ECE 692 Advanced Topics on Power System Stability
2 – Power System Modeling
Spring 2016
Instructor: Kai Sun
1
Outline
•Modeling of synchronous generators for Stability Studies
•Modeling of loads
•Modeling of frequency and voltage regulation systems
•Materials
– Kundur’s Chapters 3‐5 and 7‐9 – Chapter 12 of “Power System Analysis”(3rd Ed.) by Saadat
2
2.1 Modeling of Synchronous Generators for Stability Studies
3
Synchronous Generators
Cylindrical/round rotor
Salient-pole rotor
Field winding
Armature winding
Stator
16 poles salient-pole rotor (12 MW)
Round rotor generator under construction
4
(Source: http://emadrlc.blogspot.com)
Stator and Rotor Windings
•
Consider a reference frame rotating synchronously
with d and q-axes at speed r (assumed to be
along with the axis of a-a’ at t=0)
•
 is the displacement of d-axis from the axis of a-a’
•
 is the displacement of q-axis from the rotating
reference axis
Reference axis
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
• Round‐rotor: 3+4
r
Direct or
d axis
    r t 

2

Quadrature
or q axis
ANSI/IEEE standard 100-1977 defines the q-axis to lead the d-axis by 900
5
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
lb b
lac
lb c
laF
lb F
laD
lb D
l cb
l cc
l cF
l cD
lFb
lDb
lFc
lDc
lFF
lDF
lFD
lDD
lQ b
lQ c
lQ F
lQ D
L SR   i abc 


L RR  i FDQ 
l a Q    ia 


l b Q    ib 
l cQ    ic 

 
l F Q   iF 
l D Q   iD 

 
l Q Q   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   l F a
   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
6
Self‐ and Mutual Inductances
Each of windings 1 & 2 is stationary (on the
stator) or rotating (on the rotor)

N1
N2
l12 =N1N2 P12 = l21 (Symmetric)
 a   l a a l a b
  
bb
 b   l b a LlSS
 c   l ca
l cb

 
 F   l F a Ll F b
RS
   l
lDb
Da
D

 
 Q   l Q a l Q b
lac
lb c
laF
lb F
laD
lb D
l cc
l cF
l cD
lFc
lDc
lFF
lDF
lFD
LRR
lDD
lQ c
lQ F
lQ D
LSR
l a Q    ia 


l b Q    ib 
l cQ    ic 

 
l F Q   iF 
l D Q   iD 

 
l Q Q   iQ 
P12 - permeance of the mutual flux path
(mainly influenced by the air gap)
•
Stator self/muual inductances (e.g. laa and lab):
P12 between stator windings is a function of 
and reaches its maximum twice per cycle
P12  P0+P2cos2(+)  l12 =l0+l2cos2(+)
•
Stator to Rotor Mutual Inductances (e.g. laF):
P12 between stator and rotor windings is almost
constant but N1N2 is a function of  and
reaches its maximum once per cycle; flux
leakage can be ignored (l0=0)
N1N2  N0cos(+) 
•
l12 =l1cos(+)
Rotor Inductances are all constant
F
R
0
R
D
0
0
0
Q
Using a reference revolving
with the rotor will lead to a
constant inductance matrix
7
Park’s Transformation
i 0dq  Pi abc
ψabc   LSS LSR 
ψ   

 FDQ  LRS LRR 
e abc   R abc
e    0
 FDQ  
 e0   R a
e  
 d 0
 eF   0
 
0   0
e   0
 q 
 0   0
P
1/ 2

2 / 3  cos 
  sin 

i abc 
i
 ψ 0 dq  Pψ abc
 FDQ 
0   i abc  d  ψ abc 

R FDQ  i FDQ  dt  ψ FDQ 
0
Ra
0
0
0
0
0
0
0
RF
0
0
0
0
RD
0
0
0
0
Ra
0
0
0
0
0    i0 


0    id 
0   iF 

 
0   iD 
0    iq 

 
RQ   iQ 


1/ 2
cos(  2 / 3)
 sin(  2 / 3)

1/ 2

cos(  2 / 3) 
 sin(  2 / 3) 
0  L0 0
0
0
0
0  i0 
  
 i 
0
0
0
L
kM
kM
d
d
F
D
  
  d
F   0 kMF LF MR 0
0  iF 
  
 
0
0  iD 
D  0 kMD MR LD
   0 0
Lq kMQ iq 
0
0
 q 
 
0
0 kMQ LQ  iQ 
Q   0 0
 
L0  Ls  2M s
3
Ld  Ls  M s  Lm
2
3
Lq  Ls  M s  Lm
2
k  3/ 2
e0 dq  Pe abc
 L0
0

0

0
0

 0
0
Ld
0
kM F
0
kM D
0
0
kM F
LF
MR
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 


 0 
8
Per Unit Representation
• 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
S3 base (VA)
= es base×is base
Zs base ( )
=es base/is base
Ls base (H) =Zs base/base
base (elec. rad/s) =2fbase
tbase(s)
s base (Wbturns) =Ls base×is base= es base/base
Tbase (Nm)
=
p.u.
2
2 If f=fbase
=1/ base =1/(2fbase)
s base×is base
• Lad-Laq based per unit system: assume all per unit mutual inductances between
in q-axis
the stator and rotor circuits are all equal to
in d-axis or
F
D
Q
0
0
0
0
0  i0  Base d q 0
0  L0
 
  
1
fbase
0
0
0

L
L
L
L
l
ad
ad
ad
 id 
 d 
S3 base
2
es base
q   0
0
0 Laq  iq 
Ll  Laq 0
 
 
iF base
iD base
iQ base
is base
3

0
0
0
L
L
M
ad
F
R
  iF 
 F 
 D   0
0
Lad
MR LD 0   iD 
   iFbase, iDbase and iQbase enable a symmetric
  

0
0
0
0
L
L
  iQ  per-unit inductance matrix
 Q  
aq
Q
9
Equivalent Circuits
e0   Ra
e  
 d 0
eF   0
 
0   0
e   0
 q 
0   0
 0   L0
   0
 d 
 F   0
p   
 D   0
 q   0
  
 Q   0
0
Ra
0
0
0
RF
0
0
0
0
0
0
0
0
0
0
0
0
RD
0
0
0
Ra
0





0
0

RQ 
0
0
0
i0   L0
i  
 d 0
iF   0
 
iD   0
i   0
 q 
iQ   0
0
0
0
0
Ll  Lad
Lad
Lad
0
Lad
LF
MR
0
Lad
0
MR
0
LD
0
0
Ll  Laq
0
0
0
Laq
0
Ll  Lad
Lad
0
Lad
LF
0
Lad
MR
0
0
0
Lad
0
0
MR
0
0
LD
0
0
0
Ll  Laq
Laq
0    i0 
 
0    id 
0   iF 
 p 
0   iD 
Laq    iq 
  
LQ   iQ 
 
 i0   0 
 i  

  d  r q 
 iF   0 
 p 

0  iD   0 
Laq  iq   r d 
   

LQ  iQ   0 
 
0
0
0
They follow Faraday’s law
(differential operator p=d/dt)
iD+iF
ed
10
Equivalent Circuits with Multiple Damper Windings
Usually ignored
MR-Lad 0
L1d =LD - MR
Lfd=LF -MR
Rfd=RF
R1d =RD
efd=eF
L1q=LQ – Laq L2q=LG – Laq
R1q=RQ
R2q=RG
EPRI Report EL-1424-V2, “Determination of Synchronous
Machine Stability Study Constants, Volume 2”, 1980
11
Lad=KsdLadu
Ksd = at/at0 = at/(at+ I) = I0 / I
12
Steady‐state Analysis
All derivatives are zero:
pr=0  r=1 and L=X in p.u.
pfd=0  efd= Rfdifd
p1d=0  i1d=0 d= -Ldid+Ladifd
p1q=0  i1q=0 q= -Lqiq
pd=0  ed =rLqiq -Raid
= Xqiq -Raid
pq=0  eq = -rLdid +rLadifd -Raiq
= -Xdid
+Xadifd -Raiq
Terminal voltage & current phasors:
=ed+jeq
=
=id+jiq
– (Ra+ jXq
where
=j[Xadifd-(Xd-Xq)id]
If saliency is neglected
Xd=Xq=Xs (synchronous reactance)
Eq=Xadifd
13
Computing per‐unit steady‐state values
r  1p.u.
•
Active and Reactive Powers
S  Et It*
 (ed  j eq )(id  j iq )
 (ed id  eqiq )  j(eqid  ed iq )
ed  r q  Raid
eq  r d  Raiq
Pt  ed id  eqiq  rTe  Ra (id2  iq2 )
Qt  eqid  ed iq
•
Air-gap torque (or electric torque)
Te   d iq  q id =Pt  Ra (id2  iq2 )
14
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
15
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
16
Transient and sub‐transient parameters
+
pfd
-
+
p1d
-
R1d>>Rfd
+
p1q
-
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 Kundur’s 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).
17
18
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)
19
Xd ≥ Xq ≥ X’q ≥ X’d ≥ X”q ≥ X”d >Xl
T’d0 > T’d > >T”d0 > T”d
T’q0 > T’q > >T”q0 > T”q
20
Swing equations
2 H d 2
 Tm  Te - K D r
2
0 dt
2H
(s)
d (r )
 Tm  Te - K D r
dt
1 d
= r
0 dt
Some references define M=2H, called the mechanical starting time, i.e. the time required
for rated torque to accelerate the rotor from standstill to rated speed
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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
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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)
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Neglect of Stator p terms
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Simplified Models
pΨ  (R L1  Ω)  Ψ  e
• [d fd 1d q 1q r ]T or
[d fd 1d q 1q 2q r ]T
2 H  pr  Tm  Te
p  r  1
Let pd=pq=0 and r=1 pu
=q= -Lqiq
• [fd 1d 1q r ]T or
[fd 1d 1q 2q r ]T
– Inertia
2H ~ pr
– Transient
T’d0 ~ pfd
– Sub-transient T”d0 ~p1d
T’q0 ~ p1q
T”q0 ~ p2q
Let p1d =p1q=0 (neglect
damper windings)
[fd r
=d= -Ldiq
]T
– Inertia
– Transient
2H ~ pr
T’d0~ pfd
Constant flux linkages
• [r ]T (classic model)
– Inertia
2H ~ pr
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E’q= fd Lad/(Lad+Lfd)
Efd =efd Lad/Rfd
T’d0 ~pE’q(fd )
T”d0 ~p’d (1d )
T”q0 ~ p”q (1q)
Source: J. Weber, “Description of
Machine Models GENROU,
GENSAL, GENTPF and GENTPJ,”
PowerWorld, Oct 2015
26
27
Phasor diagram
=jXadifd
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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
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Comparison of PSS/E Generator Models
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