ece 422 system operations planning 2 –synchronous machine...
TRANSCRIPT
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Spring 2015Instructor: Kai Sun
ECE 422 Power System Operations & Planning
2 – Synchronous Machine Modeling
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Outline
•Synchronous Machine Modeling•Simplified Models for Stability Studies•Materials– Saadat’s Chapter 8– Kundur’s Chapters 3‐5
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Synchronous Generators
Stator
Salient-pole rotor Cylindrical/round rotor
Field current
Armature winding
Field winding
4
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
• 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
16 poles salient-pole rotor (12 MW)
(Source: http://emadrlc.blogspot.com)
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Under Steady‐State Conditions
•MMFs: Fsr =Fr+Fs Fr=Fsr -Fs
• EMFs: Esr=Ea+Ear Ea=Esr -Ear
Esr =V+(Ra+jXl)Ia
• For a round rotor, defineEar =-jXarIa
where Xs=Xl+Xar (synchronous reactance)
d
Axis of coil a
(reference)
N
S
n
m
ea
Fr
Fs
Fsr
Fs
Fsr
Fr
( ) ( )
a a ar a
a s a
E V R jX jX IV R jX I
Ea
EsrEar
Ea
Load
r
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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
Stator and Rotor Windings
Note: ANSI/IEEE standard 100-1977 defines the quadrature (q) axis to lead the d-axis by 900
2rt
• 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
Quadrature or q axis
Direct or d axis
Reference axis
r
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Voltage and Flux Equations (Salient‐pole machine)
0 0 0 0 00 0 0 0 00 0 0 0 0
0 0 0 0 00 0 0 0 000 0 0 0 00
a a aa
b b bb
c c cc
FF F F
DD D D
QQ Q Q
e iRe iRe iR d
Re i dtRe i
Re i
0
0abc abc abcabc
FDQFDQ FDQ FDQ
ddt
e i ψRRe i ψ
a aaa ab ac aF aD aQ
b bba bb bc bF bD bQ
c cca cb cc cF cD cQ
F a F b F c F F F D F QF F
D a D b D c D F D D D QD D
Q a Q b Q c Q F Q D Q QQ Q
il l l l l lil l l l l lil l l l l l
l l l l l l il l l l l l il l l l l l i
abc abcSS SR
FDQ FDQRS RR
ψ iL Lψ iL L
• Model windings as a group of magnetically coupled circuits with inductances depending on
• Rotor self‐inductances (lFF, lDD, lQQ)
• Rotor mutual inductances (lFD, lDQ, lFQ)
• Rotor self‐inductances (lFF, lDD, lQQ)
• Rotor mutual inductances (lFD, lDQ, lFQ)
• Stator self‐inductances (laa, lbb, lcc)
• Stator mutual inductances (lab, lbc, lac)
• Stator self‐inductances (laa, lbb, lcc)
• Stator mutual inductances (lab, lbc, lac)
• Stator‐to‐rotor mutual inductances (laF, lbD, laQ)• Stator‐to‐rotor mutual inductances (laF, lbD, laQ)
A main objective of synchronous machine modeling is to find constants for simplification of voltage and flux equations
Rc
Ra
Rb
aF
D
Q
c
b
RF
RD
RQ
ea
ec
eb
eF
eQ=0
eD=0
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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= N1N2 P12 = N12P12 = 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(+)
• 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(+)
a aaa ab ac aF aD aQ
b bba bb bc bF bD bQ
c cca cb cc cF cD cQ
F a F b F c F F F D F QF F
D a D b D c D F D D D QD D
Q a Q b Q c Q F Q D Q QQ Q
il l l l l lil l l l l lil l l l l l
l l l l l l il l l l l l il l l l l l i
Assume a sinusoidal space distribution of each MMF
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Stator self‐inductances (laa, lbb, lcc) and Stator Mutual Inductances (lab, lbc, lac)• 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
laa= Ls + Lm cos2
• 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)
laa=Ls + Lm cos2
lbb=Ls + Lm cos2(-2/3)
lcc=Ls + Lm cos2(+2/3)
Ls >Lm 0
lab = -Ms - Lmcos2(+/6)
lbc = -Ms - Lmcos2(-/2)
lca = -Ms - Lmcos2(+5/6)
Ms Ls/2
q axis
d axis
Reference axis
b aq
d
= -30o
b a
q
d
=150o
S N
N S
c
c
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Stator to Rotor Mutual Inductances(LSR: laF~lcF, laD~lcD, laQ~lcQ)
• The rotor sees a constant permeance if neglecting variations in the air gap due to stator slots
• 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
• d‐axis
• q‐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)
LaQ = LQa = -MQ sin
LbQ = LQb = -MQ sin(-2/3)
LcQ = LQc = -MQ sin(+2/3)
baq (Q)
d (F, D)
c
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For Salient‐pole Rotors
Which of the curves will be different for round rotors? No 2nd harmonic
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Rotor Inductances (LRR: lFF, lDD, lQQ, lFD, lFQ, lDQ)
•They are all constant– Rotor self inductances
– Rotor mutual inductances
≜
lFF ≜ LF
lDD ≜ LD
lQQ ≜ LQ
lFD = lDF ≜ MR
lFQ = lQF = 0
lDQ = lQD = 0
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LRS = LTSR
F R 0R D 00 0 Q
Summary Observations:• Only LRR is constant• LSS and LSR are or time dependent• How to simplify LSS and LSR?
– Diagonalization– Remove or time dependency
What if we define q-axis lagging d-axis by 90o?
abc abcSS SR
FDQ FDQRS RR
ψ iL Lψ iL L
How to remove or time dependency?
• 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
FDQ = - LSR iabc + LRR iFDQ
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.
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FDQ = - LSR iabc + LRR iFDQ
F = – 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
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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) ]
For balanced steady-state conditions:ia= Im sinstib= Im sin(st - 2/3)ic= Im sin(st + 2/3)
id= kd Im sin(st-)×3/2
iq= -kq Im cos(st-)×3/2
=rt+-/2rs
id= kd Im sin(/2-)×3/2
iq= -kq Im cos (/2-)×3/2Constant currents!
• Define
i0= k0(ia + ib + ic)
00 0 0
cos cos( 2 / 3) cos( 2 / 3)sin sin( 2 / 3) sin( 2 3)
/
d d d
q
a
d b
q cq q
k k kk k k
i ii
kii ik k
0dq abci Pi
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Park’s (dq0) Transformation (cont’d)
P-TP-1=U P-1= PT or PTP=U (P is an orthogonal matrix)
1
1/ 2 cos sin
2 / 3 1/ 2 cos( 2 / 3) sin( 2 / 3)
1/ 2 cos( 2 / 3) sin( 2 / 3)
P
1 / 2 1 / 2 1 / 22 / 3 cos cos( 2 / 3) cos( 2 / 3)
sin sin( 2 / 3) sin( 2 / 3)
P
kd=kq= and k0=
P3 =edid+eqiq+e0i0If we expect the transformation to be power invariant:
What if we define q-axis lagging d-axis by 90o?
3T
a a b b c c abc abcP e i e i e i e i 1 10 0( )T
dq dq P e P i 1
0 0T T
dq dq e P P i
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Flux Equations after Park’s Transformation
0 adq bc Pψψ 0 dq
FD QQ
abc
FD
ψψ ψψ P 0
0 U
abc abcSS SR
FDQ FDQRS RR
ψ iL Lψ iL L
01 1
0 SS SR
RS R
dq dq
FDQF
RDQ
P 0 P 00 U 0 U
L
iL LψLψ i
01
dq
F
abc
DF QDQ
ψ
ψψP 0
U
ψ
0
0 0dq dqSS SR
RS RRFDQFDQ
ψ iL LL Lψ i
0 00 0 0 0 0 00 0 00 0 0 0
0 0 00 0 00 0 0 0
d dd F D
q qq Q
F F RF F
D R DD D
Q QQ Q
iLiL kM kMiL kM
kM L M ikM M L i
kM L i
032 2
3 3 / 22
s s d s s m
q s s m
L L M L L M L
L L M L k
0 00 0 0 0 0 00 0 00 0 0
0 0 00 0 0 00 0 0 0
d dd F D
F FF F R
D DD R D
q qq Q
Q QQ Q
iLiL kM kMikM L M
ikM M LiL kM
kM L i
10 00
0SS SR
RS RR
dq dq
FDQFDQ
P P 0U 0 ULψ
L
iLψ iL
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Voltage Equations after Park’s Transformation0
1
dq
F
abc
DF QDQ
e
eeP 0
U
e
00 adq bc Pee
0 01 1
dq abc
FDQ
dq
FDQFDQ
P 0 P 00 U 0 U
e i
e i
R 00 R
10 dq
FDQ
ddt
ψ
ψP 00 U
10 0abc
FD
dq dq
FDQQ QFD
P 0 P 0e iie 0 0 U
00 RU
R 10 dq
FDQ
ddt
P 0 P 00 U 0 ψU
ψ
0 0 0 001
dq dq dp dqdq
FDQFDQ FDQ FDQFDQ
dd
dtt
d
Pe i ψ ψP 0
0 U
R 00 Re ψ ψi
Note: P is NOT constant
1 1 1r
d d d ddt dt d d
P P P P P P1/ 2 1/ 2 1/ 2
2 / 3 cos cos( 2 / 3) cos( 2 / 3)sin sin( 2 / 3) sin( 2 / 3)
0 sin cos0 sin( 2 /3) cos( 2 /3)0 sin( 2 /3) cos( 2 /3)
0 0 00 0 10 1 0
r
10
0 0( ) ( )
( )dq r q r q q r Q Q
r d r d d r F F r D D
d L i kM idt
L i kM i kM i
P P
0
0abc abc abcabc
FDQFDQ FDQ FDQ
ddt
e i ψRRe i ψ
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Voltage Equations after Park’s Transformation (cont’d)0 00 0 0 0 0 0 0 0 0 0
0 0 0 0 0 00 0 0 0 0 0
0 0 0 0 0 0 0 00 0 0 0 0 0 000 0 0 0 00
a a
db a r q r Q d F D
qc r d a r F r D q Q
F F F RF F
D D RD D
QQ Q
ie R Lie R L kM L kM kMie L R kM kM L kM
R kM L Me iR kM Me i
Re i
0
00 0 0 0
d
q
F
D D
Q Q Q
iiid
idtL i
kM L i
00 00 0 0 0 0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0 0 0
0 0 0 0 0 0 0 000 0 0 0 0 00 0 0 0 0 0 0 00
a
dd a r q r Q d F D
FF F F RF
DD D R D
qr d r F r D a q Qq
Q Q
ie R Lie R L kM L kM kM
iR kM L MeiR kM M L
iL kM kM R L kMeR i
0
0
d
F
D
q
Q Q Q
ii
ididt
ikM L i
00 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 0
0 0 0 0 0 0 0 000 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00
a
dd a d F D
FF F F RF
DD D R D
qa q Qq
Q Q QQ
ie R Lie R L kM kM
iR kM L MeiR kM M L
iR L kMeR kM Li
0 0
00
0
d r q
F
D
q r d
Q
ii
ididt
i
i
( ) ( ) = ( )r q q r Q Q r q r d d r F F r D D r dL i kM i L i kM i kM i
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Winding Circuits after Park’s Transformation00 00 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 00 0 0 0 0 0 0 0
0 0 0 0 0 0 0 000 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00
a
dd a d F D
FF F F RF
DD D R D
qa q Qq
Q Q QQ
ie R Lie R L kM kM
iR kM L MeiR kM M L
iR L kMeR kM Li
0 0
00
0
d r q
F
D
q r d
Q
ii
ididt
i
i
• d‐axis flux causes a speed voltage rd inthe q‐axis winding
• q‐axis flux causes a speed voltage ‐rq in the d‐axis winding
kMF
kMD
MR
kMR
L0
r i0
id
iq
Ld
Lq
e0
ed
eq
rq
rd
r
r
eF
eQ=0
eD=0
RF
RQ
RD
Ra
Ra
Ra
d
q
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Alternative Park’s Transformation
• If kd=kq=2/3 and k0=1/3, a unit‐to‐unit relationship holds between abc and dq0 variables.
ia= Im sinst
ib= Im sin(st - 2/3)
ic= Im sin(st + 2/3)
id= kd Im sin(st-)×3/2
iq= -kq Im cos(st-)×3/2
i0= k0(ia + ib + ic)
P
• By defining proper base inductances, the matrix may become symmetric in per unit
1/ 2 1/ 2 1/ 22 cos cos( 2 / 3) cos( 2 / 3)3
sin sin( 2 / 3) sin( 2 / 3)
P 1
1 cos sin1 cos( 2 / 3) sin( 2 / 3)1 cos( 2 / 3) sin( 2 / 3)
P
0 00 0 0 0 0 00 0 00 0 0 00 0 00 0 00 0 0 0
d dd F D
q qq Q
F FF F R
D DD R D
Q QQ Q
iLiL M MiL M
iM L MiM M LiM L
0
0 0
0 0 0 0 00 0 00 0 0 0
30 0 0 230 0 02
30 0 0 02
d F D
q Qd d
q qF F R
F F
D DD R D
Q Q
Q Q
LL M M i
L M iiM L M
iiM M Li
M L
Lad-Laq based per unit system: assume that all the per unit mutual inductances between the stator and rotor circuits in each axis are equal ( or )
0 0 0 0 0 00 0 00 0 0 00 0 00 0 00 0 0 0
l ad ad ad
l aq aq
ad F R
ad R D
aq Q
LL L L L
L L LL L ML M L
L L
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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
• 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) =2fbase
– mbase (mech. rad/s) =base×(2/pf) – tbase (s) =1/ base =1/(2fbase)– s base (Wbturns) =Ls base×is base= es base/base
– Tbase (Nm) = S3 base / mbase = s base×is base
p.u.
Base d q 0 F D Q1 fbase
2 es base S3 base
3 is base iF base iD base iQ base
Selection of rotor based quantities:
iFbase, iDbase and iQbase should enable a
symmetric per-unit inductance matrix
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If f=fbase
23
Equivalent Circuits
iD+iF
00 00 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 0
0 0 0 0 0 0 0 000 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00
a
dd a l ad ad ad
FF ad F RF
DD ad R D
qa l aq aqq
Q aq QQ
ie R Lie R L L L L
iR L L MeiR L M LiR L L Le
R L Li
0 0
00
0
d r q
F
D
q r d
Q
ii
ip i
i
i
ed= -(Ll+Lad)×pid+Lad×piF +Lad×piD-rq -Raid
=Lad×p(-id+iF+iD) -Ll×pid -rq -Raid
pd= -(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
eq= -(Ll+Laq)×piq +Laq×piQ +rd -Raiq
=Laq×p(-iq+iQ) -Ll×piq +rd -Raiq
eQ=0= -Laq×piq + LQ×piQ +RQiQ
= Laq×p(-iq+iQ) +(LQ-Laq)×piQ +RQiQ
ed
00 0 0 0 0 0 00 0 00 0 00 0 00 0 0 00 0 0 0
dd l ad ad ad
FF ad F R
DD ad R D
qq l aq aq
Q aq Q Q
iLiL L L L
iL L Mp p iL M L
iL L LL L i
Define differential operator p=d/dt
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Equivalent Circuits with Multiple Damper Windings (e.g. Round‐rotor Machines)
L1d =LD - MR
R1d =RD
Lfd=LF - MR
Rfd=RF
efd=eF
L1q=LQ – Laq
R1q=RQ
L2q=LG – Laq
R2q=RG
Subscript Notations:( )fd ~ field winding quantities( )kd ~ k-th d-axis damper winding quantities( )kq ~ k-th q-axis damper winding quantities
• d axis:Lfd≜LF - MR Rfd≜RF
fd≜ F efd≜eF
L1d ≜LD - MR R1d ≜RD
1d ≜ D
MR-Lad 0 (named Lfkd1 in some literature 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 R1q≜RQ
1q ≜ Q
L2q≜LG – Laq R2q≜RG
2q ≜ G
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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 testsUsually ignored
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Lad=KsdLadu
Ksd = at/at0 = at/(at+ I) = I0 / I
27
Steady‐state AnalysisAll flux linkages, voltages and currents are constant:p1d=0 R1di1d=0 i1d=0p1q=0 R1qi1q=0 i1q=0(Damper winding currents are all zero due tono change in the magnetic field)
d= -( Ll+Lad)id+Ladifd = -Ldid+Ladifd
q= -( Ll+Laq)iq = -Lqiq
pd=0 ed = -rq -Raid =rLqiq -Raid
pq=0 eq =rd -Raiq = -rLdid +rLadifd -Raiq
pfd=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?
28
ea= Em cos(st+)eb= Em cos(st +-2/3)ec= Em cos(st ++2/3)
ed= Em cos( -0)= Et cos( -0)eq= Em sin( -0) = Et sin( -0)where Et is the per unit RMS value of the armature terminal voltage, which equals the peak value Em in per unit.
= Xqiq –Raid – jXdid +jXadifd –jRaiq
= – Ra +Xqiq –jXdid +jXadifd
Compared to = –Ra –jXS
=rt+0
E
LoadXS Xq
=j[Xadifd – (Xd – Xq)id]≝
= – Ra – jXq
Terminal voltage & current phasors:
=ed+jeq =id+jiq
where ed=Xqiq –Raid
eq= – Xdid +Xadifd –Raiq
1/ 2 1/ 2 1/ 22 cos cos( 2 / 3) cos( 2 / 3)3
sin sin( 2 / 3) sin( 2 / 3)
P
= –Ra –jXS (id+jiq)
=–Ra +XSiq –jXSid+
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Steady‐state equivalent circuit
• Under no‐load or open‐circuit conditions, id=iq=0. =jXadifd
i=0 (load angle)
• For round rotor machines or machines with neglected saliency
Xd=Xq=Xs (synchronous reactance)= – (Ra+jXs)
Eq=Xadifd
=j[Xadifd-(Xd-Xq)id]
30
Computing per‐unit steady‐state values
*
( j )( j )
( ) j( )
t t
d q d q
d d q q q d d q
S E Ie e i i
e i e i e i e i
t d d q q
t q d d q
P e i e i
Q e i e i
d r q a d
q r d a q
e R i
e R i
2 2
2 2
( ) ( )
( )t r d q q d a d q
r e a d q
P i i R i i
T R i i
2 2
2 2
= / ( )
= ( ) 1p.u.
e d q q d
t r a d q
t a d q r
T i i
P R i i
P R i i
• Active and Reactive Powers
• Air-gap torque (or electric torque)
31
Example 3.2 in Kundur’s Book
MR
L1q/R1q>>L2q/R2q
R1d>>Rfd
Lfd/Rfd>>L1d/R1d
Figure 3.22
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33
39.1o
1.565pu
34
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
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• Study transient behavior of a simple RL circuit
• Apply Laplace Transform
• Apply Inverse Laplace Transform
• Example 8.1R=0.125, L=10mH, Vm=151V=L/R=0.08s
Transient Phenomena
Unit step
i(t)
Steady-state (sinusoidal component)
dc offset (transient component)
( ) sin( )) (mv t V t u t
( )( ) ( )di tRi t L v tdt
(s) [ ( ) (0)] ( )RI L sI s i V s 2 2
( ) 1 sin cos( ) 1 ( )(1 )/
mVV s sI sR s R sL Rs
/sin( )( ( )) sintmm I eti It ⁄ , ⁄ , tan ⁄
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Short‐circuit and open circuit time constantsConsider 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 )
• Time constant or 0 equals the division of the total inductance and resistance (L/R) with the effective circuit
00
1( )1
fd
dd d
d e
sL s Li s
)1)(1()1)(1()(00 dd
dddd TsTs
TsTsLsL
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Transient and sub‐transient parameters
d axis circuit q axis circuit
Considered rotor windings
Only fieldWinding
Add the damper winding Only 1st damper winding Add the 2nd damper winding
Time constant(open circuit)
Time constant(short circuit)
Inductance(Reactance)Ld(s) and Lq(s)
L1q /R1q>>L2q /R2qR1d>>Rfd Lfd /Rfd>>L1d /R1d
8.07(s) 1.00(s)
T’d0=
T’d=
L’d= L’’d=
T’’d0= T’q0= T’’q0=
T’’d= T’q= T’’q=
0.03(s) 0.07(s)
L’q= L’’q=0.30(pu) 0.23(pu) 0.65(pu) 0.25(pu)
Based on the parameters of Example 3.2
+
//
Ll+Lad//Lfd
//
// //
Ll+Lad//Lfd//L1d
+
//
Ll+Laq//L1q
//
// //
Ll+Laq//L1q//L2q
• 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).
38
39
Synchronous, Transient and Sub‐transient Inductances
• Under steady‐state condition: s=0 (t)Ld(0)=Ld
• During a rapid transient: s
•Without the damper winding :s>>1/T’d and 1/T’d0 but << 1/T”d and 1/T”d0
)1)(1()1)(1()(00 dd
dddd TsTs
TsTsLsL
1
0 0 1 1
( ) ad fd dd dd d d l
d d ad fd ad d fd d
L L LT TL L L LT T L L L L L L
0
( ) ad fddd d d l
d ad fd
L LTL L L LT L L
(d-axis sub-transient inductance)
(d-axis synchronous inductance)
(d-axis transient inductance)
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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
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Parameter Estimation by Frequency Response Tests
T’d0 > T’d > >T”d0 > T”d> Tkd
-20dB/decade
Bode Plot
1/T’d0 <1/ T’d << 1/T”d0 <1/T”d<1/Tkd)1)(1(
)1)(1()(00 dd
dddd TsTs
TsTsLsL
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Swing Equations
• Define per unit inertia constant
emam TTT
dtdJ J combined moment of inertia of generator
and turbine, kgm2
m angular velocity of the rotor, mech. rad/st time, sTa accelerating torque in N.mTm mechanical torque in N.mTe electromagnetic torque in N.m
0
212 VA
m
base
JH
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
basem
HJ VA220
(s) (s)
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0
0 0
0
2
2 VA
2VA /
2
2 (in per unit)
m
m m
m
ma m e
mbase m e
m m e
base
m m e
base
rm e
dJ T T Tdt
H d T Tdt
d T THdt
d T THdt T
dH T Tdt
where r (in per unit) =
0 0 0
//
r fm r
m f
pp
Angular position of the rotor in electrical radian with respect to a synchronously rotating reference
0 0rt t in rad
0r rddt in rad/s
2
2
( )r rd dddt dt dt
in rad/s2
2
0 02
( ) ( )r rd dddt dt dt
in rad/s2
0 0
1 rr
ddt
2
20
2 m eH d T T
dt
2
20 0
2 - = - Dm e D r m e
H d K dT T K T Tdt dt
If adding a damping term
proportional to speed deviation:
where
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Block diagram representation of swing equations
( )2 -rm e D r
dH T T Kdt
0
1 = rddt
2
20 0
2 - = - Dm e D r m e
H d K dT T K T Tdt dt
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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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• Consider 5 windings: d, q, F (fd), D (1d) and Q (1q)– Voltage and flux equations:
– Swing equations:
• Define state vector x=[d fd 1d q 1q r ]T
Thus, the state‐space model:
• 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)
State Space Model on a Salient‐pole Machine
0 00
1 1
1 1
0
0, , ,
00
00
d dd r q
fd fdfd
d d
q q r dq
q q
ieie
iei
iei
e i Ψ ω ΩΨ
ddt
e R i L i ΩΨ Ψ L i
1( )ddt
Ψ R L Ω Ψ e
0
2 ( )
1
rm e m d q q d
r r
dH T T T i idt
ddt
( , , , , )fd m d qe T e ex f x
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Simplified Models• [d fd 1d q 1q r ]T
• [fd 1d 1q r ]T
– Inertia 2H ~ pr
– Transient T’d0 ~ pfd
– Sub-transient T”d0 ~p1d
T”q0 ~ p1q
• [fd r ]T
– Inertia 2H ~ pr
– Transient T’d0~ pfd
• [r ]T
– Inertia 2H ~ pr
Neglect damper windings, i.e. p1dand p1q
Neglect pd, pq and variations of r(i.e. r=1 pu) in the voltage equations
Constant flux linkage assumption
(classic model)
=q= -Lqiq
=d= -Ldiq
1( )p Ψ R L Ω Ψ e
21
r m e
r
H p T Tp
48
Neglect of Stator p terms
49
Classic Model
•Eliminate the differential equations on flux linkages (swing equations are the only differential equations left)
•Assume X’d=X’q
0 0( )t a d tE E R jX I
E’ is constant and can be estimated by computing its pre-disturbance value
( )t a d tE E R jX I
21
r m e
r
H p T Tp
50
Simplified models neglecting p and saliency effects
Used in short-circuit analysis
Used for stability studies
Used for steady-state analysis
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Comparison of PSS/E Generator Models