ECE 422 Power System Operations & Planning
2 – Synchronous Machine Modeling
Spring 2015
Instructor: Kai Sun
1
Outline
•Synchronous Machine Modeling
•Simplified Models for Stability Studies
•Materials
– Saadat’s Chapter 8
– Kundur’s Chapters 3‐5
2
Synchronous Generators
Cylindrical/round rotor
Salient-pole rotor
Field winding
Armature winding
Stator
Field current
3
Types of Rotors
• Salient pole rotors
– Concentrated windings on poles and non‐uniform air gap
– Short axial length and large diameter
– Hydraulic turbines operated at low speeds (large number of poles)
– Have damper/amortisseur windings to help damp out speed oscillations
16 poles salient-pole rotor (12 MW)
• Round rotors
– 70% of large synchronous generators (150~1500MVA)
– Distributed winding and uniform air gap
– Large axial length and small diameter to limit the centrifugal forces
– Steam and gas turbines, operated at high speeds, typically 3600 or 1800rpm (2 or 4‐pole)
– No special damper windings but eddy in the solid steal rotor gives damping effects
Round rotor generator under construction
(Source: http://emadrlc.blogspot.com)
4
d
Under Steady‐State Conditions
• MMFs: Fsr =Fr+Fs  Fr=Fsr -Fs
Esr=Ea+Ear  Ea=Esr -Ear
Esr =V+(Ra+jXl)Ia
• For a round rotor, define
Ear =-jXarIa
Fr
r
n
Fsr
N
• EMFs:

 ea
S
Axis of
coil a
(reference)
Fs
m
Ea  V  ( Ra  jX   jX ar ) I a
 V  ( Ra  jX s ) I a
Fr
where Xs=Xl+Xar (synchronous reactance)
Fsr

Ea
Ea
Esr Ear
Load
Fs
5
Stator and Rotor Windings
Armature windings: • a‐a’, b‐b’ and c‐c’ windings
Rotor windings:
• Field windings
– Field winding F‐F’ produces a flux on the d‐axis. • Damper windings
– Two damper windings D‐D’ and Q‐Q’ respectively on d‐ and q‐axes
– For a round‐rotor machine, consider a second damper winding G‐G’ on the q‐
axis (two windings on each axis)
Total number of windings:
• Salient pole: 3+3 (discussed here)
• Round‐rotor: 3+4
•
Assume a synchronously rotating reference frame
at speed r (along with the axis of a-a’ at t=0)
•
Assume  to be the displacement of q-axis from
the reference axis
  r t   

2
Reference
axis
Direct or
d axis
r
Quadrature
or q axis
Note: ANSI/IEEE standard 100-1977 defines the
quadrature (q) axis to lead the d-axis by 900
6
Voltage and Flux Equations (Salient‐pole machine)
•
Model windings as a group of magnetically coupled circuits with inductances depending on 
RF
 ea
  Ra
e
 
b

 0
 ec
 0


e
 F
 0
e  0   0
 D
 
 eQ  0   0
0
Rb
0
0
0
0
0
Rc
0
0
0
0
0
RF
0
0
0
0
0
RD
0
0
0
0
 ψ abc   L SS
ψ   
 FDQ   L RS
 ia 
 i 
 b
 ic  d
 
iF  dt
i 
D 
iQ 
 a 
 
 b
 c 
 
 F 
 
 D
 Q 
eF
lab
lac
laF
laD
lb b
l cb
lb c
l cc
lb F
l cF
lb D
l cD
lFb
lFc
lFF
lFD
lDb
lDc
lDF
lDD
lQ b
lQ c
lQ F
lQ D
L SR   i abc 


L RR  i FDQ 
laQ 
l b Q 
l cQ 

lFQ 
lDQ 

l Q Q 
  ia 
i 
 b
  ic 


i
 F 
i 
 D 
 iQ 
Rc
F
RD
eD=0
c
b
D
eQ=0
ec
a
Ra
Q
0   i abc  d  ψ abc 

R FDQ  i FDQ  dt  ψ FDQ 
e abc   R abc
e    0
 FDQ  
 a   l a a
  
 b   lb a
 c   l ca

 

 F   lFa
   l
 D   Da
 Q   l Q a








RQ 
0
0
0
0
0
ea
Rb
eb
RQ
•
Stator self‐inductances (laa, lbb, lcc)
•
Stator mutual inductances (lab, lbc, lac)
•
Stator‐to‐rotor mutual inductances (laF, lbD, laQ)
•
Rotor self‐inductances (lFF, lDD, lQQ)
•
Rotor mutual inductances (lFD, lDQ, lFQ)
A main objective of synchronous machine modeling is to find
constants for simplification of voltage and flux equations
7
 a   l a a
  
 b   lb a
 c   l ca

 
 F   l F a
   l
 D   Da
 Q   l Q a
lab
lac
laF
laD
lb b
l cb
lb c
l cc
lb F
l cF
lb D
l cD
lFb
lFc
lFF
lFD
lDb
lDc
lDF
lDD
lQ b
lQ c
lQ F
lQ D
laQ 
l b Q 
l cQ 

lFQ 
lDQ 

l Q Q 
  ia 
i 
 b
  ic 


 iF 
i 
 D 
 iQ 
The matrix is symmetric because the mutual inductance by definition is the flux
linkage with one winding per unit current in the other winding, i.e.
l12 ≝ N1  / i2 = N1 N2  / (N2i2)= N1 N2  / MMF2= N1N2 P12 = N12P12 = l21
P12 ~ permeance of the mutual flux path
N12 ~ effective coupling between two windings
A salient pole machine has significantly different permeances in d and q axes:
• P12 between stator windings (e.g. Paa and Pab) is a function of the rotor
position  and reaches the maximum twice in 0o~360o
P12  P0+P2cos2(+) 
l12 =l0+l2cos2(+)
Assume a sinusoidal space
distribution of each MMF
• P12 between stator and rotor windings (e.g. PaF) may be treated as
constant but effective N12 is a function of the rotor position  and reaches
the maximum once in 0o~360o
N12  N0cos(+) 
l12 =l1cos(+)
8
Reference axis
Stator self‐inductances (laa, lbb, lcc) and Stator Mutual Inductances (lab, lbc, lac)
d axis
• Assume a sinusoidal space distribution of MMF Fa
• laa is equal to the ratio of flux linking phase a winding to current ia, with zero currents in all other circuits, which can be approximated as
Ls >Lm 0
laa= Ls + Lm cos2
q axis
b
laa=Ls + Lm cos2
q
a
lbb=Ls + Lm cos2(-2/3)
= -30o
S
lcc=Ls + Lm cos2(+2/3)
• lab <0 since windings a and b have 120o (>90o) displacement • Has the maximum absolute value when  = ‐300 or 1500.
lab = lba = -Ms - Lmcos2(+/6)
lab = -Ms - Lmcos2(+/6)
lbc = -Ms - Lmcos2(-/2)
lca = -Ms - Lmcos2(+5/6)
Ms  Ls/2
N
d
c
a
b
d
N
S
=150o
q
c
9
Stator to Rotor Mutual Inductances
(LSR: laF~lcF, laD~lcD, laQ~lcQ)
b
• The rotor sees a constant permeance if neglecting variations in the air gap due to stator slots
d (F, D)
q (Q)
a

• When the flux linking a stator winding and a rotor winding reaches the maximum when they aligns with each other and is 0 when they are displaced by 90o
c
• d‐axis
laF = lFa = MF cos
laD = lDa = MD cos
lbF = lFb = MF cos(-2/3)
lbD = lDb = MD cos(-2/3)
lcF = lFc = MF cos(+2/3)
lcD = lDc = MD cos(+2/3)
• q‐axis
LaQ = LQa = -MQ sin
LbQ = LQb = -MQ sin(-2/3)
LcQ = LQc = -MQ sin(+2/3)
10
For Salient‐pole Rotors
Which of the curves will be different for round rotors?
No 2nd harmonic
11
Rotor Inductances (LRR: lFF, lDD, lQQ, lFD, lFQ, lDQ)
•They are all constant
– Rotor self inductances
lFF ≜ LF
lDD ≜ LD
lQQ ≜ LQ
– Rotor mutual inductances lFD = lDF ≜ MR
lFQ = lQF = 0
lDQ = lQD = 0
12
Summary
 ψ abc   L SS
ψ   
 FDQ  L RS
L SR   i abc 


L RR   i FDQ 
Observations:
• Only LRR is constant
• LSS and LSR are  or time dependent
• How to simplify LSS and LSR?
– Diagonalization
– Remove  or time dependency
How to remove  or time dependency?
LRS =
The 1st harmonic in LSS and 2nd harmonic in LSR are due to the rotor rotating relative to a, b and c to cause variations in P12 or N12
•
Constant LRR doesn’t have harmonic terms because it is in a reference frame rotating with the rotor
What if we define q-axis
lagging d-axis by 90o?
LTSR
F
R
R
D
0
•
0
FDQ = - LSR iabc + LRR iFDQ
0
0
Q
LSR iabc = -FDQ + LRR iFDQ
•
LSR iabc may be represented by functions independent of  if we represent stator currents and flux linkages also in a reference frame rotating with the rotor.
13
FDQ
F
= - LSR iabc + LRR iFDQ
= – laFia – lbFib – lcFic +lFFiF + lFDiD + lFQiQ
= – MFcos ia – MFcos(-2/3) ib – MF cos(+2/3)ic +LFiF + MRiD +0
= – MF [ia cos + ib cos(-2/3) + ic cos(+2/3)]
+ LFiF + MRiD
D
= – laDia – lbDib – lcDic +lDFiF + lDDiD + lDQiQ
= – MD cos ia - MD cos(-2/3) ib - MD cos(+2/3)ic +MRiF + LDiD + 0
= – MD [ia cos + ib cos(-2/3) + ic cos(+2/3)]
+ MRiF + LDiD
Q
= – laQia – lbQib – lcQic
+lQFiF + lQDiD + lQQiQ
= MQ sin ia + MQ sin(-2/3) ib + MQ sin(+2/3) ic
+0 + 0 + LQiQ
= MQ [ia sin + ib sin(-2/3) + ic sin(+2/3) ]
+LQiQ
14
Park’s (dq0) Transformation
• Define
id= kd [ia cos +ib cos(-2/3) +ic cos(+2/3)]
iq= -kq [ia sin +ib sin(-2/3) +ic sin(+2/3) ]
=rt+-/2
rs
For balanced steadystate conditions:
ia= Im sinst
ib= Im sin(st - 2/3)
ic= Im sin(st + 2/3)
id= kd Im sin(/2-)×3/2
iq= -kq Im cos (/2-)×3/2
id= kd Im sin(st-)×3/2
iq= -kq Im cos(st-)×3/2
Constant currents!
• Define
i0= k0(ia + ib + ic)
i0   k0

k0
k0
  




i
k
k
k
cos

cos(

2

/
3)
cos(

2

/
3)
d
d
d  d

iq  kq sin  kq sin(  2 / 3) kq sin(  2 / 3)

  
ia 
i 
 b
ic 
i 0dq  Pi abc
15
Park’s (dq0) Transformation (cont’d)
P3  eaia  ebib  ecic  eTabc i abc  (P 1e0 dq )T P 1i 0 dq  eT0 dq P T P 1i 0 dq
If we expect the transformation to be power invariant:
P3 =edid+eqiq+e0i0
P-TP-1=U P-1= PT or PTP=U (P is an orthogonal matrix)
kd=kq=
P
and k0=
1/ 2

2 / 3  cos 
  sin 

1/ 2

P 1  2 / 3 1/ 2

1/ 2
1/ 2
cos(  2 / 3)
 sin(  2 / 3)

1/ 2

cos(  2 / 3) 
 sin(  2 / 3) 


cos(  2 / 3)  sin(  2 / 3) 

cos(  2 / 3)  sin(  2 / 3) 
cos 
 sin 
What if we define q-axis lagging
d-axis by 90o?
16
Flux Equations after Park’s Transformation
ψ 0 dq  Pψ abc
 ψ abc   L SS
ψ   
 FDQ  L RS
 ψ abc   P 1 0   ψ 0 dq 
 ψ 0 dq   P 0   ψ abc 

ψ   
 


ψ 

ψ
0
U
F
DQ

  FDQ 

 
 ψ FDQ   0 U   FDQ 


1
1

ψ
i


L
L




 SS
P
0
P
0  0dq 
0dq
L SR   i abc 
SR 





 

 



L RR   i FDQ 
 0 U ψFDQ  LRS LRR   0 U i 
 FDQ 
 ψ 0 dq   P 0   L SS



ψ
 FDQ   0 U  L RS
L SR   P

L RR   0
1
 0   L0
0
0
0
0
0  i0 
  
 i 
0
0
0
L
kM
kM
d
d
F
D
  
  d
 q   0
0
0
0 kMQ  iq 
Lq
 
  
0
0  iF 
LF
MR
 F   0 kM F
   0 kM
0
0  iD 
MR
LD
D
 D 
  
0
0
0
0
kM
L
 Q  
Q
Q 
 iQ 
L0  Ls  2 M s
3
Ld  Ls  M s  Lm
2
3
Lq  Ls  M s  Lm
2
k  3/ 2


0   i 0 dq 


U 
i
 FDQ 
 ψ 0 dq   LSS


ψ
 FDQ  LRS


LSR   i 0 dq 

LRR  
i
 FDQ 
 0  L0
0
0
0
0
0  i0 
  
 i 
0
0
0
L
kM
kM
d
d
F
D
  
  d
 F   0 kMF LF
0
0  iF 
MR
 
  

0
0  iD 
LD
 D   0 kMD MR
   0
0
0
0
Lq kMQ  iq 
q
  
  
0
0
0 kMQ LQ  iQ 
 Q   0
 
17
Voltage Equations after Park’s Transformation
e0 dq  Pe abc
e abc   P 1
e   
 FDQ   0
e abc   R abc
e    0
 FDQ  
0   i abc  d  ψ abc 

R FDQ  i FDQ  dt  ψ FDQ 
e0 dq 


e FDQ 
0  e0 dq   R abc


U  e FDQ   0
0   P 1
R FDQ   0
e0 dq   P 0   R abc



e FDQ   0 U   0
0   P 1
R FDQ   0
 P 1

 0
e0 dq   R 0 dq


e
 FDQ   0
P
0

U


d
0   i 0 dq   P P 1
  dt




R FDQ 
i FDQ   0
d 1
d d 1
d
P P
P  r P P1
dt
dt d
d
0 0 0 
 r 0 0 1
0 1 0 


1
0   i 0 dq  d 
P


 dt   0
U 
 
i FDQ 
0

U
 ψ 0 dq  


ψ
 FDQ  
0   i 0 dq   P 0  d   P 1 0 

   0 U  dt  
i
0
U
U   FDQ  
  


0

U
 ψ 0 dp  d  ψ 0 dq 

 

ψ
 FDQ  dt  ψ FDQ 
 1/ 2

1/ 2
1/ 2


 2 / 3  cos cos(  2 / 3) cos(  2 / 3) 
 sin   sin(  2 / 3)  sin(  2 / 3)


 ψ 0 dq  


ψ
 FDQ  
Note: P is NOT
constant
sin
 cos
0

0 sin(  2 / 3)  cos(  2 / 3)


0 sin(  2 / 3)  cos(  2 / 3)
0
 0  

d 1

P P  0 dq   r q   
(r Lq )  (iq )  r kM Q iQ

dt
 r d  r Ld  (id )  r kM F iF  r kM D iD 
18
Voltage Equations after Park’s Transformation (cont’d)
 ea
  Ra
e
 
 b
 0
 ec
 0


 eF
 0
e  0   0
 D
 
 eQ  0   0
0
0
0
0
Ra
 r Ld
0
0
  r Lq
Ra
0
0
0
 r kM F
RF
0
0
 r kM D
0
RD
0
0
0
0
0
e0   Ra
e  
Ra
 d 0
eF   0
0

  
0
0   0
e   0 r Ld
 q 
0
0   0
0
0
RF
0
r kM F
0
0
0
RD
r kM D
0
r Lq
0
0
Ra
0
0
0

 r kM Q 

0

0


0

RQ

0
0 
r kM Q 
0 

0 
0 

RQ 
(   r Lq )  (  iq )   r kM Q iQ =   r q
 e0   R a
e  
 d 0
 eF   0
 
0   0
e   0
 q 
 0   0
0
0
0
0
Ra
0
0
RF
0
0
0
0
0
0
RD
0
0
0
0
Ra
0
0
0
0
0 
0 
0 

0 
0 

RQ 
  i0 
 i 
 d
 iF 


i
D 
 i 
 q
 iQ 
 L0
0

0

0
0

 0
  i0   L 0
 i  
 d 0
  iq   0


 iF   0
i   0
D  
 iQ   0
 i0   L0
 i  
 d 0
iF   0
 
iD   0
 i   0
 q 
iQ   0
0
0
0
0
Ld
0
kM F
kM D
0
Lq
0
0
kM F
0
LF
MR
kM D
0
MR
LD
0
kM Q
0
0
0
Ld
kM F
kM D
0
kM F
LF
MR
0
kM D
MR
LD
0
0
0
0
0
0
0
0
0
0
Lq
kM Q
0    i0 


0    id 
kM Q  d   iq 

 
0  dt  iF 
0   iD 

 
LQ   iQ 


0   i0 
 
0   id 
0  d iF 
  
0  dt iD 
kM Q   iq 
  
LQ  iQ 
 
 r Ld  (  id )   r kM F i F   r kM D iD   r d
0
0
0
0
Ld
kM F
kM F
LF
kM D
MR
0
0
kM D
MR
LD
0
0
0
0
Lq
0
0
0
kM Q
  i0 
0 
 i 
0 
 d
0  d  iF 



0  dt  i D 
 i 
kM Q 
 q

LQ 
 iQ 
 0 
   
 r q
 0 


 0 
  r d 


19
0


Winding Circuits after Park’s Transformation
 e0   R a
e  
 d 0
 eF   0
 
0   0
e   0
 q 
 0   0
0
Ra
0
0
0
0
0
0
0
RF
0
0
0
0
0
0
RD
0
0
Ra
0
0
0
0
0    i0   L 0


0    id   0
0   iF   0

 
0   iD   0
0    iq   0
 
 
RQ   iQ   0


r
Ra
RF
L0
e0
eF
Rr
a
kMF
RD
MR
kMD
eD=0
Ld
0
kM F
0
kM D
0
0
kM F
LF
MR
0
kM D
0
MR
0
LD
0
0
Lq
0
0
0
kM Q
0    i0   0 


0    id    r q 


0  d  iF   0 

 

0  dt  iD   0 
kM Q    iq    r d 
 
 

LQ   iQ   0 


• d‐axis flux causes a speed voltage rd in
the q‐axis winding
• q‐axis flux causes a speed voltage ‐rq in the d‐axis winding
id
d
ed
rq
Rra
kMR
eQ=0
i0
0
Ld
RQ
q
iq
eq
Lq
rd
20
Alternative Park’s Transformation
ia= Im sinst
id= kd Im sin(st-)×3/2
P
ib= Im sin(st - 2/3)
iq= -kq Im cos(st-)×3/2
ic= Im sin(st + 2/3)
i0= k0(ia + ib + ic)
• If kd=kq=2/3 and k0=1/3, a unit‐to‐unit relationship holds between abc and dq0 variables.  1/ 2
2
P   cos 
3
  sin 
1/ 2

cos(  2 / 3) 
 sin(  2 / 3)  sin(  2 / 3) 
1/ 2
cos(  2 / 3)
cos
 sin
1

P1  1 cos(  2 / 3)  sin(  2 / 3)
1 cos(  2 / 3)  sin(  2 / 3)
• By defining proper base inductances, the matrix may become symmetric in per unit
L0

 0   0
   0
 d 
 q  
 0
 F  
   0
 D 
 Q  
0

0
Ld
0
3
MF
2
3
MD
2
0
0
0
Lq
0
0
3
MQ
2
0
0
0 
MF MD 0 
0
0 MQ 

LF MR 0 


MR LD
0 


0
0 LQ 

i0 
i 
 d
iq 
 
iF 
i 
D 
iQ 
 0 
 
 d
 q 
 
 F 
 D 
 
 Q 
0 0 0 0 0 0 00    i0 
LL0 0
0




i
0
0
0
L
M
M

0
L
L
0
L
L
0
d
d
F
D
l
ad
ad
ad



00
   iq 
0
Llq  Laq 0 0 0 0 MLaq
Q



0
0
0
L
L
M
i
0
0
0
M
L
M
adF

F F
R R
 F 
00 M
LadD
0 0 M RM R LDLD 00   iD 



0
0
0
0
L
L
 0
0
M Q aq 0
LQQ   iQ 
Lad-Laq based per unit system: assume that all the per unit mutual inductances between the
or
)
stator and rotor circuits in each axis are equal (
21
Per Unit Representation
p.u.
• Using the machine ratings as the base values
–
es base (V)
peak value of rated line-to-neutral voltage
–
is base (A)
peak value of rated line current
–
fbase (Hz)
rated frequency
• Accordingly:
– S3 base (VA)
= 3ERMS base× IRMS base = 3(es base/ 2)×(is base/ 2)= es base×is base
– Zs base ( )
=es base/is base
– Ls base (H)
=Zs base/base
– base (elec. rad/s)
=2fbase
– mbase (mech. rad/s)
=base×(2/pf)
– tbase (s)
=1/ base =1/(2fbase)
– s base (Wbturns)
=Ls base×is base= es base/base
– Tbase (Nm)
= S3 base / mbase =
Base d q
1
0
F
D
Q
fbase
2
es base
3
is base
iD base
If f=fbase
s base×is base
Selection of rotor based quantities:
iFbase, iDbase and iQbase should enable a
S3 base
iF base
2
2 iQ base
symmetric per-unit inductance matrix
22
Equivalent Circuits
Define differential operator p=d/dt
e0  Ra 0 0 0 0 0 
e  

 d   0 Ra 0 0 0 0 
eF   0 0 RF 0 0 0 
 

0   0 0 0 RD 0 0 
e   0 0 0 0 Ra 0 
 q 

0   0 0 0 0 0 RQ 
i0  L0
0
0
0
0
0  i0   0 
0
 0   L0
i  
i  





0
0   d  
 d   0 Ll  Lad Lad Lad
 d   0 Ll  Lad
 r q
iF   0
 F   0
Lad
LF MR
0
0  iF   0 
Lad
 
p
 
 p  

Lad
Lad
MR LD
0
0  iD   0 
iD   0
 D   0
i   0
 q   0

0
0
0
0 Ll  Laq Laq  iq   
 q 
  
    r d
0
Laq
LQ  iQ   0 
0
0
0
iQ   0
 Q   0
 
ed= -(Ll+Lad)×pid+Lad×piF +Lad×piD-rq -Raid
=Lad×p(-id+iF+iD) -Ll×pid -rq -Raid
pd= -(Ll+Lad)×pid +Lad×piF +Lad×piD
= Lad×p(-id+iF+iD) -Ll×pid
eF= -Lad×pid +LF×piF +MR×piD +RFiF
= Lad×p(-id+iF+iD) +(MR- Lad)×p(iD+iF)
+(LF - MR)×piF + RFiF
eD=0= -Lad×pid +MR×piF +LD×piD+ RDiD
= Lad×p(-id+iF+iD) +(MR- Lad)×p(iD+iF)
+(LD- MR)×piD+ RDiD
0
Lad
0
Lad
0
0
LF
MR
0
MR
LD
0
0
0
0
0
Ll  Laq
Laq
0  i0 
 
0  id 
0  iF 
 p 
0  iD 
Laq  iq 
  
LQ  iQ 
 
iD+iF
ed
eq= -(Ll+Laq)×piq +Laq×piQ +rd -Raiq
=Laq×p(-iq+iQ) -Ll×piq +rd -Raiq
eQ=0= -Laq×piq + LQ×piQ +RQiQ
= Laq×p(-iq+iQ) +(LQ-Laq)×piQ +RQiQ
23
Equivalent Circuits with Multiple Damper Windings (e.g. Round‐rotor Machines) Subscript Notations:
• d axis:
Lfd ≜LF - MR
fd ≜ F
L1d ≜LD - MR
1d ≜ D
Rfd ≜RF
efd ≜eF
( )fd ~ field winding quantities
( )kd ~ k-th d-axis damper winding quantities
( )kq ~ k-th q-axis damper winding quantities
R1d ≜RD
L1d =LD - MR
Rfd=RF
MR-Lad 0 (named Lfkd1 in some literature R1d =RD
to model rotor mutual flux leakage, i.e. the flux linking the rotor’s field and damper windings but not stator windings)
• q axis:
L1q≜LQ – Laq
1q ≜ Q
L2q≜LG – Laq
2q ≜ G
Lfd=LF - MR
R1q≜RQ
efd=eF
L1q=LQ – Laq L2q=LG – Laq
R1q=RQ
R2q=RG
R2q≜RG
24
Example: a model with 3 rotor windings in each of d‐ and q‐axis equivalent circuits • EPRI Report EL‐1424‐V2, “Determination of Synchronous Machine Stability Study Constants, Volume 2”, 1980
– The proposed equivalent circuits are expected to contain sufficient details to model all machines
– Parameters are estimated by frequency response tests
Usually ignored
25
Lad=KsdLadu
Ksd = at/at0 = at/(at+ I) = I0 / I
26
Steady‐state Analysis
All flux linkages, voltages and currents are constant:
p1d=0  R1di1d=0 i1d=0
p1q=0  R1qi1q=0 i1q=0
(Damper winding currents are all zero due to
no change in the magnetic field)
d= -( Ll+Lad)id+Ladifd = -Ldid+Ladifd
q= -( Ll+Laq)iq
= -Lqiq
pd=0 ed = -rq -Raid =rLqiq -Raid
pq=0 eq =rd -Raiq = -rLdid +rLadifd -Raiq
pfd=0 efd= Rfdifd
r=1 and L=X in p.u.
ed=Xqiq -Raid
eq= -Xdid +Xadifd -Raiq
efd= Rfdifd
Can we have a single equivalent circuit
representing both d and q axes circuits under
balanced steady-state conditions?
27
ea= Em cos(st+)
eb= Em cos(st +-2/3)
ec= Em cos(st ++2/3)
 1/ 2
2
P   cos 
3
  sin 
1/ 2
1/ 2

cos(  2 / 3) cos(  2 / 3) 
 sin(  2 / 3)  sin(  2 / 3) 
=rt+0
Terminal voltage & current phasors:
ed= Em cos( -0)= Et cos( -0)
eq= Em sin( -0) = Et sin( -0)
=ed+jeq
where Et is the per unit RMS value of the armature terminal voltage, which equals the peak value Em in per unit.
=id+jiq
where ed=Xqiq –Raid
eq= – Xdid +Xadifd –Raiq
= Xqiq –Raid – jXdid +jXadifd –jRaiq
= – Ra +Xqiq –jXdid +jXadifd
Compared to = –Ra –jXS
=
–Ra –jXS (id+jiq)
=–Ra +XSiq –jXSid+
XS Xq
E
Load
=j[Xadifd – (Xd – Xq)id]≝
=
– Ra – jXq
28
Steady‐state equivalent circuit
• Under no‐load or open‐circuit conditions, id=iq=0. =jXadifd
i=0 (load angle)
=j[Xadifd-(Xd-Xq)id]
• For round rotor machines or machines with neglected saliency
Xd=Xq=Xs (synchronous reactance)
=
– (Ra+jXs)
Eq=Xadifd
29
Computing per‐unit steady‐state values
•
Active and Reactive Powers
S  Et It*
 (ed  j eq )(id  j iq )
 (ed id  eqiq )  j(eqid  ed iq )
Pt  ed id  eqiq
Qt  eqid  ed iq
ed  r q  Raid
eq  r d  Raiq
Pt  r ( d iq  q id )  Ra (id2  iq2 )
 rTe  Ra (id2  iq2 )
•
Air-gap torque (or electric torque)
Te   d iq  q id
=Pt / r  Ra (id2  iq2 )
=Pt  Ra (id2  iq2 )
r  1p.u.
30
Example 3.2 in Kundur’s Book
R1d>>Rfd
Lfd/Rfd>>L1d/R1d
MR
L1q/R1q>>L2q/R2q
Figure 3.22
31
32
39.1o
1.565pu
33
Sub‐transient and Transient Analysis
• Following a disturbance, currents are induced in rotor circuits. Some of these induced rotor currents decay more rapidly than others. – Sub‐transient parameters: influencing rapidly decaying (cycles) components
– Transient parameters: influencing the slowly decaying (seconds) components
– Synchronous parameters: influencing sustained (steady state) components
34
Transient Phenomena
i(t)
• Study transient behavior of a simple RL circuit
v(t )  Vm sin(t   )  u (t )
Ri (t )  L
Unit step
di (t )
 v(t )
dt
• Apply Laplace Transform
RI (s)  L[sI (s)  i(0)]  V (s)
I ( s) 
V ( s) 1

R 1  s

Vm s sin    cos 
 2
R (s   2 )(1  sL / R)
• Apply Inverse Laplace Transform
i(t )  I m sin(t     )  I met / sin(   )
Steady-state
(sinusoidal component)
⁄ ,
⁄
,
tan
⁄
dc offset
(transient component)
• Example 8.1
R=0.125, L=10mH, Vm=151V
=L/R=0.08s
35
Short‐circuit and open circuit time constants
Consider the d-axis network
• Short-circuit time constant 
– Instantaneous change on d
– Delayed change on id ( through
)

• Open-circuit time constant 0
– Instantaneous change on id
– Delayed change on  d ( through
Ld ( s)  
 d
id
 Ld
e fd  0
1  s
1  s 0

)
(1  sTd )(1  sTd)
Ld ( s )  Ld
(1  sTd0 )(1  sTd0 )
• Time constant  or 0 equals the division of the total inductance and
resistance (L/R) with the effective circuit
36
Transient and sub‐transient parameters
R1d>>Rfd
L1q /R1q>>L2q /R2q
Lfd /Rfd>>L1d /R1d
d axis circuit
Considered rotor windings
Time constant
(open circuit)
Only field
Winding
T’d0=
+
8.07(s)
Time constant
(short circuit)
Inductance
(Reactance)
Ld(s) and Lq(s)
T’d=
q axis circuit
Add the damper winding
T’’d0=
//
L’d= Ll+Lad//Lfd
0.30(pu)
T’’d=
//
//
0.23(pu)
T’q=
L’q=
Add the 2nd damper winding
T’’q0=
//
0.07(s)
1.00(s)
L’’d= Ll+Lad//Lfd//L1d
Based on the parameters of Example 3.2
+
T’q0=
0.03(s)
//
Only 1st damper winding
//
Ll+Laq//L1q
0.65(pu)
T’’q=
//
//
L’’q= Ll+Laq//L1q//L2q
0.25(pu)
• Note: time constants are all in p.u. To be converted to seconds, they have to be multiplied by tbase=1/base (i.e. 1/377 for 60Hz).
37
38
Synchronous, Transient and Sub‐
transient Inductances
(1  sTd )(1  sTd)
Ld ( s )  Ld
(1  sTd0 )(1  sTd0 )
• Under steady‐state condition: s=0 (t)
Ld(0)=Ld
(d-axis synchronous inductance)
• During a rapid transient: s
Lad L fd L1d
TdTd
Ld  Ld ()  Ld
 Ll 
Td0Td0
Lad L fd  Lad L1d  L fd L1d
(d-axis sub-transient
inductance)
• Without the damper winding :s>>1/T’d and 1/T’d0 but << 1/T”d and 1/T”d0
Lad L fd
Td

Ld  Ld ()  Ld
 Ll 
Td0
Lad  L fd
(d-axis transient inductance)
39
Xd ≥ Xq ≥ X’q ≥ X’d ≥ X”q ≥ X”d
T’d0 > T’d > >T”d0 > T”d
T’q0 > T’q > >T”q0 > T”q
40
Parameter Estimation by Frequency Response Tests
Ld ( s )  Ld
(1  sTd )(1  sTd)
(1  sTd0 )(1  sTd0 )
T’d0 > T’d > >T”d0 > T”d> Tkd
1/T’d0 <1/ T’d << 1/T”d0 <1/T”d<1/Tkd
Bode Plot
-20dB/decade
41
Swing Equations
dm
J
 Ta  Tm  Te
dt
J
m
t
Ta
Tm
Te
combined moment of inertia of generator
and turbine, kgm2
angular velocity of the rotor, mech. rad/s
time, s
accelerating torque in N.m
mechanical torque in N.m
electromagnetic torque in N.m
• Define per unit inertia constant
2
1 J 0 m
H
(s)
2 VAbase
J
2H

2
(s)
VAbase
0m
Some references define TM or M=2H, called the mechanical starting time, i.e. the time
required for rated torque to accelerate the rotor from standstill to rated speed
42
d m
 Ta  Tm  Te
dt
2H
d m
 Tm  Te
VA
base
 02m
dt
J
Angular position of the rotor in electrical radian
with respect to a synchronously rotating reference
  r t  0t   0
in rad
d  m
2H 
dt   0 m

Tm  Te
 
 VA base /  0 m
d
  r   0    r in rad/s
dt
d 
2H  m
dt   0 m
 Tm  Te
 
Tbase

d 2 d  r d (  r )
in rad/s2


dt 2
dt
dt
2H
d r
 Tm  Te
dt
(in per unit)
where
m r / p f r


r (in per unit) =
0m 0 / p f 0
d ( r )
d (  r )
d 2
2
in
rad/s




0
0
dt 2
dt
dt
where  r 
 r
0

1 d
0 dt
2 H d 2
 Tm  Te
2
0 dt
If adding a damping term
proportional to speed deviation:
2 H d 2
K D d
T
T
K

=
T
T




m
e
D
r
m
e
0 dt 2
0 dt
43
Block diagram representation of swing equations
2 H d 2
K D d
T
T
K

=
T
T




m
e
D
r
m
e
0 dt 2
0 dt
2H
d (r )
 Tm  Te - K D r
dt
1 d
=  r
0 dt
44
State‐Space Representation of a Synchronous Machine
So far, we modeled all critical dynamics about a synchronous machine:
• State variables (pX): – stator and rotor voltages, currents or flux linkages
– swing equations (rotor angle and speed)
• Time constants:
– Inertia: 2H
– Sub‐transient and transient time constants, e.g. T’d0 and T”d0
• Other parameters
– Stator and rotor self‐ or mutual‐inductances and resistances
– Rotor mechanical torque Tm and stator electromagnetic torque Te
45
State Space Model on a Salient‐pole Machine
• Consider 5 windings: d, q, F (fd), D (1d) and Q (1q)
d
– Voltage and flux equations:
e  R  i  L  i  ΩΨ
e0 
e 
d 
e fd 
e   ,
0 
e 
q 
0 
i0 
i 
 d
i fd 
i   , Ψ 
i1d 
i 
 q
i1q 
 
 0 
 
 d 
 fd 
 ,
1d 
 
 q 
1q 
 
dt
 0 
 
 r q
 0 
ω  ΩΨ  

 0 
 

r d


0


– Swing equations:
• Define state vector x=[d fd 1d q 1q r ]T
Thus, the state‐space model:
Ψ  Li
d
Ψ  (R L1  Ω)  Ψ  e
dt
2H
d r
 Tm  Te  Tm  ( d iq  q id )
dt
d
 r  0  r  1
dt
x  f (x, e fd , Tm , ed , eq )
• efd and Tm are usually known but ed and eq are related to its loading conditions (the grid), so algebraic power‐flow equations should be introduced.
• The grid model is a set of Differential‐Algebraic Equations (DAEs)
46
Simplified Models
• [d fd 1d q 1q r ]T
Neglect pd, pq and variations of r
(i.e. r=1 pu) in the voltage equations
pΨ  (R L1  Ω)  Ψ  e
2 H  pr  Tm  Te
p  r  1
=q= -Lqiq
• [fd 1d 1q r ]T
– Inertia
2H ~ pr
– Transient
T’d0 ~ pfd
– Sub-transient T”d0 ~p1d
T”q0 ~ p1q
Neglect damper windings, i.e. p1d
and p1q
=d= -Ldiq
• [fd r ]T
– Inertia
– Transient
2H ~ pr
T’d0~ pfd
Constant flux linkage assumption
• [r ]T (classic model)
– Inertia
2H ~ pr
47
Neglect of Stator p terms
48
Classic Model
•Eliminate the differential equations on flux linkages (swing equations are the only differential equations left)
•Assume X’d=X’q
2 H  pr  Tm  Te
p  r  1
Et  E   ( Ra  jX d ) It
E’ is constant and can be estimated by
computing its pre-disturbance value
E   Et 0  ( Ra  jX d ) It 0
49
Simplified models neglecting p and saliency effects
Used in short-circuit analysis
Used for stability studies
Used for steady-state analysis
50
Comparison of PSS/E Generator Models
51