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TRANSMISSION & DISTRIBUTIONA Division of Global Power
POWER SYSTEM STABILITY CALCULATION TRAININGD 2 T i t St bilitDay 2 - Transient Stability
July5,2013Prepared by: Peter Anderson
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OUTLINE
2
OUTLINE
The Swing Equation
Application to Synchronous Machines
Step-by-Step Solution Method eBoo
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THE SWING EQUATION
3
THE SWING EQUATIONPower-Angle Relationship
d2 = d = 0.(Pm Pe)dt2 dt 2Hdt2 dt 2HH = Inertia Constant = J.2/2S in MW-s/MVA or seconds
Metric: GD2 in kg-m2 622
10.kVA
)RPM()GD(48163.5=H
Imperial: WR2 in lb.ft2 622
10.kVA
)RPM()WR(231.0=H
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ANALYSIS OF THE SWING EQUATION
4
ANALYSIS OF THE SWING EQUATIONd = 0.(Pm Pe)dt 2Hdt 2H
In terms of short-term transient stability studiesIn terms of short-term transient stability studies,
Pm is constantm
Rotor acceleration is a function of the Electrical Power
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APPLICATION TO SYNCHRONOUS
5
MACHINESIncrease in Mechanical Power InputPm increased Rotor accelerates from 25 deg. to new operating point at 40 deg. 1.2
1.4
gRotor overshoots to 60 deg, where area above Pm equals the area below PmNow Pe >Pm and rotor
0.8
1
Now Pe Pm and rotor decelerates towards =40 deg.Rotor will oscillate around =40 deg. 0.2
0.4
0.6
40 deg.If the overshoot reaches =140 deg. The rotor will not be able to return to the new operating point and will slip
0
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180
operating point and will slip to the next pole position
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STEP BY STEP SOLUTION METHOD
6
STEP-BY-STEP SOLUTION METHODPm Increased
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12)T Pm Pe Pa Acceleration t1 t2
(3)(4) (5)*k (6)*(7) deg/s (9)*(10) (2)+(11)0.0 0 Pm0 Pe0 Pm0Pe0 0 0 0=0 +0.0 0 Pm1 Pe0 Pm1Pe0 0 t/2 1 0+1=1 t 1 10.5 1 Pm1 Pe1 Pm1Pe1 1 t 2 1+2=2 t 2 21 2 Pm1 Pe2 Pm1Pe2 2 t 3 2+3=3 t 3 31.5 3 Pm1 Pe3 Pm1Pe3 3 t 4 3+4=4 t 4 42 4 Pm1 Pe4 Pm1Pe4 4 t 5 4+5=5 t 5 5
0 8
1
1.2
1.4
0.2
0.4
0.6
0.8
0
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180
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STEP BY STEP SOLUTION METHOD
7
STEP-BY-STEP SOLUTION METHODPm Increased from 0.42 to 0.8 pu
1.00
1.20
0.60
0.80
0.20
0.40
0.00
0 30 60 90 120 150 180
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STEP BY STEP SOLUTION METHOD
8
STEP-BY-STEP SOLUTION METHODT Pm Pe Pa t1 t2 0 0 25 0 0 42 0 42 0 0 0 0 0 00.0 25.0 0.42 0.42 0 0 0 0 0.00 25.0 0.80 0.42 0.38 10.87 0.25 2.72 2.7 0.5 1.4 26.40.5 26.4 0.80 0.44 0.36 10.25 0.5 5.13 7.8 0.5 3.9 30.31 30.3 0.80 0.50 0.30 8.52 0.5 4.26 12.1 0.5 6.1 36.31.5 36.3 0.80 0.59 0.21 5.98 0.5 2.99 15.1 0.5 7.5 43.92 43.9 0.80 0.69 0.11 3.08 0.5 1.54 16.6 0.5 8.3 52.22.5 52.2 0.80 0.79 0.01 0.29 0.5 0.14 16.8 0.5 8.4 60.63 60.6 0.80 0.87 0.07 2.05 0.5 1.02 15.8 0.5 7.9 68.53.5 68.5 0.80 0.93 0.13 3.75 0.5 1.87 13.9 0.5 6.9 75.44 75.4 0.80 0.97 0.17 4.83 0.5 2.41 11.5 0.5 5.7 81.14.5 81.1 0.80 0.99 0.19 5.42 0.5 2.71 8.8 0.5 4.4 85.55 85.5 0.80 1.00 0.20 5.67 0.5 2.84 5.9 0.5 3.0 88.55.5 88.5 0.80 1.00 0.20 5.75 0.5 2.87 3.0 0.5 1.5 90.06 90.0 0.80 1.00 0.20 5.76 0.5 2.88 0.2 0.5 0.1 90.16.5 90.1 0.80 1.00 0.20 5.76 0.5 2.88 2.7 0.5 1.4 88.77 88.7 0.80 1.00 0.20 5.75 0.5 2.88 5.6 0.5 2.8 85.97 5 85 9 0 80 1 00 0 20 5 69 0 5 2 84 8 4 0 5 4 2 81 77.5 85.9 0.80 1.00 0.20 5.69 0.5 2.84 8.4 0.5 4.2 81.78 81.7 0.80 0.99 0.19 5.46 0.5 2.73 11.2 0.5 5.6 76.18.5 76.1 0.80 0.97 0.17 4.92 0.5 2.46 13.6 0.5 6.8 69.39 69.3 0.80 0.94 0.14 3.90 0.5 1.95 15.6 0.5 7.8 61.59.5 61.5 0.80 0.88 0.08 2.27 0.5 1.14 16.7 0.5 8.4 53.210 53.2 0.80 0.80 0.00 0.01 0.5 0.00 16.7 0.5 8.4 44.8
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STEP BY STEP SOLUTION METHOD
9
STEP-BY-STEP SOLUTION METHODPm Increased from 0.42 to 0.8 pu
80 0
90.0
100.0
50.0
60.0
70.0
80.0
A
n
g
l
e
(
d
e
g
)
10 0
20.0
30.0
40.0
M
a
c
h
i
n
e
0.0
10.0
0 2.5 5 7.5 10 12.5 15
Time(s)
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TRANSMISSION & DISTRIBUTIONA Division of Global Power
POWER SYSTEM STABILITY CALCULATION TRAININGD 3 T i t St bilitDay 3 - Transient Stability
July8,2013Prepared by: Peter Anderson
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CASE STUDY: ANALYSIS OF A FAULT CASE
2
CASE STUDY: ANALYSIS OF A FAULT CASEL1
L2
3 h f lt li L2 l t t b3-phase fault on line L2, close to generator busFrequency = 60 HzInertia Constant = 3.0 sP f lt P A l C i /0 8Pre-fault Power Angle Curve = sin/0.8Post-fault Power Angle Curve = sin/1.1Fault duration = 0.1 sTime Step = 0.02 s
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CASE STUDY: ANALYSIS OF A FAULT CASE
3
CASE STUDY: ANALYSIS OF A FAULT CASE
Step 1: Construct Power-Angle Curvesp g
1.40
1.00
1.20
0 40
0.60
0.80
0.00
0.20
0.40
0 30 60 90 120 150 180
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CASE STUDY: ANALYSIS OF A FAULT CASE
4
CASE STUDY: ANALYSIS OF A FAULT CASEStep 1: Construct Step-by-Step Table
K 3600 =180*f/(H*S)/( )1 2 3 4 5 6 7 8 9 10 11 12
Pa t1 t2 (3)(4) (5)*K (6)*(7) (9)N1+(8)N (9)*(10) (2)+(11)
s deg pu pu pu deg/s/s s deg/s deg/s s deg deg0.0 39.8 0.80 0.80
T Pm Pe
0.0 39.8 0.80 0.800 39.8 0.80 0.00 0.80 2880 0.01 28.80 28.8 0.02 0.6 40.4
0.02 40.4 0.80 0.00 0.80 2880 0.02 57.60 86.4 0.02 1.7 42.10.04 42.1 0.80 0.00 0.80 2880 0.02 57.60 144.0 0.02 2.9 45.00.06 45.0 0.80 0.00 0.80 2880 0.02 57.60 201.6 0.02 4.0 49.00.08 49.0 0.80 0.00 0.80 2880 0.02 57.60 259.2 0.02 5.2 54.20.08 49.0 0.80 0.00 0.80 2880 0.02 57.60 259.2 0.02 5.2 54.20.1 54.2 0.80 0.00 0.80 2880 0.01 28.80 288.00.1 54.2 0.80 0.76 0.04 152 0.01 1.52 289.5 0.02 5.79 60.00.12 60.0 0.80 0.81 0.01 52 0.02 1.04 200.6 0.02 4.0 64.00.14 64.0 0.80 0.85 0.05 179 0.02 3.58 197.0 0.02 3.9 67.90.16 67.9 0.80 0.88 0.08 291 0.02 5.83 191.2 0.02 3.8 71.80.16 67.9 0.80 0.88 0.08 291 0.02 5.83 191.2 0.02 3.8 71.80.18 71.8 0.80 0.91 0.11 389 0.02 7.78 183.4 0.02 3.7 75.40.2 75.4 0.80 0.93 0.13 471 0.02 9.42 174.0 0.02 3.5 78.90.22 78.9 0.80 0.95 0.15 539 0.02 10.77 163.2 0.02 3.26 82.20.24 82.2 0.80 0.96 0.16 592 0.02 11.85 151.3 0.02 3.03 85.20.26 85.2 0.80 0.98 0.18 634 0.02 12.68 138.7 0.02 2.77 88.00.26 85.2 0.80 0.98 0.18 634 0.02 12.68 138.7 0.02 2.77 88.00.28 88.0 0.80 0.98 0.18 665 0.02 13.30 125.4 0.02 2.51 90.50.3 90.5 0.80 0.99 0.19 687 0.02 13.74 111.6 0.02 2.23 92.7
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CASE STUDY: ANALYSIS OF A FAULT CASE
5
CASE STUDY: ANALYSIS OF A FAULT CASEStep 1: Plot Results
100.0
120.0
60.0
80.0
h
i
n
e
A
n
g
l
e
(
d
e
g
)
20.0
40.0
M
a
c
h
0.0
0 0.1 0.2 0.3 0.4 0.5 0.6
Time(s)
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THREE PHASE SHORT CIRCUIT
6
THREE PHASE SHORT-CIRCUITSteady-state:Leakage Reactance of the machine (X )Leakage Reactance of the machine (Xl) Armature Reaction to the Fault Current (Xa)Synchronous Reactance Xd = Xl + Xay d l a
Xl Xa
E
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THREE PHASE SHORT CIRCUIT
7
THREE PHASE SHORT-CIRCUITAt the instant of the Fault:
Leakage Reactance of the machine (Xl)Armature Reaction to the Fault Current (Xa)Since the Air Gap flux cannot change instantaneously, currents are induced in the field and damper windings (Xdw and Xf)f)Sub-transient Reactance Xd = Xl + 1/(1/Xa +1/Xdw + 1/Xf)
Xl Xa
XdwE Xf
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THREE PHASE SHORT CIRCUIT
8
THREE PHASE SHORT-CIRCUITShortly after the instant of the Fault:Leakage Reactance of the machine (Xl)Leakage Reactance of the machine (Xl)Armature Reaction to the Fault Current (Xa)Damper winding currents (low X/R ratio) die out quicklyq yTransient Reactance Xd = Xl + 1/(1/Xa + 1/Xf)
E
Xl Xa
E Xf
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THREE PHASE SHORT CIRCUIT
9
THREE PHASE SHORT-CIRCUIT
6
2
4
(
p
u
)
Subtransient
4
2
0
C
u
r
r
e
n
t
Transient
Synchronous
6
4
0 0.05 0.1 0.15 0.2 0.25 0.3Time(s)
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SYNCHRONOUS MACHINE MODELS
10
SYNCHRONOUS MACHINE MODELSSingle Phase Equivalent of a 3-phase Generator
jXd I
E U~
Im
EjXd IjXd.I
Re U
II
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SYNCHRONOUS MACHINE MODELS
11
SYNCHRONOUS MACHINE MODELSThree Possible Generator Models:Synchronous Model (Constant voltage behind Xd)Synchronous Model (Constant voltage behind Xd)Transient Model (Constant voltage behind Xd)Sub-transient Model (Constant/Variable voltage behind Xd))
Sub-transient model allows exciter effects to be explicitly represented
For each model, the prime mover can be represented as a fixed power model or a variable power model under the control of governor p gaction
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MODEL APPLICATION
12
MODEL APPLICATIONUse of Mixed Generator Models:Complex models used for machines of interestSimpler models used for remote machines
Requires less dataRequires less dataSignificantly reduces the computing burden
With the removal of computer limitations, it is recommended to use the sub-transient model for all generators
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TRANSMISSION & DISTRIBUTIONA Division of Global Power
POWER SYSTEM STABILITY CALCULATION TRAININGD 4 T i t St bilitDay 4 - Transient Stability
July9,2013Prepared by: Peter Anderson
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OUTLINE
2
OUTLINE Machine Differential Equations
Exciter Differential Equations
G Diff ti l E ti Governor Differential Equations
Solution of Differential Equations
Network Solution
Sample Cases
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BASIC MODELS IN STABILITY STUDIES
3
BASIC MODELS IN STABILITY STUDIESSynchronous MachinesSwing Equations:Swing Equations:
( )aem0 p.DPP.H
=p
0=p
H
Flux Linkage Differential Equations:
( )[ ] ( )[ ]'dq'qq'd"dq"qq"d EIXX1=pEEIXX1=pE ( )[ ] ( )[ ]
( )[ ] ( )[ ]'qd'ddfd''q"qd"d'd'q""q
dqqq'0q
ddqqq"0q
d
EIXXE1
=pEEIXXE1
=pE
EIXXT
pEEIXXT
pE
( )[ ] ( )[ ]qdddfd'0d
qqdddq"0d
qT
pT
p
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BASIC MODELS IN STABILITY STUDIES
4
BASIC MODELS IN STABILITY STUDIESSynchronous MachinesAlgebraic Model:Algebraic Model:
VE.Y=II
.XR
=VE dqadd
( )FrameferenceReSystemtoFrameferenceReMachinefromtionTransforma
VE.YII
.RXVE qadqq
( ) ( )( )2/jexpI+I=jI+I qdMRI
YY.E
Y.V V
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BASIC MODELS IN STABILITY STUDIES
5
BASIC MODELS IN STABILITY STUDIESExciter
( )[ ]EVVVK1=pE fdfbtsEfd ( )[ ]
( )VpEKT1
=pV
EVVVKT
pE
fbfdff
fb
fdfbtsEE
fd
maxDpE;maxEVminE;V+KE
=V fdE0tE
0fds
f
Vs Exciter Limiter RateLimiter+
Vt KE EfdEmax, Dmax1+TEp
,Emin Dmax
Vfb pKf1+Tfp
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BASIC MODELS IN STABILITY STUDIES
6
BASIC MODELS IN STABILITY STUDIESGovernor
( ) ( )[ ]1 ( ) ( )[ ]
[ ]mons
m
on00sc
on
PCT1
=pP
CR/PT1
=pC
limM
s
GP
T
0 Regulator Limiter Control Valve Turbine
+ 1 1 Con 1 Pm0.R 1+Tcp 1+Tsp
Ps
Glim
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SOLUTION OF DIFFERENTIAL EQUATIONS
7
SOLUTION OF DIFFERENTIAL EQUATIONSMethod of Integration:Implicit Trapezoidal Rule 70
80
90
Implicit Trapezoidal Rule( ) ( ) ( )( )
VariableIntegrable=Y
pY+pY.2h
+Y=Y th+tt)h+t(
20
30
40
50
60
Y
LengthStepnIntegratio=h
VariableIntegrableNon=X
g
I i T f i f Diff i l E i
0
10
20
2 4 6 8 10
X
Iterative Transformation of Differential Equations to Algebraic Equations:
( )h+t)h+t( X.YX+YC=Y ( )
( ) ( )( ) ( )( )[ ]
( )[ ]
ttt
h+t)h+t(
h
XF,YF.2h
+Y=ttanCons=YC
X.YX+YCY
( ) ( )( )[ ]h+th+t XF.2h
=X.YX
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SOLUTION OF DIFFERENTIAL EQUATIONS
8
SOLUTION OF DIFFERENTIAL EQUATIONSSolution Algorithm:
1 Establish Initial Conditions by Solving for Y and X at t=01. Establish Initial Conditions by Solving for Y and X at t=02. Solve for Y(t+h) using values of X(t) by means of Trapezoidal
Rule3 Solve for X using values of Y from above3. Solve for X(t+h) using values of Y(t+h) from above4. Resolve for Y(t+h) using values of X(t+h)5. Repeat Steps 2, 3 & 4 until convergence is achieved6 Proceed to next time step6. Proceed to next time step
Numerically StableAllows Larger Time Steps to be used than other methodsgIn Step 2 in practice, an extrapolated value of X(t) is used except immediately after a discontinuity usingX = 2X XX(t+h) = 2X(t) X(t-h)
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NETWORK SOLUTION
9
NETWORK SOLUTION1. Calculate Integrable Variables2 T f f M hi R f F t S t2. Transform from Machine Reference Frame to System
Reference Frame3. Calculate Norton equivalent Nodal Injected Current4. Solve for V using I = Y.V where Y = Nodal Admittance4. Solve for V using I Y.V where Y Nodal Admittance
Matrix5. Use New values for V to move to next time step6. Network Solution carried out simultaneously with
Integration ProcessIntegration Process eBook
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SAMPLE CASES
10
SAMPLE CASES4x500MWST 250MWH
GBUS1 GBUS2
600MVA 0 975pf
4x590MVA,21/400kV 1x295MVA,21/400kV
600MVA,0.975pf
100km
HVBUS1
400km
100km 50km
GBUS3
HVBUS31x590MVA,21/400kV
600MVA,0.975pf1200MVA,0.975pf
500MWST
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FAULT POSITION 1
11
FAULT POSITION-1Line Fault close to Generator HV Bus
BaseFaultatGBus1 BaseFaultatGBus2
0
30
60
90
e
(
d
e
g
r
e
e
s
w
r
t
G
1
)
0
30
60
90
e
(
d
e
g
r
e
e
s
w
r
t
G
1
)
90
60
30
0 50 100 150 200
M
a
c
h
i
n
e
A
n
g
l
e
90
60
30
0 50 100 150 200
M
a
c
h
i
n
e
A
n
g
l
e
Time(cycles) G2 G3 Time(cycles) G2 G3
60
90
w
r
t
G
1
)
BaseFaultatGBus3
60
30
0
30
c
h
i
n
e
A
n
g
l
e
(
d
e
g
r
e
e
s
w
90
60
0 50 100 150 200
M
a
c
Time(cycles) G2 G3
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FAULT POSITION 2
12
FAULT POSITION-2Line Fault close to System HV Bus
90BaseFaultatHVBus1
90BaseFaultatHVBus3
0
30
60
g
l
e
(
d
e
g
r
e
e
s
w
r
t
G
1
)
0
30
60
n
g
l
e
(
d
e
g
r
e
e
s
w
r
t
G
1
)
90
60
30
0 50 100 150 200
M
a
c
h
i
n
e
A
n
g
90
60
30
0 50 100 150 200
M
a
c
h
i
n
e
A
n
Time (cycles) G2 G3Time(cycles) G2 G3 Time(cycles)G2 G3
10
20
s
w
r
t
G
1
)
BaseAllFaults
40
30
20
10
0
a
c
h
i
n
e
A
n
g
l
e
(
d
e
g
r
e
e
s
50
0 10 20 30 40 50
M
a
Time(cycles)
HVBus1 HVBus3 GBus1 GBus2 GBus3
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GENRATOR INERTIA
13
GENRATOR INERTIALine Fault close to Generator HV Bus (GBUS-3_
10
20
w
r
t
G
1
)
G3Hincreased
20
10
0
g
l
e
(
d
e
g
r
e
e
s
w
40
30
20
0 50 100 150 200
M
a
c
h
i
n
e
A
n
g
0 50 100 150 200
Time(cycles)
G3 G30
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GENERATOR EXCITATION CONTROL
14
GENERATOR EXCITATION CONTROLLine Fault close to Generator HV Bus (GBUS-3_
10
20
r
t
G
1
)
Excitersadded
20
10
0
l
e
(
d
e
g
r
e
e
s
w
r
60
50
40
30
M
a
c
h
i
n
e
A
n
g
l
60
0 50 100 150 200
Time(cycles) G30 G3
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GENERATOR REACTANCE
15
GENERATOR REACTANCELine Fault close to Generator HV Bus (GBUS-3_
60
90
t
G
1
)
G3Xdreduced
0
30
60
e
(
d
e
g
r
e
e
s
w
r
t
60
30
M
a
c
h
i
n
e
A
n
g
l
e
90
0 50 100 150 200
Time(cycles) G3 G30
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GENERATOR MODEL TYPE
16
GENERATOR MODEL TYPELine Fault close to Generator HV Bus (GBUS-3_
60
90
r
t
G
1
)
G2FixedV/Xd'
0
30
g
l
e
(
d
e
g
r
e
e
s
w
r
90
60
30
M
a
c
h
i
n
e
A
n
g
0 50 100 150 200Time(cycles)G2 G3 G20 G30
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FAULT CLEARING TIME
17
FAULT CLEARING TIMELine Fault close to Generator HV Bus (GBUS-3)
120150180
w
r
t
G
1
)
BaseAllFaults
306090
120
l
e
(
d
e
g
r
e
e
s
w
9060300
M
a
c
h
i
n
e
A
n
g
90
0 10 20 30 40 50Time(cycles)
5cycles 7cycles 10cycles 12cycles
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LOAD REPRESENTATION
18
LOAD REPRESENTATIONLine Fault close to Generator HV Bus (GBUS-3_
60
90
r
t
G
1
)
ConstantPowerAllFaults
0
30
60
e
(
d
e
g
r
e
e
s
w
r
30
0
M
a
c
h
i
n
e
A
n
g
l
e
60
0 10 20 30 40 50
M
Time(cycles)
3cycles 4cycles 5cycles3 cycles 4 cycles 5 cycles
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SUMMARY OF CASES
19
SUMMARY OF CASESParameter Stability
Increase DecreaseIncrease DecreaseFaultPosition SystemDependentIncreasedInertia Y GeneratorExcitationControl Y DecreasedSynchronousReactance Y Si lifi d G t M d l YSimplifiedGeneratorModel Y
IncreasedFaultClearingTime Y
Decreased VoltageDependencyofLoad Y
Fault Clearing Time & Voltage Dependency of Load have the most significant impacts on stabilitythe most significant impacts on stability
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