• From now, we ignore the superbar “-” with variables in per unit.
ψ 0 
ψ 
 d
ψ F 
 
ψ D 
ψ q 
 
ψ Q 
 L0








Ll + Lad
Lad
Lad
LF
Lad
MR
Lad
MR
LD
Ll + Laq
Laq
41
  −i0 
  −i 
  d
  iF 
× 
  iD 
Laq   −iq 
  
LQ   iQ 
Equivalent Circuits for d- and q-axes
ed = -ωrψq -Raid + pψd
= -ωrψq -Raid +Lad×d(-id+iF+iD)/dt -Ll×did/dt
eF = RFiF + pψF
= RFiF -Lad×did/dt +LF×diF/dt +MR×diD/dt
eD=0= RDiD+pψD
=RDiD -Lad×did/dt +MR×diF/dt +LD×diD/dt
Ll+Lad
eq= ωrψd -Raiq + pψq
=ωrψd -Raiq +Laq×d(-iq+iQ)/dt -Ll×diq/dt
eQ=0=RQiQ +pψQ
=RQiQ -Laq×diq/dt + LQ×diQ/dt
Ll+Laq
42
Equivalent Circuits for d- and q-axes
ed= Lad×d(-id+iF+iD)/dt -Ll×did/dt -ωrψq -Raid
iD+iF
eF= -Lad×did/dt +LF×diF/dt +MR×diD/dt +RFiF
= Lad×d(-id+iF+iD)/dt +(MR- Lad)×d(iD+iF) /dt
+(LF- MR)×diF/dt + RFiF
eD=0= -Lad×did/dt +MR×diF/dt +LD×diD/dt+ RDiD
= Lad×d(-id+iF+iD)/dt +(MR- Lad)×d(iD+iF)
+(LD- MR)×diD/dt+ RDiD
eq= Laq×d(-iq+iQ)/dt -Ll×diq/dt +ωrψd -Raiq
eQ=0= -Laq×diq/dt + LQ×diQ/dt +RQiQ
= Laq×d(-iq+iQ)/dt +(LQ-Laq)×diQ/dt +RQiQ
43
Equivalent Circuits with Multiple Damper
Windings (e.g. Round-rotor Machines)
d axis:
Lfd ≜LF - MR
Rfd ≜RF
ψfd ≜ψ F
efd ≜eF
L1d (or Lkd1) ≜LD - MR
R1d (or Rkd1) ≜RD
MR-Lad ≈0
L1d =LD - MR
Lfd=LF - MR
Rfd=RF
R1d =RD
efd=eF
(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(or Lkq1)≜LQ – Laq
R1q(or Rkq1)≜RQ
L2q(or Lkq2)≜LG - Laq
R2q(or Rkq2)≜RG
L1q=LQ – Laq L2q=LG – Laq
R1q=RQ
44
R2q=RG
Example: a model with 3 rotor windings in each of dand 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
45
46
Example 3.1 (Kundur’s book)
A 555 MVA, 24kV, 0.9p.f., 60Hz, 3 phase, 2 pole synchronous generator has the following
inductances an resistances associated with the stator and field windings:
mH
lab=-1.6379-0.0458 cos(2θ+π/3) mH
laa=3.2758+0.0458 cos(2θ)
mH
LF=576.92
mH
laF=40.0 cosθ
Ω
RF=0.0715
Ω
Ra=0.0031
a. Determine Ld and Lq in H
b. If the stator leakage inductance Ll is 0.4129mH, determine Lad and Laq in H
c. Using the machine rated values as the base values for the stator quantities, determine the per
unit values of the following in the Lad base reciprocal per unit system (assuming Lad=MF=MR
in per unit): Ll, Lad, Laq, Ld, Lq, MF, LF, Lfd, Ra and RF
Solution:
=3.2758+0.0458 cos2θ mH
a. laa =Ls + Lm cos2θ
lab = -Ms - Lmcos(2θ+π/3) = -1.6379-0.0458 cos(2θ+π/3) mH
=40.0 cosθ mH
laF = MF cosθ
Ld = Ls + Ms + 3Lm/2
Lq = Ls + Ms - 3Lm/2
b. Lad = Ld – Ll
Laq = Lq – Ll
3
2
3
1.6379+ ×0.0458
2
=3.2758+1.6379+ ×0.0458
=3.2758-
=4.9825 – 0.4129
=4. 4851 – 0.4129
47
=4.9825 mH
=4.8451 mH
=4.5696 mH
=4.432 mH
c.
3-phase VAbase = 555 MVA
ERMS base
es base (peak)
IRMS base
is base (peak)
=24/ 3
=13.856 kV
= 2× IRMS base = 2×13.3512×103
=18.8815×103A
= 2× ERMS base = 2× 13.856
=19.596 kV
=3-phaseVAbase / (3× ERMS base)=555×106/(3×13.856×103)=13.3512×103A
=1.03784 Ω
=es base/is base
=19.596×103/(18.8815×103)
=2π×60
=377 elec. rad/s
Ls base
=Zs base/ωbase
=1.03784/377×103
=2.753
mH
iF base
=Lad/MF× is base
=4.5696/40×18.8815×103
=2158.0
A
eF base
=3-phaseVAbase/iFbase = 555×106/2158
ZF base
= eF base / iF base
=257.183×103/2158
=119.18
Ω
LF base
= ZF base / ωbase
=119.18×103/377
=316.12
mH
MF base
= LS base × iS base / iF base =2.753×188815/2158
Zs base
ωbase
=257.183 kV
=241 mH
Then per unit values are:
Ll =0.4129/2.753
=0.15 pu
MF=MF/MF base=400/241
=1.66 pu
Lad=4.5696/2.753
=1.66 pu
LF=576.92/316.12
=1.825 pu
Laq=4.432/2.753
=1.61pu
Lfd = LF-MR =LF-Lad=1.825 - 1.66 =0.165 pu
Ld= Ll + Lad=0.15+1.66
=1.81 pu
Ra=0.0031/1.03784
=0.003 pu
Lq= Ll + Laq=0.15+1.61
=1.76 pu
RF=0.0715/119.18
=0.0006 pu
48
Steady-state Analysis
All flux linkages, voltages and currents are
constant:
pψ1d=0 → R1di1d=0→ i1d=0
pψ1q=0 → R1qi1q=0→ i1q=0
(Damper winding currents are all zero due to
no change in the magnetic field)
ψd= -( Ll+Lad)id+Ladifd+Ladi1d = -Ldid+Ladifd
ψq= -( Ll+Laq)iq+Laqi1q
= -Lqiq
pψd=0→ ed = -ωrψq -Raid =ωrLqiq -Raid
pψq=0→ eq =ωrψd -Raiq =-ωrLdid +ωrLadifd -Raiq
pψfd=0→ efd= Rfdifd
ωr=1 and L=X in p.u.
ed=Xqiq -Raid
eq=-Xdid +Xadifd -Raiq
Can we have a single equivalent circuit
representing both d and q axes circuits under
balanced steady-state conditions?
ifd =(eq+ Raiq+Xdid)/Xad
49
ea= Em cos(ωst+α)
eb= Em cos(ωst +α-2π/3)
ec= Em cos(ωst +α+2π/3)
θ=ωrt+θ0
ed= Em cos(α -θ0)= Et cos(α -θ0)
eq= Em sin(α -θ0) = Et sin(α -θ0)
𝐸�𝑡 =ed+jeq
𝐼̃𝑡 =id+jiq
where Et is the per unit RMS value of the
armature terminal voltage, which equals the peak
value Em in per unit.
𝐸�𝑡 = Xqiq -Raid+-jXdid +jXadifd -jRaiq
= -Ra𝐼̃𝑡 +Xqiq -jXdid +jXadifd
𝐸�𝑡 = 𝐸� –Ra𝐼̃𝑡 –jXS 𝐼̃𝑡
XS=Xq
𝐸�E
𝐸� =j[Xadifd-(Xd-Xq)id] ≝ 𝐸�𝑞
𝐸�𝑡 =𝐸�𝑞 – Ra𝐼̃𝑡 – jXq𝐼̃𝑡
50
ed=Xqiq -Raid
eq=-Xdid +Xadifd -Raiq
𝐼̃𝑡
𝐸�𝑡
Load
Steady-state equivalent circuit
• Under no-load or open-circuit conditions, id=iq=0.
𝐸�𝑡 = 𝐸�𝑞 =jXadifd
δi=0 (load angle)
• For round rotor machines or machines with
neglected saliency
Xd=Xq=Xs (synchronous reactance)
𝐸�𝑡 =𝐸�𝑞 – (Ra+jXs)𝐼̃𝑡
Eq=Xadifd
51
𝐸�𝑞 =j[Xadifd-(Xd-Xq)id]
Computing steady-state values
52
Representation of Magnetic Saturation
• Assumptions for stability studies
– The leakage fluxes are not significantly affected by saturation of
the iron portion, so Ll is constant and only Lad and Laq saturate in
equivalent circuits
Ladu→Lad
Ladu and Laqu denote the
unsaturated values
Laqu→ Laq
– The leakage fluxes do not contribute to the iron saturation. Thus,
saturation is determined only by the air-gap flux linkage
ψat = ψ2𝑎𝑎 + ψ2𝑎𝑎
– The saturation relationship ψat ~ ifd or ψat ~ MMF under loaded
conditions is the same as under no-load conditions. This allows
the open-circuit characteristic (OCC) to be considered
– No magnetic coupling between d and q axes, such that their
effects of saturation can be modeled individually
53
Air-gap flux and voltage
=
ψ at ψ ad
+ jψ aq
For steady state conditions:
=ψ d + Ll id + jψ q + jLl iq
ed = -ψq -Raid → ψq = - ed -Raid
=eq + Ra iq − jed − jRa id + Ll It
eq = ψd - Raiq → ψd = eq+ Raiq
=
− jE t − jRa It + Ll It
ψ at =Eat =| E t + ( Ra + jX l ) It |
∆
=
− j[ E t + ( Ra + jX l ) It ] =
− jE at
54
Estimating Saturation Factors Ksd and Ksd
Lad=KsdLadu
Laq=KsqLaqu
• Salient pole machines
– The path for q-axis flux is largely
in air, so Laq does not vary
significantly with saturation of the
iron portion of the path
– Assume Ksq=1.0 for all loading
conditions.
• Round rotor machines
– There is a magnetic saturation in
both axes, but q axis saturation
data is usually not available
– Assume Ksq =Ksd
Ksd = Lad/Ladu=ψat/ψat0 = ψat/(ψat+ ψI)
• Thus, we focus on estimating Ksd
= I0 / I
(See Kundur’s Example 3.3 on Estimating Ksd
for different loading conditions)
55
Estimating the Saturation Characteristic
• Modeled by 3 approximate functions
– Segment I (ψat<ψT1):
ψI=ψat0 - ψat=0
ψI
– Segment II (ψT1<ψat<ψT2):
ψ = 𝐴 𝑒 𝐵𝑠𝑠𝑠(ψ𝑎𝑎−ψ𝑇𝑇)
𝐼
𝑠𝑠𝑠
– Segment III (ψat>ψT2):
linear
nonlinear
ψI=ψG2+Lratio(ψat-ψT2)- ψat
• Note that segments I and II are not
connected since when ψat=ψT1,
ψI=Asat≠ 0 (usually small)
linear
• Assume that segments II and III are
connected at ψat=ψT2 to solve ψG2
𝐴𝑠𝑠𝑠 𝑒 𝐵𝑠𝑠𝑠(ψ𝑇𝑇 −ψ𝑇𝑇 ) =ψG2+Lratio(ψT2-ψT2)- ψT2
ψG2 = ψT2+𝐴𝑠𝑠𝑠 𝑒 𝐵𝑠𝑠𝑠(ψ𝑇𝑇 −ψ𝑇𝑇 )
56
We need to know Asat, Bsat,
ψT1, ψT2 and Lratio
57
P186 of Anderson’s book
S(1.0)
S(1.2)
ψ𝐼 = 𝐴𝑠𝑠𝑠 𝑒 𝐵𝑠𝑠𝑠 (ψ𝑎𝑎 −ψ𝑇𝑇 ) ~
58
SG = AG e BG (Vt −0.8)
Example 3.2 in Kundur’s Book
R1d>>Rfd
Lfd/Rfd>>L1d/R1d
L1q/R1q>>L2q/R2q
MR
59
60
≈39.1o
≈1.565pu
61
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
62
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= 𝐿𝑎𝑎 +𝐿𝑓𝑓
8.07(s)
Time constant
(short circuit)
Inductance
(Reactance)
𝑅𝑓𝑓
q axis circuit
Add the damper
winding
T’’d0=
𝐿𝑎𝑎 //𝐿𝑓𝑓 + 𝐿1𝑑
𝑅1𝑑
0.03(s)
Only 1st damper
winding
T’q0= 𝐿𝑎𝑎 +𝐿1𝑞
1.00(s)
T’d= 𝐿𝑎𝑎 //𝐿𝑙 + 𝐿𝑓𝑓 T’’d= 𝐿𝑎𝑎 //𝐿𝑓𝑓 //𝐿𝑙 + 𝐿1𝑑 T’ =
q
𝑅𝑓𝑓
L’d= Ll+Lad//Lfd
0.30(pu)
𝑅1𝑑
L’’d= Ll+Lad//Lfd//L1d
0.23(pu)
Based on the parameters of Example 3.2
𝑅1𝑞
L’q=
𝐿𝑎𝑎 //𝐿𝑙 + 𝐿1𝑞
𝑅1𝑞
Ll+Laq//L1q
0.65(pu)
63
Add the 2nd damper
winding
T’’q0= 𝐿𝑎𝑎 //𝐿1𝑞 + 𝐿2𝑞
0.07(s)
𝑅2𝑞
T’’q= 𝐿𝑎𝑎 //𝐿1𝑞 //𝐿𝑙 + 𝐿2𝑞
𝑅2𝑞
L’’q= Ll+Laq//L1q//L2q
0.25(pu)
Identifying Terminal Quantities
• Apply Laplace transform to the incremental
forms of flux linkage equations (see Kundur’s
Chapter 4.1 for details):
∆ψd(s) = G(s) ∆efd(s) – Ld(s) ∆id(s)
∆ψq(s) = –Lq(s) ∆iq(s)
≈ Ld
(1 + sT4 )(1 + sT6 )
(1 + sT1 )(1 + sT3 )
≈ G0
(1 + sTkd )
(1 + sT1 )(1 + sT3 )
Since R1d>>Rfd
T2, T3<<T1,
T5,T6<<T4
T1+T2≈ T1+T3, T4+T5≈ T4+T6
>>
>>
>>
(1 + sTd′ )(1 + sTd′′) ≈ (1 + sT4 )(1 + sT6 )
(1 + sTd′0 )(1 + sTd′′0 ) ≈ (1 + sT1 )(1 + sT3 )
64
(Classical Expressions)
65
Summary of Transient and Sub-transient
Parameters (Classical Expressions)
d axis circuit
OC Time
Constant
SC Time
Constant
Inductance
(Reactance)
T’d0=
𝐿𝑎𝑎 +𝐿𝑓𝑓
𝑅𝑓𝑓
q axis circuit
T’’d0=
𝐿𝑎𝑎 //𝐿𝑓𝑓 + 𝐿1𝑑
𝑅1𝑑
T’q0= 𝐿𝑎𝑎 +𝐿1𝑞
𝑅1𝑞
T’d=
T’’d=
T’q= 𝐿 //𝐿 + 𝐿
𝑎𝑎
𝑙
1𝑞
𝐿𝑎𝑎 //𝐿𝑙 + 𝐿𝑓𝑓
𝐿𝑎𝑎 //𝐿𝑓𝑓 //𝐿𝑙 + 𝐿1𝑑
𝑅1𝑞
𝑅𝑓𝑓
𝑅1𝑑
L’d=
Ll+Lad//Lfd
L’’d=
Ll+Lad//Lfd//L1d
L’q=
Ll+Laq//L1q
T’’q0=
𝐿𝑎𝑎 //𝐿1𝑞 + 𝐿2𝑞
𝑅2𝑞
T’’q=
𝐿𝑎𝑎 //𝐿1𝑞 //𝐿𝑙 + 𝐿2
𝑅2𝑞
L’’q=
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/ωbase (i.e. 1/377 for 60Hz).
66
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′′
= Ll +
Ld′′ = Ld (∞) = Ld
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
67
(d-axis transient inductance)
Xd ≥ Xq ≥ X’q ≥ X’d ≥ X”q ≥ X”d
68
T’d0 > T’d > >T”d0 > T”d
T’q0 > T’q > >T”q0 > T”q
Parameter Estimation by Frequency Response Tests
∆ψd(s) = G(s) ∆efd(s) – Ld(s) ∆id(s)
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
69
70
Swing Equations
dωm
J
= Ta = Tm − Te
dt
J
ωm
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 Jω0m
H=
(s)
2 VA base
J=
2H
ω
2
(s)
VA base
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
71
d ωm
= T=
Tm − Te
a
dt
d ωm
2H
= Tm − Te
VA
base
ω02m
dt
Angular position of the rotor in electrical radian
with respect to a synchronously rotating reference
J
δ = ω r t − ω0 t + δ 0
in rad
d  ωm
2H 
dt  ω 0 m

Tm − Te
 =
 VA base / ω 0 m
dδ
ω r − ω0 =
=
∆ωr in rad/s
dt
d ω
2H  m
dt  ω 0 m
 Tm − Te
 =
Tbase

d 2δ d ωr d (∆ωr )
in rad/s2
=
=
dt 2
dt
dt
2H
d ωr
= Tm − Te
dt
(in per unit)
where
ωm ωr / p f ωr
=
=
ωr (in per unit) =
ω0 m ω0 / p f ω 0
d (ωr )
d (∆ωr )
d 2δ
2
in
rad/s
=
ω
=
ω
0
0
dt 2
dt
dt
∆ωr=
∆ωr
=
ω0
1 dδ
ω0 dt
2 H d 2δ
= Tm − Te
2
ω0 dt
If adding a damping term
proportional to speed deviation:
2 H d 2δ
K D dδ
=
T
T
K
ω
T
T
=−
∆
−
m
e
D
r
m
e
ω0 dt 2
ω0 dt
72
Block diagram representation of swing equations
2 H d 2δ
K D dδ
=
T
T
K
T
T
ω
=−
∆
−
m
e
D
r
m
e
ω0 dt 2
ω0 dt
2H
d (∆ωr )
Tm Te - K D ∆ωr
=−
dt
1 dδ
∆ωr =
ω0 dt
73
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
74
Differential Equations on a Salient-pole Machine
• Consider 5 windings (d, F, D, q and Q)
– Voltage equations:
e=R×i+L×di/dt+ωrN×I
pi = -
L-1(R+ωrN)
i
Since ψ=L×i,
+L-1e
here is pψ= - (R+ωrN)L-1ψ+e
– Swing equations:
𝑇𝑒 = ψ𝑑 𝑖𝑞 − ψ𝑞 𝑖𝑑
τ pωr =Tm -Te (where τ=2H/ω0)
pδ=ωr - ω0=ωr - 1
ψ𝑑 = −𝐿𝑑 𝑖𝑑 + 𝑘𝑀𝐹 𝑖𝐹 + 𝑘𝑀𝐷 𝑖𝐷 ,
ψ𝑞 = −𝐿𝑞 𝑖𝑞 + 𝑘𝑀𝑄 𝑖𝑄
75
𝑇𝑒 = −𝐿𝑑 𝑖𝑞 ⋅𝑖𝑑 + 𝑘𝑀𝐹 𝑖𝑞 𝑖𝐹 + 𝑘𝑀𝐷 𝑖𝑞 𝑖𝐷 +
𝐿𝑞 𝑖𝑑 ⋅𝑖𝑞 − 𝑘𝑀𝑄 𝑖𝑑 𝑖𝑄
State-space Model
d
 i 
 d 
 i 
 F 
i 
 D 
 i 
 q  / dt


i
Q




ωr 


 δ 



















0
0
−L−1  R +ω N 
r 

0
0
0
(T’d0, T”d0, T’q0, T”q0)
Ld iq
−kM F iq
−kM Diq
− Lqid
kM Q id
τ
τ
τ
τ
τ
0
0
0
0
0
0
1
0 
 i
0   d

  iF
0 
  iD
0   iq


0  iQ













 ω 
0 r


 δ 
0   




 L−1e

+ 



 Tm /τ

 −1














Thus, the state-space model of a synchronous machine:
dx/dt=f(x, ed, eq and efd, Tm)
where x=[id iFd i1d iq i1q ωr δ]T or [ψd ψfd ψ1d ψq ψ1q ωr δ]T is the state vector.
(See the section 4.12 in Anderson’s book).
efd and Tm are usually known but ed and eq are related to state variables, so
additional equations (on the machine outside, e.g. loading conditions) should
be introduced
76
Simplified Models
• [ψd ψfd ψ1d ψq ψ1q ωr δ]T
Neglect pψd, pψq and variations of ωr
(i.e. ωr=1 pu) in the voltage equations
=ψq= -Lqiq
• [ψfd ψ1d ψ1q ωr δ]T
– Inertia
τ ~ pωr
– Transient
T’d0 ~ pψfd
– Sub-transient T”d0 ~pψ1d
T”q0 ~ pψ1q
Neglect damper windings, i.e. pψ1d
and pψ1q
=ψd= -Ldiq
• [ψfd ωr δ]T
– Inertia
– Transient
τ ~ pωr
T’d0~ pψfd
Constant flux linkage assumption
• [ωr δ]T (classic model)
– Inertia
τ ~ pωr
77
Neglect of Stator pψ terms
78
Neglecting Damper Windings
ψd= - Ldid+ Ladifd
ψq= - Lqiq
ψfd= - Ladid+ (Lad+Lfd) ifd
pψfd= -efd - Rfdifd
(only differential equation left for the equivalent circuits)
T’d0pE’q= Efd –E’q–(Ld– L’d)id
T’q0pE’d=
–E’d–(Lq– L’q)iq
𝐿𝑎𝑎
ψ
𝐿𝐹 fd
where E’q=
𝐿𝑎𝑎
e
𝑅𝑓𝑓 fd
and Efd=
(added for round-rotor machines to give the so called
“Two-Axis Model”)
X’d=Ll+Lad//Lfd
Xq=Ll+Laq
79
Classic Model
• Eliminate the differential equations on flux linkages (swing
equations are the only differential equations left)
• Assume X’d=X’q
2𝐻
pωr
𝜔0
=Tm -Te
pδ=ωr -1
~
~ ~
Et = E ′ − ( Ra + jX d′ ) I t
E ′ = E t 0 + ( Ra + jX d′ ) It 0
E’ is constant and can be estimated by
computing its pre-disturbance value
80
Simplified models neglecting pψ and saliency effects
Used in short-circuit analysis
Used for stability studies
Used for steady-state analysis
81
Comparison of PSS/E Generator Models
82