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Double-fed electric machines –
steady state analysis
Set 1
J. McCalley
Four configurations
We will study
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 inject a voltage signal via the converter to control it.
3
Basic concepts
 s  m
slip  s 
;
s
ns  nm

;
ns
We can manipulate to get:
 s  377 rad/sec ;
# of pole
pairs
m  p m
60 f s
ns 
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:
Two 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Supersynchronous operation
4
Per-phase steady-state model
The underlining
of a parameter
indicates that it
is a phasor.
X r
Rr I r
Ers
V s  E s  ( Rs  jXs ) I s
STATOR VOLTAGE EQUATION:
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
We neglect magnetizing
inductance Lm in this
figure, for simplicity,
but it will be added back
V r in later. It is modeled in
parallel with Es.
at fs
This is major departure
from SCIG where V’’r=0.
These quantities
are referred to
stator side.
X s =stator leakage reactance
ROTOR VOLTAGE EQUATION:
V r  Ers  ( Rr jX r ) I r at fr These quantities
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 fr
Rr =rotor resistance
X r=rotor leakage reactance=r Lr
are referred to
rotor side, as
indicated by
double-prime
notation.
Source: G. Abad, J. Lopez, M. Rodriguez, L. Marroyo, & G Iwanski, “Doubly fed induction machine: modeling and control for wind energy generation,” Wiley 2011.
5
Referring quantities
We know (see *) the coils on the armature and rotor will see a flux linkage proportional
to the magnetizing flux φm and the number of turns N. Application of Faraday’s law
results in the stator emf and the induced rotor voltage being expressed as:
Ks, Kr: stator and rotor winding factors, respectively,
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:
E s  2 K s N s f s  m
Ks Ns f s
Es
Ers
Es
m 




E rs Kr Nr f r
2 K s N s f s
2 K r N r f r
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 rs s
a
Nr
(A)
* A. Fitzgerald, C. Kingsley, and A. Kusko, “Electric Machinery,” 3rd edition, McGraw-Hill, 1971, pg. 148.
6
Referring quantities
E rs 
We just derived that:
sE s
a
(A)
At a locked rotor condition (s=1), the device is simply a static transformer,
and we have:
Es
E rs 
a
 E s  aE rs
This tells us (for locked rotor) if 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
Es
3
3
Ir
Rotor quantities are referred
to the stator-side, indicated
by unprimed quantities (we
could also use single prime
quantities as notation here).
Vr
This is locked rotor condition
(s=1), therefore ωr=ωs and
Ers=Es
L r  Lr a 2
3
3
Is
Rr  Rra 2
3
3 Ers
Rr
We can account for other
slip conditions using ωr=sωs
and from (A), Ers=aE’’rs=sEs.
7
Referring quantities
Rs
Is
jsωsLσr
jωsLσs
3
3
V s  E s  ( Rs  jXs ) I s
or using the nomenclature
of this figure
3
3
Es
Vs
Rr
Ir
3
3
From slide 5, the statorside voltage equation
(referred to stator) is:
3
3 Ers=sEs
Vr
V s  E s  ( Rs  js Ls )I s
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. At the same time,
we model the magnetizing
inductance jωsLm.
8
Referring quantities
Is
Rs
jωsLσs
jωsLσr
Es
Ir
3
3
3
3
Vs
Rr/s
3
3 jωsLm
Vr/s
9
Power relations
Is
Rs
jωsLσs
jωsLσr
Es
Ir
3
3
3
3
Vs
Rr/s
3
3 jωsLm
Vr/s
Modify the above circuit to separate slip-dependent and slip-independent 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….
10
Power relations
Veq=Vr(1-s)/s
Is
Rs
jωsLσs
jωsLσr
Ir
+-
3
3
3
3
Es
Vs
Rr
3
3 jωsLm
Req=
Rr(1-s)/s
Vr
The mechanical power out of the machine is the power into the machine less the
stator and rotor copper losses (as indicated in the figure at the bottom), i.e.,
Pmech  Ps  Pr  Ploss, s  Ploss,r
Power balance relation:
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.




Pmech  3Re V s I s  3Re V r I r  3 I s Rs  3 I r Rr
*
Here, the first two terms are given by:
   3V I cos(
P  3Re V I   3V I cos(
Ps  3Re V s I
r
r
*
s
*
r
s s
r r
v,s
v ,r
 i , s )
 i , r )
*
2
2
Ps
Ps>0 it receives power via stator
Ps<0 it delivers power via stator
Pr>0 it receives power via rotor
Pr<0 it delivers power via rotor
Pmech>0 is motor; <0 is gen
Pr
DFIM
Ploss,s+Ploss,r
Pmech
11
Power relations
Veq=Vr(1-s)/s
Rs
Is
jωsLσs
Ir
3
3
3
3
+-
3
3 jωsLm
Es
Vs
Rr
jωsLσr
Req=
Rr(1-s)/s
Vr
The mechanical power out of the machine is the power into the machine less the
stator and rotor copper losses, i.e.,
Pmech  Ps  Pr  Ploss, s  Ploss,r
Power balance relation:
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.




Pmech  3Re V s I s  3Re V r I r  3 I s Rs  3 I r Rr
*
*
2
2
From a cct analysis point of view, if we include remaining terms, we should get 0, i.e.,

  
 3V I   3V I   3 I


3 V s I s  3 V r I r  3 I s Rs  3 I r Rr  3 I r2  Req   3Re V eq I r  0
*
 Pmech
*
s
*
s
2
r
*
r
2
2
s
*

Rs  3 I r Rr  3 I r2  Req   3Re V eq I r
2
*

This shows the slip-dependent terms are those responsible for mechanical power.
12
Power relations
Rs
Is
jωsLσs
jωsLσr
3
3
3
3
Ps
Pmech
Ploss,s
Es
Vs
3
3 jωsLm
Veq=
Vr(1-s)/s I
Rr
r
+Ploss,r R =
eq
Rr(1-s)/s
Vr
Introduction of
voltage source
in rotor circuit
provides ability
to affect Pmech.
Pr
Ps
Ps
Pr
DFIM
Pmech
Pr
Ploss,s+Ploss,r
Pmech
13
Power relations
Is
Rs
jωsLσs
jωsLσr
3
3
3
3
Es
Vs

Pmech  3 I r2  Req   3Re V eq I r
*

Pmech
3
3 jωsLm
Veq=
Vr(1-s)/s I
Rr
r
+Req=
Rr(1-s)/s
Vr
With Pmech>0 (see below), motor convention
implies a positive value of Req represents a positive
mechanical load (as it did for SCIM), so Req term
should be positive on RHS. For Veq term, rotor
current (Ir) direction is out of positive side of
voltage source; therefore voltage source supplies
power to circuit and should be subtracted from the
mechanical loading represented by the Req term.
If Pmech>0the machine is delivering power through the shaft: MOTOR!
If Pmech<0the machine is receiving power through the shaft: GEN!
Terms are always same sign;
If 0<s<1(sub)Req term is pos; Veq term is pos.
but Pmech takes their difference
If 0>s>-1(super)Req term is neg; Veq term is neg.
So we observe that in either mode (sub or super), machine may motor or
generate, depending on relative magnitudes of Req and Veq terms.
14
Power relations
Rs
Is
jωsLσs
3
3
3
3

Veq=
Vr(1-s)/s I
Rr
r
+Req=
Rr(1-s)/s
Vr
jωsLσr
3
3 jωsLm
Es
Vs

Pmech  3 I r2  Req   3Re V eq I r 
*
Pmech
Is Pslip equal to the power associated
  1 s  * 
 R (1  s) 
with the elements in the dotted box?
 3 I r2  r

3Re
V r 
Ir

s


  s  
No! Pmech is equal to the power
associated with the elements in
 1 s  2
 1 s 
 1 s 
*
*
2
 3
I
R

3
Re
V
I

3
I
R

3Re
V
I




r r
r r
r
r
 r r




the dotted box, and we see from
 s 
 s 
 s 
the derived equation that Pmech
 1 s 
 1 s 
 1 s 

and Pslip differ by –(1-s)/s.
  Ploss ,r  Pr    
  Pr  Ploss ,r    
  Pslip 
s
s
s






If the power at the rotor terminals
1 s 
does not equal Pmech, what
Pmech  
 Pslip
makes up the difference?
 s 
Power from the stator
What is Pslip?
that crosses the airgap!
It is Pr-Ploss,r = what is provided at the rotor terminals not lost
We will return to this!
in the winding losses. Where does it go? To or from the shaft!
15

 

A first torque expression
Rs
Is
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

 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 4: s 
s 
and
m   s (1  s)  1  s 
Therefore:

m
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 ,r  i ,r 
16
A second (equivalent) torque expression
Is
Rs
jωsLσs
jωsLσr
3
3
3
3
Stator power:
3
3 jωsLm
Es
Vs
 
Ps  3 Re V s I
Veq=
Vr(1-s)/s I
Rr
r
+Req=
Rr(1-s)/s
Vr
Idea here is to compute Pmech from:
*
s
Pmech  Ps  Pr  Ploss, s  Ploss,r
using stator & rotor voltage eqts.
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
17
A second (equivalent) torque expression
Rs
Is
jωsLσs
jωsLσr
3
3
3
3
Rotor power:
3
3 jωsLm
Es
Vs

Pr  3 Re V r I r
*
Veq=
Vr(1-s)/s I
Rr
r
+Req=
Rr(1-s)/s
Vr

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

 3 ReI
P  3 Re V I

 3 Re R I
R
 js L
  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
18
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
*
*

19
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 (applying to first term only):

 
 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
20
Two equivalent torque expressions
Tem 
3 p I r2 Rr
r

3p
r
Vr I r cosv  i 
Torque expression #1: Need ωr, 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….
21
Additional equivalent torque expressions
This will be very similar to HW3, problem 1.
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: Recall L=λ/ieach self inductance is comprised of mutual and leakage
The flux seen by the stator winding from the
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
stator current is the same as the flux seen by
the rotor winding from the stator current plus
the flux from the stator current that leaks.
 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
22
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




Lm
*
 3p
Im I s  r  Lm I s2
Lr
Purely
real
 3p
 
Lm
*
Im I s  r
Lr


Purely
real
23
Airgap and slip power
On slides 17 and 18, 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
*



Pr  3Rr I r2  3 Re jss Lm I s I r
This quantity is power flowing from
stator terminals to rotor shaft (neg for
generator operation). In other words, it
is power across the airgap (rotor losses
accounted for in slip power). Therefore:

Pairgap  Ps  3R I  3 Re js Lm I r I
2
s s
*
*
s

*

This quantity is power transferred
from the grid to the rotor shaft
through the converter (neg 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
2
s s
 3 Im L
s
m
Ir
I 
*
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
24
Airgap and slip power
Pslip   sPairgap
Slip power (the power flowing to the shaft from the
rotor circuit) is –s times the airgap power (the power
flowing to the shaft from the stator circuit). Therefore:
•
•
In subsynchronous mode (0<s<1), Pairgap and Pslip have
opposite signs (only one of them is into the rotor shaft).
Because |s| is always small, |Pslip|=|s||Pairgap| is much smaller
than |Pairgap|. So the sign of Pairgap determines the operation
(Pairgap>0 is motor, Pairgap<0 is generator).
In supersynchronous mode (-1<s<0), Pairgap and Pslip have
same signs - either both are positive and therefore into the
rotor shaft (motor operation) or both are negative and
therefore out of the rotor shaft (generator operation).
25
Torque from 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 
1
Substituting: Pslip   sPairgap  Pairgap 
Pslip
s
(**)
Aside: From (*) and (**), we may derive the below
relation, which was also derived on slide 13.
1 s 
Pmech  
Pslip 
s
26
Tem 
r
s
s
p
Pslip
s s
s  p
T

Pslip
 em
r  s
p
T

Pslip
 em
r
Approximate relations between active powers
On slides 17 and 18, 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
Recall: 1  s  
s

*


Pr   sPs
Pmech  Ps  Pr  Ploss, s  Ploss,r
Pmech  Ps  Pr
m
Ps
 Pmech 

p m
p
ps
Tem  Pmech

Ps

Ps
m s m  s
*
27
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 
P
 P  3R I  3 Rej L I I  1  s  

P  P  3R I  3 Rejs L I I 
1 s
Approximate


Ps  3Rs I s2  3 Re js Lm I r I s
2
r r
r
airgap
slip
s
2
s s
s
r
m
*
r
s
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
1 s 
Pmech  
Pslip 
s


p
Tem 
Pairgap
s
Tem 
p
r
Pslip
r
r
m
m
*
s
s
*
r
s
s
s
Tem  Pmech
*


I 
Pairgap  Ps  3 Re js Lm I r I s
m
m 


Ps  3 Re js Lm I r I s
*
Pr  3 Re jss Lm I s I r
*
r
p
m

Pslip  Pr  3 Re jss Lm I s
*
*
r
Pr   sPs
Pmech  Ps  Pr
Pmech  (1  s) Ps

Pmech  m Ps
s
1 s 
Pmech  
Pr 
 s 
p
Tem 
Ps
s
Tem 
p
r
Pr
28
Active power relations - summary
Exact
Approximate
Pslip   sPairgap
Pr   sPs
Pmech  1  s Pairgap
Pmech  (1  s) Ps
 Pairgap  sPairgap  Pairgap  Pslip
Mechanical power is the airgap power
• less the slip power, in subsynchronous
mode (when Pairgap and Pslip are
opposite sign);
• plus the slip power, in
supersynchronous mode (when Pairgap
and Pslip are same sign).
 Ps  sPs  Ps  Pr
Mechanical power is the stator power
• less the rotor power, in
subsynchronous mode (when Ps and
Pr are opposite sign);
• plus the rotor power, in
supersynchronous mode (when Ps and
Pr are same sign).
29
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).
30
Generator modes
Yellow is supersynchronism; red is subsynchronism, bold box is generator operation.
Mode
Slip and speed
Pmech
Ps
Pr
1. Motor
(Tem>0)
s<0, ωm>ωs
(suprsynchrnsm)
>0 (shft delivers
mech pwr)
>0 (shft receives >0 (shft receives
power frm stator) power frm rotor)
2. Generator
(Tem<0)
s<0, ωm>ωs
(suprsynchrnsm)
<0 (shft receives
mech pwr)
<0 (shft delivers
power to stator)
<0 (shft delivers
power to rotor)
3. Generator
(Tem<0)
s>0, ωm<ωs
(subsynchrnsm)
<0 (shft receives
mech pwr)
<0 (shft delivers
power to stator)
>0 (shft receives
power frm rotor)
4. Motor
(Tem>0)
s>0, ωm<ωs
(subsynchrnsm)
>0 (shft delivers
mech pwr)
>0 (shft receives <0 (shft delivers
power frm stator) power to rotor)
Operation (motor/generator) is determined by the sign of Tem or equivalently, the sign of
Ps≈Pairgap (see slide 25).
For each mode, we may use the two relations to track the signs of Ps and Pr from the
signs of Tem and s. For example, for mode 2, Tem<0Ps<0; Ps<0, s<0 Pr<0
Ps 
s
p
Tem
Pr   sPs
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.
31
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.
32
Generator modes
What is the fundamental
difference between DFIG &
SCIG that allows DFIG to
operate as a gen in both subsynch & super-synch?
Control over rotor power!
Mode 2
Pm= Pmech
m   s
Mode 3
m   s
These figures show actual flow
direction for generator operation.
They also neglect losses.
Unlike the SCIG, we can
make the rotor cct inject or
withdraw as much or as little
power as we like, independent of conditions. Thus,
• when power supplied by
turbine is small, inject
power via rotor into shaft –
subsynchronous!
• When power supplied by
turbine is large, withdraw
power via rotor from the
shaft – supersynchronous!
33
This figure assumes proper
sign convention (power
flowing to the rotor or into
the stator is positive).
A question on rating
Pmech  Ps  Pr Pgrid  Ps  Pr
P
 sPmech
Ps  mech
Pr 
1 s
1 s
Pgrid
Ps
Without losses
Pairgap
Assume an operating condition
such that Pmech=PWTrating. Then
PW Trating
1 s
Pr 
Pr
Pmech
Pmech  Pgrid  PWTrating
Ps 
Pslip
 sPW Trating
1 s
For example, consider Pmech=PWTrating=-2 MW. In supersynchronous mode, with s=-0.3,
2
Ps 
 1.5385MW.
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. The maximum power in subsynchronous mode at 30% slip is:
Pmech  Ps (1  s)  1.5385(1  0.3)  1.0769MW
34
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
35
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
d 



   Vt 
 Justificat ion : v 
dt
t


• Three-phase
base
base base
power base:
S
 3V
I
• Base inductance:
base
Lbase 
I base
• Base speed:
base
m,base 
p
• Base torque:
Sbase
Tbase 
m,base
36
Per-unitization – stator quantities
Once all base quantities are obtained, then per-unitization is easy:
• Stator voltage in pu:
• Stator current in pu:
• Stator flux in pu:
Vs
Vbase
Is
I s , pu 
I base
V s , pu 
s
 s , pu 
base
 
*
s s
• Stator active power in pu:
Re V I
P s , pu  3
Sbase
• Stator reactive power in pu:
Im V s I s
Q s , pu  3
Sbase
 
*
As usual, only the magnitude is transformed (angle remains unchanged).
37
Per-unitization – rotor quantities
• Rotor voltage in pu:
• Rotor current in pu:
• Rotor flux in pu:
• Rotor active power in pu:
• Rotor reactive power in pu:
Vr
Vbase
Ir
I r , pu 
I base
V r , pu 
r
 r , pu 
base
P r , pu  3
Q r , pu
For rotor quantities, we
use the same base
quantities as for the stator
quantities (with actual
rotor quantities referred to
the stator side).
 
*
r r
Re V I
Sbase
Im V I 
3
*
r r
Sbase
As usual, only the magnitude is transformed (angle remains unchanged).
38
Per-unitization – torque, speed, R, L
• Torque in pu:
• Speed in pu:
• Resistances in pu:
• Inductance in pu:
Tem
Tbase
m
m, pu 
m,base
Tem , pu 
Rr
r pu 
Z base
l pu 
L
Lbase
For rotor quantities, we
use the same base
quantities as for the stator
quantities (with actual
rotor quantities referred to
the stator side).
Note that the
resistances and
inductances when
expressed in pu are
lower case.
39
Voltage equations expressed in per unit
From slides 17, 18, 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 22, 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
Replace voltages, currents, flux linkages, w/ product of pu value & base quantity:
V s , puVbase  I s , pu I basers Z base  j s , pubase  s , pubase
V r , puVbase  I r , pu I baserr Z base  js s , pubase  r , pubase
Divide by base quantities (note from slide 36 that Vbase=ZbaseIbase=λbaseωbase
and so this last step is a division by same value)
V s , pu  I s , pu rs  j s , pu  s , pu
V r , pu  I r , pu rr  js s , pu  r , pu
V s , pu  I s , pu rs  j  s , pu
V r , pu  I r , pu rr  js r , pu
ωs,pu=1
40
Voltage equations expressed in per unit
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:
 s , pu  ls I s , pu  lm I r , pu
 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
*
Tbase 
Sbase 3 Vbase I base


m,base m,base / p
Tem, pu
 
 
as follows:


3 p Im  r I *r 3 p Im  r I *r
r
I *r 


 Im
 Im  r , pu I *r , pu

3 Vbase I base
Tbase
Vbase / m,base I 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 )
Notation Police!!!! .
I have used
θv,s, θi,s, θv,r, and θi,r previously (see slides 11 and 16).
Here I am using
γv, γi,, φv, and φi, respectively.
41
Homework #3 (also posted to website)
Homework #3: This homework is due Monday September 26, 2016. It is
also posted to the website.
A. Using previous relations provided in the “Set 1” 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 30 in the “Set 1” slides.
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.
42
Homework
Per-unit
values
u (or a)
Rs
Lσs
Lm
R’r
Lσr
Rr
Lσr
Ls
Lr
Vbase
Ibase
rs
lσs
lm
rr
lσr
43