TRANSMISSION &
DISTRIBUTION
A Division of Global Power
POWER SYSTEM STABILITY CALCULATION TRAINING
D2
T i t St bilit
D
ay
2
-
T
rans
i
en
t

St
a
bilit
y
July5,2013
Prepared by: Peter Anderson
eBook for You
OUTLINE
2
OUTLINE
• The Swing Equation
• Application to Synchronous Machines
• Step-by-Step Solution Method
eBook for You
THE SWING EQUATION

=H
Imperial: WR
2
in lb.ft
2
6
22
10.kVA
)RPM(×)WR(×231.0
=H
eBook for You
ANALYSIS OF THE SWING EQUATION
4
ANALYSIS

OF

THE

SWING

EQUATION
dω = ω0.(Pm – Pe)
dt
2H
dt
2H
In terms of short
-
term transient stability studies

at 40 de
g
.
1.2
1.4
g
Rotor overshoots to 60 deg,
where area above Pm equals
the area below Pm
Now
Pe
>
Pm and rotor
0.8
1
Now

Pe
Pm

and

rotor

decelerates towards θ=40
deg.
Rotor will oscillate around
θ
=
40 deg.

-
BY
-
STEP

SOLUTION

METHOD
Pm Increased
(1 ) (2) (3 ) (4 ) (5 ) (6 ) (7 ) (8 ) (9 ) (1 0) (1 1) (12)
T θ Pm Pe Pa Ac c eleration ∆t1 ∆ω ω ∆t2 ∆θ θ
(3)‐(4) (5)*k (6 )*(7) deg/s (9 )*(10) (2)+(11)
‐0.0 θ0 Pm0 Pe0 Pm0‐Pe0 0 ‐ 0 ω0=0 ‐
+0.0 θ0 Pm1 Pe0 Pm1‐Pe0 α0 ∆t/2 ∆ω1 ω0+∆ω1=ω1 ∆t ∆θ1 θ1
0.5 θ1 Pm1 Pe1 Pm1‐Pe1 α1 ∆t ∆ω2 ω1+∆ω2=ω2 ∆t ∆θ2 θ2
1 θ2 Pm1 Pe2 Pm1‐Pe2 α2 ∆t ∆ω3 ω2+∆ω3=ω3 ∆t ∆θ3 θ3
1.5 θ3 Pm1 Pe3 Pm1‐Pe3 α3 ∆t ∆ω4 ω3+∆ω4=ω4 ∆t ∆θ4 θ4
2 θ4 Pm1 Pe4 Pm1‐Pe4 α4 ∆t ∆ω5 ω4+∆ω5=ω5 ∆t ∆θ5 θ5
08
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 17 0 180

BY
-
STEP

SOLUTION

METHOD
T θ Pm Pe Pa α∆t1 ∆ω ω ∆t2 ∆θ θ
00
25 0
042
042
0
0
0
0
00
‐
0
.
0
25
.
0
0
.
42
0
.
42

84
05
42
81 7
7
.
5
85
.
9
0
.
80
1
.
00
‐
0
.
20
‐
5
.
69
0
.
5
‐
2
.


SOLUTION

METHOD
Pm Increased from 0.42 to 0.8 pu
80 0
90.0
100.0
50.0
60.0
70.0
80
.
0
Angle(deg)
10 0
20.0
30.0
40.0
Machine
0.0
10
.
0
0 2.5 5 7.5 10 12.5 15
Time(s)
eBook for You
TRANSMISSION &
DISTRIBUTION
A Division of Global Power

A

FAULT

CASE
L1
L2
3
h f lt li L2 l t t b
3
-p
h
ase
f
au
lt
on
li
ne
L2
, c
l
ose
t
o genera
t
or
b
us
Frequency = 60 Hz

3
CASE

STUDY:

ANALYSIS

OF

A

FAULT

CASE
Ste
p
1: Construct Power-
A
n
g
le Curves
p
g
1.40
1.00
1.20
040
0.60
0.80
0.00

N‐1
+(8)
N
(9 )*(10 ) (2)+(11 )
s deg pu pu pu deg/s/s s deg/s deg/s s deg deg
‐
0.0
39.8
0.80
0.80
T θ Pm Pe
0.0
39.8
0.80
0.80
0 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 42.1
0.04 42.1 0.80 0.00 0.80 2880 0.02 57.60 144.0 0.02 2.9 45.0
0.06 45.0 0.80 0.00 0.80 2880 0.02 57.60 201.6 0.02 4.0 49.0
0.08
49.0
0.80
0.00
0.80
2880
0.02
57.60
259.2
0.02
5.2

3.8
71.8
0.16
67.9
0.80
0.88
0.08
291
0.02
5.83
191.2
0.02
3.8
71.8
0.18 71.8 0.80 0.91 ‐0.11 ‐389 0.02 ‐7.78 183.4 0.02 3.7 75.4
0.2 75.4 0.80 0.93 ‐0.13 ‐471 0.02 ‐9.42 174.0 0.02 3.5 78.9
0.22 78.9 0.80 0.95 ‐0.15 ‐539 0.02 ‐10.77 163.2 0.02 3.26 82.2
0.24 82.2 0.80 0.96 ‐0.16 ‐592 0.02 ‐11.85 151.3 0.02 3.03 85.2
0.26
85.2
0.80
0.98
‐
0.18
‐
634
0.02
‐
12.68
138.7

FAULT

CASE
Step 1: Plot Results
100.0
120.0
60.0
80.0
h
ineAngle (deg)
20.0
40.0
Mac
h
0.0
0 0.1 0.2 0.3 0.4 0.5 0.6
Time(s)
eBook for You
THREE PHASE SHORT
CIRCUIT
6
THREE

PHASE

SHORT
-
CIRCUIT
Steady-state:


l
a
X
l
X
a
E
eBook for You
THREE PHASE SHORT
CIRCUIT
7
THREE

PHASE

SHORT
-
CIRCUIT
At the instant of the Fault:
Leakage Reactance of the machine (X
l
)
Armature Reaction to the Fault Current (X
a
)
Since the Air Gap flux cannot change instantaneously,
currents are induced in the field and damper windings (X
dw
and
X

PHASE

SHORT
-
CIRCUIT
Shortly after the instant of the Fault:

Leakage Reactance of the machine (X
l
)

Leakage

Reactance

of

the

machine

(X
l
)
Armature Reaction to the Fault Current (X
a
)
Damper winding currents (low X/R ratio) die out
q
uickl

6
2
4
(pu)
Sub‐transie n
t
‐
4
‐2
0
Current
Transient
Sync hronous
‐6
4
0 0.05 0.1 0.15 0.2 0.25 0.3
Time(s)
eBook for You
SYNCHRONOUS MACHINE MODELS
10
SYNCHRONOUS

MACHINE

MODELS
Single Phase Equivalent of a 3-phase Generator
jXd I
EU
~
Im


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
p
ower model under the control of
g
overnor
pg
action
eBook for You
MODEL APPLICATION
12
MODEL

APPLICATION
Use of Mixed Generator Models:
Complex models used for machines of interest
Simpler models used for remote machines
•
Requires less data
•
Requires

Prepared by: Peter Anderson
eBook for You
OUTLINE
2
OUTLINE
• Machine Differential Equations
• Exciter Differential Equations
GDifftilEti
•
G
overnor
Diff
eren
ti
a
l

E
qua
ti
ons
• Solution of Differential Equations
• Network Solution
• Sample Cases
eBook for You
BASIC MODELS IN STABILITY STUDIES
3
BASIC

MODELS

]
'
d
q
'
q
q
'
d
"
d
q
"
q
q
"
d
E
‐
I
X
‐
X
1
=
pE
E
‐
I
X

q
"
q
d
"
d
'
d
'
q
"
"
q
d
q
q
q
'
0q
d
d
q
q
q
"
0q
d
E‐IX‐X‐E
1
=

fd
'
0d
q
q
d
d
d
q
"
0d
q
T
p
T
p
eBook for You
BASIC MODELS IN STABILITY STUDIES
4
BASIC

MODELS

IN

STABILITY

STUDIES
Synchronous Machines
Algebraic Model:

RXV‐E
qadqq
(
)
()()
∂‐2/πj‐expI+I=jI+I
qdMR
I
Y
Y. E
Y.V
V
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