165 .9

Vr 
 300 .6  165 .9
Actual rotor voltage:
43
a
0.34
Example Problem
The 2 MW DFIG given by the data on slide 38 is delivering, from the stator, rated
load (2 MW) at rated voltage with zero stator reactive power in a 50 Hz grid. The
slip is s=-0.25 (super-synchronous). Compute:
(h) Rotor real power
(d) Stator flux
(a) Synchronous speed
(e) Rotor current (i) Rotor reactive power
(j) Total real power generated
(b) Line-to-neutral voltage
(f) Rotor flux
(c) Line current
(g) Rotor voltage (k) Tem
 
(h) Rotor real power Pr  3 Re V r I r
*


Pr  3 Re 102.2 165.9  (1807.4 16.5)*  0.55 MW
 
(i) Rotor reactive power Qr  3 Im V r I r
*


Qr  3 Im 102.2 165.9  (1807.4 16.5)*  23.4kVAR
(j) Total real power generated
Comments:
Ps  Pr  2  0.55  2.55MW 1. P must be larger in magnitude to supply losses
m
Pmech  Ps  Pr  Ploss ,s  Ploss ,r
2. This wind turbine’s rating should be 2.55 MW.
3. The DFIG stator winding is rated for 2MW.
44
Example Problem
The 2 MW DFIG given by the data on slide 38 is delivering, from the stator, rated
load (2 MW) at rated voltage with zero stator reactive power in a 50 Hz grid. The
slip is s=-0.25 (super-synchronous). Compute:
(h) Rotor real power
(d) Stator flux
(a) Synchronous speed
(e) Rotor current (i) Rotor reactive power
(j) Total real power generated
(b) Line-to-neutral voltage
(f) Rotor flux
(c) Line current
(g) Rotor voltage (k) Tem
(k) Tem
Tem  3 p
 
Lm
*
Im  s I r
Ls


2.5 103
*


Tem  3  2
Im
1
.
28

90

1807
.
4


16
.
5

 12.9kNm
3
2.58710
45
Wind turbine control levels
Rotor-side converter (RSC) is
controlled so that it provides
independent control of Tem
and Qs. Let’s study the
steady-state actions of this
particular control function.
Level I: Regulates power flow
between grid and generator.
Level II: Controls the amount
of energy extracted from the
wind by wind turbine rotor.
Level III: Responds to windfarm or grid-central control
commands for MW dispatch,
voltage, frequency, or inertial
46
control.
Level 1 control
We achieve
control objectives
by controlling
rotor-side
voltage.
This (openloop) control
not heavily
used for DFIGs
Assume DC bus voltage is
controlled by grid-side
converter (GSC) to a predetermined value for
proper operation of both
GSC and RSC.
We control rotor
voltage to achieve a
specified torque and
stator reactive power.
47
Level 1 control
Our objective here is, for a fixed stator voltage (fixed by the grid), and a
desired torque Tem,ref and a desired stator reactive power Qs,ref, we want to
determine the rotor voltage to make it so. We are also interested in the
stator flux, stator current, rotor current, and rotor flux, and stator real power,
as shown in the diagram below.
48
Level 1 control
We draw the phasor diagram with stator flux as the reference (0 degrees). Here,
the stator flux, denoted by ψs (instead of λs), is specified as the reference. We
have identified particular angles in this phasor diagram. It is operating as a
motor (current is almost in phase with voltage), and the stator is absorbing
reactive power (Is has a negative angle relative to Vs, so Zmotor=Vs/Is has a
positive angle, indicating it is inductive and therefore absorbing.
49
Level 1 control: Qs equation
V s  I s Rs  js  s
From voltage equation (slide 35):
If we neglect drop across the stator resistance (it is typically very small), then:
V s  js  s
 

Substitute into the stator reactive power equation: Qs  3 Im V s I s  3 Im js  s I s

Use Im(ja)=Re(a): Qs  3 Re s  s I s
*

*
*

From previous slide, note that ɣi is the angle by which Is leads λs , i.e.,
 s  s 0;
Substituting:
I s  I s  i
Qs  3 Res s 0I s    i   3s s I s Re   i 
 3s s I s Recos i  j sin  i   3s s I s cos i
Final equation for Qs: Qs  3s s I s cos i
50
Level 1 control: Tem equation
From HW3 (see slide 37):
T  3 p Im , I 
em
*
s
s
Again (from phasor diagram), note that ɣi is the angle by which Is leads λs , i.e.,
 s  s 0;
Substituting:
I s  I s  i
Tem  3 p Ims 0I s  i   3 ps I s Im i 
 3 ps I s Imcos i  j sin  i   3 ps I s sin  i
Final torque equation:
Tem  3 ps I s sin  i
51
Level 1 control: Is equation
From phasor diagram:
I s  I s cos i  jI s sin  i
But recall our Qs and Tem equations:
Qs  3s s I s cos i
I s cos i 
I s sin  i 
Tem  3 ps I s sin  i
Qs
3s s
Tem
3 ps
Substituting into Is equation:
Is 
Qs
3s s
j
Tem
3 ps
V s  js  s  Vs  s s
Q
T
Substituting into Is equation: I s  s  j s em
3Vs
3 pVs
Recall from slide 50:
52
Level 1 control: λr equation
From slide 20:
L
1
s  m r
Ls
Ls Lr
L
1
I r  m s 
r
Ls Lr
Lr
Is 
 s  Ls I s  Lm I r
 r  Lm I s  Lr I r
L2m
  1
Ls Lr
Using these relations, together with:
V s  js  s
Qs
sTem
Is 
j
3Vs
3 pVs
we may derive:
 Vs Lr Qs Ls Lr   sTem Ls Lr 
r  

  j


L
3
V
L
3
pV
L
s
m 
s
m 
 s m

 V 1 Qs Ls   sTem Ls 
Ir   s

  j


L
3
V
L
3
pV
L
s
m
s
m
 s m

53
Level 1 control: λr equation
Now use the rotor flux equation derived on the previous slide
together with the rotor voltage equation (slide 35):
V L
Q Ls Lr 
r   s r  s

 s Lm 3Vs Lm 
  T L L 
j  s em s r 
 3 pVs Lm 
V r  I r Rr  jss  r
Neglecting the voltage drop in the rotor resistance, we may derive:
  T L L 
V L
Q Ls Lr 
V r  r  s em s r   jr  s r  s

3
pV
L

L
3
V
L
s
m 
s
m 

 s m
54
Level 1 control: summary
Qs
T
 j s em
3Vs
3 pVs
 Vs Lr Qs Ls Lr   sTem Ls Lr 
r  

  j


L
3
V
L
3
pV
L
s
m 
s
m 
 s m

 Vs 1 Qs Ls   sTem Ls 
Ir  

  j


L
3
V
L
3
pV
L
s
m
s
m
 s m

  T L L 
V L
Q Ls Lr 
V r  r  s em s r   jr  s r  s

3
pV
L

L
3
V
L
s
m
s
m
s
m




Is 
s 
Vs
s
Also, we have stator and rotor powers as a function of Tem:
Ps 
s
p
Tem  Pr 
 r
Tem
p
55
Level 1 control: magnitudes
Magnitudes
are attractive
because then we can plot them.
2
2
Q 
 T 
 f Is (Vs , Qs , Tem )
I s2   s    s em 
 3Vs 
 3 pVs 
 Vs
2
r  
 s
V
s  s
s
V
I r2   s
 s
2
Lr Qs Ls Lr   sTem Ls Lr 

 

Lm 3Vs Lm   3 pVs Lm 
2
 T
Q L 
1
 s s    s em
Lm 3Vs Lm 
 3 pVs
2



T

L
L
2  Vs
s r
Vr2  r2  s em


r 

3
pV
L
s
m


 s
Pr 
r
p
Tem
Ls 

Lm 
2
 f r (Vs , Qs , Tem )
 f s (Vs )
2
Q Ls Lr 
Lr
 s

Lm 3Vs Lm 
 f Ir (Vs , Qs , Tem )
2
 fVr (Vs , Qs , Tem , r )
 f Pr (Tem , r )
And this shows that these terms are functions of our desired reference quantities.
The above relations are given as a function of ωr, but it may be more intuitive
to plot them as a function of rotor speed, ωm, where we can compute
ωr =sωm/(1-s). You can think of the rotor speed as ωm=(1-s) ωs which shows
that for low positive slips, rotor speed is just below synchronous speed, and
for low negative slips, rotor speed is just above synchronous speed.
56
Level 1 control
2
Q 
 T 
I s2   s    s em 
 3Vs 
 3 pVs 
Fixed Qs=0
2
Fixed Tem=-1
• Is is independent of ωm but increases with |Tem| and with |Qs|
• Is is the same independent of whether machine is absorbing or supplying vars.
• Above equation indicates Is should be the same for Tem=1, Tem=-1. However,
above equation neglected stator resistance Rs. Assuming fixed Vs, in motor
mode (Tem=1), Rs causes voltage across rotor circuit to be less, and so Ir must
be greater to deliver same torque. In gen mode, Rs causes voltage across
rotor circuit to be more, and so Ir must be less to deliver same torque.
57
Level 1 control
2
V 1
 T L 
Q L 
I r2   s
 s s    s em s 
  s Lm 3Vs Lm 
 3 pVs Lm 
2
Fixed torque implies fixed rotor
current if stator flux is fixed.
Tem  3 p
 
Lm
*
Im  s I r
Ls
Because Tem=Pmechp/ωm, Pmech
must decrease as ωm increases.
Fixed Tem=-1
• Ir is independent of ωm for fixed torque but increases as Qs moves from + (absorbing)
to – (supplying).
58
Level 1 control
Both rotor current and stator current equations have real part determined by Qs
and imaginary part determined by Tem (Vs is at 90° so real part of currents is in
quadrature with Vs)
 V 1 Qs Ls 
Ir   s


 s Lm 3Vs Lm 
Qs
sTem  Vs 1 Qs Ls   sTem
Im  Is  Ir 
j


  j
3Vs
3 pVs  s Lm 3Vs Lm   3 pVs
Qs
sTem
Is 
j
3Vs
3 pVs
Add them to
obtain
magnetizing
current
 Qs Vs 1 Qs Ls 




 3Vs s Lm 3Vs Lm 
Magnetizing component.
Qs=0 (no stator reactive power):
 sTem sTem Ls 
j


3
pV
3
pV
L
s
s
m

Very close to zero since Ls~Lm.
 Vs 1 



L
 s m
0<Qs<3Vs2/Lsωs (reactive power into stator, abs)
Qs=3Vs2/Lsωs (reactive power into stator, abs)  Q
Magnetized from rotor current
Magnetized from both currents.


 3Vs 
Qs<0 (reactive power from stator, sup):
 T L 
j  s em s 
 3 pVs Lm 
Ls 

Lm 
s
Magnetized from stator current.
Magnetized from both currents.
59
Level 1 control
Pr 
r
p
Tem
Fixed Qs=0
Fixed Tem=-1
• Pr linearly decreases w/ ωm for –Tem (gen) and linearly increases w/ ωm for +Tem(mot).
• Pr is independent of whether machine is absorbing or supplying vars.
Remember: ωm=(1-s)ωs,
ωr=sωs.
60
Level 1 control
2
  T Ls Lr 
Qs Ls Lr 
2  Vs Lr
Vr2  r2  s em



r 


3
pV
L

L
3
V
L
s
m
s
m


 s m

Fixed Qs=0
2
Fixed Tem=-1
• Vr is linearly decreasing with ωm to ωm=ωs and then linearly increasing with ωm.
• Vr depends mainly on speed of machine.
• Vr does not change much with Tem or with Qs because VsLr/ωsLm tends to dominate.
Remember: ωm=(1-s)ωs,
ωr=sωs.
61
Level 1 control
Fixed Qs=0
Fixed Tem=-1
• Efficiency increases with ωm under all conditions (see next slide):
• In the subsynchronous mode, stator windings carry |Pmech|+|Pr|.
• In the supersynchronous mode, stator windings carry |Pmech|-|Pr|.
• Efficiency decreases as |Qs| increases (most efficient for unity power factor).
• More efficient when absorbing (magnetized from stator) than supplying
(magnetized from rotor)
62
Generator modes
Mode 2
Pm= Pmech
m  s
Mode 3
m  s
63
Representing RSC with impedance
It can be convenient in analyzing the steady-state performance of the DFIG to
represent the RSC as an equivalent impedance, as indicated in the below figure.
We can follow our earlier development (see slide 9), but with our RSC
equivalent impedance represented:
Rr
Rs
jsωsLσr
Is
jωsLσs
Ir
3
3
3
3
Es
Vs
3
3
Req
3
3 Ers=sEs
Vr
In slide 9:
V r  s E s  ( Rr  jss Lr ) I r
Now:
 I r ( Req  jss Leq )  s E s  ( Rr  jss Lr ) I r
Divide by s
Divide by s
Vr
R
 E s  ( r  j s Lr ) I r
s
s
3 jωrLeq
3
=jsωs Leq
 I r ( Req  jss Leq )
s
 Ir(
Req
s
Rr
 E s  (  js Lr ) I r
s
 js Leq )  E s  (
Rr
 js Lr ) I r
s
64
Representing RSC with impedance
Is
Rs
jωsLσs
Rr/s
Ir
3
3
3
3
Vs
jωsLσr
Req/s
Vm 3
3 jωsLm
Equivalent RSC impedance is:
Represent it in the circuit with:
Vr/s
3 jωs Leq
3
Zeq  Req  jr Leq
Z eq Req jr Leq Req



 js Leq
s
s
s
s
Let’s assume the DFIG operates at unity power factor. Then Qs=0, and for Vs=Vs/_0°,

 

Pairgap  Ps  Ploss ,s  3 Vs I s  Rs I s2  3 Vs  Rs I s I s
Question: Do we need to specify motor or generator operation in the above equation?
Answer: Not for the relation Pairgap= Ps-Ploss,s (see slide 30). For motor op, Ps>0 and
losses subtract so that Pairgap is smaller than Ps, consistent with the fact that power
flows from stator to rotor. For gen op, Ps<0 and losses add so that Pairgap is larger than
Ps, consistent with the fact that power flows from rotor to stator.
However, the relation on the right assumes that Is is a magnitude (positive), and so it
is correct for motor op. For gen op, we must use a negative magnitude to get the sign
of VsIs correct. We could correct this by writing the RHS as VsIs-RsIs2= (Vs-RsIs)Is, i.e.
use phasor notation for the current instead of just magnitude.
65
Representing RSC with impedance
Is
Rs
jωsLσs
Vm
Ir
3
3
3
3
Vs
Rr/s
jωsLσr
Req/s
3
3 jωsLm
Vr/s

3 jωs Leq
3

Pairgap  Ps  Ploss ,s  3 Vs  Rs I s I s
From slide 25, we know for the model (with losses) that Tem 
Equating the two airgap expressions:
s
p

2
4 Rss
Tem
3p
2 Rs
Vs  Vs2 
Is 
s
Pairgap  Pairgap 
s
p
Tem
Tem  3 Vs  Rs I s I s
Rewriting, we find a quadratic in Is: Rs I s  Vs I s 
Obtain roots:

p
s
3p
Tem  0
Could be positive (motor) or
negative (generator)
With stator current calculated, we can use the circuit to find Vr and Ir….
66
Representing RSC with impedance
Rs
Is
jωsLσs
jωsLσr
Rr/s
Ir
3
3
3
3
Im
Vs
Req/s
Vm 3
3 jωsLm
From KVL we can compute Vm:
Vr/s
V m  V s  I s Rs  js Ls 
Then compute the magnetizing current Im:
Then compute the rotor current Ir:
3 jωs Leq=jXeq/s
3
Im 
Vm
V  I s Rs  js Ls 
 s
js Lm
js Lm
Ir  Im Is 
Then compute the rotor voltage Vr:
We can now obtain Zeq/s or Zeq:

R

 V m  I r  r  js Lr  
Req
V /s
 s

Z eq / s 
 js Leq  r  


s
Ir
Ir




where Ir is computed from above relations.
(Xeq= ωrLeq)
V s  I s Rs  js Ls 
Is
js Lm
R

V r / s  V m  I r  r  js Lr 
 s

R

 V s  I s Rs  js Ls   I r  r  js Lr 
 s

Z eq  Req  jss Leq 
V r  sV m  I r Rr  jss Lr  


I r 
Ir

67
Representing RSC with impedance
Tem  3 p
 
Lm
*
Im  s I r
Ls
Tem is increasing here.
Req<0converter
transfers active
power to rotor.
Req>0rotor
delivers active
power to the
converter.
68
Homework #4
Consider a 1.5 MW, 690 v, 50 Hz 1750 rpm DFIG wind energy system. The
parameters of the generator are given on the next slide. The generator operates
with a maximum power point tracking (MPPT) system so that its mechanical
torque Tem is proportional to the square of the rotor speed. The stator power factor
is unity. For each of the following speeds: 1750, 1650, 1500, 1350, and 1200 rpm,
compute:
• Slip
• Tem (kN-m)
• Vr (volts)
• Ir (amps)
• Req (ohms)
• Xeq (ohms)
What kind of machine is this at 1500 rpm?
69
Homework #4
70
Homework #4
Converter equivalent impedance at 1500 rpm:
m 
1500rev 2rad min
 157.0796rad / sec
min
rev 60sec
m  pm  2 *157.0796 314.1592rad / sec
f m  m / 2  314.1592/ 2  50Hz
So 1500 rpm is synchronous speed!
71
Homework #4
There is another solution which has very
large current and is clearly not realistic.
Be careful here because this
solution assumed the direction
of current Ir opposite to what we
have assumed.
Observe that slip=0. This implies that a DC current flows through the rotor circuit
from the converter and the rotor leakage reactance and equivalent reactance
are zero. The DFIG is operating like a synchronous machine where the rotor flux
is produced by a DC current through a DC exciter.
72
SCIG Torque-slip characteristic
You may recall, from EE 303 or your undergraduate course on electric machines that
the torque-slip characteristic of the squirrel-cage induction generator (SCIG) appears
as below. One observes that the SCIG operates as a generator only when it is in
supersynchronous mode and a motor only when it is in subsynchronous mode.
Motoring
Generating
Subsynchronous
Supersynchronous
Let’s see how we obtain this curve for SCIG, and let’s also compare what
we do to what we need to do to obtain the analogous curves for the DFIG.
73
Comparison of equivalent circuits: SCIG vs DFIG
Is
Rs
jωsLσs
jωsLσr
Rr/s
Ir
3
3
3
3
Im
Vm 3
3 jωsLm
Vs
SCIG
The difference between the machines in terms of steady-state models is
the ability to electrically absorb or supply complex power S via the rotor.
Is
Rs
jωsLσs
jωsLσr
Rr/s
Ir
3
3
3
3
Im
Vs
Vm 3
3 jωsLm
Req/s
Vr/s
3 jωs Leq=jXeq/s
3
DFIG
Where do we see rotor losses in these circuits? … (next slide)
(Xeq= ωrLeq)
74
Comparison of equivalent circuits: SCIG vs DFIG
Is
Rs
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Im
Vm 3
3 jωsLm
Vs
SCIG
Split up the R/s terms in each circuit as R+R(1-s)/s
and the rotor losses become immediately apparent.
Is
Rs
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vs
Vm 3
3 jωsLm
DFIG
Vr/s
Req
3 jωs Leq=jXeq/s
3
(Xeq= ωrLeq)
Where do we see mechanical power in these circuits? … (next slide)
75
Comparison of equivalent circuits: SCIG vs DFIG
Is
Rs
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Im
Vm 3
3 jωsLm
Vs
SCIG
The mechanical power is represented
by the slip-dependent resistances.
Is
Rs
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vs
Vm 3
3 jωsLm
DFIG
Vr/s
Req
3 jωs Leq=jXeq/s
3
(Xeq= ωrLeq)
But what do the other two terms in the DFIG circuit represent? … (next slide)
76
Comparison of equivalent circuits: SCIG vs DFIG
Is
Rs
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Im
Vm 3
3 jωsLm
Vs
SCIG
These terms represent the real and reactive power exchange between the rotor
and the RSC. As we saw on slide 68, these terms, Req and Xeq can be pos (rotor
transfers power to RSC) or neg (RSC transfers power to rotor).
Is
Rs
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vs
Vm 3
3 jωsLm
DFIG
Vr/s
Req
3 jωs Leq=jXeq/s
3
(Xeq= ωrLeq)
How to compute torque in for these machines? … (next two slides)
77
Comparison of equivalent circuits: SCIG vs DFIG
78
Torque equation for SCIG
Is
Rs
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Im
Vm 3
3 jωsLm
Vs
SCIG
Rr (1  s) Note that the “s” on the denominator provides
that Pmech is positive for s>0, motor action,
Pmech  3I
s
and negative for s<0, generator action.
1
p
p 2 Rr (1  s)
Tem 
Pmech 
Pmech  3
Ir
m
m
m
s
2
r
p
p 2 Rr
2 Rr (1  s )
3
Ir
 3 Ir
s (1  s)
s
s
s
How to obtain Ir? …. (next slide)
79
Torque equation for SCIG
Zs=Rs+jXσs
Is
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Im
Vm 3
3 Zm=jωsLm
Vs
Find Thevenin looking in here.
Zth
Is
jωsLσr
Rr
Vth  V s
Ir Rr(1-s)/s
Zm
Zs  Zm
Z th 
Zs Zm
Zs  Zm
3
3
3
3
Comment: Zm>>ZS, so
Vth≈Vs, Zth=Zs is not a
bad approximation.
Vth
Tem  3
p
s
I r2
3 pVth Rr / ss
Rr

2
s 
Rr 
2


R


X

X
 th

th
r
s 

Ir 
 V th
Z th  ( Rr / s)  jX r
80
SCIG Torque-slip characteristic
You may recall, from EE 303 or your undergraduate course on electric machines that
the torque-slip characteristic of the squirrel-cage induction generator (SCIG) appears
as below. One observes that the SCIG operates as a generator only when it is in
supersynchronous mode and a motor only when it is in subsynchronous mode.
Motoring
Generating
Subsynchronous
Supersynchronous
Now let’s take a look at the torque-speed curves for the DFIG…. (next slide)
81
Torque equation for DFIG
Rs
Is
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vm 3
3 jωsLm
Vs
Vr/s
Req
3 jωs Leq=jXeq/s
3
 ( Rr  Req )(1  s) 

Pmech  3I 
s


1
p
p 2  ( Rr  Req )(1  s) 

Tem 
Pmech 
Pmech  3
I r 
m
m
m 
s

( Rr  Req )(1  s) 
p
p 2  Rr  Req 
2
  3 I r 

3
I r 
s (1  s) 
s
s  s 

2
r
How to obtain Ir? …. (next slide)
82
Comparison of equivalent circuits: SCIG vs DFIG
Rs
Is
jωsLσs
jωsLσr
Rr
3
3
3
3
Req(1-s)/s
Im
Vm 3
3 Zm=jωsLm
Vs
Ir Rr(1-s)/s
Vr/s
Vth  V s
Find Thevenin looking in here.
Zth
Is
jωsLσr
Rr
Ir Rr(1-s)/s
Comment: Zm>>ZS, so
Vth≈Vs, Zth=Zs is not a
bad approximation.
Req(1-s)/s
Vr/s
3 pVth2 ( Rr  Req ) / ss
 Rr  Req 
 
Tem  3 I 
2
2
s 
s
Rr  Req  
X eq 
 
 Rth 
   X th  X r 

s
s 

 
2
r
Zm
ZZ
Z th  s m
Zs  Zm
Zs  Zm
3
3
3
3
Vth
p
Req
3 jωs Leq=jXeq/s
3
Ir 
Req
3 jωs Leq=jXeq/s
3
 V th
Z th  ( Rr  Req ) / s  j ( X r  X eq / s)
83
Torque-slip characteristic for DFIG
So how do we obtain the torque-slip characteristic for the DFIG?
1. Develop values of Zeq for various values of torque-speed control point (slides 66-67):
4 Rss
Tem
3p
2 Rs
Vs  Vs2 
Is 
V  I s Rs  js Ls 
Ir  Im Is  s
Is
js Lm
V m  V s  I s Rs  js Ls 
Z eq  Req  jss Leq 
V r  sV m  I r Rr  jss Lr  


I r 
Ir

Aside: The above points result from the turbine control characteristic. This characteristic
originates from the maximum power extracted from the wind, which is given by the
power curve, described by Pmech~ωm3. But Pmech=Temωm therefore Tem~ ωm2.
2. For each value of Zeq, express Tem as a function of s (or ωm= ωs(1-s)) for various
values of s. torque-speed control point (slides 66-67):
3 pVth2 ( Rr  Req ) / ss
 Rr  Req 
 
Tem  3 I 
2
2
s 
s
Rr  Req  
X eq 
 
 Rth 
   X th  X r 

s
s

 

p
2
r
84
Torque-slip characteristic for DFIG
The sign of Req and Xeq
are for rotor current
direction defined out of
the rotor. These signs
reverse for rotor
current direction into
the rotor as we have
done.
85
Efficiency
Consider our HW assignment, at a speed of 1750 rpm and unity power factor.
Compute the efficiency of the DFIG.
Is
Rs
jωsLσs
Ir Rr(1-s)/s
3
3
3
3
Vs
jωsLσr
Im
Vm 3
3 Zm=jωsLm
Req(1-s)/s
Vr/s
Req
3 jωs Leq=jXeq/s
3
At 1750, the slip is s=(1500-1750)/1500=-0.1667
From your homework, you should compute that Is=1068.2 amperes
Ir=1125.6 amperes, Req=0.05375 ohms, Xeq=0.02751 ohms.
The mechanical power supplied to the generator
Pmech  3I r2 ( Req  Rr )(1  s) / s
 3(1125.6) 2 (0.05375 0.00263)(1  0.01667) /(0.1667)
 1500kW
86
Efficiency
Consider our HW assignment, at a speed of 1750 rpm and unity power factor.
Compute the efficiency of the DFIG.
Is
Rs
jωsLσs
jωsLσr
3
3
3
3
Im
Vm 3
3 Zm=jωsLm
Vs
Ir Rr(1-s)/s
Req(1-s)/s
Vr/s
Req
3 jωs Leq=jXeq/s
3
From your homework, you should compute that Is=1068.2 amperes
Ir=1125.6 amperes, Req=-0.05375 ohms, Xeq=-0.02751 ohms.
The rotor power is
Pr  3I r2 Req
 3(1125.6) 2 (0.05375)
 204.29kW
This power is negative (because Req is negative); it is supersynchronous, therefore it
is flowing out of the rotor to the RSC.
87
Efficiency
Consider our HW assignment, at a speed of 1750 rpm and unity power factor.
Compute the efficiency of the DFIG.
Is
Rs
jωsLσs
jωsLσr
3
3
3
3
Im
Vm 3
3 Zm=jωsLm
Vs
Ir Rr(1-s)/s
Req(1-s)/s
Vr/s
Req
3 jωs Leq=jXeq/s
3
From your homework, you should compute that Is=1068.2 amperes
Ir=1125.6 amperes, Req=0.05375 ohms, Xeq=0.02751 ohms.
The rotor and stator winding losses are
Plosses ,r  3I r2 Rr
Plosses , s  3I s2 Rs
 3(1125.6) 2 (0.00263)
 10.0kW
 3(1068.2) 2 (0.00265)
 9.07kW
88
Efficiency
Consider our HW assignment, at a speed of 1750 rpm and unity power factor.
Compute the efficiency of the DFIG.
Is
Rs
jωsLσs
Ir Rr(1-s)/s
3
3
3
3
Vs
jωsLσr
Im
Vm 3
3 Zm=jωsLm
Req(1-s)/s
Vr/s
Req
3 jωs Leq=jXeq/s
3
From your homework, you should compute that Is=1068.2 amperes
Ir=1125.6 amperes, Req=0.05375 ohms, Xeq=0.02751 ohms.
The stator active power is
Ps  3Vs I s coss 
690
1068.2  cos(180)  1276.64kW
3
89
Efficiency
Consider our HW assignment, at a speed of 1750 rpm and unity power factor.
Compute the efficiency of the DFIG.
Is
Rs
jωsLσs
jωsLσr
3
3
3
3
Vs
Ir Rr(1-s)/s
Im
Vm 3
3 Zm=jωsLm
Req(1-s)/s
Vr/s
Req
3 jωs Leq=jXeq/s
3
From your homework, you should compute that Is=1068.2 amperes
Ir=1125.6 amperes, Req=0.05375 ohms, Xeq=0.02751 ohms.
The total power delivered to the grid is
Pg | Ps |  | Pr | 1276.64  204.29  1480.93
The difference between Pm and Pg is the losses on the stator and rotor windings:
| Pm |  | Pg | Plosses ,r  Plosses ,s  15001480.93  19.07
Efficiency is:

Pg
Pm

1480.93
 98.7%
1500
90
DFIG for non-unity power factor
“FERC 661-A [1] specifies that large wind farms must maintain a power factor within the
range of 0.95 leading to 0.95 lagging, measured at the POI as defined in the Large
Generator Interconnect Agreement (LGIA) if the Transmission Provider shows, in the
system impact study that they are needed to ensure the safety or reliability of the
transmission system..”
[1] Order for Wind Energy, Order No. 661-A, 18 CFR Part 35 (December 12, 2005). See also Interconnection for Wind Energy,
Order No. 661, 70 FR 34993 (June 16, 2005), FERC Stats. & Regs. ¶ 31,186 (2005) (Final Rule); see also Order Granting
Extension of Effective Date and Extending Compliance Date, 70 FR 47093 (Aug. 12, 2005), 112 FERC ¶ 61,173 (2005).
“The Electrical System Operator (IESO) of Ontario essentially requires reactive power
capabilities for large wind farms that are equivalent to that for synchronous generators,
taking into consideration an equivalent impedance between the generator terminals
and the POI [2]. The requirements include:… Supplying full active power continuously
while operating at a generator terminal voltage ranging from 0.95 pu to 1.05 pu of the
generator’s rated terminal voltage.”
“The Alberta Electric System Operator’s requirements [4] include: The wind farm’s
continuous reactive capability shall meet or exceed 0.9 power factor (pf) lagging to 0.95
pf leading at the collector bus based on the wind farm aggregated MW output.”
E. Camm and C. Edwards, “Reactive Compensation Systems for Large Wind Farms,” IEEE Transmission and Distribution
Conference and Exposition, 2008.
91
DFIG for non-unity power factor
E. Camm and C. Edwards, “Reactive Compensation
Systems for Large Wind Farms,” IEEE Transmission
and Distribution Conference and Exposition, 2008.
92
DFIG for non-unity power factor
“Along with the evolution of wind turbine technology, technical standards of wind
generation interconnections become more restrictive. For example, unity power factor
has been required for wind generation interconnections in many utilities or control
areas in earlier years. Recently, the more restrict requirement with 0.95 lead and lag
power factor has been under discussion since the DFIG and full converter wind turbine
technology has become mainstream of wind generation interconnection requests.”
I. Green and Y. Zhang, “California ISO experience with wind farm
modeling,” IEEE Power and Energy Society General Meeting, 2011.
93
DFIG for non-unity power factor
Rs
Is
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vs
Vm 3
3 jωsLm
Vr/s
Req
3 jωs Leq=jXeq/s
3
Define: φ as power factor angle: -180<φ<180
Ps
Ps  3Vs I s cos  I s 
3Vs cos
Identify the current phasor as
I s  I s (cos  j sin  )
Therefore: I s 
Ps is negative for gen; then cosφ is also negative;
Ps is positive for motor; then cosφ is also positive;
so Is is always positive.
Ps
P
sin 
(cos  j sin  )  s (1  j
)
3Vs cos
3Vs
cos
Recalling sin 2   cos 2   1  sin   1  cos 2  , we may write
1  cos 2 
Ps
Is 
(1  j
)
3Vs
cos 
We have just made the numerator
positive for all values of φ.
94
DFIG for non-unity power factor
Rs
Is
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vs
Vm 3
3 jωsLm
Vr/s
Req
3 jωs Leq=jXeq/s
3
1  cos 2 
Ps
Is 
(1  j
)
3Vs
cos 
The sign of Ps determines the sign of the real part of the current.
Ps is negative if machine is in generating mode (supplying real power).
In this case, cosφ is negative because φ is in quadrant 2 or 3.
If machine is supplying Q, then sign of Qs should be negative, sign of Im{Is*} should
be negative, and therefore sign of Im{Is} should be positive. Given cos φ is negative:
1  cos 2 
Ps
Is 
(1  j
)
3Vs
cos 
If machine is absorbing Q, then sign of Qs should be positive, sign of Im{Is*} should
be positive, and therefore sign of Im{Is} should be negative. Given cos φ is negative:
1  cos 2 
Ps
Is 
(1  j
)
3Vs
cos 
95
DFIG for non-unity power factor
Rs
Is
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vs
Vm 3
3 jωsLm
Define the magnetizing current factor:
From the circuit, KCL requires:
Vr/s
Km 
Ir  Im Is
Im
I s ,rated
But the magnetizing current is entirely imaginary: I m
Req
3 jωs Leq=jXeq/s
3
 I m  K m I s ,rated
 jI m or I m  jKm I s,rated
Substitution of the Im expression into the rotor current expression yields:
I r  jKm I s,rated  I s
1  cos 2 
Ps
(1  j
) . Substituting into I :
If the machine is absorbing Q, then I s 
r
3Vs
cos 
I r  jK m I s ,rated
1  cos 2 
Ps

(1  j
)
3Vs
cos 
96
DFIG for non-unity power factor
Rs
Is
jωsLσs
Rr
jωsLσr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vm 3
3 jωsLm
Vs
I r  jK m I s ,rated
Vr/s
Req
3 jωs Leq=jXeq/s
3
1  cos 2 
Ps

(1  j
)
3Vs
cos 
Assume the machine is operated at rated power, Ps,rated, and recall
I r  jK m I s ,rated 
Ps ,rated
3V
1  cos 2 
(1  j
)
cos 
P
s
Ps
s , rated
I


I

Recall from slide 94: s
s , rated
3Vs cos
3Vs cos
and the substitute into previous expression :
I r  jK m
Ps ,rated
3Vs cos 

Ps ,rated
3Vs
1  cos 2 
(1  j
)
cos 
Factor out the Ps,rated/3Vs….(next slide):
97
DFIG for non-unity power factor
Rs
Is
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vm 3
3 jωsLm
Vs
I r  jK m
Ps ,rated
3Vs cos 

Ps ,rated
3Vs
Vr/s
Req
3 jωs Leq=jXeq/s
3
1  cos 2 
(1  j
)
cos 
 Ps ,rated   jK m
1  cos2  
 (1  j
)

Factor out the -Ps,rated/3Vs  I r 
3Vs  cos
cos

Combine terms with “j”
Simplify


2
K

 Ps ,rated 
1

cos

m

1  j
Ir 

 cos

3Vs 
cos




 K  1  cos2  
 Ps ,rated 

1  j m
Ir 


3Vs 
cos



98
DFIG for non-unity power factor
Rs
Is
jωsLσs
jωsLσr
Rr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vs
Vm 3
3 jωsLm
Vr/s
Req
3 jωs Leq=jXeq/s
3
So this is for absorbing
(underexcited operation)
 Ps ,rated 
1 
Ir 
3Vs 

 K  1  cos2 
j m

cos





If we repeat the exercise for
supplying (overexcited operation),
we will obtain this:
 Ps ,rated 
1 
Ir 
3Vs 

 K  1  cos2 
j m

cos





The difference in sign on the square root term indicates higher rotor current is required
for overexcited operation than for underexcited operation. No big surprise there!
And so the rotor winding should be rated for the overexcited operation, at
rated stator active power output. This would be…. (next slide)
99
DFIG for non-unity power factor
Rs
Is
jωsLσs
Rr
jωsLσr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vs
Vm 3
3 jωsLm
Rotor current for rated stator
active power and reactive power
generation
Vr/s
Ir 
K rs 
Ir
I s ,rated
 K  1  cos2 
1  m

cos

Ps ,rated
3Vs
It is interesting to see the relative magnitude
between Ir and Is. Again, from slide 94:
Req
3 jωs Leq=jXeq/s
3
Is 
2
P
Ps
 I s ,rated  s ,rated
3Vs cos
3Vs cos
 K  1  cos  
3Vs cos Ps ,rated


1  m


Ps ,rated
3Vs
cos


2




2
 K  1  cos2 
 K rs  cos 1   m

cos





2
100
DFIG for non-unity power factor
Rs
Is
jωsLσs
Rr
jωsLσr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vs
Vm 3
3 jωsLm
Rotor current for rated stator
active power and reactive power
generation
Rotor current for rated stator
active power and stator unity
power factor
Ratio of rotor current required for
a given stator power factor when
supplying Q and that required for
unity stator power factor, all at
rated stator active power
Vr/s
Ir 
I
Ps ,rated
3Vs
pf 1.0
r

Req
3 jωs Leq=jXeq/s
3
 K  1  cos2 
1  m

cos

Ps ,rated
3Vs
I rpf 1.0
2
1  K m2
 K  1  cos  

1  m


cos




1  K m2
2
Ir




2
101
DFIG for non-unity power factor
Rs
Is
jωsLσs
Rr
jωsLσr
Ir Rr(1-s)/s
3
3
3
3
Req(1-s)/s
Im
Vs
Vm 3
3 jωsLm
Krs factor for a given stator power
factor at rated stator active power.
Vr/s
 K  1  cos2 
K rs  cos 1   m

cos





2
K rspf 1.0  1  K m2
Krs factor for unity stator power
factor at rated stator active power.
Ratio of Krs factor for a given
power factor to Krs factor for unity
power factor, for rated stator
active power.
Req
3 jωs Leq=jXeq/s
3
 K  1  cos  

1  m


cos



 cos
1  K m2
2
K rs
K rspf 1.0
2
102
DFIG for non-unity power factor
Ir
Krs factor for a given stator power
K

factor at rated stator active power. rs I
s , rated
Km 
Im
I s ,rated
 K  1  cos2 
 cos 1   m

cos





2
 I m  K m I s ,rated
The prime notation on the Krs
at the top of the graph
indicates the values have been
referred to the rotor for an
“a” of about 0.3.
103
DFIG for non-unity power factor
 K  1  cos  

1  m


cos



 cos
1  K m2
2
Ratio of Krs factor for a given
power factor to Krs factor for unity
power factor, for rated stator
active power.
K rs 
104
Ir
I s ,rated
K rs
K rspf 1.0
2